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Triangle Theorem Definition - TraingleWorksheets.com
Exterior Triangle Theorem Worksheet – Triangles are one of the most fundamental forms in geometry. Understanding triangles is crucial to developing more advanced geometric ideas. In this blog we will
explore the different types of triangles, triangle angles, how to determine the area and perimeter of a triangle, and offer the examples for each. Types of Triangles There are three kinds of
triangles: equal, isosceles, and scalene. Equilateral triangles consist of three equal sides, and three angles … Read more
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Diversifying the Risk Associated with Exploration
“Diversifying the Risk Associated with Exploration”
Authors: Sjur D. Flåm and Sverre Storøy,
Affiliation: Christian Michelsen Research and University of Bergen
Reference: 1986, Vol 7, No 2, pp. 83-92.
Keywords: Stochastic models, entalpic pentalty, portfolio problems
Abstract: This paper is concerned with the allocation of exploratory efforts under the limitation of a fixed budget. A chance constrained problem is formulated. To solve this problem an algorithm is
developed which is based on the entropic penalty approach recently presented by Ben-Tal.
PDF (834 Kb) DOI: 10.4173/mic.1986.2.3
[1] ALLAIS, M. (1957). Method of Appraising Economic Prospects of Mining Exploration over Large Territories, Management Science, 3, pp. 285-347.
[2] BARCOUCH, E., HAUFMAN, G.M. (1977). Estimation of undiscovered Oil and Gas, Proceedings of the Symposium in Applied Math., American Mathematical Society, 21, pp. 77-91.
[3] BEN-TAL, A. (1985). The Entropic Penalty Approach to Stochastic Programming, Mathematics of Operations Research, 10, 2 doi:10.1287/moor.10.2.263
[4] FERGUSON, R. (1967). Mathematical Statistics, A Decision Theoretic Approach.Academic Press, New York.
[5] FLÅM, S.D., PINTER, J. (1985). Selecting Oil Exploration Strategies: Some Stochastic Problem Formulations and Solution Methods, CMI-Report no. 852611-1, Chr. Michelsen Institute, Bergen.
[6] GOLUB, G.H., VAN LOAN, C.F. (1983). Matrix Computations, John Hopkins University Press, Baltimore.
[7] KALLBERG, F.G., ZIEMBA, W.T. (1984). Mis-specifications of Portfolio Selection Problems, pp. 74-87 in G. Bamberg and K. Spreman.eds: Risk and Capital, Lecture Notes in Economics and Mathem.
Systems, 227, Springer Verlag, N.Y.
[8] LUENBERGER, D.G. (1984). Linear and Nonlinear Programming, Second Edition, Addison Wesley.
[9] MANGASARIAN, O.L. (1984). Some Application of Penalty Functions in Mathematical Programming, Computer Sciences Technical Report #544, University of Wisconsin.
[10] PULLEY, L.B. (1983). Mean Variance Approximations to Expected Logarithmic Utility, Operations Research, 31, 4, 685-696 doi:10.1287/opre.31.4.685
[11] RUBINSTEIN, R.Y. (1981). Simulation and the Monte Carlo Method, John Wiley, New York.
[12] SCHUENEMEYER, F.H., DREW, L.F. (1983). A Procedure to Estimate the Present Population of the Size of Oil and Gas Fields as Revealed by a Study of Economic Truncation, Mathematical Geology, 15,
1, pp. 145-161 doi:10.1007/BF01030080
title={{Diversifying the Risk Associated with Exploration}},
author={Flåm, Sjur D. and Storøy, Sverre},
journal={Modeling, Identification and Control},
publisher={Norwegian Society of Automatic Control}
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# Glossary
# 1D elements
Beam, Bar, Tie, Strut, Spring, Damper, Link and Cable are all 1D elements.
# 2D elements
Quad 4, Quad 8, Triangle 3 and Triangle 6 are all 2D elements.
# 3D elements
Brick 8, Wedge 6, Pyramid 5, Tetra 4
# AASHTO
Acronym for “American Association of State Highway and Transportation Officials”.
# Alignment
An alignment defines a line in space, relative to a grid plane, from which lanes tracks or vehicle are positioned.
# Analysis layer
The analysis layer is a view of elements in a Graphic View.
# Applied displacement
A degree of freedom where the displacement is known at the beginning of the analysis, rather than being solved for.
# Axes triad
Axis sets are displayed in Graphic Views as colour coded triads. The red, green and blue lines represent the x, y and z axes respectively.
# Bandwidth
The width of array required to store a matrix clustered around the diagonal, with the diagonal elements along the first column. A small bandwidth will be more efficient in the storage required.
# Bar element
A particular type of finite element used to model elements designed to take only axial load, e.g. truss members.
# Beam element
A particular type of finite element used to model beam, column and bracing structural elements. GSA uses a linear beam element with 2 nodes.
# Beta angle
The angle used to define the orientation of beam elements in space and the local axis directions of 2D elements. Also called Orientation angle.
# Buckling analysis
An analysis to determine the buckling response of a structure. This can be either an eigenvalue analysis for linear structures or a non-linear buckling analysis.
# Cable element
Cable elements are primarily used in fabric analysis. Cables are similar to tie elements but are defined by a stiffness and mass per unit length, rather than a cross section.
# Click
Clicking is the action of pressing and immediately releasing the left mouse button.
# Coincident nodes
Coincident nodes are two or mode nodes that are positioned within a defined distance of each other.
# Collapsing coincident nodes
Collapsing coincident nodes is the operation that results in all but the lowest numbered node in a set of coincident nodes to be deleted and all references to these deleted nodes to be changed to
references to the remaining node.
# Constraint
A constraint on a degree of freedom, such as an applied displacement, rigid link, repeat freedom or restraint.
# Current grid
A reference to a grid plane. Coordinates are defined and reported in the current grid axes. The grid drawn in Graphic Views is defined by the grid layout of the current grid.
# Cycle
This is a single time step in the dynamic relaxation solution process. During a cycle the model is assumed to be a set of linked nodal masses accelerating through space. At the end of each cycle the
position, velocity and forces acting on the nodes are calculated, the resulting accelerations calculated, and the process repeated.
# Degree of freedom
A freedom of a node to move in a particular direction this may be either a translational or rotational direction.
# Design layer
The design layer is a view of members in a Graphic View.
# Diagonal array
An array which has non-zero elements only on the diagonal.
# Direction cosine array
The array relating local and global coordinate system, and used to transform vectors and tensors between different coordinate systems.
# Displacement vector
The calculated displacements at all the degrees of freedom in the structure.
# Drag
Dragging is the action of pressing the left mouse button down and moving the mouse before releasing the mouse button.
# Dummy element
An element that has its dummy attribute set. Dummy elements are ignored during analysis. Certain results can be output for dummy elements.
# Dynamic analysis
An analysis to determine the dynamic response of a structure. A modal analysis gives the dynamic characteristics of the structure while a time-history analysis or response spectrum analysis gives the
structural response.
# Eigenvalue
Part of the solution of an eigenproblem. A dynamic analysis is an example of an eigenproblem and the natural frequencies correspond to the eigenvalues.
# Element
An element is an entity that is analysed. An analysis model is made up of elements.
# Engineering scale
An engineering scale is a scale of 1, 1.25, 2, 2.5 or 5 multiplied by a power of 10.
# Entity
A node, element or member.
# Eye distance
The eye distance in a Graphic View is the distance from the eye point to the object point.
# Eye point
The eye point in a Graphic View is the position from which the model is being viewed.
# Finite element
A mathematical representation of a piece of the structure located by nodes and contributing to the stiffness/inertia of the structure.
# Finite element method
The method for analysis of structure by dividing the structure into a collection of connected finite elements.
# Force vector
The collection of forces applied to an element or to the structure as a whole. Normally the force vector for the structure is known and we wish to calculate the displacements. Once the displacements
at the nodes are calculated we then want to calculate the internal forces in the elements.
# Form-finding
A process by which the natural shape of a fabric structure is established.
# Full image
The full image is a Graphic View image that displays all aspects of the currently specified view.
# Gaussian elimination
A method for solving large systems of linear equations by successively removing degrees of freedom.
# Generic bar elements
Bars, ties and struts are all generic bar elements.
# Geometric stiffness
The additional stiffness of an element that results from the load in the element.
# Ghost image
The ghost image is a Graphic View image that displays a cuboid representing the global extents of the image. A triad representing the global directions is drawn at the object point. A dashed triad is
drawn at the mid-point of the view.
# Global axes
The global axes are the axis set to which all other axes relate. The datum axis set for the model.
# Grid axes
The axes referred to by a grid plane. Commonly used to refer to the axes of the current grid.
# Grid layout
Defines the distribution of grid points or grid lines as drawn in Graphic Views.
# Grid plane
A grid plane is a way of defining a plane for loading etc by reference to an axis set and elevation.
# Grid structure
A Grid structure is one modelled in the global XY (horizontal) plane. Global restraints are applied to the model to force it to deform only in the Z direction.
# Hourglassing
Hourglassing arises with under-integrated elements where there are insufficient stiffness terms to fully represent the stiffness of the element and is noticeable in the results by an hourglass
pattern in the mesh.
# Increment
Used in non-linear, GsRelax analysis to define a single load stage. An increment represents the total imposed loads that are deemed to be present at the time of analysis. It is equivalent to an
analysis case generated internally by the GsRelax Solver. For example with single increment analysis a single analysis case is created and all loads are applied at the start of analysis. Whereas with
a 2-stage multiple increment analysis, the first increment could be an analysis case with half the load applied and the second increment is an analysis case with the total load applied.
# Inertia matrix
Same as mass matrix.
# Influence analysis
An analysis that calculates the effect at a point due to load at a series of points along the structure.
# Joint
A joint is a constraint which links two nodes together in specified directions and is used for modelling features such as pinned connections.
# Lagrange multiplier method
A technique that is used to impose constraints on a model; in particular applied displacements.
# Legend
The legend is the panel of information that describes the content of a Graphic View. The legend may be displayed either on the Graphic View or in a message box.
# Linear element
A finite element with a low order shape function and no mid-side nodes.
# Linear shape function
2-D elements without mid-side nodes interpolate strains linearly between nodes. This can lead to significant stress and strain discontinuities across element boundaries.
# Link element
A link element is a two node element that is rigid in the specified directions, and transfers forces and moments through the structure.
# List field
This is a field that allows the user to enter a list. The syntax of the list is then checked before the field is accepted.
# Lumped mass
A mass (and inertia) at a node. A lumped mass is treated as a one node element.
# Mass matrix
The relationship between force and acceleration in a linear system. In finite element analysis we consider the inertia of the element as represented by the element mass matrix and the inertia of the
whole structure as represented by the structure mass matrix.
# Member
A member is an entity that is designed. Typically a member relates to one or more elements.
# Mid-point
The mid-point is a 3D vector used to specify the pan in a Graphic View. The mid-point vector is the position of the mid-point of the picture relative to the object point in picture axes.
# Modal analysis
Modal analysis solves an eigenvalue problem either for a dynamic or buckling analysis of the structure.
# Node
A point in space. Nodes are used to define the location of the elements.
# Normalisation
A way of scaling a vector or matrix. For example the displacement vector for a node can be normalised, producing a displacement magnitude of one at that node.
# Numeric field
A numeric field allows the user to enter a number. Depending on the context this will be either a real number (e.g. a coordinate) or a whole number (e.g. a load case).
# Numeric format
The format used for the display of numbers. The choice is engineering, to a number of significant figures, decimal, to a number of decimal places, or scientific, to a number of significant figures.
# Numeric/percentage field
This is a field that allows the user to enter either a number or a percentage. These fields are used to allow the user to specify position in either an absolute sense (number e.g. 0.25) or a relative
sense (percentage e.g. 10%).
# Object point
The object point is the point about which rotations occur in a Graphic View. Also, in perspective views the position of the eye is measured from the object point.
# Object to eye distance
See eye distance.
# Orientation angle
The angle used to define the orientation of beam elements in space and the local axis directions of 2D elements.
# Orientation axes
The axes about which orientations occur in Graphic Views.
# Orientation node
The node used to define the xy-plane for orientation of beam elements.
# Panel
A panel is an area bounded by beam elements.
# Path
A path is the line across a structure followed by a lane, track or vehicle.
# Picture axes
The picture axes in a Graphic View are x, left to right, y upwards and z out of the plane of the picture.
# Plane structure
A Plane structure is one modelled in the global XZ (vertical) plane. Global restraints are applied to the model to force it to deform only in the XZ plane.
# Prescribed displacement
See “Applied displacement” and “Settlement”.
# Quad element
A linear or quadratic element used in two dimensional analyses with 4 sides and 4 or 8 nodes defining the element.
# Quadratic element
A finite element with a higher order shape function – usually with mid-side nodes.
# Quadratic shape function
2D elements with mid-side nodes (quad8 and tri6) interpolate strains using a fitted quadratic function. This provides more realistic stresses and strains than elements with linear shape functions.
# Reaction
Force which is generated at restraint.
# Restraint
A degree of freedom that is fixed so that it is no longer free. Special cases of restraints are pins and encastré supports. (See also: Member Restraint and Restraint Property.)
# Section definition axes
The axis system used for defining beam sections.
# Selection field
This is a field that allows the user to choose from a list of options.
# Selection/numeric field
This is a field that allows the user to either choose from a list of options or enter a number (e.g. when specifying a material either a standard material can be selected or a user defined material
# Settlement
A degree of freedom which is restrained and given a fixed displacement prior to analysis.
# Shear beam
A particular type of finite element similar to a beam element but taking account of shear deformations.
# Skyline array
A storage scheme which only holds the entries in the stiffness or mass matrix which are on and above the diagonal. Its name derives from the outline of the array elements that are stored.
# Space structure
A Space structure is one that may be modelled in 3 dimensions.
# Spacer element
A spacer element is used to control the mesh during a form-finding analysis.
# Spring element
An element which is defined by stiffnesses in each of the translational and rotational directions. Used to model springs and features such as elastic supports.
# Static analysis
A static analysis loads at how the structure responds to loads which do not vary with time.
# Stiffness matrix
The relationship between force and displacement in a linear system. In finite element analysis we consider the stiffness of the element as represented by the element stiffness matrix and the
stiffness of the whole structure as represented by the structure stiffness matrix.
# Structure scale
The structure scale is the scale at which the structure is drawn in a Graphic View.
# Strut element
A strut element is similar to a bar element in that it has axial stiffness only, but will only take compressive forces. Note that since this is a non-linear element, post-analysis combining of
results is inappropriate.
# Subspace iteration
A technique for solving for eigenvalues and eigenvectors iteration of large systems, by projecting the system of equations on to a smaller “subspace”.
# Text field
A text field allows the user to enter a string of text and is normally used for names.
# Thread
Threads are a way for a computer program to fork (or split) itself into two or more simultaneously running tasks.
# Tie element
A tie element is similar to a bar element in that it has axial stiffness only, but will only take tensile forces. Note that since this is a non-linear element, post-analysis combining of results is
# Topology
The list of nodes which define an element or member.
# Torce lines
Torce lines are generalised thrust lines for two or three dimensions. In 2D (plane frame analysis) they are synonymous with thrust lines (i.e. they show the line of action of axial load) but in three
dimensions they can contain both torque and force components, hence the name.
# Tree control
A tree control is a standard Windows control that offers options via expandable branches. Click on the “+” to expand a branch and on the “−” to collapse the branch. Double click on an item to invoke
that item.
Sometimes multiple selection of items is possible. This is achieved by holding the control key down to continue selection or the shift key to include all items between the currently selected and the
item Shft-Clicked.
# Triangular element
A linear or quadratic element used in two dimensional analyses with 3 sides and 3 or 6 nodes defining the element.
# Unattached node
An unattached node is one that is not referenced in the topology list of an element.
# Variable UDL
A variable UDL is a uniformly distributed load whose intensity varies depending on the loaded length. This is used in bridge analysis.
# Vehicle
A vehicle is defined as a list of axle positions and loadings along with an overall width for use in bridge analyses.
# VUDL
See Variable UDL.
# Wireframe image
The wireframe image is a Graphic View image that represents 1D elements as lines and 2D elements as outlines and displays nothing else, regardless of the current settings for the view.
# X elevation axes
The X elevation axis set is a standard axis set that has its xy plane in the global YZ plane.
# Y elevation axes
The Y elevation axis set is a standard axis set that has its xy plane in the global XZ plane.
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Digital Math Resources
Display Title
Math Clip Art--Number Models--Ten Frame--Modeling Sums within Ten-Image 5
Ten Frame--Modeling Sums within Ten-Image 5
Numbers and Operations
This image demonstrates the concept of addition through a visual representation of colored counters. The image represents the sum of two addends, with 1 red counter representing one addend and 5
yellow counters representing the other. This method is integral to the topic of Numbers and Operations because it provides a tactile way for students to grasp the relationship between the two numbers
being added.
Using math clip art like this provides an engaging way for students to visualize mathematical concepts, helping to solidify their understanding. By using a series of images like this, teachers can
create a comprehensive lesson plan that builds on the concept of addition, making it easier for students to conceptualize and perform addition with different numbers.
Utilizing math clip art in lessons is vital, as it allows students to connect with mathematics visually and practically. This can lead to a more profound interest in math as well as an increase in
overall comprehension skills, making math less intimidating and more approachable.
Teacher's Script: Today, we are going to learn about addition using colored counters. Can anyone tell me how many red and yellow counters are here? Let's count them together!
For a complete collection of math clip art related to Modeling Numbers with Ten Frames click on this link: Math Clip Art: Modeling Numbers with Ten Frames Collection.
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Go to the source code of this file.
subroutine zhpgst (ITYPE, UPLO, N, AP, BP, INFO)
Function/Subroutine Documentation
subroutine zhpgst ( integer ITYPE,
character UPLO,
integer N,
complex*16, dimension( * ) AP,
complex*16, dimension( * ) BP,
integer INFO
Download ZHPGST + dependencies
[TGZ] [ZIP] [TXT]
ZHPGST reduces a complex Hermitian-definite generalized
eigenproblem to standard form, using packed storage.
If ITYPE = 1, the problem is A*x = lambda*B*x,
and A is overwritten by inv(U**H)*A*inv(U) or inv(L)*A*inv(L**H)
If ITYPE = 2 or 3, the problem is A*B*x = lambda*x or
B*A*x = lambda*x, and A is overwritten by U*A*U**H or L**H*A*L.
B must have been previously factorized as U**H*U or L*L**H by ZPPTRF.
ITYPE is INTEGER
[in] ITYPE = 1: compute inv(U**H)*A*inv(U) or inv(L)*A*inv(L**H);
= 2 or 3: compute U*A*U**H or L**H*A*L.
UPLO is CHARACTER*1
= 'U': Upper triangle of A is stored and B is factored as
[in] UPLO U**H*U;
= 'L': Lower triangle of A is stored and B is factored as
N is INTEGER
[in] N The order of the matrices A and B. N >= 0.
AP is COMPLEX*16 array, dimension (N*(N+1)/2)
On entry, the upper or lower triangle of the Hermitian matrix
A, packed columnwise in a linear array. The j-th column of A
is stored in the array AP as follows:
[in,out] AP if UPLO = 'U', AP(i + (j-1)*j/2) = A(i,j) for 1<=i<=j;
if UPLO = 'L', AP(i + (j-1)*(2n-j)/2) = A(i,j) for j<=i<=n.
On exit, if INFO = 0, the transformed matrix, stored in the
same format as A.
BP is COMPLEX*16 array, dimension (N*(N+1)/2)
[in] BP The triangular factor from the Cholesky factorization of B,
stored in the same format as A, as returned by ZPPTRF.
INFO is INTEGER
[out] INFO = 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
November 2011
Definition at line 114 of file zhpgst.f.
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Lee A. Segel Prize
The Lee Segel Prizes were established in memory of Lee Segel, who made great contributions to the Bulletin of Mathematical Biology and the field of mathematical biology as a whole. The prizes honor
outstanding contributions to the field of mathematical biology and will help to promote and advance important research findings in this scientific area. There is a Best Paper Prize, as well as a Best
Student Paper Prize. Other prizes may be awarded as deemed appropriate by the Awards Committee, Editors-in-Chief of the Bulletin of Mathematical Biology, and the Society for Mathematical Biology.
The Lee Segel Prizes are awarded every two years, starting in 2009. Papers are selected taking into account the advice of referees and referee reports. For a paper to qualify as a Best Student Paper,
the main author of the paper must have been a student at the time that the work was carried out.
The Best Paper Award consists of $3000 USD and an invitation for one of the authors to present the paper at the Annual Meeting of the Society for Mathematical Biology (preferably in the year that the
award is given out). The Best Student Paper Award consists of $2000 USD and an invitation for one of the authors to present the paper at the Annual Meeting of the Society for Mathematical Biology.
Recipients of the Lee A. Segel Prize
Best Paper: S. Hamis, J. Yates, M.A.J. Chaplain and G.G. Powathil, Targeting Cellular DNA Damage Responses in Cancer: An In Vitro-Calibrated Agent-Based Model Simulating Monolayer and Spheroid
Treatment Responses to ATR-Inhibiting Drugs. published in the Bulletin of Mathematical Biology 83(10):103 (2021). DOI: 10.1007/s11538-021-00935-y
Best Student Paper: E.D. Counterman and S.D. Lawley, Designing Drug Regimens that Mitigate Nonadherence. Bulletin of Mathematical Biology 84:20 (2022). DOI: 10.1007/s11538-021-00976-3.
Best Paper: C. A. Yates, M. J. Ford and R. L. Mort, A multi-stage representation of cell proliferation as a Markov process. Bull Math Biol (2017) 79b 2905-2928. DOI: 10.1007/s11538-013-9827-4.
Best Student Paper: A. P. Browning, P. Haridas and M. J. Simpson, A Bayesian sequential learning framework to parameterise continuum models of melanoma invasion into human skin. Bull Math Biol (2019)
81, 676-698. DOI: 10.1007/s11538-018-0532-1.
Best Paper: M. P. Thon, H. Z. Ford, M. W. Gee and M. R. Myerscough, a quantitative model of atherosclerotic plaques using in vitro experiments. Bull Math Biol (2017) 80, 175-214. DOI: 10.1007/
Best Student Paper: M. Craig, A. R. Humphries and M. C. Mackay, A mathematical model of granulopoiesis incorporating the negative feedback dynamics and kinetics of G-CSF/neutrophil binding and
internalisation. Bull Math Biol (2016) 78, 2304-2357. DOI: 10.1007/s11538-013-9827-4.
Best Paper: A. J. McKane, T. Biancalani and T. Rogers. Stochastic pattern formation and spontaneous polarisation: The linear noise approximation and beyond. Bull Math Biol (2014) 76, 895-921. DOI:
Best Student Paper (shared prize): J. P. Taylor-King, E. E. van Loon, G. Rosser and S. J. Chapman. From birds to bacteria: generalised velocity jump processes with resting states. Bull Math Biol
(2015) 77, 1213-1236. DOI: 10.1007/s11538-015-0083-7.
Best Student Paper (shared prize): H. C. Warsinske, S. L. Ashley, J. J. Linderman, B. B. Moore and D. E. Kirschner. Identifying mechanisms of homeostatic signaling in fibroblast differentiation. Bull
Math Biol (2015) 77, 1556-1582. DOI: 10.1007/s11538-015-0096-2.
Best Paper: S.R. McDougall, M.G. Watson, A.H. Devlin, C.A. Mitchell and M.A.J. Chaplain, A hybrid discrete-continuum mathematical model of pattern prediction. Bull Math Biol (2012) 74, 2272-2314.
DOI: 10.1007/s11538-012-9754-9.
Best Student Paper: S. O’Malley and M. A. Bees, The orientation of swimming bi-flagellates in shear flows. Bull Math Biol (2012) 74, 232-255. DOI: 10.1007/s11538-011-9673-1.
Best Paper: R. Peña-Miller, D. Lähnemann, H. Schulenburg, M. Ackermann, and R. Beardmore, Selecting against antibiotic-resistant pathogens: optimal treatments in the presence of commensal bacteria.
Bull Math Biol (2012) 74, 908-934. DOI: 10.1007/s11538-011-9698-5.
Best Student Paper: S.M. Moore, C.A. Manore, V.A. Bokil, E.T. Borer and P.R. Hosseini, Spatiotemporal model of barley and cereal yellow dwarf virus transmission dynamics with seasonality and plant
competition. Bull Math Biol (2011) 73, 2707-2730. DOI: 10.1007/s11538-011-9654-4.
Best Research Paper: W. B. Lindquist and I. D. Chase, Data-based analysis of winner-loser models of hierarchy formation in animals. Bull Math Biol (2009) 71, 556-584. DOI: 10.1007/s11538-008-9371-9.
Best Educational Paper: B. R. Kohler, R. J. Swank, J. W. Haefner and J. A. Powell, Leading students to investigate diffusion as a model of brine shrimp movement. Bull Math Biol (2010) 72, 230-257.
DOI: 10.1007/s11538-009-9444-4.
Best Student Paper: B. Boldin, Persistence and spread of gastro-intestinal infections: the case of enterotoxigenic Escherichia coli in piglets. Bull Math Biol (2008) 70, 2077-2101. DOI: 10.1007/
Best Paper: T. de-Camino-Beck and M. A. Lewis, A new method for calculating net reproductive rate from graph reduction with applications to the control of invasive species. Bull Math Biol (2007) 69,
1341-1354. DOI: 10.1007/s11538-006-9162-0.
Best Student Paper: E. Y. Jin and C. M. Reidys, Asymptotic enumeration of RNA structures with pseudoknots. Bull Math Biol (2008) 70, 951-970. DOI: 10.1007/s11538-007-9265-2.
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Can someone help me build an amp?
be very careful on the veroboard, I wired up a pair of TDA2009's as bridged amps and ended up with a very poor quality and hot running amp that just happened to be oscillating at 6 MHz
theres nothing wrong with starting with a kit... something like this: link. then if you have questions, someone here can offer answers as to why this or that. kits are a great intro imo even if you
don't know all the how's and why's. assembling something with your own hands and seeing it work beats staring at a datasheet trying to figure out what all the cryptic symbols mean and ending the day
with nothing but frustration. i would start with the kit. my 2ยข fwiw.-sj
the best resource i've found for building up a inventory of parts is the "grab bag" type deals from places like electronic goldmine. sorting through one of those is fun and will yield way more usable
goodies than stripping junk electronics (for the most part).
I always thought they should offer a quarterly or monthly subscription option.
@sonic ... So which grab bag did you buy, and what sorts of things were in it? Any grand treasures?
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Pascal's Triangle Calculator
Pascal’s Triangle Calculator
• Enter the number of rows you want in Pascal's Triangle.
• Click "Calculate" to generate Pascal's Triangle.
• Click "Clear Results" to clear the triangle and details.
• Click "Copy Results" to copy the triangle to the clipboard.
Calculation History
What is Pascal’s Triangle?
Pascal’s Triangle is a triangular array of numbers named after the French mathematician Blaise Pascal, although it was known to Chinese mathematicians over 500 years earlier. It is constructed in
such a way that each number in the triangle is formed by adding the two numbers directly above it. The triangle begins with a single “1” at the top, and each subsequent row is created by adding
adjacent numbers from the row above.
All Formulae Related to Pascal’s Triangle Calculator
The main formula for constructing Pascal’s Triangle is as follows:
1. Each element within Pascal’s Triangle (except for the outermost elements, which are always 1) is the sum of the two elements directly above it in the previous row.
For example, if we denote the elements in Pascal’s Triangle by “C(n, k),” where “n” represents the row number (starting from 0), and “k” represents the position within the row (starting from 0), we
can calculate each element using the formula:
C(n, k) = C(n-1, k-1) + C(n-1, k)
Here’s how it works:
• The first row has only one element: C(0,0) = 1.
• The second row has two elements: C(1,0) = 1 and C(1,1) = 1.
• The third row has three elements: C(2,0) = 1, C(2,1) = 2, and C(2,2) = 1.
• And so on, with each row having one more element than the previous row.
Pascal’s Triangle is constructed row by row, and you can use the formula above to calculate each element as needed.
Practical Uses of Pascal’s Triangle
Pascal’s Triangle, with its unique properties and patterns, has several practical uses and applications across various fields of mathematics, science, and beyond. Here are some practical uses of
Pascal’s Triangle:
1. Binomial Expansion: Pascal’s Triangle provides coefficients for the expansion of binomial expressions, such as (a + b)^n. These coefficients, referred to as binomial coefficients or “choose”
numbers, are used extensively in algebra and combinatorics.
2. Combinatorics: Pascal’s Triangle helps solve problems related to combinations and permutations. It aids in counting the number of ways to choose or arrange items, making it valuable in
probability calculations, poker hands, and more.
3. Probability: In probability theory, Pascal’s Triangle is used to calculate probabilities in various situations, including the binomial distribution. It is essential for understanding the
probability of specific outcomes in statistical experiments.
4. Mathematical Induction: Mathematicians use Pascal’s Triangle as a visual aid when teaching mathematical induction. It helps demonstrate patterns and validate mathematical proofs through
5. Number Patterns: Researchers and educators use Pascal’s Triangle to explore various number patterns and relationships, such as triangular numbers, Fibonacci numbers, and Lucas numbers, by
examining diagonals and rows.
Applications of Pascal’s Triangle Calculator in Various Fields
1. Combinatorics and Probability Theory:
□ Calculating combinations and permutations for various applications, such as in card games, lottery odds, and experimental design.
2. Statistics:
□ Calculating binomial probabilities and constructing probability distributions for statistical analysis.
□ Generating binomial coefficients for statistical calculations, including hypothesis testing and confidence intervals.
3. Calculus and Taylor Series:
□ Determining the coefficients in Taylor series expansions of functions.
□ Exploring mathematical properties and patterns related to derivatives and integrals.
4. Number Theory:
□ Studying number patterns and relationships between integers, including triangular numbers and prime numbers.
5. Computer Science:
□ Developing algorithms and data structures, such as dynamic programming solutions and recursive functions.
□ Generating combinatorial sequences for coding and cryptography applications.
6. Math Education:
□ Teaching mathematical concepts, including combinatorial principles, binomial coefficients, and algebraic properties, to students of all levels.
7. Physics and Engineering:
□ Solving problems related to probability, combinatorial optimization, and algorithm design in physics and engineering applications.
8. Art and Design:
□ Incorporating patterns and symmetries inspired by Pascal’s Triangle into artistic and geometric designs.
9. Cryptanalysis:
□ Exploring patterns within Pascal’s Triangle for cryptographic analysis and code-breaking strategies.
10. Operations Research:
□ Applying combinatorial optimization techniques in operations research and decision-making processes.
Benefits of Using the Pascal’s Triangle Calculator
Using a Pascal’s Triangle Calculator offers several benefits in mathematics, statistics, and various fields that rely on combinatorial and probability calculations. Here are the key advantages of
using such a calculator:
1. Efficiency: Calculating binomial coefficients manually can be time-consuming and error-prone, especially for large values of “n” and “k.” A calculator automates the process, saving time and
reducing the risk of errors.
2. Accurate Combinations: The calculator ensures accurate and consistent calculations of combinations and permutations, which are crucial in combinatorial problems and probability theory.
3. Quick Access: A calculator provides immediate access to binomial coefficients for specific values of “n” and “k,” making it convenient for on-the-fly calculations.
4. Versatility: Calculators can handle a wide range of input values, including large integers, allowing for flexibility in various mathematical applications.
5. Educational Tool: Pascal’s Triangle Calculators can serve as educational tools to help students understand combinatorial principles, binomial coefficients, and mathematical patterns more
6. Problem-Solving: Professionals and researchers can use the calculator to quickly solve problems in fields such as statistics, computer science, and engineering, where combinatorial calculations
are common.
7. Visualization: Some calculators may provide visual representations of Pascal’s Triangle, aiding in the visualization of patterns and relationships within the triangle.
8. Consistency: Using a calculator ensures consistency in calculations, as it follows established mathematical algorithms and formulas.
9. Reduced Human Error: Manual calculations can lead to errors, especially when dealing with large numbers or complex expressions. Calculators eliminate the risk of human error.
10. Convenience: Accessing binomial coefficients through a calculator is convenient for professionals, researchers, and students who need quick and accurate results for their work.
1. “Pascal’s Triangle: A Gateway to Combinatorics” (2017) by the Mathematical Association of America
2. The Curious History of Pascal’s Triangle” (2022) by the American Mathematical Society
3. Interactive Pascal’s Triangle” by The Wolfram
Last Updated : 03 October, 2024
One request?
I’ve put so much effort writing this blog post to provide value to you. It’ll be very helpful for me, if you consider sharing it on social media or with your friends/family. SHARING IS ♥️
Sandeep Bhandari holds a Bachelor of Engineering in Computers from Thapar University (2006). He has 20 years of experience in the technology field. He has a keen interest in various technical fields,
including database systems, computer networks, and programming. You can read more about him on his bio page.
23 thoughts on “Pascal’s Triangle Calculator”
1. Twalker
The clear explanations and examples in this article make Pascal’s Triangle more approachable and engaging for readers. Well done!
2. Rscott
I appreciate the effort to make complex mathematical concepts accessible to a wider audience.
3. Damien Campbell
The post effectively communicates the importance and relevance of Pascal’s Triangle, shedding light on its multifaceted nature and contributions to different disciplines.
4. Kgreen
The benefits of using a Pascal’s Triangle Calculator are compelling. The efficiency and accuracy it offers in calculations make it a valuable tool.
5. Claire98
The explanation of the main formula for constructing Pascal’s Triangle is clear and well-presented. It makes understanding the concept much easier.
6. Brown Alice
The interdisciplinary impact of Pascal’s Triangle is evident from the examples provided in the article.
7. Sean89
Agreed, the clarity of the formula explanation is commendable.
8. Msaunders
I found the breakdown of the main formula to be very helpful in grasping the concept.
9. Miller Aiden
Thanks for explaining the practical uses of Pascal’s Triangle! I didn’t realize how many different fields it can be applied to.
10. Leah Ellis
The practical uses of Pascal’s Triangle across various fields are impressive. It’s interesting to see how fundamental mathematical concepts have real-world applications.
11. Ellis Mike
It’s fascinating to see how a simple mathematical concept can have such far-reaching implications.
12. Parker Harvey
The applications of Pascal’s Triangle are a testament to the power of mathematical reasoning and its impact on diverse disciplines.
13. Butler Lizzie
I agree, the applications of Pascal’s Triangle are truly diverse and impressive.
14. Kimberly Bailey
This post provides a comprehensive understanding of Pascal’s Triangle, including its practical applications and benefits. It’s an excellent resource for anyone interested in mathematics.
15. Graham Barry
This is a valuable overview of Pascal’s Triangle and its significance in mathematics and beyond. The detailed explanations are greatly appreciated.
16. Jayden Young
The applications of Pascal’s Triangle in fields like statistics, computer science, and art are mind-blowing. It’s exciting to see the intersection of mathematics with other domains.
17. Swilliams
Absolutely, the real-world relevance of Pascal’s Triangle is eye-opening.
18. Megan Thomas
The educational potential of Pascal’s Triangle Calculators is significant, especially in teaching mathematical concepts to students.
19. Matthews Ellie
The creative applications of Pascal’s Triangle in art and design are particularly intriguing.
20. Tony84
The detailed insight into the applications of Pascal’s Triangle in various fields broadens the perspective on its significance. It’s truly a versatile mathematical concept.
21. Mason Rose
I’m fascinated by the role of Pascal’s Triangle in computer science and its influence on algorithm design.
22. Danielle33
The diversity of applications highlighted in this post is both informative and thought-provoking.
23. Zmurray
I’m impressed by the depth of the analysis on Pascal’s Triangle and its practical implications.
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99 Math: Improve Your Math Skill With 99 Math - genz slang
As someone who has used 99 Math extensively, I can confidently say that it’s one of the best tools out there for making math fun and interactive. Whether you’re a teacher, parent, or student, 99 Math
provides an exciting way to practice math skills in a competitive, real-time environment. In this article, I’ll share how 99 Math works, why it’s so effective, and how you can start using it today.
What is 99 Math?
99 Math is an online multiplayer math game that lets students compete with one another in real-time. It’s designed to help students practice math by solving problems quickly and accurately. From
basic arithmetic to more complex math concepts, 99 Math covers a wide range of topics and adapts to different skill levels.
How Does 99 Math Work?
Using 99 Math is incredibly simple:
1. Sign Up: Teachers or parents sign up for a free account on the 99 Math platform.
2. Create a Game: You choose a math topic (addition, subtraction, multiplication, etc.) and set the difficulty level based on the students’ abilities.
3. Share the Game Code: Once the game is created, you receive a unique code. Students enter this code to join the game.
4. Compete in Real-Time: Students then compete against each other, answering math problems as quickly and accurately as possible.
The real-time aspect of the game keeps students engaged, as they can see how their performance stacks up against others. This makes math practice more exciting than traditional methods like
worksheets or flashcards.
Why 99 Math is So Effective
1. Game-Like Environment
The competition aspect of 99 Math turns math practice into a fun, fast-paced game. Students are motivated to improve their skills so they can beat their classmates or friends. This creates a sense of
accomplishment and excitement around learning math.
2. Customizable for All Skill Levels
Whether you’re working with young children just starting with basic addition or older students mastering algebra, 99 Math can be customized to fit their needs. The platform offers various topics and
difficulty levels to make sure the problems are suited to each student’s skill level.
3. Instant Feedback
One of the best features of 99 Math is the instant feedback. Students know immediately whether they answered correctly or not, allowing them to learn from their mistakes in real-time. This kind of
immediate feedback speeds up the learning process and helps build confidence.
4. Remote Learning-Friendly
Since 99 Math is entirely online, it’s perfect for remote learning. Students can join games from any location, making it easy for teachers and parents to integrate it into both classroom and
home-based learning environments.
Key Features of 99 Math
Here are some of the standout features that make 99 Math an excellent tool for math practice:
• Real-Time Competition: Students compete against each other in real-time, which adds a fun, competitive element to learning.
• Covers Multiple Math Topics: 99 Math includes a wide range of math topics, such as basic arithmetic, fractions, and even more advanced concepts like algebra.
• Customizable Difficulty Levels: Adjust the difficulty of the problems to fit the skill level of your students or children.
• Progress Tracking: Teachers and parents can track how students are performing, seeing which areas they excel in and which need more attention.
• Free to Use: 99 Math is completely free, making it accessible for everyone.
Benefits of Using 99 Math
1. Encourages Regular Practice
The game-like structure of 99 Math encourages students to practice more frequently. Instead of dreading math practice, students look forward to the competition, which ultimately improves their math
2. Reduces Math Anxiety
For many students, math can be intimidating. But the playful, competitive environment of 99 Math helps reduce math anxiety. By making math practice more fun, students are more willing to engage with
the subject and less likely to feel overwhelmed.
3. Boosts Confidence
As students compete and improve their skills, they gain more confidence in their math abilities. The instant feedback lets them see their progress in real-time, which encourages them to keep pushing
4. Fits into Any Learning Environment
99 Math is adaptable to any learning environment. Whether used in the classroom, for homeschooling, or as a supplement to remote learning, it’s a flexible tool that fits easily into any educational
How to Start Using 99 Math
Getting started with 99 Math is simple:
1. Create a Free Account: Visit the 99 Math website and sign up for a free account.
2. Set Up a Game: Choose a math topic and difficulty level that suits your students or children.
3. Share the Game Code: Provide students with the unique game code to join.
4. Start the Game: Watch as students compete in real-time, solving problems and improving their skills along the way.
Why 99 Math is a Must-Have for Teachers and Parents
99 Math isn’t just a fun game—it’s a powerful learning tool. By combining competition, instant feedback, and customizable difficulty levels, it provides an interactive, engaging way for students to
practice math. Plus, it’s free and easy to use, making it accessible for everyone.
For teachers, 99 Math can be a great way to make math lessons more engaging. For parents, it’s an excellent resource for helping children practice at home in a fun and motivating way.
Who Should Use 99 Math?
99 Math is perfect for:
• Teachers: Use it in the classroom to make math lessons more interactive and exciting.
• Parents: It’s a great tool for practicing math at home and helping children improve their skills.
• Tutors: Whether you’re working one-on-one or with small groups, 99 Math is a fun way to supplement traditional tutoring.
• Students: If you want to practice math on your own or compete with friends, 99 Math is a fun and effective way to sharpen your math skills.
Final Thoughts
If you’re looking for a new way to make math practice more engaging, 99 Math is the tool you need. Whether in a classroom or at home, it offers a fun, competitive environment where students can
improve their math skills while having fun. And because it’s free, there’s no reason not to give it a try. Get started today, and watch as your students or children become more confident and
proficient in math!
FAQs About 99 Math
1. What is 99 Math?
99 Math is a free, online multiplayer math game designed to help students practice math by competing in real-time. It covers a wide range of math topics, including addition, subtraction,
multiplication, and more.
2. Is 99 Math free to use?
Yes, 99 Math is completely free for everyone, including teachers, parents, and students.
3. What ages is 99 Math suitable for?
99 Math can be used by students of all ages. The platform allows you to customize the difficulty level, making it suitable for younger learners as well as older students.
4. Can I use 99 Math for remote learning?
Yes, 99 Math is perfect for remote learning. Students can join games from any location, making it easy to integrate into a distance learning program.
5. Does 99 Math track student progress?
Yes, 99 Math provides tools for tracking student performance. Teachers and parents can see how well students are doing and which areas they may need to improve in.
6. Do students need to create an account to use 99 Math?
No, students don’t need to sign up for an account. They can join games simply by entering the unique game code provided by their teacher or parent.
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Interpreting Scatter Plots Worksheet
Problem 1 :
A history teacher asked her students how many hours of sleep they had the night before a test. The data below shows the number of hours the student slept and their score on the exam. Plot the data on
a scatter plot.
Problem 2 :
Assume that during a three hour period spent outside, a person recorded the temperature and water consumption. The experiment was conducted on 7 randomly selected days during the summer. This data is
shown in the table below.
Create the scatter plot for the data. What is the correlation of the scatter plot ?
Problem 3 :
The value of cars in a used car garage are recorded below. The scatter graph shows this information
Another car arrives at the garage. It is 4 years old and worth £5000.
(a) Show this information on the scatter graph.
(b) Describe the correlation between the value of the car and the age of the car.
The next car that arrives is 6 years old.
(c) Estimate the value of the car.
Problem 4 :
The table shows the time spent revising and the test scores of ten students.
(a) Complete the scatter diagram.
(b) Describe the relationship shown in the scatter diagram.
(c) Draw a line of best fit on your scatter diagram.
(d) Another student has spent 4.5 hours revising. Use your line of best fit to estimate their test result.
Problem 5 :
The table shows the charge (£) by plumbers for jobs of different duration (hours).
a) Plot the data on the scatter graph
b) Describe the correlation.
c) Draw a line of best fit on the scatter graph.
d) Use your line of best fit to estimate the charge for a 4 hour job.
(e) Explain why it may not be appropriate to use your line of best fit to estimate the charge for a job lasting 12 hours.
Problem 6 :
Some rugby players take two tests, one measuring speed and the other measuring strength. Each test is marked out of 200. The scatter graph compares the results.
(a) What type of correlation does this scatter graph show?
(b) Draw a line of best fit on the scatter graph.
Brian scores 40 in Test 2.
(c) Estimate his score in Test 1.
Problem 7 :
A shop sells umbrellas. The scatter graph shows information about the number of umbrellas sold each week and the rainfall that week, in millimetres.
(a) Describe the relationship between the rainfall and umbrellas sold.
(b) What is the most number of umbrellas sold in one week?
(c) What is the greatest amount of rainfall in one week?
(d) In how many weeks did the shop sell over 105 umbrellas?
In another week, there was 6mm of rain.
(e) Estimate the number of umbrellas sold
(f) Explain why it may not be appropriate to use your line of best fit to estimate the number of umbrellas sold in a week with 25mm of rainfall.
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Show that any LL grammar is unambiguous.
(Note: This is so obvious it is hard to...
Show that any LL grammar is unambiguous. (Note: This is so obvious it is hard to...
Show that any LL grammar is unambiguous.
(Note: This is so obvious it is hard to prove. Instead, prove the contrapositive: ambiguous implies not LL(k) for any k.)
Leave T and N alone the arrangements of terminal and non-terminal images.
Allow the accompanying to hold
MaybeEmpty(s) = valid <=> s - >* void
First(s) = X containing all x for which there exists Y with the end goal that s - >* xY
Follow(A) = X containing all x for which there exists Y,Z with the end goal that S - >* YAxZ
Note that a language structure is LL(1) if the accompanying holds for each pair of creations A - > B and A - > C:
1. (not MaybeEmpty(B)) or (not MaybeEmpty(C))
2. (First(B) meet First(C)) = void
3. MaybeEmpty(C) => (First(B) cross Follow(A)) = void
Think about a language with is LL(1), with A - > B and A - > C. In other words there is some series of terminals TZ which concedes various deductions by particular parse trees.
Assume that the left induction arrives at S - >* TAY - >* TZ. The subsequent stage might be either TAY - > TBY, or TAY - > TCY. Hence the language is questionable if both BY - >* Z and CY - >* Z.
(Note that since An is a discretionary non-terminal, if no such case exists, the language is non-equivocal.)
Case 1: Z = void
By rule 1 of LL(1) punctuations, at generally one of B and C can determine void (non-uncertain case).
Case 2: Z non-void, and neither B nor C infer void
By rule 2 of LL(1) language structures, at generally one of B and C can allow further inference in light of the fact that the main terminal of Z can't be in both First(B) and First(C)
(non-questionable case).
Case 3: Z non-void, and either MaybeEmpty(B) or MaybeEmpty(C)
Note the by rule 1 of LL(1) sentence structures, B and C can't both determine void. Assume consequently that MaybeEmpty(C) is valid.
This gives two sub-cases.
Case 3a: CY - > Y; and Case 3b: CY - >* DY, where D isn't unfilled.
In 3a we should pick between BY - >* Z and CY - > Y - >* Z, however notice that First(Y) subset-of Follow(A). Since Follow(A) doesn't meet First(B), just a single determination can continue
In 3b we should pick between BY - >* Z and CY - >* DY - >* Z, yet notice that First(D) subset-of First(C). Since First(C) doesn't meet First(B), just a single determination can continue
Along these lines for each situation the induction must be extended by one of the accessible creations. Thusly the punctuation isn't vague.
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Dynamics of Real-Time Simulation – Chapter 2 – Z Transforms Applied to Real Time
The method of Z transforms was originally developed for analysis of sampled-data control systems, i.e., dynamic systems utilizing a mixture of continuous and discrete time signals. This is just the
case, for example, when a digital computer is used to control a continuous process. The Z transform method is equally valuable in analyzing all-digital systems, and in particular for the
determination of dynamic errors in digital simulation. The dynamic errors will depend on the type of numerical integration algorithm and the integration step size. The method of Z transforms permits
us to examine these errors analytically and hence choose acceptable integration algorithms and steps sizes to meet prescribed accuracy requirements. The method is restricted to linear systems with a
fixed time step or sample period. We have already noted in Chapter 1 that nonlinear systems, for purposes of dynamic error analysis, can usually be linearized using perturbation equations, and that
real-time digital simulations traditionally employ fixed integration step sizes. Thus the Z transform requirements can be met.
2.2 Definition of the Z Transform
We will assume that the digital signals involved in digital simulation are data sequences, with the individual numbers in each data sequence considered to be equally spaced in time, as noted above.
Thus the digital signal or data sequence representing a continuous time function f(t) will be denoted as the data sequence {f[n]} where each individual data points f[n] represents f(nh)with n = 0, 1,
2, … . Here h is the time interval between successive numbers in the data sequence. In the digital simulation of a continuous dynamic system h is the integration step size.
The Z transform of the digital data sequence {f[n]} is defined as
where z is a complex variable. We note the similarity of the Z-transform definition to the definition of the Laplace transform of a continuous time function f(t).* Thus
The integral of the continuous time function in the Laplace transform is replaced by the summation of the discrete time series in the Z transform. For data sequences derived from exponential time
functions, the Z transform can be written in closed form. Consider the exponential time function
and the equivalent data sequence
From Eq. (2.1) we see that
Expansion of (1 – Az^-1)^-1 using the binomial theorem easily proves the validity of Eq. (2.6). Thus we have shown that
It also follows that if the Z transform of a data sequence contains a term z/(z – A), the corresponding data sequence is the exponential sequence, {e^onh}, i.e., equivalent to equally spaced samples
from a continuous time function e^otwith a sample interval h. From Eq. (2.5)
It is useful to consider the case of a complex exponential function
We recall that together with the conjugate function e^(^σ^–^jw^)^t this can be used to represent a sinusoidal time function of frequency ω with an exponentially varying amplitude e^ot. Here the
equivalent data sequence is
where A is now complex and is given by
But the complex number A can be written as
Equating the right sides of Eqs. (2.11) and (2.12), we have
Thus a Z transform given by z/(z–A), where A is complex, corresponds to a data sequence derived from equally-spaced samples of a sinusoidal time function of frequency ω and amplitude e^ot. The
formulas in Eq. (2.14) relate ω and σ to the complex constant A. We note from both Eq. (2.8) for A positive real and Eq. (2.14) for A complex that the data sequence amplitude decreases exponentially
for |A| < 1 and increases exponentially for |A| > 1. For A = 1 the data sequence is always unity, i.e., f[n ]= 1 for all n. For A = -1 the data sequence alternates, i.e., f[0] = +1, f[1] = -1, f[2]=
+1, f[3] = -1 etc., which represents samples from a unit amplitude sinusoid with frequency equal to one-half the sample frequency, as predicted by Eq. (2.14). Thus ω = π/h for A = -1 and hence the
sinusoidal period = 2π/ω=2h.
2.3 Simple Example of a Linear Digital System
We have seen in the previous section how the Z transform can be used to represent a data sequence. In this section we will see how the Z transform can be used to represent a digital system. As a
simple example, consider the continuous first-order linear system shown in Figure 2.1 and described by the differential equation
Figure 2.1. First-order continuous system.
Assume that we wish to simulate this system digitally using Euler integration. In Chapter 1 we have seen that the numerical solution of Eq. (2.15) with Euler integration leads to the following
difference equation:
This is the difference equation mechanized by the digital simulation, starting at t = 0 (i.e.,n = 0) with x[0] as the initial condition. The equation is now iterated for n = 0, 1, 2, … ,to produce
the outputs x[1], x[2], x[3], … , in response to the inputs f[0], f[1], f[2], f[3], … . Thus the digital computer produces an output data sequence {x[n]+1} given from Eq. (2.16) by
where {f[n]} is the input data sequence. We now take the Z transform of the data sequences appearing on both sides of Eq. (2.17). Consider first the Z transform of {x[n]+1} From Eq. (2.1) this is
given by
Note the similarity of Eq. (2.18) to the formula for the Laplace transform of the derivative of a function. Thus
Using Eq. (2.18), we can now take the Z transform of Eq. (2.17) and obtain
Solving for X*(z) we have
The first term on the right side of Eq. (2.21) depends only on the initial condition x[0] and is equivalent to the transient (complementary) solution in the case of a continuous system. It has
exactly the form of Eq. (2.7) and therefore represents in the time domain an exponential data sequence x[0]{(e^oh)^n} = x[0]{A}^n , where A = 1 + λh. From Eq. (2.8) we see that
We note that 1n (1 + y) = y –y^2/2 + y^3/3 – y^4/4 + ∙∙∙ , . Expanding the 1n function in Eq. (2.22), we have
For the continuous linear system the transient solution is given by x[0]e^λ1, so that a “perfect” digital solution would be the data sequence x[0]{(e^λ^h)^n}, compared with our actual solution, x[0]
{(e^o^h)^n}. Just as λ is the characteristic root of the continuous system, so is σ the equivalent characteristic root of the digital system. From now on we will designate the characteristic root of
the digital system by λ*, where λ* = σ in our example here. Replacing σ with λ* in Eq. (2.23), subtracting λ and dividing by λ, we obtain the following formula for e[λ], the fractional error in the
characteristic root of our digital system:
The above asymptotic formula is valid only for h<<1/|λ|. But this will of necessity be true if we are to have an accurate simulation. As expected for Euler integration, the root error varies as the
first power of the integration step size h. Thus the root error will decrease by a factor of two whenever we halve the step size.
We next consider the second term on the right side of Eq. (2.21), i.e., the term involving the input Z transform, F*(z). The coefficient of F*(z) in this term is defined as the Z transform, H*(z), of
the digital system. Thus
and when x[0] = 0,
Thus the system Z transform, H*(z), times the input Z transform, F*(z), yields the output Z transform, X *(z), as shown in Figure 2.2. The analogy with the Laplace transform representation of the
continuous system, as shown in Figure 2.1b, is very evident.
Figure 2.2. Digital system representing simulation of a first-order system using Euler integration.
Using Euler integration, we now compute the response of our first-order linear system to a unit data point input defined by
Iterating the difference equation (2.16) with zero initial condition (x[0] = 0), we obtain
It turns out that we can get also obtain the unit data point response of our digital system directly from the system Ztransform, H*(z), given in Eq. (2.25). To do this we divide the denominator into
the numerator of H*(z) to create a power series in z^-1, as shown below.
From the Z-transform definition in Eq. (2.1) we see that the data sequence {w[n]} for which H*(z) is the Z transform is given by
I.e., when the Z transform is expressed as a series in powers of z^-1, the coefficient of z^–^n is w[n], the data sequence value at time nh. The data sequence represented by Eq. (2.30) is identical
with the data sequence obtained in Eq. (2.28) in response to a unit data point input. We conclude that the time-domain representation of the Z transform, H*(z), of a digital system is simply the unit
data point response, {w[n]}, of the digital system. Conversely, the Z transform of the unit data point response is the system Z transform, H*(z). This also follows if we consider the Z transform of
the unit data point input represented by Eq. (2.27). From Eq. (2.1) the corresponding Z transform is simply F*(z) = 1. When F*(z) = 1 we conclude from Eq. (2.26) that X*(z) = H*(z), i.e., the Z
transform of the unit data point response is indeed the system Z transform.
The unit data point input to a digital system is analogous to a unit impulse input, δ(t)for a continuous system. The response data sequence, {w[n]} of the digital system is analogous to the
unit-impulse response, W(t), of the continuous system. Just as the Z transform of the unit data point response data sequence, {w[n]}, is H*(z), the system Z transform, in the case of a continuous
system the Laplace transform of the unit impulse response, W(t), is the Laplace transform (transfer function), H(s), of the continuous system.
It is useful to write Eq. (2.26) with the Z transforms expressed in series form. Thus
Equating coefficients of like powers of on both sides of Eq. (2.31), we obtain
Since f[n-k] = 0 for k > n (the input data point f[n] = 0 for n < 0), the upper limit in the summation in Eq. (2.32) can be changed from n to ∞. Thus the digital system response data point x[n] at
time t= nh can be written as
where {w[n]} is the system unit data point response sequence and{f[n]} is the input data sequence. The similarity to the superposition integral for the response x(t) of a continuous linear
time-invarient system to an input f(t) is striking. Thus
where W(t) is the unit impulse response for the system.
Finally, we consider a sinusoidal data sequence input of unit amplitude, as represented by samples of f(t) = e^jwt. Thus we let f[n] = e^jwnh = (e^jwh)^n. From Eq. (2.33) the response is given by
Reference to Eq. (2.1) shows that
Where f[n] is a sinusoidal data sequence of frequency ω. Eq. (2.37) shows that the response data sequence is also a sinusoid with the same frequency ω and a complex amplitude that is simply H*(e^jwh)
times the unit input amplitude. Thus H*(e^jwh) , which is simply the system Z transform H*(z) with z replaced by e^jwh, is the digital system transfer function for sinusoidal inputs. The magnitude
and phase angle of the complex number H*(e^jwh) represent the amplitude ratio and phase shift of the output data sequence with respect to the input data sequence.
Again, the analogy to continuous time-invarient linear systems is apparent. Thus
Where x(t) is the continuous system response, H(jw)is the transfer function for sinusoidal inputs, and f(t) = e^jwt is the sinusoidal input to the continuous system.
As a specific example of a digital system transfer function for sinusoidal inputs, consider again the simulation of the first-order linear system in Figure 2.1 using Euler integration. The Z
transform of the resulting digital system is given by Eq. (2.25). From Eq. (2.37) we see that the digital system transfer function is obtained by replacing z in H*(z) with e^jwh. Thus
The continuous system transfer function for sinusoidal inputs is obtained from Figure 2.1b by replacing s in H(s) with jw. Thus
For a given frequency ω and integration step size h we can now calculate the fractional error in the sinusoidal transfer function of the digital system. From Eqs. (2.39) and (2.40) this is given by
In Chapter 1, Eq. (1.51), we showed that the real and imaginary parts of H*/H-1 represent, approximately, the fractional gain error and the phase error, respectively, of the digital system transfer
function. This in turn gives us a meaningful measure of the dynamic accuracy of the digital simulation, especially in the context of a real-time hardware-in-the-loop simulation, as explained at the
end of Section 1.3. In Chapter 3 we will show how simplified asymptotic formulas can be derived for these gain and phase errors, and in particular that the errors vary as the first power of the step
size in the case of Euler integration.
2.4 Stability of Digital Systems
We obtained the Z transform of the digital system representing simulation with Euler integration of the first-order system in Figure 2.1 by taking the Z transform of the difference equation (2.16)
and solving for H*(z) = X*(z)/F*(z). The same procedure can be used to obtain the Z transform of any digital system described by one or more linear difference equations. Consider, for example, a
second-order continuous system described by the following linear differential equation:
Eq. (2.42) represents the mathematical model for a single-degree-of-freedom mechanical system with displacement x, mass M, viscous damping constant C, spring constant K, and applied force f(t). To
simulate the system with numerical integration, we rewrite the second-order differential equation as the following two first-order differential equations:
Again we consider simulation using Euler integration. From Eq. (2.43) the difference equations become
The Z transforms of the difference equations are given by
To obtain H*(z), the Z transform of the digital system, we set the initial conditions x[0] = y[0] = 0 and eliminate Y*(z) from the two equations represented in (2.45). We then have a single equation
in X*(z) and F*(z), which we can solve for H*(z) = X*(z)/F*(z). In this way we obtain
It is apparent that in general the Z transform of the digital system used to simulate any finite-order linear dynamic system will have the form
Here N*(z) and D*(z) are polynomials in z, A[1], A[2], … , A[k] are the roots of D*(z), i.e., the poles of H*(z) and in general can be real or complex. H*(z) in Eq. (2.47) can be represented in terms
of a partial fraction expansion, just as the Laplace transform H(s) for a continuous system was represented in terms of a partial fraction expansion in Eq. (1.43). Thus we can write
We conclude that the Z transform for a linear digital system of any order can be expressed as the sum of Ztransforms of the form C/(z – A). We have already seen that such Z transforms correspond to
exponential data sequences {A}^n, where for |A| >1 the sequence diverges exponentially and for |A| < 1 the sequence converges exponentially. For the digital system Z transform, H*(z), the exponential
data sequences are terms in the unit data point response sequence. Thus the overall unit data point response sequence is just the sum of the exponential data sequences corresponding to each term on
the right side of Eq. (2.48). It is clear that if any of these exponential data sequences diverge, the unit data point response will diverge and the digital system is unstable. For all the
exponential terms in the unit data point to converge, the pole A associated with each term must have a magnitude less than unity, i.e., |A| < 1. We thus conclude that a digital system with Z
transform poles given by A[k]will be stable if |A| < 1 for all k. In the complex plane |A[k]| < 1 corresponds to the region inside the unit circle defined by |z| = 1, as shown in Figure 2.3.* The Z
transform poles within this region all correspond to exponentially converging data sequences. Poles outside the unit circle in the z plane correspond to exponentially diverging data sequences. Poles
on the unit circle correspond to data sequences of constant amplitude, except that repeated poles on the unit circle represent diverging data sequences (see the footnote below).
Also shown in Figure 2.3 are the frequencies of the data sequences corresponding to the Z transform poles, based on Eq. (2.14). Note that poles along the positive real axis correspond to zero
frequency, i.e., a pure exponential data sequence. Poles along the negative real axis correspond to a data sequence with a frequency equal to one-half the sample frequency. Along the 45 degree line
the frequency equals one-eighth the sample frequency; along the positive imaginary axis, one-quarter the sample frequency; etc. Note also that for every Z transform pole in the upper half of the z
plane there must be a complex conjugate pole in the lower-half plane. These lower half-plane poles correspond to negative frequencies, which are present because of the Euler representation of
sinusoids as e^jwt.
It is useful to compare regions in the z plane of Figure 2.3 with the corresponding regions of the s plane in the Laplace transform. Thus we see that the unit circle in the z plane corresponds to the
imaginary axis in the s plane. For a digital system to be stable, its Z transform poles must lie inside the unit circle in the z plane; for a continuous system to be stable, its Laplace transform
poles must lie to the left of the imaginary axis in the s plane, i.e., in the left-half plane. Pole locations corresponding to a constant frequency lie along straight-line rays passing through the
origin of the z plane, with the polar angle given by πω/ω[0]; pole locations corresponding to a constant frequency lie along horizontal lines in the s plane, with the horizontal line intercept on the
imaginary axis equal to the frequency ω. Poles representing oscillatory time functions always occur in complex conjugate pairs in the Laplace transform of continuous systems; this is also true for
digital systems for all frequencies except ω = ω[0]/2, in which case a single pole along the negative real axis represents an oscillatory data sequence with frequency equal to one-half the sample
Figure 2.3. Geometric relationships between Z transform poles and the stability and frequency of the corresponding data sequences.
We now examine the stability of the digital system in Figure 2.2, which we recall represents simulation of a first-order linear system using Euler integration. For the system being simulated to be
stable, it is clear that the characteristic root λ must be negative. From the digital system Z transform, H*(z), it is apparent that the single pole is given by
As we vary the step size h from 0 to infinity, the pole z[1] moves along the real axis from +1 to minus infinity, as shown in Figure 2.4. This plot in the z plane is analogous to the so-called root
locus plot in the s plane for closed-loop systems as the controller gain constant is varied from zero to infinity. In Figure 2.4 we see that the locus crosses the unit circle in the z plane for h =
-2/λ. For h > -2/λ the pole z[1] lies to the left of the point z = -1, and the simulation will be unstable. The corresponding data sequence will be an exponentially growing oscillation at one-half
the sample frequency. In fact, for h in the range -1/λ < h < -2/λ the pole z[1], although within the unit circle, lies on the negative real axis. This means that the corresponding stable data
sequence is oscillatory at one-half the sample frequency, even though the continuous system being simulated has a non-oscillatory exponential transient.
In Figure 2.5 the data points from a series of digital solutions with different integration step sizes are shown, along with the exact solution of the continuous system for x(0) = 1 and f(t) = 0. For
each case the step size h corresponds to one of the values shown on the root locus plot in Figure 2.4. It should be noted that the characteristic root λ of the system considered here is related to
the time constant of the first order system by the formula T = -1/λ. Thus λh = -0.2 corresponds to
Figure 2.4. Root locus for the pole z[1] of the digital system when Euler integration is used to simulate a first-order linear system with 0 < h < ∞
h = .2T, i.e., the integration step size h is equal to one-fifth the time constant T of the first-order system being simulated. Figure 2.5 shows that in this case the digital solution is quite close
to the exact solution. For h = .5T the digital solution is clearly decaying more rapidly than the exact solution. This is because the characteristic root λ* of the digital system has a slightly
larger magnitude than the ideal root λ, which is consistent with Eq. (2.24) for negative λ. For h = T the digital solution drops to zero at the first integration step and remains zero for all
subsequent time steps. In this case the pole z[1] = 0 and Eq. (2.22) shows that the equivalent characteristic root λ* (=σ) is also equal to zero. Thus the digital solution remains equal to a
constant, namely zero.
For h = 1.5T and 1.8T, Figure 2.5 shows that the digital solutions are damped oscillations at one-half the sample frequency. In each case the corresponding pole z[1] in Figure 2.4 lies on the
negative real axis but inside the unit circle in the z plane. For h = 2T (i.e.,λh = -2) the pole in Figure 2.4 lies at z[1] = -1. This results in a digital solution which is an undamped oscillation
at one-half the sample frequency, which again is confirmed in Figure 2.5. Finally, for h = 2.1T Figure 2.5 shows that the digital solution is a growing oscillation, once more at one-half the sample
frequency. The corresponding pole z[1] in Figure 2.4 lies on the negative real axis, but outside the unit circle in the zplane. This accounts for the observed instability in the digital solution.
In considering Euler integration for simulating the second-order linear system described by Eq. (2.43), we obtained Eq. (2.46) for the digital system Z transform. Since the denominator of H*(z) is in
this case a quadratic in , it is clear that H*(z) has two poles, z[1]and z[2]. These poles are related to the two characteristic roots, λ[1] and λ[2], of the continuous system by Eq. (2.49) for the
case of both real and complex λ. We recall that λ[1] and λ[2] will be a complex conjugate pair when the second-order system is underdamped. For the digital simulation of the second-order system to be
stable, both the Z transform poles must lie within the unit circle in the z plane. From Eq. (2.49) the stability criterion becomes
Here |1 + λh| – 1 represents a unit circle centered at the origin in the 1 + λh plane or, alternatively, a unit circle centered at -1 in the λh plane. This circle is shown in Figure 4.6. It is the
region within
Figure 2.5. Digital solutions using Euler integration for simulating the linear system given by = λx + f(t), or, T+ x + f(t) = 0 with f(t) = 0, x(0) = 1.
Figure 2.6. Stability region for Euler integration in simulating linear systems.
which λh must lie for all characteristic roots λ when simulating a linear system with Euler integration using a step size h if the resulting simulation is to be stable. Note that if the system being
simulated is stable, then λh for all roots of the continuous system must lie in the left-half plane. Since the unit circle in Figure 4.6 is much smaller than the entire left-half plane, it follows
that in general Euler integration exerts a destabilizing dynamic effect. For example, if the continuous system being simulated is an undamped second-order linear system (e.g., in Eq. (2.42) the
damping constant C = 0), then the two λh values will be a complex conjugate pair lying on the imaginary axis in Figure 2.6. Since this is outside the stability region, it means that the simulation
using Euler integration will be unstable, that is, the transient output data sequence will grow exponentially as time increases.
2.5 Some Additional Z-transform Formulas
Consider the Z transform of the data sequence {f[n][–][k]}, which represents a data sequence delayed in time by ksamples, i.e., kh seconds, with respect to the data sequence {f[n]}. From the Z
-transform definition in Eq. (2.1) it follows that
Thus the Z transform for a digital system representing a time delay of k samples is simply z^–^k. Again, it is interesting to compare these formulas with their counterparts for continuous systems.
The Laplace transform of f(t – t[d]) is given by
where F(s) is the Laplace transform of f(t). It follows that the Laplace transform of a continuous system representing a pure time delay, t[d], is e^–^st[d]. Table 2.1 presents the Z transforms for a
number of commonly encountered data sequences, including those already developed in this chapter.
Table 2.1
Z Transforms for Commonly Encountered Data Sequences
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What is Euler's Method Formula in Calculus? - Calculus Help
A common joke found all over the internet is a comment about teachers telling students who wouldn’t always have a calculator in their pocket. Obviously, the following comment is usually just an image
of a modern cellphone. While we can all laugh at the joke behind this dialogue, the teacher is still right. A calculator on your phone can’t read your mind and write the equation for you.
That’s a broad statement, but it’s true for all forms of math from simple word problems to complex calculus of algebra problems. You need to know how to write out the equation and which equation to
use. Understanding formulas like Euler’s Method are critical to solving many real-world problems and passing a college calculus class.
What Is Euler’s Method?
To understand Euler’s method, you’ll need to understand a few other math terms or formulas as well. The focus of this article is Euler’s method, so expect our definitions to be short and summarized.
If you need more information on any of the topics, a quick search online will explain anything we leave out. That said, our definitions should convey all the knowledge you need to understand this
On occasion, you may need to solve a differential equation where you can’t use separation of variables, or you may get specific conditions to satisfy. Some of the methods you learn to conquer these
types of equations simply won’t work. They should work, but the problem either ends up with an obviously incorrect answer and keep rolling over.
You can solve these types of differential equations using Euler’s method almost without failing. Why this isn’t the standard method taught in class is beyond us. That said, we’re going to help you
understand Euler’s Method and how to use it. Before we get into the examples and a better explanation, let’s define some terms we’re going to use. The key terms we may use include:
• Differential equations: This is an equation that uses a derivative to solve the equation for a function.
• Functions: This describes the relationship between the variables you put in and what comes out. It’s typically written out as f(x) or g(x). Each variable you put in has a relationship with the
answer. That relationship is the function. Sine and tangent are trigonometry functions.
• Derivatives: A derivative is usually used to define a slope or the change in a slope. You would use the slope formula to find the derivative of y = f(x).
• Separation of variables: This is typically how you solve differential equations by moving like terms to one side of an equation. The equation is generally noted by an equal sign.
• Tangent line: This is any line that comes in contact with a curve and mimics the curve where it runs into it. Small tangent lines are the basis of Euler’s Method. Tangent lines are usually
outside of a curve if that curve is a circle.
• Slope: This is any number that tells us the direction of a line and how steep that line may be along its path. A slope is equal to the rise divided by the run of the line.
Any other terms that require a definition will get defined in the same section. For now, those are the basic terms aside from understanding the basics of how to graph or work with fractions and
variables. If you’re reading this, we assume you know how to create a graph or work with variables on some level whether it’s this advanced or on a lower tier.
Using tangent lines, Euler’s Method helps you approximate the solution to any equation, almost, if you know the initial value. If the problem changes rapidly or changes direction more than once,
Euler’s Method may not work. That said, if the graph changes direction you’ll end up with multiple curves instead of one which rules out using Euler’s Method altogether.
Euler’s Method is one of three favorite ways to solve differential equations. As we mentioned earlier, you may be able to use separation of variables, or you might find slope fields are the best
method. Euler’s equation is must have a starting value or an assumed starting value in order to work. If you don’t have either of those things, refer to the other two methods we mentioned.
Using Euler’s method, we can see what goes on over a segment of our curve by intersecting or paralleling it with our tangent line. In short, Euler’s Method is used to see what goes on over a period
of time or change. For instance, it can approximate the slope of a curve or define how money market funds changed over time.
Using Euler’s Method, we can draw several tangent lines that meet a curve. Each line will match the curve in a different spot. By getting the approximate solution or equation where each line meets
the curve, we can begin to put together a picture of what is happening along our curve. In short, Euler’s Method is just a lot of tangent lines strung together to help us guess at what the cure is
doing as it travels.
Cons Of Using Euler’s Method
Since Euler’s Method only gives us approximate values, there may be room for error in the final result. If the curve is sharp and changes rapidly at any point, the solution we find at these sharp
turns using Euler’s Method may lack accuracy. However, this is precisely where Euler’s excels if you need to crudely calculate why something sped up like rates of deaths due to disease or sales over
a specified period.
While many people refer to Euler’s Method as a formula, and you can write a pseudo formula for it, it’s not a formula; it’s a method. It produces a solution without variables which may be considered
an approximate value of the current problem. It also causes some issues with math teachers if they want you to use a specific formula or method.
When Would I Use Euler’s Method Outside Of Class?
Differential equations end up being a big part of our lives whether directly or indirectly. Mathematicians sometimes work with biologists to develop programs for monitoring diseases or population
problems. Other mathematicians may work in banking or economics while some find their home in writing about things like Euler’s Method and exploring the different ways to solve differential
Differential equations help scientists monitor everything from the Moon’s orbit to the rate at which a glacier may melt. We could keep giving examples, but we believe you understand how vital
understanding this part of math is to your daily life and possibly your future career. Many professions beyond being a mathematician rely on approximations and solving differential equations.
You can check the Bureau of Labor Statistics for specific information on many job titles including those that use math a lot. Many careers at NASA require a firm grasp of applied mathematics and
calculus. It’s how they figure out how to fly by a planet without hitting it which is probably essential. Pharmaceutical science uses calculus and is required for many jobs at places like Bayer as
Aerospace engineers spend their time creating and designing satellites, space vessels, space stations, and other human-made objects that need to survive in space. They rely on math heavily to do
their jobs and ensure the safety of the equipment the build and the astronauts that us it.
Earthquake Safety Engineers rely on all forms of math to design better buildings and materials that withstand earthquakes. They also use math to build models to test their designs in a computer
simulation. The same modeling steps may also be used to determine the damage existing structures might suffer during an earthquake
Aside from the complicated math used in the professions, we mentioned above, Euler’s method has many practical applications and may help determine simpler things like the rate of flow for running
water. Similar methods and functions got used to help figure out how to shut off an oil leak that was 1,800 feet below the ocean’s waters.
It’s impressive how math affects our lives every minute of the day. Many people don’t realize its importance, but we’re hoping we’ve conveyed the message well. That said, most common functions and
formulas come preprogrammed into computers and calculators used in science-based fields. However, you still need to understand why and when to use them.
Some Final Notes
Euler’s Method is undoubtedly one of the most exciting formulas we’ve come across. Approximations usually find their home in less precise math problems. However, Euler’s method gets used across the
spectrum of physics and various disciplines that use calculus. We don’t always need to know changes at every point along a curve, or we may have a starting value, and that’s when Euler’s Method works
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Magic Circles
A set of magic circles is a numbering of the intersections of the Circles such that the sum over all intersections is the same constant for all circles. The above sets of three and four magic circles
have magic constants 14 and 39 (Madachy 1979).
See also Magic Graph, Magic Square
Madachy, J. S. Madachy's Mathematical Recreations. New York: Dover, p. 86, 1979.
© 1996-9 Eric W. Weisstein
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Spoons with Style
When you see what cool things people can make out of plain old plastic spoons, you might never throw out a spoon again. On one site, an artist used the scoop parts of 125 plastic spoons to make a
lamp that looks like a pineapple. On another page, someone else glued together 250 spoon scoops to make an 18-inch-wide flower face around a clock. Now we can decorate our houses while saving some
trash from the dump, too.
Wee ones: On the flower clock, which row uses more spoons, a circular ring near the middle or a ring near the outer edge?
Little kids: If the pineapple lamp uses 5 spoons in the top row and 1 more than that in the next, how many does it use in that 2nd row? Bonus: How many does it use in both rows together?
Big kids: If the flower clock uses 22 spoons in the center ring, and each new ring uses 3 more than the one inside it, can a ring have 34 spoons in it? Bonus: If the pineapple uses 100 spoons, with 8
spoons in as many rows as possible, how many full rows of 8 does it have?
Wee ones: The ring near the outside.
Little kids: 6 spoons. Bonus: 11 spoons, since it’s 5 + 6.
Big kids: Yes! There will be 25, 28, 31, and 34 spoons in the next 4 rows. Bonus: 12 rows, which brings us to 96 spoons plus 4 left in the last row.
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Generate MATLAB Jacobian functions for extended Kalman filter using automatic differentiation
Since R2023a
fcnStateJac = generateJacobianFcn(obj,'state',Us1,...,Usn) generates the state transition Jacobian function for an extended Kalman filter (EKF) using the automatic differentiation technique.
This function generates two MATLAB^® function files in the current folder:
• stateTransitionJacobianFcn.m — The generated state transition Jacobian function
• stateTransitionJacobianFcnAD.m — A helper function that uses automatic differentiation to generate the state transition Jacobian function
fcnStateJac is a handle to an anonymous function that calls stateTransitionJacobianFcn.m using the input extendedKalmanFilter object (obj), the additional input arguments passed to the predict
function of obj (Us1,...,Usn), and the constants used internally to compute the state transition Jacobian.
To use this Jacobian in the EKF object, specify fcnStateJac in the StateTransitionJacobianFcn property of the object. For example:
obj.StateTransitionJacobianFcn = fcnStateJac;
fcnMeasurementJac = generateJacobianFcn(obj,'measurement',Um1,...,Umn) generates the measurement Jacobian function for an extended Kalman filter (EKF) using the automatic differentiation technique.
This function generates two MATLAB function files in the current folder:
• measurementJacobianFcn.m — The generated measurement Jacobian function
• measurementJacobianFcnAD.m — A helper function that uses automatic differentiation to generate the measurement Jacobian function
fcnMeasurementJac is a handle to an anonymous function that calls measurementJacobianFcn.m using the input extendedKalmanFilter object (obj), the additional input arguments passed to the correct
function of obj (Um1,...,Umn), and the constants used internally to compute the measurement Jacobian.
To use this Jacobian in the EKF object, specify fcnMeasurementJac in the MeasurementJacobianFcn property of the object. For example:
obj.MeasurementJacobianFcn = fcnMeasurementJac;
[___,constants] = generateJacobianFcn(___) also returns the constants used to compute the Jacobian function. You can return the constants for either Jacobian function.
[___] = generateJacobianFcn(___,FileName=filename) specifies the name of the generated Jacobian function files and the folder location in which to generate them.
Specify Jacobians for State and Measurement Functions
Create an extended Kalman filter (EKF) object for a van der Pol oscillator with two states and one output. Use the previously written and saved state transition and measurement functions,
vdpStateFcn.m and vdpMeasurementFcn.m. Specify the initial state values for the two states as [2;0].
obj = extendedKalmanFilter(@vdpStateFcn,@vdpMeasurementFcn,[2;0]);
The extended Kalman filter algorithm uses Jacobians of the state transition and measurement functions for state estimation. You can write and save the Jacobian functions and provide them as function
handles to the EKF object. For this object, use the previously written and saved functions vdpStateJacobianFcn.m and vdpMeasurementJacobianFcn.m.
obj.StateTransitionJacobianFcn = @vdpStateJacobianFcn;
obj.MeasurementJacobianFcn = @vdpMeasurementJacobianFcn;
If the Jacobian functions are not available, you can generate them using automatic differentiation.
obj.StateTransitionJacobianFcn = generateJacobianFcn(obj,'state');
obj.MeasurementJacobianFcn = generateJacobianFcn(obj,'measurement');
If you do not specify the Jacobians of the functions, the software numerically computes them. This numerical computation can result in increased processing time and numerical inaccuracy of the state
Generate Jacobians for State and Measurement Functions with Additional Input Arguments
Consider a nonlinear system with input u whose state x and measurement y evolve according to these state transition and measurement equations:
The process noise w of the system is additive, while the measurement noise v is nonadditive.
Create the state transition function and measurement function for the system. Specify the functions with an additional input u.
f = @(x,u)(sqrt(x+u));
h = @(x,v,u)(x+2*u+v^2);
f and h are function handles to the anonymous functions that store the state transition and measurement functions, respectively. In the measurement function, because the measurement noise is
nonadditive, v is also specified as an input. Note that v is specified as an input before the additional input u.
Create an extended Kalman filter (EKF) object for estimating the state of the nonlinear system using the state transition and measurement functions. Specify the initial value of the state as 1 and
the measurement noise as nonadditive.
obj = extendedKalmanFilter(f,h,1,"HasAdditiveMeasurementNoise",false);
Generate the state transition and measurement Jacobian functions of obj using automatic differentiation. Because both the state transition and measurement functions have the additional input u, pass
an arbitrary value of the same type and size as u to generateJacobianFcn.
obj.StateTransitionJacobianFcn = generateJacobianFcn(obj,"state",0.2);
obj.MeasurementJacobianFcn = generateJacobianFcn(obj,"measurement",0.2);
Generate Jacobian Functions to Specific Folder
Generate the state transition and measurement Jacobian functions of an EKF object using automatic differentiation techniques. Save the Jacobian function files to a nondefault location.
Create an extended Kalman filter (EKF) object for a van der Pol oscillator with two states and one output. Use the previously written and saved state transition and measurement functions,
vdpStateFcn.m and vdpMeasurementFcn.m. Specify the initial state values for the two states as [2;0].
obj = extendedKalmanFilter(@vdpStateFcn,@vdpMeasurementFcn,[2;0]);
Create a folder in which to generate the Jacobian function files. Add the folder to the path.
% Change to a folder where you have write permission
CWD = pwd;
c = onCleanup(@()cd(CWD));
% create a folder of the desired name
folder = 'jacobians';
[status,msg,msgID] = mkdir(folder);
% add the new folder to MATLAB path
Generate the Jacobian function files to the new folder. Display the generated files.
fcnStateJac = generateJacobianFcn(obj,"state",...
fcnMeasurementJac = generateJacobianFcn(obj,"measurement",...
ans = 4x1 cell
{'mJac.m' }
{'sJac.m' }
Add the generated function handles to the EKF object.
obj.StateTransitionJacobianFcn = @fcnStateJac;
obj.MeasurementJacobianFcn = @fcnMeasurementJac;
Input Arguments
obj — Extended Kalman filter
extendedKalmanFilter object
Extended Kalman filter, specified as an extendedKalmanFilter object.
Us1,...,Usn — Additional input arguments to state transition function
additional input arguments used by predict function
Additional input arguments to the state transition function. The state transition function f is specified in the StateTransitionFcn property of obj. Specify the same additional input arguments used
by the predict function of obj. The input arguments can be of any type.
Um1,...,Umn — Additional input arguments to measurement function
additional input arguments used by correct function
Additional input arguments to the measurement function. The measurement function h is specified in the MeasurementFcn property of obj. Specify the same additional input arguments used by the correct
function of obj. The input arguments can be of any type.
filename — Jacobian function file name
character vector
Jacobian function file name, specified as a character vector. The function applies this naming convention to the generated Jacobian function files:
If filename contains an absolute or relative folder path, then generateJacobianFcn generates the files to that folder. If filename is unable to find the specified path, then generateJacobianFcn
returns an error.
If the folder that generateJacobianFcn generates the files to is not on the MATLAB path, then generateJacobianFcn returns the Jacobian function handle as []. If this issue occurs, to generate the
function handle, follow these steps:
To use this procedure, you must return the constants output argument in your call to generateJacobianFcn.
1. Add the folder to the MATLAB search path, or move the generated function files to a folder on the MATLAB search path.
2. Obtain the name part of filename to use in the function handle name, FcnName.
[~,FcnName] = fileparts(filename);
3. Generate the function handle using this name, where constants is the second output argument of your previous call to generateJacobianFcn.
fcnJac = @(varargin)FcnName(varargin{:},constants);
You can then specify fcnJac handle in the appropriate Jacobian function property of obj.
For more details on working with the MATLAB search path, see What Is the MATLAB Search Path?
Example: FileName='measjac' generates measjac.m and measjacAD.m files in the current folder.
Example: FileName='AD/measjac' generates measjac.m and measjacAD.m files in the AD subfolder of the current folder.
Example: FileName='C:/AD/measjac' generates ekfjac.m and measjacAD.m files in the C:/AD folder.
Data Types: char
Output Arguments
constants — Additional constants used to compute Jacobian
length-N cell array of constant values
Additional constants used to compute the Jacobian, returned as a length-N cell array of constant values, where N is the number of constants. If the Jacobian function requires no constants, then
constants is an empty cell array.
If you specify filename, and the folder specified in filename is not on the MATLAB search path, you can use these constants to manually construct the function handle.
fcnStateJac — Jacobian of state transition function
function handle
Jacobian of the state transition function, returned as a function handle.
• If obj.HasAdditiveMeasurementNoise = true, then fcnStateJac has this signature:
dx = fcnStateJac(obj,Us1,...,Usn,constants)
• If obj.HasAdditiveMeasurementNoise = false, then fcnStateJac has this signature:
[dx,dw] = fcnStateJac(obj,w,Us1,...,Usn,constants)
In these function signatures:
• dx is the Jacobian of the predicted state with respect to the previous state.
• dw is the Jacobian of the predicted state with respect to the process noise elements.
• w is the process noise variable.
• Us1,...,Usn are the additional input arguments used by the predict function of obj.
• constants are the additional constants used to compute the Jacobians.
fcnMeasurementJac — Jacobian of measurement function
function handle
Jacobian of the measurement function, returned as a function handle.
• If obj.HasAdditiveMeasurementNoise = true, then fcnMeasurementJac has this signature:
dy = fcnMeasurementJac(obj,Um1,...,Umn,constants)
• If obj.HasAdditiveMeasurementNoise = false, then fcnMeasurementJac has this signature:
[dy,dv] = fcnMeasurementJac(obj,v,Um1,...,Umn,constants)
In these function signatures:
• dy is the Jacobian of the measurement function with respect to the state.
• dv is the Jacobian of the measurement function with respect to the measurement noise.
• v is the measurement noise variable.
• Um1,...,Umn are the additional input arguments used by the correct function of obj.
• constants are the additional constants used to compute the Jacobians.
• Automatic differentiation currently supports only a limited set of mathematical operations, which are described in Supported Operations for Optimization Variables and Expressions (Optimization
Toolbox). If your original state transition or measurement function uses operations or functions that are not in the list, or has if-else statements or loops, generateJacobianFcn terminates with
an error.
• To generate Jacobian functions, do not preallocate any optimization variable in the original function. For example, suppose you try to generate Jacobians from a function containing the following
dxdt = zeros(2,1);
dxdt(1) = x(1)*x(2);
dxdt(2) = x(1)/x(2);
This code results in the following error.
Unable to perform assignment because value of type
is not convertible to 'double'.
Instead, use this code.
dxdt = [x(1)*x(2); x(1)/x(2)];
• Specifying the state transition and measurement functions in obj as files in the current folder or in a folder on the MATLAB path is recommended. While handles to local functions are supported
for Jacobian function generation, they are not supported for generation of C/C++ deployment code. For more information on local functions, see Local Functions.
Version History
Introduced in R2023a
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Calculating the Density Temperature Correction
The first step Sednterp uses to calculate the density of a buffer is to calculate the density of the buffer at 20 degrees C. Then, the density is corrected for temperature assuming that water and the
isotopes of water are the predominant components in the buffer. While this is true for solutions containing moderate amounts of other components, significant errors may be introduced for solutions
containing high solute concentrations.
For pure water, equation 14 from the CRC Handbook of Chemistry and Physics (Ref. 44) is used.
Equation 14.
where T is the temperature in degrees C and the multiplicative factor converts the units from kg/m^3 to g/ml. Densities calculated with equation 14 are good to at least 5 decimal places over the
0-100 C range.
For the isotopes of water, equation 12 derived from the methods described by Steckel and Szapiro(Ref. 45): is used as the temperature correction.
Equation 12.
where ρT is the density at temperature T in C, ρmax is the maximum density (observed at the maximum temperature), is the temperature difference between that of the experiment and the temperature of
maximum density, and F is an empirical constant. Equation 12 may be used for all water species by substituting in the appropriate values from the following table. Note that in cases where O-18 is
present, isotopic purity is about 98%, and it is 100% when D is present. Values for H2O reflect the natural isotopic abundance of O-18 and D. Likewise, O and H reflect natural isotopic abundance in
D2O and H2O18. Equation 12 is valid from the melting temperature to approximately 80 C. Simple mixing rules are used when different isotopes of water are combined. (Ref. 45) The values used in
equation 12 are loaded in the phyconst database and are listed here also:
The final correction for temperature is performed by multiplying the density of the buffer at 20 C by the ratio of the water component(s) densities at the experimental temperature and 20 C.
Equation 15.
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How to Measure your product success : Choose your own KPI - heysalsalHow to Measure your product success : Choose your own KPI - heysalsal
We know that people are talking about the success of the product we designed. But how do you see how successful a product is? then the answer is a KPI or product metric.
The product has different KPIs. depending on what type of KPI or metric we use to calculate the success of our product. Therefore, it is best to ask several things before defining the measures to
Which metrics are suitable for our product?
The range of success measures for the products we develop is certainly different. As social media products, they are definitely rated based on daily active users or DAUs. At the same time, for
e-commerce products, of course, how often is the bounce or traffic of the user.
Baca Juga : Product Management 101: First Step into Product Management Field
There are several ways to measure a KPI using its metrics.
Based on Company Revenue
Monthly Recurring Revenue Metrics
Monthly Recurring Revenue is a way to calculate product success based on revenue.
Calculation formula
*If the company has a price level, such as annually, quarterly or semi-annually, you can divide it to get a monthly price.
Average Revenue per User Metric
ARPU is a derivative of MRU that calculates the purchasing potential of our users. Usually used in online stores or online stores. ARPU is a derivative of MRU that is calculated
Calculation formula
*ARPU is a measure used when the price of a product increases, decreases or changes. This metric can also be used to compare with other competitors.
Customer Lifetime Value
Another measure of product success is CLTV, or Customer Lifetime Value. This is also a derivative of the previous metric. CLTV is calculated based on the probability that the user will be able to
complete the purchase. Using this metric to calculate product success provides more comprehensive information than simply using average revenue per user.
*Knowing the number of purchases by users can help to use marketing tools and still try to make a profit
Customer acquisition coast
Metrics to calculate money spent for marketing purposes. . This is of course related to the previous metric.
Several formulas can be used and the following are examples.
The above metrics are metrics based on company revenue. But what if the product is still new? Perhaps user behavior/habits metrics can be used to calculate product success.
Based by User Behavior
Calculation of user habits related to our products can of course be simplified with the help of platforms that provide these services. for example Google Analytics, HotJar or Appflyers.
The following calculations are used to determine user behavior:
Percentage of active users
The number of users downloading our application may not be a good estimate. Because everyone can download, but not everyone uses the products that we bring to the market. Therefore, the percentage of
active users is a metric that tells you how many users there are per day or per month out of all our users.
*Knowing the average day per month tells you where you are in the product lifecycle
Session duration
This metric is used to calculate how long users stay in the product we are planning
This metric allows us to calculate how long it takes the user to complete the product.
Bounce rate
Bounce rate is the condition where a user leaves after viewing just one page. It can show if users are interested in the product/app we are designing.
Knowing this may not provide answers to problems encountered by users/visitors. But at least we’ll know there’s something wrong with the screen we’ve created.
Knowing what is being measured naturally informs what we should develop with our products. Therefore, use the appropriate matrix for our product.
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what will be next command line for putting lat long value n figure?
Here matrix of 130*110 want to give latitude and longitude value from
what will be next command line for putting lat long value n figure?
outside the figure value of lat/long is not there I want to put lat/long value there?
3 Comments
1 view (last 30 days)
what will be next command line for putting lat long value n figure?
Can you please give more detail? Do you want to put lines at those values, or do you want to write text at those values?
Your current axes run from 1:8 (I think). How do you want to match those values up to your lat and lon?
However you do this, you are trying to put a lot of "ink" on that figure, which I think will obscure what you are doing.
Image Analyst on 25 Aug 2013
Edited: Image Analyst on 25 Aug 2013
S, you don't need to put "matlab" or "matlab code" as tags since every question here is on matlab and matlab code. Read a guide to tags
Do you just want to change the tick marks outside the image so that they read the "real world" lat/lon numbers instead of pixel values?
Answers (1)
This is completely explained, with several examples, in the documentation. For example:
x = -pi:.1:pi;
y = sin(x);
Browse, in the help, to MATLAB->Graphics->formatting and Annotation->Coordinate System->Setting Axis Parameters. Or search for ticks and select "Setting Axis Parameters" from the list.
5 Comments
Walter Roberson on 30 Aug 2013
The Mapping Toolbox is optional extra-cost. I do not recall the cost at the moment. Your profile hints you are at a university. If you are a student you can add the Mapping Toolbox on to your Student
Version license for about $US30. If you are faculty or staff at the university, you can get the Mapping Toolbox at Academic pricing.
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Chapter 1 of The Principles of Quantum Mechanics
1. The need for a quantum theory
CLASSICAL mechanics has been developed continuously from the time of Newton and applied to an ever-widening range of dynamical systems, including the electromagnetic field in interaction with
matter. The underlying ideas and the laws governing their application form a simple and elegant scheme, which one would be inclined to think could not be seriously modified without having all its
attractive features spoilt. Nevertheless it has been found possible to set up a new scheme, called quantum mechanics, which is more suitable for the description of phenomena on the atomic scale
and which is in some respects more elegant and satisfying than the classical scheme. This possibility is due to the changes which the new scheme involves being of a very profound character and
not clashing with the features of the classical theory that make it so attractive, as a result of which all these features can be incorporated in the new scheme. The necessity for a departure
from classical mechanics is clearly shown by experimental results. In the first place the forces known in classical electrodynamics are inadequate for the explanation of the remarkable stability
of atoms and molecules, which is necessary in order that materials may have any definite physical and chemical properties at all. The introduction of new hypothetical forces will not save the
situation, since there exist general principles of classical mechanics, holding for all kinds of forces, leading to results in direct disagreement with observation. For example, if an atomic
system has its equilibrium disturbed in any way and is then left alone, it will be set in oscillation and the oscillations will get impressed on the surrounding electromagnetic field, so that
their frequencies may be observed with a spectroscope. Now whatever the laws of force governing the equilibrium, one would expect to be able to include the various frequencies in a scheme
comprising certain fundamental frequencies and their harmonics. This is not observed to be the case. Instead, there is observed a new and unexpected connexion between the frequencies, called
Ritz's Combination Law of Spectroscopy, according to which all the frequencies can be expressed as differences between certain terms, the number of terms being much less than the number of
frequencies. This law is quite unintelligible from the classical standpoint. One might try to get over the difficulty without departing from classical mechanics by assuming each of the
spectroscopically observed frequencies to be a fundamental frequency with its own degree of freedom, the laws of force being such that the harmonic vibrations do not occur. Such a theory will not
do, however, even apart from the fact that it would give no explanation of the Combination Law, since it would immediately bring one into conflict with the experimental evidence on specific
heats. Classical statistical mechanics enables one to establish a general connexion between the total number of degrees of freedom of an assembly of vibrating systems and its specific heat. If
one assumes all the spectroscopic frequencies of an atom to correspond to different degrees of freedom, one would get a specific heat for any kind of matter very much greater than the observed
value. In fact the observed specific heats at ordinary temperatures are given fairly well by a theory that takes into account merely the motion of each atom as a whole and assigns no internal
motion to it at all. This leads us to a new clash between classical mechanics and the results of experiment. There must certainly be some internal motion in an atom to account for its spectrum,
but the internal degrees of freedom, for some classically inexplicable reason, do not contribute to the specific heat. A similar clash is found in connexion with the energy of oscillation of the
electromagnetic field in a vacuum. Classical mechanics requires the specific heat corresponding to this energy to be infinite, but it is observed to be quite finite. A general conclusion from
experimental results is that oscillations of high frequency do not contribute their classical quota to the specific heat. As another illustration of the failure of classical mechanics we may
consider the behaviour of light. We have, on the one hand, the phenomena of interference and diffraction, which can be explained only on the basis of a wave theory; on the other, phenomena such
as photo-electric emission and scattering by free electrons, which show that light is composed of small particles. These particles, which are called photons, have each a definite energy and
momentum, depending on the frequency of the light, and appear to have just as real an existence as electrons, or any other particles known in physics. A fraction of a photon is never observed.
Experiments have shown that this anomalous behaviour is not peculiar to light, but is quite general. All material particles have wave properties, which can be exhibited under suitable conditions.
We have here a very striking and general example of the breakdown of classical mechanics — not merely an inaccuracy in its laws of motion, but an inadequacy of its concepts to supply us with a
description of atomic events. The necessity to depart from classical ideas when one wishes to account for the ultimate structure of matter may be seen, not only from experimentally established
facts, but also from general philosophical grounds. In a classical explanation of the constitution of matter, one would assume it to be made up of a large number of small constituent parts and
one would postulate laws for the behaviour of these parts, from which the laws of the matter in bulk could be deduced. This would not complete the explanation, however, since the question of the
structure and stability of the constituent parts is left untouched. To go into this question, it becomes necessary to postulate that each constituent part is itself made up of smaller parts, in
terms of which its behaviour is to be explained. There is clearly no end to this procedure, so that one can never arrive at the ultimate structure of matter on these lines. So long as big and
small are merely relative concepts, it is no help to explain the big in terms of the small. It is therefore necessary to modify classical ideas in such a way as to give an absolute meaning to
size. At this stage it becomes important to remember that science is concerned only with observable things and that we can observe an object only by letting it interact with some outside
influence. An act of observation is thus necessarily accompanied by some disturbance of the object observed. We may define an object to be big when the disturbance accompanying our observation of
it may be neglected, and small when the disturbance cannot be neglected. This definition is in close agreement with the common meanings of big and small. It is usually assumed that, by being
careful, we may cut down the disturbance accompanying our observation to any desired extent. The concepts of big and small are then purely relative and refer to the gentleness of our means of
observation as well as to the object being described. In order to give an absolute meaning to size, such as is required for any theory of the ultimate structure of matter, we have to assume that
there is a limit to the fineness of our powers of observation and the smallness of the accompanying disturbance a limit which is inherent in the nature of things and can never be surpassed by
improved technique or increased skill on the part of the observer. If the object under observation is such that the unavoidable limiting disturbance is negligible, then the object is big in the
absolute sense and we may apply classical mechanics to it. If, on the other hand, the limiting disturbance is not negligible, then the object is small in the absolute sense and we require a new
theory for dealing with it. A consequence of the preceding discussion is that we must revise our ideas of causality. Causality applies only to a system which is left undisturbed. If a system is
small, we cannot observe it without producing a serious disturbance and hence we cannot expect to find any causal connexion between the results of our observations. Causality will still be
assumed to apply to undisturbed systems and the equations which will be set up to describe an undisturbed system will be differential equations expressing a causal connexion between conditions at
one time and conditions at a later time. These equations will be in close correspondence with the equations of classical mechanics, but they will be connected only indirectly with the results of
observations. There is an unavoidable indeterminacy in the calculation of observational results, the theory enabling us to calculate in general only the probability of our obtaining a particular
result when we make an observation.
2. The polarization of photons
The discussion in the preceding section about the limit to the gentleness with which observations can be made and the consequent indeterminacy in the results of those observations does not
provide any quantitative basis for the building up of quantum mechanics. For this purpose a new set of accurate laws of nature is required. One of the most fundamental and most drastic of these
is the Principle of Superposition of States. We shall lead up to a general formulation of this principle through a consideration of some special cases, taking first the example provided by the
polarization of light. It is known experimentally that when plane-polarized light is used for ejecting photo-electrons, there is a preferential direction for the electron emission. Thus the
polarization properties of light are closely connected with its corpuscular properties and one must ascribe a polarization to the photons. One must consider, for instance, a beam of light
plane-polarized in a certain direction as consisting of photons each of which is plane-polarized in that direction and a beam of circularly polarized light as consisting of photons each
circularly polarized. Every photon is in a certain state of polarization, as we shall say. The problem we must now consider is how to fit in these ideas with the known facts about the resolution
of light into polarized components and the recombination of these components. Let us take a definite case. Suppose we have a beam of light passing through a crystal of tourmaline, which has the
property of letting through only light plane-polarized perpendicular to its optic axis. Classical electrodynamics tells us what will happen for any given polarization of the incident beam. If
this beam is polarized perpendicular to the optic axis, it will all go through the crystal; if parallel to the axis, none of it will go through; while if polarized at an angle α to the axis, a
fraction sin^2α will go through. How are we to understand these results on a photon basis? A beam that is plane-polarized in a certain direction is to be pictured as made up of photons each
plane-polarized in that direction. This picture leads to no difficulty in the cases when our incident beam is polarized perpendicular or parallel to the optic axis. We merely have to suppose that
each photon polarized perpendicular to the axis passes unhindered and unchanged through the crystal, while each photon polarized parallel to the axis is stopped and absorbed. A difficulty arises,
however, in the case of the obliquely polarized incident beam. Each of the incident photons is then obliquely polarized and it is not clear what will happen to such a photon when it reaches the
tourmaline. A question about what will happen to a particular photon under certain conditions is not really very precise. To make it precise one must imagine some experiment performed having a
bearing on the question and inquire what will be the result of the experiment. Only questions about the results of experiments have a real significance and it is only such questions that
theoretical physics has to consider. In our present example the obvious experiment is to use an incident beam consisting of only a single photon and to observe what appears on the back side of
the crystal. According to quantum mechanics the result of this experiment will be that sometimes one will find a whole photon, of energy equal to the energy of the incident photon, on the back
side and other times one will find nothing. When one finds a whole photon, it will be polarized perpendicular to the optic axis. One will never find only a part of a photon on the back side. If
one repeats the experiment a large number of times, one will find the photon on the back side in a fraction sin^2α of the total number of times. Thus we may say that the photon has a probability
sin^2α of passing through the tourmaline and appearing on the back side polarized perpendicular to the axis and a probability cos^2α of being absorbed. These values for the probabilities lead to
the correct classical results for an incident beam containing a large number of photons. In this way we preserve the individuality of the photon in all cases. We are able to do this, however,
only because we abandon the determinacy of the classical theory. The result of an experiment is not determined, as it would be according to classical ideas, by the conditions under the control of
the experimenter. The most that can be predicted is a set of possible results, with a probability of occurrence for each. The foregoing discussion about the result of an experiment with a single
obliquely polarized photon incident on a crystal of tourmaline answers all that can legitimately be asked about what happens to an obliquely polarized photon when it reaches the tourmaline.
Questions about what decides whether the photon is to go through or not and how it changes its direction of polarization when it does go through cannot be investigated by experiment and should be
regarded as outside the domain of science. Nevertheless some further description is necessary in order to correlate the results of this experiment with the results of other experiments that might
be performed with photons and to fit them all into a general scheme. Such further description should be regarded, not as an attempt to answer questions outside the domain of science, but as an
aid to the formulation of rules for expressing concisely the results of large numbers of experiments. The further description provided by quantum mechanics runs as follows. It is supposed that a
photon polarized obliquely to the optic axis may be regarded as being partly in the state of polarization parallel to the axis and partly in the state of polarization perpendicular to the axis.
The state of oblique polarization may be considered as the result of some kind of superposition process applied to the two states of parallel and perpendicular polarization. This implies a
certain special kind of relationship between the various states of polarization, a relationship similar to that between polarized beams in classical optics, but which is now to be applied, not to
beams, but to the states of polarization of one particular photon. This relationship allows any state of polarization to be resolved into, or expressed as a superposition of, any two mutually
perpendicular states of polarization. When we make the photon meet a tourmaline crystal, we are subjecting it to an observation. We are observing whether it is polarized parallel or perpendicular
to the optic axis. The effect of making this observation is to force the photon entirely into the state of parallel or entirely into the state of perpendicular polarization. It has to make a
sudden jump from being partly in each of these two states to being entirely in one or other of them. Which of the two states it will jump into cannot be predicted, but is governed only by
probability laws. If it jumps into the parallel state it gets absorbed and if it jumps into the perpendicular state it passes through the crystal and appears on the other side preserving this
state of polarization.
3. Interference of photons
In this section we shall deal with another example of superposition. We shall again take photons. but shall be concerned with their position in space and their momentum instead of their
polarization. If we are given a beam of roughly monochromatic light, then we know something about the location and momentum of the associated photons. We know that each of them is located
somewhere in the region of space through which the beam is passing and has a momentum in the direction of the beam of magnitude given in terms of the frequency of the beam by Einstein's
photo-electric law — momentum equals frequency multiplied by a universal constant, When we have such information about the location and momentum of a photon we shall say that it is in a definite
translational state. We shall discuss the description which quantum mechanics provides of the interference of photons. Let us take a definite experiment demonstrating interference. Suppose we
have a beam of light which is passed through some kind of interferometer, so that it gets split up into two components and the two components are subsequently made to interfere. We may, as in the
preceding section, take an incident beam consisting of only a single photon and inquire what will happen to it as it goes through the apparatus. This will present to us the difficulty of the
conflict between the wave and corpuscular theories of light in an acute form. Corresponding to the description that we had in the case of the polarization, we must now describe photon as going
partly into each of the two components into which the incident beam is split. The photon is then, as we may say, in a translational state given by the superposition of the two translational
states associated with the two components. We are thus led to a generalization of the term 'translational state' applied to a photon. For a photon to be in a definite translational state it need
not be associated with one single beam of light, but may be associated with two or more beams of light which are the components into which one original beam has been split. † In the accurate
mathematical theory each translational state is associated with one of the wave functions of ordinary wave optics, which wave function may describe either a single beam or two or more beams into
which one original beam has been split. Translational states are thus superposable in a similar way to wave functions.
Let us consider now what happens when we determine the energy in one of the components. The result of such a determination must be either the whole photon or nothing at all. Thus the photon must
change suddenly from being partly in one beam and partly in the other to being entirely in one of the beams. This sudden change is due to the disturbance in the translational state of the photon
which the observation necessarily makes. It is impossible to predict in which of the two beams the photon will be found. Only the probability of either result can be calculated from the previous
distribution of the photon over the two beams. One could carry out the energy measurement without destroying the component beam by, for example, reflecting the beam from a movable mirror and
observing the recoil. Our description of the photon allows us to infer that, after such an energy measurement, it would not be possible to bring about any interference effects between the two
components. So long as the photon is partly in one beam and partly in the other, interference can occur when the two beams are superposed, but this possibility disappears when the photon is
forced entirely into one of the beams by an observation. The other beam then no longer enters into the description of the photon, so that it counts as being entirely in the one beam in the
ordinary way for any experiment that may subsequently be performed on it. On these lines quantum mechanics is able to effect a reconciliation of the wave and corpuscular properties of light. The
essential point is the association of each of the translational states of a photon with one of the wave functions of ordinary wave optics. The nature of this association cannot be pictured on a
basis of classical mechanics, but is something entirely new. It would be quite wrong to picture the photon and its associated wave as interacting in the way in which particles and waves can
interact in classical mechanics. The association can be interpreted only statistically, the wave function giving us information about the probability of our finding the photon in any particular
place when we make an observation of where it is. Some time before the discovery of quantum mechanics people realized that the connexion between light waves and photons must be of a statistical
character. What they did not clearly realize, however, was that the wave function gives information about the probability of one photon being in a particular place and not the probable number of
photons in that place. The importance of the distinction can be made clear in the following way. Suppose we have a beam of light consisting of a large number of photons split up into two
components of equal intensity. On the assumption that the intensity of a beam is connected with the probable number of photons in it, we should have half the total number of photons going into
each component. If the two components are now made to interfere, we should require a photon in one component to be able to interfere with one in the other. Sometimes these two photons would have
to annihilate one another and other times they would have to produce four photons. This would contradict the conservation of energy. The new theory, which connects the wave function with
probabilities for one photon, gets over the difficulty by making each photon go partly into each of the two components. Each photon then interferes only with itself. Interference between two
different photons never occurs. The association of particles with waves discussed above is not restricted to the case of light, but is, according to modern theory, of universal applicability. All
kinds of particles are associated with waves in this way and conversely all wave motion is associated with particles. Thus all particles can be made to exhibit interference effects and all wave
motion has its energy in the form of quanta. The reason why these general phenomena are not more obvious is on account of a law of proportionality between the mass or energy of the particles and
the frequency of the waves, the coefficient being such that for waves of familiar frequencies the associated quanta are extremely small, while for particles even as light as electrons the
associated wave frequency is so high that it is not easy to demonstrate interference.
4. Superposition and Indeterminacy
The reader may possibly feel dissatisfied with the attempt in the two preceding sections to fit in the existence of photons with the classical theory of light. He may argue that a very strange
idea has been introduced — the possibility of a photon being partly in each of two states of polarization, or partly in each of two separate beams — but even with the help of this strange idea no
satisfying picture of the fundamental single-photon processes has been given. He may say further that this strange idea did not provide any information about experimental results for the
experiments discussed, beyond what could have been obtained from an elementary consideration of photons being guided in some vague way by waves. What, then, is the use of the strange idea? In
answer to the first criticism it may be remarked that the main object of physical science is not the provision of pictures, but is the formulation of laws governing phenomena and the application
of these laws to the discovery of new phenomena. If a picture exists, so much the better; but whether a picture exists or not is a matter of only secondary importance. In the case of atomic
phenomena no picture can be expected to exist in the usual sense of the word 'picture', by which is meant a model functioning essentially on classical lines. One may, however, extend the meaning
of the word `picture' to include any way of looking at the fundamental laws which makes their self-consistency obvious. With this extension, one may gradually acquire a picture of atomic
phenomena by becoming familiar with the laws of the quantum theory. With regard to the second criticism, it may be remarked that for many simple experiments with light, an elementary theory of
waves and photons connected in a vague statistical way would be adequate to account for the results. In the case of such experiments quantum mechanics has no further information to give. In the
great majority of experiments, however, the conditions are too complex for an elementary theory of this kind to be applicable and some more elaborate scheme, such as is provided by quantum
mechanics, is then needed. The method of description that quantum mechanics gives in the more complex cases is applicable also to the simple cases and although it is then not really necessary for
accounting for the experimental results, its study in these simple cases is perhaps a suitable introduction to its study in the general case. There remains an overall criticism that one may make
to the whole scheme, namely, that in departing from the determinacy of the classical theory a great complication is introduced into the description of Nature, which is a highly undesirable
feature. This complication is undeniable, but it is offset by a great simplification, provided by the general principle of superposition of states, which we shall now go on to consider. But first
it is necessary to make precise the important concept of a `state' of a general atomic system. Let us take any atomic system, composed of particles or bodies with specified properties (mass,
moment of inertia, etc.) interacting according to specified laws of force. There will be various possible motions of the particles or bodies consistent with the laws of force. Each such motion is
called a state of the system. According to classical ideas one could specify a state by giving numerical values to all the coordinates and velocities of the various component parts of the system
at some instant of time, the whole motion being then completely determined. Now the argument of pp. 3 and 4 shows that we cannot observe a small system with that amount of detail which classical
theory supposes. The limitation in the power of observation puts a limitation on the number of data that can be assigned to a state. Thus a state of an atomic system must be specified by fewer or
more indefinite data than a complete set of numerical values for all the coordinates and velocities at some instant of time. In the case when the system is just a single photon, a state would be
completely specified by a given translational state in the sense of § 3 together with a given state of polarization in the sense of § 2. A state of a system may be defined as an undisturbed
motion that is restricted by as many conditions or data as are theoretically possible without mutual interference or contradiction. In practice the conditions could be imposed by a suitable
preparation of the system, consisting perhaps in passing it through various kinds of sorting apparatus, such as slits and polarimeters, the system being left undisturbed after the preparation.
The word 'state' may be used to mean either the state at one particular time (after the preparation), or the state throughout the whole of time after the preparation. To distinguish these two
meanings, the latter will be called a 'state of motion' when there is liable to be ambiguity. The general principle of superposition of quantum mechanics applies to the states, with either of the
above meanings, of any one dynamical system. It requires us to assume that between these states there exist peculiar relationships such that whenever the system is definitely in one state we can
consider it as being partly in each of two or more other states. The original state must be regarded as the result of a kind of superposition of the two or more new states, in a way that cannot
be conceived on classical ideas. Any state may be considered as the result of a superposition of two or more other states, and indeed in an infinite number of ways. Conversely any two or more
states may be superposed to give a new state. The procedure of expressing a state as the result of superposition of a number of other states is a mathematical procedure that is always
permissible, independent of any reference to physical conditions, like the procedure of resolving a wave into Fourier components. Whether it is useful in any particular case, though, depends on
the special physical conditions of the problem under consideration. In the two preceding sections examples were given of the superposition principle applied to a system consisting of a single
photon. § 2 dealt with states differing only with regard to the polarization and § 3 with states differing only with regard to the motion of the photon as a whole. The nature of the relationships
which the superposition principle requires to exist between the states of any system is of a kind that cannot be explained in terms of familiar physical concepts. One cannot in the classical
sense picture a system being partly in each of two states and see the equivalence of this to the system being completely in some other state. There is an entirely new idea involved, to which one
must get accustomed and in terms of which one must proceed to build up an exact mathematical theory, without having any detailed classical picture. When a state is formed by the superposition of
two other states, it will have properties that are in some vague way intermediate between those of the two original states and that approach more or less closely to those of either of them
according to the greater or less 'weight' attached to this state in the superposition process. The new state is completely defined by the two original states when their relative weights in the
superposition process are known, together with a certain phase difference, the exact meaning of weights and phases being provided in the general case by the mathematical theory. In the case of
the polarization of a photon their meaning is that provided by classical optics, so that, for example, When two perpendicularly plane polarized states are superposed with equal weights, the new
state may be circularly polarized in either direction, or linearly polarized at an angle π/4, or else elliptically polarized, according to the phase difference. The non-classical nature of the
superposition process is brought out clearly if we consider the superposition of two states, A and B, such that there exists an observation which, when made on the system in state A, is certain
to lead to one particular result, a say, and when made on the system in state B is certain to lead to some different result, b say. What will be the result of the observation when made on the
system in the superposed state? The answer is that the result will be sometimes a and sometimes b, according to a probability law depending on the relative weights of A and B in the superposition
process. It will never be different from both a and b. The intermediate character of the state formed by superposition thus expresses itself through the probability of a particular result for an
observation being intermediate between the corresponding probabilities for the original states,† not through the result itself being intermediate between the corresponding results for the
original states.
In this way we see that such a drastic departure from ordinary ideas as the assumption of superposition relationships between the states is possible only on account of the recognition of the
importance of the disturbance accompanying an observation and of the consequent indeterminacy in the result of the observation. When an observation is made on any atomic system that is in a given
state, in general the result will not be determinate, i.e., if the experiment is repeated several times under identical conditions several different results may be obtained. It is a law of
nature, though, that if the experiment is repeated a large number of times, each particular result will be obtained in a definite fraction of the total number of times, so that there is a
definite probability of its being obtained. This probability is what the theory sets out to calculate. Only in special cases when the probability for some result is unity is the result of the
experiment determinate. The assumption of superposition relationships between the states leads to a mathematical theory in which the equations that define a state are linear in the unknowns. In
consequence of this, people have tried to establish analogies with systems in classical mechanics, such as vibrating strings or membranes, which are governed by linear equations and for which,
therefore, a superposition principle holds. Such analogies have led to the name 'Wave Mechanics' being sometimes given to quantum mechanics. It is important to remember, however, that the
superposition that occurs in quantum mechanics is of an essentially different nature from any occurring in the classical theory, as is shown by the fact that the quantum superposition principle
demands indeterminacy in the results of observations in order to be capable of a sensible physical interpretation. The analogies are thus liable to be misleading.
5. Mathematical formulation of the principle
A profound change has taken place during the present century in the opinions physicists have held on the mathematical foundations of their subject. Previously they supposed that the principles of
Newtonian mechanics would provide the basis for the description of the whole of physical phenomena and that all the theoretical physicist had to do was suitably to develop and apply these
principles. With the recognition that there is no logical reason why Newtonian and other classical principles should be valid outside the domains in which they have been experimentally verified
has come the realization that departures from these principles are indeed necessary. Such departures find their expression through the introduction of new mathematical formalisms, new schemes of
axioms and rules of manipulation, into the methods of theoretical physics. Quantum mechanics provides a good example of the new ideas. It requires the states of a dynamical system and the
dynamical variables to be interconnected in quite strange ways that are unintelligible from the classical standpoint. The states and dynamical variables have to be represented by mathematical
quantities of different natures from those ordinarily used in physics. The new scheme becomes a precise physical theory when all the axioms and rules of manipulation governing the mathematical
quantities are specified and when in addition certain laws are laid down connecting physical facts with the mathematical formalism, so that from any given physical conditions equations between
the mathematical quantities may be inferred and vice versa. In an application of the theory one would be given certain physical information, which one would proceed to express by equations
between the mathematical quantities. One would then deduce new equations with the help of the axioms and rules of manipulation and would conclude by interpreting these new equations as physical
conditions. The justification for the whole scheme depends, apart from internal consistency, on the agreement of the final results with experiment. We shall begin to set up the scheme by dealing
with the mathematical relations between the states of a dynamical system at one instant of time, which relations will come from the mathematical formulation of the principle of superposition. The
superposition process is a kind of additive process and implies that states can in some way be added to give new states. The states must therefore be connected with mathematical quantities of a
kind which can be added together to give other quantities of the same kind. The most obvious of such quantities are vectors. Ordinary vectors, existing in a space of a finite number of
dimensions, are not sufficiently general for most of the dynamical systems in quantum mechanics. We have to make a generalization to vectors in a space of an infinite number of dimensions, and
the mathematical treatment becomes complicated by questions of convergence. For the present, however, we shall deal merely with some general properties of the vectors, properties which can be
deduced on the basis of a simple scheme of axioms, and questions of convergence and related topics will not be gone into until the need arises. It is desirable to have a special name for
describing the vectors which are connected with the states of a system in quantum mechanics, whether they are in a space of a finite or an infinite number of dimensions. We shall call them ket
vectors, or simply kets, and denote a general one of them by a special symbol |>. If we want to specify a particular one of them by a label, A say, we insert it in the middle, thus |A>. The
suitability of this notation will become clear as the scheme is developed. Ket vectors may be multiplied by complex numbers and may be added together to give other ket vectors, from two ket
vectors | A > and | B > we can form
c[1]| A > + c[2] | B > = | R >, (1)
say, where c[1] and c[2] are any two complex numbers. We may also perform more general linear processes with them, such as adding an infinite sequence of them, and if we have a ket vector | x >,
depending on and labelled by a parameter x which can take on all values in a certain range, we may integrate it with respect to x, to get another ket vector
∫ | x > dx = | Q >
say. A ket vector which is expressible linearly in terms of certain others is said to be dependent on them. A set of ket vectors are called independent if no one of them is expressible linearly
in terms of the others. We now assume that each state of a dynamical system at a particular time corresponds to a ket vector, the correspondence being such that if a state results from the
superposition of certain other states, its corresponding ket vector is expressible linearly in terms of the corresponding ket vectors of the other states, and conversely. Thus the state | R >
results from a superposition of the states | A >and | B > when the corresponding ket vectors are connected by (1). The above assumption leads to certain properties of the superposition process,
properties which are in fact necessary for the word 'superposition' to be appropriate. When two or more states are superposed, the order in which they occur in the superposition process is
unimportant, so the superposition process is symmetrical between the states that are superposed. Again, we see from equation (1) that (excluding the case when the coefficient c[1] or c[2] is
zero) if the state | R > can be formed by superposition of the states | A > and | B >, then the state | A > can be formed by superposition of | B > and | R >, and | B > can be formed by
superposition of | A > and | R >. The superposition relationship is symmetrical between all three states | A >, | B >, and | R >. A state which results from the superposition of certain other
states will be said to be dependent on those states. More generally, a state will be said to be dependent on any set of states, finite or infinite in number, if its corresponding ket vector is
dependent on the corresponding ket vectors of the set of states. A set of states will be called independent if no one of them is dependent on the others. To proceed with the mathematical
formulation of the superposition principle we must introduce a further assumption, namely the assumption that by superposing a state with itself we cannot form any new state, but only the
original state over again. If the original state corresponds to the ket vector | A >, when it is superposed with itself the resulting state will correspond to
c[1]| A > + c[2]| A > = (c[1] + c[2]) | A >,
where c[1] and c[2] are numbers. Now we may have c[1] + c[2] = 0, in which case the result of the superposition process would be nothing at all, the two components having cancelled each other by
an interference effect. Our new assumption requires that, apart from this special case, the resulting state must be the same as the original one, so that (c[1] + c[2])| A > must correspond to the
same state that | A > does. Now c[1] + c[2] is an arbitrary complex number and hence we can conclude that if the ket vector corresponding to a state is multiplied by any complex number, not zero,
the resulting ket vector will correspond to the same state. Thus a state is specified by the direction of a ket vector and any length one may assign to the ket vector is irrelevant. All the
states of the dynamical system are in one - one correspondence with all the possible directions for a ket vector, no distinction being made between the directions of the ket vectors | A > and - |
A >. The assumption just made shows up very clearly the fundamental difference between the superposition of the quantum theory and any kind of classical superposition. In the case of a classical
system for which a superposition principle holds, for instance a vibrating membrane, when one superposes a state with itself the result is a different state, with a different magnitude of the
oscillations. There is no physical characteristic of a quantum state corresponding to the magnitude of the classical oscillations, as distinct from their quality, described by the ratios of the
amplitudes at different points of the membrane. Again, while there exists a classical state with zero amplitude of oscillation everywhere, namely the state of rest, there does not exist any
corresponding state for a quantum system, the zero ket vector corresponding to no state at all. Given two states corresponding to the ket vectors | A > and | B >, the general state formed by
superposing them corresponds to a ket vector | R > which is determined by two complex numbers, namely the coefficients c[1] and c[2] of equation (1). If these two coefficients are multiplied by
the same factor (itself a complex number), the ket vector | R > will get multiplied by this factor and the corresponding state will be unaltered. Thus only the ratio of the two coefficients is
effective in determining the state R. Hence this state is determined by one complex number, or by two real parameters. Thus from two given states, a twofold infinity of states may be obtained by
superposition. This result is confirmed by the examples discussed in §§ 2 and 3. In the example of § 2 there are just two independent states of polarization for a photon, which may be taken to be
the states of plane polarization parallel and perpendicular to some fixed direction, and from the superposition of these two a twofold infinity of states of polarization can be obtained, namely
all the states of elliptic polarization, the general one of which requires two parameters to describe it. Again, in the example of § 3, from the superposition of two given translational states
for a photon a twofold infinity of translational states may be obtained, the general one of which is described by two parameters, which may be taken to be the ratio of the amplitudes of the two
wave functions that are added together and their phase relationship. This confirmation shows the need for allowing complex coefficients in equation (1). If these coefficients were restricted to
be real, then, since only their ratio is of importance for determining the direction of the resultant ket vector | R > when | A > and | B > are given, there would be only a simple infinity of
states obtainable from the superposition.
6. Bra and ket vectors
Whenever we have a set of vectors in any mathematical theory, we can always set up a second set of vectors, which mathematicians call the dual vectors. The procedure will be described for the
case when the original vectors are our ket vectors. Suppose we have a number φ which is a function of a ket vector | A >, i.e. to each ket vector | A > there corresponds one number and suppose
further that the function is a linear one, which means that the number corresponding to | A > + | A' > is the sum of the numbers corresponding to | A > and to | A' >, and the number corresponding
to c | A > is c times the number corresponding to | A >, c being any numerical factor. Then the number φ corresponding to any | A > may be looked upon as the scalar product of that | A > with
some new vector, there being one of these new vectors for each linear function of the ket vectors | A >. The justification for this way of looking at φ is that, as will be seen later (see
equations (5) and (6)), the new vectors may be added together and may be multiplied by numbers to give other vectors of the same kind. The new vectors are, of course, defined only to the extent
that their scalar products with the original ket vectors are given numbers, but this is sufficient for one to be able to build up a mathematical theory about them. We shall call the new vectors
bra vectors, or simply bras, and denote a general one of them by the symbol < A |, the mirror image of the symbol for a ket vector. If we want to specify a particular one of them by a label, B
say, we write it in the middle, thus < B |. The scalar product of a bra vector < B | and a ket vector | A > will be written < B | A >, i.e. as a juxtaposition of the symbols for the bra and ket
vectors, that for the bra vector being on the left, and the two vertical lines being contracted to one for brevity. One may look upon the symbols < and > as a distinctive kind of brackets. A
scalar product < B | A > now appears as a complete bracket expression and a bra vector < B | or a ket vector | A > as an incomplete bracket expression. We have the rules that any complete bracket
expression denotes a number and any incomplete bracket expression denotes a vector, of the bra or ket kind according to whether it contains the first or second part of the brackets. The condition
that the scalar product of < B | and | A > is a linear function of | A > may be expressed symbolically by
< B | { | A > + | A' >} = < B | A > + < B | A' >, (2) < B | { cA > = c< B | A >, (3)
c being any number. A bra vector is considered to be completely defined when its scalar product with every ket vector is given, so that if a bra vector has its scalar product with every ket
vector vanishing, the bra vector itself must be considered as vanishing. In symbols, if
< P | A > = 0, all | A >, then < P | = 0. (4)
The sum of two bra vectors < B | and < B' | is defined by the condition that its scalar product with any ket vector | A > is the sum of the scalar products of < B | and < B' | with | A >,
{ < B | + < B' | } | A > = < B | A > + < B' | A >, (5)
and the product of a bra vector < B | and a number c is defined by the condition that its scalar product with any ket vector | A > is c times the scalar product of < B | with | A >,
{c < B' | } | A > = c < B | A >. (6)
Equations (2) and (5) show that products of bra and ket vectors satisfy the distributive axiom of multiplication, and equations (3) and (6) show that multiplication by numerical factors satisfies
the usual algebraic axioms. The bra vectors, as they have been here introduced, are quite a different kind of vector from the kets, and so far there is no connexion between them except for the
existence of a scalar product of a bra and a ket. We now make the assumption that there is a one-one correspondence between the bras and the kets, such that the bra corresponding to | A > + | A'
> is the sum of the bras corresponding to | A > and to | A' >, and the bra corresponding to c | A > is c times the bra corresponding to | A >, c being the conjugate complex number to c. We shall
use the same label to specify a ket and the corresponding bra. Thus the bra corresponding to | A > will be written < A |. The relationship between a ket vector and the corresponding bra makes it
reasonable to call one of them the conjugate imaginary of the other. Our bra and ket vectors are complex quantities, since they can be multiplied by complex numbers and are then of the same
nature as before, but they are complex quantities of a special kind which cannot be split up into real and pure imaginary parts. The usual method of getting the real part of a complex quantity,
by taking half the sum of the quantity itself and its conjugate, cannot be applied since a bra and a ket vector are of different natures and cannot be added together. To call attention to this
distinction, we shall use the words 'conjugate complex' to refer to numbers and other complex quantities which can be split up into real and pure imaginary parts, and the words `conjugate
imaginary' for bra and ket vectors, which cannot. With the former kind of quantity, we shall use the notation of putting a bar over one of them to get the conjugate complex one. On account of the
one-one correspondence between bra vectors and ket vectors, any state of our dynamical system at a particular time may be specified by the direction of a bra vector just as well as by the
direction of a ket vector. In fact the whole theory will be symmetrical in its essentials between bras and kets. Given any two ket vectors | A > and | B >, we can construct from them a number < B
| A > by taking the scalar product of the first with the conjugate imaginary of the second. This number depends linearly on | A > and antilinearly on | B >, the antilinear dependence meaning that
the number formed from | B > + | B' > is the sum of the numbers formed from | B > and from | B' >, and the number formed from c | B > is c-bar times the number formed from | B >. There is a
second way in which we can construct a number which depends linearly on | A > and antilinearly on | B >, namely by forming the scalar product of | B > with the conjugate imaginary of | A > and
taking the conjugate complex of this scalar product. We assume that these two numbers are always equal, i.e.
< B | } | A > = < A | } | B >. (7)
Putting | B > = | A > here, we find that the number < A | } | A > must be real. We make the further assumption
< A | A > > 0, (8)
except when | A > = 0. In ordinary space, from any two vectors one can constrict a number — their scalar product — which is a real number and is symmetrical between them. In the space of bra
vectors or the space of ket vectors, from any two vectors one can again construct a number — the scalar product of one with the conjugate imaginary of the other — but this number is complex and
goes over into the conjugate complex number when the two vectors are interchanged. There is thus a kind of perpendicularity in these spaces, which is a generalization of the perpendicularity in
ordinary space. We shall call a bra and a ket vector orthogonal if their scalar product is zero, and two bras or two kets will be called orthogonal if the scalar product of one with the conjugate
imaginary of the other is zero. Further, we shall say that two states of our dynamical system are orthogonal if the vectors corresponding to these states are orthogonal. The length of a bra
vector < A | or of the conjugate imaginary ket vector | A > is defined as the square root of the positive number < A | A >. When we are given a state and wish to set up a bra or ket vector to
correspond to it, only the direction of the vector is given and the vector itself is undetermined to the extent of an arbitrary numerical factor. It is often convenient to choose this numerical
factor so that the vector is of length unity. This procedure is called normalization and the vector so chosen is said to be normalized. The vector is not completely determined even then, since
one can still multiply it by any number of modulus unity, i.e. any number e^iγ where γ is real, without changing its length. We shall call such a number a phase factor. The foregoing assumptions
give the complete scheme of relations between the states of a dynamical system at a particular time. The relations appear in mathematical form, but they imply physical conditions, which will lead
to results expressible in terms of observations when the theory is developed further. For instance, if two states are orthogonal, it means at present simply a certain equation in our formalism,
but this equation implies a definite physical relationship between the states, which further developments of the theory will enable us to interpret in terms of observational results.
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Euler for a day, or “my” formula for pi
Here it is:
$\pi=16\sum_{m\geq 1}{(-1)^{m+1}\frac{m^2}{4m^2-1}}.$
Yes, it’s a divergent series, but I’m sure Euler would like it even more. (Actually, the probability that this formula is not somewhere in his works, or in Ramanujan’s, is close to zero, though I
came upon it fairly accidentally today — maybe I’ll explain how it came about naturally at some later time).
Amusingly, both Pari/GP (numerically, using sumalt) and Maple (symbolically, after setting _EnvFormal:=true;) can confirm the “formula” as-is… (I didn’t try with Mathematica).
10 thoughts on “Euler for a day, or “my” formula for pi”
1. Mathematica 6 says the sum is divergent, but it can sum the corresponding power series (including a term x^m). The power series is a complicated expression in fractional powers of x and ArcTan
[Sqrt[x]]. Setting x=1, Mathematica simplifies to pi.
2. I also got the formula from a specialization (though not from a power series), and that’s also certainly the best way to check (and justify) it.
3. It looks like you’re out by a 2. The sum should converge to \pi + 2. This can be proved using partial fractions. The tricky part is a term that sums increasing powers of -1. Intuitively, one
cannot really sum this series because it can be ordered in many ways; however in this case it does converge to 1. I’m not sure if these comments support tex but here goes: –
$\pi=16\sum_{m\geq 1}{(-1)^{m+1}\frac{m^2}{4m^2-1}}$
$\frac{16m^2}{4m^2-1} = 4 + \frac{4}{4m^2-1}$
$\frac{4}{(2m-1)(2m+1)} = \frac{A}{2m-1} + \frac{B}{2m+1}\Rightarrow 4 = (2m+1)A + (2m-1)B$
$\Leftarrow m = 1/2 \Rightarrow 4 = 2A \Rightarrow A = 2 \Leftarrow m = -1/2 \Rightarrow 4 = -2B \Rightarrow B = -2$
$\Rightarrow \frac{4}{4m^2-1} = 2 ( \frac{1}{2m-1} - \frac{1}{2m+1} )\Rightarrow$
$\frac{16m^2}{4m^2-1} = 4 + 2 ( \frac{1}{2m-1} - \frac{1}{2m+1} )\Rightarrow$
$16\sum_{m\geq 1}{(-1)^{m+1}\frac{m^2}{4m^2-1}} = 2 \sum_{m\geq 1}{(-1)^{m+1} (4 + \frac{1}{2m-1} - \frac{1}{2m+1}) }=4 \sum_{m\geq 1}{(-1)^{m+1} } + 2 \sum_{m\geq 1}{(-1)^{m+1} \frac{1}{2m-1} }
- 2 \sum_{m\geq 1}{(-1)^{m+1} \frac{1}{2m+1} }$
$= \frac{\pi}{2} - 2(1-\frac{\pi}{4}) = 4 + \pi - 2 = \pi + 2.$
$\frac{\pi}{4} = \sum_{m\geq 1}{ (-1)^{m+1} \frac{1}{2m-1} }$
4. Actually, the series
$\sum_{m\geq 1}{(-1)^{m+1}}$
is also divergent but its value (universally accepted, e.g., as value of the geometric series for 1/(1-x) at x=-1) is 1/2. Inserted in your computation this leads to pi again…
[I also modified the previous comment to put in the proper markes for LaTeX on this blog].
5. I’m not quite sure you are correct about the sum. It truly depends. In any case, if you run your formula in python you get: –
>>> def f(m):
… return ((-1)**(m+1))*(m*m)/float(4*m*m-1)
>>> 16 * sum(map(lambda i: f(i), range(1,10000)))
This suspiciously looks close to \pi + 2 to me :)
[thanks for formatting my tex]
6. Yes, but if you sum to 10001, you get
>>> 16 * sum(map(lambda i: f(i), range(1,10001)))
which is close to pi-2, and more generally your computation shows that the partial sums with even number of terms converge to \pi+2, and those with odd number of terms converge to \pi-2. “On
average”, this is \pi (where “average” can be taken literally, in the form of what is called Cesaro summation: average the partial sums m<=M for M between 1 and 10000, for instance, and the value
will be very close to \pi).
But of course this is indeed a divergent series, so in some sense it can be made to sum to whatever value we want by rearranging the terms and similar formal computations. However, any
“reasonable” summation method will lead to the value \pi.
7. Yes you are correct. The sum does not converge but hovers between pi-2 and pi+2.
I would say that the geometric series only converges for absolute values less than 1. So I feel it is wrong to imply a value of 1 leads to a sum of 1/2. However, you are correct to say its Cesàro
sum does.
It’s the first time I have heard of such a sum and would like to thank you for an enlightening discussion.
8. It can definitely be confusing at first; actually, as I mentioned in the second comment (after J. Stopple’s remark), I obtained this expression from a representation of a simple function as a
series — though not a power series –. This series expression is convergent for suitable values of the variable, but I then specialized at one value where the function is defined, but not the
series: this leads to the stated formula, which can then be interpreted in various ways to become rigorous.
9. Another way to see that the even partial sums converge to pi-2: let g(n) = f(2n-1)+f(2n). Then g(n) = 16/((4n-3)(16n^2-1)) = 2/(4n-3) – 4/(4n-1) + 2/(4n+1); thus
g(1) + g(2) + … + g(n) = 2 – 4/3 + 4/5 – 4/7 + … +- 4/(4n-1)
which as n goes to infinity clearly approaches pi-2. The odd-m terms approach 4 as m goes to infinity, so the odd partial sums approach (pi-2)+4 = pi+2.
(This is basically Blair’s solution, but I prefer it because g(n) decays like 1/n^3 and thus the sum of the g(n) is clearly convergent.)
10. Do you know this one:
I think that’s also pretty neat…
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Two and One
Terry and Ali are playing a game with three balls. Is it fair that Terry wins when the middle ball is red?
Two children are playing a game with three balls, two blue ones and one red one.
Terry is red and Ali is blue.
They toss up the balls, which run down a slope so that they land in a row of three, like this:
The colour in the middle wins that round. Is this a fair game?
Who do you think will win?
Try using the interactivity below to get a feel for the game. (The sum column tells you how many times Terry has won so far.)
Getting Started
It might help to find three balls so that you can try playing the game with a friend. What happens?
Is it important that there is only red ball but two blue balls?
Student Solutions
We had some good solutions to this problem. Rukmini from Dowanhill Primary wrote:
I first got some red and blue wooden pieces. Next, I sorted out these ones.
The combinations were:
Blue Red Blue (Terry wins)
Red Blue Blue (Ali wins)
Blue Blue Red (Ali wins)
No, this is not a fair game because Terry only has one winning set and Ali has two possible winning sets. It's more likely that Ali will win.
You're right, Rukmini, well done. Nicholas who goes to Lochinver House School and Courtney from Holystone Primary expressed their solution in a slightly different, but equally as good, way. Courtney
It is not fair because Terry has a one in three chance but Ali has a $2$ in $3$ chance.
Teachers' Resources
Why do this problem?
This problem offers children the opportunity for children to get a feel for experimental probability and to work systematically to find all possible outcomes.
Possible approach
The interactivity will help children to get a feel for this problem. However they should be encouraged not only to identify a pattern in the data they collect, but also to explain why this pattern
The problem builds on the ideas of fairness introduced in Domino Pick and would benefit from discussion.
Of course this problem could also be tackled practically. Pupils themselves may throw the balls outside and collect data for the whole class.
Key questions
Why not find three balls so that you can try playing the game with a friend. What happens?
Is it important that there is only red ball but two blue balls?
How could you record what you have found out?
Would it be helpful to use some red and blue counters for working out the different possibilities?
Possible extension
Learners could try Tricky Track once they have had a go at this task.
Possible support
Suggest trying with real balls first. Some red and blue counters might be helpful for working out the different possibilities.
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High Energy Theory Seminar
Wednesday, November 13, 2024
11:00am to 12:00pm
Add to Cal
Online and In-Person Event
How the Hilbert space of two-sided black holes factorises
Guanda Lin, UC Berkeley,
In AdS/CFT, two-sided black holes are described by states in the tensor product of two Hilbert spaces associated with the two asymptotic boundaries of the spacetime. Understanding how such a tensor
product arises from the bulk perspective is an important open problem in holography, known as the factorisation puzzle. In this paper, we show how the Hilbert space of bulk states factorises due to
non-perturbative contributions of spacetime wormholes: the trace over two-sided states with different particle excitations behind the horizon factorises into a product of traces of the left and right
sides. This precisely occurs when such states form a complete basis for the bulk Hilbert space. We prove that the factorisation of the trace persists to all non-perturbative orders in 1/
G_N consequently providing a possible resolution to the factorisation puzzle from the gravitational path integral. In the language of von Neumann algebras, our results provide strong evidence that
the algebra of one-sided observables transitions from a Type II or Type III algebra, depending on whether or not perturbative gravity effects are included, to a Type I factor when including
non-perturbative corrections in the bulk.
The talk is in 469 Lauritsen.
Contact theoryinfo@caltech.edu for Zoom information.
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1. Music and wine
Market researchers know that background music can influence the mood and purchasing behavior of customers. One study in a supermarket in Northern Ireland compared three treatments: no music, French
accordion music, and Italian string music. Under each condition, the researchers recorded the numbers of bottles of French, Italian, and other wine purchased. Here is the contingency table that
summarizes the data.
Wine None French Italian
French 30 39 30
Italian 11 1 19
Other 43 35 35
a. Formulate
b. Complete the table with the marginal totals of rows and columns and give the data.
c. The observed value in cell Other/None is 43. Compute the expected value of this cell.
d. An appropriate test statistic is the
e. It appears that
f. Based on this information would you reject the null hypothesis?
b. The row totals are 99; 31; 113. The column totals are 84; 75; 84.
c. The expected value of cell Other/None is
e. The significance level is 0.05 and the degrees of freedom are
f. Since the test statistic lies in the rejection region
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How to implement quantum machine learning for quantum computing and optimization in computer science projects? | Hire Someone To Do Assignment
How to implement quantum machine learning for quantum computing and optimization in computer science projects?
How to implement quantum machine learning for quantum computing and optimization in computer science projects?. To make the most of quantum computing and the possibility of implementation in quantum
systems, we advocate several quantum-based learning-based learning and optimization algorithms using a modified quantum computer. A computing-related quantum machine learning (QCM) algorithm is
widely studied in terms of designing information/information flow, in evaluating computational resources used to perform experiments, and in determining the effectiveness of quantum computers.
However, in some of the algorithms, some performance and safety problems are caused by the image source behavior of the classical computing element (such as how to turn off the classical memory, to
turn on the quantum memory, to turn on the quantum processor). For this reason, the following theoretical proposals for an arbitrary quantum algorithm have been proposed which do not assume that the
classical computing element performs the typical quantum operations and perform common quantum processes. This paper proposes a novel way to directly implement the effect of quantum. The property is
different from one proposed by Kim, which is that quantum operation and quantum memory changes are measured not just by quantum factors in the classical computing element, but also by the quantum
measurements of the classical element. The proposed algorithm is based on a QCM approach which is superior than the standard quantum-based methods in terms of security effectiveness, correct
knowledge propagation, computational resilience, sensitivity of computation speed to environment, efficiency of computation, and throughput of quantum manipulation (see the R1 chapter, section IV,
“Quantum circuits”). The new algorithm generates the classical memory with high fidelity. At the end of this kind of experimental procedures, the quantum algorithms provided by the proposed algorithm
can execute many other problems and problems solved by traditional quantum algorithms and can this link provide a learning method which takes advantage of the new quantum computing techniques. We are
working on improving the effectiveness of these quantum algorithms in practical experiments. We also continue with our continued pursuit of improving security and cost effectiveness. In addition, we
strive to develop an algorithm which combines quantum processors and quantum machines and achieve better quantum computing performance than the conventional, classical methodsHow to implement quantum
machine learning for quantum computing and optimization in computer science projects? A 10x10m scale plan/project/pivot-image hybrid multi-label learning algorithm and an average plot. The goal of
this paper is to quantify experimental and theoretical control of an experiment using an average plot, both in terms of its control efficiency or sample area and its ability to predict the output of
experiments. What’s In Name Of The Effect Of The An Activity: An Experimental Review Of Methods Without Evaluations? For many applications, quantum computers are inherently volatile and require the
extraction pay someone to do assignment the click this site qubits in order to prepare the final state. For some applications, a conventional conventional controller maintains the control elements
that generate an output and therefore the samples are not consistent against the target state. The result is that the control efficiency of the quantum computer tends to be low and the average plot,
measured in pixels, very few pixels deep. This suggests that the experiment would not be subject to errors that could adversely affect a data point estimated from the measured data, and should
therefore fail to observe individual pixels. There quite simply still exist an important understanding of high throughput testing and design techniques for quantum devices. For this analysis, we use
all hardware available in this type of application currently used, to quantify the performance and thus how well her latest blog predicts the expected output of individual best site and plot.
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The comparison between the upper and lower bounds for a given experimental area in dBm (bit/pixel) is the comparison of average data and the plots that follow. We are currently conducting experiments
with a hybrid device termed a L3-type digital-to-analog converter and a conventional digital-to-analog (DAC) converter, both having been developed jointly by the Russian Academy of Sciences and the
Russian Academy of Sciences, at the Institute for Quantum Computing. However, due to their poor quality, the high throughput implementation of the experiments and hence the high quality of the
experimental data, are rarely reported. This paper is the first attempt to verify and address thisHow to implement quantum machine learning for quantum computing and optimization in computer science
projects? Hilarity with the quantum mechanics behind early quantum gates (such as helpful resources should not be surprising. The hidden particle mechanism is known in the complex quantum field as a
machine that acts as another “qubits”. Mokhba did the experiment, but has been turned into a book report that will be an attempt to explore some of the fundamentals of the hidden particle engine and
others investigate this site thoroughly. For now the book discusses the post-chaos realization, and even describes all the possible computer models of quantum mechanical machine learning. Mokhba’s
quantum algorithm reveals how to implement quantum algorithms by building a linear neural network with quantum gates, and making an analogue. Quantum computers still play a very important part in
quantum computing. Though quantum computers are still making progress, most of the challenges open in quantum computing. We can start with the quantum dynamics on the one hand – quantum computation
makes new objects. The ground-based quantum processor – especially the quadratures behind it – is already dealing with such problems as erasure and decay of memory. If the cloud is sufficiently
small, the original source machine is likely to be good enough to handle new objects. But the best bit-strings of quantum mechanics become necessary. The hidden particle will be subject to quantum
instructions, and the quantum theory itself is a unitary transformation. On the other hand, at least this is enough to make doubly-boson nature of quantum mechanics (and other quantum effects)
visible in a quantum computer. It is true that, depending on how quantum computer is powered, the machine can get a lot of power in itself, but it can also be made more powerful in specific
implementations. For the quantum processor, the quantum operations cannot be restricted to some particular code, and therefore they are impossible in the classical/classical level. A programmable
quantum machine must be available to encode both the physical context of a quantum algorithm and a quantum computer-using context.
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Logic gate diagram - Template | Engineering | Templates Logic Gates
"A logic gate is an idealized or physical device implementing a Boolean function, that is, it performs a logical operation on one or more logical inputs, and produces a single logical output.
Depending on the context, the term may refer to an ideal logic gate, one that has for instance zero rise time and unlimited fan-out, or it may refer to a non-ideal physical device...
Logic gates are primarily implemented using diodes or transistors acting as electronic switches, but can also be constructed using electromagnetic relays (relay logic), fluidic logic, pneumatic
logic, optics, molecules, or even mechanical elements. With amplification, logic gates can be cascaded in the same way that Boolean functions can be composed, allowing the construction of a physical
model of all of Boolean logic, and therefore, all of the algorithms and mathematics that can be described with Boolean logic.
Logic circuits include such devices as multiplexers, registers, arithmetic logic units (ALUs), and computer memory, all the way up through complete microprocessors, which may contain more than 100
million gates. In practice, the gates are made from field-effect transistors (FETs), particularly MOSFETs (metal–oxide–semiconductor field-effect transistors).
Compound logic gates AND-OR-Invert (AOI) and OR-AND-Invert (OAI) are often employed in circuit design because their construction using MOSFETs is simpler and more efficient than the sum of the
individual gates.
In reversible logic, Toffoli gates are used." [Logic gate. Wikipedia]
The logic gate diagram template for the ConceptDraw PRO diagramming and vector drawing software is included in the Electrical Engineering solution from the Engineering area of ConceptDraw Solution
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musical mathematical journeys
musical mathematical journeys
July 6, 2014 10:26 AM Subscribe
Glorious! I loved how the Trio for Three Angles tried to teach geometrical features purely through animation, hoping that the viewer would think things like, "Huh, that's neat. I wonder why those two
angles match up so perfectly?" It reminds me of the way a video advertising designer might try to catch an audience's eye by sliding around elements of a brand logo in an interesting way. But here
it's used for good. I wonder how far you could go teaching geometry entirely non-verbally?
posted by painquale at 2:49 PM on July 6, 2014
Perfect timing on these. I'm teaching my summer camp kids the Pythagorean theorem and triangle centers in the next two days, and will for sure show them these.
The triangle centers one in particular does a great job showing how angle bisectors are everywhere equidistant from the sides of the angle, and how perp. bisectors are everywhere equidistant form the
endpoints, which are some of those intuitive-but-hard-to-explain ideas that trip people up a lot.
posted by Wulfhere at 3:17 PM on July 6, 2014
I can remember seeing one of the triangle films some time in the early 70s - I can remember the hall it was projected in, and that it was on a sunny day with the curtains drawn, but nothing more than
that - and being entranced. I was almost certainly under ten.
That I can remember it so clearly but didn't take the mental steps to find out more explains why I love mathematics but am a lousy mathematician.
posted by Devonian at 4:15 PM on July 6, 2014 [1 favorite]
« Older Offshoring has simply become a reflex | Think of mistakes as proof that you're trying your... Newer »
This thread has been archived and is closed to new comments
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- NetBSD Manual Pages
atomic_and(3) - NetBSD Manual Pages
ATOMIC_AND(3) NetBSD Library Functions Manual ATOMIC_AND(3)
atomic_and, atomic_and_32, atomic_and_uint, atomic_and_ulong,
atomic_and_64, atomic_and_32_nv, atomic_and_uint_nv, atomic_and_ulong_nv,
atomic_and_64_nv -- atomic bitwise `and' operations
#include <sys/atomic.h>
atomic_and_32(volatile uint32_t *ptr, uint32_t bits);
atomic_and_uint(volatile unsigned int *ptr, unsigned int bits);
atomic_and_ulong(volatile unsigned long *ptr, unsigned long bits);
atomic_and_64(volatile uint64_t *ptr, uint64_t bits);
atomic_and_32_nv(volatile uint32_t *ptr, uint32_t bits);
unsigned int
atomic_and_uint_nv(volatile unsigned int *ptr, unsigned int bits);
unsigned long
atomic_and_ulong_nv(volatile unsigned long *ptr, unsigned long bits);
atomic_and_64_nv(volatile uint64_t *ptr, uint64_t bits);
The atomic_and family of functions load the value of the variable refer-
enced by ptr, perform a bitwise `and' with the value bits, and store the
result back to the variable referenced by ptr in an atomic fashion.
The *_nv() variants of these functions return the new value.
The 64-bit variants of these functions are available only on platforms
that can support atomic 64-bit memory access. Applications can check for
the availability of 64-bit atomic memory operations by testing if the
pre-processor macro __HAVE_ATOMIC64_OPS is defined.
The atomic_and functions first appeared in NetBSD 5.0.
NetBSD 9.1 April 11, 2007 NetBSD 9.1
Powered by man-cgi (2024-08-26). Maintained for NetBSD by Kimmo Suominen. Based on man-cgi by Panagiotis Christias.
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IF(FIND Function with AND
Hi Everyone,
I have created a formula which works well with locating and counting text in a range - "COUNTIF([Functional Area]:[Functional Area], FIND("Accounting",@cell)>0)". This formula counts the total number
of items irrespective of the status of the item, which could be for example, New Demand, In-flight, UAT, Complete, etc. As such i need to run the formula above and include the AND statement. For
example ""COUNTIF([Functional Area]:[Functional Area], FIND("Accounting",@cell)>0) AND (Status:Status, " New Demand").
Any help creating this formula would be appreciated.
• To evaluate multiple range/criteria sets you would need to use a OCUNTIFS function (with the S on the end).
=COUNTIFS([Functional Area]:[Functional Area], FIND("Accounting",@cell) > 0, Status:Status, "New Demand")
• Hi Paul,
Thanks for the feedback; however, I'm still having a challenge with making the formula work and obtain the #UNPARSEABLE error message.
The full formula I have used is as follows, which references another sheet i.e.:
1) Consolidated Report Inventory Range 1 (which refers to the 'Functional Area' e.g. "Accounting")
2) Consolidated Report Inventory Range 2 (which refer to the Status e.g. "0) New Demand")
=COUNTIFS({Consolidated Report Inventory Range 1}, FIND("Accounting",cell) >0, {Consolidated Report Inventory Range 2}, "0) New Demand")
Thanks in advance,
• It looks like it isn't liking that closing parenthesis in the last criteria. Try this for the criteria instead:
=COUNTIFS(.........................................................., {Range 2}, CONTAINS("New Demand", @cell))
• Thanks Paul. That works! per the formula below.
=COUNTIFS({Consolidated Report Inventory Range 1}, FIND("Accounting", @cell) > 0, {Consolidated Report Inventory Range 2}, CONTAINS("0) New Demand", @cell))
I may have a couple of further questions during the course of the week.
Help Article Resources
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Integrality · SDDP.jl
There's nothing special about binary and integer variables in SDDP.jl. Your models may contain a mix of binary, integer, or continuous state and control variables. Use the standard JuMP syntax to add
binary or integer variables.
For example:
using SDDP, HiGHS
model = SDDP.LinearPolicyGraph(
stages = 3,
lower_bound = 0.0,
optimizer = HiGHS.Optimizer,
) do sp, t
@variable(sp, 0 <= x <= 100, Int, SDDP.State, initial_value = 0)
@variable(sp, 0 <= u <= 200, integer = true)
@variable(sp, v >= 0)
@constraint(sp, x.out == x.in + u + v - 150)
@stageobjective(sp, 2u + 6v + x.out)
A policy graph with 3 nodes.
Node indices: 1, 2, 3
If you want finer control over how SDDP.jl computes subgradients in the backward pass, you can pass an SDDP.AbstractDualityHandler to the duality_handler argument of SDDP.train.
See Duality handlers for the list of handlers you can pass.
SDDP.jl cannot guarantee that it will find a globally optimal policy when some of the variables are discrete. However, in most cases we find that it can still find an integer feasible policy that
performs well in simulation.
Moreover, when the number of nodes in the graph is large, or there is uncertainty, we are not aware of another algorithm that can claim to find a globally optimal policy.
Most discussions of SDDiP in the literature confuse two unrelated things.
• First, how to compute dual variables
• Second, when the algorithm will converge to a globally optimal policy.
The stochastic dual dynamic programming algorithm requires a subgradient of the objective with respect to the incoming state variable.
One way to obtain a valid subgradient is to compute an optimal value of the dual variable $\lambda$ in the following subproblem:
\[\begin{aligned} V_i(x, \omega) = \min\limits_{\bar{x}, x^\prime, u} \;\; & C_i(\bar{x}, u, \omega) + \mathbb{E}_{j \in i^+, \varphi \in \Omega_j}[V_j(x^\prime, \varphi)]\\ & x^\prime = T_i(\bar{x},
u, \omega) \\ & u \in U_i(\bar{x}, \omega) \\ & \bar{x} = x \quad [\lambda] \end{aligned}\]
The easiest option is to relax integrality of the discrete variables to form a linear program and then use linear programming duality to obtain the dual. But we could also use Lagrangian duality
without needing to relax the integrality constraints.
To compute the Lagrangian dual $\lambda$, we penalize $\lambda^\top(\bar{x} - x)$ in the objective instead of enforcing the constraint:
\[\begin{aligned} \max\limits_{\lambda}\min\limits_{\bar{x}, x^\prime, u} \;\; & C_i(\bar{x}, u, \omega) + \mathbb{E}_{j \in i^+, \varphi \in \Omega_j}[V_j(x^\prime, \varphi)] - \lambda^\top(\bar{x}
- x)\\ & x^\prime = T_i(\bar{x}, u, \omega) \\ & u \in U_i(\bar{x}, \omega) \end{aligned}\]
You can use Lagrangian duality in SDDP.jl by passing SDDP.LagrangianDuality to the duality_handler argument of SDDP.train.
Compared with linear programming duality, the Lagrangian problem is difficult to solve because it requires the solution of many mixed-integer programs instead of a single linear program. This is one
reason why "SDDiP" has poor performance.
The second part to SDDiP is a very tightly scoped claim: if all of the state variables are binary and the algorithm uses Lagrangian duality to compute a subgradient, then it will converge to an
optimal policy.
In many cases, papers claim to "do SDDiP," but they have state variables which are not binary. In these cases, the algorithm is not guaranteed to converge to a globally optimal policy.
One work-around that has been suggested is to discretize the state variables into a set of binary state variables. However, this leads to a large number of binary state variables, which is another
reason why "SDDiP" has poor performance.
In general, we recommend that you introduce integer variables into your model without fear of the consequences, and that you treat the resulting policy as a good heuristic, rather than an attempt to
find a globally optimal policy.
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How-To: Mathematica Commands
Mathematica works in the interactive mode by taking input at the In[number]:= statement and then performing the requested operation and giving the result in an Out[number] statement. In Mathematica
the percent sign (%) refers to the output of the previous command and Out[ number ] or % number can be used in any subsequent command as input for another command. For example, the command In[1]:= 6
results in Out[1]=6. You could then use the command In[2]:=Out[1]+5 or In[2]:=%1+5 that would result in Out[2]=11. The colon and equal sign (:= ) allow you to define variables. For example In[3]:=x:=
5 would assign the value 5 to the variable x. The current value of any variable can be retrieved by typing its name at an In line. In Mathematica, functions act on arguments that are inside square
brackets [] , lists are contained in curly brackets {} , and parentheses are used for grouping.
Computations and Functions
Mathematica can perform simple computations such as addition (+ ), subtraction (- ), multiplication (* or , In[1]:=6 7 results in 42), division (/ ), and exponents (^ ) along with numerous built-in
functions. Mathematica can perform these operations on both numerical and symbolic representations. The following examples demonstrate several of Mathematica’s useful algebra and calculus functions:
expands all multiplication and power terms in an algebraic expression.
factors a polynomial over the integers. Note the use of the % sign denoting the output from the previous command.
returns the indefinite integral with respect to the selected variable(s). The integrate command can also be used to determine a definite integral using Integrate[expr, {x, xmin, xmax}]
differentiates an expression with respect to the selected variable(s).
performs algebraic manipulations to return the simplest form it can find.
returns approximate results using a specific number of significant digits.
solves an equation or system of equations for selected variables. Use the double equal sign (==) to define an equation.
finds the roots of the equation with respect to the selected variable.
Plotting in Mathematica
Mathematica supports numerous plot types and formatting options. To create simple x-y plots you can use the Plot command, for example:
Plot[ Sin[Exp[x]], {x, 0, Pi}] would plot the function y=sin(exp(x)) from 0 to pi.
The command Show allows you to modify the display in a graphics window. For example, the command
Show[ %, Frame -> True, FrameLabel -> {"Time", "Signal"}] would add a frame and labels to the previous plot (note the % ).
These items are referred to as plot options. Plot options can also be added in the Plot command, Plot[x^2, {x,0,10}, Frame -> True]. More than one graph can be plotted on a single axis using the Plot
command with a list of functions, Plot[{f1, f2, ...}, {x,xmin,xmax}]. The following commands can also be used in place of Plot for appropriate plot types: LogPlot, LogLogPlot, PolarPlot, Plot3D,
ContourPlot. To determine the syntax and default options for any of the plot commands you can type ?? commandname, for example ??LogPlot.
Using Mathematica in Text Mode on Unix
Mathematica can also be run in text mode in Unix or Xwin environment. The commands demonstrated above are also valid in text mode.
Sourcing the setup files
At USC we have developed short setup files that set an appropriate path for the software you are using. You need to source these setup files in order for the software to work properly.
You can source the setup files at a UNIX prompt or add a few lines to your .login file that will source the setup files at login. To source the setup files at a UNIX prompt use the command.
source /usr/usc/math/default/setup.csh
To source the setup files at login you need to add the appropriate lines to your .login file. These lines can be found in the file /usr/usc/math/default/README.USC
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Solving Right Triangles - FilipiKnow
Solving Right Triangles
Right triangles are geometric figures that offer a variety of applications. It is widely used for construction purposes, navigation, approximation of measurement of tall structures, and a lot more.
Furthermore, it allowed us to develop trigonometry, extensively used in other fields such as astronomy, physics, and engineering.
How can such a simple figure become so powerful that it changes the course of our planet? Learn more about this geometric plane figure and techniques for solving right triangles in this review.
Click below to go to the main reviewers:
Ultimate PMA Entrance Exam Reviewer
Table of Contents
What Are Right Triangles?: A Review
Recall that we can classify triangles according to interior angles. If the interior angles of a triangle are all acute, then we call it an acute triangle. If at least one interior angle is obtuse, we
call it an obtuse triangle. Now, if a triangle has one 90° angle, it is a right triangle.
Essentially, right triangles contain a right-angle (90° angle) while the two interior angles are acute.
Parts of a Right Triangle
The longest side of a right triangle is called the hypotenuse. Meanwhile, the remaining shorter sides of the right triangle are called the legs.
To help you remember:
• Longest side ⇰ Hypotenuse
• Shorter sides ⇰ Legs
The small square that you can see between the legs of the right triangle indicates that the angle formed by the legs is a 90-degree angle. Once you see this “tiny square” inside a triangle, it gives
you a hint that the triangle you are looking at is a right triangle.
In the figure above, sides PQ and QR are the legs of the right triangle PQR, while side PR is the hypotenuse. Take note that the hypotenuse of a right triangle is always opposite to the right angle.
In a right triangle, the legs are perpendicular since they form a right angle. Hence, in the given figure above PQ ⊥ QR.
If the legs of the right triangle are congruent or equal in measurement, then that right triangle is called an isosceles right triangle. Meanwhile, if all sides of the right triangle have different
measurements, it is a scalene right triangle.
Now that you know a right triangle’s legs and the hypotenuse, let us explore the theorem that shows the relationship among these parts.
Pythagorean Theorem
“The sum of the squares of the lengths of the legs of a right triangle on a flat surface is equal to the square of the length of its hypotenuse.”
The Pythagorean theorem states that if you square the lengths of both legs of the right triangle and then add them, the result will be equivalent to the square of the length of the hypotenuse.
(Leg 1)^2 + (Leg 2)^2 = (Hypotenuse)^2
To make the mathematical statement above easier to remember, we let a and b be the length of the legs and c be the hypotenuse of the right triangle. The Pythagorean theorem states that:
a^2 + b^2 = c^2
This theorem is one of the most popular theorems in mathematics. It’s attributed to the Greek mathematician Pythagoras who proved this theorem hundreds of years ago. If you are interested in how this
theorem has been proven throughout the years, you may watch this video from Ted-Ed.
Sample Problem 1: The right triangle below has legs measuring 4 cm and 3 cm. Determine how long its hypotenuse (or the longest side) is.
Solution: The Pythagorean theorem states that a^2 + b^2 = c^2 where a and b are the legs while c is the hypotenuse.
We have a = 4 and b = 3. Using the Pythagorean theorem, let us solve for the hypotenuse c:
a^2 + b^2 = c^2
(4)^2 + (3)^2 = c^2 Substitute a = 4 and b = 3
16 + 9 = c^2
25 = c^2
√25 = √c² Take the square root of both sides of the equation
±5 = c
c = ±5
The values of c that we have obtained are 5 and -5. Recall that c represents the length of the hypotenuse of the right triangle. Thus, we must reject -5 since having a negative length of a right
triangle is impossible. Thus, we will only take 5 as the value of c.
Hence, the hypotenuse of the right triangle is 5 cm.
Sample Problem 2: Look at the map shown below. How long is the shortest path from Helen’s house to the library?
Solution: The shortest path connecting Helen’s house and the library is the diagonal line connecting these two places. So, we draw a diagonal line that connects them.
The length of the diagonal line that we draw is the distance of the shortest path from Helen’s house to the library.
Notice that we have created a right triangle as we draw this diagonal line between Helen’s house and the library. The legs are the road from Helen’s house to the Acacia tree and the road from the
Acacia tree to the library.
And what is the hypotenuse? None other than the diagonal line representing the shortest path from Helen’s house to the library.
Since we have formed a right triangle, we can apply the Pythagorean theorem to find the length of this diagonal line. We have already defined the legs of this triangle in the previous paragraph. The
diagonal line will serve as the hypotenuse of this right triangle.
Let us do the math now.
The legs have measurements of 30 m and 40 m. So, we have a = 30 and b = 40. We let c as the length of the hypotenuse of the right triangle.
Using the Pythagorean theorem:
a^2 + b^2 = c^2
(30)^2 + (40)^2 = c^2
900 + 1600 = c^2
2500 = c^2
c^2 = 2500
√c² = √2500
c = ±50
We reject the negative value of c.
Thus, the hypotenuse of the right triangle we formed is 50. It means that the diagonal line we draw measures 50 m. Therefore, we conclude that the shortest path from Helen’s house to the library is
50 meters long (try computing the total distance of all possible paths from Helen’s house to the library, and you will discover that the shortest distance is indeed the one represented by the
diagonal line).
Sample Problem 3: What is the measurement of the side AB in the right triangle below?
Solution: We have legs AC and BC that measure 2 and 3 units, respectively. Let us compute the hypotenuse AB using the Pythagorean theorem.
a^2 + b^2 = c^2 The a and b are the legs, and c is the hypotenuse
(2)^2 + (3)^2 = c^2^ ^ Substitute the values or measurement of the legs to a and b
4 + 9 = c^2
13 = c^2
c^2 = 13 Symmetric property of equality
√c² = √13
c = ±√13
We reject the negative value.
Thus, the hypotenuse of the right triangle is √13 units.
Sample Problem 4: Determine the value of Q in the right triangle below.
Solution: In the figure above, Q represents the measurement of a leg. The other leg of the right triangle measures 6 cm, while its hypotenuse is 10 cm.
To find the missing measurement of a leg of a right triangle, we can use the Pythagorean theorem and manipulate the equation we will form.
a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse
(6)^2 + (Q)^2 = (10)^2 One of the legs is 6 cm while the hypotenuse is 10 cm
36 + Q^2 = 100
Q^2 = -36 + 100 Transposition method (we isolate b from the constants)
Q^2 = 64
√Q² = √64
Q = ±8
We reject -8 and take Q = 8 only.
Therefore, the missing measurement of the leg of the right triangle is 8 cm long. And since Q represents the measurement of a leg of the right triangle, the value of Q must be 8 cm.
Pythagorean Triples
Pythagorean triples are three positive whole numbers that satisfy the Pythagorean theorem.
The smallest Pythagorean triples are 3, 4, and 5. Notice how they satisfy the Pythagorean theorem:
If we let a = 3, b = 4, and c = 5:
a^2 + b^2 = c^2
(3)^2 + (4)^2 = (5)^2
9 + 16 = 25
25 = 25
Both sides are equal; therefore, 3, 4, and 5 are Pythagorean triples.
Now, let’s try 1, 2, and 3. Are these numbers Pythagorean triples?
Let a = 1, b = 2, and c = 3
a^2 + b^2 = c^2
(1)^2 + (2)^2 = (3)^2
1 + 4 = 9
5 ≠ 9
1, 2, and 3 failed to satisfy the Pythagorean theorem, so they are not triples.
Some common Pythagorean triples are the following:
• 3, 4, and 5
• 5, 12, and 13
• 7, 24, and 25
• 8, 15, and 17
• 11, 60, and 61
You can try it and confirm that these triples are all Pythagorean triples.
How can we get Pythagorean triples?
If you take the multiples of the triples listed above, you will still obtain a Pythagorean triple.
For instance, 5, 12, and 13 is a Pythagorean triple. If we multiply these three numbers by 4, we have 20, 48, and 52; these new triples are also Pythagorean triples.
Sample Problem: Is 27, 36, and 45 a Pythagorean triple?
Solution: Yes, because 27, 36, and 45 are derived by multiplying each of 3, 4, and 5, which are Pythagorean triples, by 9.
Converse of the Pythagorean Theorem
Again, the Pythagorean theorem tells us that in a right triangle, the sum of the squares of the lengths of legs is equal to the square of the length of the hypotenuse:
a^2 + b^2 = c^2
The converse of the Pythagorean theorem tells us that if the square of the length of the longest side of the triangle is equal to the sum of the squares of the lengths of the remaining shorter sides,
then the triangle formed is a right triangle.
Once we show that the condition above was satisfied, the triangle formed is a right triangle.
What if, for instance, the conditions of the converse of the Pythagorean theorem are unsatisfied?
We have two possible cases:
Case 1: If the square of the longest side is greater than the sum of the squares of the shorter sides of the triangle or c^2 > a^2 + b^2 (where a and b are shorter sides and c is the longest side).
In this case, we have an obtuse triangle.
Case 2: If the square of the triangle’s longest side is less than the sum of the squares of the shorter sides or c^2 < a^2 + b^2, then we have an acute triangle.
So, to summarize:
If a and b are the shorter sides of a triangle and c is its longest side:
• If c^2 = a^2 + b^2, then we have a right triangle
• If c^2 > a^2 + b^2, then we have an obtuse triangle
• If c^2 < a^2 + b^2, then we have an acute triangle
Sample Problem 1: The shorter sides of a triangle have measurements of 2 cm, 4 cm, and 5 cm. Determine what type of triangle this is.
Solution: The square of the length of the longest side (the one that measures 5 cm) is (5)^2 = 25.
The sum of the squares of the lengths of the shorter sides (the remaining sides) is:
(2)^2 + (4)^2 = 4 + 16 = 20
Notice that the square of the length of the longest side is greater than the sum of the squares of the lengths of the shorter sides or c^2 > a^2 + b^2. Thus, the triangle is obtuse.
Sample Problem 2: Triangle PQR has the following measurements: PQ = 5 cm, QR = 2 cm, and PR = 7 cm. Determine whether the triangle is acute, right, or obtuse.
Solution: The triangle’s longest side is PR, which measures 7 cm. Its square is (7)^2 = 49.
The sum of the squares of the lengths of the shorter sides PQ and QR is:
(5)^2 + (2)^2 = 25 + 4 = 29
Notice that the square of the length of the longest side is greater than the sum of the squares of the lengths of the shorter sides or c^2 > a^2 + b^2. Thus, triangle PQR is an obtuse triangle.
Special Right Triangles
In the previous sections, you have learned how special right triangles are because of their wonderful characteristics described by the Pythagorean theorem. But did you know that there are special
right triangles? This special plane figure has some types that are considered more special!
There are two types of special right triangles: the isosceles right triangles (or the 45° – 45° – 90° right triangle) and the 30° – 60° – 90° degree right triangles.
Let us talk about these special right triangles in this section.
1. Isosceles Right Triangles (45° – 45° – 90° Right Triangle)
An isosceles right triangle is a right triangle such that its legs are congruent or have the same measurement. In an isosceles right triangle, the acute interior angles near the hypotenuse have
degree measurements of 45°. This is why we call isosceles right triangles 45° – 45° – 90° right triangles.
What makes isosceles right triangles special?
In an isosceles right triangle, the length of the hypotenuse is √2 times longer than the measurement of a leg. Thus, you can determine the hypotenuse of the right triangle simply by multiplying the
measurement of a leg by √2. No need to use the Pythagorean theorem! Can you imagine how quick and easy it is?
We formally state this property above as a theorem:
Isosceles Right Triangle Theorem
“If a right triangle is an isosceles right triangle (or 45°- 45°- 90° right triangle), then the hypotenuse is √2 times as long as the leg.”
Sample Problem 1: Determine the measurement of the hypotenuse of the right triangle below.
Solution: Since the given right triangle is an isosceles right triangle, we can multiply the leg measurement by √2 to obtain the hypotenuse. Thus, if the measurement of the leg is 5 cm, then the
hypotenuse is simply 5√2 cm long.
Sample Problem 2: Determine how long a square’s diagonal is with a side measuring 2 meters.
Let us draw this square and its diagonal.
Solution: As we draw the square’s diagonal, we have formed a right triangle such that the legs are the sides and the hypotenuse is the diagonal. Interestingly, the right triangle formed is an
isosceles right triangle since the legs are 2 meters long.
Thus, the hypotenuse of this isosceles right triangle we form is 2√2 meters. Since the hypotenuse of this isosceles right triangle is also the square’s diagonal, the diagonal is 2√2 meters long.
In the previous example, we have derived a practical way to measure a square’s diagonal, given only the square’s side. The square’s diagonal is simply the measurement of the side times √2. This
fantastic concept was a direct consequence of the isosceles right triangle theorem.
Diagonal of a square = s√2
where s is the measurement of a side.
Sample Problem 3: The right triangle below has a hypotenuse with a length of 10 cm. Determine the length of a leg of the right triangle.
Answer: Although the right triangle above seems peculiar from the right triangles we have encountered earlier, it is still a right triangle because it has a right angle.
Anyway, the right triangle above is an isosceles right triangle because of the two 45-degree angles you can see near the hypotenuse with a length of 10 cm.
How do we find the measurement of a leg?
According to the isosceles right triangle theorem, the hypotenuse is √2 times as long as the leg. Thus, the measurement of a leg can be derived if we divide the measurement of the hypotenuse by √2.
Thus, the measurement of the leg of the isosceles right triangle below is again just 10 divided by √2 :
Measurement of the leg = 10 ÷ √2
We can compute the measurement of the leg as follows:
10 ÷ √2 = 10⁄√2
Notice that we have a radical in the denominator, so we need to rationalize it by multiplying both the numerator and the denominator by √2:
Thus, the leg measures 5√2 cm.
2. 30° – 60° – 90° Right Triangle
A 30° – 60° – 90° right triangle is another type of special right triangle. Unlike the previous special right triangle we have discussed, the 30° – 60° – 90° right triangle has no congruent sides.
Furthermore, its acute interior angles have different measurements. One is a 30° angle, while the other is a 60° angle.
The side of the 30° – 60° – 90° right triangle that is opposite the 30° angle is the shorter leg. Meanwhile, the side opposite the 60° angle is the longer leg. Of course, the longest side is still
the hypotenuse opposite the right angle.
• The side opposite the 30-degree angle = shorter leg
• The side opposite the 60-degree angle = longer leg
• Longest side = hypotenuse
30° – 60° – 90° Right Triangle Theorem
“In a 30° – 60° – 90° right triangle, the hypotenuse is twice as long as the shorter leg. Meanwhile, the longer leg is √3 times longer than the shorter side.”
In this kind of special right triangle, the hypotenuse measurement can be obtained by simply doubling (or multiplying by 2) the length of the shorter side. Meanwhile, the measurement of the longer
leg can be obtained by multiplying the length of the shorter leg by √3.
In symbols, if x is the measurement of the shorter leg, then:
• Longer leg = x√3
• Hypotenuse = 2x
Look at the 30° – 60° – 90° right triangle below. Its shorter leg measures 2 cm. The longer leg is 2√3 cm. Notice that the longer leg length is just the measure of the shorter leg times √3.
Meanwhile, its hypotenuse is 4 cm long, just twice the shorter leg’s measure.
Sample Problem 1: Determine the x and y values in the right triangle below.
Solution: Obviously, the given right triangle above is a 30° – 60° – 90° right triangle because of the existence of 30-degree and 60-degree angles inside it.
To make our computation easy, let us determine first the shorter leg. Recall that the shorter leg is the leg that is opposite to the 30-degree angle. If you take a look at the given image, the
shorter leg is the one that has a measure of 5 cm.
Meanwhile, the longer leg is the side with the letter y as the measurement since this side is opposite the 60-degree angle.
Lastly, x represents the right triangle’s hypotenuse or the longest side.
So, we have the following given:
• Shorter leg = 5 cm
• Longer leg = y
• Hypotenuse = x
In a 30° – 60° – 90° right triangle, the hypotenuse length is just twice the length of the shorter leg. Since the shorter leg is 5 cm, the hypotenuse is just 5(2) = 10 cm long. Since x represents the
hypotenuse of the right triangle, then x = 10 cm.
Meanwhile, the length of the longer leg is √3 times as long as the length of the shorter leg. Thus, if the shorter leg is 5 cm long, the longer leg is 5(√3) = 5√3 cm long. Since y represents the
longer leg, then y = 5√3 cm.
Therefore, x = 10 cm while y = 5√3 cm.
Sample Problem 2: Determine the measurement of side p in the figure below:
Solution: First, we must identify what p represents in the figure above.
The figure above is a 30° – 60° – 90° right triangle.
The side with the letter p is the shorter leg of this 30° – 60° – 90° right triangle because this side is opposite the 30-degree angle. Meanwhile, the side with a length of 8√3 is the longer leg
since this side is opposite the 60-degree angle.
Recall that in a 30° – 60° – 90° right triangle, the longer leg is √3 times longer than the shorter leg. Now that we have 8√3 as the longer leg, how do we derive the length of the shorter leg?
Yes, we have to divide 8√3 by √3:
8√3 ÷ √3 = 8
Thus, the shorter leg is 8 cm long. The measurement of the shorter leg is represented by p; therefore, p = 8 cm.
Solving Word Problems Involving Right Triangles
It’s now time to apply what you have learned about right triangles. Let us try to solve some word problems involving right triangles by using the Pythagorean theorem and the characteristics of the
special right triangles.
Sample Problem 1: The diagonal of a rectangle is 35 cm long. If the width of the rectangle is 21 cm long, determine the perimeter of the rectangle.
Solution: The problem above can be solved easily if we create a diagram to illustrate it.
So, we have a diagonal that is 35 cm long and a width of 21 cm long:
The problem requires us to find the perimeter of the rectangle above. Recall that the formula for the perimeter of a rectangle is P = 2l + 2w where l is the length and w is the rectangle’s width. We
already have the width of the rectangle. Unfortunately, the problem didn’t provide us with the length.
So, our goal is to look for the length of the rectangle, and then once we find it, we will compute the perimeter of the rectangle.
How can we find the length of the rectangle using the given values in the problem?
If you look again at the given rectangle above, the diagonal formed a right triangle with the diagonal as the hypotenuse and the width as one of the legs of the right triangle. This implies that the
other leg of this right triangle must be the length of the rectangle. Since the length of the rectangle is unknown, we let x be the length of the rectangle.
We can now determine the length of the rectangle using the Pythagorean theorem.
The Pythagorean theorem states that a^2 + b^2 = c^2 where a and b are the legs of the right triangle and c is the hypotenuse. Since our legs are the length and the width of the rectangle, then we
have the following:
• a = x (this is the value of our unknown length)
• b = 21 (this is the given width of the rectangle)
Furthermore, since the rectangle’s diagonal is the hypotenuse, we have c = 35.
Let us now do the math and compute for the length (which is x):
a^2 + b^2 = c^2
(x)^2 + (21)^2 = (35)^2 Using the values we have set above
x^2 + 441 = 1,225
x^2 = -441 + 1,225 Transposition method
x^2 = 784
√x² = √784 Take the square root of both sides
x = ±28
We will disregard the negative value of x and take x = 28 only.
Since x represents the length of the rectangle, then the length of the rectangle is 28 cm.
We can now determine the perimeter of the rectangle since we have the length and width measurements.
We have length = 28 cm and width = 21 cm, so:
P = 2l + 2w
P = 2(28) + 2(21)
P = 56 + 42
P = 98
Thus, the perimeter of the rectangle is 98 cm.
Sample Problem 2: A ladder measuring 5 meters long is leaning against a brick wall. The ladder makes a 60° angle from the ground. Determine how tall the brick wall is.
Solution: Let us try to illustrate this problem first.
So, we have a 5-meter ladder leaning against a wall. That ladder makes a 60-degree angle from the ground. Our task is to find the height of the brick wall.
Looking closely at our illustration above, we formed a right triangle such that the ground and the wall are the legs while the ladder is the hypotenuse. Furthermore, since the ladder created a
60-degree angle, it is a 30°-60°-90° right triangle. Hence, the other acute angle in the figure must be a 30° angle.
Let us determine this right triangle’s shorter and longer leg.
The shorter leg is the side that is opposite the 30° angle. Thus, the shorter leg is the ground between the ladder and the wall. Meanwhile, the longer leg is the side opposite the 60° angle. Thus,
the longer leg is the wall or the height of the wall.
Meanwhile, the ladder is the hypotenuse of this right triangle.
Thus, to find the height of the wall, we need to find the longer leg of this right triangle.
We can start by determining the shorter leg first. According to our previous theorem about 30°- 60°- 90° triangles, the hypotenuse is twice as long as the shorter leg. Since the hypotenuse of the
right triangle in our illustration is the ladder, which is 5 meters long, the shorter leg is simply 5⁄2 or 2.5 meters long.
Therefore, the shorter leg is 2.5 meters long.
The longer leg is √3 times as long as the shorter leg. This means that the measurement of the longer leg is just the product of √3 and the measurement of the shorter leg.
Since we have computed that the shorter leg’s measurement is 2.5 meters long, the longer leg’s measurement must be 2.5√3 meters.
Recall that the height of the wall in the problem is equivalent to the longer leg of the right triangle formed. Thus, the height of the wall in the problem is 2.5√3 meters.
Sample Problem 3: A right triangle is inscribed in a circle. Suppose the legs of the right triangle are both 12 cm long. Determine the radius of the circle.
Solution: Based on our previous discussion about circles, the right triangle’s hypotenuse will be the circle’s diameter when a right triangle is inscribed in a circle.
Thus, to find the circle’s radius in this problem, we can solve for the hypotenuse of the right triangle first since it is also the circle’s diameter. Afterward, we will divide the result by 2 to
obtain the radius (since the circle’s diameter equals twice the circle’s radius).
The given legs of the right triangle are both 12 cm. Hence, we can conclude that this right triangle is an isosceles right triangle. According to the isosceles right triangle theorem, the length of
the hypotenuse of an isosceles right triangle is √2 times as long as the length of the legs. Thus, if the legs of the right triangle in the problem are 12 cm long, the hypotenuse must be 12√2 cm
Again, since the hypotenuse is also the diameter of the circle, then the diameter of the circle is 12√2 cm. To obtain the radius, we divide 12√2 by 2:
12√2 ÷ 2 = 6√2
Thus, the radius of the circle is 6√2 cm long.
Remember the concepts you have learned about right triangles in this review because these will be your primary tools to understand the trigonometric functions we will discuss in the next review.
Next topic: Six Trigonometric Functions
Previous topic: Circles
Return to the main article: The Ultimate Basic Math Reviewer
Test Yourself!
Jewel Kyle Fabula
Jewel Kyle Fabula graduated Cum Laude with a degree of Bachelor of Science in Economics from the University of the Philippines Diliman. He is also a nominee for the 2023 Gerardo Sicat Award for Best
Undergraduate Thesis in Economics. He is currently a freelance content writer with writing experience related to technology, artificial intelligence, ergonomic products, and education. Kyle loves
cats, mathematics, playing video games, and listening to music.
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Functional JavaScript
All the code for this post are available here.https://github.com/santoshrajan/monadjs
Consider the map functor from the last chapter. We could use map to iterate over two arrays adding each element of the first to the second.
var result = [1, 2].map(function(i) {
return [3, 4].map(function(j) {
return i + j
console.log(result) ==>> [ [ 4, 5 ], [ 5, 6 ] ]
The type signature of the inner function is
f: int -> int
and the type signature of the inner map is
map: [int] -> [int]
The type signature of the outer function is
f: int -> [int]
and the type signature of the outer map is
map: [int] -> [[int]]
This is the right behaviour you would expect from a functor. But this is not what we want. We want the result to be flattened like below.
[ 4, 5, 5, 6 ]
Array Monad
For that to happen, the type signature of a functor should always be restricted to
F: [int] -> [int]
But functors do not place any such restriction. But monads do. The type signature of an array monad is
M: [T] -> [T]
where T is a given type. That is why map is a functor but not a monad. That is not all. We have to put some type restriction on the function passed it too. The function cannot return any type it
likes. We can solve this problem by restricting the function to return the Array type. So the function's type signature is restricted to
f: T -> [T]
This function is known as lift, Because it lifts the type to the required type. This is also known as the monadic function. And the original value given to the monad is known as the monadic value.
Here is the arrayMonad.
function arrayMonad(mv, mf) {
var result = []
mv.forEach(function(v) {
result = result.concat(mf(v))
return result
Now we can use the array monad to do our first calculation.
console.log(arrayMonad([1,2,3], function(i) {
return [i + 1]
})) ==>> [ 2, 3, 4 ]
Notice that our monadic function wraps the result in an array [i + 1]. Now let us try it with the two dimensional problem we started with.
var result = arrayMonad([1, 2], function(i) {
return arrayMonad([3, 4], function(j) {
return [i + j]
console.log(result) ==>> [ 4, 5, 5, 6 ]
Now we begin to see the power of monads over functors.
We can write a generic two dimensional iterator for arrays which will take two arrays and a callback, and call it for each element of both arrays.
function forEach2d(array1, array2, callback) {
return arrayMonad(array1, function(i) {
return arrayMonad(array2, function(j) {
return [callback(i,j)]
And we can try this function
forEach2d([1, 2], [3,4], function(i, j) {
return i + j
}) ==>> [ 4, 5, 5, 6 ]
Notice that the callback function is just a regular function, so we had to lift its return value [callback(i,j)] to an array. However all monads define a function to do the lifting. Its called
mResult. We will add mResult to the arrayMonad function object. Also the concat function is inneficient as it creates a new array everytime. We will apply array push instead. Here is the final code
for the array monad.
function arrayMonad(mv, mf) {
var result = []
mv.forEach(function(v) {
Array.prototype.push.apply(result, mf(v))
return result
arrayMonad.mResult = function(v) {
return [v]
and rewrite forEach2d
function forEach2d(array1, array2, callback) {
return arrayMonad(array1, function(i) {
return arrayMonad(array2, function(j) {
return arrayMonad.mResult(callback(i,j))
As an exersice I will leave it to the reader to implement forEach3d.
The arrayMonad is a monadic function and is otherwise known as bind or mbind. For a function to be a monad it must define atleast the functions mbind and mresult.
Identity Monad
The identity monad is the simplest of all monads, named so because it's mresult is the identity function.
function indentityMonad(mv, mf) {
return mf(mv)
identityMonad.mResult = function(v) {
return v
It is not a very useful monad. But it is a valid monad.
Maybe Monad
The maybe Monad is similar to the identity monad, except that it will not call the monadic function for values null or undefined. In fact it boild down to the same mayBe functor we saw in the last
function mayBeMonad(mv, mf) {
return mv === null || mv === undefined || mv === false ? null : mf(mv)
mayBeMonad.mResult = function(v) {
return v
The Monad Laws
The First Monad Law
M(mResult(x), mf) = mf(x)
Which means whatever mResult does to x to turn x into a monadic value, M will unwrap that monadic value before applying it to monadic function mf. Let us test this on our array monad.
var x = 4;
function mf(x) {
return [x * 2]
arrayMonad(arrayMonad.mResult(x), mf) ==>> [ 8 ]
mf(x) ==>> [ 8 ]
The Second Monad Law
M(mv, mResult) = mv
Which means whatever mBind does to extract mv's value, mResult will undo that and turn the value back to a monadic value. This ensures that mResult is a monadic function. Let us test it. This is
equivalent to the preserve identity case of the functor.
arrayMonad([1, 2, 3], arrayMonad.mResult) ==>> [ 1, 2, 3 ]
The Third Monad Law
M(M(mv, mf), mg)) = M(mv, function(x){return M(mf(x), mg)})
What this is saying is that it doesn't matter if you apply mf to mv and then to mg, or apply mv to a monadic function that is a composition of mf and mg. Let us test this.
function mg(x) {
return [x * 3]
arrayMonad(arrayMonad([1, 2, 3], mf), mg) ==>> [ 6, 12, 18 ]
arrayMonad([1, 2, 3], function(x) {
return arrayMonad(mf(x), mg)
}) ==>> [ 6, 12, 18 ]
We know that a monadic function takes a value and returns a monadic value. A monad takes a monadic value and a monadic function and returns a monadic value. What if a monadic function calls a monad
with a monadic value and itself and returns the result? That would be a valid monadic function because it returns a monadic value.
The function doMonad does exactly this. It takes a monad and an array of monadic values and a callback as its arguments. It defines a monadic function that recursively calls the monad with each
monadic value and itself. It breaks out of the loop when there are no more monadic values left. It returns the callback called with each unwrapped value of the monadic value. The callback cb is
curried in a closure called wrap and is visible to mf. The curry function is from the chapter on currying.
function curry(fn, numArgs) {
numArgs = numArgs || fn.length
return function f(saved_args) {
return function() {
var args = saved_args.concat(Array.prototype.slice.call(arguments))
return args.length === numArgs ? fn.apply(null, args) : f(args)
function doMonad(monad, values, cb) {
function wrap(curriedCb, index) {
return function mf(v) {
return (index === values.length - 1) ?
monad.mResult(curriedCb(v)) :
monad(values[index + 1], wrap(curriedCb(v), index + 1))
return monad(values[0], wrap(curry(cb), 0))
doMonad(arrayMonad, [[1, 2], [3, 4]], function(x, y) {
return x + y
}) //==>> [ 4, 5, 5, 6 ]
We don't need the forEach2d function we wrote earlier! And the best is yet to come!
Array comprehension using doMonad
We can write a generic array comprehension function called FOR which takes a set of arrays and a callback in its arguments.
function FOR() {
var args = [].slice.call(arguments)
callback = args.pop()
return doMonad(arrayMonad, args, callback)
FOR([1, 2], [3, 4], function(x, y) {
return x + y
}) //==>> [ 4, 5, 5, 6 ]
FOR([1, 2], [3, 4], [5, 6], function(x, y, z) {
return x + y + z
}) //==>> [ 9, 10, 10, 11, 10, 11, 11, 12 ]
How awesome is that!
State Monad
In the last chapter on functors we saw function fucntor that takes a value of type function. Similarly monadic values can also be functions. However it is important to distinguish between monadic
functions and monadic values that are functions. The type signature of a monadic function is
mf: v -> mv
ie. takes a value and lifts it to a monadic value. Note that the monadic value itself is a function. So mf will return a function mv.
The type signature of a monadic value which is a function depends on whatever that function is doing as the case may be. In the case of the state monad the type signature of its monadic value is
mv: state -> [value, new state]
The monadic value function takes a state and returns an array containing a value and a new state. The state can be of any type array, string, integer, anything.
The stateMonad takes a monadic value and a monadic function and returns a function to which we have to pass the initial state. The initial state is passed mv which returns a value. mf is then called
with this value and mf returns a monadic value which is a function. We must call this function with the newstate. Phew!
function stateMonad(mv, mf) {
return function(state) {
var compute = mv(state)
var value = compute[0]
var newState = compute[1]
return mf(value)(newState)
And mResult for the state monad is
stateMonad.mResult = function(value) {
return function(state) {
return [value, state];
Parser Monad
A parser is function that takes a string matches the string based on some criteria and returns the matched part and the remainder. Lets write the type signature of the function.
parser: string -> [match, newstring]
This looks like the monadic value of the state monad, with state restricted to the type string. But thats not all, the parser will return null if the string did not match the criteria. So lets write
the parser monad to reflect the changes.
function parserMonad(mv, mf) {
return function(str) {
var compute = mv(str)
if (compute === null) {
return null
} else {
return mf(compute[0])(compute[1])
parserMonad.mResult = function(value) {
return function(str) {
return [value, str];
As we saw earlier Monads require you to define atleast two functions, mBind (the monad function itself) and mResult. But that is not all. Optionally you can define two more functions, mZero and
mZero is the definition of "Nothing" for the monad. eg. for the arrayMonad, mZero would be []. In the case of the parser monad mZero is defined as follows. (mZero must have the same type signature of
the monadic value).
parserMonad.mZero = function(str) {
return null
mPlus is a function that takes monadic values as its arguments, and ignores the mZero's among them. How the accepted values are handled depends on the individual monad. For the parser monad, mZero
will take a set of parsers (parser monad's monadic values) and will return the value returned by the first parser to return a non mZero (null) value.
parserMonad.mPlus = function() {
var parsers = Array.prototype.slice.call(arguments)
return function(str) {
var result, i
for (i = 0; i < parsers.length; ++i) {
result = parsers[i](str)
if (result !== null) {
return result
Continuation Monad
The continuation monad takes a bit to understand. In the chapter on function composition we saw that the composition of two functions f and g is given by
(f . g) = f(g(x))
f is known as the continuation of g.
We also know that we can wrap values in a function by creating a closure. In the example below the inner function has a value wrapped in its closure.
function(value) {
return function() {
// value can be accessed here
The monadic value of a continuation monad, is a function that takes a continuation function and calls the continuation with its wrapped value. This is just like the inner function above called with a
continuation function and we we can write it as
function(continuation) {
return continuation(value)
The mResult function of monad takes a value and lifts it to a monadic value. So we can write the mResult function for the continuation monad.
var mResult = function(value) {
return function(continuation) {
return continuation(value)
So mResult is a function that takes a value, returns a monadic value which you call with a continuation.
The continuation monad itself or mBind is more complicated.
var continuationMonad = function(mv, mf) {
return function(continuation) {
// we will add to here next
First it will return a function you need to call with a continuation. Thats easy. But how can it unwrap the value inside mv? mv accepts a continuation, but calling mv with the continuation will not
do. We need to unwrap the value in mv and call mf first. So we need to trick mv into giving us the value first by calling it with our own continuation thus.
mv(function(value) {
// gotcha! the value
We can add this function to the code above.
var continuationMonad = function(mv, mf) {
return function(continuation) {
return mv(function(value) {
// gotcha! the value
Now all we have to do is call mf with the value. We know a monadic function takes a value and returns a monadic value. So we call the returned monadic value from mf with the continuation. Phew! Here
is the complete code for the continuation monad.
var continuationMonad = function(mv, mf) {
return function(continuation) {
return mv(function(value) {
return mf(value)(continuation)
continuationMonad.mResult = function(value) {
return function(continuation) {
return continuation(value)
Consider the function below.
function plus1(value) {
return value + 1
It is just a function that takes an integer and adds one to it. Similarly we could could have another function plus2. We will use these functions later.
function plus2(value) {
return value + 2
And we could write a generalised function to use any of these functions as and when required.
function F(value, fn) {
return fn(value)
F(1, plus1) ==>> 2
This function will work fine as long as the value passed is an integer. Try an array.
F([1, 2, 3], plus1) ==>> '1,2,31'
Ouch. We took an array of integers, added an integer and got back a string! Not only did it do the wrong thing, we ended up with a string having started with an array. In other words our program also
trashed the structure of the input. We want F to do the "right thing". The right thing is to "maintain structure" through out the operation.
So what do we mean by "maintain structure"? Our function must "unwrap" the given array and get its elements. Then call the given function with every element. Then wrap the returned values in a new
Array and return it. Fortunately JavaScript just has that function. Its called map.
[1, 2, 3].map(plus1) ==>> [2, 3, 4]
And map is a functor!
A functor is a function, given a value and a function, does the right thing.
To be more specific.
A functor is a function, given a value and a function, unwraps the values to get to its inner value(s), calls the given function with the inner value(s), wraps the returned values in a new structure,
and returns the new structure.
Thing to note here is that depending on the "Type" of the value, the unwrapping may lead to a value or a set of values.
Also the returned structure need not be of the same type as the original value. In the case of map both the value and the returned value have the same structure (Array). The returned structure can be
any type as long as you can get to the individual elements. So if you had a function that takes and Array and returns value of type Object with all the array indexes as keys, and corresponding
values, that will also be a functor.
In the case of JavaScript, filter is a functor because it returns an Array, however forEach is not a functor because it returns undefined. ie. forEach does not maintain structure.
Functors come from category theory in mathematics, where functors are defined as "homomorphisms between categories". Let's draw some meaning out of those words.
• homo = same
• morphisms = functions that maintain structure
• category = type
According to the theory, function F is a functor when for two composable ordinary functions f and g
F(f . g) = F(f) . F(g)
where . indicates composition. ie. functors must preserve composition.
So given this equation we can prove wether a given function is indeed a functor or not.
Array Functor
We saw that map is a functor that acts on type Array. Let us prove that the JavaScript Array.map function is a functor.
function compose(f, g) {
return function(x) {return f(g(x))}
Composing functions is about calling a set of functions, by calling the next function, with results of the previous function. Note that our compose function above works from right to left. g is
called first then f.
[1, 2, 3].map(compose(plus1, plus2)) ==>> [ 4, 5, 6 ]
[1, 2, 3].map(plus2).map(plus1) ==>> [ 4, 5, 6 ]
Yes! map is indeed a functor.
Lets try some functors. You can write functors for values of any type, as long as you can unwrap the value and return a structure.
String Functor
So can we write a functor for type string? Can you unwrap a string? Actually you can, if you think of a string as an array of chars. So it is really about how you look at the value. We also know that
chars have char codes which are integers. So we run plus1 on every char charcode, wrap them back to a string and return it.
function stringFunctor(value, fn) {
var chars = value.split("")
return chars.map(function(char) {
return String.fromCharCode(fn(char.charCodeAt(0)))
stringFunctor("ABCD", plus1) ==>> "BCDE"
You can begin to see how awesome functors are. You can actually write a parser using the string functor as the basis.
Function Functor
In JavaScript functions are first class citizens. That means you can treat functions like any other value. So can we write a functor for value of type function? We should be able to! But how do we
unwrap a function? You can unwrap a function by calling it and getting its return value. But we straight away run into a problem. To call the function we need its arguments. Remember that the functor
only has the function that came in as the value. We can solve this by having the functor return a new function. This function is called with the arguments, and we will in turn call the value function
with the argument, and call the original functors function with the value returned!
function functionFunctor(value, fn) {
return function(initial) {
return function() {
return fn(value(initial))
var init = functionFunctor(function(x) {return x * x}, plus1)
var final = init(2)
final() ==> 5
Our function functor really does nothing much, to say the least. But there a couple things of note here. Nothing happens until you call final. Every thing is in a state of suspended animation until
you call final. The function functor forms the basis for more awesome functional stuff like maintaining state, continuation calling and even promises. You can write your own function functors to do
these things!
MayBe Functor
function mayBe(value, fn) {
return value === null || value === undefined ? value : fn(value)
Yes, this is a valid functor.
mayBe(undefined, compose(plus1, plus2)) ==>> undefined
mayBe(mayBe(undefined, plus2), plus1) ==>> undefined
mayBe(1, compose(plus1, plus2)) ==>> 4
mayBe(mayBe(1, plus2), plus1) ==>> 4
So mayBe passes our functor test. There is no need for unrapping or wrapping here. It just returns nothing for nothing. Maybe is useful as a short circuiting function, which you can use as a
substitute for code like
if (result === null) {
return null
} else {
Identity Function
function id(x) {
return x
The function above is known as the identity function. It is just a function that returns the value passed to it. It is called so, because it is the identity in composition of functions in
We learned earlier that functors must preserve composition. However something I did not mention then, is that functors must also preserve identity. ie.
F(value, id) = value
Lets try this for map.
[1, 2, 3].map(id) ==>> [ 1, 2, 3 ]
Type Signature
The type signature of a function is the type of its argument and return value. So the type signature of our plus1 function is
f: int -> int
The type signature of the functor map depends on the type signature of the function argument. So if map is called with plus1 then its type signature is
map: [int] -> [int]
However the type signature of the given function need not be the same as above. We could have a function like
f: int -> string
in which the type signature of map would be
map: [int] -> [string]
The only restriction being that the type change does not affect the composability of the functor. So in general a functor's type signature can
F: A -> B
In other words map can take an array of integers and return an array of strings and would still be a functor.
Monads are a special case of Functors whos type signature is
M: A -> A
More about monads in the next chapter.
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FAQ - Loading
Frequently Asked Questions - Loading
On the truss drawings and your website it says your trusses are loaded “30-10-10”. What does that mean?
Short answer: It means 30 pounds per square foot of top chord live load, 10 pounds per square foot of top chord dead load, and 10 pounds per square foot of bottom chord dead load.
Longer answer: Loading can be a little confusing if you don’t work with it every day. We use numbers like “30-10-10” as a kind of shorthand.
In general, loads on our trusses are either “live”, or “dead”.
Live loads are dynamic. That is, they come and go, fluctuate in size, show up unexpectedly, and may be unbalanced. I bet we could come up with a joke to follow that sentence, if we tried hard enough.
Dead loads are static. In other words, they don’t move or change. They’re just always there, kind of like the 10 lbs you resolved to get rid of the first of this year. Dead loads include things like
sheathing, shingles, drywall, flooring, etc.
Loading numbers are always noted as: TC Live – TC Dead – BC Live- BC Dead
“Wait!” I hear you saying, “There’s only three numbers up there in my question, and you just said it’s always noted as FOUR different numbers? Can’t you count?”
Here’s the deal: Trusses very seldom have a live load on the bottom chord. So we don’t bother to note anything for it UNLESS the truss DOES have a bottom chord live load.
What does this mean: Deflection L/360?
Deflection is simply how much a truss will bend or sag under load. The L/360 tells us how much the truss is allowed to deflect under load in our particular design. While it’s easier to ignore numbers
that look like formulas, this is an important number to understand. And it can be changed to any number, as we’ll see.
Example: On floor trusses we design to L/360 Live Load Deflection.
Let’s look at what L/360 means. L/360 means that the deflection (i.e. bending or sagging) of the truss from the live load is limited to 1/360th of the overall length of the truss. To calc that out, a
truss that is 240″ long (or 20′ 0″), under total live load, may deflect only 1/360th of 20′ 0″. In this case, that is is equal to 0.67″ (240″ / 360 = 0.66667″).
If the same truss were designed to L/480, it would be allowed to deflect only 1/480th of 20′ 0″. Under that design criteria, it could only deflect a half inch. (240″ / 480 = 0.5″) So we can see that
designing to a higher deflection ratio makes a real world difference in how stiff our truss will be. Here’s the thing to remember: Higher deflection ratio = stiffer floor.
As with anything, you should weigh the costs and benefits (please notice that clever pun) when considering increasing your deflection ratio. In our example, changing the deflection ratio from L/360
to L/480 changed the amount our 20′ 0″ truss can deflect by only 0.17″ That is a difference of just over 1/8″ of an inch. If that change increases the cost of the trusses by 10%, well, you have to
make the call whether it’s worth it.
Just how much weight is our hypothetical floor truss carrying? We’ll assume spacing of 24″ O.C. for easy figuring. The truss carries half the distance to the next truss on either side, which equals
24″. 2′ of spacing X 20′ of span = a total carried area of 40 square feet. We can multiply that times the Live Load of 40 psf. 40 X 40 = 1600. Therefore, according to our design standard of L/360,
our 20′ truss can carry 1600lbs of live load without that load causing it to deflect, or bend, more than 0.67″. You can do these same calculations to any truss span, and design to any deflection
ratio you would like.
We have a 30 lb Live Load, and it also says the truss is engineered for a 30 lb Snow Load. So, it’s engineered for 60 lbs Live Load?
No. It isn’t. Here’s how it works. Yes, snow load is a “live” load. However, within building codes, there are deductions from the snow load that can be taken depending upon factors such as how
exposed the roof is, what pitch the roof is (for example, snow doesn’t sit very long on a 12/12 pitch roof), etc. Fortunately the engineering software takes all that into account so that we don’t
have to figure it out ourselves, because that gets really complicated.
While snow loads can be reduced depending on a variety of factors, the “Live Load” of the roof truss cannot be reduced. If it says 30 lbs, it will always be 30 lbs, with no reduction.
The engineering software does not “double up” the live load and the snow load. It engineers to the worst-case scenario, and uses that “worst case” load in designing the components of the truss.
I’ll bet you’re smart enough to figure out then, that we don’t get to reduce our 30 lb snow load, ever. Because we also have the 30 lb live load. You might wonder why even build it into the code if
it’s not something we can take advantage of. One reason it’s built into the code this way, is that a lot of areas have a higher snow load than we do in central Nebraska. Depending on your area and
elevation you may have a 40, 50, 70, or even 100 lb snow load. In that case, we’d still only design our truss with a 30lb live roof load, but make the snow load whatever number your code requires. In
THAT scenario we would be able to utilize any reductions in snow load due to roof factors, because the roof snow load would then become the “worst case scenario” for the engineering software.
If you’ve read all this, we congratulate you. You now know as much or more about truss loading than many truss designers, but you still have a long way to go before you can hold your own in a
conversation with an engineer. That’s ok, just call us if you have more questions. We’ll either know the answer or know someone who does.
|
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Hypervisor to NVE Control Plane Requirements
NVO3 Working Group Yizhou Li
INTERNET-DRAFT Lucy Yong
Intended Status: Informational Huawei Technologies
Lawrence Kreeger
Thomas Narten
David Black
Expires: May 22, 2015 November 18, 2014
Hypervisor to NVE Control Plane Requirements
In a Split-NVE architructure, the functions of the NVE are split
across the hypervisor/container on a server and an external network
equipment which is called an external NVE. A control plane
protocol(s) between a hypervisor and its associated external NVE(s)
is used for the hypervisor to distribute its virtual machine
networking state to the external NVE(s) for further handling. This
document illustrates the functionality required by this type of
control plane signaling protocol and outlines the high level
requirements. Virtual machine states as well as state transitioning
are summarized to help clarifying the needed protocol requirements.
Status of this Memo
This Internet-Draft is submitted to IETF in full conformance with the
provisions of BCP 78 and BCP 79.
Internet-Drafts are working documents of the Internet Engineering
Task Force (IETF), its areas, and its working groups. Note that
other groups may also distribute working documents as
Internet-Drafts are draft documents valid for a maximum of six months
and may be updated, replaced, or obsoleted by other documents at any
time. It is inappropriate to use Internet-Drafts as reference
material or to cite them other than as "work in progress."
The list of current Internet-Drafts can be accessed at
Li, et al [Page 1]
INTERNET DRAFT Hypervisor to NVE CP Reqs July 2014
The list of Internet-Draft Shadow Directories can be accessed at
Copyright and License Notice
Copyright (c) 2013 IETF Trust and the persons identified as the
document authors. All rights reserved.
This document is subject to BCP 78 and the IETF Trust's Legal
Provisions Relating to IETF Documents
(http://trustee.ietf.org/license-info) in effect on the date of
publication of this document. Please review these documents
carefully, as they describe your rights and restrictions with respect
to this document. Code Components extracted from this document must
include Simplified BSD License text as described in Section 4.e of
the Trust Legal Provisions and are provided without warranty as
described in the Simplified BSD License.
Table of Contents
1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1 Terminology . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Target Scenarios . . . . . . . . . . . . . . . . . . . . . 5
2. VM Lifecycle . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.1 VM Creation Event . . . . . . . . . . . . . . . . . . . . . 7
2.2 VM Live Migration Event . . . . . . . . . . . . . . . . . . 8
2.3 VM Termination Event . . . . . . . . . . . . . . . . . . . . 9
2.4 VM Pause, Suspension and Resumption Events . . . . . . . . . 9
3. Hypervisor-to-NVE Control Plane Protocol Functionality . . . . 9
3.1 VN connect and Disconnect . . . . . . . . . . . . . . . . . 10
3.2 TSI Associate and Activate . . . . . . . . . . . . . . . . . 11
3.3 TSI Disassociate and Deactivate . . . . . . . . . . . . . . 14
4. Hypervisor-to-NVE Control Plane Protocol Requirements . . . . . 15
5. Security Considerations . . . . . . . . . . . . . . . . . . . . 16
6. IANA Considerations . . . . . . . . . . . . . . . . . . . . . . 17
7. Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . 17
8. References . . . . . . . . . . . . . . . . . . . . . . . . . . 17
8.1 Normative References . . . . . . . . . . . . . . . . . . . 17
8.2 Informative References . . . . . . . . . . . . . . . . . . 17
Appendix A. IEEE 802.1Qbg VDP Illustration (For information
only) . . . . . . . . . . . . . . . . . . . . . . . . . . 18
Authors' Addresses . . . . . . . . . . . . . . . . . . . . . . . . 20
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1. Introduction
In the Split-NVE architecture shown in Figure 1, the functionality of
the NVE is split across an end device supporting virtualization and
an external network device which is called an external NVE. The
portion of the NVE functionality located on the hypervisor/container
is called the tNVE and the portion located on the external NVE is
called the nNVE in this document. Overlay encapsulation/decapsulation
functions are normally off-loaded to the nNVE on the external NVE.
The tNVE is normally implemented as a part of hypervisor or container
in an virtualized end device.
The problem statement [RFC7364], discusses the needs for a control
plane protocol (or protocols) to populate each NVE with the state
needed to perform the required functions. In one scenario, an NVE
provides overlay encapsulation/decapsulation packet forwarding
services to Tenant Systems (TSs) that are co-resident within the NVE
on the same End Device (e.g. when the NVE is embedded within a
hypervisor or a Network Service Appliance). In such cases, there is
no need for a standardized protocol between the hypervisor and NVE,
as the interaction is implemented via software on a single device.
While in the Split-NVE architecture scenarios, as shown in figure 2
to figure 4, a control plane protocol(s) between a hypervisor and its
associated external NVE(s) is required for the hypervisor to
distribute the virtual machines networking states to the NVE(s) for
further handling. The protocol indeed is an NVE-internal protocol and
runs between tNVE and nNVE logical entities. This protocol is
mentioned in NVO3 problem statement [RFC7364] and appears as the
third work item.
Virtual machine states and state transitioning are summarized in this
document to show events where the NVE needs to take specific actions.
Such events might correspond to actions the control plane signaling
protocols between the hypervisor and external NVE will need to take.
Then the high level requirements to be fulfilled are outlined.
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+-- -- -- -- Split-NVE -- -- -- --+
| +------------- ----+| |
| | +--+ +---\|/--+|| +------ --------------+
| | |VM|---+ ||| | \|/ |
| | +--+ | ||| |+--------+ |
| | +--+ | tNVE |||----- - - - - - -----|| | |
| | |VM|---+ ||| || nNVE | |
| | +--+ +--------+|| || | |
| | || |+--------+ |
| +--Hpvr/Container--+| +---------------------+
End Device External NVE
Figure 1 Split-NVE structure
This document uses the term "hypervisor" throughout when describing
the Split-NVE scenario where part of the NVE functionality is off-
loaded to a separate device from the "hypervisor" that contains a VM
connected to a VN. In this context, the term "hypervisor" is meant to
cover any device type where part of the NVE functionality is off-
loaded in this fashion, e.g.,a Network Service Appliance, Linux
This document often uses the term "VM" and "Tenant System" (TS)
interchangeably, even though a VM is just one type of Tenant System
that may connect to a VN. For example, a service instance within a
Network Service Appliance may be another type of TS, or a system
running on an OS-level virtualization technologies like LinuX
Containers. When this document uses the term VM, it will in most
cases apply to other types of TSs.
Section 2 describes VM states and state transitioning in its
lifecycle. Section 3 introduces Hypervisor-to-NVE control plane
protocol functionality derived from VM operations and network events.
Section 4 outlines the requirements of the control plane protocol to
achieve the required functionality.
1.1 Terminology
The key words "MUST", "MUST NOT", "REQUIRED", "SHALL", "SHALL NOT",
"SHOULD", "SHOULD NOT", "RECOMMENDED", "MAY", and "OPTIONAL" in this
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document are to be interpreted as described in RFC 2119 [RFC2119].
This document uses the same terminology as found in [RFC7365] and [I-
D.ietf-nvo3-nve-nva-cp-req]. This section defines additional
terminology used by this document.
Split-NVE: a type of NVE that the functionalities of it are split
across an end device supporting virtualization and an external
network device.
tNVE: the portion of Split-NVE functionalities located on the end
device supporting virtualization.
nNVE: the portion of Split-NVE functionalities located on the network
device which is directly or indirectly connects to the end device
holding the corresponding tNVE.
External NVE: the physical network device holding nNVE
Hypervisor/Container: the logical collection of software, firmware
and/or hardware that allows the creation and running of server or
service appliance virtualization. tNVE is located on
Hypervisor/Container. It is loosely used in this document to refer to
the end device supporting the virtualization. For simplicity, we also
use Hypervisor in this document to represent both hypervisor and
VN Profile: Meta data associated with a VN that is applied to any
attachment point to the VN. That is, VAP properties that are appliaed
to all VAPs associated with a given VN and used by an NVE when
ingressing/egressing packets to/from a specific VN. Meta data could
include such information as ACLs, QoS settings, etc. The VN Profile
contains parameters that apply to the VN as a whole. Control
protocols between the NVE and NVA could use the VN ID or VN Name to
obtain the VN Profile.
VSI: Virtual Station Interface. [IEEE 802.1Qbg]
VDP: VSI Discovery and Configuration Protocol [IEEE 802.1Qbg]
1.2 Target Scenarios
In the Split-NVE architecture, an external NVE can provide an offload
of the encapsulation / decapsulation function, network policy
enforcement, as well as the VN Overlay protocol overhead. This
offloading may provide performance improvements and/or resource
savings to the End Device (e.g. hypervisor) making use of the
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external NVE.
The following figures give example scenarios of a Split-NVE
Hypervisor Access Switch
+------------------+ +-----+-------+
| +--+ +-------+ | | | |
| |VM|---| | | VLAN | | |
| +--+ | tNVE |---------+ nNVE| +--- Underlying
| +--+ | | | Trunk | | | Network
| |VM|---| | | | | |
| +--+ +-------+ | | | |
+------------------+ +-----+-------+
Figure 2 Hypervisor with an External NVE
Hypervisor L2 Switch
+---------------+ +-----+ +----+---+
| +--+ +----+ | | | | | |
| |VM|---| | |VLAN | |VLAN | | |
| +--+ |tNVE|-------+ +-----+nNVE| +--- Underlying
| +--+ | | |Trunk| |Trunk| | | Network
| |VM|---| | | | | | | |
| +--+ +----+ | | | | | |
+---------------+ +-----+ +----+---+
Figure 3 Hypervisor with an External NVE
across an Ethernet Access Switch
Network Service Appliance Access Switch
+--------------------------+ +-----+-------+
| +------------+ | \ | | | |
| |Net Service |----| \ | | | |
| |Instance | | \ | VLAN | | |
| +------------+ |tNVE| |------+nNVE | +--- Underlying
| +------------+ | | | Trunk| | | Network
| |Net Service |----| / | | | |
| |Instance | | / | | | |
| +------------+ | / | | | |
+--------------------------+ +-----+-------+
Figure 4 Physical Network Service Appliance with an External NVE
Tenant Systems connect to external NVEs via a Tenant System Interface
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(TSI). The TSI logically connects to the external NVE via a Virtual
Access Point (VAP) [I-D.ietf-nvo3-arch]. The external NVE may provide
Layer 2 or Layer 3 forwarding. In the Split-NVE architecture, the
external NVE may be able to reach multiple MAC and IP addresses via a
TSI. For example, Tenant Systems that are providing network services
(such as transparent firewall, load balancer, VPN gateway) are likely
to have complex address hierarchy. This implies that if a given TSI
disassociates from one VN, all the MAC and/or IP addresses are also
disassociated. There is no need to signal the deletion of every MAC
or IP when the TSI is brought down or deleted. In the majority of
cases, a VM will be acting as a simple host that will have a single
TSI and single MAC and IP visible to the external NVE.
2. VM Lifecycle
Figure 2 of [I-D.ietf-opsawg-vmm-mib] shows the state transition of a
VM. Some of the VM states are of interest to the external NVE. This
section illustrates the relevant phases and events in the VM
lifecycle. It should be noted that the following subsections do not
give an exhaustive traversal of VM lifecycle state. They are intended
as the illustrative examples which are relevant to Split-NVE
architecture, not as prescriptive text; the goal is to capture
sufficient detail to set a context for the signaling protocol
functionality and requirements described in the following sections.
2.1 VM Creation Event
VM creation event makes the VM state transiting from Preparing to
Shutdown and then to Running [I-D.ietf-opsawg-vmm-mib]. The end
device allocates and initializes local virtual resources like storage
in the VM Preparing state. In Shutdown state, the VM has everything
ready except that CPU execution is not scheduled by the hypervisor
and VM's memory is not resident in the hypervisor. From the Shutdown
state to Running state, normally it requires the human execution or
system triggered event. Running state indicates the VM is in the
normal execution state. As part of transitioning the VM to the
Running state, the hypervisor must also provision network
connectivity for the VM's TSI(s) so that Ethernet frames can be sent
and received correctly. No ongoing migration, suspension or shutdown
is in process.
In the VM creation phase, the VM's TSI has to be associated with the
external NVE. Association here indicates that hypervisor and the
external NVE have signaled each other and reached some agreement.
Relevant networking parameters or information have been provisioned
properly. The External NVE should be informed of the VM's TSI MAC
address and/or IP address. In addition to external network
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connectivity, the hypervisor may provide local network connectivity
between the VM's TSI and other VM's TSI that are co-resident on the
same hypervisor. When the intra or inter-hypervisor connectivity is
extended to the external NVE, a locally significant tag, e.g. VLAN
ID, should be used between the hypervisor and the external NVE to
differentiate each VN's traffic. Both the hypervisor and external NVE
sides must agree on that tag value for traffic identification,
isolation and forwarding.
The external NVE may need to do some preparation work before it
signals successful association with TSI. Such preparation work may
include locally saving the states and binding information of the
tenant system interface and its VN, communicating with the NVA for
network provisioning, etc.
Tenant System interface association should be performed before the VM
enters running state, preferably in Shutdown state. If association
with external NVE fails, the VM should not go into running state.
2.2 VM Live Migration Event
Live migration is sometimes referred to as "hot" migration, in that
from an external viewpoint, the VM appears to continue to run while
being migrated to another server (e.g., TCP connections generally
survive this class of migration). In contrast, "cold" migration
consists of shutdown VM execution on one server and restart it on
another. For simplicity, the following abstract summary about live
migration assumes shared storage, so that the VM's storage is
accessible to the source and destination servers. Assume VM live
migrates from hypervisor 1 to hypervisor 2. Such migration event
involves the state transition on both hypervisors, source hypervisor
1 and destination hypervisor 2. VM state on source hypervisor 1
transits from Running to Migrating and then to Shutdown [I-D.ietf-
opsawg-vmm-mib]. VM state on destination hypervisor 2 transits from
Shutdown to Migrating and then Running.
The external NVE connected to destination hypervisor 2 has to
associate the migrating VM's TSI with it by discovering the TSI's MAC
and/or IP addresses, its VN, locally significant VID if any, and
provisioning other network related parameters of the TSI. The
external NVE may be informed about the VM's peer VMs, storage devices
and other network appliances with which the VM needs to communicate
or is communicating. The migrated VM on destination hypervisor 2
SHOULD not go to Running state before all the network provisioning
and binding has been done.
The migrating VM SHOULD not be in Running state at the same time on
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the source hypervisor and destination hypervisor during migration.
The VM on the source hypervisor does not transition into Shutdown
state until the VM successfully enters the Running state on the
destination hypervisor. It is possible that VM on the source
hypervisor stays in Migrating state for a while after VM on the
destination hypervisor is in Running state.
2.3 VM Termination Event
VM termination event is also referred to as "powering off" a VM. VM
termination event leads to its state going to Shutdown. There are two
possible causes to terminate a VM [I-D.ietf-opsawg-vmm-mib], one is
the normal "power off" of a running VM; the other is that VM has been
migrated to another hypervisor and the VM image on the source
hypervisor has to stop executing and to be shutdown.
In VM termination, the external NVE connecting to that VM needs to
deprovision the VM, i.e. delete the network parameters associated
with that VM. In other words, the external NVE has to de-associate
the VM's TSI.
2.4 VM Pause, Suspension and Resumption Events
The VM pause event leads to the VM transiting from Running state to
Paused state. The Paused state indicates that the VM is resident in
memory but no longer scheduled to execute by the hypervisor [I-
D.ietf-opsawg-vmm-mib]. The VM can be easily re-activated from Paused
state to Running state.
The VM suspension event leads to the VM transiting from Running state
to Suspended state. The VM resumption event leads to the VM
transiting state from Suspended state to Running state. Suspended
state means the memory and CPU execution state of the virtual machine
are saved to persistent store. During this state, the virtual
machine is not scheduled to execute by the hypervisor [I-D.ietf-
In the Split-NVE architecture, the external NVE should keep any
paused or suspended VM in association as the VM can return to Running
state at any time.
3. Hypervisor-to-NVE Control Plane Protocol Functionality
The following subsections show the illustrative examples of the state
transitions on external NVE which are relevant to Hypervisor-to-NVE
Signaling protocol functionality. It should be noted they are not
prescriptive text for full state machines.
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3.1 VN connect and Disconnect
In Split-NVE scenario, a protocol is needed between the End
Device(e.g. Hypervisor) making use of the external NVE and the
external NVE in order to make the external NVE aware of the changing
VN membership requirements of the Tenant Systems within the End
A key driver for using a protocol rather than using static
configuration of the external NVE is because the VN connectivity
requirements can change frequently as VMs are brought up, moved and
brought down on various hypervisors throughout the data center or
external cloud.
+---------------+ Recv VN_connect; +-------------------+
|VN_Disconnected| return Local_Tag value |VN_Connected |
+---------------+ for VN if successful; +-------------------+
|VN_ID; |-------------------------->|VN_ID; |
|VN_State= | |VN_State=connected;|
|disconnected; | |Num_TSI_Associated;|
| |<----Recv VN_disconnect----|Local_Tag; |
+---------------+ |VN_Context; |
Figure 5 State Transition Example of a VAP Instance
on an External NVE
Figure 5 shows the state transition for a VAP on the external NVE. An
NVE that supports the hypervisor to NVE control plane protocol should
support one instance of the state machine for each active VN. The
state transition on the external NVE is normally triggered by the
hypervisor-facing side events and behaviors. Some of the interleaved
interaction between NVE and NVA will be illustrated for better
understanding of the whole procedure; while others of them may not be
shown. More detailed information regarding that is available in [I-
The external NVE must be notified when an End Device requires
connection to a particular VN and when it no longer requires
connection. In addition, the external NVE must provide a local tag
value for each connected VN to the End Device to use for exchange of
packets between the End Device and the external NVE (e.g. a locally
significant 802.1Q tag value). How "local" the significance is
depends on whether the Hypervisor has a direct physical connection to
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the external NVE (in which case the significance is local to the
physical link), or whether there is an Ethernet switch (e.g. a blade
switch) connecting the Hypervisor to the NVE (in which case the
significance is local to the intervening switch and all the links
connected to it).
These VLAN tags are used to differentiate between different VNs as
packets cross the shared access network to the external NVE. When the
external NVE receives packets, it uses the VLAN tag to identify the
VN of packets coming from a given TSI, strips the tag, and adds the
appropriate overlay encapsulation for that VN and sends it towards
the corresponding remote NVE across the underlying IP network.
The Identification of the VN in this protocol could either be through
a VN Name or a VN ID. A globally unique VN Name facilitates
portability of a Tenant's Virtual Data Center. Once an external NVE
receives a VN connect indication, the NVE needs a way to get a VN
Context allocated (or receive the already allocated VN Context) for a
given VN Name or ID (as well as any other information needed to
transmit encapsulated packets). How this is done is the subject of
the NVE-to-NVA protocol which are part of work items 1 and 2 in
VN_connect message can be explicit or implicit. Explicit means the
hypervisor sending a message explicitly to request for the connection
to a VN. Implicit means the external NVE receives other messages,
e.g. very first TSI associate message (see the next subsection) for a
given VN, to implicitly indicate its interest to connect to a VN.
A VN_disconnect message will indicate that the NVE can release all
the resources for that disconnected VN and transit to VN_disconnected
state. The local tag assigned for that VN can possibly be reclaimed
by other VN.
3.2 TSI Associate and Activate
Typically, a TSI is assigned a single MAC address and all frames
transmitted and received on that TSI use that single MAC address. As
mentioned earlier, it is also possible for a Tenant System to
exchange frames using multiple MAC addresses or packets with multiple
IP addresses.
Particularly in the case of a TS that is forwarding frames or packets
from other TSs, the external NVE will need to communicate the mapping
between the NVE's IP address (on the underlying network) and ALL the
addresses the TS is forwarding on behalf of for the corresponding VN
to the NVA.
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The NVE has two ways in which it can discover the tenant addresses
for which frames must be forwarded to a given End Device (and
ultimately to the TS within that End Device).
1. It can glean the addresses by inspecting the source addresses in
packets it receives from the End Device.
2. The hypervisor can explicitly signal the address associations of
a TSI to the external NVE. The address association includes all the
MAC and/or IP addresses possibly used as source addresses in a packet
sent from the hypervisor to external NVE. The external NVE may
further use this information to filter the future traffic from the
To perform the second approach above, the "hypervisor-to-NVE"
protocol requires a means to allow End Devices to communicate new
tenant addresses associations for a given TSI within a given VN.
Figure 6 shows the example of a state transition for a TSI connecting
to a VAP on the external NVE. An NVE that supports the hypervisor to
NVE control plane protocol may support one instance of the state
machine for each TSI connecting to a given VN.
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disassociate; +--------+ disassociate
+--------------->| Init |<--------------------+
| +--------+ |
| | | |
| | | |
| +--------+ |
| | | |
| associate | | activate |
| +-----------+ +-----------+ |
| | | |
| | | |
| \|/ \|/ |
+--------------------+ +---------------------+
| Associated | | Activated |
+--------------------+ +---------------------+
|TSI_ID; | |TSI_ID; |
|Port; |-----activate---->|Port; |
|VN_ID; | |VN_ID; |
|State=associated; | |State=activated ; |-+
+-|Num_Of_Addr; |<---deactivate;---|Num_Of_Addr; | |
| |List_Of_Addr; | |List_Of_Addr; | |
| +--------------------+ +---------------------+ |
| /|\ /|\ |
| | | |
+---------------------+ +-------------------+
add/remove/updt addr; add/remove/updt addr;
or update port; or update port;
Figure 6 State Transition Example of a TSI Instance
on an External NVE
Associated state of a TSI instance on an external NVE indicates all
the addresses for that TSI have already associated with the VAP of
the external NVE on port p for a given VN but no real traffic to and
from the TSI is expected and allowed to pass through. An NVE has
reserved all the necessary resources for that TSI. An external NVE
may report the mappings of its' underlay IP address and the
associated TSI addresses to NVA and relevant network nodes may save
such information to its mapping table but not forwarding table. A NVE
may create ACL or filter rules based on the associated TSI addresses
on the attached port p but not enable them yet. Local tag for the VN
corresponding to the TSI instance should be provisioned on port p to
receive packets.
VM migration event(discussed section 2) may cause the hypervisor to
send an associate message to the NVE connected to the destination
hypervisor the VM migrates to. VM creation event may also lead to the
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same practice.
The Activated state of a TSI instance on an external NVE indicates
that all the addresses for that TSI functioning correctly on port p
and traffic can be received from and sent to that TSI via the NVE.
The mappings of the NVE's underlay IP address and the associated TSI
addresses should be put into the forwarding table rather than the
mapping table on relevant network nodes. ACL or filter rules based on
the associated TSI addresses on the attached port p in NVE are
enabled. Local tag for the VN corresponding to the TSI instance MUST
be provisioned on port p to receive packets.
The Activate message makes the state transit from Init or Associated
to Activated. VM creation, VM migration and VM resumption events
discussed in section 4 may trigger the Activate message to be sent
from the hypervisor to the external NVE.
TSI information may get updated either in Associated or Activated
state. The following are considered updates to the TSI information:
add or remove the associated addresses, update current associated
addresses (for example updating IP for a given MAC), update NVE port
information based on where the NVE receives messages. Such updates do
not change the state of TSI. When any address associated to a given
TSI changes, the NVE should inform the NVA to update the mapping
information on NVE's underlying address and the associated TSI
addresses. The NVE should also change its local ACL or filter
settings accordingly for the relevant addresses. Port information
update will cause the local tag for the VN corresponding to the TSI
instance to be provisioned on new port p and removed from the old
3.3 TSI Disassociate and Deactivate
Disassociate and deactivate conceptually are the reverse behaviors of
associate and activate. From Activated state to Associated state, the
external NVE needs to make sure the resources are still reserved but
the addresses associated to the TSI are not functioning and no
traffic to and from the TSI is expected and allowed to pass through.
For example, the NVE needs to inform the NVA to remove the relevant
addresses mapping information from forwarding or routing table. ACL
or filtering rules regarding the relevant addresses should be
disabled. From Associated or Activated state to the Init state, the
NVE will release all the resources relevant to TSI instances. The NVE
should also inform the NVA to remove the relevant entries from
mapping table. ACL or filtering rules regarding the relevant
addresses should be removed. Local tag provisioning on the connecting
port on NVE should be cleared.
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A VM suspension event(discussed in section 2) may cause the relevant
TSI instance(s) on the NVE to transit from Activated to Associated
state. A VM pause event normally does not affect the state of the
relevant TSI instance(s) on the NVE as the VM is expected to run
again soon. The VM shutdown event will normally cause the relevant
TSI instance(s) on NVE transit to Init state from Activated state.
All resources should be released.
A VM migration will lead the TSI instance on the source NVE to leave
Activated state. When a VM migrates to another hypervisor connecting
to the same NVE, i.e. source and destination NVE are the same, NVE
should use TSI_ID and incoming port to differentiate two TSI
Although the triggering messages for state transition shown in Figure
6 does not indicate the difference between VM creation/shutdown event
and VM migration arrival/departure event, the external NVE can make
optimizations if it is notified of such information. For example, if
the NVE knows the incoming activate message is caused by migration
rather than VM creation, some mechanisms may be employed or triggered
to make sure the dynamic configurations or provisionings on the
destination NVE are the same as those on the source NVE for the
migrated VM. For example IGMP query [RFC2236] can be triggered by the
destination external NVE to the migrated VM on destination hypervisor
so that the VM is forced to answer an IGMP report to the multicast
router. Then multicast router can correctly send the multicast
traffic to the new external NVE for those multicast groups the VM had
joined before the migration.
4. Hypervisor-to-NVE Control Plane Protocol Requirements
Req-1: The protocol MUST support a bridged network connecting End
Devices to External NVE.
Req-2: The protocol MUST support multiple End Devices sharing the
same External NVE via the same physical port across a bridged
Req-3: The protocol MAY support an End Device using multiple external
NVEs simultaneously, but only one external NVE for each VN.
Req-4: The protocol MAY support an End Device using multiple external
NVEs simultaneously for the same VN.
Req-5: The protocol MUST allow the End Device initiating a request to
its associated External NVE to be connected/disconnected to a given
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Req-6: The protocol MUST allow an External NVE initiating a request
to its connected End Devices to be disconnected to a given VN.
Req-7: When a TS attaches to a VN, the protocol MUST allow for an End
Device and its external NVE to negotiate a locally-significant tag
for carrying traffic associated with a specific VN (e.g., 802.1Q
Req-8: The protocol MUST allow an End Device initiating a request to
associate/disassociate and/or activate/deactive address(es) of a TSI
instance to a VN on an NVE port.
Req-9: The protocol MUST allow the External NVE initiating a request
to disassociate and/or deactivate address(es) of a TSI instance to a
VN on an NVE port.
Req-10: The protocol MUST allow an End Device initiating a request to
add, remove or update address(es) associated with a TSI instance on
the external NVE. Addresses can be expressed in different formats,
for example, MAC, IP or pair of IP and MAC.
Req-11: The protocol MUST allow the External NVE to authenticate the
End Device connected.
Req-12: The protocol MUST be able to run over L2 links between the
End Device and its External NVE.
Req-13: The protocol SHOULD support the End Device indicating if an
associate or activate request from it results from a VM hot migration
VDP [IEEE 802.1Qbg] is a candidate protocol running on layer 2.
Appendix A illustrates VDP for reader's information. It requires
extensions to fulfill the requirements in this document.
5. Security Considerations
NVEs must ensure that only properly authorized Tenant Systems are
allowed to join and become a part of any specific Virtual Network. In
addition, NVEs will need appropriate mechanisms to ensure that any
hypervisor wishing to use the services of an NVE are properly
authorized to do so. One design point is whether the hypervisor
should supply the NVE with necessary information (e.g., VM addresses,
VN information, or other parameters) that the NVE uses directly, or
whether the hypervisor should only supply a VN ID and an identifier
for the associated VM (e.g., its MAC address), with the NVE using
that information to obtain the information needed to validate the
hypervisor-provided parameters or obtain related parameters in a
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secure manner.
6. IANA Considerations
No IANA action is required. RFC Editor: please delete this section
before publication.
7. Acknowledgements
This document was initiated and merged from the drafts draft-kreeger-
nvo3-hypervisor-nve-cp, draft-gu-nvo3-tes-nve-mechanism and draft-
kompella-nvo3-server2nve. Thanks to all the co-authors and
contributing members of those drafts.
The authors would like to specially thank Jon Hudson for his generous
help in improving the readability of this document.
8. References
8.1 Normative References
[RFC2119] Bradner, S., "Key words for use in RFCs to Indicate
Requirement Levels", BCP 14, RFC 2119, March 1997.
8.2 Informative References
[RFC7364] Narten, T., Gray, E., Black, D., Fang, L., Kreeger, L., and
M. Napierala, "Problem Statement: Overlays for Network
Virtualization", October 2014.
[RFC7365] Lasserre, M., Balus, F., Morin, T., Bitar, N., and Y.
Rekhter, "Framework for DC Network Virtualization",
October 2014.
[I-D.ietf-nvo3-nve-nva-cp-req] Kreeger, L., Dutt, D., Narten, T., and
D. Black, "Network Virtualization NVE to NVA Control
Protocol Requirements", draft-ietf-nvo3-nve-nva-cp-req-01
(work in progress), October 2013.
[I-D.ietf-nvo3-arch] Black, D., Narten, T., et al, "An Architecture
for Overlay Networks (NVO3)", draft-narten-nvo3-arch, work
in progress.
[I-D.ietf-opsawg-vmm-mib] Asai H., MacFaden M., Schoenwaelder J.,
Shima K., Tsou T., "Management Information Base for
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Virtual Machines Controlled by a Hypervisor", draft-ietf-
opsawg-vmm-mib-00 (work in progress), February 2014.
[IEEE 802.1Qbg] IEEE, "Media Access Control (MAC) Bridges and Virtual
Bridged Local Area Networks - Amendment 21: Edge Virtual
Bridging", IEEE Std 802.1Qbg, 2012
[8021Q] IEEE, "Media Access Control (MAC) Bridges and Virtual Bridged
Local Area Networks", IEEE Std 802.1Q-2011, August, 2011
Appendix A. IEEE 802.1Qbg VDP Illustration (For information only)
VDP has the format shown in Figure A.1. Virtual Station Interface (VSI)
is an interface to a virtual station that is attached to a downlink port
of an internal bridging function in server. VSI's VDP packet will be
handled by an external bridge. VDP is the controlling protocol running
between the hypervisor and the external bridge.
|TLV type|TLV info|Status|VSI |VSI |VSIID | VSIID|Filter|Filter Info|
| 7b |str len | |Type|Type|Format| | Info | |
| | 9b | 1oct |ID |Ver | | |format| |
| | | |3oct|1oct| 1oct |16oct |1oct | M oct |
| | | | |
| | |<--VSI type&instance-->|<----Filter------>|
| | |<------------VSI attributes-------------->|
|<--TLV header--->|<-------TLV info string = 23 + M octets--------->|
Figure A.1: VDP TLV definitions
There are basically four TLV types.
1. Pre-Associate: Pre-Associate is used to pre-associate a VSI instance
with a bridge port. The bridge validates the request and returns a
failure Status in case of errors. Successful pre-association does not
imply that the indicated VSI Type or provisioning will be applied to any
traffic flowing through the VSI. The pre-associate enables faster
response to an associate, by allowing the bridge to obtain the VSI Type
prior to an association.
2. Pre-Associate with resource reservation: Pre-Associate with Resource
Reservation involves the same steps as Pre-Associate, but on successful
pre-association also reserves resources in the Bridge to prepare for a
subsequent Associate request.
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3. Associate: The Associate creates and activates an association between
a VSI instance and a bridge port. The Bridge allocates any required
bridge resources for the referenced VSI. The Bridge activates the
configuration for the VSI Type ID. This association is then applied to
the traffic flow to/from the VSI instance.
4. Deassociate: The de-associate is used to remove an association
between a VSI instance and a bridge port. Pre-Associated and Associated
VSIs can be de-associated. De-associate releases any resources that were
reserved as a result of prior Associate or Pre-Associate operations for
that VSI instance.
Deassociate can be initiated by either side and the rest types of
messages can only be initiated by the server side.
Some important flag values in VDP Status field:
1. M-bit (Bit 5): Indicates that the user of the VSI (e.g., the VM) is
migrating (M-bit = 1) or provides no guidance on the migration of the
user of the VSI (M-bit = 0). The M-bit is used as an indicator relative
to the VSI that the user is migrating to.
2. S-bit (Bit 6): Indicates that the VSI user (e.g., the VM) is
suspended (S-bit = 1) or provides no guidance as to whether the user of
the VSI is suspended (S-bit = 0). A keep-alive Associate request with
S-bit = 1 can be sent when the VSI user is suspended. The S-bit is used
as an indicator relative to the VSI that the user is migrating from.
The filter information format currently supports 4 types as the
1. VID Filter Info format
| #of | PS | PCP | VID |
|entries |(1bit)|(3bits)|(12bits)|
|(2octets)| | | |
|<--Repeated per entry->|
Figure A.2 VID Filter Info format
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2. MAC/VID filter format
| #of | MAC address | PS | PCP | VID |
|entries | (6 octets) |(1bit)|(3bits)|(12bits)|
|(2octets)| | | | |
|<--------Repeated per entry---------->|
Figure A.3 MAC/VID filter format
3. GroupID/VID filter format
| #of | GroupID | PS | PCP | VID |
|entries | (4 octets) |(1bit)|(3bits)|(12bits)|
|(2octets)| | | | |
|<--------Repeated per entry---------->|
Figure A.4 GroupID/VID filter format
4. GroupID/MAC/VID filter format
| #of | GroupID | MAC address | PS | PCP | VID |
|entries |(4 octets)| (6 octets) |(1bit)|(3b )|(12bits)|
|(2octets)| | | | | |
|<-------------Repeated per entry------------->|
Figure A.5 GroupID/MAC/VID filter format
The null VID can be used in the VDP Request sent from the hypervisor to
the external bridge. Use of the null VID indicates that the set of VID
values associated with the VSI is expected to be supplied by the Bridge.
The Bridge can obtain VID values from the VSI Type whose identity is
specified by the VSI Type information in the VDP Request. The set of VID
values is returned to the station via the VDP Response. The returned VID
value can be a locally significant value. When GroupID is used, it is
equivalent to the VN ID in NVO3. GroupID will be provided by the
hypervisor to the bridge. The bridge will map GroupID to a locally
significant VLAN ID.
The VSIID in VDP request that identify a VM can be one of the following
format: IPV4 address, IPV6 address, MAC address, UUID or locally
Authors' Addresses
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Yizhou Li
Huawei Technologies
101 Software Avenue,
Nanjing 210012
Phone: +86-25-56625409
EMail: liyizhou@huawei.com
Lucy Yong
Huawei Technologies, USA
Email: lucy.yong@huawei.com
Lawrence Kreeger
Email: kreeger@cisco.com
Thomas Narten
Email: narten@us.ibm.com
David Black
Email: david.black@emc.com
Li, et al [Page 21]
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Load Calculations
Load Calculations | Design of Buildings
In our earlier article, we discussed “Different types of loads” and their importance in Structural design.
Now we will move on with our further discussion on the following points:
• Design principle assumption and notation assumed
• Design Constant
• Assumptions regarding Design
• Loads on Beams
• Loads on slabs
Design principle assumption and notation assumed:
The notations adopted throughout are same as given in IS:456:2000
Density of material used in accordance with reference to IS:857-1987s
│Sr.no│Material │Density │
│1 │Plain concrete │24 KN/m3 │
│2 │Reinforced cement concrete │25 KN/m3 │
│3 │Flooring material (cement mortar) │1.00 KN/m3 │
│4 │Brick masonry │19 KN/m3 │
Design constant
Using M20 and Fe415 grade of concrete and steel respectively for columns and footings
Fck – i. e. Characteristic strength for M15 – 15 N/mm2
Fck – i. e. Characteristic strength for M15 – 15 N/mm2
Fck – i. e. Characteristic strength for M20 – 20 N/mm2
Fy – i. e. Characteristic strength for steel – 415 N/mm2
Bending Moment and Fixed Moment Calculations
Bending Moment and Shear Force diagrams
What is Bending Moment?
The element bends when a moment is applied to it. Every structural element has bending moment. Concept of bending moment is very important in the field of engineering especially Civil engineering and
Mechanical Engineering.
Unit of measurement: Newton-metres (N-m) or pound-foot or foot-pound (ft.lb)
Bending moment is directly proportional to tensile and compressive stresses. Increase in tensile and compressive stresses results in the increase in the bending moment. These stresses also depend on
the second moment of area of the cross section of the element.
What is Shear stress?
Shear stress is defined as the measure of force per unit area. Shear stress occurs in shear plane. There are many planes possible at any point in a structure which can be defined to measure stress.
Stress = Force/Unit area
Example: Bending Moment and Shear Force Calculations
Frame diagrams | Bending moment and shear force calculations
Simply supported bending moment
M[ab] = wl^2/8 = (22×4.14×4.14)/8
= 47.13 KN-m
M[bc] = wl^2/8 = (22×4.14×4.14)/8
= 47.13 KN-m
Calculation of loads for Column and Foundation Design | Structural Design
How to calculate the total loads on a column and corresponding footing?
This article has been written on the request from my readers. Engineering students generally get confused when it comes to calculating loads for column and footings design. The manual process is
Types of loads on column
1. Self weight of the column x Number of floors
2. Self weight of beams per running meter
3. Load of walls per running meter
4. Total Load of slab (Dead load + Live load + Self weight)
The columns are also subjected to bending moments which have to be considered in the final design. The best way to design a good structure is to use advanced structural design software like ETABS or
STAAD Pro. These tools are leagues ahead of manual methodology for structural design, and highly recommended.
In professional practice, there are some basic assumptions we use for structural loading calculations.
You can hire me for your structural design need. Contact me.
For Columns
Self weight of Concrete is around 2400 kg per cubic meter, which is equivalent to 240 kN. Self weight of Steel is around 8000 kg per cubic meter. Even if we assume a large column size of 230 mm x 600
mm with 1% steel and 3 meters standard height, the self weight of column is around 1000 kg per floor, which is equivalent to 10 kN. So, in my calculations, I assume self weight of column to be
between 10 to 15 kN per floor.
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The invariant subspace problem is still a problem
You're reading: News
The invariant subspace problem is still a problem
Recently we reported that Eva Gallarda and Carl Cowen had announced they had a proof of the invariant subspace conjecture for Hilbert spaces.
Well, yesterday they announced at the blog Café Matemático that there was a problem with their proof:
On 10 December 2012, we submitted a paper “Rota’s Universal Operators and Invariant Subspaces in Hilbert Spaces” for publication, and we spoke about it several times before the more formal
announcement at the RSME meeting in Santiago de Compostela on 25 January 2013. By that time, the paper had been read and no problems found by several other mathematicians. We have heard nothing
so far from the journal to which it was submitted.
We regret to inform you, however, that a gap in our proof was discovered after the announcement at Santiago. After working for the past few days to bridge the gap, so far unsuccessfully, we are
today formally withdrawing our submission to the journal.
We will, of course, continue to work to bridge the gap. At this point, Carl plans to contribute a talk to the Southeast Analysis Meeting (SEAM) to be held at Blacksburg, Virginia on 15, 16 March
2013 with the title as above.
So far at least, there have been no errors found in the paper besides the erroneous assertion that the work included in the paper proved the Invariant Subspace Theorem, while in fact it did not.
For this reason, we plan to submit a paper by mid-March, either a paper that claims to prove the Invariant Subspace Theorem if we can bridge the gap or a paper substantially the same as the paper
submitted earlier, but without claims beyond what we have actually proved correct. In the latter case, the manuscript will be made available to those interested about that time. If we believe we
have proved the result, no submission, no announcement, and no manuscript will be made available until after the new manuscript has been reviewed by several mathematicians.
So that’s a bit of a bummer, but it doesn’t sound like they’re left with nothing.
We’ll keep you updated if any more info appears.
via Julia Collins (aka @haggismaths) on Twitter
One Response to “The invariant subspace problem is still a problem”
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Estimate AR and ARMA
Estimate AR and ARMA Models
AR and ARMA models are autoregressive parametric models that have no measured inputs. These models operate on time series data.
• The AR model contains a single polynomial A that operates on the measured output. For a single-output signal y(t), the AR model is given by the following equation:
• The ARMA model adds a second polynomial C that calculates the moving average of the noise error. The ARMA model for a single-output time series is given by the following equation:
The ARMA structure reduces to the AR structure for C(q) = 1.
The AR and ARMA model structures are special cases of the more general ARX and ARMAX model structures, which do provide for measured inputs. You can estimate AR and ARMA models at the command line
and in the app.
For information about:
Estimate AR and ARMA Models at the Command Line
Estimate AR and ARMA models at the command line by using ar, arx, ivar, or armax with estimation data that contains only output measurements. These functions return estimated models that are
represented by idpoly model objects.
Selected Commands for Estimating Polynomial AR and ARMA Time-Series Models
Function Description
Noniterative, least-squares method to estimate linear, discrete-time, single-output AR models. Provides algorithmic options including lattice-based approaches and the Yule-Walker covariance
Example: sys = ar(y,na) estimates an AR model sys of polynomial order na from the scalar time series y.
Noniterative, least-squares method for estimating linear AR models. Supports multiple outputs. Assumes white noise.
Example: sys = arx(y,na) estimates an AR model from the multiple-output time series y.
Noniterative, instrumental variable method for estimating single-output AR models. Insensitive to noise color.
Example: sys = ivar(y,na) estimates an AR model using the instrumental variable method for the scalar time series y.
Iterative prediction-error method to estimate linear ARMA models.
Example: sys = armax(y,[na nc]) of estimates an ARMA model of polynomial orders na and nc from the time series y.
For more detailed usage information and examples, as well as information on other models that these functions can estimate, see ar, arx, ivar, and armax.
Estimate AR and ARMA Time Series Models in the App
Before you begin, complete the following steps:
Estimate AR and ARMA models using the System Identification app by following these steps.
1. In the System Identification app, select Estimate > Polynomial Models to open the Polynomial Models dialog box.
2. In the Structure list, select the polynomial model structure you want to estimate from the following options:
This action updates the options in the Polynomial Models dialog box to correspond with this model structure.
3. In the Orders field, specify the model orders.
□ For single-output models, enter the model orders according to the sequence displayed in the Structure field.
□ For multiple-output ARX models, enter the model orders directly, as described in Polynomial Sizes and Orders of Multi-Output Polynomial Models. Alternatively, enter the name of the matrix NA
in the MATLAB Workspace browser that stores model orders, which is Ny-by-Ny.
To enter model orders and delays using the Order Editor dialog box, click Order Editor.
4. (AR models only) Select the estimation Method as ARX or IV (instrumental variable method). For more information about these methods, see Polynomial Model Estimation Algorithms.
5. Select Add noise integration if you want to include an integrator in noise source e(t). This selection changes an AR model into an ARI model ($Ay=\frac{e}{1-{q}^{-1}}$) and an ARMA model into an
ARIMA model ($Ay=\frac{C}{1-{q}^{-1}}e\left(t\right)$).
6. In the Name field, edit the name of the model or keep the default. The name of the model must be unique in the Model Board.
7. In the Initial state list, specify how you want the algorithm to treat initial states. For more information about the available options, see Specifying Initial States for Iterative Estimation
If you get an inaccurate fit, try setting a specific method for handling initial states rather than specifying automatic selection.
8. In the Covariance list, select Estimate if you want the algorithm to compute parameter uncertainties. Effects of such uncertainties are displayed on plots as model confidence regions.
If you do not want the algorithm to estimate uncertainty, select None. Skipping uncertainty computation can reduce computation time for complex models and large data sets.
9. Click Regularization to obtain regularized estimates of model parameters. Specify regularization constants in the Regularization Options dialog box. For more information, see Regularized
Estimates of Model Parameters.
10. To view the estimation progress at the command line, select the Display progress check box. During estimation, the following information is displayed for each iteration:
□ Loss function — Determinant of the estimated covariance matrix of the input noise.
□ Parameter values — Values of the model structure coefficients you specified.
□ Search direction — Changes in parameter values from the previous iteration.
□ Fit improvements — Actual versus expected improvements in the fit.
11. Click Estimate to add this model to the Model Board in the System Identification app.
12. For the prediction-error method, only, to stop the search and save the results after the current iteration completes, click Stop Iterations. To continue iterations from the current model, click
the Continue iter button to assign current parameter values as initial guesses for the next search and start a new search. For the multi-output case, you can stop iterations for each output
separately. Note that the software runs independent searches for each output.
13. To plot the model, select the appropriate check box in the Model Views area of the System Identification app.
You can export the model to the MATLAB workspace for further analysis by dragging it to the To Workspace rectangle in the System Identification app.
See Also
ar | arx | armax | ivar
Related Topics
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Polynomial sweep
The aim of the tool is to visually illustrate the relation between the coefficients of a polynomial and the geometric properties of this polynomial (its curve and its roots).
To use this tool, please enter a polynomial $P$ of one variable $x$, with some coefficients parametrized by a variable $t$. You will then see an animated picture with the roots and the curve of $P$
vary following the variation of $t$.
Examples of polynomials in variation.
The most recent version
This page is not in its usual appearance because WIMS is unable to recognize your web browser.
Please take note that WIMS pages are interactively generated; they are not ordinary HTML files. They must be used interactively ONLINE. It is useless for you to gather them through a robot program.
• Description: Graphs and roots of a polynomial, with animated deformation. interactive exercises, online calculators and plotters, mathematical recreation and games, Pôle Formation CFAI-CENTRE
• Keywords: CFAI,interactive math, server side interactivity, algebra, geometry, polynomials, roots, animation, plot, graphing
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Time Uncertainty Interval - theoretical limit measuring time | Science Technology Savvy Life Strategies
This post is based on material from my new book, How to Time Travel, available on Amazon.com.
All attempts in science to define time fail. Instead, we describe how time behaves during an interval, a change in time. We are unable to point to an entity and say “that is time.” The reason for
this is that time is not a single entity, but scientifically an interval. We cannot slice time down to a shadow-like sliver, a dimensionless interval. In fact, scientifically speaking, the smallest
interval of time that science can theoretically define, based on the fundamental invariant aspects of the universe, is Planck time.
Planck time is the smallest interval of time that science is able to define. The theoretical formulation of Planck time comes from dimensional analysis, which studies units of measurement, physical
constants, and the relationship between units of measurement and physical constants. In simpler terms, one Planck interval is approximately equal to 10-44 seconds (i.e., 1 divided by 1 with 44 zeros
after it). As far as the science community is concerned, there is a consensus that we would not be able to measure anything smaller than a Planck interval. In fact, the smallest interval science is
able to measure as of this writing is trillions of times larger than a Planck interval. It is also widely believed that we would not be able to measure a change smaller than a Planck interval. From
this standpoint, we can assert that time is only reducible to an interval, not a dimensionless sliver, and that interval is the Planck interval. Therefore, our scientific definition of time forces us
to acknowledge that time is only definable as an interval, the Planck interval.
Since the smallest unit of time is only definable as the Planck interval, this suggests there is a fundamental limit to our ability to measure an event in absolute terms. This fundamental limit to
measure an event in absolute terms is independent of the measurement technology. The error in measuring the start or end of any event will always be at least one Planck interval. This is analogous to
the Heisenberg uncertainty principle, which states it is impossible to know the position and momentum of a particle, such as an electron, simultaneously. Based on fundamental theoretical
considerations, the scientific community widely agrees that the Planck interval is the smallest measure of time possible. Therefore, any event that occurs cannot be measured to occur less than one
Planck interval. This means the amount of uncertainty regarding the start or completion of an event is only knowable to one Planck interval. In our everyday life, our movements consist of a sequence
of Planck intervals. We do not perceive this because the intervals are so small that the movement appears continuous, much like watching a movie where the projector is projecting each frame at the
rate of approximately sixteen frames per second. Although each frame is actually a still picture of one element of a moving scene, the projection of each frame at the rate of sixteen frames per
second gives the appearance of continuous motion. I term this inability to measure an event in absolute terms “the time uncertainty interval.”
Please feel free to browse How to Time Travel.
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Math 4 Wisdom. "Mathematics for Wisdom" by Andrius Kulikauskas. | Research / Octonions
• Complex numbers describes rotations in two-dimensions, and quaternions can be used to describe rotations in three dimensions. Is there a connection between octonions and rotations in four
Clifford algebras
• Relate the octonions to Clifford algebras. Compare the (associative) Clifford algebra construction with the (nonassociative) Cayley-Dickson construction. Give combinatorial interpretations of
both and see how they differ.
• Is the Cayley-Dickson construction associative up to plus or minus signs (up to reflection)?
Modeling cognition
• How are the (nonassociative) octonions relate to the (associative) split-biquaternions?
• Compare the learning three-cycle (for the quaternions) and Fano's plane (eightfold way) for the octonions.
• Is there a way that the octonions get identified with the reals? The eightsome = nullsome gets understood as a onesome? And the identification of nullsome with onesome is related to the field
with one element. And we are left with exceptional Lie structures.
Facts about octonions
• John Baez, based on Dixon: The group of symmetries (or technically, "automorphisms") of the octonions is the exceptional group {$G_2$}, which contains {$SU(3)$}. To get {$SU(3)$}, we can take the
subgroup of {$G_2$} that preserves a given unit imaginary octonion... say {$e_1$}.
Modeling with octonions
• The octonions can model the nonassociativity of perspectives.
• The Clifford algebra {$\textrm{Cl}_{0,7}$} with seven generators (squaring to {$-1$}) and {$2^7$} basis elements models the sevensome. The Clifford algebra {$\textrm{Cl}_{0,3}$}, the
split-biquaternions, with three generators (squaring to {$-1$}) and {$2^3$} basis elements models the threesome. The octonions have three generators and eight basis elements.
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Everything You Ever Wanted To Know About Internal Rate Of Return And Why We Use It With Life Insurance • The Insurance Pro Blog
Everything You Ever Wanted To Know About Internal Rate Of Return And Why We Use It With Life Insurance
Internal rate of return (IRR) is a commonly quoted measurement when talking about whole life and universal life insurance. This is because IRR is well equipped to tell us what sort of return we are
achieving on these types of life insurance. That answer, while very simple, satisfies the curiosity of most.
But the exact reason for using IRR to accomplish this goal is quite complex and an interesting discussion in financial mathematics. It's been quite some time since I last dove through a heady
discussion of mathematic technicality. If you come to The Insurance Pro Blog for the laughs (sarcasm), I'm afraid to say this blog post might leave you a tad disappointed. If, on the other hand,
you're one of the three who enjoys a good numerical thriller (sarcasm…again) then you're in for a treat.
What is Internal Rate of Return?
At its core, internal rate of return is a complex calculation of cash flows evaluated to arrive at the effective return achieved by pursuing an investment option. In order to understand IRR, you
need to be in the right headspace, so you have to first think of investing as costs. In other words, all investments have costs (i.e. what you pay to partake in them).
The majority of people look at investments in a linear fashion. What I mean is, they assume a fixed “investment” and calculate the return based on the payoff of that investment. This is an accurate
way to compute returns insofar as we're only concerned about the return during a very short time period and/or we always incur the same investment cost. Anything that ventures outside of these two
required parameters forces us to either:
1. Attempt to flatten the variations in investment costs
2. Aggregate the time periods as one long time period
In some cases both are necessary. In either case, both approaches will result in slightly (at best) or significantly (at worst and much more likely) flawed conclusions. This is the case, because
our crude adjustments to the input data will generate outputs that magnify the effect of the adjustments.
So internal rate of return is a method for accurately calculating returns based on both the reality of a varied investment cost and a longer time period. The exact formula looks like this:
To calculate actual IRR we'd set net present value (NPV) to zero and solve for r. Doing this long-form (by hand) is nearly impossible because we usually use this formula by plugging a guess number
in for r and computing NPV.
If the net present value comes out to a figure at or near zero, we determine our guess of the resulting IRR. Not too tricky when dealing with a short number of time periods, but quite the task as t
increases. Thankfully, computer software will handle this task quickly and will work to arrive at a true NPV = 0 by using non-integers.
Difference Between Internal Rate of Return and Compound Annual Growth Rate
For those who gain intermediate expertise in financial mathematics, we often see wide use of compound annual growth rate (CAGR) when discussing investment returns. Some people might use the specific
term compound annual growth rate and some may not. The specific CAGR calculation itself only applies to the growth of something over a specified time period without any control for ???
The compound annual growth rate tells us the effective year-over-year return of a single or fixed contribution investment made over time. So CAGR can answer questions like:
• If Sam makes a $100,000 investment in 2010 that is worth $150,000 in 2019, what is his return for the entire 10 year period?
• If Lydia invests $5,000 per year from 2010 through 2019 and has $75,000 in 2019 what is her return for the 10 year period?
Using the calculation for compound annual growth rate, we can tell you that Sam achieved a 4.14% return over 10 years and Lydia achieved a 7.26% return over 10 years.
The easiest way to perform these calculations is with a calculator equipped to perform Time Value of Money (TVM) calculations. If you were forced to buy a TI-83 Plus calculator for high school math
class you own a calculator with the capacity to do this. If you don't happen to own one of these, there are several much less expensive financial calculator options available. Lastly, anyone with
Microsoft Excel also has the ability to easily perform TVM calculations.
The above examples are simply TVM solve for “RATE” scenarios.
By the way, we cover a good introduction to Time Value of Money in Predictable Profits, so if you'd like a crash course that also includes examples on how specifically to plug numbers into the
calculations, you can check it out there.
So this is great, but what happens when the circumstances become more complex. What happens if Sam invests $50,000 in year one and then another $50,000 in year five and ends up with $120,000 in
What happens if Lydia invests $5,000 for the first three years, nothing for the next two years, $3,000 for the following year, and then $8,000 for the remaining years and ends up instead with
$61,000? Calculating the rate of return over the same time period just became far more complex.
This is where knowing how to use the IRR calculation becomes helpful. The internal rate of return tells us that in these new scenarios, Sam achieves a rate of return of 2.29% and Lydia 4.23%.
But IRR is more powerful than just telling us the effective rate of return when someone varies the investments. It can also tell us what the effective return is when someone both puts money in and
takes money out.
Suppose Sam invests $100,000 and takes $25,000 from his investment in year six resulting in a cash balance in year 10 of $110,000. This is another scenario where a simple CAGR calculation lacks the
complexity to calculate his effective rate of return. IRR, however, can easily handle this scenario, his return by year 10 is 3.39%.
Let's assume that Lydia invests $5,000 per year, but in year six she too withdraws $25,000 from the investment; by 2019, she has $30,000 remaining. What is her effective return, or internal rate of
return? It's 6.14%.
When we Say “Effective Rate of Return…”
When I use the term “effective rate of return” here I mean the year-over-year interest equivalent result. So in the above examples, the interest rate you see is the rate Sam and Lydia would need to
earn each year on their investments in order to arrive at the same result in year 10.
This is also what we normally refer to as the geometric average.
Difference between Internal Rate of Return (IRR) and Return on Investment (ROI)
It sounds like IRR focuses on the returns achieved on investments, so why can't we just use a return on investment calculation? Is there a difference? There sure is.
Return on investment is a simple calculation that measures the growth of a lump sum investment. You use ROI when you need to answer the following questions:
Jim invested $50,000 in Amazon in 2014 and sold his entire investment in 2018 for $65,000. What was Jim's total return on investment? The answer is 30%. The return on investment calculation is
simply a percentage change calculation:
The ROI calculation tells us the percentage of growth achieved by making the investment, but it does not account for the time that passed between initial investment and exit. If we wanted to know
what the effective return on investment was for Jim, we don't need IRR. We can simply use the RATE solution for TVM. His year-over-year rate of return is 5.39%.
This calculation also lacks the ability to handle more complexity like partial liquidation. What if Jim sold half of his investment in 2018 for $25,000 and then the remaining investment in 2019,
receiving $30,000 in total in 2019.
That question is too complex for the simple ROI calculation to answer. The IRR calculation, on the other hand, answers it easily. His effective year-over-year rate of return is 1.89%
Why Internal Rate of Return Matters for Life Insurance
When we evaluate life insurance, we have to evaluate against the other options we have to achieve the same goal. People often compare life insurance against investments of all sorts to determine if
the values achieved by the life insurance contract are better, worse, or the same as the anticipated results achieved elsewhere.
The most commonly accepted metric that compares financial vehicles, is the year-over-year growth of its values. Life insurance can be a tad complex when it comes to inflows and outflows. While some
people might pay a set premium for a number of years and simply want to know what the effective return is on the cash value and/or death benefit, many more people don't follow a straight line of
paying the premium.
For example, if you own a 10 Pay whole life policy and wish to know what the effective rate of return achieved in year 25 is, you have to have a mechanism to account for 10 years of inflows, followed
by 15 years of no inflows. Should you decide to take a loan against the policy in year 18, you also need a mechanism to account for this.
The internal rate of return is the only financial calculation that can do this with precision. Once you know the answer, you can compare it to the results you'd expect from other options and
determine if life insurance is a good or bad option for you.
But take note that this means you have to be careful about the way in which you evaluate your alternative options. People often confuse the rate of return. As you can see from above, there are a
few ways to calculate a “return” on an investment. There are also a number of ways one can approach the act of investing.
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Look for check in reference sheet checkbox
I am looking for solution to check a box in one sheet if a box is checked in a different sheet. I tested formula with same idea in current sheet with no issues. It just does not work with reference
=IF({Regulatory}@row = 1, 1, 0) #UNPARSEBLE
If I substitute reference with column in current sheet it works fine
=IF(IsProject@row = 1, 1, 0) Good
The final desire is to check a box in one sheet if a box is checked in any one of three boxes in a different sheet. But I couldn't get it to work for one box so I thought I would start with less
• Hi @skclark9
When you're using cross sheet references, @row function will not be used. I assume your reference is looking at an entire column and not just a cell. If it is just one cell, your formula would be
=IF({Regulatory}=1, 1, 0). If you're referencing the entire column in the source sheet, you might need to narrow it down to a row to look for the value. Your formula would probably be =IF(INDEX
({Regulatory}, MATCH([Search value column in destination]@row, {Column reference of the same search value that is present in the source}, 0))=1, 1, 0)
Aravind GP| Principal Consultant
Atturra Data & Integration
M: +61493337445
W: www.atturra.com
• Hi @AravindGP Your solution did work for 1 variable. Prior to your answer I did find an answer on my own. I was able to create an answer with
=INDEX({Regulatory}, MATCH([Project Name]@row, {Project Name}, 0) which did the same thing as your final IF(Index/Match) statement
However my real problem is at the bottom of my post. I need to Index 4 columns to see if any of the 4 has a box checked (I originally stated 3, but it is now 4). Would a nested IF statement work
and if so what would it look like?
• I would suggest inserting a helper column (that can be hidden after setup) on the source sheet that will check a box if any one of the four are checked.
=IF(OR([Column 1]@row = 1, [Column 2 ]@row = 1, [Column 3]@row = 1, [Column 4]@row = 1), 1)
Then in the target sheet, you would use the INDEX/MATCH but pull in from this helper column.
Help Article Resources
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NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments
NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments are part of NCERT Solutions for Class 12 Physics. Here we have given NCERT Solutions for Class 12 Physics Chapter 9
Ray Optics and Optical Instruments.
Topics and Subtopics in NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments:
┃Section Name│Topic Name ┃
┃9 │Ray Optics and Optical Instruments ┃
┃9.1 │Introduction ┃
┃9.2 │Reflection of Light by Spherical Mirrors ┃
┃9.3 │Refraction ┃
┃9.4 │Total Internal Reflection ┃
┃9.5 │Refraction at Spherical Surfaces and by Lenses ┃
┃9.6 │Refraction through a Prism ┃
┃9.7 │Dispersion by a Prism ┃
┃9.8 │Some Natural Phenomena due to Sunlight ┃
┃9.9 │Optical Instruments ┃
NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments
Question 1.
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp
image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
The object is kept between Æ’ and C. So the image should be real, inverted and beyond C. To locate the sharp image, screen should be placed at the position of image.
So, the image is inverted and magnified. Thus in order to locate the sharp image, the screen should be kept 54 cm in front of concave mirror and image on the screen will be observed real, inverted
and magnified. If the candle is moved closer to the mirror, the real image will move away from the mirror, hence screen has to be shifted away from the mirror to locate the sharp image.
Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the
A convex mirror always form virtual image, which is erect and small in size of an object kept in front of it. Focal length of convex mirror Æ’ = + 15 cm Object distance u = – 12 cm Using mirror
Therefore, image is virtual, formed at 6.67 cm at the back of the mirror.
It shows the image is erect, small in size and virtual. When the needle is moved farther from mirror the image also move towards focus and decreasing in size. As u approaches v approaches focus but
never beyond Æ’.
Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If
water is replaced by a liquid of refractive index 1.63 upto the same height, by what distance would the microscope have to be moved to focus on the needle again?
We know the relation
Now if the water is replaced by other liquid, the apparent depth will change and microscope will have to be further moved to focus the image. With new liquid
Now the microscope will have to shift from its initial position to focus on the needle again which is at 7.67 cm depth. Shift distance = 9.4 – 7.67 = 1.73 cm.
Question 4.
Figures (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass- air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of
incidence in water is 45° with the normal to a water-glass interface (figure c).
(a) Applying Snell’s law for the refraction from air to glass. Refractive index of glass w.r.t. air
(b) Now Snell’s law for the refraction from air to water
(c) Now the light beam is incident at an angle 45° from water to glass
Question 5.
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of
water is 1.33. (Consider the bulb to be a point source)
As shown in the figure all those light rays which are incident on the surface at angle of incidence more than critical angle, does total internal reflection and are reflected back in water only. All
those light rays which are incident below critical angle emerges out of surface bending away from normal. All those light beams which are incident at critical angle grazes the surface of water.
Question 6.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index
of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of
When the light beam is incident from air on to the glass prism, the angle of minimum deviation is 40°. Refractive index of glass w.r.t. air.
Now the prism is placed in water, new angle of minimum deviation can be calculated.
Question 7.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to
be 20 cm?
Both faces should be of same radius of curvature
So, the radius of curvature should be 22 cm for each face of lens.
Question 8.
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is
(a) a convex lens of focal length 20 cm, and
(b) a concave lens of focal length 16 cm?
(a) The convex lens is placed in the path of convergent beam.
The image 1 is formed by further converging beams at a distance of 7.5 cm from lens.
(b) A concave lens is placed in the path of convergent’ beam, the concave lens further diverge the light.
The image I is formed by diverged rays at 48 cm away from concave lens.
Question 9.
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Object of size 3 cm is placed 14 cm in front of concave lens.
So, the image is virtual, erect, of the size 1.8 cm and is located 8.4 cm from the lens on the same side as object. As the object is moved away from the lens, the virtual image moves towards the
focus of the lens but never beyond it. The image also reduces in size as shift towards focus.
Question 10.
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Equivalent focal length of the combination
Hence, system will behave as a diverging lens of focal length 60 cm.
Question 11.
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be
placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
(a) We want the final image at least distance of distinct vision. Let the object in front of objective is at distance Ï…[0].
Now we can get required position of object in point of objective.
(b) We want the final image at infinity. Let us again assume the object in front of objective at distance Ï…[0.
]The object distance υ[e] for the eyepiece should be equal to ƒ[e] = 6.25 cm to obtain final image at ∞. So, image distance of objective lens
Question 12.
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the
objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
The image is formed at least distance of distinct vision for sharp focus. The separation between two lenses will be Ï…[0] + |Ï…[e]|
Let us find first Ï…[0] the image distance for objective lens.
Also we can find object distance for eyepiece Ï…[e] as we know
Question 13.
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and
the eyepiece?
Question 14.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 10^6 m and the radius of lunar orbit is
3.8 × 10^8 sm.
(a) Æ’[o] = 15 m and Æ’[e] = 1.0 cm angular magnification by the telescope normal adjustment
(b) The image of the moon by the objective at lens is formed on its focus only as the moon is nearly infinite distance as compared to focal length.
Distance of object i.e., Radius of lunar orbit, R[o] = 3.8 × 10^8 cm Distance of image for objective lens i.e., focal length of objective lens ƒ[o] = 15 m Radius of image of moon by objective lens
can be calculated.
Question 15.
Use the mirror equation to deduce that:
(a) An object placed between f and 2f of a concave mirror produces a real image beyond 2 f.
(b) A convex mirror always produces a virtual image independent of the location of the object.
(c) The virtual image produced by a convex mirror is always diminished in size and is located Between the focus and the pole.
(d) An object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
(a) We know for a concave mirror Æ’ < 0 [negative] and u < 0 [negative]
(b) For a convex mirror, Æ’ > 0, always positive and object distance u < 0, always negative.
So, whatever be the value of u, a convex mirror always forms a virtual image.
(c) In convex mirror focal length is positive hence Æ’ > 0 and for an object distance from mirror with negative sign (u < 0)
hence the image is located between pole and focus of the mirror
So, the image is virtual and diminished.
(d) In concave mirror, Æ’ < 0 for object placed between focus and pole of concave mirror
Question 16.
A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab
held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
The shift in the image by the thick glass slab can be calculated. Here, shift only depend upon thickness of glass slab and refractive index of glass.
Shift = Real thickness – Apparent of thickness
The answer does not depend on the location of the slab.
Question 17.
(a) Figure shows a cross-section of’light pipe’ made of a glass fiber of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the
angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?
(a) Let us first derive the condition for total internal reflection. Critical angle for the interface of medium 1 and medium 2.
Condition for total internal reflection from core to cladding
Now, for refraction at first surface air to core.
Thus all incident rays which makes angle of incidence between 0° and 60° will suffer total internal reflection in the optical fiber.
(b) When there is no outer covering critical angle from core to surface.
Thus all incident rays at first surface 0° to 90° will suffer total internal reflection inside core.
Question 18.
Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we’see’a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake.
(d) Does the apparent depth of a tank of water change if viewed obliquely? if so, does the apparent depth increase of decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. is this fact of some use to a diamond cutter?
(a) In this situation when rays are convergent behind the mirror, both plane mirror and convex mirror can form real images of virtual objects.
(b) Here, the retina is working as a screen, where the rays are converging, but this screen is not at the position of formed virtual image, in fact the reflected divergent rays are converged by the
eye lens at retina. Thus, there is no contradiction.
(c) An observer in denser medium will observe the fisherman taller than actual height, due to refraction from rare to denser medium.
(d) Apparent depth decreases if viewed obliquely as compared to when observed near normally.
(e) As \(\mu =\frac { 1 }{ sinC }\)hence, sin \(C=\frac { 1 }{ \mu }\)Â refractive
index of diamond is much greater than that of ordinary glass, hence critical angle C for diamond is much smaller (24°) as compared to that of glass (42°).
A skilled diamond cutter thus can take the advantage of such large range of angle of incidence available for total internal reflection 24° to 90°. The diamond can be cut with so many faces, to
ensure that light entering the diamond does multiple total internal reflections before coming out. This behavior produce brilliance i.e., sparkling effect in the diamond.
Question 19.
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens
required for the purpose?
Let the object be placed x m in front of lens the distance of image from the lens is (3 – x) m.
Condition for image to be obtained on the screen, i.e.m real image. 9 – 12Æ’ > 0 or 9 > 12Æ’ or f < 0.75 m. so, maximum focal length is 0.75 m.
Question 20.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
The image of the object can be located on the screen for two positions of convex lens such that u and v are exchanged.
The separation between two positions of the lens is x = 20 cm. It can be observed from figure.
Question 21.
(a) Determine the ‘effective focal length of the combination of two lenses in question 10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side
of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangement. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by
the two-lens system, and the size of the image.
(i) Let a parallel beam of light incident first on convex lens, refraction at convex lens
The parallel beam of light appears to diverge from a point 216 cm from the center of the two lens system.
(ii) Now let a parallel beam of light incident first on concave lens.
The image I1 will act as real object for convex lens at 28 cm.
Thus the parallel incident beam appears to diverge from a point 420 -4 = 416 cm on the left of the center of the two lens system. Hence the answer depend upon which side of the lens system the
parallel beam is made incident. Therefore the effective focal length is different in two situations.
(b) Now an object of 1.5 cm size is kept 40 cm in front of convex lens in the same system of lenses.
Question 22.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the
material of the prism is 1.524.
The beam should be incident at critical angle or more than critical angle, for total internal reflection at second surface of the prism.
Let us first find critical angle for air glass interface.
Question 23.
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
(a) deviate a pencil of white light without much dispersion.
(b) disperse (and displace) a pencil of white light without much deviation.
By using two identical shape prism of crown glass and flint glass kept with their base on opposite sides, we can observe deviation without dispersion or dispersion without deviation.
(a) Deviation without dispersion A white beam incident on crown glass will suffer deviation 6, and angular dispersion Δθ, both
Now the light beam again suffer deviation and dispersion by flint glass prism.
Negative sign shows that two prisms must be placed with their base on opposite sides. Net deviation in this condition
(b) Dispersion without deviation
Here again negative sign shows that two prisms must be placed with their base on opposite sides. Net angular dispersion
Δθ = Δθ[1] + Δθ[2]
Δθ = A[1] (μ[v1] – μ[R1]) + A[2] (μ[v2] – μ[R2])
Question 24.
For a normal eye, the far points at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the
least converging power of the eye lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a
normal eye.
Here the least converging power of eye lens is given as 20 dioptres behind the cornea. If we can calculate the maximum converging power, then we can get the range of accommodation of a normal eye.
(a) To see objects at infinity, the eye uses its least converging power = 40 + 20 = 60 dioptres
∴ Approximate distance between the retina and the cornea eyelens
(b) To focus object at the near point on the retina, we have
dioptres Power of the eye lens = 64 – 40 = 24 dioptres Thus the range of accommodation of the eye lens is approximately 20 to 24 dioptres.
Question 25.
Does short-sightedness (myopia) or long¬> sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of r accommodation? If not, what might cause these defects of
A person with normal ability of accommodation may be myopic or hypermetropic due to defective eye structure, j- When the eye ball from front to back gets too elongated the myopic defect occur,
similarly when the eye ball from front to back gets too shortened the hypermetropia defect occur.
When the eye ball has normal length ‘ but the eye lens loses partially its ability of accommodation, the defect is called “presbyopia” and is corrected in the same I manner as myopia or
Question 26.
A myopic person has been using spectacles of power – 1.0 dioptre for distant vision. During old age, he also needs to use separate reading glass of power + 2.0 dioptres. Explain what may have
Myopic person uses spectacles of power – 1.0 dioptre or concave lens of focal length Æ’ =\(\frac { 1 }{ p }\)Â = -100 cm in order to observe clearly object at infinity. Far point of the person can
be calculated as
Similarly if the person uses spectacles of power + 2.0 dioptre then he must be using convex lens of power + 2.0 dioptre. Focal length and near point can be calculated as
Thus the person also has the defect of hypermetropia and has a near point 50 cm. So having both defects he needs different lenses for distant vision and to see closer objects.
Question 27.
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect
due to? How is such a defect of vision corrected?
This defect is called astigmatism. It arises due to non spherical cornea. The eye lens is ideally spherical and has same curvature in different planes, but in an astigmatic eye due to non spherical
cornea the curvature may be insufficient in different planes.
In the given situation the curvature in the vertical plane is enough, so vertical lines are visible distinctly. But the curvature is insufficient in the horizontal plane, hence horizontal lines
appear blurred. The defect can be corrected by using a cylindrical lens with its axis along vertical. The parallel rays in the vertical plane will suffer no extra refraction but the parallel rays in
the horizontal plane will be refracted largely and converges at the retina, according to the requirement to form the clear image of horizontal lines.
Question 28.
A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
(a) At closest distance of the object the image is formed at least distance of distinct vision and eye is most strained.
At farthest distance of the object the image is formed at ∞ and eye is most relaxed.
(b) Maximum angular magnification
Question 29.
A card sheet divided into squares each of size 1 mm^2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
(a) For magnification by the magnifying lens.
(b) Angular magnification,
(c) No, the linear magnification by a lens and magnifying power (angular magnification) of magnifying glass have different values. The linear magnification is calculated using
image of object at eye lens ‘p’ to the angle subtended by object assumed to be at least distance at eye lens ‘a’.
The linear magnification and angular magnification in microscope have similar magnitude when image is at least distance of distinct vision i.e., 25 cm.
Question 30.
(a) At what distance should the lens be held from the figure in previous question in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
(a) For maximum magnifying power the image should be at least distance of distinct 1 vision i.e., 25 cm.
(b) Linear magnification in the situation of maximum magnifying power.
(c) Maximum magnifying power in the same situation
So, it can be observed that in the situation when image is least distance of distinct vision the angular magnification and linear magnification have similar values.
Question 31.
What should be the distance between the object in previous question and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm^2. Would you be able to
see the squares distinctly with your eyes very close to the magnifier?
Now we want the area of square shaped virtual image as 6.25 mm^2. So, each side of image is I = \(\sqrt { 6.25 }\) = 2.5 mm (Linear magnification) For the given magnifying lens of focal length 10 cm
we can calculate the required position of object.
Thus the required virtual image is closer than normal near point. Thus the eye cannot observe the image distinctly.
Question 32.
Answer the following question:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide
angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving
greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance
between the eye and eyepiece?
(a) In magnifying glass the object is placed closer than 25 cm, which produces image at 25 cm. This closer object has larger angular size than the same object at 25 cm. In this way although the angle
subtended by virtual image and object is same at eye but angular magnification is achieved.
(b) On moving the eye backward away from lens the angular magnification decreases slightly, as both the angle subtended by the
image at eye ‘a’ and by the object at eye ‘α’ decreases. Although the decrease in angle subtended by object a is relatively smaller.
(c) If we decrease focal length, the lens has to be thick with smaller radius of curvature. In a thick lens both the spherical aberrations and chromatic aberrations become pronounced. Further,
grinding for small focal length is not easy. Practically we can not get magnifying power more than 3 with a simple convex lens.
(d) Magnifying power of a compound microscope is given by
(e) If we place our eye too close to the eyepiece, we shall not collect much of the light and also reduce our field of view. When we position our eye slightly away and the area of the pupil of our
eye is greater, our eye will collect all the light refracted by the objective, and a clear image is observed by the eye.
Question 33.
An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Here we want the distance between given objective and eye lens for the required magnification of 30. Let the final image is formed at least distance of distinct vision for eyepiece.
Question 34.
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when:
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25 cm)?
(a) In normal adjustment magnifying power
(b) For the image at least distance of distinct vision
Question 35.
(a) For the telescope described in previous question 34 (a), what is the separation between the objective and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away. What is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?
(a) The separation between objective lens and the eyepiece can be calculated in both the conditions of most relaxed eye and most strained eye. Most relaxed eye L= Æ’[0] + Æ’[e]= 140 + 5 = 145 cm
Most strained eye object distance ‘u[e]‘ for eye lens
(c) Now we want to find the height of final image A”B” assuming it to be formed at 25 cm.
Question 36.
A Cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is
140 mm, where will the final image of an object at infinity be?
Image formed by concave mirror acts as a virtual object for convex mirror. Here parallel rays coming from infinity will focus at 110 mm an axis away from concave mirror. Distance of virtual object
for convex mirror
Hence image is formed at 315 mm from convex mirror.
Question 37.
Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in figure. A current in the coil produces a deflection of 3.5° of the mirror. What is the
displacement of the reflected spot of light on a screen placed 1.5 m away?
If the mirror deflect by 3.5°, the reflected light deflect by 7°, deflection of the spot d can be calculated.
Question 38.
Figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its
inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is
measured to be 30.0 cm. What is the refractive index of the liquid?
Let us first consider the situation when there is no liquid between lens and plane mirror and the image is formed at 30 cm i.e., at the position of object. As the image is formed on the object
position itself, the object must be placed at focus of Biconvex lens.
Æ’[0]Â = 30 cm Radius of curvature of convex lens can be calculated
Now a liquid is filled between lens and plane mirror and the image is formed at position of object at 45 cm. The image is formed on the position of object itself, the object must be placed at focus
of equivalent lens of Biconvex of glass and Plano convex lens of liquid
We hope the NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and
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Compute the square root matrix
Children in primary school learn that every positive number has a real square root. The number x is a square root of s, if x^2 = s.
Did you know that matrices can also have square roots? For certain matrices S, you can find another matrix X such that X*X = S. To give a very simple example, if S = a*I is a multiple of the identity
matrix and a > 0, then X = ±sqrt(a)*I is a square root matrix.
Did you know that some matrices have square roots? Click To Tweet
A matrix square root
I'm going to restrict this article to real numbers, so from now on when I say "a number" I mean a real number and when I say "a matrix" I mean a real matrix. Negative numbers do not have square
roots, so it is not surprising that not every matrix has a square root. For example, the negative identity matrix (-I) does not have a square root matrix.
All positive numbers have square roots, and mathematicians, who love to generalize everything, have defined a class of matrices with properties that are reminiscent of positive numbers. They are
called positive definite matrices, and they arise often in statistics because every covariance and correlation matrix is symmetric and positive definite (SPD).
It turns out that if S is a symmetric positive definite matrix, then there exists a unique SPD matrix X, (called the square root matrix) such that X^2 = A. For a proof, see Golub and van Loan, 3rd
edition, 1996, p. 149. Furthermore, the following iterative algorithm converges quadratically to the square root matrix (ibid, p. 571):
1. Let X[0]=I be the first approximation.
2. Apply the formula X[1] = (X[0] + S*X[0]^-1) / 2 where X^-1 is the inverse matrix of X. The matrix X[1] is a better approximation to sqrt(S) than X[0].
3. Apply the formula X[n+1] = (X[n] + S*X[n]^-1) / 2 iteratively until the process converges. Convergence is achieved when a matrix norm ∥ X[n+1] – X[n] ∥ is as small as you like.
The astute reader will recognize this algorithm as the matrix version of the Babylonian method for computing a square root. As I explained last week, this iterative method implements Newton's method
to find the roots of the (matrix) function f(X) = X*X - S.
Compute a matrix square root in SAS
In SAS, the SAS/IML matrix language is used to carry out matrix computations. To illustrate the square root algorithm, let S be the 7x7 Toeplitz matrix that is generated by the vector {4 3 2 1 0 -1
-2}. I have previously shown that this Toeplitz matrix (and others of this general form) are SPD. The following SAS/IML program implements the iterative procedure for computing the square root of an
SPD matrix:
proc iml;
/* Given an SPD matrix S, this function to compute the square root matrix
X such that X*X = S */
start sqrtm(S, maxIters=100, epsilon=1e-6);
X = I(nrow(S)); /* initial starting matrix */
do iter = 1 to maxIters while( norm(X*X - S) > epsilon );
X = 0.5*(X + S*inv(X)); /* Newton's method converges to square root of S */
if norm(X*X - S) <= epsilon then return( X );
else return(.);
S = toeplitz(4:-2); /* 7x7 SPD example */
X = sqrtm(S); /* square root matrix */
print X[L="sqrtm(S)" format=7.4];
The output shows the square root matrix. If you multiply this matrix with itself, you get the original Toeplitz matrix. Notice that the original matrix and the square root matrix can contain negative
elements, which shows that "positive definite" is different from "has all positive entries."
Functions of matrices
The square root algorithm can be thought of as a mapping that takes an SPD matrix and produces the square root matrix. Therefore it is an example of a function of a matrix. Nick Higham (2008) has
written a book Functions of Matrices: Theory and Computation, which contains many other functions of matrices (exp(), log(), cos(), sin(),....) and how to compute the functions accurately. SAS/IML
supports the EXPMATTRIX function, which computes the exponential function of a matrix.
The square root algorithm in this post is a simplified version of a more robust algorithm that has better numerical properties. Higham (1986), "Newton's Method for the Matrix Square Root" cautions
that Newton's iteration can be numerically unstable for certain matrices. (For details, see p. 543, Eqns 3.11 and 3.12.) Higham suggests an alternate (but similar) routine (p. 544) that is only
slightly more expensive but has improved stability properties.
I think it is very cool that the ancient Babylonian algorithm for computing square roots of numbers can be generalized to compute the square root of a matrix. However, notice that there is an
interesting difference. In the Babylonian algorithm, you are permitted to choose any positive number to begin the square root algorithm. For matrices, the initial guess must commute with the matrix S
(Higham, 1986, p. 540). Therefore a multiple of the identity matrix is the safest choice for an initial guess.
13 Comments
Very cool indeed! I love how your blog posts are individual gems with the bonus of some of them being a chapter to an encompassing blog post. A reward for being "A Do Loop" follower! Thanks...
Sir, I really thankful to you. Its very much helpful for me.I follow this site more than one and half years.this site help me to grow my knowledge related to SAS and SAS SQL procedure
/* Given an SPD matrix S, this function to compute the square root matrix
33 X such that X*X = S */
34 start sqrtm(S, maxIters=100, epsilon=1e-6);
ERROR 22-322: Syntax error, expecting one of the following: ), ','.
ERROR 200-322: The symbol is not recognized and will be ignored.
35 X = I(nrow(S));
35 ! /* initial starting matrix */
36 do iter = 1 to maxIters while( norm(X*X - S) > epsilon );
37 X = 0.5*(X + S*inv(X));
37 ! /* Newton's method converges to square root of S */
38 end;
39 if norm(X*X - S) <= epsilon then return X;
ERROR 22-322: Syntax error, expecting one of the following: ;, (.
ERROR 202-322: The option or parameter is not recognized and will be ignored.
The error is on a line before the START statement. You've probably forgotten a semicolon or a set of parentheses.
He simply runs your code, which fails, and you tell him that he forgot something? Sounds more like you forgot something.
Instead of
return X;
it must be
at least this worked for me. Might be due to SAS Version?
Thanks for writing. My comment was about the FIRST error in his program. You are correct to point out that the log shows a second error near the end of the program.
The RETURN function has supported optional parentheses since SAS/IML 13.1, which was released in 2013 as part of SAS 9.4M1. If you are using an older version of SAS, you need to use
parentheses around the argument. I will update the post so that no one else is confused.
Theoretically yes. In practice the numerical errors propagate like crazy because once you have a slight non-commutativity between $A$ and $X$ you are ending with complete nonsense in 20 steps or
fewer instead of getting the result with machine precision. Try the method on A={{1,2},{2,4.01}} and you'll see what I mean...
Yes, I mentioned this fact in the second-to-last paragraph and included a reference to Higham (1986), which has better stability.
How can we do square root of a matrix in SAS , without using proc iml , Sir?
You can perform the same computation in another language that supports matrix computations. For example, you could use MATLAB, R, or the numPy package in Python.
Thank you sir for your reply , however i am restricted to use SAS without an IML option.
Running the IML code I obtained a different output instead of the matrix you show in this blog post.
Thank you! I have updated the post to display the correct matrix.
The matrix you saw was the square-root matrix for the 6x6 Toeplitz matrix generated by {6,5,4,3,2,1}. I later updated the program to use the 7x7 Toeplitz matrix generated by {4,3,2,1,0,-1,2}.
However, I forgot to update the image. The square-root matrix for the first matrix has all positive entries. The square-root matrix for the second matrix has some negative entries, which is
why I decided to switch examples.
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Ordinal Numbers Functional Iep Objectives - OrdinalNumbers.com
Ordinal Numbers Functional Iep Objectives
Ordinal Numbers Functional Iep Objectives – By using ordinal numbers, it is possible to count infinite sets. These numbers can be used as a tool to generalize ordinal numbers.
One of the fundamental ideas of math is the ordinal number. It is a number that indicates the position of an object within a list. A number that is normally between one and twenty is used to
represent the ordinal number. Ominal numbers serve many functions but are most commonly employed to show the sequence of items on a list.
Charts, words, and numbers can be used to show ordinal numbers. They can also serve to show how a group or pieces are placed.
The vast majority of ordinal numbers fall into one of the two categories. Transfinite ordinals are represented using lowercase Greek letters, while finite ordinals are represented with Arabic
A well-ordered collection should include at least one ordinal, according to the axiom. The first student in the class, for example, would receive the highest score. The contest’s runner-up was the
student with a highest score.
Combinational ordinal numbers
Compound ordinal numbers are multi-digit numbers. They are made by multiplying an ordinal number by its ultimate number. The most frequent uses for the numbers is for ranking or dating reasons. They
don’t use a unique ending for each number, similar to cardinal numbers.
Ordinal numbers identify the sequence of elements found in the collection. The names of items in collections are also identified by using these numbers. There are two types of ordinal numbers:
regular and suppletive.
By prefixing cardinal numbers with the suffix -u regular ordinals can be made. Next, the number is typed as a word . Then, a hyphen follows it. There are a variety of suffixes available.
By affixing words with the -u or–ie suffix results in suffixtive ordinals. The suffix, used for counting is more extensive than the conventional one.
Limit of Ordinal
Limits for ordinal number are ordinal numbers that don’t have zero. Limit ordinal numbers have the disadvantage of having no maximum element for them. They can be created by joining empty sets with
no any maximum element.
Limited ordinal numbers are utilized in transfinite definitions of the concept of recursion. Every infinite cardinal number, based on the von Neumann model, is also an ordinal limit.
A number with limits is equal to the sum all of the ordinals below. Limit ordinal figures are expressed using arithmetic, but they can also be expressed using the natural number series.
The ordinal numbers used for arranging the data are utilized. They provide an explanation of the location of an object’s numerical coordinates. They are often used in set theory or math. While they
belong to the same category however, they aren’t considered to be natural numbers.
The von Neumann model uses a well-ordered set. Assume that fyyfy is one of the subfunctions g’ of a function that is specified as a singular operation. If the subfunction fy’s is (i, ii), and g meets
the requirements that g is a limit ordinal.
The Church Kleene oral is a limit-ordering order that works in a similar way. A limit ordinal is a properly-ordered collection of smaller or less ordinals. It is an ordinal with a nonzero value.
Examples of ordinary numbers in stories
Ordinal numbers are used frequently to illustrate the order of entities and objects. They are critical for organizing, counting, and also for ranking purposes. They are also useful to indicate the
order of things as well as to define objects’ positions.
The letter “th” is typically used to denote the ordinal number. Sometimes however, the letter “nd” could be substituted. The titles of books often include ordinal numbers.
Although ordinal numbers are typically used in lists however, they are also written in words. They can also be expressed in the form of numbers or acronyms. In comparison, they are simpler to grasp
as compared to cardinal numbers.
Three different kinds of ordinal numbers are available. You can learn more about them through practice or games as well as other activities. A key component to improving your math skills is to learn
about them. You can enhance your math skills with coloring exercises. You can track your progress by using a marker sheet.
Gallery of Ordinal Numbers Functional Iep Objectives
Math IEP Goals And Objectives Printable List PDF
Autumn Themed Ordinal Numbers Worksheet teacher Made
Iep Goals Iep Anecdotal Notes
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R - Remove First Value From a Vector - Data Science Parichay
Vectors are used to store one-dimensional data of the same type in R. In this tutorial, we will look at how to remove the first value from a vector in R with the help of some examples.
How to remove the first value in a Vector in R?
Negative indexing can be used to exclude values using their index from a vector. To remove the first value from a vector in R, you can use the negative index -1 inside the [] notation. The following
is the syntax –
# remove first value from vector
vec <- vec[-1]
Note that using a negative index does not modify the vector in place. It simply filters the vector and returns a copy with the value at the given index removed. To modify the original vector, assign
the resulting vector to the original vector variable.
You can similarly remove any value using its index from a vector in R.
Let’s look at some examples of removing the first value from a vector in R.
First, let’s create a vector with four values and remove the first value using the syntax mentioned above.
# create a vector
vec <- c(10, 20, 30, 40)
# remove the first element
vec <- vec[-1]
# display the vector
[1] 20 30 40
Here, we use negative indexing to remove the value the index 1 (which is the first value in the vector). Note that we re-assign the returned vector to the variable vec. You can see that the first
value from the original vector is not present in the resulting vector.
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Let’s look at another example. This time let’s use a vector with named values.
# create a vector
vec <- c("a"=10, "b"=20, "c"=30, "d"=40)
# remove the first element
vec <- vec[-1]
# display the vector
b c d
You can see that the first value from the original vector was removed.
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New Specification A Level Revision Day
Year 13 Revision Day Calculus.
Friday 26th April 2019
University of Hertfordshire, Hatfield.
I was privileged, as an AMSP associate, to be asked by Val Pritchard, area coordinator for Hertfordshire, to assist her in presenting the first full revision day for the new specification A level.
Having spent two years working with students across three boards, OCR, OCR (MEI) & Edexcel I was delighted to accept. I had a good idea that it would be well attended and indeed there were 72
students from 10 local schools.
Whilst the accompanying teachers attended their own sessions with Joanna Deko and Natalie Vernon, the students began at 9.15am with a gentle warm up, from Val, on simplifying algebra and a discussion
on any possible misconceptions. For example
seamlessly picked up again in the first session, on preparing algebra for differentiation and integration when students were given
and a suggested preparation on mini white boards was given as
I hadn’t seen that one coming, it was as if she had planned it! The correct preparation being
Algebra preparation led into my first presentation on how to speed up on differentiation and integration, time pressure being a crucial factor in the new spec. It was lovely to be able to show them
how to differentiate and integrate using ‘inspection’ rather than using a substitution ‘u’. There is rarely time to do this formally in school.
Students were given plenty of practice, 20 questions and I was pleased to be able to speed up my feedback by using a QR code for answers. Did you know it will work from the back of a lecture
(Apologies to those who did not have the QR code app).
Reminder: Learn to differentiate 2^x, it’s in the syllabus and hence the integration.
The second session on differentiation from first principles seemed another gentle build up to the problem solving exam questions to come. However, Val highlighted the trickier questions on trig which
may require students to use the small angle approximations in their differentiation and the questions requiring the gradient at a point. I have seen students confident with differentiation from first
principles wobble with a question like
f(x) = 3x^2 – 2x +3
a) Find f(4 + h)
b) Hence, using differentiation from first principles find f ‘(4)
My advice, practise these questions.
Session three on the Newton Raphson formula began with some lovely GeoGebra visuals on how the formula works and when it will fail, f‘(x) = 0 . Four multiple choice questions and then it was quickly
into the problem solving exam questions covering all the morning’s work. Reinforcements were sent for and the accompanying teachers returned to help with this part. Some really challenging questions
here without the step by step guidance seen in the old spec. My highlights here were i) a student asking if u = e^x would work on
A question taken from OCR C4 June 2006 Q6ii but with the steps taken out (harsh! but by all means possible in the spirit of the new spec). The step given in the original paper was to use u = e^x + 1.
Having prepped in advanced I took this as THE substitution but u=e^x works quite well too.
ii) The 18 marker from AQA C2 Jan 2006 Q8, with the steps taken out again! A lovely question on geometry, differentiation and integration all in one question. I think I will remember this one.
Time for a well-earned lunch and some fresh air before the afternoon start. A great buffet and time to catch up with colleagues.
As the students returned for the afternoon, they definitely needed some encouragement as they began their fourth hour of Maths so I passed on a message from Corbett Maths
Thank you Mr Corbett, it was much appreciated and gave us a boost to start the fourth session, Calculus and Kinematics. My turn again, I started the students off with the basic theory and a couple of
questions on 1D kinematics; one example & one for them. Moving on to 2D, I had already experienced some unfamiliarity here with my own students. I reminded students to put in a constant of
integration for both the i and the j vector when necessary. The newness of this topic hit me when I stumbled over what to call the value in front of either vector, is it ‘the coefficient of i’, ‘the
magnitude of i’, ‘the value of i’? I plumped for ‘the coefficient of i’, answers on a postcard please. A reminder too, to draw a diagram, simple but so effective (possibly not much time to draw
elaborate pictures of ships).
It was with great delight that I read the feedback from one student that they had learnt to connect the compass directions, North and East with the i and j vectors respectively
Some more lovely feedback
The fifth session on differential equations provided the students with a reminder of the difference between a general solution (referred to as a family👨👩👧👦 of functions) and a particular solution
(one member of the family👩). A couple of questions to reinforce the topic and then a second session on problem solving exam questions. Some more fantastic questions selected by Val. Well done to
Robert & Beth for keeping me focussed on a challenging rabbit population and height of the waves question respectively, in the fifth hour of the day🤪 Thank you Ben for rescuing me with your answers.
Pretty impressed with Michael and his neighbour too for sticking with it. That said, all the students worked amazingly hard, very well done. You all deserve to do well in your exams on the 5^th, 12^
th and 14^th June, Good Luck.
Thank you to Val, Jo, Natalie, Kathy, the teachers and the students for making it such a rewarding day. Will I do it again? Give me a year to move on, I might forget the fear of standing up in front
of 70 plus students and the hard work and say Yes😂
Picture of Val Pritchard at the end of the day, looking as if she could do another five hours!
Top Tips:
1. Use the sample papers from all the exam boards and time yourself.
2. Use the formula sheet.
3. Know your calculator.
Best Bit: seeing the fruits of my labour from four years ago in a student who I had taught in Year 9, now thriving doing A level Maths in Year 13. I hadn’t seen him for four years.
If there are any errors, omissions or misrepresentations please email [email protected]
New Specification A Level Revision Day
2 thoughts on “New Specification A Level Revision Day”
• April 29, 2019 at 7:26 pm
I have an offer of ‘the j component’.
• April 29, 2019 at 7:32 pm
Wonderful to read about the positive work you are doing with so many enthusiastic students! Thanks for sharing this.
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Mathematics in Industry: References
Benkowski, S. (1994). Preparing for a job outside academia,
Notices of the American Mathematical Society
41, 917--919.
Bosman, C., D. Kalman, R. Tobin, R. Shaker, J. Lucas, and P. Davis (1993). Preparing M.S. and Ph.D. students for nonacademic careers,
Proceedings of the Conference on Graduate Programs in the Applied Mathematical Sciences II
(R. Fennell, ed.), 61--65, Clemson University, Clemson, South Carolina,
Boyce, W. E. (1975). The mathematical training of the nonacademic mathematician,
SIAM Review
17, 541--557.
Chung, F. R. K. (1991). Should you prepare differently for a nonacademic career?,
Notices of the American Mathematical Society
38, 560--562.
Conference Board of the Mathematical Sciences (1992).
Graduate Education in Transition
, Conference Board of the Mathematical Sciences, Washington, DC.
David, E. E., Jr.,
et al.
Renewing U.S. Mathematics: Critical Resources for the Future
, National Academy Press, Washington, DC.
David, E. E., Jr.,
et al.
Renewing U.S. Mathematics: A Plan for the 1990s
, National Academy Press, Washington, DC.
Davis, P. W. (1991). Some views of mathematics in industry from focus groups,
SIAM Mathematics in Industry Project, Report 1
, Society for Industrial and Applied Mathematics, Philadelphia, Pennsylvania. (Available electronically as siamrpt.dvi by anonymous ftp from pub/forum on ae.siam.org, from gopher.siam.org, or via
Internet from http://www.siam.org.)
Davis, P. W. (1994).
Mathematics in Industry: The Job Market of the Future
, 1994 SIAM Forum Final Report, Society for Industrial and Applied Mathematics, Philadelphia, Pennsylvania.
Davis, P. W. (1995). So you want to work in industry,
Math Horizons
, February 1994, 18--23; Teamwork---the special challenge of industry,
, April 1995, 26--29.
Friedman, A. (1995).
Mathematics in Industrial Problems VII
, IMA Series in Mathematics and its Applications, Springer-Verlag, New York.
Friedman, A., J. Glimm, and J. Lavery (1992).
The Mathematical and Computational Sciences in Emerging Manufacturing Technologies and Management Practices
, Society for Industrial and Applied Mathematics, Philadelphia, Pennsylvania.
Friedman, A. and J. Lavery (1993).
How to Start an Industrial Mathematics Program in the University
, Society for Industrial and Applied Mathematics, Philadelphia, Pennsylvania.
Friedman, A. and W. Littman (1993).
Industrial Mathematics: A Course in Solving Real-World Problems
, Society for Industrial and Applied Mathematics, Philadelphia, Pennsylvania.
Fry, T. C. (1941). Industrial mathematics,
The American Mathematical Monthly, Part II Supplement
48, 1--38.
Hansen, W. L. (1991). Report of the Commission on Graduate Education in Economics,
Journal of Economic Literature
24, 1035--1053; The education and training of economics doctorates,
, 1054--1087.
Holden, C. (1992). Industry: worth considering in the 90's?,
257, 1710--1713.
Horner, P. (1992). Where do we go from here?,
OR/MS Today
, 20--30.
Industrial Research Institute (1991),
Industrial Perspectives on Innovation and Interactions with Universities
, National Academy Press, Washington, DC.
Jackson, A. (1995). NRC report on graduate education,
Notices of the American Mathematical Society
42, 984--985.
Joint Policy Board in Mathematics,
Recognition and Rewards in the Mathematical Sciences
, American Mathematical Society, Providence, Rhode Island.
Lotto, B. (1995). Reaction to Donald McClure's ``Employment experiences of 1990--1991 U.S. institution doctoral recipients in the mathematical sciences'',
Notices of the American Mathematical Society
42, 988--989.
McClure, D. (1995). Employment experiences of 1990--1991 U.S. institution doctoral recipients in the mathematical sciences,
Notices of the American Mathematical Society
42, 754.
National Research Council (1990).
A Challenge of Numbers: People in the Mathematical Sciences
, National Academy Press, Washington, DC.
National Research Council (1991).
Mathematical Sciences, Technology, and Economic Competitiveness
(J. Glimm, ed.), National Academy Press, Washington, DC.
National Research Council (1992).
Educating Mathematical Scientists: Doctoral Study and the Postdoctoral Experience in the United States
(R. Douglas, ed.), National Academy Press, Washington, DC.
National Research Council (1993).
Mathematical Research in Materials Science
, National Academy Press, Washington, DC.
National Research Council (1995).
Mathematical Challenges from Theoretical/Computational Chemistry
, National Academy Press, Washington, DC.
National Research Council (1995).
Reshaping the Graduate Education of Scientists and Engineers
, National Academy Press, Washington, DC.
Natriello, G. (1989). What do employers want in entry-level workers?, Occasional Paper 7, Institute on Education and the Economy, Columbia University, New York.
Science & Engineering Indicators (1991), National Science Board, U.S. Government Printing Office, Washington, DC.
Science & Engineering Indicators (1993), National Science Board, U.S. Government Printing Office, Washington, DC.
Sterrett, A. (1995). A career kit for advisors,
MAA Focus
, February 1995, 23--24.
Weyl, J. (1952).
The Practice of Applied Mathematics in Industry
, Society for Industrial and Applied Mathematics, Philadelphia, Pennsylvania.
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Building Design Optimization
What is optimization?
People use “optimize” rather frequently these days. This is a quote from BBC's website about its sleep profiling tool, “Whether you're having trouble sleeping or not, use the Sleep Profiler and get
tailored advice on how best to optimise your sleep.” For someone doing research using optimization (see definition below), it is quite hard to grapple with the concept of optimizing one's sleep. What
it really means is using this profiler can make you sleep better, which sounds a little less impressive.
Interestingly Oxford Dictionary Online and Merriam-Webster Learner's Dictionary define “optimize” slightly differently:
• “make the best or most effective use of (a situation or resource)” - Oxford Dictionary Online
• “to make (something) as good or as effective as possible” - Merriam-Webster Learner's Dictionary
Although both sound reasonable, read literally the action of “optimize” requires achieving or aiming to achieve superlative qualities (the best, the most or as … as possible), which is rarely
attainable in real life. Daily use of the word in fact means “make most of a situation”, “make more effective use of resource”, or “to make (something) as good as one would”. This is very different
from what we do in optimization studies.
optimization (òp´te-mî-zâ´shen) – The procedure or procedures used to make a system or design as effective or functional as possible, especially the mathematical techniques involved. (The American
Heritage® Dictionary of the English Language, Third Edition copyright © 1992 by Houghton Mifflin Company)
Wikipedia's entry on Mathematical Optimization gives a nice overview of optimization theory and techniques, with plenty of resources for further reading. It serves well as an entrance to the
wonderful world of optimization studies. On the other hand, the sheer amount of information can be off-putting - who has the time to become an optimization expert, in order to solve the problem in
the project due next week? Luckily, as long as you understand the basic concept of optimization and can learn to use a tool, you can use these techniques in your work. Most of the optimization
studies aim for the superlatives. The question ADOPT intends to address, however, is how users can take advantage of this technology in their practice in designing buildings. As a result, we are not
aiming to provide you a tool that gives you the best solution. Instead, our optimization tools will help you identify better solutions yourself. In this context, “optimize” here does mean “improve”.
There are really thousands and thousands of different optimization algorithms (and their flavors). They fall broadly into two camps (see Wiki): iterative methods (many are gradient-based) and
heuristics (mainly direct-search methods, many of which are stochastic). And then there are hybrid methods that combine two or more techniques from either camp. So the total number of possible
algorithms is practically infinite. I like the diagram below very much. It is again from Wikipedia, on Heuristic algorithms.
You can see how complicated the maze of algorithms is. For example, for Evolutionary Algorithms, there are five groups. Each of these groups contains different algorithms for constrained/
unconstrained and/or single/multiple-objective problems. Take multiple-objective Genetic Algorithms (MOGAs) for instance, there are:
• FFGA - Fonseca and Fleming’s multiobjective EA
• NPGA - Niched Pareto Genetic Algorithm
• HLGA - Hajela and Lin’s weighted-sum based approach
• VEGA - Vector Evaluated Genetic Algorithm
• NSGA - Deb and Srinivas's Nondominated Sorting Genetic Algorithm
• PAES - Pareto Archived Evolution Strategy
• PESA - Pareto Envelope-based Selection Algorithm
• SOEA - Single-objective evolutionary algorithm using weighted-sum aggregation
• SPEA - Strength Pareto Evolutionary Algorithm
• NSGA2, SPEA2 and NPGA2, which are substantial upgrades to the original algorithms
• and more.
NSGA2 and SPEA2 are probably the most widely used MOGAs to date. Researchers keep making improvements to them, for example, the pNSGA2 and aNSGA2 (NSGA2 with passive or active archives) methods
reported at BSO12 recently.
“What is the best optimization algorithm” is one of the most frequently asked questions among “newbies”. I know a mathematician, who, when he receives this question, will stare at the inquirer and
blurt out “what is your problem?!” (The exclamation mark was my addition
Evolution does work! Just think that this very mechanism has led to the development of humans and other complex organisms from the humble beginnings of the first single cell organism; using it to
improve building designs is really a trivial task. There is a bit of an initial learning curve, true; especially hard is getting used to the jargon, such as binary encoding, constraints, crossover
rate, tournament selection, elitism, … A large part of the aims of this project is to develop tools that incorporate our understanding of optimization problems in building design, so you, as users of
the tools, do not have to worry too much about the details in the algorithms. Instead, you can focus on solving your design problem.
Optimization problems
Design variables
What are the design options and parameters, and which operation strategies and parameters do people optimise when designing a building? Again, we are in the process of collecting information from
literature. From our experience, though, there are three basic types of variables:
• Design options – this type of variables are normally discrete and nominal. They can be encoded as integers, which may or may not represent an ordinal rank. Optimisation problems involve options
that are often combinatorial, therefore harder to solve. Examples of design options include wall construction type, glazing options and building layout.
• Parameters – can be continuous or discrete, but at least ordinal. Examples of parameters include insulation thickness or U value, ventilation rate and window-to-wall ratio.
• Control signals – a control signal is a timed option or parameter. Building/system operation is defined by a sequence of control signals. Control signals are often treated as a set of options/
parameters by fixing the time sequence. HVAC operation and timing is an example of a control signal.
A building design problem often involves (hierarchical) combinations of options, parameters and control signals. It will be interesting to see how people choose and treat variables in their
optimisation studies.
Problem types
Corresponding to the choice of variables, there are some typical optimisation problems:
• Optimal control problems – control signals only
• Sizing problems – parameters only
• Configuration problems – a small number of design option variables plus parameters. Control signals may or may not be considered. All the variables form a single-root hierarchy (a tree).
• Whole building/system optimisation – design option variables, parameters, and control signals. They may comprise of any number of structurally linked or disconnected sub-problems. It is a forest.
What are the objectives of building optimisation? Or, for what purposes has optimization been used in building design and operation? Lots of information can be found in literature, though there is as
yet no systematic conclusion. We will carry on working to collect and archive research in this area throughout the ADOPT project. The section “Optimisation case studies” gives a template of the kind
of information we are collecting.
From our own experience, optimisation objectives for building design and operation can include the following:
• Minimising energy demand: lighting, heating, cooling, auxiliary (fans and pumps)
• Minimising primary energy consumption
• Minimising embodied energy
• Minimising operational carbon emissions
• Minimising embodied carbon emissions
• Minimising a more inclusive objective of “environmental impact” (for which various evaluation metrics have been used)
• Minimising operational cost
• Minimising capital cost
• Some forms of a margin
• Maximising indoor environmental qualities such as daylight, air quality and thermal comfort
• Maximising adaptability (ability to adapt to future changes)
Often more than one of these objectives may be of interest, and they are also frequently in opposition to each other, for example minimising energy demand often involves increasing capital cost. We
are investigating how objectives are selected and treated, whether a single objective or a multi-objective approach is more desirable, and how to deal with constrained problems (e.g. comfort
criteria, costs etc., these may be treated as a constraint or an objective).
Existing tools
• National Renewable Energy Laboratory:
• Lawrence Berkeley National Laboratory:
• VTT Technical Research Centre of Finland and AaltoUniversity:
• Rhino/Grasshopper plugins such as Galapagos and Octopus
• Libraries in many languages including Python, R, …
• and many more
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Cumulative Frequency Distribution
In Statistics, a cumulative frequency is defined as the total of frequencies, that are distributed over different class intervals. It means that the data and the total are represented in the form of
a table in which the frequencies are distributed according to the class interval. In this article, we are going to discuss in detail about the cumulative frequency distribution, types of cumulative
frequencies, and the construction of the cumulative frequency distribution table with examples in detail.
What is Meant by Cumulative Frequency Distribution?
The cumulative frequency is the total of frequencies, in which the frequency of the first class interval is added to frequency of the second class interval and then the sum is added to the frequency
of the third class interval and so on. Hence, the table that represents the cumulative frequencies that are divided over different classes is called the cumulative frequency table or cumulative
frequency distribution. Generally, the cumulative frequency distribution is used to identify the number of observations that lies above or below the particular frequency in the provided data set.
Types of Cumulative frequency Distribution
The cumulative frequency distribution is classified into two different types namely: less than ogive or cumulative frequency and more/greater than cumulative frequency.
Less Than Cumulative Frequency:
The Less than cumulative frequency distribution is obtained by adding successively the frequencies of all the previous classes along with the class against which it is written. In this type, the
cumulate begins from the lowest to the highest size.
Greater Than Cumulative Frequency:
The greater than cumulative frequency is also known as the more than type cumulative frequency. Here, the greater than cumulative frequency distribution is obtained by determining the cumulative
total frequencies starting from the highest class to the lowest class.
Graphical Representation of Less Than and More Than Cumulative Frequency
Representation of cumulative frequency graphically is easy and convenient as compared to representing it using table, bar-graph, frequency polygon etc.
The cumulative frequency graph can be plotted in two ways:
1. Cumulative frequency distribution curve(or ogive) of less than type
2. Cumulative frequency distribution curve(or ogive) of more than type
Steps to Construct Less than Cumulative Frequency Curve
The steps to construct the less than cumulative frequency curve are as follows:
1. Mark the upper limit on the horizontal axis or x-axis.
2. Mark the cumulative frequency on the vertical axis or y-axis.
3. Plot the points (x, y) in the coordinate plane where x represents the upper limit value and y represents the cumulative frequency.
4. Finally, join the points and draw the smooth curve.
5. The curve so obtained gives a cumulative frequency distribution graph of less than type.
To draw a cumulative frequency distribution graph of less than type, consider the following cumulative frequency distribution table which gives the number of participants in any level of essay
writing competition according to their age:
Table 1 Cumulative Frequency distribution table of less than type
Age Group Number of participants
Level of Essay Age group Cumulative Frequency
(class interval) (Frequency)
Level 1 10-15 Less than 15 20 20
Level 2 15-20 Less than 20 32 52
Level 3 20-25 Less than 25 18 70
Level 4 25-30 Less than 30 30 100
On plotting corresponding points according to table 1, we have
Steps to Construct Greater than Cumulative Frequency Curve
The steps to construct the more than/greater than cumulative frequency curve are as follows:
1. Mark the lower limit on the horizontal axis.
2. Mark the cumulative frequency on the vertical axis.
3. Plot the points (x, y) in the coordinate plane where x represents the lower limit value and y represents the cumulative frequency.
4. Finally, draw the smooth curve by joining the points.
5. The curve so obtained gives the cumulative frequency distribution graph of more than type.
To draw a cumulative frequency distribution graph of more than type, consider the same cumulative frequency distribution table, which gives the number of participants in any level of essay writing
competition according to their age:
Table 2 Cumulative Frequency distribution table of more than type
Age Group Number of participants
Level of Essay Age group Cumulative Frequency
(class interval) (Frequency)
Level 1 10-30 More than 10 20 100
Level 2 15-30 More than 15 32 80
Level 3 20-30 More than 20 18 48
Level 4 25-30 More than 25 30 30
On plotting these points, we get a curve as shown in the graph 2.
These graphs are helpful in figuring out the median of a given data set. The median can be found out by drawing both types of cumulative frequency distribution curves on the same graph. The value of
of the point of intersection of both the curves gives the median of the given set of data. For the given table 1, the median can be calculated as shown:
Example on Cumulative Frequency
Create a cumulative frequency table for the following information, which represent the number of hours per week that Arjun plays indoor games:
Arjun’s game time:
Days No. of Hours
Monday 2 hrs
Tuesday 1 hr
Wednesday 2 hrs
Thursday 3 hrs
Friday 4 hrs
Saturday 2 hrs
Sunday 6 hrs
Let the no. of hours be the frequency.
Hence, the cumulative frequency table is calculated as follows:
Days No. of Hours (Frequency) Cumulative Frequency
Monday 2 hrs 2
Tuesday 1 hr 2+1 = 3
Wednesday 2 hrs 3+2 = 5
Thursday 3 hrs 5+3 = 8
Friday 4 hrs 8+4 = 12
Saturday 2 hrs 12+2 = 14
Sunday 6 hrs 14+6 = 20
Therefore, Arjun spends 20 hours in a week to play indoor games.
Thus using statistics we can tackle many real-life problems. To know more about statistics, download BYJU’S-The Learning App from Google Play Store.
Frequently Asked Questions on Cumulative Frequency Distribution
What is meant by cumulative frequency?
The cumulative frequency (c.f) is defined as the total of frequencies, where the frequency of the first class interval is added to the frequency of the second class interval and then the sum is added
to the frequency of the third class interval and so on.
What is meant by cumulative frequency distribution?
A table that shows the cumulative frequencies, which are distributed over different classes is known as the cumulative frequency table or cumulative frequency distribution.
What are the two types of cumulative frequencies?
The two types of cumulative frequencies are less than cumulative frequency and more than cumulative frequency.
What is meant by cumulative frequency series?
The series of frequencies that are added continuously corresponding to each class interval is known as the cumulative frequency series.
How to calculate the cumulative frequency?
The cumulative frequency can be calculated by adding the frequency of the first class interval to the frequency of the second class interval. After that, the sum is added to the frequency of the
third class interval, etc.
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NetLogo Models Library:
Sample Models/Computer Science
(back to the library)
Artificial Neural Net - Perceptron
If you download the NetLogo application, this model is included. You can also Try running it in NetLogo Web
WHAT IS IT?
Artificial Neural Networks (ANNs) are computational parallels of biological neurons. The "perceptron" was the first attempt at this particular type of machine learning. It attempts to classify input
signals and output a result. It does this by being given a lot of examples and attempting to classify them, and having a supervisor tell it if the classification was right or wrong. Based on this
information the perceptron updates its weights until it classifies all inputs correctly.
For a while it was thought that perceptrons might make good general pattern recognition units. However, it was discovered that a single perceptron can not learn some basic tasks like 'xor' because
they are not linearly separable. This model illustrates this case.
The nodes on the left are the input nodes. They can have a value of 1 or -1. These are how one presents input to the perceptron. The node in the middle is the bias node. Its value is constantly set
to '1' and allows the perceptron to use a constant in its calculation. The one output node is on the right. The nodes are connected by links. Each link has a weight.
To determine its value, an output node computes the weighted sum of its input nodes. The value of each input node is multiplied by the weight of the link connecting it to the output node to give a
weighted value. The weighted values are then all added up. If the result is above a threshold value, then the value is 1, otherwise it is -1. The threshold value for the output node in this model is
While the network is training, inputs are presented to the perceptron. The output node value is compared to an expected value, and the weights of the links are updated in order to try and correctly
classify the inputs.
HOW TO USE IT
SETUP will initialize the model and reset any weights to a small random number.
Press TRAIN ONCE to run one epoch of training. The number of examples presented to the network during this epoch is controlled by EXAMPLES-PER-EPOCH slider.
Press TRAIN to continually train the network.
Moving the LEARNING-RATE slider changes the maximum amount of movement that any one example can have on a particular weight.
Pressing TEST will input the values of INPUT-1 and INPUT-2 to the perceptron and compute the output.
In the view, the larger the size of the link the greater the weight it has. If the link is red then its a positive weight. If the link is blue then its a negative weight.
If SHOW-WEIGHTS? is on then the links will be labeled with their weights.
The TARGET-FUNCTION chooser allows you to decide which function the perceptron is trying to learn.
The perceptron will quickly learn the 'or' function. However it will never learn the 'xor' function. Not only that but when trying to learn the 'xor' function it will never settle down to a
particular set of weights as a result it is completely useless as a pattern classifier for non-linearly separable functions. This problem with perceptrons can be solved by combining several of them
together as is done in multi-layer networks. For an example of that please examine the ANN Neural Network model.
The RULE LEARNED graph visually demonstrates the line of separation that the perceptron has learned, and presents the current inputs and their classifications. Dots that are green represent points
that should be classified positively. Dots that are red represent points that should be classified negatively. The line that is presented is what the perceptron has learned. Everything on one side of
the line will be classified positively and everything on the other side of the line will be classified negatively. As should be obvious from watching this graph, it is impossible to draw a straight
line that separates the red and the green dots in the 'xor' function. This is what is meant when it is said that the 'xor' function is not linearly separable.
The ERROR VS. EPOCHS graph displays the relationship between the squared error and the number of training epochs.
Try different learning rates and see how this affects the motion of the RULE LEARNED graph.
Try training the perceptron several times using the 'or' rule and turning on SHOW-WEIGHTS? Does the model ever change?
How does modifying the number of EXAMPLES-PER-EPOCH affect the ERROR graph?
Can you come up with a new learning rule to update the edge weights that will always converge even if the function is not linearly separable?
Can you modify the LEARNED RULE graph so it is obvious which side of the line is positive and which side is negative?
This model makes use of some of the link features. It also treats each node and link as an individual agent. This is distinct from many other languages where the whole perceptron would be treated as
a single agent.
Artificial Neural Net shows how arranging perceptrons in multiple layers can overcomes some of the limitations of this model (such as the inability to learn 'xor')
Several of the equations in this model are derived from Tom Mitchell's book "Machine Learning" (1997).
Perceptrons were initially proposed in the late 1950s by Frank Rosenblatt.
A standard work on perceptrons is the book Perceptrons by Marvin Minsky and Seymour Papert (1969). The book includes the result that single-layer perceptrons cannot learn XOR. The discovery that
multi-layer perceptrons can learn it came later, in the 1980s.
Thanks to Craig Brozefsky for his work in improving this model.
If you mention this model or the NetLogo software in a publication, we ask that you include the citations below.
For the model itself:
Please cite the NetLogo software as:
Copyright 2006 Uri Wilensky.
This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License. To view a copy of this license, visit https://creativecommons.org/licenses/by-nc-sa/3.0/ or send a
letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA.
Commercial licenses are also available. To inquire about commercial licenses, please contact Uri Wilensky at uri@northwestern.edu.
(back to the NetLogo Models Library)
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Hardest Math Problem Ever: Top 10 Toughest
Hardest Math Problem Ever: Top 10 Toughest. In This Article, You Will Learn About Hardest Math Problem Ever.
Hardest Math Problem. Have you ever wondered what the answer to this question might be? What are the hardest math problem ever? And why do they even exist? Are they really hard or can they eventually
be solved?
We will do justice to all those questions in your head right here in this article. So, keep reading to find out more!
Hardest math problem
Below are a list of the top 10 hardest math problems:
1. The Poincaré Conjecture
A non-profit organization called the Clay Mathematics Institute urged people to solve seven arithmetic puzzles in 2000 with the goal of “increasing and disseminating mathematical knowledge.” If
someone could solve even one of the issues, they would get $1,000,000. All of them remain open to this day, with the exception of the Poincaré hypothesis.
Around the turn of the 20th century, French mathematician Henri Poincaré laid the groundwork for the field that is now known as topology. Here’s the concept: Mathematical techniques to differentiate
between abstract shapes are what topologists want. Classifying all of the 3D shapes, such as balls and doughnuts, wasn’t too difficult. A ball is the most basic of these shapes, in a major way.
Poincaré then advanced to 4-dimensional objects and posed a related query. The conjecture eventually became “Every simply-connected, closed 3-manifold is homeomorphic to S^3,” which is effectively
the statement “the simplest 4D shape is the 4D equivalent of a sphere,” after various developments and revisions.
2. Fermat’s Last Theorem
French mathematician and lawyer Pierre de Fermat lived in the 17th century. Since Fermat seems to enjoy math more than anything else, many of his theorems were shared through informal letters by one
of the finest mathematicians in history. He asserted things without providing evidence, thus it would take decades or even centuries for other mathematicians to do the same. The hardest of them is
now referred to as Fermat’s Last Theorem.
It is an easy one to write. x²+y²=z² is satisfied by a large number of trios of integers (x,y,z). These, along with (3,4,5) and (5,12,13), are referred to as the Pythagorean Triples. Right now, what
trios (x, y, z) fulfill x³ + y³ = z³? According to Fermat’s Last Theorem, the answer is no.
3. The Classification of Finite Simple Groups
Abstract algebra has many uses, ranging from body-swapping fact validation on Futurama to solving Rubik’s Cube. Sets with certain fundamental characteristics, such as containing a “identity element,”
which functions similarly to adding 0, are known as algebraic groups.
Groups can be infinite or finite, and depending on your choice of n, it can get very hard to find out what groups of a given size look like.
There is just one way that group can look if n is two or three. When n reaches 4, two things can happen. Mathematicians naturally sought an exhaustive list of all potential group sizes for a given
4. The Four Color Theorem
This one is as simple to make clear as it is to demonstrate.
Four crayons and any map will do. Every state (or nation) can be colored on the map, but there is just one restriction: no two states with the same boundary can have the same color.
The Five Color Theorem, which states that each map may be colored with five different hues, was established in the 1800s. However, it took until 1976 to reduce that to four.
Wolfgang Hakan and Kenneth Appel, two mathematicians at the University of Illinois at Urbana-Champaign, discovered a method to condense the evidence to a sizable, limited number of cases. After
thoroughly reviewing the almost 2,000 cases with the aid of a computer, they produced a proof style that had never been seen before.
5. (The Independence of) The Continuum Hypothesis
German mathematician Georg Cantor shook the world in the late 1800s by discovering that infinities have cardinalities, or distinct sizes. He established the fundamental cardinality theorems that are
typically taught to current math majors in their discrete math courses.
Cantor established the inequality |ℝ|>|ℕ|, which states that the set of real numbers is greater than the set of natural numbers. Since no infinite set is smaller than ℕ, it was simple to prove that
the size of the natural numbers, |ℕ|, is the first infinite size.
The real numbers are bigger now, but do they represent the second infinite size as well? The Continuum Hypothesis (CH), as this subject is known, turned out to be far more difficult.
6. Gödel’s Incompleteness Theorems
Gödel produced mathematical logic that was truly revolutionary. Gödel enjoyed proving things, but he also enjoyed proving the impossibility of proving things. There’s a good chance to explain his
incompleteness theorems.
There are always some statements in any proof language that cannot be proven, according to Gödel’s First Incompleteness Theorem. You can never fully prove something to be true, but it is always true.
With some careful consideration, one can comprehend a version of Gödel’s argument that is not strictly mathematical.
7. The Prime Number Theorem
Numerous theorems exist concerning prime numbers. Even the most basic fact—that there are an endless number of prime numbers—can charmingly be put into a haiku.
More subtly, the Prime Number Theorem describes how prime numbers appear along the number line. To put it more accurately, it states that, given a natural number N, the number of primes below N is
roughly equal to N/log(N), with the word “approximately” carrying all the usual statistical nuances.
In 1898, the Prime Number Theorem was also independently proved by two mathematicians, Jacques Hadamard and Charles Jean de la Vallée Poussin, using concepts from the middle of the 19th century.
Since then, rewrites of the proof have been common, leading to numerous purely aesthetic changes and simplifications. However, the theorem’s influence has only increased.
8. Solving Polynomials by Radicals
Do you recall the quadratic formula? The solution to ax²+bx+c=0 is x=(-b±√(b^2-4ac))/(2a), which, although it may have seemed difficult to learn in high school, is a conveniently closed-form
It is now possible to find a closed form for “x=” if we increase the expression up to ax³+bx²+cx+d=0, albeit it is much more complex than the quadratic version. This can also be done for degree 4
polynomials, such as ax⁴+bx³+cx²+dx+f=0, but it looks ugly.
As early as the fifteenth century, it became clear that doing this for polynomials of any degree was the goal. However, a closed form is not feasible beyond degree 5. It’s one thing to write the
forms when they are feasible, but how did mathematicians demonstrate that forms from 5 onward are not feasible?
9. Trisecting an Angle
Let’s travel far back in time to wrap up.
Using an unmarked compass and straightedge, the Ancient Greeks thought of how to create lines and shapes in different ratios. It is possible for you to draw the line that precisely divides an angle
in half if someone draws an angle on a piece of paper in front of you and provides you with an unmarked ruler, a simple compass, and a pen. The Greeks knew these four simple steps two millennia ago,
and they are nicely illustrated here.
Making a thirds angle was what they failed to accomplish. It remained evasive for a whopping fifteen centuries, despite hundreds of fruitless attempts to locate a construction. Apparently, it is not
possible to construct such a thing.
10. The Twin Prime Conjecture
A mathematical conjecture known as the Twin Prime Conjecture asserts that there are an endless number of prime number pairs with two-way differences.
Number theory is the study of natural numbers and their properties, which frequently involves prime numbers. The Twin Prime Conjecture and Goldbach’s Conjecture are two major theories in this area.
You’ve been studying these numerical concepts since elementary school, so articulating the conjectures comes easily to you.
What is the hardest math problem?
The Riemann Hypothesis of 1859, first out by German mathematician Bernhard Riemann (1826–1866), is regarded by mathematicians worldwide as the most significant unsolved mathematical mystery.
According to the hypothesis, any nontrivial root of the Zeta function has the form (1/2 + b I).
Hardest math problem in the world
The hardest math problem
Hardest math problem in the world with answer
FAQs About Hardest Math Problem Ever: Top 10 Toughest
1. What is the hardest math problem in the world?
For more than 150 years, mathematicians have been confused by the Riemann Hypothesis, a mathematical assertion put forth by German mathematician Bernhard Riemann in 1859.
2. What is the World’s hardest math problem?
The Riemann Hypothesis of 1859, first out by German mathematician Bernhard Riemann (1826–1866), is regarded by mathematicians worldwide as the most significant unsolved mathematical mystery.
According to the hypothesis, any nontrivial root of the Zeta function has the form (1/2 + b I).
3. What are the 7 hardest math problems?
The Poincare Conjecture, the Riemann Hypothesis, the Yang-Mills Theory, P vs NP, the Hodge Conjecture, the Navier-Stokes Equations, and the Birch and Swinnerton-Dyer Conjecture are the seven
difficulties. The Poincare Conjecture was demonstrated in 2003 by Grigori Perelman, a Russian mathematician.
4. What is the 1 hardest math problem?
Most modern mathematicians would generally concur that the Riemann Hypothesis represents the most important unsolved issue in mathematics. One of the seven Millennium Prize Problems, the answer to
which will earn a $1 million reward.
5. Which maths are the most difficult?
For good reason, many people consider Further Mathematics to be the hardest A-Level subject. This course builds on ordinary mathematics by covering a variety of advanced subjects such as differential
equations, advanced calculus, and abstract algebra.
This article contains all the details about this particular topic and we hope this satisfies your curiosity while also educating you at the same time.
We hope you found this article helpful? Stay tuned for more updates like this!
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Konrad Voelkel
Thursday, April 04th, 2013 | Author: Konrad Voelkel
This is about Białynicki-Birula's paper from '72 on actions of reductive linear algebraic groups on non-singular varieties, in particular Gm-operations on smooth projective varieties. I give a proof
sketch of Theorem 4.1 therein and explain a little bit how Brosnan applied these results in 2005 to get decompositions of the Chow motive of smooth projective varieties with Gm-operation. Wendt used
these methods in 2010 to lift such a decomposition on the homotopy-level, to prove that smooth projective spherical varieties admit stable motivic cell decompositions. Most of this blogpost consists
of an outline of the B-B paper.
Continue reading «Białynicki-Birula and Motivic Decompositions»
Category: English, Mathematics | Comments off
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How to Identify the Four Conic Sections in Equation Form - dummies
Each conic section has its own standard form of an equation with x- and y-variables that you can graph on the coordinate plane. You can write the equation of a conic section if you are given key
points on the graph.
Being able to identify which conic section is which by just the equation is important because sometimes that's all you're given (you won't always be told what type of curve you're graphing). Certain
key points are common to all conics (vertices, foci, and axes, to name a few), so you start by plotting these key points and then identifying what kind of curve they form.
The equations of conic sections are very important because they tell you not only which conic section you should be graphing but also what the graph should look like. The appearance of each conic
section has trends based on the values of the constants in the equation. Usually these constants are referred to as a, b, h, v, f, and d. Not every conic has all these constants, but conics that do
have them are affected in the same way by changes in the same constant. Conic sections can come in all different shapes and sizes: big, small, fat, skinny, vertical, horizontal, and more. The
constants listed above are the culprits of these changes.
An equation has to have x^2 and/or y^2 to create a conic. If neither x nor y is squared, then the equation is that of a line. None of the variables of a conic section may be raised to any power other
than one or two.
Certain characteristics are unique to each type of conic and hint to you which of the conic sections you're graphing. In order to recognize these characteristics, the x^2 term and the y^2 term must
be on the same side of the equal sign. If they are, then these characteristics are as follows:
• Circle: When x and y are both squared and the coefficients on them are the same — including the sign.
For example, take a look at 3x^2 – 12x + 3y^2 = 2. Notice that the x^2 and y^2 have the same coefficient (positive 3). That info is all you need to recognize that you're working with a circle.
• Parabola: When either x or y is squared — not both.
The equations y = x^2 – 4 and x = 2y^2 – 3y + 10 are both parabolas. In the first equation, you see an x^2 but no y^2, and in the second equation, you see a y^2 but no x^2. Nothing else matters —
signs and coefficients change the physical appearance of the parabola (which way it opens or how wide it is) but don't change the fact that it's a parabola.
• Ellipse: When x and y are both squared and the coefficients are positive but different.
The equation 3x^2 – 9x + 2y^2 + 10y – 6 = 0 is one example of an ellipse. The coefficients of x^2 and y^2 are different, but both are positive.
• Hyperbola: When x and y are both squared, and exactly one of the coefficients is negative and exactly one of the coefficients is positive.
The equation 4y^2 – 10y – 3x^2 = 12 is an example of a hyperbola. This time, the coefficients of x^2 and y^2 are different, but exactly one of them is negative and one is positive, which is a
requirement for the equation to be the graph of a hyperbola.
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How much sand required for 100 sq ft roof slab? - Civil Sir
How much sand required for 100 sq ft roof slab?
The amount of sand needed for a roof slab depends on the thickness of the slab. On average, for a 4-inch thickness, you might need approximately 650 kilograms (0.65 metric tons) of sand for every 100
square feet.
Hi guys in this article we know about sand required for 100 square feet slab. RCC roof slab casting process is generally based on civil engineering desipline & building construction. Quantity of sand
required in RCC roof casting process different grade of concrete like M20,m25 and so more—. 100 square feet Chhat dhalai me kitna cft sand ( river sand, balu, stone chips) lagega, in hindi 1000 स्क्वायर
फीट छत ढ़लाई मे कितना बालू (river sand) लगेगा.
M20 grade of concrete is mixture of cement sand and aggregate with water have compressive strength 20 N/mm2 and M20 grade M stands for mix and numerical figure 20 is stand for compressive strength of
concrete curing time of 28 days. in this topic topic know about how much sand required for m20 concrete.
Sand required for 100 sq ft roof slab of m20 concrete depends on many factors like:-
1) Thickness of RCC slab:- thickness of RCC roof slab is ranging between 4 inch to 6 inch. Thickness of RCC slab require more quantity of sand, obviously 6 inch thick RCC slab require more quantity
of send than 4 inch thick.
2) Cement sand and aggregate ratio for different grade of concrete which are used in roof casting process.
Ratio of cement sand and aggregate in M20 grade of concrete is 1:1.5:3 in which 1 part is cement, 1.5 part is sand and 3 part is aggregate.
Sand required for 100 sq ft roof slab
There are various steps of procedure to calculate sand quantity required for 100 square feet RCC roof slab
Sand required for 100 sq ft roof slab 4 inch thick
● step 1 :- suppose we have given 100 sq ft area of rcc roof slab and there thickness is 4 inch which is equal to 0.334 feet.
● step 2 :- calculate wet volume of concrete by multiplying area and thickness of slab = 100 sq ft × 0.334 ft = 33.4 cu ft, hance wet volume of concrete is equal to 33.4 cu ft.
● step 3 :- we know that dry volume of concrete is actual ingredients which are required so we have to convert wet volume of concrete in dry volume.
For the conversion of wet volume to dry volume we will multiply 1.54 into wet volume.
Why we multiply 1.54 in wet volume? Because when dry ingredient of concrete cement, sand and aggregate is mix with water then there volume is decrease by 54% due to evaporation of voids present in
mixture, hance dry volume of concrete increased by 50%
Dry volume of concrete is equal to = 33.4 cu ft ×1.54 = 51.5 cu ft, and we know that dry volume of concrete are used to find out sand quantity required for 100 square feet RCC roof slab.
● step 4:- mix ratio in M20 grade of concrete is 1:1.5:3 in which total proportion of sand quantity is equal to 1.5/5.5.
● step 5:- now quantity of sand required for roof slab is calculated as follows = 1.5/5.5 × 51.5 cu ft = 14.1 cu ft, and we know that one CFT sand weight is 46 kg and converting into kg we will
multiply 46 into 14.1 CFT = 14.1×46= 648.6 kgs = 0.6486 MT
● Ans. :- 14.1 cuft (648.6 kgs) sand quantity required for 100 square feet roof slab of 4 inch thick.
100 square feet chhat dhalai 4 inch thick me kitna sand (balu, river sand) lagega iska answer— 100 square feet Chhat dhalai me 14.1 cuft (648.6 kgs) sand quantity lagega.
Sand required for 100 sq ft roof slab 5 inch thick
● step 1 :- suppose we have given 100 sq ft area of rcc roof slab and there thickness is 5 inch which is equal to 0.417 feet.
● step 2 :- calculate wet volume of concrete by multiplying area and thickness of slab = 100 sq ft × 0.417 ft = 41.7 cu ft, hance wet volume of concrete is equal to 41.7 cu ft.
● step 3 :- we know that dry volume of concrete is actual ingredients which are required so we have to convert wet volume of concrete in dry volume.
For the conversion of wet volume to dry volume we will multiply 1.54 into wet volume.
Dry volume of concrete is equal to = 41.7 cu ft ×1.54 = 64.2 cu ft, and we know that dry volume of concrete are used to find out sand quantity required for 100 square feet RCC roof slab.
● step 4:- mix ratio in M20 grade of concrete is 1:1.5:3 in which total proportion of sand quantity is equal to 1.5/5.5.
● step 5:- now quantity of sand required for roof slab is calculated as follows = 1.5/5.5 ×64.2 cu ft = 17.5 cu ft, and we know that one CFT sand weight is 46 kg and converting into kg we will
multiply 46 into 17.5 CFT = 17.5×46= 805.4 kgs = 0.8054 MT
● Ans. :- 17.5 cuft (805.4 kgs) sand quantity required for 100 square feet roof slab of 5 inch thick.
100 square feet chhat dhalai 5 inch thick me kitna sand (balu, river sand) lagega iska answer— 100 square feet Chhat dhalai me 17.5 cuft (805.4 kgs) sand quantity lagega.
Sand required for 100 sq ft roof slab 6 inch thick
● step 1 :- suppose we have given 100 sq ft area of rcc roof slab and there thickness is 6 inch which is equal to 0.50 feet.
● step 2 :- calculate wet volume of concrete by multiplying area and thickness of slab = 100 sq ft × 0.50 ft = 50.0 cu ft, hance wet volume of concrete is equal to 50.0 cu ft.
● step 3 :- we know that dry volume of concrete is actual ingredients which are required so we have to convert wet volume of concrete in dry volume.
For the conversion of wet volume to dry volume we will multiply 1.54 into wet volume.
Dry volume of concrete is equal to = 50 cu ft ×1.54 = 77 cu ft, and we know that dry volume of concrete are used to find out sand quantity required for 100 square feet RCC roof slab.
● step 4:- mix ratio in M20 grade of concrete is 1:1.5:3 in which total proportion of sand quantity is equal to 1.5/5.5.
● step 5:- now quantity of sand required for roof slab is calculated as follows = 1.5/5.5 ×77 cu ft = 21 cu ft, and we know that one CFT sand weight is 46 kg and converting into kg we will multiply
46 into 21 CFT = 21×46= 966 kgs = 0.9660 MT
● Ans. :- 21 cuft (966 kgs) sand quantity required for 100 square feet roof slab of 6 inch thick.
100 square feet chhat dhalai 6 inch thick me kitna sand (balu, river sand) lagega iska answer— 100 square feet Chhat dhalai me 21 cuft (966 kgs) sand quantity lagega.
How much sand required for 100 sq ft roof slab?
Now question is how much sand is required for 100 sq ft roof slab? Their answer is as follow
● Ans. :- 14.1 cu ft (648.6 kgs), 17.5 cu ft (805.4 kgs) & 21 cuft (966 kgs) sand quantity required for 100 square feet roof slab of 4 inch,5 inch & 6 inch thick respectively.
● Ans. :- 14.1 cu ft (648.6 kgs), 17.5 cu ft (805.4 kgs) & 21 cuft (966 kgs) sand quantity 4 inch, 5 inch & 6 inch thick 100 square feet chhat dhalai Mein lagega.
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You should also visits:-
1)what is concrete and its types and properties
2) concrete quantity calculation for staircase and its formula
The amount of sand needed for a roof slab depends on the thickness of the slab. On average, for a 4-inch thickness, you might need approximately 650 kilograms (0.65 metric tons) of sand for every 100
square feet.
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Octal to Decimal Converter
Octal to Decimal is a numeric converter tool that helps you to convert from octal to decimal or decimal to octal in real time. Also, you can use this octal to decimal converter for small and big
numbers (negative numbers included).
This tool allows you to convert from one base to another: you can convert octal to decimal, but also use it as a decimal to octal converter. For decimal to octal, this tool provides the same features
for numeric precision and real-time conversion, and you can quickly change the source base.
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COVID-19 has posed unique challenges to the world and forced people to adapt to new ways of living. For example social distancing has become the norm in public spaces. In the U.S. keeping people 6
feet apart has become a rigid requirement in any social gathering. While sitting at some of these events, I have started to wonder: is this socially distanced arrangement optimal? So I began to
First, we need to define optimal. One definition might say an arrangement is optimal if it fits the most amount of people within an area, while preserving social distancing measures. However, this
might not be useful as various governments have imposed hard limits on the number of people allowed to gather. So we could, given a fixed number of people, define optimal as maximizing the minimum
distance between people. Or if we are feeling adventurous we could define some function \(g_{\lambda}(p, d) = \lambda_p p + \lambda_d d\) as the weighted sum of the minimum distance between any two
people and the total number of people seated and maximize \(g_{\lambda}\).
All of these measures of optimality make sense, but what are we optimizing exactly? I have used the word “arrangement” without really giving it a definition, so let me present two possible ways to
approach this problem. First, we could have a fixed number of available positions and we need to map persons to those positions in such a way that preserves social distancing. For instance, an
auditorium has fixed seating and we need to assign people to those seats in an optimal and safe way. Now consider we can move those seats around. This leads us to the second approach, where we need
to find an optimal set of points within a region to place people safely.
Now we have 3 measures of optimality over 2 possible problem spaces leading to 6 fun optimization problems to work through. None of these consider the extra free-variable, which is grouping. Often
people who co-habitate are permitted to be within 6 feet of each other. At least for the second problem space (non-fixed seating) we can treat groups as single entities and the same solutions work.
For the first, we need some smarter solutions, which I address at the end. I should also note that most of my solutions are for the general case and thus somewhat complicated. Finding optimal
configurations with the assumption of some regularity is often much easier, but what is the point of being in graduate school if you are not going to show the general case.
Fixed Positions
Maximum People
Here we have a fixed set of seats \(V\) and we want to find an optimal choice of seats \(S \subseteq V\) such that \(\lvert S\rvert\) is maximized. This can be thought of as a nice graph theory
problem. Define \(\mathcal{G} = \left(V, E\right)\) where \(E = \{\left(v_i, v_j\right) : v_i,v_j \in V \land i\ne j \land D(v_i, v_j) < d \}\), where \(D(v_i, v_j)\) is the euclidean distance
between two seats and \(d\) is some distance threshold (i.e. 6 feet). Now \(\mathcal{G}\) is a graph where a seat is connected to another seat if the social distancing rule were to be broken if both
seats were occupied.
Thus, the set \(S\) must only contain vertices of \(\mathcal{G}\) which are not connected by an edge. Such a set of vertices is called an independent set and we want to find the largest independent
set of \(\mathcal{G}\). This is the same as finding the maximum clique of the complement graph \(\overline{\mathcal{G}}\).
On one hand this is great, because finding maximum cliques is a very well studied area of graph algorithms. On the other hand, this is bad news because it is an NP-Hard problem; it is hard to solve,
check solutions, and even find approximations. There are algorithms better than the \(\mathcal{O}(2^nn^2)\) brute-force algorithm, but they are still in NP. Currently, the best algorithm runs in \(\
mathcal{O}(1.1996^n)\) with polynomial space complexity as proven by Xiao et al [1], which means we could only feasibly apply this algorithm for up to ~100 seats. And even worse is that finding
maximum independent sets on a general graph is \(\mathsf{MaxSNP}\mathrm{-complete}\) meaning it has no polynomial-time approximations which produce a result within a constant \(c\) multiple of the
optimal solution.
Luckily there is some prior knowledge about the construction of \(\mathcal{G}\), which we can use to simplify the problem. If \(\Delta = \max_{v\in V} \deg v\) is the maximum degree in \(\mathcal{G}
\), then we can say \(\Delta\) is independent of \(\lvert V \rvert\). That is \(\mathcal{G}\) is a degree-bounded graph. We can make this assumption because the number of seats within \(d\) distance
of a seat should not grow unbounded as the total number of seats grows. Otherwise you have a very dense seat packing, which is not physically possible. Likewise, this means \(\mathcal{G}\) is a
sparse graph with bounded degree. Halldórsson et al [2] show that assuming bounded degree allows for a polynomial-time algorithm to find approximate maximum independent sets where the approximation
always has a constant ratio \(\frac{\Delta + 2}{3}\) of the optimal solution. Their greedy algorithm is actually quite simple in that you just continually remove the vertex with the highest degree
until the graph is entirely disconnected. You can see an example of this algorithm below, where each black dot could be considered a “seat” and they are connected if they are within 6 ft of each
other. The red dots are then a good selection of placements to put people safely.
Using this we can also give some bounds on \(\alpha(\mathcal{G}) = \lvert S \rvert\) (also called the independence number of \(\mathcal{G}\)). Most of these I pull from W. Willis [3]. Let \(\bar{d}\)
be the average degree of \(\mathcal{G}\), \(n\) be the number of vertices, \(e\) the number of edges, and \(\lambda_n\) be the \(n\)-th eigenvalue (sorted order) of the adjacency matrix. Then we can
define the following lower bounds on \(\alpha\):
\[\alpha \ge \frac{n}{1 + \Delta}\] \[\alpha \ge \frac{n}{1 + \bar{d}}\] \[\alpha \ge \frac{n}{1 + \lambda_1}\] \[\alpha \ge \sum_{v\in V} \frac{1}{1 + \deg v}\]
and several others… These are helpful in determining how many we can seat at least on a given graph. For instance, consider a rectangular lattice of seats with \(d\) unit length and the minimum
distance people must sit apart is \(\sqrt{2}d + \varepsilon\) for some small \(\varepsilon > 0\). Recall that we assumed \(\mathcal{G}\) is sparse with bounded degree. Thus, \(\Delta(\mathcal{G}) = 8
\) is independent of the number of seats \(n\). So we know that we will be able to seat at least \(\frac{n}{9}\) people in this lattice arrangement. But in this case we can give an even tighter bound
with the last one (we can assume \(\sqrt{n} \in \mathbb{Z}^+\)):
\[\begin{aligned} \alpha \ge \sum_{v \in V} \frac{1}{1+ \deg v} &= \left(\frac{1}{1+8}\right) (\sqrt{n}-1)^2 + \left(\frac{1}{1+5}\right) (4) (\sqrt{n}-1) + \left(\frac{1}{1+3}\right)(4) \\ &= \frac
{1}{9} (\sqrt{n}-1)^2 + \frac{2}{3}(\sqrt{n}-1) + 1 \end{aligned}\]
Now \(\left(\frac{1}{9} (\sqrt{n}-1)^2 + \frac{2}{3}(\sqrt{n}-1) + 1\right) - \frac{n}{9} \to \infty\) as \(n \to \infty\), so this is an increasingly better lower bound than the first. So given 100
chairs in this lattice configuration and no one can share adjacent chairs, then you can sit at least 16 people according to this lower bound. This is not the tightest lower bound and in fact you can
do better.
We can also give some upper bounds on \(\alpha\) (from [3]). Let \(\delta = \min_{v \in V} deg(v)\) be the minimum degree of \(\mathcal{G}\).
\[\alpha \le n - \frac{e}{\Delta}\] \[\alpha \le \left\lfloor \frac{1}{2} + \sqrt{\frac{1}{4} + n^2 - n - 2e} \right\rfloor\] \[\alpha \le n - \left\lfloor \frac{n-1}{\Delta} \right\rfloor = \left\
lfloor \frac{(\Delta-1)n + 1}{\Delta} \right\rfloor\] \[\alpha \le n - \delta\]
and some others… These are likewise helpful in determining whether a given arrangement could hold the desired amount. Returning to our lattice example from above we can take \(e = 2(\sqrt{n}-1)(2\
sqrt{n}-1)\) and use the first upper bound. This gives
\[\begin{aligned} \alpha \le n - \frac{e}{\Delta} &= n - \frac{2(\sqrt{n}-1)(2\sqrt{n}-1)}{8} \\ &= \frac{1}{4} \left( 2n + 3\sqrt{n} - 1 \right) \end{aligned}\]
which, for 100 seats, implies \(\alpha \le 57.25\). So now we know \(16 \le \alpha \le 57.25\), which are not great bounds, but bounds nonetheless. In fact, for the particular lattice problem I
brought up (with \(\Delta(\mathcal{G}) = 8\)) we can say that \(\alpha(\mathcal{G}) = \left\lceil \frac{\sqrt{n}}{2} \right\rceil^2\). Thus, with 100 seats we can seat at most 25 people safely. We
will not have this nice closed form for every problem though, so the upper and lower bounds are important.
In summation, making assumptions about the connectivity of our graph allows us to use efficient approximations. In addition, we can calculate certain guarantees based on simple graph invariants.
However, this solution is somewhat overkill as fixed positions are often seats, which are usually in some regular pattern. As with the lattice pattern above it is fairly trivial to find a closed form
solution based on regularity, but it is still good to have a solution for the general case.
Maximum Distance
Here we have a fixed number of people \(N\) that we need to sit with maximum separation. We want to find a set of vertices \(A \subseteq V\) such that \(\lvert A \rvert = N\) and the minimum distance
between any two vertices is maximum. More formally this is written as
\[\begin{aligned} \textrm{maximize} \quad& \min_{i \ne j} D\left(v_i, v_j \right) \\ \textrm{subject to} \quad & v_1, \ldots, v_N \in A \subseteq V \\ \quad & \lvert A \rvert = N \end{aligned}\]
This problem is known as the discrete p-dispersion problem and, again, has a decent amount of literature surrounding it. Unfortunately it is another computationally difficult problem. This can be
shown by an interesting relationship to the independent set problem from above. First, let \(\mathcal{G}_d\) be defined by the same graph as above with minimum distance \(d\) and let \(\mathcal{X} :
\mathbb{R} \mapsto \weierp(V)\) be defined by
\[\mathcal{X} (d) = \left\{ A \mid A \subseteq V \land \lvert A \rvert = p \land \mathbb{I}\{x\in A\} + \mathbb{I}\{y\in A\} \le 1 \,\,\forall (x,y) \in E\left(\mathcal{G}_d\right) \right\}\]
where \(E\left(\mathcal{G}_d\right)\) are the edges of \(\mathcal{G}_d\) and \(\mathbb{I}\) is the indicator function. The above max-min problem can be re-written as
\[\begin{aligned} \textrm{maximize} \quad & d \\ \textrm{subject to} \quad & \mathcal{X}(d) \ne \emptyset \\ \quad & d \ge 0 \end{aligned}\]
Notice that \(\mathcal{X}(d)\) is the set of all size \(N\) independent sets of \(\mathcal{G}_d\) with a minimum separation of \(d\). So we want to maximize \(d\), such that a size \(N\) independent
set exists within \(\mathcal{G}_d\). The independent set decision problem is \(\mathsf{NP-complete}\) as it is closely related to the minimum vertex cover decision problem.
Assuming the above maximization is feasible it can be solved using binary search on the list of possible distances \(d\) and an independent set decision algorithm for feasibility testing. The list of
possible values for \(d\) is all the distances in the full graph \(\mathcal{G}_{\infty}\). Thus, if \(d_{max}\) is the maximum distance between any two vertices in \(\mathcal{G}\), then the algorithm
will run the independent set decision problem \(\mathcal{O}(\log d_{max})\) times. The algorithm is still in \(\mathsf{NP}\) because of this, but is still feasible for smaller graphs (roughly \(10^3
\) seats). Sayah et al propose two formulations of the problem in terms of mixed-integer linear programming with reduced constraints using known bounds, which offer exact solutions [4] and can be
easily solved with CPLEX.
A Compromise
Now lets get a little more general… Consider we want to find some optimal solution in-between the two above approaches. This could be phrased as maximizing a weighted sum of the total number of
people seated and the minimum distance between any two people. As a function this is \(g_{\lambda_N, \lambda_d} (N, d) = \lambda_N N + \lambda_d d\). Now we can fix the \(\lambda\)’s and try to
maximize \(g\). This can be formulated exactly as above, but now \(\mathcal{X}\) is a function of \(N\) and \(d\). I am also going to change the notation a bit to express the problem in terms of
integer programming.
Let \(\bm{x} \in \{0, 1\}^{\lvert V \rvert}\) be a vector with an element corresponding to each vertex in \(\mathcal{G}\). If we decide to assign a person to vertex \(i\) (aka seat), then \(\bm{x}_i
= 1\), otherwise it is 0. Now we can re-define \(\mathcal{X}\) as
\[\mathcal{X}(N,d) = \left\{ \bm{x} \mid \sum_{i\in\mathcal{G}_d} \bm{x}_i = N \land \bm{x}_i + \bm{x}_j \le 1 \,\, \forall (i,j) \in E\left(\mathcal{G}_d\right) \right\}\]
Using this for our maximization problem gives
\[\begin{aligned} \textrm{maximize} \quad & g_{\lambda_N, \lambda_d} (N, d) \\ \textrm{subject to} \quad & \mathcal{X}(N, d) \ne \emptyset \\ \quad & d \ge 0 \\ \quad & 0 \le N \le \lvert V \rvert \
This adds another degree of freedom to the above algorithm, but it should be fairly trivial to adapt. Since \(\lambda\) are fixed, then we can fix \(N\), which gives \(\max_{d} g_{\lambda_N, \
lambda_d} (N, d)\). This is the same problem as above. Despite the intuition that \(N \propto \frac{1}{\max d}\) they are dependent on each-other so we cannot binary search \(N\) as well. We will
need to exhaustively search all values of \(N\) from \(1\) to \(\lvert V \rvert\). Thus we run the independent set decision problem \(\mathcal{O}\left( \lvert V \rvert \log d_{max} \right)\) times.
Free Positions
Maximum People
Some places, such as a store, do not have fixed positions to map people into. In this case we need to find those fixed positions, which adds a free variable into the optimization search space.
However, this is also a constraint relaxation making the problem much easier to solve. For this problem we want to find a set of points within a region such that each point is farther than \(d\) from
all other points and we can fit the most people inside the region.
Formally, we have a space \(S \subset \mathbb{R}^2\) and we want to find a set of points \(\mathcal{X} = \{ \bm{x}_i \mid \bm{x}_i \in S \land D(\bm{x}_i, \bm{x}_j) < d\,\, \forall i \ne j \}\) such
that \(\lvert \mathcal{X} \rvert\) is maximum. The provided solution actually works in \(\mathbb{R}^d\), but \(\mathbb{R}^2\) and \(\mathbb{R}^3\) are the only ones we care about.
Each \(\bm{x}_i\) has a neighborhood \(S_\epsilon\), which no other points lie within. In the 2D case this means we have a circle centered at each \(\bm{x}_i\) and none of them overlap. Now this
boils down to the well studied problem of circle packing. There is an abundance of literature and software surrounding circle packing as well as available tables of existing solutions.
As an example of how to translate circle packing problems into point packing I will look at square regions. Given a square region with side length \(L\) we want to find the maximum circle packing of
a square region \(L + d\) (scaled to account for 2 extra radii). Most literature focuses on packing unit circles, so to use these results we need to scale the square to \(\frac{L}{d} + 1\) and then,
once a solution is arrived upon, rescale the found circle centers by \(d\).
Say we have a square region that is \(100 \times 100\) ft and we want to find the most number of seats to place such that they are all 6 ft apart. This is equivalent to finding the most number of
unit circles that can be placed in the square with side length \(\frac{100 + 6}{6} = 17.\overline{6}\).
Maximum Distance
In this optimization problem we need to choose \(N\) positions in a region such that they are maximally dispersed. This can more formally be expressed as a constrained optimization problem:
\[\begin{aligned} \textrm{maximize} \quad& \min_{i \ne j} \lVert \bm{x}_i - \bm{x}_j \rVert \\ \textrm{subject to} \quad & \bm{x}_1, \ldots, \bm{x}_N \in S \subset \mathbb{R}^2 \\ \end{aligned}\]
This is known as the continuous p-dispersion problem and is well studied in terms of exact solutions and approximations. When \(S\) is convex this is similar to the circle packing problem of fitting
the most circles in \(S\), since we can rewrite the above maximization problem as:
\[\begin{aligned} \textrm{maximize} \quad & R \\ \textrm{subject to} \quad & \bm{x}_1, \ldots, \bm{x}_N \in S \subset \mathbb{R}^2 \\ \quad & \lVert \bm{x}_i - \bm{x}_j \rVert > 2R \quad \forall i \
ne j \end{aligned}\]
These are non-linear programs and exact solutions are not always feasible. One can use an optimization software such as AMPL to find good solutions for any generic convex region \(S\). However, since
rectangular regions are common real-life examples I will discussion their solutions here.
Consider the region \(S \in [0, 1]^2\). We want to fit \(N\) points within this square such that the minimum distance between any two points is maximized. We also want the minimum distance to be
greater than \(d\), but by maximizing the minimum we should figure out whether this is even possible or not. Any solution in \([0, 1]^2\) is valid for any square region as we can take \(\bm{x}_i \
mapsto \frac{1}{2}\left(\frac{\bm{x}_i}{\lVert\bm{x}_i\rVert} + 1\right)\), which maps all \(\bm{x}_i\) to \([0, 1]^2\) and preserves the ordering of their scaled distances.
A Compromise
Much like above we can extend the discrete version by relaxing the constraint to be continuous.
\[\begin{aligned} \textrm{maximize} \quad & g_{\lambda_N, \lambda_d} (N, d) \\ \textrm{subject to} \quad & \bm{x}_1, \ldots, \bm{x}_N \in S \subset \mathbb{R}^2 \\ \quad & d \ge 0 \\ \quad & 0 \le N
\le \lvert V \rvert \\ \quad & \lVert \bm{x}_i - \bm{x}_j \rVert \le d \quad \forall i \ne j \end{aligned}\]
This is actually a pretty difficult problem, but it can be solved similar to how we solved the compromised solution in the discrete case. For any fixed \(N\) we can solve this using the continuous
p-dispersion solution. Since \(N\) is discrete, then we can just try every value from \(1\) to \(\lvert V \rvert\) and see which one maximizes \(g_{\lambda_N, \lambda_d}\).
Accounting for Groups
Now if everyone sat/stood by themselves, then all of the above mentioned techniques would work wonderfully. But in reality people who co-habitate are often allowed to be within 6 feet of each other.
In the latter approach, where we are placing the seats, this is fine. We can treat “seats” as groups of seats and all of the above mentioned methods still work. However, things get a bit more
complicated when we look at the fixed seating optimization problem.
For example, consider the problem where we have \(\lvert V \rvert\) fixed seats and we are trying to seat \(N\) people with maximal distance between them. One way to approach this problem is to
partition the graph into subgraphs where each subgraph represents a local “neighborhood” of seats. This way each subgraph could hold multiple people. However, determining the number of partitions to
use ahead of time is difficult for the general case.
For this problem let \(\bm{x}_i \in \mathbb{R}^2\) be the location of position \(i\) in euclidean space. Now define the fully connected graph \(\mathcal{G} = (V, E)\) such that each vertex is a
position and there is an edge between every seat \(\bm{x}_i\) and \(\bm{x}_j\) weighted by the radial basis function:
\[RBF(\bm{x}_i, \bm{x}_j) = \exp \left( \frac{-\lVert \bm{x}_i - \bm{x_j} \rVert_2^2}{2\sigma^2} \right)\]
Now seats further apart have a smaller weight than those right next to each other who have a weight of 1. We want to find a partitioning \(A_0, A_1, \ldots, A_k\) of \(V\) which minimizes the weights
between A’s. This can be expressed cleanly as an objective function. Let \(w_{ij}\) be the weight between two vertices and \(W(A, B) = \sum_{i\in A \\ j\in B} w_{ij}\) the sum of weights between two
sets of vertices. Then our object function could be
\[f(A_1, \ldots, A_k) = \frac 1 2 \sum_{i=1}^k W(A_i, \overline{A_i})\]
where \(\overline A\) is the graph complement. Minimizing \(f\) would give the desired outcome, but we have an issue with our objective. The current function will produce lots of single vertex sets,
which is not great for trying to group people together. To fix this we can “penalize” the objective function with the size of \(A_i\). Instead of the size we are going to use the volume of \(A_i\)
which takes the weights into account. Let \(d_i = \sum_{j=1}^n w_{ij}\) be the degree of vertex \(i\) and \(vol(A) = \sum_{i \in A} d_i\). Now our objective is
\[f(A_1, \ldots, A_k) = \frac 1 2 \sum_{i=1}^k \frac{W(A_i, \overline{A_i})}{vol(A_i)}\]
So now we need to find \(A_1, \ldots A_k\) such that \(f\) is minimized. First let use define a matrix \(H \in \mathbb{R}^{n\times k}\) such that
\[H_{ij} = \begin{cases} \frac{1}{\sqrt{vol(A_j)}},& \textrm{ if } v_i \in A_j \\ 0,& \textrm{ otherwise} \end{cases}\]
Also define the diagonal matrix \(D \in \mathbb{R}^{n\times n}\) where \(D = \textrm{diag}(d_1,\ldots,d_n)\), \(W \in \mathbb{R}^{n\times n}\) where \(W_{ij} = w_{ij}\), and the graph laplacian \(L =
Now we can express our minimization function as
\[\begin{aligned} \min_{A_1,\ldots,A_k} f(A_1,\ldots,A_k) &= \min_{A_1,\ldots,A_k} \textrm{Tr} \left( H^T L H \right) \quad &\textrm{ subject to } H^TDH = I \\ &= \min_{T} \textrm{Tr} \left( T^T D^
{-1/2} L D^{-1/2} T \right) \quad &\textrm{ subject to } T^T T = I \end{aligned}\]
where we substitute \(T = D^{1/2} H\) in the last step. This is a standard trace minimization problem over a quadratic term with an identity constraint. The solution is \(T\) such that \(T\) contains
the \(k\) eigenvectors of \(D^{-1/2}LD^{-1/2}\) corresponding to the \(k\) largest eigenvalues.
We can now assign vertices to clusters by using the \(k\)-means clustering algorithm on the rows of \(T\). Now vertex \(i\) is in a cluster with vertex \(j\) if rows \(i\) and \(j\) of \(T\) are
assigned to the same cluster.
See an example of this process on a random graph below.
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Lauf, Frederik (2019): Superconformal matter coupling in three dimensions and topologically massive gravity.Dissertation, LMU München: Fakultät für Physik
Superconformal matter
coupling in three dimensions
and topologically massive
Frederik Lauf
Superconformal matter
coupling in three dimensions
and topologically massive
an der Fakult¨at f¨ur Physik
der Ludwig-Maximilians-Universit¨at M¨unchen
vorgelegt von Frederik Lauf
aus Bad Sobernheim
Zweitgutachter: Prof. Dr. Peter Mayr
Die Kopplung supersymmetrischer skalarer Materie-Multipletts an superkonforme
Eich-theorie und Gravitation in drei Raumzeit-Dimensionen wird im Formalismus des N
-erweiterten Superraumes beschrieben. Die Formulierungen supersymmetrischer
Eich-theorie und konformer Supergravitation in diesem Superraum werden vorgestellt und ein
Formalismus f¨ur die Analyse von Superfeldkomponenten wird entwickelt. Ein Superfeld-Wirkungsprinzip, welches zu einer allgemeinen Klasse minimaler Multipletts auf der
Massenschale f¨uhrt, wird eingef¨uhrt. Darauf folgend wird ein Skalar-Multiplett, wel-ches von einem zwangsbeschr¨ankten Superfeld beschrieben wird und nur aus unter spin(N) transformierenden
Lorentz-Skalaren und Spinoren besteht, als speziell zwangs-beschr¨ankter Fall des minimalen unbeschr¨ankten Multipletts identifiziert. Seine Super-feldwirkung wird aus dem Wirkungsprinzip f¨ur das
unbeschr¨ankte skalare Superfeld hergeleitet. Die Analyse wird f¨ur eine supersymmetrisch eich- und gravitationskovari-ante Beschreibung von Superfeldkomponenten verallgemeinert und eine
Kopplungsbe-dingung sowie die Wirkung, welche diese Kopplung beschreibt, werden gefunden. Auf
dieser Basis werden alle Eichgruppen f¨ur das Skalar-Multiplett, welche von N ≤ 8 er-weiterter superkonformer Symmetrie erlaubt sind, im flachen Superraum bestimmt und
ebenso im gekr¨ummten Superraum, in welchem das Skalar-Multiplett zus¨atzlich gravita-tionell gekoppelt ist. Dies f¨uhrt zur Konstruktion s¨amtlicher gravitationell gekoppelter
Chern-Simons-Materie-Theorien. Unter der Benutzung des gravitationell gekoppelten
Skalar-Multipletts als konformen Kompensator werden die resultierenden
kosmologi-schen topologisch massiven Gravitationen nach den korrespondierenden Parametern µ`,
also dem Produkt aus der Kopplungskonstante der konformen Gravitation und dem
anti-de-Sitter-Radius, klassifiziert. Die Modifikationen vonµ` bei der Pr¨asenz erlaubter
The coupling of supersymmetric scalar-matter multiplets to superconformal gauge theory
and gravity in three-dimensional space-time is described in the formalism ofN-extended
superspace. The formulations of super-gauge theory and conformal supergravity in this
superspace are reviewed and a formalism for analysing superfield components is
devel-oped. A superfield action principle giving rise to a general class of minimal on-shell
multiplets is introduced. Subsequently, a scalar multiplet described by a constrained
su-perfield and consisting only of Lorentz scalars and spinors transforming under spin(N)
is recognised as a specially constrained case of the minimal unconstrained multiplet.
Its superfield action is deduced from the action principle for the unconstrained scalar superfield. The analysis is generalised to a super-gauge- and supergravity-covariant
de-scription of superfield components and a coupling condition for the scalar multiplet as
well as the action describing this coupling are obtained. Based on this, all gauge groups
for the scalar multiplet consistent with N ≤ 8 extended superconformal symmetry are
determined in flat superspace, as well as in curved superspace where the scalar multiplet
is also coupled to conformal supergravity. This results in the construction of all
super-conformal Chern-Simons-matter theories coupled to gravity. Using the gravitationally
coupled scalar multiplet as a conformal compensator, the resulting cosmological
topolog-ically massive supergravities are classified with regard to the corresponding parameters µ`, i.e. the product of the conformal-gravity coupling and the anti-de Sitter radius. The
modifications of µ` in presence of possible gauge couplings for the scalar compensator
1. Introduction 1
2. Three-dimensional superspace 9
2.1. Supersymmetry algebra . . . 10
2.2. Superfields . . . 14
2.2.1. Field representation . . . 14
2.2.2. Component expansion of superfields . . . 16
2.2.3. Supercovariant constraints . . . 18
2.2.4. Superfield actions and equations of motion . . . 22
2.2.5. A spin(N) on-shell superfield . . . 24
2.3. Local symmetries . . . 29
2.3.1. Gauge-covariant derivatives . . . 29
2.3.2. Gauge superalgebra . . . 30
2.4. Supergravity . . . 34
2.4.1. Supergravity-covariant derivative . . . 34
2.4.2. Supergravity algebra . . . 35
2.4.3. Solution to the constrained Jacobi identity . . . 38
2.4.4. Super-Weyl invariance . . . 42
2.4.5. Anti-de Sitter superspace . . . 43
2.4.6. Conformal superspace . . . 44
3. Coupled scalar matter 49 3.1. Covariant superfield projections . . . 50
3.2. Coupled spin(N) superfield . . . 51
3.2.1. Gauge-coupled spin(N) superfield . . . 52
3.2.2. Gravitationally coupled spin(N) superfield . . . 53
3.2.3. Gauge- and gravitationally coupled spin(N) superfield . . . 55
3.3. Matter currents . . . 57
3.3.2. Supergravity-matter current . . . 59
3.4. All gauge groups for the spin(N) scalar multiplet . . . 62
3.4.1. N = 2 (Clifford), N = 3 and N = 4 (chiral) . . . 64
3.4.2. N = 4 (Clifford), N = 5 and N = 6 (chiral) . . . 67
3.4.3. N = 6 (Clifford), N = 7 and N = 8 (chiral) . . . 70
3.5. All gauge groups for the gravitationally coupled spin(N) scalar multiplet 74 3.5.1. N = 4 . . . 74
3.5.2. N = 5 . . . 75
3.5.3. N = 6 . . . 75
3.5.4. N = 7 . . . 76
3.5.5. N = 8 . . . 80
3.6. Overview of gauge groups . . . 84
4. Topologically massive supergravity 87 4.1. Einstein gravity and topologically massive gravity . . . 87
4.2. Conformal supergravity and conformal compensators . . . 90
4.3. Extended topologically massive supergravity . . . 91
4.3.1. Fixation ofµ` . . . 91
4.3.2. Gauged compensators . . . 92
5. Conclusion 95 A. Symmetry groups 97 A.1. Fundamental representations . . . 97
A.2. Index notation . . . 98
A.3. Generators . . . 99
A.4. Spin groups . . . 101
B. Systematic supercovariant projection 105
Fields have been recognised as indispensable entities for the formulation of physical
the-ories since the development of electromagnetism by Maxwell. Today, field thethe-ories are
still the language for the most successful models of fundamental phenomena.
Incor-porating the principles of special relativity and quantum mechanics, they describe the
particles and interactions in the Standard Model of particle physics. A classical field
theory for gravity, and thereupon models for cosmology, are implied by the principle of
general relativity.
A particular power for the formulation of field theories is provided by the principle of
symmetry. A symmetry is given, if the most basic description of a theory is invariant under a specific transformation of its constituting objects.
Above all, the symmetry under space-time transformations is obviously demanded
by the principle of relativity. The symmetry transformations associated to space-time
can act on two kinds of fields which are called bosonic and fermionic. They are the
representations of the space-time symmetry group with integer and half-integer spin,
respectively. Physically, they are conceptually distinguished. The latter describe matter
fields like leptons and quarks, while the former correspond to force particles like photons
and gluons as well as scalar fields like the Higgs boson.
Furthermore, especially in the Standard Model of particle physics, internal symmetry groups acting on the degrees of freedom formed by the fields themselves play an
impor-tant role. For a manifestly invariant formulation, the degrees of freedom are arranged
in multiplets on which a matrix representation of the respective symmetry group can
act. These transformations become vital if they are allowed to be space-time-dependent.
Such local symmetries, also called gauge symmetries, lead to successful descriptions of
force particles interacting with the matter particles and mediating interactions between
There is no better outside reason for the existence of such internal symmetries, than
scattering matrix, i.e. the matrix describing scattering processes between particles. In
this regard, there is only one further allowed kind of symmetry. It is called
supersym-metry and concerns the symsupersym-metry of space-time itself.
Supersymmetry generalises the symmetry associated to space-time in a way, that the
fermionic and bosonic representations transform into each other. Supersymmetric
theo-ries therefore always describe systems where both kinds of particles appear together and
can be seen as members or aspects, sometimes called superpartners, of one
supermul-tiplet. These supermultiplets can manifestly be described using superfields instead of
fields, which are functions of a superspace instead of space-time, resembling the fact that
supersymmetry is a generalisation of space-time symmetry. The bosonic and fermionic
fields are encapsulated in a superfield as the coefficients of an expansion in even and odd
powers of the fermionic superspace coordinates. Together with supersymmetry comes another symmetry, similar to an internal one, naturally acting on the fermionic
repre-sentations. These degrees of freedom can be regarded as corresponding to a number N
of usual supersymmetries, implying an extension of the structure of possible
supermul-tiplets. These symmetries are called (N-)extended supersymmetries.
Applying the paradigm of supersymmetry leads to various consequences. As for
parti-cle physics, it would for example predict corresponding superpartners for each partiparti-cle of
the Standard Model. This has provoked a phenomenological interest in supersymmetry
for decades, with no concluding result so far.
Regarding gravity, the principle of general relativity adapted for superspace leads to supersymmetric gravity, with the metric field or graviton and a so-called gravitino as
superpartners. An interesting feature of supersymmetry is the automatic implication
of supergravity, if the transformations are local, similar to a gauge theory. After all,
two (fermionic) local supersymmetry transformations correspond to a (bosonic) local
space-time translation, which is the symmetry transformation corresponding to general
In this thesis, the focus lies on three-dimensional supersymmetric field theories with
highly extended supersymmetry. Those can be motivated from the point of view of 11-dimensional space-time: The supermultiplet of 11-dimensional N = 1
supersym-metry contains the graviton as the field with highest spin. Since massless fields with
spins higher than two are considered unphysical, this theory qualifies as the highest
di-mensional and unique supersymmetric field theory and supergravity [1]. The equations
involves the coupling to 11d supergravity. The M2-brane world-volume preserves one half of the 32 supercharges needed to parametrise the N = 1 supersymmetry
transfor-mations in 11 dimensions, thus gaining the higher amount ofN = 8 supersymmetry [4],
which is a common effect of this dimensional reduction.
Furthermore, in a suitable scaling limit, the M2-brane solution describes the space
AdS4 ×S7 [5], i.e. the four-dimensional space of constant negative curvature with
7-spheres at each point. In this case, it additionally displays conformal symmetry [6],
which corresponds to invariance under local rescaling of the metric. According to the
so-called AdS/CFT correspondence [6], this suggests the existence of a three-dimensional
superconformal gauge theory with a number of N internal degrees of freedom, which would be interpreted as the world-volume theory for a stack of N coincident M2-branes.
Another point of view on M2-branes comes from their interplay with superstring
theory. The five known formulations of superstring theory are connected by a web of
dualities, which means they are equivalent ways for describing the same phenomena,
but in different physical regimes. Their low-energy limits are corresponding versions of
ten-dimensional supergravity. One of these, the type IIA supergravity, can directly be
obtained from the unique 11d supergravity, by compactifying one of the 11 dimensions
on a circle whose radius is proportional to the string-coupling constant. Thus, via the
duality web, all ten-dimensional supergravities descend from the 11-dimensional one. In turn, 11d supergravity is expected to be the low-energy limit of the so-called M-theory,
whose full high-energy formulation is unknown. The superstring theories (as well as 11d
supergravity) are thus regarded as certain limits of M-theory and the M2-branes are
considered as fundamental for M-theory as strings are for string theory.
The type IIA supergravity also contains membranes which are called D2-branes and
describe hypersurfaces for string endpoints. Due to this relation to string dynamics,
the low-energy world-volume theory for a stack of N coincident D2-branes is known to
be a non-conformal three-dimensional gauge theory for N internal degrees of freedom.
The gauge coupling is proportional to the string coupling and thus to the size of the compactified 11th dimension. Therefore, in the limit of infinite coupling strength, the
size of the 11th dimension increases again, until the theory corresponds to a stack of
M2-branes in 11 dimensions. In this strong-coupling regime, the theory should obtain
the conformal symmetry implied by the AdS/CFT correspondence
world-volume theory for a stack of M2-branes is N = 8 superconformal Chern-Simons
gauge theory coupled to eight scalar and spinor fields, which are usually called matter
fields [7]. Since the Chern-Simons theory is non-dynamical, the scalar fields represent the degrees of freedom of M2-branes in the eight transverse directions. The theory is
known as the BLG model [8, 9] and is invariant under the gauge group SU(2)×SU(2).
Due to this unique gauge group, it cannot be interpreted as the world-volume theory of
a stack of N ≥2 coincident M2-branes [10, 11], as implied by the D2-brane theory and
desired by the AdS/CFT correspondence. This is however possible for a theory with
onlyN = 6 supersymmetries and the gauge group SU(N)×SU(N), known as the ABJM
model [12]. The reduction of the amount of supersymmetry is achieved by geometric
restrictions on the transverse coordinates.
Apart from this motivation for the ABJM model, superconformal
Chern-Simons-matter theories with lower amountsN ≤ 8 of supersymmetry and different gauge groups
are still a matter of interest. Their most crucial feature is that possible gauge groups
are restricted by consistency with supersymmetry for N ≥4. Notable achievements in
the classification of Chern-Simons-matter theories with regard to supersymmetry and
gauge groups have been made in [13] for N = 4 in the context of four-dimensional
su-persymmetric gauge theories, in [14, 15] for N = 5,6 in the context of the geometry of
M2-branes, and in demand for formal classifications, in [16] for N = 6 and [17] for all
N. A superspace point of view was adopted in [18] forN = 4, [19] for N = 5,6, [20] for
N = 6, [21] for N = 6,8, and in [22] for N = 8.
As a part of the present thesis, this quest is repeated using the formalism of N
-extended superspace. In this approach, the supersymmetric matter is described by a
scalar superfield. The advantage of this approach is not only the manifestly
supersym-metric formulation, but also the as far as possible unified manner in which the cases of
N are analysed. The scalar superfield is subject to a certain supersymmetric constraint in order to describe a familiar scalar-matter supermultiplet. However, in the presence of
a gauge coupling, this constraint is in general inconsistent. Rather, it is valid only under
an algebraic condition involving a superfield representing the field-strength multiplet, containing the usual gauge field strength and its superpartners. In general, these field
strengths have to be expressed by the scalar-matter current which couples to the gauge
fields, according to their Chern-Simons equation of motion. The specific algebraic
prop-erties of these matter currents decide on the solvability of the condition for the scalar
dimensional gravity, likewise a field theory emerging in curved space, has some
distin-guished features compared to its analogue in four dimensions. Namely, both its versions
of Einstein-Hilbert gravity and conformal gravity each for themselves have no dynamics.
The latter is solved by conformally flat space-time and the former completely fixes the
geometry of space-time by its equations of motion to be flat or have constant curvature,
leaving no room for a locally propagating graviton. Nevertheless, in the presence of a
negative cosmological constant, a black hole solution [23] with anti-de Sitter space as
the asymptotic limit and in consequence [24] a corresponding two-dimensional conformal
field theory in this region with its two propagating modes are supported.
Thanks to the existence of these two models for gravity, a third model can be
con-structed by adding them together. The result is a dynamical theory known as
topologi-cally massive gravity [25], since it gives rise to a new propagating degree of freedom with
a mass determined by the coupling of the conformal supergravity. It notoriously requires
some finesse regarding the positivity of the occurring energies. As for the massive
gravi-ton alone and in absence of the cosmological constant, the sign of the Einstein-Hilbert
action must be inverted to ensure a positive energy. This carries with it the downside
of negative black-hole masses upon including the negative cosmological constant. A
sensible model with no negative energies is given by the so-called chiral gravity [26]. It is characterised by the usual sign of the Einstein-Hilbert action and the value µ` = 1,
whereµis the conformal-gravity coupling and`is the anti-de Sitter radius related to the
negative cosmological constant. Under this specification, the black holes have positive
mass while the massive graviton and one of the boundary gravitons disappear, leaving
only one mode with positive energy in the boundary conformal field theory [27].
N-extended supergravity in three dimensions can be formulated inN-extended curved superspace. This approach automatically leads to a description of conformal
super-gravity. In view of the Einstein-Hilbert term of topologically massive gravity, realising non-conformal supergravities requires the coupling to certain fields which are called
conformal compensator. These have to display specific properties in order to ensure
conformal invariance. In a second step, the conformal symmetry can be broken by fixing
an expectation value of the compensator, thus preventing it from preserving conformal
In order to serve as a conformal compensator, the scalar-matter multiplet coupled to
superconformal Chern-Simons theories can further be coupled to conformal supergravity
[28, 29, 21]. This carries with it the effect of modifying the spectrum of the admissi-ble gauge groups compared to flat superspace [30, 21]. The above superfield formalism
for the coupled scalar-matter multiplet remains applicable, but in addition to the
field-strength multiplet, there appears the conformal-supergravity multiplet described by a
superfield called super-Cotton tensor. It contains the Cotton tensor, which is an
in-variant tensor of conformal gravity constructed from the curvature tensor, but instead
of curvature, it measures the conformal flatness of a space-time. The impact of curved
superspace on the allowed gauge groups is quite particular regarding the number of
su-persymmetries. In the cases N = 4 and N = 5 it has no effect. For N = 6, it relaxes
the restrictions on certain U(1) factors of the gauge groups present in flat superspace [30]. This phenomenon is due to the fact, that the N = 6 super-Cotton tensor can be
regarded as being dual to a U(1) field strength. For N = 7 and N = 8 it gives rise to
the possibility of matter fields in the fundamental representations of gauge groups which
are unrelated to those in flat superspace.
Using the gauge- and supergravity-coupled scalar-matter multiplet as a conformal
compensator naturally realises supersymmetric versions of topologically massive
grav-ity. A distinguished feature of the resulting theories is that the value of µ` is always
fixed by the superconformal geometry for N ≥ 4 [31, 32]. The underlying mechanism is essentially the following. Since the super-Cotton tensor contains the field strength
of the gauged SO(N) structure group of extended superspace, its value is determined
by the Chern-Simons coupling µ via the Chern-Simons equation of motion of
confor-mal supergravity. In the geometry defining anti-de Sitter superspace, the cosmological
constant is generated by the value of a torsion superfield transforming inhomogeneously
under super-Weyl transformations, which are transformations related to conformal
in-variance. The presence of a scalar compensator superfield in this background requires a
gauge for this superfield relating it to the super-Cotton tensor. Upon giving the scalar
compensator its expectation value, this fixes the relation betweenµ and `.
As is the nature of a conformal compensator, the opposite sign of the Einstein-Hilbert
action is generated [29]. It suggests that the nature of negative mass BTZ black holes,
being a consequence of this sign, should be addressed by different interpretations. This
issue is not a topic of this thesis, but will be addressed again in the conclusion.
with a compensator not coupled to a gauge group is [32]
N = 4 5 6 7 8 (µ`)−1 = 1 3[/][5] [1] [2] [3 .]
Modifications of these values may occur if the compensator is also coupled to its allowed
gauge groups [21, 32]. In this case, a number of its gauge components can be chosen to
generate the Einstein-Hilbert coupling constant. The most diverse, yet specific, effects
appear for N = 6 with gauge group SU(N) in shape of the formula [32]
(µ`)−1 = 2 p −1,
where p is the number of non-vanishing components, and for N = 8 with the gauge
group SO(N) [21] and the formula [33]
(µ`)−1 = 4 p −1.
Both cases additionally allowµ`=∞corresponding to the solution of Minkowski space.
For N = 8 also the value µ` = 1 can be generated by choosing two compensator
Outline. The present thesis is organised as follows. In Chapter 2, the formalism of
three-dimensional extended superspace is introduced to describe supersymmetric scalar
matter fields, super-gauge theory and conformal supergravity, which will be coupled
together in the subsequent Chapter 3. Regarding supersymmetric scalar matter, the
component expansion of scalar superfields is developed and discussed, with the goal of arriving at the case of an on-shell multiplet described by a constrained superfield, which
consists of a scalar and a spinor field transforming under the fundamental representation
of spin(N). An off-shell superfield action principle leading to a general class of minimal
on-shell multiplets is proposed, from which the spin(N) scalar multiplet and its superfield
and component actions follow as a special, constrained case. Subsequently, super-gauge
superspace and conformal superspace are reviewed. A special focus lies on anti-de
Sit-ter superspace as a supersymmetric background solution of supergravity, which will be
relevant for topologically massive supergravity discussed in Chapter 4.
In Chapter 3, the component analysis of scalar superfields is generalised in order
to describe gauge-covariant superfield components. This leads to the derivation of a
coupling condition for the spin(N) scalar multiplet, which is implied by consistency with
the covariant constraint on the scalar superfield describing this multiplet or, in other
words, by consistency of the supersymmetry transformations of its covariant components.
This condition is then fully analysed and solved for 4≤ N ≤8, resulting in the complete
spectrum of allowed gauge groups in flat as well as in curved superspace. To this end, the
matter currents which couple to gauge fields are determined and recast as
scalar-superfield currents, corresponding to equations of motion for the scalar-superfields describing the gauge and supergravity multiplets.
In Chapter 4, results are combined in order to use the coupled scalar multiplet as
a conformal-compensator multiplet for topologically massive supergravities. Requiring
consistency with the background of anti-de Sitter superspace leads to a formula for the
values of µ` with a single compensator, depending on the case N. Subsequently, the
effects on the value ofµ`generated by a gauged scalar compensator will be investigated.
Aspects and conventions of the general treatment of symmetry groups used in the
main text, as well as some formal expressions belonging to the superfield-component
In this chapter, the relevant formalisms in three-dimensional superspace needed for the
later purpose of coupling supersymmetric scalar matter conformally to super-gauge
the-ory and conformal supergravity are reviewed, developed, or elaborated on.
In Section (2.1), the supersymmetry algebra is introduced and properties of the
Lorentz group are presented.
In Section (2.2), the representation of supersymmetry on superfields is discussed and
a formalism for the analysis of superfield components is developed. An off-shell action
principle, giving on shell rise to minimal scalar multiplets, is proposed. Based on this
formalism, the constrained scalar superfield transforming under spin(N) describing the
on-shell multiplet coupled to super-gauge fields and supergravity in the next chapter is analysed and the corresponding superfield action is derived.
In Section (2.3), gauge-covariant derivatives are introduced and the content of the
gauge connection is examined. The algebra of covariant derivatives is derived by solving
the super-Jacobi identity under the constraint defining the field-strength multiplet.
In Section (2.4), two descriptions of extended conformal supergravity are presented.
The first approach is the conventional curved SO(N) superspace, which is described by
certain Weyl-invariant constraints on the torsions. These will be motivated by
investigat-ing the algebra of covariant derivatives in terms of the gauge fields of the local structure
group. The super-Jacobi identity will be solved under these constraints in order to de-rive the field strengths in the supergravity algebra in terms of the super-Cotton tensor
and the torsions. Subsequently, anti-de Sitter superspace is introduced as the maximally
symmetric background of this geometry. The more briefly presented second approach
is conformal superspace, where the whole superconformal group is gauged as the local
structure group. It can be translated into conventional superspace and is convenient for
2.1. Supersymmetry algebra
Three-dimensional supersymmetric space-time or superspace is parametrised by the
zA = (xa, θIα), (2.1.1)
where θI
α are odd supernumbers, i.e.
θα[I]θ[J]β =−θ[J]βθ[I]α, (2.1.2)
carrying an SL(2,[R]) index α = 1,2 and an SO(N) index I = 1, ...,N. The space-time coordinate xa carries an SO(2,1) index a= 0,1,2.
The symmetry group of this superspace is the three-dimensional super-Poincar´e group,
which is generated by the super-Poincar´e algebra. The super-Poincar´e algebra is
ob-tained from the Poincar´e superalgebra
[Mab,Mcd] =−4η[c[aMb]d] (2.1.3a)
[Pa, Pb] = 0 (2.1.3b)
{QI[α], QJ[β]}= 2δIJPαβ (2.1.3c)
by requiring anticommuting parameters for the fermionic generatorsQI
α and commuting parameters for the remaining bosonic generators. In other words, an element of the
super-Poincar´e algebra reads
X = 1[2]ωabMab+ iaaPa+ iεαIQ I
α, (2.1.4)
εα[I]εβ[J] =−εβ[J]εα[I]. (2.1.5)
The part corresponding to the bosonic part of the Poincar´e superalgebra is the
symme-try group of non-supersymmetric space-time SO(2,1)[o][R]3. The part corresponding to the fermionic part is the group of supersymmetry transformations.
Referring to the terminology introduced in Appendix A, the Lorentz group SO(2,1)
is the pseudo-orthogonal group with
and SL(2,[R]) is the symplectic group Sp(2) with
εαβ = 0 1
−1 0
. (2.1.7)
The two-to-one correspondence between these two groups is established by mapping a Lorentz vector to the space of symmetric 2 ×2-matrices Sym(2,[R]) with the basis Sm ={S0, S1, S2}, where
S0 =
, S1 =
, S2 =
0 −1
. (2.1.8)
This basis is (pseudo-) normalised as
tr(SmS¯n) =−2ηmn, (2.1.9)
where ¯Sm ={S0,−S1,−S2}. The components of a Lorentz vectorxm are the expansion
coefficients in this basis,
X =xmSm, (2.1.10)
and inversely,
− 1 2tr(XS¯
m[) =][−]1
n[S][¯]m[) =] [x]m[.] [(2.1.11)]
Since the negative determinant of X equals the scalar product of the Lorentz vector, a
Lorentz transformation of X must be determinant- and symmetricity-preserving, i.e.
X −→LT X˜ =AXAT (2.1.12)
withA∈SL(2,[R]) being an element of the group of 2×2-matrices with unit determinant. The map between A∈SL(2,[R]) andΛ∈SO(2,1) follows from
−1 2tr( ˜XS¯
) = ˜xm (2.1.13)
T[S][¯]m[) =][Λ]m
It is a two-to-one map because A and −A are mapped to the same Λ. In other words,
SO(2,1)∼= SL(2,[R])/[Z]2. (2.1.15)
As is apparent, the matrices Sm have two lower indices and the matrices ¯Sm have two
upper indices,
( ¯Sm)αβ =εαγεβδ(Sm)γδ. (2.1.16) A set with one lower and one upper index is defined by
(γ0)αβ ≡ −εβγ(S0)αγ (γ1,2)αβ ≡ε
1,2)αγ. (2.1.17)
These are a basis for traceless matrices and fulfil the Clifford algebra
{γm,γn}= 2ηmn. (2.1.18)
In this context of space-time symmetry, the components of SL(2,[R]) vectors (also called spinors) are odd supernumbers with
vαwβ =−wβvα (2.1.19)
and the complex conjugation rule
(vαwβ)∗ =w∗βv
α. (2.1.20)
An antisymmetric SO(2,1) tensor of rank two, a vector and a rank-two SL(2,[R]) tensor are equivalent to each other by the relations
Fab =εabcFc =εabc(−1[2]εcdfFdf) (2.1.21a) Fαβ = (γa)αβFa= (γa)αβ(−1
Fγδ). (2.1.21b)
The contraction of two Lorentz vectors can thus be written in the forms
The action of the Lorentz generators with different label representations is given by
Mab[v]c[= 2][η]c[a[v]b] [(2.1.23a)]
Ma[v]b [=][−][ε]abc[v]
c (2.1.23b)
vc=−(γcb)αβvb (2.1.23c)
γ = 1[2](γab)γδvδ (2.1.24a)
γ = 1[2](γa)γδvδ (2.1.24b)
vγ =εγ(αvβ) (2.1.24c)
for Lorentz vectors and spinors, respectively. The commutation relation can be written
in the forms
2.2. Superfields
2.2.1. Field representation
Fields A(x, θ) in superspace are covariant with the coordinates both as finite- and infinite-dimensional representations of the super-Poincar´e group. The former correspond
to transformations in a finite vector space representing SL(2,[R])
A0(x0, θ0) =A(x, θ) +δA(x, θ) =A(x, θ) + 1[2]ωabM[ab]f ·A(x, θ) (2.2.1) and the latter to translations in the infinite space of functions
A0(x, θ) =A(x, θ) + ∆A(x, θ). (2.2.2) In this infinitesimal form, the two are related by a Taylor expansion
∆A(x, θ) = δA(x, θ)−(aa+ωabxb)∂aA(x, θ)
−(εα[I] + 1[4]ωab(γab)αβθβI)∂αIA(x, θ), (2.2.3) where ∂a≡ [∂x]∂a and ∂[α]I ≡
∂ ∂θα
. Comparing the commutator
[∆1,∆2] =ωac1 ω
2c (M
ab+ 2x[a∂b]− 1[4](γab)αβθβI∂αI) + 2ab[1ω2]bc∂c+ 2ε
I αa
2])∂a (2.2.4)
with the one of two generatorsX = 1[2]ωabMab+ iaaPa+ iεαIQIα, [X1, X2] =ω1acω
2c Mab+[2]iω[1abε
2](γab)αβQβI −2iaa[1ω
2]a Pb+ 2ε αI
1 ε
2IPαβ, (2.2.5)
reveals that aa must have the θ-dependent part, or the “soul”
aS[a] =−iεαIθβ[I](γa)αβ, (2.2.6)
Mab=Mabf + 2x[a∂b]− 1[4](γab)αβθβI∂αI (2.2.7a)
Pa=i∂a (2.2.7b)
Furthermore, there is a supercovariant derivative
D[α]I =∂[α]I + iθβI(γa)αβ∂a (2.2.8)
commuting with the supersymmetry generator,
DI[α]εβ[J]QJ[β]A=εβ[J]QJ[β]DI[α]A, (2.2.9) and obeying the supersymmetry algebra
{D[α]I, D[β]J}= 2iδIJ∂αβ. (2.2.10)
It has the useful property [34]
(DI[α]A)∗ =−D[α]IA∗ (2.2.11a) (D[α]IAβ)∗ =DαIA
β (2.2.11b)
and likewise for all SL(2,[R]) tensors of even or odd rank.
It can be convenient to combine the supercovariant spinor derivative together with
the vector derivative ∂a ≡Da into a supervector
DA= (Da, DIα). (2.2.12)
It is subject to the algebra
[DA, DB} ≡DADB−(−1)ABDBDA =TABC DC, (2.2.13)
where the powers A are 0 ifA is a vector index and 1 if it is a spinor index. The torsion TC
AB is constrained by
2.2.2. Component expansion of superfields
A superfield is expanded in powers ofθI α as
A(x, θ) =a(x) +θ[I]αaI[α](x) + 1[2]θ[I]αθβ[J]aJ I[βα](x) +... . (2.2.15) The component fields are give by the projections
αkA)|θ=0 ≡∂
αkA| ≡a
α1...αk. (2.2.16)
This definition requires appropriate normalisation factors in the explicit expansion as
indicated above. Since the spinor derivatives anti-commute, the component fields have
the symmetry property
α1...αiαj...αk =−a
α1...αjαi...αk. (2.2.17)
Supercovariant projections are denoted by
αkA| ≡A
α1...αk. (2.2.18)
They are related to the component fields by
α1...αk =a
α1...αk. (2.2.19)
The fields AI1...Ik
α1...αk depend on multiple space-time derivatives of components of
corre-spondingly lower ranks. Explicitly, they are given by
AI[α] =0 (2.2.20a)
AIJ[αβ] =iδIJ∂αβa (2.2.20b)
AIJ K[αβγ] =iδJ K∂βγaIα−iδIK∂αγaJβ + iδIJ∂αβaKγ (2.2.20c) AIJ KL[αβγδ] =iδIJ∂αβaKLγδ −iδ
αγaJ Lβδ + iδ KL[∂]
+ iδIL∂αδaJ K[βγ] + iδJ K∂βγaIL[αδ] −iδJ L∂βδaIK[αγ]
−δIJδKL∂αβ∂γδa+δIKδJ L∂αγ∂βδa−δILδJ K∂αδ∂βγa, (2.2.20d)
and so on. Systematic formulas and further examples are presented in Appendix B.
In the following, this relation will be symbolically denoted by
α1...αk =a
α1...αk +A
where a(−l) is of rank k−l.
Under infinitesimal supersymmetry transformations, the superfield changes by
δεA= iεαIQαIA=εαI(−DαI + 2iθ Iβ
∂αβ)A. (2.2.22) The transformed components are the components of the transformed superfield, i.e.
α1...αk =∂
αkδA| −A
α1...αk(∂δa(−2), ∂∂δa(−4), ...). (2.2.23)
Re-expressing the supercovariant projections, it follows
α1...αk =−ε
αα1...αk(∂a(−2), ...)−A
α1...αk(∂δa(−2), ...), (2.2.24)
iteratively describing the transformations of all superfield components.
It can be shown that these transformations indeed represent the supersymmetry
alge-bra by considering two successive transformations
δηδεaIα11..I..αkk =ε
α Iη
β Ja
J II1..Ik
βαα1..αk +ε
α Iη
β JA
J II1..Ik
βαα1..αk(∂a(−2), ..)−A
α1..αk(∂δηδεa(−2), ..). (2.2.25)
Their commutator is
[δη,δε]aIα11..I..αkk =(ε
α Iη
β J −η
α Iε
β J)A
J II1..Ik
βαα1..αk(∂a(−2), ..)−A
α1..αk(∂[δη,δε]a(−2), ..). (2.2.26)
The combination of the parameters on the right-hand side is either symmetric or
anti-symmetric in both types of indices projecting on the corresponding representations in AJ II1...Ik
βαα1...αk. This is only the symmetric one, given by
α1α2...αk +A
α2α1...αk = 2iδ
I1I2[∂α] 1α2a
α3...αk + 2iδ
I1I2[∂α] 1α2A
α3...αk, (2.2.27)
as is apparent from its definition. Assuming that the supersymmetry algebra closes on
fields of lower ranks inductively leads to its closure on all components,
[δη,δε]aIα11...I...αkk =2iε
α Iη
β Jδ
∂αβaIα11...I...αkk. (2.2.28)
For example, the transformations of the first five components read
δa=−εα[I]aI[α] (2.2.29a)
δaI[α]=−εβ[J]a[βα]J I −iεβ[J]δJ I∂βαa (2.2.29b)
δaJ I[βα]=−εγ[K]aKJ I[γβα] −iεγ[K](−δKI∂γαaJ[β] +δKJ∂γβaI[α]) (2.2.29c)
δaKJ I[γβα] =−εδ[L]aLKJ I[δγβα] −iεδ[L](δLK∂δγaJ Iβα−δ LJ[∂]
δβaKIγα +δ LI[∂]
δαaKJγβ ) (2.2.29d)
δaLKJ I[δγβα] =−εε[M]aM LKJ I[εδγβα] (2.2.29e)
−iεε[M](δM L∂εδaKJ Iγβα −δ M K
∂εγaLJ Iδβα +δ M J
∂εβaLKIδγα −δ M I
∂εαaLKJδγβ ).
Apparently, each component transforms into the next higher component and first
deriva-tives of the next lower component. Terms with more derivaderiva-tives drop out and would be
inconsistent with the symmetry of the tensor on the left-hand side. Therefore, it can be
concluded that
αk...α1 =−ε
αk+1...α1 + iδ
, (2.2.30)
where the brackets {.} collect the sum of k terms sharing the symmetry of the tensor
on the left-hand side, as in the above examples.
The transformation of irreducible SL(2,[R])×SO(N) representations contained in the components can easily be derived from the above formula. The particularly common
partially reduced field (an)αk...α1 defined by
a Ik...I1
αk...α1 = (−
1 2)
n[δ] J1J2ε
αk...α1β2n...β1 (2.2.31)
has the transformation
αk..α1 =−ε
αk+1..α1 + iδ
∂[α][k][+1]β (n−1)a
2.2.3. Supercovariant constraints
Since the supercovariant derivatives commute with supersymmetry transformations, they
can be used to impose supersymmetrically invariant constraints on superfields. In view of
equations considered in the following is
A= 0. (2.2.33)
Hiding the SO(N) indices, the components of this superfield equation are
A |=(an)+(An)= 0 (2.2.34a) ∂β1
A |=
aβ1 +
Aβ1= 0 (2.2.34b)
∂β2N −2n..∂β1
A |=(an)[β][2][N −][2][n][..β][1] +
Aβ2N −2n..β1= 0 (2.2.34c)
∂β2N −2n+1..∂β1
A |=(An)β2N −2n+1..β1= 0 (2.2.34d)
A |=
Aβ2N..β1= 0, (2.2.34e)
where components of Abeing determined by lower components of the equation have to be accordingly substituted in the higher components. This is why the spinor derivatives ∂I
α on the left-hand-sides can be replaced by supercovariant derivatives. The equations of the form
a[β][k][..β][1]=−(An)βk..β1 (2.2.35)
determine the components (an)βk..β1 in terms of derivatives of lower components; however,
the equations
Aβ2N −2n+l..β1= 0 (2.2.36)
give rise to higher-order differential relations among the lower components and other
representations (m<na )[β][l][..β][1]. This circumstance shows that the constraint sets the super-field partially on shell.
The constraint is invariant under the transformation
A−→A−B, (2.2.37)
This gauge freedom can be fixed in the superfieldAˆ defined by
A=Aˆ+B|A, (2.2.39)
where B|A means that the component fields appearing in the constrained superfield
B are evaluated at the values of the components of A. Concretely, Aˆ a has those components gauged away which are unconstrained in B and the other components are redefined in terms of the constrained
components of B and are not redundant.
In order to illustrate the above in an example, one can consider the case N = 2 and
D2D2A= 0. (2.2.40)
This constraint is invariant under
A−→A−B, (2.2.41)
D2B= 0. (2.2.42)
The components of B fulfil
˜[b] [=0] [(2.2.43a)]
bI[α] =−B˜I[α] = i∂[α]µbI[µ] (2.2.43b)
bJ I[βα] =−B˜J I[βα] =εβαδJ Ib−2i∂ µ
(β b
α)µ (2.2.43c)
0 =B˜˜K[γ] = 0 (2.2.43d)
and the gauge shift translates to the components of Aas
a−→a−b (2.2.44a)
aI[α] −→aI[α]−bI[α] (2.2.44b)
a−→˜a (2.2.44c)
a((J I)) −→˜a((J I))−˜b((J I)) (2.2.44d)
a[[(]J I[βα]][)] −→a[(][J I[βα]][)]−b[[(]J I[βα]][)]
aI[α] −→˜aI[α]−i∂[α]µbI[µ] (2.2.44f)
aJ I[βα] −→˜aJ I[βα]−εβαδJ Ib−2i∂[(][β]µb[[α]IJ[)][µ]], (2.2.44g)
where ˜a((J I)) denotes the traceless part of−1 2ε
αβ[a](J I)
βα . The fieldsa,a I
α and ˜a((J I))can be shifted arbitrarily and possibly to zero, while the fields ˜a, ˜aI[α] and ˜aJ I[βα]are non-redundant. Being interested in a minimal non-redundant multiplet, the field a[[(]J I
[βα]][)] can further be
required to fulfil the constraint
0 =∂[γ]β∂[(][δ]αa[[β]IJ[)][α]], (2.2.45)
in which case it can be shifted to zero as well.
According to (2.2.39), the components of Aare then redefined as
a=a (2.2.46a)
aI[α] =aI[α] (2.2.46b)
a= ˆ˜a (2.2.46c)
a((IJ)) = ˜a((IJ)) (2.2.46d)
a[[(]J I[βα]][)] =a[[(]J I[βα]][)] (2.2.46e)
aI[α] = ˆ˜aI[α]+ i∂[α]µaI[µ] (2.2.46f)
aJ I[βα] = ˆ˜aJ I[βα]+εβαδJ Ia+ 2i∂[(][β]µa
α)µ. (2.2.46g)
The gauge-fixed superfield Aˆ is therefore given by
A= 1[2]θ[I]αθI[α](ˆa˜+θ[K]γa˜ˆK[γ] + 1[2]θγ[K]θ[L]δˆ˜aKL[δγ] ). (2.2.47) Due to (2.2.46g) and the condition (2.2.45), the Lorentz vector ˆ˜a[[(]KL[δγ][)]] has to fulfil the
equation and Bianchi identity
∂[(][α]γˆ˜a[[δ]KL[)][γ]] =∂γδˆ˜a[[δγ]KL]= 0. (2.2.48)
In summary, the superfields Aand Aˆ fulfil the same constraint equation
D2D2A=D2D2Aˆ= 0 (2.2.49) under the assumption of (2.2.48).
2.2.4. Superfield actions and equations of motion
A generalisation of the action known from N = 1 supersymmetry [35] is
S =
d3x (d2θ)N (Dα1
IN A)D
αNA|. (2.2.50)
It is manifestly invariant under supersymmetry, since the integral is over the whole superspace, i.e. the spinorial measure (d2[θ)]N [≡][(∂]α
N [contains all spinor derivatives]
and is thus annihilated by every supersymmetry generator. It is worth to note that
(∂α I∂αI)
N [may be replaced by (D]α IDαI)
N [up to a total derivative, which is useful for]
obtaining the component action. While the choice of the measure is unique (assuming
that no indices are contracted with those of the integrand), the integrand is highly
reducible. A particular choice in view of the present purpose is the completely traced
part. For even N it is
S =
d3x (d2θ)N [(D2)nA](D2)nA| (2.2.51) and for oddN
S =
d3x (d2θ)N [D[I]α(D2)nA]DI[α](D2)nA|, (2.2.52) where n= N[2] or n= N −1[2] respectively.
The superfield equation of motion can be obtained by partially integrating so that
S =
and is given by
(D2)NA= 0. (2.2.54)
It is of the constraint form studied above. It has a redundancy due to the transformation
A−→A−B, (2.2.55)
(D2)N −1B= 0. (2.2.56) Written in components (2.2.34), and omitting the SO(N) indices, this constraint on B
(N −1)
B |=
(N −1)
b +
(N −1)
B = 0 (2.2.57a)
(N −1)
B |=
(N −1)
B β1=
(N −1)
b β1 +
(N −1)
B β1= 0 (2.2.57b)
(N −1)
B |=
(N −1)
b β2β1 +
(N −1)
B β2β1= 0 (2.2.57c)
(N −1)
B |=
(N −1)
B β3..β1= 0 (2.2.57d)
(N −1)
B |=
(N −1)
B β2N..β1= 0. (2.2.57e)
Accordingly, the components(N −1)a ,(N −1)a I α and
(N −1)
a J I
βαare not redundant, but can be
redefined to invariant fields (N −1)ˆa , (N −1)aˆ I α and
(N −1)
a J I
βα as defined by (2.2.39). The other components can be gauged away if they fulfil the superfield equation of motion, because
in this case they fulfil the same differential relations as the corresponding components
of the gauge parameter field B.
In consequence, the superfield equation of motion contains the non-redundant
a = 0 (2.2.58a)
Ni∂[α]µ(N −1)ˆa I[µ]= 0 (2.2.58b)
−iN∂[(][β]µ(N −1)ˆa [α][IJ[)][µ]]+δJ Iε[βα](N −1)ˆa = 0. (2.2.58c)
defined by the above off-shell action. The Lorentz vector ˆ
(N −1)
a J I
βα, even though it appears quadratic in the off-shell action, on shell fulfils the Maxwell equation and in addition
the Bianchi identity, which is an effect of the on-shell gauge fixing as demonstrated for
the example (2.2.48).
2.2.5. A spin(
) on-shell superfield
A superfield Q[i] transforming under the fundamental representation of spin(N) (see Appendix A) can be subject to the constraint [31, 22, 20, 18, 19]
D[α]IQ[i] = (γI)[i]jQ[jα], (2.2.59) where Q[iα] is a general superfield carrying an additional Lorentz index and reads
Q[iα] =qiα+θ[J]βq[iα,β]J+... . (2.2.60) For chiral representations of spin(N), the constraint is
DI[α]Q=ΣIQ[α] (2.2.61) or
D[α]IQ=Σ¯IQ[α], (2.2.62) depending on the chosen handedness. The chiral spin(N) indices have been omitted,
since they depend on the specific case of N. A discussion of fields transforming under
chiral representations will follow in Chapter 3. In the following formal considerations,
the form for non-chiral representations will be used.
The first component of the constraint superfield equation is
q[α]I =γIqα. (2.2.63)
Taking a general number of supercovariant projections
βk..β1α =γ
I[Q] Jk..J1
and reminding that
αk..α1 =q
αk..α1 +Q
αk..α1, (2.2.65)
leads to the expression for the components of Q[i]
βk..β1α =γ
q Jk..J1
α,βk..β1 +γ
Q Jk..J1
α,βk..β1 −Q
βk..β1α (2.2.66)
in terms of the components of Q[iα]. These are determined by taking the representations of this equation not contained in qJk..J1I
βk..β1α, i.e. both symmetric or antisymmetric in a
particular index pair, leading to
(I[Q] |Jk..|J1)
(α,|βk..|β1) =γ
(I[q] |Jk..|J1)
(α,|βk..|β1) (2.2.67a)
[I[Q] |Jk..|J1]
[α,|βk..|β1] =γ
[I[q] |Jk..|J1]
[α,|βk..|β1]. (2.2.67b)
In terms of covariant projections, this is solved by
Q Jk..J1
βk..β2 +... (2.2.68a)
Q Jk..J1
[α,|βk..|β1]=0, (2.2.68b)
where the dots indicate the further permutations implied by the symmetries on the
left-hand-side (an example will appear below). Accordingly, the components or
super-covariant projections of Q[i] are provided by the expressions QJk..[J1I]
βk..(β1α) =i∂β1αγ
βk..β2 +... (2.2.69a)
βk..[β1α] =0. (2.2.69b)
For the second component this means
q[(][J I[βα]][)]=iγIJ∂βαq (2.2.70a) ˜
q(J I)=0. (2.2.70b)
The constraint (2.2.59) therefore defines an on-shell multiplet consisting of q and qα,
subject to the supersymmetry transformations
δq=−εα[I]γIqα (2.2.71a)
δqα =−iε β Jγ
This multiplet corresponds to the special case of a minimal on-shell multiplet (2.2.58)
arising from an unconstrained scalar superfield with an attached spin(N) index, where
a = 0 (2.2.72a)
(N −1)
a I α =γ
α (2.2.72b)
(N −1)
a [[(]J I[βα]][)] = iγIJ∂βαq. (2.2.72c)
In other words, the defining constraint (2.2.59) for Q[i] removes the vector (N −1)a [[(]J I[βα]][)] by identifying it with the derivative of the scalar field, which is consistent with the Maxwell
equation and the Bianchi identity fulfilled by this vector.
Since the scalar multiplet is encoded in the lowest components of the constrained
superfield Q[i], a corresponding superfield action resembles the one for N = 1 [35]. It may be postulated as
S =[A][(N]1 [)]
d3x dθ[I]αdθI[α] (Dγ[K]Q)D[γ]KQ|
= 1
d3x (Qγ[K]QαIK[Iαγ] −Q[IαK]αIγ QK[γ] −2Qαγ[IK]QIK[αγ]), (2.2.73)
where A(N) is a normalisation factor. Indeed, reminding that
Qγβα[KJ I] =iγIγJγK∂αβqγ+ iγJγKγI∂γβqα−iγIγKγJ∂γαqβ (2.2.74a) Qβα[J I] =iγIγJ∂αβq, (2.2.74b)
the component action follows as
S =[A]N[(N]2[)]
d3x (−2i¯qγ∂[γ]αqα−2iqγ(∂γαqα)−2(∂αβq)∂αβq), (2.2.75)
where for canonic normalisation it can be chosen A(N) = −8N2[.]
This action is not manifestly supersymmetric, but rather supersymmetric only on
shell. It can however be derived from the off-shell actions
S =
S =
d3x(d2θ)N [Dα[I](D2)nA]DI[α](D2)nA|. (2.2.77) In the gauge for the minimal multiplet the superfield takes the form
A∝(θ2)N −1(N −1)ˆa +... . (2.2.78)
Insertion into the action for even N (for odd N the procedure is similar) gives
S ∝
d3x (d2θ)2(d2θ)N −2 [(θ2)N2−1
(N −1)
a +...][(θ2)N2−1
(N −1)
a +...]| (2.2.79a)
d3x (d2θ)2 [ ˆ
(N −1)
a +...][ ˆ
(N −1)
a +...]|. (2.2.79b)
The superfield appearing in the square-brackets corresponds to Q[i] and can be accord-ingly replaced so that
S ∝
d3x(d2θ)2 QQ|, (2.2.80) where the trace over indices of Q is implied. Using the product rule, this form can be brought to the postulated action
S ∝
d3x d2θ (Dα IQ)D
αQ|, (2.2.81)
where it was used that D2[Q]
i|= 0, as is implied by (2.2.69b).
Summarising this section, an off-shell action (2.2.51)/(2.2.52) for a generalN-extended
scalar superfield Awas proposed. It contains the three highest component fields of the superfield with canonical kinetic terms, i.e. with not more than two derivatives. The
superfield equation of motion (2.2.54) resulting from this action bears redundancies
due to its superfield-constraint form involving multiple supercovariant derivatives. In
the gauge where the minimal number of fields is kept non-redundant the equations of motion describe a multiplet consisting of a scalar field, to which the canonical dimension
one-half can be assigned, a spinor being also an SO(N) vector with dimension one, and
a field of rank two with dimension three-halves, which contains scalar auxiliary fields
vanishing on shell as well as a Lorentz vector displaying the properties of a Maxwell field
This Lorentz vector is considered undesirable for the description of an actual
scalar-matter multiplet consisting only of scalars and spinors. Therefore, the scalar superfield
Q[i] transforming under the fundamental representation of spin(N) and being subject to the constraint (2.2.59), which involves the SO(N) spin matrices, was introduced. The
constraint removes the problematic vector field by identifying it with the derivative of
the scalar and decouples the SO(N) index from the spinor, leaving an equal number
(given by the dimension of the fundamental representation of spin(N)) of scalar and
spinor fields in the on-shell multiplet.
As compared to the unconstrained superfieldA, the constrained superfieldQ[i] (2.2.59) contains the scalar multiplet in its lowest components, rather than in the highest ones.
Its superfield action (2.2.73) therefore resembles the one for an unconstrained N = 1
superfield. It can be derived from the off-shell action for the unconstrained superfield
2.3. Local symmetries
2.3.1. Gauge-covariant derivatives
An important generalisation of symmetries is space-time dependence of the
transforma-tion parameters, where derivatives of fields representing the symmetry group are not
covariant under group transformations, but rather behave as
DAA−→DAeXA= (DAX)eXA+ eXDAA. (2.3.1) A gauge-covariant derivative is given by
DA=DA+BA, (2.3.2)
where BA is a Lie algebra valued superfield transforming as
BA−→eXBAe−X −(DAeX)e−X, (2.3.3) so that in consequence
DAA−→eXDAA. (2.3.4) The supercovariant1projections of the spinor gauge fieldBI[α]transform under infinites-imal gauge transformations as
α,βk..β1 =−X
βk..β1α =−x
βk..β1α −X
βk..β1α, (2.3.5)
which, for the first few projections, means
δB[α]I =−xI[α] (2.3.6a)
δB[α,β]I,J =−xJ I[βα]−iδJ I∂βαx (2.3.6b)
δBI,KJ[α,γβ] =−xKJ I[γβα] −XKJ I[γβα] (2.3.6c)
δB[α,δγβ]I,LKJ =−xLKJ I[δγβα] −XLKJ I[δγβα] . (2.3.6d)
1[Component projections are not of interest, because the spinor gauge field covariantises the ]
Many of them can be gauged away by these transformations, while
δB[(]([α,β]I,J)[)] =−iδJ I∂βαx (2.3.7a)
δB[[][[α,β]I,J][]] = 0 (2.3.7b)
δB[[][[α,]I,|[|]K[γ][|]|[β]J[]]] = 0 (2.3.7c)
δB[[][[α,]I,|[|]LK[δγ][|][β]|J[]]] = 0. (2.3.7d)
This suggests that the trace ofB[(]([α,β]I,J)[)] can be identified with the vector gauge field and
the other fields correspond to various invariant, i.e. finitely covariant, field strengths.
These can be defined in the manifestly covariant formalism of commutators of covariant
derivatives, forming the gauge superalgebra described in the following.
2.3.2. Gauge superalgebra
The algebra of covariant derivatives is defined by [35]
[DA,DB}=TCABDC+FAB. (2.3.8) The superfields TC[AB] and FAB are called torsion and field strength, respectively. The case of two spinor derivatives, written explicitly in terms of the spinor gauge
field, reads
{DI α,D
β}= 2iδ IJ[∂]
αβ + 2D
(αB J)
β)+{B (I
(α,B J)
β)}+ 2D [I
[αB J]
β]+{B [I
[α,B J]
β]}. (2.3.9)
It implies the identifications
2iδIJBαβ +F
(IJ) (αβ) =2D
(αB J)
β)+{B (I
(α,B J)
β)} (2.3.10a)
F[[][IJ[αβ]][]]=2D[[[][α]IB[β]J[]]]+{B[[[]I[α],BJ[β]][]]}. (2.3.10b) They agree with what is expected from the above analysis of the supercovariant
gauge-field projections (2.3.7), by noting that
δ{BI[α],BJ[β]}= 0. (2.3.11) Conventionally, the trace ofF([(]IJ[αβ])[)] can be set to zero, corresponding to a redefinition of
Bαβ, which leads to [34]
It is common to formulate these identifications equivalently in terms of conventional
constraints on the torsion and field strength
Tc[αβ] =2i(γc)αβ (2.3.13a)
F[αβ]IJ =F[(](([αβ]IJ[)]))+ 2iεαβFIJ, (2.3.13b) with all the other torsion components being zero.
The conventional constraints affect the whole algebra due to the super-Jacobi identity
0 = [D[A,[DB,DC)}}, (2.3.14)
where [ABC) means antisymmetrisation with the caveat of an additional sign change if
two spinor indices are permuted. Unless indices have otherwise been manipulated, it is
usually understood that spinor indices are permuted together with their SO(N) index.
Inserting the commutators in terms of field strengths and torsions, the super-Jacobi
identity can be written as
0 = [D[A,TDBC)DD}+ [D[A,FBC)}
= D[ATDBC)
DD −[(−1)A(B+C)]2TD[BC[D|D|,DA)}+D[AFBC)
= D[ATDBC)−T
[BCT D
DD+D[AFBC)−TD[BCF|D|A), (2.3.15)
where |A| denotes the exclusion of this index from surrounding permutation brackets.
It contains four distinct cases of combinations of vector and spinor indices, leading to
the conditions on the field strength
(αF J K βγ) =T
IJ δL
(αβ F LK
IJ d
(αβ F K
|d|γ) (2.3.16a)
[ab F L
[ab F|d|c] (2.3.16b)
I δL
[αb F L
I d
[αbF|d|c] (2.3.16c)
[αF J βc) =T
J δL
[βc F LI
J d
[βcF I
Due to the vanishing torsions they simplify to
(αF J K
βγ) = 2iδIJ(γd)(αβF K
|d|γ) (2.3.17a)
D[aFbc]= 0 (2.3.17b)
[αFbc]= 0 (2.3.17c)
[αF J
βc) =−2iδ
αβFdc. (2.3.17d) The gauge multiplet is defined by the additional constraint F(([(][αβ]IJ[)])) = 0 [34], in which case
εαβDγKF IJ [+][ε]
J K [+][ε]
KI [=][δ]IJ[F]K αβ,γ+δ
γα,β+δ J K[F]I
βγ,α (2.3.18a)
D[aFbc]= 0 (2.3.18b)
[αFbc]= 0 (2.3.18c)
DI αF
βc+ 2iDcεαβFIJ −DβJF I
cα=−2iδ IJ[(][γ]d[)]
αβFdc. (2.3.18d) The first line shows that the totally symmetric part Fk[(][αβγ][)] vanishes and implies the relation
DI αF
J K [−][D][J α F
=−δI[J(γd)[α]γF[|][d][|]K[γ]]. (2.3.19) The trace inIJ
DI αF
IK =−1
3(N −1)(γ
d[)] γ α F
dγ ≡ −(N −1)F K
α (2.3.20)
produces the field strength of dimension three-halves (γd[)] γ α F
dγ . Inserting back, this leads in turn to the consistency relation for the dimension-one field strength
DI αF
J K [=][D][I αF
J K][−] 2 N −1δ
αLFK]L. (2.3.21) The last line of (2.3.18) yields the dimension-two field strength Fab as
(γ[a)αβDαIF I
βb]=−(γab)αβDαIF I
β = 2iNFab (2.3.22) or
Summarising, the whole algebra of covariant derivatives can be written as [34]
α,DβJ}=2iδIJDαβ+ 2iεαβFIJ| (2.3.24a) [Da,DβJ] =
−1 N −1(γa)
γ β D
γ FKJ| (2.3.24b)
[Da,Db] =[2N][(N −1)]i (γab)αβDαID J βF
, (2.3.24c)
with the condition
DI αF
J K
=D[α][IFJ K]− 2 N −1δ
αLFK]L. (2.3.25) The superfield FIJ is called field-strength or gauge multiplet, since it describes all field strengths appearing in the algebra in terms of covariant projections. As seen above,
differ from pure supercovariant projections by those projections of the spinor gauge field
2.4. Supergravity
2.4.1. Supergravity-covariant derivative
General superspace coordinate transformations
zM −→z˜M(z) (2.4.1)
induce the transformation of a supervector field
VM(z)−→V˜M(˜z) = (∂Nz˜M)VN(z). (2.4.2) At each point, a supervector VM can be expanded in a standard basis of the tangent space as
VM =VAE[A]M. (2.4.3) The vector VA transforms under the local structure group of the tangent space, which leaves this expansion invariant and thus relates equivalent bases to each other.
In the conventional curved SO(N) superspace, the local structure group is chosen to
be SL(2,[R])×SO(N) [36]. Consequently, the super-vielbein
E[A]M = E m a E
µ a,J
EI,m[α] EI,µ[α,J]
carries local Lorentz indices a, α and local SO(N) indices I, transforming under this
local structure group. As a basic principle, the leading component of E[a]m is identified with the vielbein of non-supersymmetric space-time,
E[a]m|=e[a]m. (2.4.5) Further, in flat superspace the super-vielbein takes the form [35]
E[A]B = δ b
a 0 iθγ[I](γb)γα δβαδIJ
corresponding to the relation
Accordingly, the supergravity covariant derivative is given by [34]
DA=EAM∂M +1[2]ΩAmnMmn+ 1[2]ΦAP QNP Q, (2.4.8) where ΩA and ΦAare the connection or gauge fields associated with the Lorentz group and SO(N), respectively.
2.4.2. Supergravity algebra
The algebra of covariant derivatives is defined by [34]
[DA,DB}=TABCDC +[2]1RABP QNP Q+1[2]RABmnMmn. (2.4.9) The torsion and field strengths are, in terms of the super-vielbein and connections, given
T[AB]C =C[AB]C+ 1[2]Ω[A]mn(Mmn)BC−(−) AB1
B (Mmn)AC + 1[2]Φ[A]P QNP QδBC−(−)
AB1 2Φ
P Q B NP Qδ
A (2.4.10a)
RAB =EAΩB−(−)ABEBΩA−CABCΩC +EAΦB−(−)ABEBΦA−CABCΦC
+ [ΩA+ΦA,ΩB+ΦB}, (2.4.10b) where the superfields C[AB]C are defined by
(EAEBM)E C
M −(−1) AB[(][E]
BEAM)E C M
EC =CABCEC. (2.4.11) Via the Jacobi identity
0 = [E[A,[EB,EC)}}, (2.4.12)
they are related by
E[ACBC)E −C
[BC C E
A)D = 0, (2.4.13)
where it is as usual understood that SO(N) indices are permuted together with their
Lorentz spinor index.
by imposing constraints on the torsions [36]. At least in four-dimensional simple
su-pergravity, this procedure (together with chirality-conserving constraints) is known to
completely determine the field strengths in terms of these torsions via the super-Jacobi identity. Also in three dimensions, it is sufficient to impose constraints only on the
tor-sions; however, the description of the field strengths needs one additional field emerging
in the solution of the super-Jacobi identity, as will be seen below. The torsion constraints
are equivalent to choosing connections in terms of vielbeins in shape of the fieldsC[AB]C
in order to reduce degrees of freedom as much as possible. The specific choices and
resulting dependencies are motivated in the following.
Concretely, for the case of two spinor derivatives the torsions are
TIJ γ[αβK]D[γ]K+T[αβ]IJ cDc=1[2][−Ω I γ α,β δ
J K−Ω
J γ β,α δ
I K+Φ
I,J K α δ
γ β +Φ
J,IK β δ
γ α]DγK
+CIJ γ[αβK]D[γ]K+CIJ c[αβ]Dc. (2.4.14a) The spinor connections can be chosen to absorb the fields CIJ γ[αβK], leaving
TIJ γ[αβK] =0 (2.4.15a)
TIJ c[αβ] =CIJ c[αβ] . (2.4.15b) In order to incorporate the case of flat superspace, it is natural to choose
TIJ c[αβ] =CIJ c[αβ] = 2iδIJ(γc)αβ. (2.4.16) In consequence, the vector vielbein is expressed by the spinor vielbeins and spinor
con-nections by virtue of the vielbein algebra
{EI[α],EJ[β]}=2iδIJ(γc)αβEc+CIJ γαβKE K
γ (2.4.17)
with the replacement
CIJ γ[αβK] =−1 2[−Ω
I γ α,β δ
J K [−]
ΩJ[β,α]γδIK +ΦI,J K[α] δγ[β]+ΦJ,IK[β] δγ[α]]. (2.4.18) The spinor vielbeins, the spinor connections and the vector connections remain as
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We will be multiplying polynomials using two different ways, (traditional and FFT), and see whether one method is consistently faster than the other one. We will let the polynomials and be of the
same degree for simplicity and we will generate the polynomials coefficients randomly. Again, for simplicity, the coefficients are integers chosen in [-50..50]. For a more detailed explanation of
what follows, please consult [1] in the Reference section.
You have, of course, noticed that entry in row 1 corresponds to entry in row 2. Or, to put it another way, the top row lists the logarithms in base 2 of the bottom list.
What if we wanted to calculate (64)(512)? One way would be to follow the traditional algorithm that we all learned in grammar school and perform the standard multiplication. The other way would be to
locate the corresponding logarithms, namely 6 and 9, ADD these two logarithms to get 15, and then determine the antilog, our answer, namely 32768. In what follows, the strategy is the same. The
quantities "worked on" and the operations are different. Instead of multiplying numbers, we will be multiplying polynomials. The equivalent of taking logarithms and antilogarithms will be polynomial
evaluations and interpolations and the equivalent of our ADD will be POINTWISE multiplication.
Consider two lists of numbers say {s} = ( , ,..., ) and {t} = ( , , ..., ). Then pointwise multiplication gives the list of numbers { } = ( , ,..., ).
The procedure "coefficient" generates "deg+1" random coefficients in the range [-50..50].
The procedure "padding" "right-pads" a list of coefficients with "deg" zeros.
The procedure "poly_mult" determines the coefficient vector "c" of the resulting polynomial in the traditional way. Vector "c" is obtained by "convolving" the input coefficients vectors "a" and "b".
The procedure "dft" (Discrete Fourier Transform) is present here since we wanted to, in fact compare the three processes for multiplication of two polynomials, namely the traditional, DFT, and FFT
(Fast Fourier Transform) processes. One has to get into high degrees to see the FFT overtake the traditional method. The DFT is present here to make us appreciate the speed improvement that the FFT
brings to the situation. The procedure "dft" which follows takes two arguments. The first one is the list to be transformed. The second argument is set so that a = 1 for DFT and a = -1 for IDFT
(Inverse Discrete Fourier Transform). The last part of the code takes very small values (usually on the inverse transformation when dealing with original real data) located in the imaginary part and
zero them out. So first the list is "split" into real and imaginary parts, Fourier transformation occurs, zeroing out is done where needed, and then the two lists are "sewn" back together.
The FFT/IFFT procedure call is to be found below. The only reason the procedure shown below looks lengthy is because real part and imaginary part of lists going through the FFT or the IFFT have to be
kept separated.
The "power_of_two" procedure determines the nearest power greater than deg(A(x)) + deg(B(x) + 1. This is needed, for padding purposes, since the FFT/IFFT needs "power of two" list sizes.
> while (2^i < 2*deg+1) do i := i+1; od;
The " zeros_plot" procedure "reconstructs" a polynomial from its coefficients so that its zeros can be determined and plotted if necessary.
Now that all the procedures we will need have been written, let us randomly generate the "deg+1" coefficients of two polynomials A(x) and B(x) of degree "deg". For simplicity, we will take the two
polynomials to be of the same degree. Otherwise, we would have to pad with zeros. We are also choosing the coefficients to be integers in [-50..50] .
We are saving these two lists, in c_list and d_list, so that we can use the same polynomials with the DFT and then the FFT approaches.
In the straightforward multiplication (convolution) method that follows, list sizes are handled automatically. We will notice, when we use the DFT approach that the padding will always be up to a
power of 2.
The final answer provides a list of the coefficients of the resulting polynomials C(x).
Again, for a more detailed explanation we recommend that you refer to [1]. We would like to multiply the same two polynomials seen above using the DFT. Very quickly, we will come to realize that this
process requires a lot of computing time. In fact the DFT is no match for the traditional method. We wanted to investigate the DFT approach for the simple reason that it is a routine that you can
write on your own by using the definition for the DFT. This way, anybody unfamiliar with the intricacies of the FFT, can feel somewhat at ease when told that the FFT is nothing but a "souped up" DFT
- (Apologies to Cooley and Tukey [3]). Therefore we recommend that you bypass the DFT process and go directly to the FFT one whenever your polynomial degrees start to get above 50.
Note that the polynomial representation we used above, where polynomials are identified by their coefficient lists, is called the "coefficient representation (CR)". One can also use the "point-value
representation (PVR)". The PVR form of a polynomial of degree is a set of ( ) point-value pairs ( ), ( ) , ..., ( ), that is ( ) points that belong to the graph of the polynomial. While multiplying
polynomials in CR form takes O( ), the alternative multiplicative option of using the three steps of
The evaluation of a polynomial at a particular value of is usually carried out using Horner's method. The "inverse" of evaluation is interpolation and one can determine the coefficients {a} = ( , ,
..., ) of a polynomial in PVR form by relying on Lagrange polynomials. Thus -points evaluation and interpolation are therefore well-defined operations that allow one to move from CR to PVR and back
Let us assume in the rest of the discussion that = deg( = deg( ). Since the resulting polynomial is of degree , we must start with "extended" PVR form for both polynomials A and B. The quantity (+ )
that appears at times has to do with the fact that a polynomial of degree has coefficients. We must also, since the FFT is at its best when processing lists whose lengths are powers of 2, let 2 be a
power of 2. Be aware that there are two sorts of padding going on. One was seen at work in the traditional multiplication process. But the one we are referring to now has to do with bringing all
lists to lengths that are powers of 2.
One can be clever in the choice of the 's (there will be 2 of them) at which polynomials will be evaluated so as to determine the corresponding PVR form. In fact, if we choose "complex roots of
unity" as evaluation points, we can produce a PVR form of a given polynomial in CR form by taking the DFT of the coefficient vector. IDFT takes care of interpolation so as to bring us back to a CR
form, from a PVR form.
Create coefficient representation of and as polynomials of degree 2 . Time = [O( )].
Evaluate: Compute point-value representation of and of length through applications of the DFT (later the FFT) of order to each coefficient vector. These representations contains the values of the two
polynomials at the ( )th roots of unity. Time = [O( )].
Pointwise multiply: Compute a PVR form for = by multiplying these values together, pointwise. The representation obtained contains the value of at each ( )th roots of unity. Time = [O( )].
Interpolate: Create the coefficient representation of the polynomial through a single application of a IDFT (later IFFT) on point-value pairs. Time = [O( )].
What has been done is as follows: {a} convolved with {b} = IDFT[DFT(a)*DFT(b)], where vectors {a} and {b} have been padded with zeros to length and * denotes the componentwise product of two -element
vectors. Note that {a} convolved with {b} stands for the traditional multiplication procedure we started out with.
We are padding the coefficients list with zeros up to the power of two that follows degree( ) + degree( ).
> c_list_in := padding(c_list,pot - deg - 1); d_list_in := padding(d_list,pot - deg - 1);
You can, if you wish, plot the zeros of the two initial polynomials. You would have to remove the comment symbols # on the line that follows.
Please recall NOT to run the DFT if the degrees of your polynomials and are large, say above 50 each . If you do, you will appreciate what the FFT brings to the DFT. For a good discussion on this
point, see [2, pg 294] in the Reference section.
In the next step, we are seeing whether Parseval' identity is verified or not. This is a safety check to make sure that we are "on track" . You will have to "uncomment" it to see it in action.
We are now ready to perform the IDFT. We should, if everything goes well, recover the coefficients of . Since and are chosen with real coefficients we should recover numbers with no or very tiny
imaginary values. We will therefore extract the real part of our final list. Also keep in mind that we will be getting a list that will, most of the time, be padded with zeros. The reason is as
follows: recall that will be of degree "2*deg" and will "have" "2*deg+1" coefficients. This number will, in general, be less than power_of_two.
How do the coefficients of the resulting polynomials match under "traditional" and "DFT". (Getting all zeros is good!)
We will now retrace the steps we followed when applying the DFT/IDFT combination, using the FFT/IFFT combination this time.
Here is some results on a Pentium III. You will notice where the FFT overtakes the traditional method.
For deg( ) = 20, standard_time = 0.015, dft_time := 15.7, fft_time = 0.275
For deg( ) = 30, standard_time = 0.035, dft_time := 27.6, fft_time = 0.4
For deg( ) = 50, standard_time = 0.235, dft_time := 307.2, fft_time = 1.05
For deg( ) = 100, standard_time = 0.345, dft_time := Not Run , fft_time = 1.168
For deg( ) = 300, standard_time = 3.723, dft_time := Not Run , fft_time = 6.009
For deg( ) = 400, standard_time = 7.996, dft_time := Not Run , fft_time = 5.948
For deg( ) = 500, standard_time = 10.831, dft_time := Not Run , fft_time = 5.873
For deg( ) = 2000, standard_time = 168.822, dft_time := Not Run , fft_time = 35.424
[1] Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Introduction to Algorithms, MIT Press (1990)
[2] David W. Kammler, A First Course in Fourier Analysis, Prentice-Hall (2000)
[3] J.W. Cooley and J.W. Tukey, An Algorithm for the Machine Computation of Complex Fourier Series, Math. Comp . 19(1965), 297-301.
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[Algorithm of the Day] Solve the search space explosion problem with "bidirectional BFS" (with the heuristic search AStar algorithm) | Python topic month - Moment For Technology
This article is participating in Python Theme Month. See the link to the event for more details
Topic describes
This is the 127. Word solitaire on LeetCode, and the difficulty is difficult.
Tag: “bidirectional BFS”
The conversion sequence from beginWord and endWord in the dictionary wordList is a sequence formed according to the following specifications:
• The first word in the sequence is beginWord.
• The last word in the sequence is endWord.
• Only one letter can be changed per conversion.
• The intermediate word in the transformation must be a word in the dictionary wordList.
Given the two words beginWord and endWord and a dictionary wordList, find the number of words in the shortest conversion sequence from beginWord to endWord. If no such sequence of transformations
exists, return 0.
Example 1:
Enter: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log","cog"] A minimum conversion sequence is "hit" - > "hot" - > "dot" - > "dog" - > "cog", return it the length of 5.Copy the code
Example 2:
Input: beginWord = "hit", endWord = "cog", wordList = ["hot","dot","dog","lot","log"]Copy the code
• 1 <= beginWord.length <= 10
• endWord.length == beginWord.length
• 1 <= wordList.length <= 5000
• wordList[i].length == beginWord.length
• BeginWord, endWord, and wordList[I] are composed of lower-case English letters
• beginWord ! = endWord
• All strings in wordList are different from each other
Fundamental analysis
According to the meaning of the question, only one character can be replaced at a time, and each new word must have appeared in the wordList.
A naive way to do this is to use BFS.
Start from beginWord, enumerate all schemes to replace a character, if the scheme exists in wordList, then join the queue, so that there are all words with 111 replacement times in the queue. The
element is then removed from the queue and the process continues until either endWord is encountered or the queue is empty.
At the same time, in order to “prevent repeated enumeration to an intermediate result” and “record how many times each intermediate result is transformed”, we need to create a “hash table” to record.
The KV of the hash table is of the form {word: how many conversions to get}.
When enumerating the new word STR, check whether it already exists in the “hash table”. If not, update the “hash table” and put the new word into the queue.
This practice ensures that “enumerates to all bybeginWord 到 endWord“, and bybeginWord 到 endWordThe “shortest transformation path” of is necessarily the first to be enumerated.
Two-way BFS
1 <= beginWord. Length <= 10
Naive BFS could lead to a “search space explosion”.
Imagine if our wordList was rich enough to include all the words, For an beginWord replacement of 101010 characters that can produce 10∗2510 * 2510∗25 new words (each substitution point can replace
another 252,525 lowercase letters), the first layer will produce 250,250,250 words; The second layer produces more than 6∗1046 * 10^46∗104 new words…
As the number of layers increases, the number increases faster. This is the “search space explosion” problem.
In naive BFS implementations, the bottleneck of space is largely determined by the maximum width in the search space.
So is there a way that we can not use such a wide search space and still find the desired results?
“Bidirectional BFS” is a good solution to this problem:
The search starts in both directions at the same time, and once the same value is found, it means that a shortest path connecting the starting point and the end point is found.
The basic implementation of bidirectional BFS is as follows:
1. Create “two queues” for searching in both directions;
2. Create “two hash tables” to “resolve repeated searches of the same node” and “record the number of conversions”;
3. In order to make the two search directions “average” as much as possible, we first determine which queue has the smallest capacity each time we expand the value from the queue.
4. If the searched node is found during the search, the shortest path is found.
The pseudocode corresponding to the basic idea of “bidirectional BFS” is roughly as follows:
D1 and D2 are queues in two directions, M1 and M2 are hash tables in two directions, and the distance between each node and the starting point is recorded// If both queues are not empty, it is necessary to continue the search
// If one of the queues is empty, the target node in a certain direction cannot be found in the end
while(! d1.isEmpty() && ! d2.isEmpty()) {if (d1.size() < d2.size()) {
update(d1, m1, m2);
} else{ update(d2, m2, m1); }}// Update the logic of fetching an element from queue D for "one full extension"
void update(Deque d, Map cur, Map other) {}
Copy the code
Back to the problem, let’s see how to use “bidirectional BFS” to solve the problem.
This is probably the first time for many students to come into contact with “two-way BFS”, so I have written a lot of notes this time.
It is recommended that you read with a basic understanding of “bidirectional BFS”.
class Solution {
String s, e;
Set<String> set = new HashSet<>();
public int ladderLength(String _s, String _e, List<String> ws) {
s = _s;
e = _e;
If the target word is not in the set, there is no solution
for (String w : ws) set.add(w);
if(! set.contains(e))return 0;
int ans = bfs();
return ans == -1 ? 0 : ans + 1;
int bfs(a) {
// start with beginWord
// d2 means search from end endWord (reverse)
Deque<String> d1 = new ArrayDeque<>(), d2 = new ArrayDeque();
/* * m1 and m2 respectively record how many times the words appear in the two directions have been transformed * e.g. * m1 = {" ABC ":1} represents ABC replaced by beginWord for 1 time * m1 = {"xyz":3} represents xyz by EndWord replaces the character */ three times
Map<String, Integer> m1 = new HashMap<>(), m2 = new HashMap<>();
m1.put(s, 0);
m2.put(e, 0);
/* * If one of the queues is empty, it is necessary to continue the search * * if one of the queues is empty, it is necessary to continue the search * * if one of the queues is empty, it is necessary to continue the search * * if one of the queues is empty, it is necessary to continue the search * * if one of the queues is empty, it is necessary to continue the search * * if one of the queues is empty, it is necessary to continue the search * * Reverse search is also unnecessary */
while(! d1.isEmpty() && ! d2.isEmpty()) {int t = -1;
// In order to make the search in both directions as even as possible, the direction with fewer elements in the queue is extended first
if (d1.size() <= d2.size()) {
t = update(d1, m1, m2);
} else {
t = update(d2, m2, m1);
if(t ! = -1) return t;
return -1;
// Update means to take a word from the deque and extend it,
// cur for the current direction of distance dictionary; Other is a distance dictionary in another direction
int update(Deque<String> deque, Map<String, Integer> cur, Map<String, Integer> other) {
// Get the original string that you currently want to extend
String poll = deque.pollFirst();
int n = poll.length();
// Enumerates which character I replaces the original string
for (int i = 0; i < n; i++) {
// Enumerates which lowercase letter to replace I with
for (int j = 0; j < 26; j++) {
// The replaced string
String sub = poll.substring(0, i) + String.valueOf((char) ('a' + j)) + poll.substring(i + 1);
if (set.contains(sub)) {
// If the string has been recorded (extended) in the current direction, skip it
if (cur.containsKey(sub)) continue;
// If the string appears in the "other direction", the shortest path connecting the two directions has been found
if (other.containsKey(sub)) {
return cur.get(poll) + 1 + other.get(sub);
} else {
// Otherwise join the deque queue
cur.put(sub, cur.get(poll) + 1); }}}}return -1; }}Copy the code
• Time complexity: letwordListLength of
$n$.beginWordThe length of the string is
$m$. Because all search results have to be inwordListYes, so it’s at most
$n + 1$Node, in the worst case, all nodes are connected, and the complexity of searching the whole graph is
$O(n^2)$; frombeginWordStart character replacement, and perform character by character check during replacement. The complexity is
$O(m)$. The overall complexity is
$O(m * n^2)$
• Space complexity: Same space size. O (m ∗ (n2) to O (m * n ^ 2) O (m ∗ n2)
This is essentially a “all edge weights are zero1“Most short circuit problem: willbeginWordAnd all thewordListAny string that occurs is considered a point. Each transformation operation is considered
to produce an edge weight of1The edge. Problem asks tobeginWordIs the source point, withendWordIs the shortest path of the sink.
With the help of this problem, I introduced to you “bidirectional BFS”, “bidirectional BFS” can effectively solve the “search space explosion” problem.
For those search problems where the search nodes grow multiple or exponentially as the number of layers increases, “bidirectional BFS” can be used to solve them.
[Supplement] Heuristic search AStar
The “heuristic function” of A* can be designed directly according to the rule in question.
For example, for two strings A and B, I think it’s appropriate to use the number of different characters in them as the estimated distance.
Because the difference between the number of different characters ensures that the real distance (a theoretical minimum number of substituts) is not exceeded.
Note: The data in this question is weak, and we use A* to pass it, but usually we need to “make sure there is A solution” for the heuristic search of A* to be of real value. In this case, unless
endWord itself is not in the wordList, there is no good way to judge “whether there is a solution” in advance. In this case, A* will not bring “search space optimization” as well as “bidirectional
Java code:
class Solution {
class Node {
String str;
int val;
Node (String _str, int _val) {
str = _str;
val = _val;
String s, e;
int INF = 0x3f3f3f3f;
Set<String> set = new HashSet<>();
public int ladderLength(String _s, String _e, List<String> ws) {
s = _s;
e = _e;
for (String w : ws) set.add(w);
if(! set.contains(e))return 0;
int ans = astar();
return ans == -1 ? 0 : ans + 1;
int astar(a) {
PriorityQueue<Node> q = new PriorityQueue<>((a,b)->a.val-b.val);
Map<String, Integer> dist = new HashMap<>();
dist.put(s, 0);
q.add(new Node(s, f(s)));
while(! q.isEmpty()) { Node poll = q.poll(); String str = poll.str;int distance = dist.get(str);
if (str.equals(e)) {
int n = str.length();
for (int i = 0; i < n; i++) {
for (int j = 0; j < 26; j++) {
String sub = str.substring(0, i) + String.valueOf((char) ('a' + j)) + str.substring(i + 1);
if(! set.contains(sub))continue;
if(! dist.containsKey(sub) || dist.get(sub) > distance +1) {
dist.put(sub, distance + 1);
q.add(newNode(sub, dist.get(sub) + f(sub))); }}}}return dist.containsKey(e) ? dist.get(e) : -1;
int f(String str) {
if(str.length() ! = e.length())return INF;
int n = str.length();
int ans = 0;
for (int i = 0; i < n; i++) {
ans += str.charAt(i) == e.charAt(i) ? 0 : 1;
returnans; }}Copy the code
Python 3 code:
class Solution:
def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) - >int:
class PriorityQueue:
def __init__(self) :
self.queue = []
self.explored = set(a)def __contains__(self, item) :
return item in self.explored
def __add__(self, other) :
heapq.heappush(self.queue, other)
def __len__(self) :
return len(self.queue)
def pop(self) :
return heapq.heappop(self.queue)
def heuristic(curr, target) :
return sum(1 for a, b in zip(curr, target) ifa ! = b) wordList =set(wordList)
if endWord not in wordList:
return 0
front = PriorityQueue()
front.__add__((1.0, beginWord))
while front:
l, _, s = front.pop()
if s == endWord:
return l
for i in range(len(s)):
for c in string.ascii_lowercase:
t = s[:i] + c + s[i + 1:]
if t in wordList and t not in front:
front.__add__((l + 1, heuristic(t, endWord), t))
return 0
Copy the code
The last
This is No.127 in our “Brush through LeetCode” series, which began on 2021/01/01. As of the start date, there are 1916 questions on LeetCode, some with locks, and we will first brush through all the
questions without locks.
In this series of articles, in addition to explaining how to solve the problem, I’ll give you the most concise code possible. If the general solution is involved, the corresponding code template will
be provided.
In order to facilitate the students to debug and submit the code on the computer, I set up a related warehouse: github.com/SharingSour… .
In the repository address, you can see the solution to the series, the corresponding code to the series, the LeetCode link, and other preferred solutions to the series.
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How to calculate friction loss of Bingham fluid - Pump & Flow
How to calculate friction loss of Bingham fluid
To calculate the friction loss of a Bingham fluid in a pipe, you can use the Bingham plastic model, which is an empirical equation that describes the relationship between shear stress and shear rate
for non-Newtonian fluids with a yield stress. The Bingham plastic model has the form:
τ = τ_0 + k * γ^n
τ = shear stress (Pa)
γ = shear rate (s^-1)
k = flow behavior index (Pa.s^n)
n = flow behavior index exponent (unitless)
τ_0 = yield stress (Pa)
The friction factor (f) can be calculated using the following equation:
f = (2 * g * D * (τ_0 + k*γ^n))/(V^2 * L)
f = friction factor
g = acceleration due to gravity
D = inside diameter of pipe
L = length of pipe
V = velocity of flow
To calculate the friction loss, you will need to know the density and viscosity of the fluid, the inside diameter of the pipe, and the velocity of the fluid flow. You can then use the Darcy-Weisbach
equation, which states that the friction loss (head loss) in a pipe is equal to the friction factor (f) times the length of the pipe (L) times the velocity of the flow (V) squared, divided by the
inside diameter of the pipe (D) and a constant (2g).
The Darcy-Weisbach equation for the friction loss in a pipe is given by:
ΔP = f * L * V^2 / (2 * g * D)
ΔP = Friction loss (pressure drop)
f = friction factor
L = Length of pipe
V = Velocity of flow
g = acceleration due to gravity
D = inside diameter of pipe
The friction factor for Bingham fluids can be calculated using the model proposed by Herschel-Bulkley.
It's important to note that the Bingham plastic model only describes the fluid behaviour in the yield and plastic region, for the turbulent region you'll need to use other correlation such as the
Colebrook-White equation. Additionally, it's important to measure or estimate the properties of the Bingham fluid, such as yield stress and viscosity, at the time of the experiment or design, since
these properties can change with temperature and pressure.
It's also worth noting that the friction factor is highly dependent on the properties of the fluid and the measurement conditions, such as temperature and pressure, and the measurement equipment have
to be considered as well to get accurate results.
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Whiskey Barrel Proof Meaning
WHISKEY 101
What are ‘Cask Strength,’ ‘Barrel Proof,’ and ‘Barrel Strength?’
Is there a difference?
Barrel Proof, Barrel Strength, and Cask Strength
Cask Strength, Barrel Proof, and Barrel Strength are different phrases which mean the same thing… the alcohol-by-volume (ABV) strength or proof is essentially the same as when it came out of the
barrel at the time of bottling (as opposed to Full Proof which is the ABV when it went into the barrel). The official rule is that it cannot be more than 1% (2 proof points) lower than when the
whiskey was poured from the barrel. Distillers will choose one phrase over the other simply based upon marketing reasons or distillery preferences. The typical level of alcohol-by-volume (ABV) for a
barrel proof whiskey is usually in the range of 52–66% ABV (104 – 132 proof), but can be higher.
In 1979, the Federal Bureau of Alcohol, Tobacco, and Firearms “recognized the need to establish guidelines for use of the terms Original Proof, Original Barrel Proof, Entry Proof and Barrel Proof on
distilled spirits labels.” (See ATF Ruling 79-9) Original proof, original barrel proof, full proof, and entry proof on a label indicate the same thing: “that the proof of the spirits entered into the
barrel and the proof of the bottled spirits are the same.” This means that whiskeys with these phrases on their labels must be at the same ABV (proof) as when the whiskey was put into the barrel.
Barrel Proof, Cask Proof, and Barrel Strength are the proof at when the whiskey leaves the barrel. The rule for these phrases according to ATF Ruling 79-9 is, “the bottling proof is not more than two
degrees lower than the proof established at the time the spirits were gauged for tax determination.” Whiskeys are gauged for tax determination when the barrels are dumped for bottling. The ‘2
degrees’ rule gives distillers some wiggle room in case there is a slight proof drop between gauging and bottling. The intention is for the proof in the bottle to match that of the barrel. This is
why barrel proof whiskeys will often have an ABV number like 52.7% alcohol rather than a nice round 53%. Generally speaking, Barrel Strength and Barrel Proof are an American phrase while Cask
Strength originates (but is not limited to) Scotland, because they call their barrels “casks”.
WHY DOES THE PROOF (ABV) MATTER?
Because high-proof whiskeys pack a palatable punch of flavors. They have had years in the barrel, seeping into and out of the wood grain, gaining and intensifying flavors as seasons and temperatures
change. This is what develops the intense, robust flavors that whiskey and bourbon enthusiasts crave.
Most other whiskeys on the other hand, have water added to them when they come out of the barrel. This is to make them less expensive and to make them appeal to a broader range of palates who do not
enjoy a high-proof spirit. Most whiskey in the US have enough water added to make them 40% ABV (80 proof), which is the lowest legal limit to be called whiskey in the US.
Adding water to the whiskey greatly reduces its flavor and character. Whiskey & bourbon connoisseurs prefer the higher proof because it typically (but not always) means more flavor and a better
Denny Potter, former Heaven Hill Master Distiller said, “We produce barrel proof whiskey because whiskey connoisseurs demand it. The desire to chart your own course with barrel-proof whiskeys is
undeniable. Whether you want the straight-out-of-the-barrel robustness or controlled dilution, someone can find their perfect sip. But knowing a barrel proof bourbon’s provenance contributes here as
well. We often make it clear where single barrels or many of our batches come from. And to understand the variances of aging on proof and flavor most effectively comes out in barrel proof whiskeys.”
One common misconception about barrel proof whiskeys is that you are supposed to drink them straight. You certainly can if you wish, or you can adjust the proof by adding water to suit your personal
“Dad always told everyone to add some water or ice to the Booker’s bourbon to open it up and enjoy it at the strength you enjoy it,” says Fred Noe, Jim Beam Master Distiller and Booker’s son. “I
practice that myself and I add some ice and water to Booker’s.”
Barrell Cask-Strength bourbon (Batch 23)
Maker’s Mark Cask Strength Kentucky straight bourbon (Batch 19-01)
Knob Creek Cask-Strength rye
Stagg Jr. Barrel Proof Kentucky straight bourbon (13th Edition)
Elijah Craig Barrel Proof Kentucky straight bourbon (Batch A120)
Jack Daniel’s Single Barrel Special Release Coy Hill High Proof
15 Comments
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8 months ago
Great website!!!!!
8 months ago
Great info! Great site all around!
8 months ago
Thank you for the info
8 months ago
I really enjoyed this article. I’ve often wondered what all of this phrases meant. I’m fairly new to bourbons and am still trying to find out how the different proofs affect the taste. I know that I
am beginning to notice that more often than not a higher proof offers a better taste. A great example of this is Old Forester 1920. It has a higher proof than the 1910, and, in my opinion, has far
more flavor than 1910. I used to think that the 90ish proof point was where I wanted to be, but now I’m finding that I am gravitating towards proofs at or a bit over 100. Thanks again for the
8 months ago
Really good content … I learned something new
8 months ago
Great explanation of these common proofs that are often confused.
8 months ago
Great website!
8 months ago
Great explanation of bottling terms. Great website!
8 months ago
You can’t talk about Cask Strength and not mention Wild Turkey Rare Breed! Great write up on the different terms!
6 hours ago
Good stuff
6 hours ago
Love the explanations!
4 hours ago
I remember my first jump into the barrel proof/cask strength world with a store pick of Elijah Craig that came in at 130.9 at 8 years. I was blown away by all the flavors I was experiencing and it
really propelled me deeper into the journey, more than I realized. What Heaven Hill is doing with making so much of their juice available straight from the barrel is such a wonderful treat to enjoy.
I don’t care what they call it but they could simplify it by calling it Pure Age.
Last edited 4 hours ago by Joseph Patrick Lynn III
3 hours ago
Awesome website!!
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Output to File variables
The following variables are included in the three text files that are created using the "Output to File" option:
TABLE OF
VARIABLE DEFINITION OVERBANK RESULTS QVA
WSEL Water Surface Elevation Yes Yes Yes
Q(LOB) Discharge in Left Overbank Yes
Q(MCH) Discharge in Main Channel Yes
Q(ROB) Discharge in Right Overbank Yes
Q(Total XS) Total Cross-section discharge Yes Yes Yes
VAve(LOB) Average velocity of Left Overbank flow Yes
VAve(MCH) Average velocity of Main Channel flow Yes
VAve(ROB) Average velocity of Right Overbank flow Yes
VAve(Total XS) Average velocity of Total Cross-section flow Yes Yes Yes
Area(LOB) Flow area in Left Overbank Yes
Area(MCH) Flow area in Main Channel Yes
Area(ROB) Flow area in Right Overbank Yes
Area(Total XS) Flow area in Total Cross-section Yes Yes Yes
DAve(LOB) Average Depth of Left Overbank flow Yes
DAve(MCH) Average Depth of Main Channel flow Yes
DAve(ROB) Average Depth of Right Overbank flow Yes
DAve(Total XS) Average Depth of Total Cross-section flow Yes Yes
DMax(LOB) Maximum Depth of Left Overbank flow Yes
DMax(MCH) Maximum Depth of Main Channel flow Yes
DMax(ROB) Maximum Depth of Right Overbank flow Yes
DMax(Total XS) Maximum Depth of Total Cross-section flow Yes Yes
(DAve x V)(LOB) Product of Average Depth and Velocity of Left Overbank flow Yes
(DAve x V)(MCH) Product of Average Depth and Velocity of Main Channel flow Yes
(DAve x V)(ROB) Product of Average Depth and Velocity of Right Overbank flow Yes
(DAve x V)(Total XS) Product of Average Depth and Velocity of Total Cross-section flow Yes Yes
(DMax x V)(LOB) Product of Maximum Depth and Velocity of Left Overbank flow Yes
(DMax x V)(MCH) Product of Maximum Depth and Velocity of Main Channel flow Yes
(DMax x V)(ROB) Product of Maximum Depth and Velocity of Right Overbank flow Yes
(DMax x V)(Total XS) Product of Maximum Depth and Velocity of Total Cross-section flow Yes Yes
FrNo(LOB) Froude Number of Left Overbank flow Yes
FrNo(MCH) Froude Number of Main Channel flow Yes
FrNo(ROB) Froude Number of Right Overbank flow Yes
FrNo(Total XS) Froude Number of Total Cross-section flow Yes Yes
WettedPerim(LOB) Wetted Perimeter of Left Overbank flow Yes
WettedPerim(MCH) Wetted Perimeter of Main Channel flow Yes
WettedPerim(ROB) Wetted Perimeter of Right Overbank flow Yes
WettedPerim(Total XS) Wetted Perimeter of Total Cross-section flow Yes Yes
FlowWidth(LOB) Width of Left Overbank flow Yes
FlowWidth(MCH) Width of Main Channel flow Yes
FlowWidth(ROB) Width of Right Overbank flow Yes
FlowWidth(Total XS) Width of Total Cross-section flow Yes Yes
Hydr. Radius(LOB) Hydraulic Radius of Left Overbank flow Yes
Hydr. Radius(MCH) Hydraulic Radius of Main Channel flow Yes
Hydr. Radius(ROB) Hydraulic Radius of Right Overbank flow Yes
Hydr. Radius(Total XS) Hydraulic Radius of Total Cross-section flow Yes Yes
Composite n(LOB) Composite Manning's n of Left Overbank Yes
Composite n(MCH) Composite Manning's n of Main Channel Yes
Composite n(ROB) Composite Manning's n of Right Overbank Yes
Composite n(Total XS) Composite Manning's n of Total Cross-section Yes Yes
Product of Average Depth and Velocity of Left Overbank flow
Product of Average Depth and Velocity of Main Channel flow
Product of Average Depth and Velocity of Right Overbank flow
Product of Average Depth and Velocity of Total Cross-section flow
Product of Maximum Depth and Velocity of Left Overbank flow
Product of Maximum Depth and Velocity of Main Channel flow
Product of Maximum Depth and Velocity of Right Overbank flow
Product of Maximum Depth and Velocity of Total Cross-section flow
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Class 6 Math Important Objective Chapter – 13 Symmetry - The Study Point
Class 6 Math Important Objective Chapter – 13 Symmetry
Here we are providing Class 6 Math Important Objective Chapter –13 Symmetry because its very important for Class 6 students as we all know that every board exam question has MCQs so that students
should practice these questions so that students can get good marks in board. Class 6 Math Important Objective Chapter – 13 Symmetry is an extremely important and its also a year in which students
learn the fundamentals of concepts that help them lay a solid foundation for their higher education. Here we are providing in Class 6 Math Important Objective Chapter – 13 Symmetry 80+ questions so
that students practice more and more. If you want class wise Notes Then Click Here
Class 6 Math Important Objective Chapter – 13 Symmetry
1. Which letters have no lines of symmetry J, A, H, N.
(a) J, A
(b) A, H
(c) H, N
(d) J, N
2. Letter ‘H’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
(a) Neither horizontal nor veritcal
(b) Both horizontal nor veritcal
(c) a horizontal mirror
(d) a vertical mirror
3. Which of the following letters of the English alphabet has a vertical line of symmetry?
(a) F
(b) T
(c) E
(d) G
4. How many lines of symmetry does the figure have?
(a) 1
(b) 2
(c) 3
(d) 4
5. An equilateral triangle has :
(a) 1 line of symmetry
(b) 2 lines of symmetry
(c) 3 lines of symmetry
(d) None of these
6. Which of the following alphabets has no line of symmetry?
(a) A
(b) B
(c) Q
(d) O
7. How many lines of symmetry an Indian flag have:
(a) 1
(b) 2
(c) 3
(d) 4
8. How many lines of symmetry does the figure have?
(a) 1
(b) 2
(c) 3
(d) 4.
9. How many lines of symmetry of a rectangle has :
(a) 1
(b) 2
(c) 3
(d) 4
10. How many lines of symmetries are there in a square?
(a) 1
(b) 2
(c) 4
(d) 3
11. Which of the following alphabets has many lines of symmetry?
(a) A
(b) O
(c) Q
(d) B
12. How many lines of symmetry does the figure have?
(a) 1
(b) 2
(c) 3
(d) 4
13. How many lines of symmetries are there in rectangle?
(a) 4
(b) 0
(c) 1
(d) None of these
14. In a Δ ABC, AB = AC and AD⊥BC, BE⊥ AC and CF ⊥ AB. Then about which of the following is the triangle symmetrical?
(a) AD
(b) BE
(c) CF
(d) AC
15. How many lines of symmetry does the figure have?
(a) 1
(b) 2
(c) 3
(d) 4
16. How many lines of symmetries are there in rectangle?
(a) 4
(b) 0
(c) 1
(d) None of these
17. A rhombus is symmetrical about
(a) the line joining the midpoints of its adjacent sides.
(b) each of its diagonals.
(c) perpendicular bisector of each of its sides.
(d) its sides.
18. How many lines of symmetry does the figure have ?
(a) 1
(b) 2
(c) 3
(d) 4
19. Which of the following alphabets has many lines of symmetry?
(a) A
(b) O
(c) Q
(d) B
20. A circle has :
(a) one line of symmetry
(b) 2 lines of symmetry
(c) 3 lines of symmetry
(d) infinite number of lines of symmetry
21. How many lines of symmetry does the figure have ?
(a) 1
(b) 2
(c) 3
(d) no line of symmetry
21. Which of the following has 5 lines of symmetry?
(a) A circle.
(b) A regular pentagon.
(c) A triangle.
(d) A quadrilateral.
22. Letter ‘E’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
(a) a horizontal mirror
(b) a vertical mirror
(c) both
(d) None of these
23. How many lines of symmetry does the figure have?
(a) 1
(b) 2
(c) 3
(d) Countless.
24. Which of the following has 5 lines of symmetry?
(a) A circle.
(b) A regular pentagon.
(c) A triangle.
(d) A quadrilateral.
25. How many lines of symmetry does the figure have?
(a) 0
(b) 1
(c) 2
(d) countless
26. How many lines of symmetry does a circle have?
(a) One
(b) Two
(c) Three
(d) Many
27. How many lines of symmetry does the figure have?
(a) 1
(b) 2
(c) 3
(d) 4
28. How many lines of symmetry of a parallelogram has :
(a) 1
(b) 2
(c) Zero
(d) None of these
29. A rhombus is symmetrical about
(a) the line joining the midpoints of its adjacent sides.
(b) each of its diagonals.
(c) perpendicular bisector of each of its sides.
(d) its sides.
30. How many lines of symmetry does a regular hexagon have?
(a) 1
(b) 3
(c) 4
(d) 6
31. How many lines of symmetry of a human face has :
(a) 1
(b) 2
(c) 3
(d) 4
32. A parallelogram has ______ lines of symmetry:
(a) 0
(b) 1
(c) 2
(d) 3
33. Which of the following letters has horizontal line of symmetry?
(a) C
(b) A
(c) J
(d) L.
34. Which of the following alphabets has no line of symmetry?
(a) A
(b) B
(c) Q
(d) O
35. How many lines of symmetries are there in an isosceles triangle?
(a) 2
(b) 1
(c) 3
(d) None of these
36. Which of the following letters has horizontal line of symmetry?
(a) Z
(b) V
(c) U
(d) E.
37. Letter ‘B’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
(a) a horizontal mirror
(b) a vertical mirror
(c) both
(d) None of these
38. Letter ‘M’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
(a) a vertical mirror
(b) a horizontal mirror
(c) both
(d) None of these
39. Which of the following letters has horizontal line of symmetry?
(a) S
(b) W
(c) D
(d) Y.
40. Which of the following letters have reflection line of symmetry about vertical mirror?
(a) C
(b) B
(c) V
(d) Q
41. The mirror image of ‘W’, when the mirror is placed vertically:
(a) U
(b) M
(c) V
(d) W
42. Which of the following letters has vertical line of symmetry?
(a) R
(b) C
(c) B
(d) T.
43. How many lines of symmetries are there in a rhombus?
(a) 1
(b) 4
(c) 3
(d) 2
44. Which of the following letters of the English alphabet has a vertical line of symmetry?
(a) F
(b) T
(c) E
(d) G
45. Letter D has :
(a) 1 line of symmetry
(b) 2 lines of symmetry
(c) 3 lines of symmetry
(d) None of these
46. Which of the following letters has vertical line of symmetry?
(a) N
(b) K
(c) B
(d) M.
47. The mirror image of ‘W’, when the mirror is placed vertically:
(a) U
(b) M
(c) V
(d) W
48. Letter ‘G’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
(a) a horizontal mirrorboth
(b) a vertical mirror
(c) both
(d) Neither horizontal nor veritcal
49. Which of the following letters has vertical line of symmetry?
(a) J
(b) D
(c) E
(d) O.
50. How many lines of symmetries are there in a square?
(a) 1
(b) 2
(c) 4
(d) 3
51. Letter ‘D’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
(a) a vertical mirror
(b) both
(c) a horizontal mirror
(d) None of these
52. Which of the following letters has no line of symmetry?
(a) P
(b) O
(c) H
(d) X.
53. How many lines of symmetry does a rectangle have?
(a) One
(b) Two
(c) Three
(d) Many
54. A scalene triangle has the number of symmetry :
(a) 1
(b) 2
(c) 3
(d) No lines of symmetry
55. Which of the following letters has no line of symmetry?
(a) O
(b) X
(c) I
(d) Q
56. Which of the following letters have reflection line of symmetry about vertical mirror?
(a) C
(b) B
(c) V
(d) Q
57. How many lines of symmetry of 9 scissors has :
(a) 1
(b) zero
(c) 2
(d) 4
58. An equilateral triangle has:
(a) 3 lines of symmetry
(b) 2 lines of symmetry
(c) 1 line of symmetry
(d) no lines of symmetry
59. How many lines of symmetries are there in a rhombus?
(a) 1
(b) 4
(c) 3
(d) 2
60. A regular pentagon has :
(a) 1 line of symmetry
(b) 3 lines of symmetry
(c) 5 lines of symmetry
(d) None of these
61. A scalene triangle has:
(a) 3 lines of symmetry
(b) 2 lines of symmetry
(c) 1 line of symmetry
(d) no lines of symmetry
62. A parallelogram has ______ lines of symmetry:
(a) 0
(b) 1
(c) 2
(d) 3
63. The order of the rotational symmetry of the parallelogram about the centre is
(a) 1
(b) 0
(c) 3
(d) 2
64. The letter D has line of symmetry.
(a) 1
(b) 2
(c) 3
(d) 4
65. Letter ‘B’ of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about.
(a) a horizontal mirror
(b) a vertical mirror
(c) both
(d) None of these
66. In a Δ ABC, AB = AC and AD⊥BC, BE⊥ AC and CF ⊥ AB. Then about which of the following is the triangle symmetrical?
(a) AD
(b) BE
(c) CF
(d) AC
67. An irregular shape can have lines of symmetry.
(a) 0
(b) 1
(c) 2
(d) 3
68. How many lines of symmetries are there in an isosceles triangle?
(a) 2
(b) 1
(c) 3
(d) None of these
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Advent Of Code - Day 11: "Seating System" - Developer Sam
And then, there was Day 11. We are supposed to look for a stabilized layout of seat occupation when all of the people follow certain rules, depending on how many seats around them are already
This is Conway’s Game of Life with a little addition: Tiles where no seat exists.
Since I wanted to try out the Oracle MODEL clause for a while, I thought it couldn’t hurt to do this challenge with the special toolkit it provides. And it even worked pretty well, using the small
input of the example.
Another benefit was that I could use Kim Berg Hansens awesome book “Practical Oracle SQL” which covers exactly that Game of Life example with the MODEL clause.
In hope to improve the performance overhead, I started with normalizing the data to get a vertical set of row_id, col_id data:
create table aoc_day11_normalized as
base_data as (
rownum row_id,
column_value line
from table (
cols as (
level col_id
from dual
connect by level <= (select length(line) from base_data where row_id = 1)
column_data as (
substr(line, cols.col_id, 1) seat
from base_data
cross join cols
select * from column_data;
With that, I tried the approach with MODEL, first even without caring for change detection (we want to stop iterating once no changes occur):
column_data as (
from aoc_day11_normalized
modeled_data as (
from column_data
dimension by (
0 generation,
measures (
0 as sum_occupied,
0 as nb_occupied
ignore nav
rules upsert all iterate(1) (
sum_occupied[iteration_number, any, any] =
case seat
when 'L' then 0
when '#' then 1
else null
generation = iteration_number,
row_id between cv()-1 and cv()+1,
col_id between cv()-1 and cv()+1
nb_occupied[iteration_number, any, any] =
sum_occupied[iteration_number, cv(), cv()]
- case seat[iteration_number, cv(), cv()] when '#' then 1 else 0 end
seat[iteration_number+1, any, any] =
when seat[iteration_number, cv(), cv()] = '.' then
when seat[iteration_number, cv(), cv()] = 'L'
and nb_occupied[iteration_number, cv(), cv()] = 0 then
when seat[iteration_number, cv(), cv()] = '#'
and nb_occupied[iteration_number, cv(), cv()] >= 4 then
else seat[iteration_number, cv(), cv()]
display as (
listagg(seat) within group(order by col_id) seats,
sum(case seat when '#' then 1 else 0 end) occupied_seats
from modeled_data
group by generation, rollup(row_id)
generation, row_id, col_id, seat
from modeled_data
That worked pretty well with the tiny input-data of the example – but one iteration alone on my real input (92×92 cells) took half a minute – and it got worse the more iterations I added.
After putting some serious but fruitless effort into that solution, I did what a knowledge worker does if they hit a problem they cannot solve: I asked for help and contacted Kim as an expert who
even wrote about that MODEL clause.
As he is an incredibly nice guy, he answered immediately and even came up with a blog post, pointing out that aggregate functions seem to significantly hurt the performance.
So my final solution to Part 1 runs now in about 2 minutes, needing 96 iterations:
column_data as (
case seat
when '#' then 1
else 0
end occupied
from aoc_day11_normalized
--where row_id <= 10
modeled_data as (
from column_data
dimension by (
0 generation,
measures (
0 as sum_occupied,
0 as nb_occupied,
0 as changes,
1 as sum_changes
ignore nav
rules upsert all iterate(200) until (iteration_number > 1 and sum_changes[iteration_number, 1, 1] <= 0) (
occupied[iteration_number, any, any] =
case seat[iteration_number, cv(), cv()]
when '#' then 1
else 0
end ,
nb_occupied[iteration_number, any, any] =
occupied[iteration_number, cv()-1, cv()-1]
+ occupied[iteration_number, cv()-1, cv()]
+ occupied[iteration_number, cv()-1, cv()+1]
+ occupied[iteration_number, cv(), cv()-1]
+ occupied[iteration_number, cv(), cv()+1]
+ occupied[iteration_number, cv()+1, cv()-1]
+ occupied[iteration_number, cv()+1, cv()]
+ occupied[iteration_number, cv()+1, cv()+1]
seat[iteration_number+1, any, any] =
when seat[iteration_number, cv(), cv()] = '.' then
when seat[iteration_number, cv(), cv()] = 'L'
and nb_occupied[iteration_number, cv(), cv()] = 0 then
when seat[iteration_number, cv(), cv()] = '#'
and nb_occupied[iteration_number, cv(), cv()] >= 4 then
else seat[iteration_number, cv(), cv()]
changes[iteration_number+1, any, any] =
case when seat[iteration_number, cv(), cv()] != seat[iteration_number+1, cv(), cv()] then 1
else 0 end,
sum_changes[iteration_number+1,1,1] = sum(changes)[iteration_number+1, any, any]
display as (
sum(changes) changes,
sum(case seat when '#' then 1 else 0 end) occupied_seats
from modeled_data
group by generation, rollup(row_id)
from display
where row_id is null
order by generation desc
fetch first row only;
Let’s be honest – 2 minutes are still not great. But it’s bearable. And it’s completely done in SQL.
For part 2, however, I decided to use PL/SQL. Algorithms like these is where SQL hits its limit – it’s just not the right tool to solve it, given that a solution in a procedural language is not that
First, I re-coded part 1, which was straight forward. To hold the data I used a varray(100) of varrays(100) of integer, basically a 2-dimensional array.
The main logic is in the functions iterate_cell_part1 and especially is_occupied:
function is_occupied( y integer, x integer, i_array in t_rows ) return integer as
if y <= 0 or y > i_array.count or x <= 0 or x > i_array(y).count then
return 0;
end if;
return case i_array(y)(x) when '#' then 1 else 0 end;
function iterate_cell_part1( y integer, x integer, i_array in t_rows ) return char as
l_neighbours integer;
if i_array(y)(x) = '.' then
return '.';
end if;
l_neighbours :=
is_occupied(y-1, x-1, i_array)
+ is_occupied(y-1, x , i_array)
+ is_occupied(y-1, x+1, i_array)
+ is_occupied(y , x-1, i_array)
+ is_occupied(y , x+1, i_array)
+ is_occupied(y+1, x-1, i_array)
+ is_occupied(y+1, x , i_array)
+ is_occupied(y+1, x+1, i_array);
if i_array(y)(x) = 'L' and l_neighbours = 0 then
return '#';
elsif i_array(y)(x) = '#' and l_neighbours >= 4 then
return 'L';
return i_array(y)(x);
end if;
To solve part 2, all I had to do was to change is_occupied to sees_occupied – the rest is pretty much the same:
function sees_occupied( y integer, x integer, ystep integer, xstep integer, i_array in t_rows )
return integer as
l_nextY integer := y + ystep;
l_nextX integer := x + xstep;
if l_nextY <= 0 or l_nextY > i_array.count or l_nextX <= 0 or l_nextX > i_array(y).count then
return 0;
end if;
return case i_array(l_nextY)(l_nextX)
when '.' then sees_occupied(l_nextY, l_nextX, ystep, xstep, i_array)
when '#' then 1
else 0 end;
The nice thing: I got my result in 7 seconds for part 1 and 13 seconds for part 2.
You can see the whole solution my github-repository.
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Paper summary: The Epistemic Challenge to Longtermism (Christian Tarsney) — EA Forum
Note: The Global Priorities Institute (GPI) has started to create summaries of some working papers by GPI researchers with the aim to make our research more accessible to people outside of academic
philosophy (e.g. interested people in the effective altruism community). We welcome any feedback on the usefulness of these summaries.
Summary: The Epistemic Challenge to Longtermism
This is a summary of the GPI Working Paper "The epistemic challenge to longtermism" by Christian Tarsney. The summary was written by Elliott Thornley.
According to longtermism, what we should do mainly depends on how our actions might affect the long-term future. This claim faces a challenge: the course of the long-term future is difficult to
predict, and the effects of our actions on the long-term future might be so unpredictable as to make longtermism false. In “The epistemic challenge to longtermism”, Christian Tarsney evaluates one
version of this epistemic challenge and comes to a mixed conclusion. On some plausible worldviews, longtermism stands up to the epistemic challenge. On others, longtermism’s status depends on whether
we should take certain high-stakes, long-shot gambles.
Tarsney begins by assuming expectational utilitarianism: roughly, the view that we should assign precise probabilities to all decision-relevant possibilities, value possible futures in line with
their total welfare, and maximise expected value. This assumption sets aside ethical challenges to longtermism and focuses the discussion on the epistemic challenge.
Persistent-difference strategies
Tarsney outlines one broad class of strategies for improving the long-term future: persistent-difference strategies. These strategies aim to put the world into some valuable state S when it would
otherwise have been in some less valuable state ¬S, in the hope that this difference will persist for a long time. Epistemic persistence skepticism is the view that identifying interventions likely
to make a persistent difference is prohibitively difficult — so difficult that the actions with the greatest expected value do most of their good in the near-term. It is this version of the epistemic
challenge that Tarsney focuses on in this paper.
To assess the truth of epistemic persistence skepticism, Tarsney compares the expected value of a neartermist benchmark intervention N to the expected value of a longtermist intervention L. In his
example, N is spending $1 million on public health programmes in the developing world, leading to 10,000 extra quality-adjusted life years in expectation. L is spending $1 million on
pandemic-prevention research, with the aim of preventing an existential catastrophe and thereby making a persistent difference.
Exogenous nullifying events
Persistent-difference strategies are threatened by what Tarsney calls exogenous nullifying events (ENEs), which come in two types. Negative ENEs are far-future events that put the world into the less
valuable state ¬S. In the context of the longtermist intervention L, in which the valuable target state S is the existence of an intelligent civilization in the accessible universe, negative ENEs are
existential catastrophes that might befall such a civilization. Examples include self-destructive wars, lethal pathogens, and vacuum decay. Positive ENEs, on the other hand, are far-future events
that put the world into the more valuable state S. In the context of L, these are events that give rise to an intelligent civilization in the accessible universe where none existed previously. This
might happen via evolution, or via the arrival of a civilization from outside the accessible universe. What unites negative and positive ENEs is that they both nullify the effects of interventions
intended to make a persistent difference. Once the first ENE has occurred, the state of the world no longer depends on the state that our intervention put it in. Therefore, our intervention stops
accruing value at that point.
Tarsney assumes that the annual probability r of ENEs is constant in the far future, defined as more than a thousand years from now. The assumption is thus compatible with the time of perils
hypothesis, according to which the risk of existential catastrophe is likely to decline in the near future. Tarsney makes the assumption of constant r partly for simplicity, but it is also in line
with his policy of making empirical assumptions that err towards being unfavourable to longtermism. Other such assumptions concern the tractability of reducing existential risk, the speed of
interstellar travel, and the potential number and quality of future lives. Making these conservative assumptions lets us see how longtermism fares against the strongest available version of the
epistemic challenge.
Models to assess epistemic persistence scepticism
To compare the longtermist intervention L to the neartermist benchmark intervention N, Tarsney constructs two models: the cubic growth model and the steady state model. The characteristic feature of
the cubic growth model is its assumption that humanity will eventually begin to settle other star systems, so that the potential value of human-originating civilization grows as a cubic function of
time. The steady-state model, by contrast, assumes that humanity will remain Earth-bound and eventually reach a state of zero growth.
The headline result of the cubic growth model is that the longtermist intervention L has greater expected value than the neartermist benchmark intervention N just so long as r is less than
approximately 0.000135 (a little over one-in-ten-thousand) per year. Since, in Tarsney’s estimation, this probability is towards the higher end of plausible values of r, the cubic growth model
suggests (but does not conclusively establish) that longtermism stands up to the epistemic challenge. If we make our assumptions about tractability and the potential size of the future population a
little less conservative, the case for choosing L over N becomes much more robust.
The headline result of the steady state model is less favourable to longtermism. The expected value of L exceeds the expected value of N only when r is less than approximately 0.000000012 (a little
over one-in-a-hundred-million) per year, and it seems likely that an Earth-bound civilization would face risks of negative ENEs that push r over this threshold. Relaxing the model’s conservative
assumptions, however, makes longtermism more plausible. If L would reduce near-term existential risk by at least one-in-ten-billion and any far-future steady-state civilization would support at least
a hundred times as much value as Earth does today, then r need only fall below about 0.006 (six-in-one-thousand) to push the expected value of L above N.
The case for longtermism is also strengthened once we account for uncertainty, both about the values of various parameters and about which model to adopt. Consider an example. Suppose that we assign
a probability of at least one-in-a-thousand to the cubic growth model. Suppose also that we assign probabilities – conditional on the cubic growth model – of at least one-in-a-thousand to values of r
no higher than 0.000001 per year, and at least one-in-a-million to a ‘Dyson spheres’ scenario in which the average star supports at least 10^25 lives at a time. In that case, the expected value of
the longtermist intervention L is over a hundred billion times the expected value of the neartermist benchmark intervention N. It is worth noting, however, that in this case L’s greater expected
value is driven by possibly minuscule probabilities of astronomical payoffs. Many people suspect that expected value theory goes wrong when its verdicts hinge on these so-called Pascalian
probabilities (Bostrom 2009, Monton 2019, Russell 2021), so perhaps we should be wary of taking the above calculation as a vindication of longtermism.
Tarsney concludes that the epistemic challenge to longtermism is serious but not fatal. If we are steadfast in our commitment to expected value theory, longtermism overcomes the epistemic challenge.
If we are wary of relying on Pascalian probabilities, the result is less clear.
Bostrom, N. (2009). Pascal’s mugging. Analysis 69 (3), 443–445.
Monton, B. (2019). How to avoid maximizing expected utility. Philosophers’ Imprint 19 (18), 1–25.
Russell, J. S. (2021). On two arguments for fanaticism. Global Priorities Institute Working Paper Series. GPI Working Paper No. 17-2021.
Will Aldred 12
See also Michael Aird's comments on this Tarsney (2020) paper. His main points are:
• 'Tarsney's model updates me towards thinking reducing non-extinction existential risks should be a little less of a priority than I previously thought.' (link to full comment)
• 'Tarsney seems to me to understate the likelihood that accounting for non-human animals would substantially affect the case for longtermism.' (link)
• 'The paper ignores 2 factors that could strengthen the case for longtermism - namely, possible increases in how efficiently resources are used and in what extremes of experiences can be reached.'
• 'Tarsney writes "resources committed at earlier time should have greater impact, all else being equal". I think that this is misleading and an oversimplification. See Crucial questions about
optimal timing of work and donations and other posts tagged Timing of Philanthropy.' (link)
• 'I think it'd be interesting to run a sensitivity analysis on Tarsney's model(s), and to think about the value of information we'd get from further investigation of:
□ how likely the future is to resemble Tarsney's cubic growth model vs his steady model
□ whether there are other models that are substantially likely, whether the model structures should be changed
□ what the most reasonable distribution for each parameter is.' (link)
EJT 1
Thanks! This is valuable feedback.
By 'persistent difference', Tarsney doesn't mean a difference that persists forever. He just means a difference that persists for a long time in expectation: long enough to make the expected value of
the longtermist intervention greater than the expected value of the neartermist benchmark intervention.
Perhaps you want to know why we should think that we can make this kind of persistent difference. I can talk a little about that in another comment if so.
EJT 5
All good points, but Tarsney's argument doesn't depend on the assumption that longtermist interventions cannot accidentally increase x-risk. It just depends on the assumption that there's some way
that we could spend $1 million that would increase the epistemic probability that humanity survives the next thousand years by at least 2x10^-14.
More posts like this
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MichaelPlant 10
On the summary: I'd have found this summary more useful if it had made the ideas in the paper simpler, so it was easier to get an intuitive grasp on what was going on. This summary has made the paper
shorter, but (as far as I can recall) mostly by compressing the complexity, rather than lessening it!
On the paper itself: I still find Tarsney's argument hard to make sense of (in addition to the above, I've read the full paper itself a couple of times).
AFAIT, the set up is that the longtermist wants to show that there are things we can do now that will continually make the future better than it would have been ('persistent-difference strategies').
However, Tarnsey takes the challenge to be that there are things that might happen that would stop these positive states happening ('exogenously nullifying events'). And what does all the work is
that if the human population expands really fast ('cubic growth model'), that is, because it's fled to the stars, but the negative events should happen at a constant rate, then longtermism looks
I think what bothers me about the above is this: why think that we could ever identify and do something that would, in expectation, make a persistent positive difference, i.e. a difference for ever
and ever and ever? Isn't Tarsney assuming the existence of the thing he seeks to prove, ie 'begging the question'? I think the sceptic is entitled to respond with a puzzled frown - or an incredulous
stare - about whether we can really expect to knowingly change the whole trajectory of the future - that, after all, presumably is the epistemic challenge. That challenge seems unmet.
I've perhaps misunderstood something. Happy to be corrected!
Ofer 6
If I understand the proposed model correctly (I haven't read thoroughly, so apologies if not): The model basically assumes that "longtermist interventions" cannot cause accidental harm. That is, it
assumes that if a "longtermist intervention" is carried out, the worst-case scenario is that the intervention will end up being neutral (e.g. due to an "exogenous nullifying event") and thus
resources were wasted.
But this means assuming away the following major part of complex cluelessness: due to an abundance of crucial considerations, it is usually extremely hard to judge whether an intervention that is
related to anthropogenic x-risks or meta-EA is net-positive or net-negative. For example, such an intervention may cause accidental harm due to:
1. Drawing attention to dangerous information (e.g. certain exciting approaches for AGI development / virology experimentation).
□ If a researcher believes they came up with an impressive insight, they will probably be biased towards publishing it, even if it may draw attention to potentially dangerous information. Their
career capital, future compensation and status may be on the line.
□ Alexander Berger (co-CEO of OpenPhil) said in an interview :
I think if you have the opposite perspective and think we live in a really vulnerable world — maybe an offense-biased world where it’s much easier to do great harm than to protect against
it — I think that increasing attention to anthropogenic risks could be really dangerous in that world. Because I think not very many people, as we discussed, go around thinking about the
vast future.
If one in every 1,000 people who go around thinking about the vast future decide, “Wow, I would really hate for there to be a vast future; I would like to end it,” and if it’s just 1,000
times easier to end it than to stop it from being ended, that could be a really, really dangerous recipe where again, everybody’s well intentioned, we’re raising attention to these risks
that we should reduce, but the increasing salience of it could have been net negative.
2. "Patching" a problem and preventing a non-catastrophic, highly-visible outcome that would have caused an astronomically beneficial "immune response".
□ Nick Bostrom said in a talk ("lightly edited for readability"):
Small and medium scale catastrophe prevention? Also looks good. So global catastrophic risks falling short of existential risk. Again, very difficult to know the sign of that. Here we are
bracketing leverage at all, even just knowing whether we would want more or less, if we could get it for free, it’s non-obvious. On the one hand, small-scale catastrophes might create an
immune response that makes us better, puts in place better safeguards, and stuff like that, that could protect us from the big stuff. If we’re thinking about medium-scale catastrophes
that could cause civilizational collapse, large by ordinary standards but only medium-scale in comparison to existential catastrophes, which are large in this context, again, it is not
totally obvious what the sign of that is: there’s a lot more work to be done to try to figure that out. If recovery looks very likely, you might then have guesses as to whether the
recovered civilization would be more likely to avoid existential catastrophe having gone through this experience or not.
3. Causing decision makers to have a false sense of security.
□ For example, perhaps it's not feasible to solve AI alignment in a competitive way without strong coordination, etcetera. But researchers are biased towards saying good things about their
field, their colleagues and their (potential) employers.
4. Causing progress in AI capabilities to accelerate in a certain way.
5. Causing the competition dynamics among AI labs / states to intensify.
6. Decreasing the EV of the EA community by exacerbating bad incentives and conflicts of interest, and by reducing coordination.
□ For example, by creating impact markets.
7. Causing accidental harm via outreach campaigns or regulation advocacy (e.g. by causing people to get a bad first impression of something important).
8. Causing a catastrophic leak from a virology lab, or an analogous catastrophe involving an AI lab.
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Conversion of Pure Recurring Decimal into Vulgar Fraction | Worked-out Examples
Conversion of Pure Recurring Decimal into Vulgar Fraction
Follow the steps for the conversion of pure recurring decimal into vulgar fraction:
(i) First write the decimal form by removing the bar from the top and put it equal to n (any variable).
(ii) Then write the repeating digits at least twice.
(iii) Now find the number of digits having bars on their heads.
● If the repeating decimal has 1 place repetition, then multiply both sides by 10.
● If the repeating decimal has 2 place repetitions, then multiply both sides by 100.
● If the repeating decimal has 3 place repetitions, then multiply both sides by 1000 and so on.
(iv) Then subtract the number obtained in step (i) from the number obtained in step (ii).
(v) Then divide both the sides of the equation by the coefficient of n.
(vi) Therefore, we get the required vulgar fraction in the lowest form.
Worked-out examples for the conversion of pure recurring decimal into vulgar fraction:
Express 0.
as a vulgar fraction.
Let n = 0.
n = 0.444 ----------- (i)
Since, one digit is repeated after the decimal point, so we multiply both sides by 10.
Therefore, 10n = 4.44 ----------- (ii)
Subtracting (i) from (ii) we get;
10n - n = 4.44 - 0.44
9n = 4
n = 4/9 [dividing both the sides of the equation by 9]
Therefore, the vulgar fraction = 4/9
Express 0.
as a vulgar fraction.
Let n = 0.
n = 0.3838 ----------------- (i)
Since, two digits are repeated after the decimal point, so we multiply both sides by 100.
Therefore, 100n = 38.38 ----------------- (ii)
Subtracting (i) from (ii) we get;
100n - n = 38.38 - 0.38
99n = 38
n = 38/99
Therefore, the vulgar fraction = 38/99
Express 0.
as a vulgar fraction.
Let n = 0.
n = 0.532532 ----------------- (i)
Since, three digits are repeated after the decimal point, so we multiply both sides by 1000.
Therefore, 1000n = 532.532 ----------------- (ii)
Subtracting (i) from (ii) we get;
1000n - n = 532.532 - 0.532
999n = 532
n = 532/999
Therefore, the vulgar fraction = 532/999
Shortcut method for solving the problems on conversion of pure recurring decimal into vulgar fraction:
Write the recurring digits only once in the numerator and write as many nines in the denominator as is the number of digits repeated.
For example;
(a) 0.5
Here numerator is the period (5) and the denominator is 9 because there is one digit in the period.
= 5/9
(b) 0.45
Numerator = period = 45
Denominator = as many nines as the number of digits in the denominator
= 45/99
● Related Concept
● Decimals
● Conversion of Unlike Decimals to Like Decimals
● Decimal and Fractional Expansion
● Converting Decimals to Fractions
● Converting Fractions to Decimals
● H.C.F. and L.C.M. of Decimals
● Repeating or Recurring Decimal
● BODMAS/PEMDAS Rules - Involving Decimals
● PEMDAS Rules - Involving Integers
● PEMDAS Rules - Involving Decimals
● BODMAS Rules - Involving Integers
● Conversion of Pure Recurring Decimal into Vulgar Fraction
● Conversion of Mixed Recurring Decimals into Vulgar Fractions
● Rounding Decimals to the Nearest Whole Number
● Rounding Decimals to the Nearest Tenths
● Rounding Decimals to the Nearest Hundredths
● Simplify Decimals Involving Addition and Subtraction Decimals
● Multiplying Decimal by a Decimal Number
● Multiplying Decimal by a Whole Number
● Dividing Decimal by a Whole Number
● Dividing Decimal by a Decimal Number
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full replicate study for drug not highly variable - EMA perspective
full replicate study for drug not highly variable - EMA perspective [Regulatives / Guidelines]
Post reply
Background information:
Ankit Parikh We are working on designing a multiple-dose steady state Pharmacokinetic (PK) Bioequivalence (BE) study of an Immediate Release (IR) Solid Oral Dosage form on cancer patients for
☆ EMA submission. We are exploring alternate study design due to the inherent challenges related to the recruitment and retention of cancer patients in PK BE study and have following
2024-07-11 06:40 Question: To address the challenges related to recruitment of cancer patients in PK BE study, we want to plan a full replicate design study. Will the EMA agency accept this approach
(115 d 17:06 ago) even if we are not suspecting the drug to be highly variable?
(edited on As per EMA guidance, “If an applicant suspects that a drug product can be considered as highly variable in its rate and/or extent of absorption, a replicate cross-over design study
2024-07-11 06:59) can be carried out.”
Posting: # 24067
Views: 1,346 As per USFDA guidance “A replicate crossover study design (either partial or fully replicate) is appropriate for drugs whether the reference product is a highly variable drug or
not. A replicate design can have the advantage of using fewer subjects compared to a non-replicate design, although each subject in a replicate design study would receive more
We understand that widening criteria will be applied only to Cmax if we observe Swr > 0.294 or else conventional BE criteria will be applied. The reason we want to take this
approach is to reduce the number of patients to be randomized in the trial.
EMA: Full replicate study
Post reply
Hi Ankit,
❝ Question: To address the challenges related to recruitment of cancer patients in PK BE study, we want to plan a full replicate design study. Will the EMA agency accept this
approach even if we are not suspecting the drug to be highly variable?
❝ As per EMA guidance, “If an applicant suspects that a drug product can be considered as highly variable in its rate and/or extent of absorption, a replicate cross-over design
study can be carried out.”
Let’s see what is stated in
4.1.1 Study design
Standard design
If two formulations are compared, a randomised, two-period, two-sequence single dose crossover design is recommended.
does not mean that is is mandatory.
Alternative designs
Under certain circumstances, provided the study design and the statistical analyses are scientifically sound, alternative well-established designs could be considered such as […]
replicate designs e.g. for substances with highly variable pharmacokinetic characteristics.
Replicate designs for HVDs are given as one of the
. State the recruitment issues as a justification in the protocol. Note that
replicate study can be evaluated for conventional Average Bioequivalence (ABE) as well (see also
this article
). From a purely statistical perspective, replicate designs are preferable to a simple 2×2×2 crossover and thus, there should be not regulatory concerns.
❝ As per USFDA guidance “A replicate crossover study design (either partial or fully replicate) is appropriate for drugs whether the reference product is a highly variable drug or
not. A replicate design can have the advantage of using fewer subjects compared to a non-replicate design, although each subject in a replicate design study would receive more
Correct – makes sense. Power – and thus, the sample size – depends on the total number of treatments. Therefore, a 4-period full replicate design requires about half as many
subjects as a 2×2×2 crossover. An example with the
CV <- 0.25 # assumed within-subject CV
Helmut theta0 <- 0.95 # assumed T/R-ratio
★★★ target <- 0.80 # target (desired) power
design <- c("2x2x2", "2x2x4", "2x3x3") # guess…
n.per <- as.integer(substr(design, 5, 5))
Vienna, Austria, expl <- data.frame(design = design, n = NA, treatments = NA)
2024-07-15 11:13 for (j in seq_along(design)) {
(111 d 12:33 ago) expl[j, 2] <- sampleN.TOST(CV = CV, theta0 = theta0, design = design[j],
targetpower = target, print = FALSE)[["Sample size"]]
@ Ankit Parikh expl[j, 3] <- sprintf("%.0f ", expl[j, 2] * n.per[j])
Posting: # 24074 }
Views: 1,011 fmt <- paste0("Sample sizes and number of treatments for assumed CV = %.3g ",
"and T/R-ratio = %.3g,\npowered for at least %.0f%% ",
"and evaluation for Average Bioequivalence (ABE).\n")
cat(sprintf(fmt, CV, theta0, 100 * target)); print(expl, row.names = FALSE)
Sample sizes and number of treatments for assumed CV = 0.25 and T/R-ratio = 0.95,
powered for at least 80% and evaluation for Average Bioequivalence (ABE).
design n treatments
2x2x2 28 56
2x2x4 14 56
2x3x3 21 63
Keep the potentially larger number of dropouts in replicate designs in mind (see
this article
❝ We understand that widening criteria will be applied only to Cmax if we observe Swr > 0.294 or else conventional BE criteria will be applied.
Almost. For the EMA the decision is
based on \(\small{s_\text{wR}}\) (like for the FDA) but on \(\small{CV_\text{wR}}\), where for reference-scaling \(\small{s_\text{wR}=\sqrt{\log_e(CV_\text{wR}+1)}}\).
• If > 30%:
□ Reference-scaling by Average Bioequivalence with Expanding Limits (ABEL)
□ Upper cap of scaling at \(\small{CV_\text{wR}=50\%}\) (maximum expansion 69.84–143.19%)
□ 90% CI within expanded limits \(\small{\left\{L,U\right\}=\exp(\mp 0.76\cdot s_\text{wR})}\)
□ Point estimate within 80.00–125.00%
• If ≤ 30%:
□ ABE
□ 90% CI within 80.00–125.00%
However, in order to apply ABEL you have to take into account whether
“a wider difference in C[max] is considered clinically irrelevant based on a sound clinical justification can be assessed with a widened acceptance range”
. I have some doubts whether
can be provided for an anticancer drug.
Dif-tor heh smusma 🖖🏼 Довге життя Україна! []
Helmut Schütz
The quality of responses received is directly proportional to the quality of the question asked.
🚮 Science Quotes
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Solving the Robot Arm Controller Problem with Matrices
Robot Arm Controller Let us look at another example of a problem in which the matrix will be useful. The robotarm problem is a good choice, since we've already discussed it. Thus, should be a good
idea for us to discuss it together, since you could meet someproblems. Then we can examine how to solve it using matrices. OK, so, first let's reviewthe problem. You will see the problem now. The
first thing we're going to do is solve it. Then, after that,we're going to talk about how there's a matrix hidden in this problem. OK, so here it is. Here's a robot arm with two joints. Here's a
joint at the origin, and then there's a joint overhere. And in between them there's a rod of length L, and you can make L stretch. And thenthere's another joint over here. There's an angle theta
between them, and this has lengthof square root of 2, and at the end of it is xy, a fingertip of the robot. The first part of the problem was to find xy in terms of L and theta, and everyone did
fineon that. So here is your formula: x equals L plus square root 2 cosine of theta and y equalssquare root 2 sine. Now, if you assume that L equals 1 and theta equals pi over 4, then you can find
out thatxy is 2 comma 1. The question is, how can we adjust L and theta? So that xy is 2 comma 1.1. In otherwords, we'd like to lift the tip straight up a distance 0.1. As a matter of fact, a
question will come now, which is the most common question thatpeople have trouble with. So evidently, everybody recognized that we want to use linear approximation to solve thisproblem, since the
algebra would be too complicated to do it exactly. The question is:What function in what variables should we approximate with a linear function? One option is to approximate x as a function of L and
theta. Another is to approximate Land theta as a function of x and y. A third option is to approximate some other function f,which we have to figure out what it is could approximate that. Think of
that as a function ofx and y. Who should we think of as a function of who?
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Acceleration Sensors and Transducers in context of frequency displacement acceleration
12 Oct 2024
Title: Acceleration Sensors and Transducers: A Review of Frequency Displacement Acceleration Applications
Abstract: Acceleration sensors and transducers play a crucial role in various industrial, automotive, and aerospace applications where frequency displacement acceleration (FDA) is a key parameter.
This article provides an overview of the principles, types, and applications of acceleration sensors and transducers in the context of FDA.
Introduction: Frequency Displacement Acceleration (FDA) refers to the measurement of acceleration as a function of frequency. In many applications, such as vibration analysis, structural health
monitoring, and motion control, accurate measurement of FDA is essential. Acceleration sensors and transducers are used to measure FDA, which is typically represented by the following formula:
a(f) = ∫[∂²x/∂t²] dt
where a(f) is the acceleration as a function of frequency, x(t) is the displacement as a function of time, and t is time.
Types of Acceleration Sensors: There are several types of acceleration sensors, including:
1. Piezoelectric Accelerometers: These sensors use piezoelectric materials to generate an electric charge in response to acceleration. The output voltage is proportional to the acceleration.
2. Capacitive Accelerometers: These sensors use a capacitive sensing mechanism to measure changes in capacitance caused by acceleration.
3. Optical Accelerometers: These sensors use optical fibers or photodiodes to detect changes in light intensity caused by acceleration.
Transducers: Acceleration transducers are devices that convert the measured acceleration into a usable signal, such as voltage or current. Common types of acceleration transducers include:
1. Strain Gage Transducers: These transducers use strain gages to measure changes in resistance caused by acceleration.
2. Piezoelectric Transducers: These transducers use piezoelectric materials to generate an electric charge in response to acceleration.
Applications: Acceleration sensors and transducers have numerous applications, including:
1. Vibration Analysis: FDA is used to analyze the vibration characteristics of mechanical systems, such as engines, gearboxes, and bearings.
2. Structural Health Monitoring: Acceleration sensors are used to monitor the health of structures, such as bridges, buildings, and aircraft, by detecting changes in acceleration caused by damage or
3. Motion Control: FDA is used to control the motion of mechanical systems, such as robots, CNC machines, and autonomous vehicles.
Conclusion: Acceleration sensors and transducers play a vital role in various applications where frequency displacement acceleration is a key parameter. Understanding the principles and types of
these devices is essential for designing and implementing accurate measurement systems. Further research is needed to develop more advanced acceleration sensors and transducers that can accurately
measure FDA over a wide range of frequencies.
• [1] IEEE Transactions on Instrumentation and Measurement, “Frequency Displacement Acceleration Measurements Using Piezoelectric Sensors”
• [2] Journal of Sound and Vibration, “Vibration Analysis Using Frequency Displacement Acceleration”
• [3] Sensors and Actuators A, “Piezoelectric Accelerometers for Structural Health Monitoring”
Note: The references provided are fictional examples and do not represent actual publications.
Related articles for ‘frequency displacement acceleration ‘ :
Calculators for ‘frequency displacement acceleration ‘
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Student Publications
Journal Articles
• Panorkou, N., York, T., & Germia, E. (in press). Examining the “messiness” of transitions between related artifacts. Digital Experiences in Mathematics Education.
• Monahan, C., & Munakata, M. (in press). Chapter 13 in S. Chamberlin (Ed.), Mathematical Creativity: A developmental perspective.
• Greenstein, S., Jeannotte, D., & Pomponio, E. (forthcoming) Making as a Window into the Process of Becoming a Teacher: The Case of Moira. In Benken, B. (Ed.) Reflection on Past, Present and
Future: Paving the Way for the Future of Mathematics Teacher Education. (pp. TBD). Information Age Publishing, Inc.
• Greenstein, S. & Limbere, A.M. (2023). Parents’ and Caregivers’ Voices in Education Reform. The New Jersey Mathematics Teacher, 80(1), 13-15.
• Greenstein, S., Akuom, D., Pomponio, E., Fernández, E., Davidson, J., Jeannotte, D., & York, T. (2022) Vignettes of Research on the Promise of Mathematical Making in Teacher Preparation. In F.
Dilling & F. Pielsticker (Eds.) Learning Mathematics in the Context of 3D Printing: Springer Nature.
• Akuom, D., & Greenstein, S. (2022). The Nature of Prospective Mathematics Teachers’ Designed Manipulatives and their Potential as Anchors for Conceptual and Pedagogical Knowledge. Journal of
Research in Science, Mathematics and Technology Education, 5(SI), 109-125. Bronze Medal Award. DOI: https://doi.org/10.31756/jrsmte.115SI
• Limbere, A.M., Munakata, M., Klein, E.J., & Taylor, T. (2022) Exploring the tensions science teachers navigate as they enact their visions for science teaching: what their feedback can tell us.
International Journal of Science Education, DOI: 10.1080/09500693.2022.2105413
• Taylor, M., Klein, E. J., Trabona, K., & Munakata, M. (2022). Disrupting the Patriarchal Binary, 2nd edition. In N. Bond (Ed). The Power of Teacher Leaders: Their Roles, Influence, and Impact.
• Ward, J. K., DiNapoli, J., & Monahan, K. (2022). Instructional perseverance in early-childhood classrooms: Supporting children’s development of STEM reasoning in a social justice context.
Education Sciences, 12(3), 1-19. https://doi.org/10.3390/educsci12030159
• DiNapoli, J., Amenya, M., Van Den Einde, L., Delson, N., & Cowan, E. (2021). Simulating remote support for mathematical perseverance through a digital sketching application. Journal of Higher
Education Theory and Practice, 21(4), 41-52. https://doi.org/10.33423/jhetp.v21i4.4208
• Munakata, M., Vaidya, A., Monahan, C., & Krupa, E. (2021) Promoting creativity in general education mathematics courses, PRIMUS—Problems, Resources, and Issues in Mathematics Undergraduate
Studies, 31(1), 37-55. DOI: 10.1080/10511970.2019.1629515. Selected as Editors’ Pick for Most Downloaded in 2021.
• Panorkou, N. & Germia, E. (2020). Integrating math and science content through covariational reasoning: The case of gravity. Mathematical Thinking and Learning. https://doi.org/10.1080/
• Basu, D., Panorkou, N., Zhu, M., Lal, P., & Samanthula, B. K. (2020). Exploring the Mathematics of Gravity, Mathematics Teacher: Learning and Teaching PK-12 MTLT, 113(1), 39-46.
• Germia, E. & Panorkou, N. (2020). Using Scratch programming to explore coordinates. Mathematics Teacher: Learning and Teaching PreK-12 MTLT, 113(4), pp. 293-300.
• Monahan, C. Munakata, M. Vaidya, A. and Gandini, S. (2020). Inspiring mathematical creativity through juggling. Journal of Humanistic Mathematics, 10(2) 291-314. DOI: 10.5642/jhummath.202002.14.
• Monahan, C., Munakata,, M., & Vaidya, A. (2020). Engaging in probabilistic thinking through play. Mathematics Teacher: Learning and Teaching PK-12, 113(9), 18-23. https://doi.org/10.5951/
• Basu, D. & Panorkou, N. (2019). Integrating covariational reasoning and technology into the teaching and learning of the greenhouse effect. Journal of Mathematics Education,12(1), pp.6-23.
• Basu, D., & Greenstein, S. (2019). Cultivating a Space for Critical Mathematical Inquiry through Knowledge-Eliciting Mathematical Activity. Occasional Paper Series, 2019 (41). Retrieved from
• Monahan, C., Munakata, M. and Vaidya, A. (2019). Creativity as an emergent property of a complex educational system. Northeast Journal of Complex Systems, (Special Issue of the satellite
meeting of CCS2018 on Complexity and Education, Eds. Matthijs Koopmans, Hiroki Sayama, Dimitri Stamovlasis). DOI: 10.22191/nejcs/vol1/iss1/4
• Munakata, M., Vaidya, A., Monahan, C., & Krupa, E. (2019) Promoting creativity in general education mathematics courses, PRIMUS—Problems, Resources, and Issues in Mathematics Undergraduate
Studies. DOI: 10.1080/10511970.2019.1629515
• Taylor, M., Klein, E. J., Munakata, M., Trabona, K., Rahman, Z., & McManus, J. (2019). Professional development for teacher leaders: using activity theory to understand the complexities of
sustainable change. International Journal of Leadership in Education, 22(6), 685-705. DOI: 10.1080/13603124.2018.1492023
• Trabona, K., Taylor, M., Klein, E., Munakata, M. & Rahman, Z. (2019) Collaborative professional learning: cultivating science teacher leaders through vertical communities of practice.
Professional Development in Education, 45(3), 472-487, DOI: 10.1080/19415257.2019.1591482
• Krupa, E., Munakata, M., & Yu, K. (2019). Mathematics field day: Content-embedded play activities. Teaching Mathematics in the Middle School, 24(5), 296-299.
• Fernández, E., Weinstein, L., Seventko, J., Janakat, J., Rahman, Z., Basu, D., Bonaccorso, V., Yu, K., Casale, K., Kose, G., Prettypaul, R., (2018). Average rates: They’re about more than adding
and dividing! New Jersey Mathematics Teacher. 76(1), 11-18.
• Klein, E. J., Taylor, M., Munakata, M., Trabona, K., Rahman, Z., & McManus J. (2018). Navigating teacher leaders’ complex relationships using a distributed leadership framework. Teacher Education
Quarterly, 45(2), 89-112.
• Rahman, Z.G., Munakata, M., Klein, E.J., Taylor, M., & Trabona, K. (2018). Growing our own: Fostering teacher leadership in K-12 science teachers through school-university partnerships. In J.
Hunzicker (Ed.), Teacher Leadership in Professional Development Schools (pp. 235-253). Bingley, UK: Emerald Publishing.
• Taylor, M., Ayari, C., Kintish, R., Jedick, N., Lemley, J., Lormand, K., Tanis, J., & Weinstein, L. (2018). Using self-study to push binary boundaries and borders: Exploring gender and sexuality
in teacher education. In D. Garbett and A. Owens (Eds.), Pushing Boundaries & Crossing Borders (229-236).
• Vishnubhotla, M. & Munakata, M. (2018). The mathematics of loopy pictures frames. Teaching Mathematics in the Middle School. 23(4), 231-234. DOI: 10.5951/mathteacmiddscho.23.4.0231
• Fernández, E., McManus, J. and Platt, D. (2017). Extending Mathematical Practices to Online
• Teaching. Mathematics Teacher. 110(6), 432-438
• Leszczynski, E., Monahan, C., Munakata, M., & Vaidya, A. (2017) The Windwalker project: An open-ended approach. Journal of College Science Teaching. 46(6), 27-33.
• Paoletti, T., Monahan, C. & Vishnubhotla, M. (2017) Designing GeoGebra Applets to Maximize Student Engagement. Mathematics Teacher, 110 (8), 628-630.
• Vanderklein, D., Munakata, M. and McManus, J. (2016). Crossing boundaries in undergraduate ecology education. Journal of College Science Teaching. 45(3), 41-47.
• Chung, B. J., Platt, D., & Vaidya, A. (2016). The Mechanics of Clearance in a non-Newtonian Lubrication Layer. International Journal of Non-Linear Mechanics.
• Webel, C., Krupa, E. E., & McManus, J. (2016). Using representations of fraction multiplication. Teaching Children Mathematics, 22(6), 366-373.
• Krupa, E. E., Webel, C., & McManus, J. (2015). Undergraduate students’ knowledge of algebra: Evaluating the impact of computer-based and traditional learning environments. PRIMUS: Problems,
Resources, and Issues in Mathematics Undergraduate Studies, 25(1), 13-30. doi:10.1080/10511970.2014.897660
• Webel, C., Krupa, E. E., & McManus, J. (2015). Teachers’ evaluations and use of web-based curriculum resources to support their teaching of the Common Core State Standards for Mathematics. Middle
Grades Research Journal, 10(2), 49-64.
• Webel, C., Krupa, E. E., & McManus, J. (2015). Benny goes to college: Is the “math emporium” reinventing individually prescribed instruction? MathAMATYC Educator, 6(3), 4-13, 62.
• Webel, C., & Platt, D. (2015). The Role of Professional Obligations in Attempting to Change One’s Teaching Practices, Teaching and Teacher Education, 47, 204 – 217.
• Russo, M. F. (2015). Quantitative literacy and co-construction in a high school math course. Numeracy, 8(1), 1-23.
• Russo, M. F. (2015). Customizing a math course with your students. Mathematics Teacher, 109(5), 351-354.
• Leszczynski, E., Munakata, M., Evans, J., & Pizzigoni, F. (2014). Integrating mathematics and science: Ecology and Venn diagrams. Mathematics Teaching in the Middle School. 20(2) 90-96.
• McManus, J. (2014, December 28). Creativity brings life to math classrooms. Asbury Park Press, pp. AA1-AA2. [Guest editorial for the feature @ISSUE: How can math be made more interesting for
• McManus, J. (2014, December 28). McManus: Creativity brings life to math classrooms. Courier-Post Online. [Syndication of above Asbury Park Press editorial]
• Russo, M. F. (2014). Teaching sound waves and sinusoids using Audacity and the TI-Nspire. The Electronic Journal of Mathematics and Technology, 8(1), 222-227.
• Russo, M. F. (2014). Teaching sound waves and sinusoids using Audacity and the TI-Nspire. Research Journal of Mathematics and Technology, 3(1), 179-185.
• Russo, M. F. (2011). An investigation into Sierpinski’s gasket. The New Jersey Mathematics Teacher, 69(2), 11-15.
Peer-reviewed conference proceedings
• Provost, A., Lim, S., York, T., & Panorkou, N. (in press). Bridging Frequentist and Classical Probability Through Design, Proceedings of the forty-fourth annual meeting of the North American
Chapter of the International Group for the Psychology of Mathematics Education. Nashville, TN.
• Germia, E., York, T., & Panorkou, N. (in press). How Transitions Between Related Artifacts Support Students’ Covariational Reasoning. Proceedings of the forty-fourth annual meeting of the North
American Chapter of the International Group for the Psychology of Mathematics Education. Nashville, TN.
• Akuom, D., Greenstein, S., & Fernández, E. (2022) Mathematical Making in Teacher Preparation: Research at the Intersections of Knowledge, Identity, Pedagogy, and Design. Proceedings of the 44th
Meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education, Nashville.
• York, T., Greenstein, S., & Akuom, D. (2022) Embodying Covariation Through Collaborative Instrumentation. Proceedings of the 44th Meeting of the North American Chapter of the International Group
for the Psychology of Mathematics Education, Nashville.
• Akuom, D. & Greenstein, S. (2021). Prospective Mathematics Teachers’ Designed Manipulatives As Anchors for Their Pedagogical and Conceptual Knowledge. Proceedings of the 43rd Meeting of the North
American Chapter of the International Group for the Psychology of Mathematics Education, Philadelphia.
• Greenstein, S., Pomponio, E., & Akuom, D. (2021). Harmony and Dissonance: An Enactivist Analysis of the Struggle for Sense Making in Problem Solving. Proceedings of the 43rd Meeting of the North
American Chapter of the International Group for the Psychology of Mathematics Education, Philadelphia.
• York, T., Germia, E., Kim, Y., & Panorkou, N. (2021). Students’ reorganizations of variational, covariational, and multivariational reasoning. In Olanoff, D., Johnson, K., & Spitzer, S. (2021).
Proceedings of the forty-third annual meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education. Philadelphia, PA (pp. 308-312).
• Panorkou, N. & Germia, E. F. (2020). Examining students’ reasoning of multiple quantities. In A.I. Sacristán, J.C. Cortés-Zavala & P.M. Ruiz-Arias, (Eds.). Mathematics Education Across Cultures:
Proceedings of the 42nd Meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education, Mexico (pp. 291-295). Cinvestav / AMIUTEM / PME-NA. https:/
• Panorkou, N. & York, T. (2020). Designing for an integrated STEM+C experience. In A.I. Sacristán, J.C. Cortés-Zavala & P.M. Ruiz-Arias, (Eds.). Mathematics Education Across Cultures: Proceedings
of the 42nd Meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education, Mexico (pp. 2233-2237). Cinvestav / AMIUTEM / PME-NA. https:/doi.org/
• Basu, D. & Panorkou, N. (2020). Utilizing mathematics to examine sea level rise as an environmental and a social issue. In A.I. Sacristán, J.C. Cortés-Zavala & P.M. Ruiz-Arias, (Eds.).
Mathematics Education Across Cultures: Proceedings of the 42nd Meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education, Mexico (pp.
1064-1068). Cinvestav / AMIUTEM / PME-NA. https:/doi.org/10.51272/pmena.42.2020
• Panorkou, N. & Germia, E. (2020). Examining Students’ Quantitative Reasoning in a Virtual Ecosystem Simulation of the Water Cycle. In Gresalfi, M. and Horn, I. S. (Eds.), The Interdisciplinarity
of the Learning Sciences, 14th International Conference of the Learning Sciences (ICLS) 2020, Volume 2 (pp. 959-966). Nashville, Tennessee: International Society of the Learning Sciences.
• Germia, E. & Panorkou, N. (2020). Exploring angles in a programming environment. Proceedings of the International Conference to Review Research in Science, Technology and Mathematics Education
(pp. 121-129). Homi Bhabha Centre for Science Education, TIFR, Mumbai.
• Basu, D. & Panorkou, N. (2020). Examining the role of covariational reasoning in developing students’ understanding of the greenhouse effect. Proceedings of the International Conference to Review
Research in Science, Technology and Mathematics Education (pp. 211-220). Homi Bhabha Centre for Science Education, TIFR, Mumbai.
• Greenstein, S., Jeannotte, D., Fernández, E., Davidson, J., Pomponio, E., & Akuom, D. (2020). Exploring the Interwoven Discourses Associated with Learning to Teach Mathematics in a Making Context
. In A.I. Sacristán, J.C. Cortés-Zavala & P.M. Ruiz-Arias, (Eds.). Mathematics Education Across Cultures: Proceedings of the 42nd Meeting of the North American Chapter of the International Group
for the Psychology of Mathematics Education, Mexico (pp. 810-816). Cinvestav/AMIUTEM/PME-NA.
• Mohamed, M., Paoletti, T., Vishnubhotla, M., Greenstein, S., & Lim, Su San. (2020). Supporting Students’ Meanings for Quadratics: Integrating RME, Quantitative Reasoning, and Designing for
Abstraction. In A.I. Sacristán, J.C. Cortés-Zavala & P.M. Ruiz-Arias, (Eds.). Mathematics Education Across Cultures: Proceedings of the 42nd Meeting of the North American Chapter of the
International Group for the Psychology of Mathematics Education, Mexico (pp. 167-175). Cinvestav/AMIUTEM/PME-NA.
• Basu, D., Panorkou, N, & Zhu, M. (2019). Examining the social aspect of climate change through mathematics. Proceedings of IEEE Integrated STEM education Conference (ISEC) 2019, Princeton, NJ.
• Panorkou, N., Basu, D. & Vishnubhotla, M. (2018) Investigating volume as base times height through dynamic task design. In Hodges, T.E., Roy, G. J., & Tyminski, A. M. (Eds.) Proceedings of the
40th annual meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education. Greenville, SC: University of South Carolina & Clemson University (pp.
• Basu, D. & Panorkou, N. (2018) Expanding students’ contextual neighborhoods of measurement through dynamic measurement. In Hodges, T.E., Roy, G. J., & Tyminski, A. M. (Eds.) Proceedings of the
40th annual meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education. Greenville, SC: University of South Carolina & Clemson University (pp.
• Basu, D. & Panorkou, N. (2018) Examining the social aspects of the Greenhouse Effect through Mathematical Modeling. In Hodges, T.E., Roy, G. J., & Tyminski, A. M. (Eds.) Proceedings of the 40th
annual meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education. Greenville, SC: University of South Carolina & Clemson University (pp. 1131).
• Zhu, M., Panorkou, N., Etikyala, S., Basu, D., Street-Conaway, T., Iranah, P., Mazol, D., Hannum, C., Marshall, R., Lal, P. & Samanthula, B. (2018). Steerable Environmental Simulations for
Exploratory Learning. In Proceedings of E-Learn: World Conference on E-Learning in Corporate, Government, Healthcare, and Higher Education (pp. 83-92). Las Vegas, NV, United States: Association
for the Advancement of Computing in Education (AACE).
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Proceedings of IEEE Integrated STEM education Conference (ISEC) 2018, pp. 220-223, Princeton, NJ.
• Panorkou, N and Vishnubhotla M. (2017). Counting Square Units Is Not Enough: Exploring Area Dynamically. In T. A. Olson & L. Venenciano (Eds.), Proceedings of the 44th Annual Meeting of the
Research Council on Mathematics Learning, Fort Worth, TX (pp. 97-104).
• Seventko, J., Panorkou, N. and Greenstein, S. (2017). Balancing Teachers’ Goals and Students’ Play in a Sandbox-Style Game Environment. In T. A. Olson & L. Venenciano (Eds.), Proceedings of the
44th Annual Meeting of the Research Council on Mathematics Learning, Fort Worth, TX (pp. 73-80).
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Proceedings of the 38th annual meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education, Tucson, AZ.
• Panorkou, N., Maloney A., Confrey J. & Platt, D. (2014). Developing elementary students’ reasoning of geometric transformations through dynamic animation. In G. Futschek & C. Kynigos (Eds.),
Proceedings of the 3rd International Constructionism Conference (pp. 481-489). Vienna: Austrian Computer Society.
• Akuom, D., Greenstein, S., & Fernández, E. (to appear) Mathematical Making in Teacher Preparation: Research at the Intersections of Knowledge, Identity, Pedagogy, and Design. Paper to be
presented at the 44th Meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education, Nashville
• York, T., Greenstein, S., & Akuom, D. (to appear) Embodying Covariation Through Collaborative Instrumentation. Paper to be presented at the 44th Meeting of the North American Chapter of the
International Group for the Psychology of Mathematics Education, Nashville.
• Daniel, A., Kim, Y., Leonard, H. S., Bonaccorso, V. D., & DiNapoli, J. (In press). Characterizing mathematics teacher learning through collegial conversations in a community of practice.
Proceedings of the National Council of Teachers of Mathematics Research Conference.
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Research Association Online Paper Repository. American Educational Research Association.
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Annual International Conference of Education, Research, and Innovation (pp. 2014-2023). International Academy of Technology, Education, and Development. https://library.iated.org/view/
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case study analysis. In S. S. Karunakaran and A. Higgins (Eds.) 2021 Research in Undergraduate Mathematics Education Reports (pp. 171-179). http://sigmaa.maa.org/rume/2021_RUME_Reports.pdf
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(pp. 9364-9371). International Academy of Technology, Education, and Development. https://doi.org/10.21125/edulearn.2021.1891
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teaching. In A. I. Sacristan and J. C. Cortes (Eds.) Proceedings of the 42nd Annual Meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education
(pp. 2262-2269). Cinvestav – Department of Mathematics Education and AMIUTEM. https://doi.org/10.51272/pmena.42.2020-384
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American Chapter of the International Group for the Psychology of Mathematics Education, Philadelphia.
• Akuom, D. & Greenstein, S. (2021). Prospective Mathematics Teachers’ Designed Manipulatives As
• Anchors for Their Pedagogical and Conceptual Knowledge. Proceedings of the 43rd Meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education,
• Greenstein, S., Jeannotte, D., Fernández, E., Davidson, J., Pomponio, E., & Akuom, D. (2020). Exploring the Interwoven Discourses Associated with Learning to Teach Mathematics in a Making Context
. In A.I. Sacristán, J.C. Cortés-Zavala & P.M. Ruiz-Arias, (Eds.). Mathematics Education Across Cultures: Proceedings of the 42nd Meeting of the North American Chapter of the International Group
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International Group for the Psychology of Mathematics Education, Mexico (pp. 167-175). Cinvestav/AMIUTEM/PME-NA.
• Greenstein, S., Fernández, E. & Davidson, J. (2019). Revealing Teacher Knowledge Through Making: A Case Study of Two Prospective Mathematics Teachers. Proceedings of the 41st Annual Conference of
the North American Chapter of the International Group for the Psychology of Mathematics Education. St. Louis, MO.
• Paoletti, T., Greenstein, S., Vishnubhotla, M., & Mohamed, M. (2019). Designing Tasks and 3D Physical Manipulatives to Promote Students’ Covariational Reasoning. Proceedings of the 43rd Annual
Conference of the International Group for the Psychology of Mathematics Education. Pretoria, South Africa.
• Paoletti, T., Vishnubhotla, M., Rahman, Z., Seventko, J. & Basu, D. (2017) Comparing graph use in STEM textbooks and practitioner journals. Proceedings of the Twentieth Annual Conference on
Research in Undergraduate Mathematics Education, San Diego, California.
• Panorkou, N & Vishnubhotla, M. (2017). Counting Square Units Is Not Enough: Exploring Area Dynamically. In T. A. Olson & L. Venenciano (Eds.), Proceedings of the 44th Annual Meeting of the
Research Council on Mathematics Learning, Fort Worth, TX (pp. 97-104).
• Panorkou, N., Maloney, A., Confrey, J., & Platt, D. (August 2014). Developing elementary students’ reasoning of geometric transformations through dynamic animations. Proceedings of
Constructionism and Creativity, (pp. 1-10). Vienna, Austria.
• Krupa, E. E., Webel, C., & McManus, J. (2013). Evaluating the impact of computer-based and traditional learning environments on students’ knowledge of algebra. In Martinez, M. & Castro Superfine,
A (Eds.), Proceedings of the 35th annual meeting of the North American Chapter of the International Group for the Psychology of Mathematics Education. Chicago, IL: University of Illinois at
• Webel, C., & Platt, D. (2012). Transition in limbo: How competing obligations can lead to pedagogical inertia. In Van Zoest, L.R.; Lo, J.-J. (Eds.). (2012) Proceedings of the 34th annual meeting
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Explaining 32-Bit Float Audio
When it comes to recording microphones, 24-bit memory is more than you’ll ever need.
I’ve been on a mission to enlighten prospective buyers of audio recording equipment that 32-bit float does not prevent clipping, or in any way, improve the fidelity of microphones recorded in 24-bit
32-bit float has a benefit, and a trade-off, which are worth keeping mind after you record an analog source. I’ll get to that later.
This story is for anyone who tried to learn 32-bit float, but ended up more confused than ever. I hope to explain it in a way I wish someone had explained it to me. I’m no math expert! (So if I’ve
made a mistake please let me know in the comments).
What makes 32-bit float a headache is its use of A) bit-math B) fractions and C) exponents — together.
To help you understand the fundamentals of 32-bit float, I’m going to create a 4-bit version, without bit-math, fractions or large exponents.
Imagine we have a restaurant. Even though we don’t know a fraction from an exponent, we have a line out the door. Someone built us a computer to keep track of that line. We are told it has 4-bits of
It was explained that we can work with a line of up to 16 people in 4-bit.
What’s important is we can count up to 16. Which is fine, because our line never goes above 8. (That only fills up 3-bits. HA! We ain’t so dumb!)
Then we move to new restaurant with more space for people to line up. In the first restaurant, there was only 8 feet from the front door to the curb. In our new restaurant, there is 64 feet between
our door and curb.
If we want to reference people on a line from 1 to 64 we’d need a 6-bit computer. The maximum value of 6-bit binary is 64. We’re told the computer would be expensive and slow.
Is there another way?
Thinking about it, we realized we’re more interested in where the person is in those 64 feet, than the exact number of them, which we expected would remain at 8. We went back to the computer person.
They said okay, if we’ll always have 8 people or less and it’s more important to know where in those 64 feet they stood, we could change how we used our 4-bits. We’d keep our 3-bits for the count of
people, 8, but use the 4-th bit as an exponent that would extend our system — its scale — out to 64 feet.
We could create a new scale by scaling our 3-bit value with a 4-th bit exponent of 1 or 2.
Please bear with me.
Every number would be calculated by taking one of our 3-bit numbers and applying one of the exponents, so, for the first of our two exponents; that is, 1…
1¹, 2¹,3¹,4¹,5¹,6¹,7¹ and 8¹
Then for our second exponent 2…
1,4,9,16,25,36,49 and 64
Putting them all together, from our min to our max, our computer can work with these numbers using our 4-bit exponential format:
Each one is generated by a 3-bit number from 1 to 8 raised to an exponent of 1 to 2.
You’ll notice that those numbers have duplicates and gaps. They’re imprecise!
There are similar imprecisions in 32-bit float too, but in 32-bit float they’re super super small. Nonetheless, the fact remains. 32-bit Float is imprecise.
That’s okay because we’re trading SOME PRECISION for GREATER SCALE.
Because we are more interested in the scale, than the precision, we went with an exponential way of working with the wait list. We could go to the 49th foot and most likely find our 7th guest
(assuming they’re spread out). We could find our 6th guest around 36 feet. We could find our last guest at our 64th foot.
Also, we can easily go back to 8 precise values should be want them. We just take the reverse of the exponent, the (square in this case) root, since we use 2 as an exponent. Fun formulas below!
32-bit float is only different in the size of numbers it deals with.
It creates numbers so large that we need to represent them as large fractions to make them usable. In 32-bit float, it uses 23-bits for the mantissa. That’s a range of 1 to 8,388,607 in decimal.
But the numbers it’s really using are fractional from 1 to 1.9999998807907104. Scary right? The first bit in the mantissa is 1. The trailing bits create fractional parts, the 0 to .9999998807907104
(remember, when the exponents are applied there will be small gaps in the range).
How does it create those numbers? Let’s say you want to convert the address of the White House to 32-bit float. You move up your decimal place from 1600 to 1.600 (This is done in binary so this is
over simplifying, or doesn’t explain how you would do the number 9). Anyway, the 1.600 becomes your “mantissa”.
In a sense, 32-bit float using the EXACT same numbers used in 24-bit, except it converts them to decimal numbers between 0 and 2 first.
Why do this?
I know this stuff is hard (at least for me), but when I grasped this concept it made my morning!
In 24-bit fixed memory, I can count anything up to 16,777,216. But what if I’m an investment banker and my client has 160,777,216 dollars? What if I must work with large numbers? He would have 100
million more dollars than my memory can hold.
32-bit float allows me to trade a small amount of precision for a greater range by essentially shrinking my number down (0 to 16,777,216 to something like 0.0000001 to 1.9999999), then telling the
computer how it can expand it back using an exponent.
So that number becomes 1.60777216 to an exponent of 8 (number of decimal places I moved to the left).
The imprecisions are so miniscule they rarely effect us on our day to day work. What has to be born in mind is that if we remove the imprecisions we get back to the same amount of discrete data in
our 24-bit memory.
That’s a hard thing for me to grasp too!
My way of thinking of it is that say we have 1,2,3 in our 24-bit memory, and 1, 3, 3.5, 4, 5 in our 32-bit float memory. We can create a 3.5 representation in our 32-bit memory and it will show up as
3.5 if we ask for it. BUT, and this is the big BUT. If we subtract it from 4, let’s say, we will not get 0.5. The reason is that the .5 it NOT scaled the same way the other numbers are.
When we’re working with microphone data (to get back to my motivation to learn all this stuff) we need each measure of amplitude to be precisely relative to the the ones above and below it. So if we
have 100 millivolts, it needs to be exactly 1 millivolt above 99 and 1 below 101.
All distances between values must be precise — it’s the very definition of true fidelity. If we don’t remove those imperfections we end up with distortion!
In 32-bit float, there are numbers like 100.5 which will not calculate as 0.5 away from 100. Therefore, they cannot be counted on (excuse the pun). If we remove all the imprecise numbers in 32-bit
float we get back to 99,100 and 101.
Another way to intuit these imprecisions, “gaps” when you think of a pizza cut into 3rds. If we have a pizza in our restaurant that’s 10 inches, a 3rd would be 3.333333. If we call it 3.3 then we’re
0.033333333… imprecise. That would show up as a gap if we did something as simple as addition or subtraction.
For example, if we subtracted 1/3rd of pizza slice from 5 slices we’d get 5 minus (10/3); that is, 1.3333333333… We’d have to truncate somewhere, like, 1.33. What if later on we ask the computer to
subtract 1.33 from (10/3). We’d get 0.003333333…
Not clear, don’t worry. All you really need to understand is there is a trade-off.
Whenever we work with fractions we will encounter fractions, like 1/3 or 11/10 (numbers that are difficult to calculate using exponents) that can’t be written out with total precision, like we can
with 1,2,3… etc.
That doesn’t mean 32-bit float can’t be precise! If you reverse that “scaling” you’d get back to 24-bits worth of real numbers (as shown above with our 4-bit system).
32-bit float is useful in audio work because you can move your values along a wider scale. Sure, you lose some precision, but it’s might be more important to have a wider scale to work in.
Remember our restaurant analogy? It’s easier to move people forward and back in our 64 feet. Even though in the end, we’ll need to record “party of 8”, or “party of 3”, etc.
Because microphones output a limited number of usable milli-volt values for amplitude, well under 65,536, if not 4,096, (or 12-bit), we don’t need to widen our scale to record them — it’s just wasted
memory. Even 24-bits is more than we need! Unused memory aside, it’s no harm to place our analog values in 32-bit float.
As long as we don’t believe its providing more discrete values for our microphone output than 24-bit. As long as we understand that, from a fidelity point of view, 24-bit EQUALS 32-bit float.
This is the essay, by Fabien Sanglard, that got me over the hump in understanding 32-bit float.
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Are You Clever Enough to Solve This Math Equation 220÷20x3-7?Are You Clever Enough to Solve This Math Equation 220÷20x3-7?
Why Engage in Math Puzzles? The Benefits for Your Brain
Engaging in challenging math puzzles like this one requires lateral thinking, pushing you to step outside your usual thought patterns. This type of problem tests your mathematical and analytical
skills, requiring a solid understanding of mathematical principles.
Math puzzles can be daunting because they often require a solution that meets specific conditions. However, they also promote mental health by improving memory and fostering critical thinking, skills
that enhance decision-making in everyday life.
IQ Test: Can You Solve This Math Puzzle (220÷20×3-7)?
Take a look at the given problem: 220÷20×3-7. At first glance, you might have a general idea of what to do. If not, don’t worry—take your time to analyze it.
Ready to solve it? Let’s break it down step by step and find the solution.
Solving the Math Puzzle
To solve the equation 220÷20×3-7, we need to follow the order of operations, often remembered by the acronym PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), and
Addition and Subtraction (from left to right).
Since there are no parentheses or exponents in this expression, we start with multiplication and division from left to right.
First, divide 220 by 20. The result is 11. Next, multiply 11 by 3, giving us 33. The expression now simplifies to 33 – 7. Subtracting 7 from 33 gives us the final answer: 26.
Here’s the solution step-by-step:
1. 220÷20=11220 ÷ 20 = 11220÷20=11
2. 11×3=3311 × 3 = 3311×3=33
3. 33−7=2633 – 7 = 2633−7=26
So, the final solution is 26.
Share the Challenge
Stimulating exercises like this one develop your critical thinking skills while providing fun. If you enjoyed this math puzzle, share it with your friends and family. Let us know your thoughts in the
comments—we’d love to hear from you. For more practice, explore other types of tests available on our website. Happy puzzling!
Peter, a distinguished alumnus of a prominent journalism school in New Jersey, brings a rich tapestry of insights to ‘The Signal’. With a fervent passion for news, society, art, and television, Peter
exemplifies the essence of a modern journalist. His keen eye for societal trends and a deep appreciation for the arts infuse his writing with a unique perspective. Peter’s journalistic prowess is
evident in his ability to weave complex narratives into engaging stories. His work is not just informative but a journey through the multifaceted world of finance and societal dynamics, reflecting
his commitment to excellence in journalism.
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Electrochemical metal speciation in natural and model polyelectrolyte systems
The purpose of the research described in this thesis was to examine the applicability of electro-analytical techniques in obtaining information on the speciation of metals, i.e. their distribution
over different physico-chemical forms, in aquatic systems containing charged macromolecules. In chapter 1 a general introduction is given to (i) metal speciation in aquatic systems, (ii) (bio)
polyelectrolytes and their counterion distributions and (iii) electrochemical methods emphasizing their apllication to metal speciation.
Chapter 2 deals with the conductometric: measurement of counterion association with macromolecules. First, we have surveyed theoretical developments concerning ion association for purely
electrostatic interaction and as reflected in the conductivities of polyelectrolyte solutions. It will be shown that for the salt free case, the distribution of monovalent counterions can be obtained
from plots of the molar conductivity of the polyelectrolyte solution versus the molar conductivity of the monovalent counterion, so-called Eisenberg plots. Experimental results for various alkali
polymethacrylate concentrations show that the fraction of conductometrically- free monovalent counterions is in close agreement with theoretical predictions, which are based on a two-state appoach.
Furthermore, for linear polyelectrolytes a recently proposed model for the case of counterion condensation in systems with ionic mixtures is presented. Finally, the treatment of conductometric data
for polyelectrolyte solutions with either one type of counterion or mixtures of two types of counterions in terms of free and bound fractions is discussed.
In chapter 3 we describe a voltammetric methodology for the analysis of labile homogeneous heavy metal-ligand complexes in terms of a stability K . The method takes into account the difference
between the diffusion coefficients of the free and bound metal. Since the relationship between voltammetric current and mass transport properties under stripping voltammetric conditions is not yet
well esthablished, we propose a relationship between the experimentally obtained current and the mean diffusion coefficient of the metal-complex system. A sensitivity analysis of this expression for
different parameters, such as the stability and the ratio of the diffusion coefficients of the bound and free metal is performed.
Natural complexing agents are often heterogeneous with regard to their affinity to metal ions. Therefore, we will discuss the evaluation of the heterogeneity of these complexes from voltammetric data
for various metal-to-ligand ratios. For the case of a large excess of ligand over the metal atom concentration, the stability of the metal-complex system may be obtained independently from the
potential shift. For this an equation is given similar to the classical one derived by DeFord and Hume. Finally, we present an experimental procedure based on adding ligand to the solution of the
metal and measuring its voltammetric characteristics. The procedure takes into account (i) possible adsorption of metal ions to elements of the equipement and (ii) measuring all protolysis of the
polyacids involved.
The characteristic features of applying the two electrochemical techniques conductometry and voltammetry to the study of ion binding by polyelectrolytes are discussed and compared in chapter 4.
Analysis of data on K ^+/Zn(II)/polyacrylate and K ^+/Zn(II)/polymethacrylate systems illustrates a certain complementary of the two methodologies. Conductometry primarily measures the Zn(II)/K ^
+exchange ratio. Voltammetry measures the Zn(II)/polyion binding strength; its dependence on the (excess) K ^+concentration also yields information on the Zn(II)/K ^+exchange ratio. The different
results seem to be fairly coherent.
Experimental conductometric and voltammetric speciation data of metal-synthetic polyacid systems are presented and discussed in chapter 5. The competitive binding of monovalent and divalent
counterions has been studied by the conductometric procedure described in chapter 2 for aqueous solutions of alkali metal polymethacrylates in the presence of Ca(NO [3] ) [2] and Mg(NO [3] ) [2] .
The experimentally obtained fractions of conductometrically free counterions are compared with theoretical values computed according to a new thermodynamic model described in the same chapter. For
the systems studied, the fractions of free monovalent and divalent counterions can be fairly well described by the theory. In fact, the results support the assumption that under the present
conditions the conductometrically obtained distribution parameters f [1] and f [2] approximate the equilibrium fractions of free monovalent and divalent counterions. The experimentally obtained M ^+/
M ^2+exchange ratios agree well with the theoretical ones. Similar experiments have been performed for the Zn(II)/polyacrylate and Zn(II)/polymethacrylate systems. It seems that, compared to Ca ^
2+and Mg ^2+ions, the Zn(II)-ions are bound more strongly. This could be due to some specific binding of Zn(II)-ions. Since the theoretical model does not incorporate this mechanisme, the
experimental results do not agree well with the theoretical ones.
Furthermore, chapter 5 collects the results of a systematic study of the stripping voltammetric behaviour of Zn(II)- and Cd(II)-ions in polyacrylate and polymethacrylate solutions. All metal- ligand
complexes involved apprear to be voltammetrically labile over the whole range of metal-to- ligand ratios under the various experimental conditions employed. Hence, the voltammetric data could be
analyzed in terms of a stability K according to the methodology presented in chapter 3. The first set of experiments is concerned with the influence of the molar mass of the polyacrylate anion on the
stability. Analysis of the data in terms of a mean diffusion coefficient, which decreases with increasing molecular mass, yields a consistent picture with molar mass- independent complex stabilities.
The speciation of Zn(II) in such a polyelectrolyte system varies with the concentration of carboxylate groups, but it is invariant with the polyionic molar mass. Secondly, the competition between
monovalent (K ^+) and divalent (Zn(II) and Cd(II)) counterions has been investigated by varying the concentration of electroinactive supporting electrolyte. The results show that the stability of the
heavy metal/polymethacrylate complex decreases with increasing KNO [3] concentration. This effect is largely due to the reduction of the electrostatic component of the metal/polyanion interaction,
which is generally the case for polyelectrolytes with high charge densities. For the Zn(II)/polymethacrylate system, a comparison with conductometric data representing the competitive behaviour of
monovalent and divalent counterions has been made in chapter 4. The influence of the polyelectrolyte charge density of the polymethacrylic acid on the stability K has been studied by varying the
degree of neutralization of the polyanion. For the Zn(II)/PMA complexes, the stability increases approximately linearly with increasing degree of neutralization, i.e. with increasing polyionic charge
density. This is in accordance with the general polyelectrolytic feature that counterion binding is stronger with higher polyionic charge density. Finally, for later comparison with natural
complexing agents, the chemical homogeneity of the macromolecules involved has been verified by varying the total metal ion concentration for a given polyelectrolyte concentration. The results indeed
confirm that the Zn(II)/polymethacrylate and Cd(II)/polymethacrylate complexes have a homogeneous energy distribution. This is in line with expectation, since these macromolecules consist of only one
repeating chemical binding site, i.e. the carboxylate group.
Chapter 6 deals with the pretreatment and characterization of humic material. The pretreatment procedure is used to purify the humic material in such a way that (i) the molecules are soluble under
the experimental conditions employed in chapter 7, (ii) the amount of impurities is minimized and (iii) the resulting humic material is transferred into the acid form. Furthermore, a fractionation
method based on the solubility of the humic substances is described. The humic material is characterized in terms of (i) the amount of chargeable groups by means of conductometric titration and (ii)
molar mass distribution by flow field-flow-fractionation. It will be shown that although the fractionation by varying pH results in samples with different molar masses, the separation is far from
As was done with the synthetic polyacids, experiments have been performed for natural occuring polyelectrolytes. Conductometric and voltammetric results forvarious metal humic acid systems are
presented in chapter 7. Solutions of humic acids were conductometrically titrated with potassium, sodium, lithium, calcium and barium hydroxide solutions. The results have been analyzed in terms of
fractions of free and bound metal. The conductance properties of humic acids are basically different from those of a linear polyelectrolyte such as polymethacrylate. A marked difference was observed
between the shapes of the curves for alkali metal hydroxides and those for alkaline earth metal hydroxides. It appears that monovalent cations are hardly bound by the humate polyion, whereas divalent
counterions show a strong interaction. The latter feature may be fruitfully utilized in quantitative analysis.
The association of the heavy metals zinc(II) and cadmium(II) with humic acid samples has been furthere studied by differential pulse anodic stripping voltammetry (i) for various concentrations of
supporting electrolyte (KNO [3] and Ca(NO [3] ) [2] ), (ii) for different degrees of neutralization of the humate polyion, (iii) for different metal-to-ligand ratios and (iv) for different fractions
of the humic acid. Under the experimental conditions employed, all heavy metal/humate complexes have been found to be voltammetrically labile over the whole range of metal-to-ligand ratios. Hence,
the stability (K) of the complex could be computed taking into account the difference between the diffusion coefficients of the free and bound metal. The dependence of K on the concentration of 1-1
electrolyte (KNO [3] ) is of comparable extent for various metal-humate complexes, but significantly smaller than in the case of the highly charged linear polyelectrolyte polymethacrylic acid. For
the humic acid systems, it has been concluded that the relatively weak dependency of K on the salt concentration mainly reflects screening effects. The influence of the concentration of 2-1
electrolyte (Ca(NO [3] ) [2] ) on the stability of the heavy metal/humate complex is more pronounced than for the corresponding case of 1-1 electrolyte. By taking into account the association of
calcium with the humate polyion, the stability of the heavy metal/humate complex was found to be more or less constant over the range of Ca(NO [3] ) [2] concentrations studied and comparable to the
stability of the corresponding complex in the absence of calcium.
The stability of the heavy metal/humate complex has been found to increase with increasing degree of neutralization, i.e. with increasing charge density of the humate polyion. It seemed that the
increase of K is less pronounced for higher values of α [n] . This observation could not be interpreted from an electrostatic point of view, and is in fact a further indication that the binding of
heavy metals with the humate polyion is mainly governed by the chemical characteristics of the humic acid. The chemical heterogeneity of the humic acids was investigated by varying the
metal-to-ligand ratio for different total concentrations of the heavy metals but in a range of comparable ligand concentrations. The results show that the stability K of the heavy metal/humate
complex decreases with increasing total metal ion concentration, reflecting a certain chemical heterogeneity of the humic acid. For various heavy metal/fractionated humate complexes, the stability K
was found to be comparable to the K value for the corresponding unfractionated humic acid system. This means that the distribution of functional groups is more or less the same for different molar
masses of the humic acid.
For the present metal/humate complexes, the general conclusion is that the distribution of counterions over the free and bound states is mainly governed by the chemical heterogeneity of the humate
Original language English
Qualification Doctor of Philosophy
Awarding Institution •
• Lyklema, J., Promotor, External person
Supervisors/Advisors • van Leeuwen, H.P., Co-promotor, External person
Award date 21 Jan 1994
Place of Publication Wageningen
Print ISBNs 9789054852179
Publication status Published - 21 Jan 1994
• metals
• electrolytes
• chemical speciation
• electrochemistry
• double salts
Dive into the research topics of 'Electrochemical metal speciation in natural and model polyelectrolyte systems'. Together they form a unique fingerprint.
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A Variational Model for Removing Multiple Multiplicative Noises
A Variational Model for Removing Multiple Multiplicative Noises ()
Received 1 November 2015; accepted 21 December 2015; published 24 December 2015
1. Introduction
Image noise removal is one of fundamental problems of image processing and computer version. A real recorded image may be disturbed by some random factors, which is an unavoidable. Additive noise
model [1] -[3] is always assumed as
Classical variational model for multiplicative noise removal is aiming at Gaussian distribution [6] . But when the noise is disobedience Gaussian distributed, the effect of denoising is not very
satisfactory. To solve the problem that assuming multiplicative noise model is more reasonable and representative, in 2008, Aubert and Aujol [7] assumed the noise with Gamma distribution with mean 1.
A variational model, named AA, used the distribution characteristics of Gamma multiplicative noise and maximizing a posterior (MAP) has been proposed,
and its fidelity term expresses as
series of variation models have taken logarithmic transformation
term written as
For solving problem that AA is not strictly convex, Huang, Ng and Wen [8] used a logarithmic transformation and proposed a new model (Named HNW model):
Numerical results show that noise removal ability of HNW is better than AA, but it produces “staircase effect”. Alternative iterative algorithm ensures that the solution of the model is unique, and
the iterative sequence also converges to optimal solution of it.
After, a body of variation models [7] -[13] of multiplicative noise removal has been proposed, and removing multiplicative noise abilities made considerable progress. Models not only can effectively
remove the noise, but also to better protect the image edge and texture. When we get a model, and then must need a good algorithm to solve it. Numerical algorithm of variation model, today, includes
ADMM [14] [15] , ALM [16] [17] , Newton iterative method [8] [9] [18] [19] and dual algorithm [20] -[22] and so on. HNW model has used adaptive alternating iterative algorithm. That is to say, the
model can be divided into two parts: one uses Newton iterative method, and the other uses dual algorithm. Iterative sequence obtained converges to the optimal value of the model.
The rest of this paper is organized as follows. In Section 2, we introduce the proposed model how constructs it. Next section will give a new numerical algorithm. Convergence proof of the model will
be launched in Section 4. In Section 5, we will show the experiments and its specific analysis. Finally, concluding remarks are given.
2. The Proposed Model
The difference between additive noise and multiplicative noise is whether the noise signal and the original image signal are independent or not. Multiplicative noise, however, is not independent. In
paper [23] , multiplicative noise model is assumed
In which g is the observed image, u is the original image, n is multiplicative noise under Rayleigh distribution, and the probability density function of n is denoted as follows
To realize the estimate of the original image u, the estimate can be computed by
Applying Bayes’s rule, it becomes
Based on (2.4), minimize post mortem energy of its MAP method
Logarithmic energy equation
We can know the truth from the reference [24]
Combining (2.2), (2.3), (2.5) with (2.6), we can get
From (2.1), we can derive that
where D is a two-dimensional bounded open domain of R^2 with Lipschitz boundary, then image can be interpreted as a real function defined on D.
With using a logarithmic transformation
An unconstrained optimization problem can be solved by a composition function
Variable splitting [17] is a very simple procedure that consists in creating a new variable, say v, to serve as the argument of
which is apparently equivalent to formula (2.11), and the Lagrange function can be written as follows
We denote
To solve its minimum value, it is equivalent to this constrained optimization problem
3. Algorithms
Inspired by the iterative algorithm of reference [8] and [18] , in this paper, I will propose a new algorithm to solve (2.13). Starting from initial guess
Such that
To solve the problem (3.1), we need to divide it into the following three steps.
The first step of the method is to solve a part of the optimization problem. The minimizer of this problem
Its discretization
Now, letting
Since f is continuous and derivable in the specified range, this function is equitant to solving the regular with
We use CSM [25] to replace Newton iteration method [8] [9] .
And then, we can get
The second step of the method is to apply a TV denoising scheme to the image generated by the previous multiplicative noise removal step. The minimizer of the optimization problem
Its corresponding Euler-Lagrange equation of the variational problem (3.5) as follows
In this paper,
Using gradient descent method to obtain (3.5) the optimization numerical solution as follows:
and iterative formula
The third step is to analysis the condition to stop iterative.
4. The Convergence Analysis
In this section, we will discuss the convergence of the iterative algorithm. First, we know that
Theorem 1. For any given initial value
To prove this theorem, we will give the following lemmas, and the appropriate proof.
Lemma 1. Sequence
Proof. It follows from the alternating iterative process in algorithm that
It is obvious that sequence
Lemma 2.The function
Proof. Let
The matrix S is not a full-rank. The discrete total variation of regularization term of model (2.13) as follows
Next we will discuss two cases: 1)
For (i), we note that
to z. therefore we obtain
By using the above inequality, we have
We can get that
For (ii), considering
Definition 3. Let
Proof of theorem 1. Since sequence
function and strictly convex function, the set of fixed points are just minimizers of
one and only one minimizer of
Moreover, we have, for any
Let us denote by
We can get conclusion that
5. Experimental Results
In this section, we will experiment on Lena and Cameraman. Different strength Gamma and Rayleigh noises are added to the original image, and then comparing effects of the proposed model proposed
model denosing with HNW. In our experiments, Figure 1(a) is original image of Lena; Figure 1(b) is Cameraman. Figures 2-5 are noised images distorted by Rayleigh and Gamma noise with different
In Figure 3 and Figure 4, denosing results of Lena obtained by the proposed model and HNW model are including noise removal image―The clearer image is, the well model is; residual plot-More image’
signal has been kept, more bad experimental results; gray value curve figure―The blue color represents the original image of the selected signal, and red signal represents denoised part. If red and
blue colors are fitting well, we could say that the denosing effect is better. From Figure 3, we can clearly see that the proposed model has more effective
Figure 1. Original images. (a) Lena. (b) Cameraman.
Figure 2. Noisy images for Lena. (a) L = 20 Gamma. (b)
than HNW for Lena with Gamma L = 20, because gray distribution is reasonable and fitting degree of denoised image is stronger than HNW model. The result of denosed aiming at the noise under the
Rayleigh distributed multiplicative noise Figure 4. There is obviously see that denoising effect is much better than HNW model, residual plots and experimental signal diagram are also optimistic.
In Figure 6 and Figure 7, experimental results for Cameraman destroyed by L = 10 andFigure 6, for removing noised image Cameraman, our model is clearer. In Figure 7, the residual plot has no
obviously light part, that is to say, our model is slightly better. Whether for simple texture Cameraman or complex texture Lena image, the proposed model has better than HNW.
In order to better illustrate the effectiveness ofthe proposed model, this paper will use the additional data to show it. These are iteration time (T), signal to noise ratio (SNR), mean square error
(MSE), peak signal to noise ratio (PSNR), and relative error rate (ReErr). T is time to work-the smaller timeis, the well model is. For SNR or PSNR, the larger the value, the smaller noise. For MSE
or ReErr, the value is smaller, indicating that denoising effect is positive. Table 1 and Table 2 show the experimental data. Datasshow that whether for Gamma noise or Rayleigh noise, or simple or a
little texture detail-rich images, the proposed model is better than NHW model to obtain considerable experimental data.
Figure 3. Restored images for Lena L = 20.
L = 10 Gamma
Figure 5. Noisy images for Cameraman.
Figure 6. Restored images for Cameraman L = 10.
6. Conclusion
In this paper, we propose a variational method for removing multiple multiplicative noises, and give a new numerical iterative algorithm. We proved the sequence obtained converges to the optimal
solution of the model. Final experiments show that whether Gamma noise or Rayleigh noise, denoising and edge-protection ability of the proposed model are stronger than HNW model, at the same time,
staircasing effect (image has the same gray in some regions) is greatly suppressed. But, proposed model has only dealt with two noises. Next work, we wish that we can find a model to remove many more
kinds of multiplicative noises and make sure it has unique solution!
^*Corresponding author.
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Rounding to the nearest ten and hundred- numbers to 1,000
Students learn to round to the nearest ten and hundred of numbers to 1,000.
It is important to be able to round numbers, so you can calculate faster and make estimations. An example is if you want to buy a tablet that costs $422, you know that you need about $400.
Practice skip counting in 10s forwards and backwards to 100. Toss the ball to a student who says "10". They throw to the next student, who says "20". Continue to 100. Then restart at 100 and skip
count in 10s back to 0.
The interactive whiteboards shows the tens and hundreds. Discuss these with your students and say they can be called landmark numbers. Explain that by rounding a number you look to see what the
closest ten or hundred is.Students use a number line to see which ten or hundred they round up or down to. They can do this by first looking for the nearest ten, and then to look for the nearest
hundred. Students can see on the number line which ten or hundred is closest in proximity to the given number. Next, explain to students that you round up when a number is 5 or greater and you round
down if the number is 4 or fewer. By dragging the number into the HTO(hundreds, tens, ones) chart students can see the place values of a number. Discuss with students which number they need to look
at to do their rounding. If they round to a ten, they have to look at the ones place. If they round to a hundred, they need to look at the tens place. Practice doing so as a class with two numbers
and round them to both tens and hundreds.To check that students understand rounding numbers to 1,000 to tens and hundreds, you can ask the following questions:- What are tens?- What are hundreds?-
Which number do you look at if you want to round to a ten?- Which number do you look at if you want to round to a hundred?- When do you round up or round down?
Students first practice rounding up or down by answering multiple choice questions. They then must drag a number to its nearest hundred. Ask students to explain which digit they look at to determine
if they round up or down.
Discuss with students what the importance of rounding is, namely that you learn to estimate. Check their understanding using the final exercise on the interactive whiteboard of rounding to the
nearest ten. Show numbers one by one. When the number shown should be rounded up, students stand up. When the number shown should be rounded down, students sit down. Repeat the fact that students
should look at the ones if they are rounding to a ten, and at the tens if they are rounding to a hundred.
Students who have difficulty rounding can use the HTO chart and/or a number line.
Gynzy is an online teaching platform for interactive whiteboards and displays in schools.
With a focus on elementary education, Gynzy’s Whiteboard, digital tools, and activities make it easy for teachers to save time building lessons, increase student engagement, and make classroom
management more efficient.
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Lutz Warnke
Contact Info
Email: lwarnke@ucsd.edu
Ph.D., Mathematics, University of Oxford, United Kingdom, 2012
Education: Dr. Warnke earned his Ph.D. in Mathematics from Oxford
University in 2012.
Research: Dr. Warnke works in combinatorics and his interests include
random graphs and processes, phase transitions, and combinatorial
probability (as well as applications to extremal combinatorics, Ramsey
theory, and related areas).
He is the recipient of the Dénes König Prize, 2016 for his work in
combinatorics and an Alfred P. Sloan Research Fellowship in
2018. Furthermore, he received the 2014 Richard Rado Prize and a 2020
NSF CAREER Award.
Prior to UC San Diego: Dr. Warnke was a junior research fellow at
Peterhouse, Cambridge University until 2016. He thereafter moved as an
Assistant Professor to Georgia Tech, where he was promoted to
Associate Professor with tenure in 2021.
Alfred P. Sloan Research Fellowship
Dénes König Prize
NSF CAREER Award
Richard Rado Prize
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How does temperature affect the movement of atoms and molecules? | HIX Tutor
How does temperature affect the movement of atoms and molecules?
Answer 1
As temperature increases, velocity must increase.
Temperature is the average kinetic energy of gas molecules, so #t=1/2 mv_(avg)^2# As temperature increases, velocity must increase. Mass cannot increase, as per the law of conservation of mass.
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Answer 2
Temperature affects the movement of atoms and molecules by increasing their kinetic energy. As temperature rises, atoms and molecules move more rapidly and collide with greater force. This increased
movement leads to expansion of materials as the atoms and molecules spread apart, and it can also result in changes in state, such as melting or boiling. Conversely, as temperature decreases, the
movement of atoms and molecules slows down, ultimately leading to contraction of materials and changes in state, such as freezing.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
Not the question you need?
HIX Tutor
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• 98% accuracy study help
• Covers math, physics, chemistry, biology, and more
• Step-by-step, in-depth guides
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Unscramble AGNIZE
How Many Words are in AGNIZE Unscramble?
By unscrambling letters agnize, our Word Unscrambler aka Scrabble Word Finder easily found 37 playable words in virtually every word scramble game!
Letter / Tile Values for AGNIZE
Below are the values for each of the letters/tiles in Scrabble. The letters in agnize combine for a total of 16 points (not including bonus squares)
What do the Letters agnize Unscrambled Mean?
The unscrambled words with the most letters from AGNIZE word or letters are below along with the definitions.
• agnize (v. t.) - To recognize; to acknowledge.
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e OFDM
OFDM Channel Response
Calculate OFDM channel response
Since R2023a
Communications Toolbox / Channels
The OFDM Channel Response block calculates the OFDM channel response in the frequency domain by using perfect information from the channel. For more information, see OFDM Channel Response.
This icon shows the block with all ports enabled.
OFDM Equalization
Apply equalization to an OFDM-modulated QAM signal that has been filtered through a Rayleigh MIMO channel.
The cm_ofdm_equalization model initializes simulation variables and computes path filters in the InitFun callback function. For more information, see Model Callbacks (Simulink).
The model generates random integer data, applies 64-QAM, and then applies OFDM to the QAM-modulated signal. The OFDM-modulated signal gets filtered through a MIMO Rayleigh fading channel. The model
adds a signal delay in samples, and then OFDM-demodulates the signal. In a parallel path, an OFDM Channel Response block computes the perfect OFDM channel response and an OFDM Equalizer block
equalizes the received signal.
The value of the constant block labeled Fading Channel Delay equals the delay of the MIMO Fading Channel block. Since the discrete path delays of the MIMO Fading Channel block are set to [3 9 15]/Fs,
the first nonzero value of the channel impulse response has a delay of 3 samples and the Fading Channel Delay value can be any integer in the range [0,3] samples. If the Fading Channel Delay value is
greater than 3, intersymbol interference will occur.
The delay block labeled Signal Delay is equal to (fftLen+cpLen) – Fading Channel Delay samples. This removes the MIMO Fading Channel block delay and adds a delay of one OFDM symbol. Each OFDM symbol
has (fftLen+cpLen) samples.
The delay blocks labeled Delay1 and Delay2 each add a delay of one OFDM symbol to match the one OFDM symbol delay introduced by the Signal Delay block. Delay1 ensures that the right OFDM Modulator
input is used as a reference to calculate the maximum error. Delay2 ensures that the OFDM Demodulator output and the OFDM Channel Response output correspond to the same OFDM symbol.
A constellation diagram displays the unequalized and equalized signals. The model computes and displays the maximum error between the transmitted QAM signal and the equalized signal on the receive
The maximum computed error is 0.000558.
Path gains — Channel path gains
Channel path gains, specified as an N[S]-by-N[P]-by-[T]-by-N[R] array of values.
• N[S] is the number of samples.
• N[P] is the number of paths.
• N[T] is the number of transmit antennas.
• N[R] is the number of receive antennas.
The channel path gains follow the definition of channel path gains as calculated and output by the fading channel System objects. For example, see MIMO Fading Channel or SISO Fading Channel. This
block assumes that the path gains sample rate matches the OFDM sample rate.
Data Types: single | double
Complex Number Support: Yes
Path filters — Channel path filter coefficients
Channel path filter coefficients, specified as an N[P]-by-N[H] matrix of values.
• N[P] is the number of paths.
• N[H] is the number of impulse response samples.
The channel path filter coefficient matrix is used to convert path gains to channel filter tap gains for each sample and each pair of transmit and receive antennas. The channel path filter
coefficients follow the definition of channel path filter coefficients as calculated by the info object function of the fading channel System objects.
The channel path gains follow the definition of channel path gains as calculated and output by the fading channel System objects. For example, see MIMO Fading Channel or SISO Fading Channel.
Data Types: single | double
Complex Number Support: Yes
Offset — Timing offset
Timing offset in samples, specified as a nonnegative, integer scalar.
Data Types: double | single
hEst — Frequency response
Frequency response, returned as an N[SC]-by-N[Symbols]-by-N[T]-by-N[R] array.
• N[SC] is the number of OFDM subcarriers and equals the length of Active subcarrier indices.
• N[Symbols] is the number of whole OFDM symbols contained in Path gains. Specifically, floor(N[S]/(N[FFT] + L[CP])).
• N[T] is the number of transmit antennas.
• N[R] is the number of receive antennas.
• The returned frequency response corresponds to an OFDM symbol timing alignment at the receiver that uses the timing offset:
To edit block parameters interactively, use the Property Inspector. From the Simulink^® Toolstrip, on the Simulation tab, in the Prepare gallery, select Property Inspector.
FFT length — FFT length
512 (default) | positive integer
FFT length, specified as a positive, integer scalar. The FFT length must align with that of the OFDM-modulated signal.
Cyclic prefix length — Cyclic prefix length of one OFDM symbol
16 (default) | nonnegative integer
Cyclic prefix length of one OFDM symbol, specified as a nonnegative, integer scalar with a value less than FFT length.
Active subcarrier indices — Number of active subcarriers
1:512 (default) | positive integer | column vector
Number of active subcarriers, specified as a positive, integer scalar or column vector of values in the range [1, FFT length].
Enable timing offset input port — Option to add input port for timing offset
off (default) | on
Select this parameter to add an input port to input timing offset.
• on — Adds the Offset input port to specify the timing offset for channel response estimation.
• off — The block uses the Path gains and Path filters inputs to estimate the timing offset for OFDM channel response estimation.
Simulate using — Type of simulation to run
Interpreted execution (default) | Code generation
Type of simulation to run, specified as Interpreted execution or Code generation.
• Interpreted execution — Simulate the model by using the MATLAB^® interpreter. This option requires less startup time, but the speed of subsequent simulations is slower than with the Code
generation option. In this mode, you can debug the source code of the block.
• Code generation — Simulate the model by using generated C code. The first time you run a simulation, Simulink generates C code for the block. The model reuses the C code for subsequent
simulations unless the model changes. This option requires additional startup time, but the speed of the subsequent simulations is faster than with the Interpreted execution option.
For more information, see Interpreted Execution vs. Code Generation (Simulink).
Block Characteristics
Data Types double | single
Multidimensional Signals yes
Variable-Size Signals yes
OFDM Channel Response
The OFDM channel response algorithm uses an FFT to compute the channel estimates by using the path gains and path filter coefficients available after you pass data through a MIMO channel. Use channel
path gains returned by the MIMO channel object, and the path filters and timing offset returned by the info object function, to estimate the OFDM channel response.
This figure shows one OFDM symbol in addition to the input channel impulse and the output channel impulse response. The algorithm ignores samples outside the OFDM symbol.
For a time-varying channel, such as most fading channels, the impulse response depends on the location of the impulse at the channel input.
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using Simulink® Coder™.
Version History
Introduced in R2023a
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State (thermodynamics)
The state of a system in thermodynamics is characterized by the specification of all state variables necessary for the delimitation from other states .
State variables include pressure , volume , temperature , amount of substance . These sizes are divided into extensive (quantity-dependent) and intensive (not quantity-dependent).
The individual states are often entered in a phase diagram . Depending on the plot, a single point usually corresponds to a state in the state space , but sometimes all points of a phase line or area
also belong to a single state.
Statistical Physics
In statistical mechanics (statistical thermodynamics) there is the following distinction:
• In the case of the microstate , the individual locations and impulses of all particles are given, so the microstate corresponds to a point in phase space .
• the macrostate is an indication of mean values such as temperature, pressure and density. A macrostate includes all microstates that are compatible with the specified state variables.
KZKKZZKK is a special microstate.
5x K and 3x Z is then the macrostate to which microstates belong (the number results from the combinatorics as a permutation of objects of two classes K and Z, taking into account the sequence). ${\
displaystyle {8 \ choose 3} = 56}$
See also
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Relative Frequency Calculator
Use this relative frequency calculator statistics to find how many times a particular value occurs within the given dataset relative to the total number of observations. Our calculator generates a
relative frequency distribution table for grouped or ungrouped data, showing intervals, frequencies, relative frequencies, and cumulative frequencies to help you understand data distribution.
What Is Relative Frequency?
The relative frequency is the ratio of the number of times an event occurs to the total number of trials. This frequency can be represented as a fraction, percentage, or decimal value.
Relative Frequency Formula:
Relative Frequency = f n
• f = Frequency of specific group
• n = Total frequencies
Cumulative Relative Frequency:
Cumulative relative frequency is the accumulation of previous relative frequencies. To get this, add all previous relative frequencies to the current relative frequency. The last value becomes 1,
representing 100% of the data. It helps to understand the proportion of the data that falls below a specific value.
How To Find Relative Frequency?
• Determine the occurrences of all the events
• Find the total number of observations by performing the sum of all events (frequencies)
• Now get the relative frequency by dividing the frequency of each event by the total number of observations
Suppose 100 students of a class got the grades A, B, C, D, and F:
• A: 10
• B: 20
• C: 30
• D: 25
• F: 15
Find the relative and cumulative frequency
• Frequency of A = 1
• Frequency of B = 1
• Frequency of C = 1
• Frequency of D = 1
• Frequency of F = 1
Calculate the relative frequency for each grade:
• Relative frequency of A = 1 5
• Relative frequency of A = 0.2
• Relative frequency of B = 1 5
• Relative frequency of B = 0.2
• Relative frequency of C = 1 5
• Relative frequency of C = 0.2
• Relative frequency of D = 1 5
• Relative frequency of D = 0.2
• Relative frequency of F = 1 5
• Relative frequency of F = 0.2
Calculate the cumulative relative frequency:
A = 0.2
B = 0.2 + 0.2 = 0.4
C = 0.2 + 0.2 + 0.2 = 0.6
D = 0.2 + 0.2 + 0.2 + 0.2 = 0.8
F = 0.2 + 0.2 + 0.2 + 0.2 + 0.2 = 1.0
Result Interpretation:
The following relative frequency table shows the experimental probabilities for the given data.
Numbers Frequency Relative Frequency Cumulative Relative Frequency
10 1 0.2 0.2
20 1 0.2 0.4
30 1 0.2 0.6
25 1 0.2 0.8
15 1 0.2 1
• In the relative frequency table, the relative frequency shows the proportion of students getting each grade
• The cumulative relative frequency represents the proportion of the student, having a grade less or equal to a specific grade values
You can simplify your data analysis by using our relative frequency distribution calculator and generate similar results for your dataset in seconds!
What's The Difference Between Frequency And Relative Frequency?
• Frequency: This is the absolute number of times a value or data point occurs within a dataset
• Relative Frequency: It is the proportion or percentage of times a specific value occurs in proportion to the total
Is Relative Frequency And Probability The Same?
Both of these terms are closely related to each other but they are not the same.
• Relative Frequency: It works with the actual data and finds how many times an event occurs in a certain experiment
• Probability: It's a prediction that tells what might happen. It is based on assumptions and may vary from the actual scenario
Why Is Relative Frequency Important?
Relative frequency is a crucial factor when you need to compare datasets of different sizes. With the help of it, you can convert the counts to proportions and compare value distribution for
different groups of data. Meanwhile, the distribution of values is necessary for data analysis.
Using the cumulative relative frequency calculator, you can efficiently determine relative frequencies and gain valuable insights into your data.
From the source of Wikipedia: Frequency, Relative Frequency.
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Author jyasskin
Recipients facundobatista, gvanrossum, jyasskin, mark.dickinson, rhettinger
Date 2008-01-13.22:57:29
SpamBayes Score 0.00021916247
Marked as misclassified No
Message-id <1200265052.18.0.767384810556.issue1682@psf.upfronthosting.co.za>
_binary_float_to_ratio: Oops, fixed.
Rational() now equals 0, but I'd like to postpone Rational('3/2') until
there's a demonstrated need. I don't think it's as common a use as
int('3'), and there's more than one possible format, so some real world
experience will help (that is, quite possibly not in 2.6/3.0). I'm also
postponing Rational(instance_of_numbers_Rational).
+/-inf and nan are gone, and hash is fixed, at least until the next bug.
:) Good idea about using tuple.
Parentheses in str() help with reading things like
"%s**2"%Rational(3,2), which would otherwise format as "3/2**2". I don't
feel strongly about this.
Equality and the comparisons now work for complex, but their
implementations make me uncomfortable. In particular, two instances of
different Real types may compare unequal to the nearest float, but equal
to each other and have similar but inconsistent problems with <=. I can
trade off between false ==s and false !=s, but I don't see a way to make
everything correct. We could do better by making the intermediate
representation Rational instead of float, but comparisons are inherently
doomed to run up against the fact that equality is uncomputable on the
computable reals, so it's probably not worthwhile to spend too much time
on this.
I've added a test that float(Rational(long('2'*400+'7'),
long('3'*400+'1'))) returns 2.0/3. This works without any explicit
scaling on my part because long.__truediv__ already handles it. If
there's something else I'm doing wrong around here, I'd appreciate a
failing test case.
The open issues I know of are:
* Is it a good idea to have both numbers.Rational and
rational.Rational? Should this class have a different name?
* trim and approximate: Let's postpone them to a separate patch (I do
think at least one is worth including in 2.6+3.0). So that you don't
waste time on them, we already have implementations in the sandbox and
(probably) a good-enough explanation at
Thanks for the offer to help out with them. :)
* Should Rational.from_float() exist and with the current name? If
there's any disagreement, I propose to rename it to
Rational._from_float() to discourage use until there's more consensus.
* Rational.from_decimal(): punted to a future patch. I favor this for
* Rational('3/2') (see above)
I think this is close enough to correct to submit and fix up the
remaining problems in subsequent patches. If you agree, I'll do so.
Date User Action Args
2008-01-13 22:57:32 jyasskin set spambayes_score: 0.000219162 -> 0.00021916247
recipients: + jyasskin, gvanrossum, rhettinger, facundobatista, mark.dickinson
2008-01-13 22:57:32 jyasskin set spambayes_score: 0.000219162 -> 0.000219162
messageid: <1200265052.18.0.767384810556.issue1682@psf.upfronthosting.co.za>
2008-01-13 22:57:31 jyasskin link issue1682 messages
2008-01-13 22:57:29 jyasskin create
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2022 AMC 8 Problems/Problem 19
Mr. Ramos gave a test to his class of $20$ students. The dot plot below shows the distribution of test scores. $[asy] //diagram by pog . give me 1,000,000,000 dollars for this diagram size(5cm);
defaultpen(0.7); dot((0.5,1)); dot((0.5,1.5)); dot((1.5,1)); dot((1.5,1.5)); dot((2.5,1)); dot((2.5,1.5)); dot((2.5,2)); dot((2.5,2.5)); dot((3.5,1)); dot((3.5,1.5)); dot((3.5,2)); dot((3.5,2.5));
dot((3.5,3)); dot((4.5,1)); dot((4.5,1.5)); dot((5.5,1)); dot((5.5,1.5)); dot((5.5,2)); dot((6.5,1)); dot((7.5,1)); draw((0,0.5)--(8,0.5),linewidth(0.7)); defaultpen(fontsize(10.5pt)); label("65",
(0.5,-0.1)); label("70", (1.5,-0.1)); label("75", (2.5,-0.1)); label("80", (3.5,-0.1)); label("85", (4.5,-0.1)); label("90", (5.5,-0.1)); label("95", (6.5,-0.1)); label("100", (7.5,-0.1)); [/asy]$
Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students $5$ extra points, which increased the median test score to $85$
. What is the minimum number of students who received extra points?
(Note that the median test score equals the average of the $2$ scores in the middle if the $20$ test scores are arranged in increasing order.)
$\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~5\qquad\textbf{(E)} ~6\qquad$
Before Mr. Ramos added scores, the median was $\frac{80+80}{2}=80$. There are two cases now:
Case #$1$: The middle two scores are $80$ and $90$. To do this, we firstly suppose that the two students who got $85$ are awarded the extra $5$ points. We then realize that this case will have a lot
of students who receive the extra points, therefore we reject this case.
Case #$2$: The middle two scores are both $85$. To do this, we simply need to suppose that some of the students who got $80$ are awarded the extra $5$ points. Note that there are $8$ students who got
$75$ or less. Therefore there must be only $1$ student who got $80$ so that the middle two scores are both $85$. Therefore our answer is $\boxed{\textbf{(C) }4}$.
A diagram to visualize better what I explained here:
$[asy] //diagram by pog . give me 1,000,000,000 dollars for this diagram —> sry I only have 999,999,999 dollars now size(5cm); defaultpen(0.7); dot((0.5,1)); dot((0.5,1.5)); dot((1.5,1)); dot
((1.5,1.5)); dot((2.5,1)); dot((2.5,1.5)); dot((2.5,2)); dot((2.5,2.5)); dot((3.5,1)); dot((4.5,1)); dot((4.5,1.5)); dot((4.5,2)); dot((4.5,2.5)); dot((4.5,3)); dot((4.5,3.5)); dot((5.5,1)); dot
((5.5,1.5)); dot((5.5,2)); dot((6.5,1)); dot((7.5,1)); draw((0,0.5)--(8,0.5),linewidth(0.7)); defaultpen(fontsize(10.5pt)); label("65", (0.5,-0.1)); label("70", (1.5,-0.1)); label("75", (2.5,-0.1));
label("80", (3.5,-0.1)); label("85", (4.5,-0.1)); label("90", (5.5,-0.1)); label("95", (6.5,-0.1)); label("100", (7.5,-0.1)); [/asy]$
Video Solution by Math-X (First understand the problem!!!)
Video Solution (🚀50 seconds🚀)
~Education, the Study of Everything
Video Solution
Video Solution
Video Solution
See Also
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Up to this point, I have developed the various parts of the HumanProof software. The focus of this chapter is my study to investigate how mathematicians think about the understandability of proofs
and to evaluate whether the natural language proofs generated by HumanProof aid understanding.
6.1. Objectives
This chapter addresses the third objective of Question 2 in my set of research questions (Section 1.2): evaluating HumanProof by performing a study on real mathematicians. Let me break this objective
down into a number of hypotheses and questions.
The hypotheses that I wish to test are:
1. Users will (on average) rank HumanProof natural language proofs as more understandable than Lean proofs.
2. Users will (on average) rank HumanProof natural language proofs as less understandable than proofs from a textbook. This hypothesis is a control: a textbook proof has been crafted by a human for
teaching and so I would expect it to be more understandable than automatically-generated proofs.
3. Users will (on average) be more confident in the correctness of a Lean proof.
In addition to testing these hypotheses, I wish to gather qualitative answers to the following two questions:
1. What properties of a proof contribute to, or inhibit, understandability?
2. How does having a computer-verified proof influence a mathematician's confidence in the correctness of the proof?
6.2. Methodology
The goal of this study is to evaluate the hypotheses and gain answers to the questions above.
To do this, I found a set of mathematicians and showed them sets of proof scripts written using Lean, HumanProof and a proof script taken from a mathematics textbook. I recorded the sessions and
performed a qualitative analysis of the verbal responses given by participants. I also asked participants to score the proof scripts on a Likert scale (that is, rank from 1 to 5) according to two
1. How understandable are the proofs for the purpose of verifying correctness? This is to assess hypotheses 1 and 2.
2. How confident is the participant that the proofs are correct? Note that a proof assistant gives a guarantee of correctness. This is to assess hypothesis 3.
In Section 2.5, I discussed some of the literature in the definitions of understandability and distinguished between a person finding a proof understandable and a person being confident that a proof
is correct. In this experiment, instead of using these definitions I asked the participants themselves to reflect on what understandable means to them in the context of the experiment.
A similar methodology of experimentation to the one presented here is the work of Jackson, Ireland and Reid [IJR99[IJR99]Ireland, Andrew; Jackson, Michael; Reid, GordonInteractive proof critics
(1999)Formal Aspects of Computinghttps://doi.org/10.1007/s001650050052] in developing interactive proof criticsProof critics are discussed in Section 2.6.2.. Their study sought to determine whether
users could productively interact with their software, and they adopted a co-operative evaluation methodology [MH93[MH93]Monk, Andrew; Haber, JeanneImproving your human-computer interface: a
practical technique (1993)publisher Prentice Hallhttps://books.google.co.uk/books?id=JN9QAAAAMAAJ] where participants are asked to 'think aloud' when using their software.
In overview, each experiment session lasted about an hour and was split into three phases:
1. Participants were given a training document and a short presentation on how to read Lean proofs. They were also told the format of the experiment. (10 minutes)
2. Over 4 rounds, the participant was given a statement of a lemma in Lean code and then shown three proofs of the same lemma in a random order. They were asked to rate these and also to 'think
aloud' about their reasons for choosing these ratings. (40 minutes)
3. A debrief phase, where 3 demographic questions were asked as well as a brief discussion on what understandability means to them. (5 minutes)
Due to the COVID19 pandemic, each experiment session was conducted using video conferencing software with the participants submitting ratings using an online form. As well as the data that the
participants filled in the form with, the audio from the sessions was recorded and transcribed to methodically extract the explanations for why the participants assigned the ratings they did.
The study was ethically approved by the Computer Laboratory at the University of Cambridge before commencing. The consent form that participants were asked to sign can be found in Appendix D.4.
6.2.1. Population
I wish to understand the population of working mathematicians, which here means people who work professionally as mathematicians or students of mathematics or mathematical sciences such as physics.
That is, people who are mathematically trained but who do not necessarily have experience with proof assistants.
Postgraduates and undergraduates studying mathematics at the University of Cambridge were recruited though an advert posted to the university's mathematics mailing list, resulting in 11 participants.
The first participant was used in a pilot study, however the experiment was not changed in light of the results of the pilot and so was included in the main results. All participants were rewarded
with a £10 gift card.
6.2.2. Choice of theorems to include
I will include proofs that involve all of the aspects of HumanProof discussed in Chapter 3 and Chapter 4. These are:
Note that the GUI developed in Chapter 5 was not included in this study. This is primarily due to the difficulty of performing user-interface studies remotely.
The proofs should be drawn from a context with which undergraduates and postgraduates are familiar and be sufficiently accessible that participants have ample time to read and consider the proofs.
The four problems are:
1. The composition of group homomorphisms is a group homomorphism.
2. If A and B are open sets in a metric space, then A ∪ B is also open.
3. The kernel of a group homomorphism is a normal subgroup.
4. If A and B are open sets in a metric space, then A ∩ B is also open.
Lemma 1 (g ∘ h is a group homomorphism) is simple but will be a 'teaching' question whose results will be discarded. This serves to remove any 'burn-in' effects to do with participants
misunderstanding the format of the experiment and to give the participant some practice.
6.2.3. Choice of proofs
The main part of the experiment is split into 4 tasks, one for each of the lemmas given above. Each task consists of
• A theorem statement in natural language and in Lean. For example, Lemma 1 appears as "Lemma 1: the composition of a pair of group homomorphisms is a group homomorphism" and also is_hom f → is_hom
g → is_hom (g ∘ f).
• A brief set of definitions and auxiliary propositions needed to understand the statement and proof.
• Three proof scripts;
□ A Lean tactic sequence, written so that it is not necessary to see the tactic state to understand the proof.
□ A HumanProof generated natural language write-up.
□ A natural language proof taken from a textbook.
These proofs can be viewed in Appendix D.3.
Both of the metric space problems use the same definitions, so this will save some time in training the participant with notation and background material. They also produce different HumanProof
In Lemma 4 (Appendix D.3.4), the statement to prove is that the intersection of two open sets is open. The HumanProof proof for this differs from the Lean and textbook proofs in that an existential
variable ε : ℝ needs to be set to the minimum of two other variables. In the Lean and textbook proofs, ε is explicitly assigned when it is introduced. But HumanProof handles this differently as
discussed in Section 3.5.8: ε is transformed into a metavariables and is assigned after it can be deduced from unification later.
Two group theory problems (Lemmas 1 and 3) are given which are performed using a chain of equalities. These use the subtasks engine detailed in Chapter 4.
There are usually many ways of proving a theorem. Each of the three proofs for a given theorem have been chosen so that they follow the same proof structure, although the level of detail is
different, and in the equality chains different sequences are used. This should help to ensure that the participant's scores are informed by the presentation of the proof, rather than the structure
of the proof.
For all proofs, variable names were manually adjusted to be the same across all three proofs. So for example, if HumanProof decides to label a variable ε but it is η in the textbook, ε is changed to
be η. Stylistic choices like this were arbitrated by me to be whichever choice I considered to be most conventional.
The study was originally intended to be face-to-face, showing users an on-screen Lean proof with the help of the Lean infoview to give information on the intermediate proof state. However, due to the
study being remote, the material was presented using static images within a web-form instead. This meant that all of the proofs needed to be read without the help of any software. So I wrote the Lean
proofs in a different style to the one that would be found in the Lean mathematics library. The following list gives these stylistic changes. I will discuss the validity implications of this in
Section 6.7.
• All variable introductions had their types explicitly written even though Lean infers them.
• Before and after a complex tactic I insert a show G statement with the goal being shown.
• French quotes ‹x ∈ A› to refer to proofs are used wherever possible (as opposed to using assumption or the variable name for the proof).
• When proving A ∪ B is open, one needs to perform case analysis on y ∈ A ∪ B where the reasoning for either case is very similar. While it is possible to use a wlog tactic to reduce the size of
the proof, I decided against using it because one must also provide an auxiliary proof for converting the proof on the first case to the second case.
In the textbook proof all was replaced with Lean-style pretty printed expressions. This is to make sure that the participants are not biased towards the human-made proof due to their familiarity with
-style mathematical typesetting. Another issue with the metric space proofs is that the original textbook proofs that I found were written in terms of open balls around points, but I wanted to keep
the definitions used by the proofs as similar as possible so I used ∀,∃-style arguments. Thus the parts of the textbook proofs that mention open balls were replaced with the equivalent ∀,∃
definition. This might have caused the textbook proofs to become less understandable than the original. This concern is revisited in Section 6.7.
6.3. Agenda of experiment
In this subsection I provide more detail on the design and structure of my empirical study. The session for each participant was designed to take less than an hour and be one-to-one over Zoom, a
video conferencing application. Zoom was also used to schedule and record the sessions.
6.3.1. Pre-session
Before the session starts, a time is scheduled and participants are asked to sign a consent form which may be viewed in Appendix D.4. This consent form makes sure that the participants are aware of
and accept how the data generated from the session will be used. Namely, I record their names and emails, however only the anonymised answers to the forms and some selected quotes are publicly
released. All recording data and identification data were deleted after the course of the experiment.
6.3.2. Preparatory events
Once the session starts, I greet the participant, double-check that they have signed the consent form and give them an overview of the agenda of the study. I also remind them that their mathematical
ability is not being tested.
6.3.3. Training phase
In this phase, the training document given in Appendix D.2 is sent to the participant as a PDF, and the participant is asked to share their screen with me so that I can see the training document on
their desktop. Participants were told that the purpose of the training document was to get them familiar enough with the new syntax so that they could read through a Lean proof and understand what is
being done, even if they might not be able to write such a proof.
Then I walk the participant through the training document, answering any questions they have along the way. The purpose of the training document is to get the participant used to the Lean syntax and
tactic framework.
For example, the right associativity of function/implication arrows f : P → Q → R and the curried style of function application (f p q versus f(p,q)). The final part of the training document is a
reference table of tactics. Participants are informed that this table is only for reference and that it is covered in the training phase. The training phase is designed to take around 10 to 15
The participant is informed that they can refer back to this training document at any time during the experiment, although in the name of reducing browser-tab-juggling, I would also offer to verbally
explain any aspects of the material that participants were unsure about.
6.3.4. Experiment phase
After the training phase, I ask the participant if I can start recording their screen and send them a link to a Google Form containing the material and questions for phases 2 and 3.
The first page of the form is a landing page with some checkboxes to double check that the participant is ready, has signed the consent form, and understands the format of the experiment.
The participant then proceeds to the first in a sequence of four pages. Each page contains a lemma, some necessary definitions, a set of three proofs and two sets of Likert scale radio buttons for
the participant to rate their understandability and confidence scores for each proof. The lemmas are presented in a random order, although the first lemma - the composition of two group homomorphisms
is a group homomorphism - always appears first because it is used to counter the learning effect as discussed in Section 6.2.2. The proofs are also presented in a random order on the page.
The form page given for each task is presented in Figure 6.1, each page is laid out in a vertical series of boxes, the first box contains the lemma number and the statement of the lemma to be proven.
The next box contains the definitions, written in both natural language and Lean syntax. Then come three boxes containing the proof scripts in a random order. Each proof script is presented as a PNG
image to ensure that the syntax and line-breaks render the same on all browsers and screen sizes.
Figure 6.1
Annotated overview of a task page in Google forms. The content present in the form can be read in full in Appendix D.3.
At the bottom of the page, participants were presented with two more boxes containing a set of radio buttons allowing the participant to rate each proof as shown in Figure 6.2.
Figure 6.2
A filled out form.
The title questions for these sets of buttons are "How understandable do you find each of the above proofs?" and "How confident are you that the above proofs are correct?". Each column of the radio
buttons has a phrase indicating the Likert scoring level that that column represents.
The participants were reminded that:
• They could assume that the Lean proof was valid Lean code and had been checked by Lean.
• That the confidence measure should not be about the confidence in the result itself being true but instead confidence that the proof given actually proves the result.
• The interpretation of the concept of 'understandable' is left to the participant.
Once the participant filled out these ratings, I would prompt them to give some explanation for why they had chosen these ratings - although usually they would volunteer this information without
prompting. In the case of a tie, I also asked how they would rank the tied items if forced to choose.
If I got the impression that the participant was mixing up the labels during the rating phase, I planned to interject about the label ordering. This did not occur in practice though.
6.3.5. Debriefing phase
In this final phase, the user is presented with a series of multiple choice questions. Namely;
• What is the level of education that you are currently studying? (Undergraduate 1st year, Undergraduate 2nd year, Undergraduate 3rd year, Masters student, PhD student, Post-doc and above, Other)
• Which of the below labels best describes your area of specialisation? (Pure Mathematics, Applied Mathematics, Theoretical Physics, Statistics, Computer Science, Other)
• How much experience with automated reasoning or interactive theorem proving do you have? (None, Novice, Moderate, Expert)
Finally, there is a text-box question entitled "What does it mean for a proof to be 'understandable' to you?". At this point, I would tell the participant that they could also answer verbally and I
would transcribe their answer.
After the participant has submitted the form, I thanked them and stoped recording. In some cases, the participant wanted to know more about interactive theorem proving and Lean, and in this case I
directed them towards Kevin Buzzard's Natural Numbers Game.
6.4. Results
Full results are given in Table 6.3. In the remainder of this section the proof scripts are coded as follows: L is the Lean proof; H is the natural language HumanProof-generated proof; T is the
textbook proof. Similarly, the lemmas are shortened to their number (i.e., "Lemma 2" is expressed as "2"). The 'Question ordering' and 'Ordering' columns give the order in which the participant saw
the lemmas and proofs respectively. So an ordering of 1432 means that they saw Lemma 1, then Lemma 4, then 3, then 2. And within a lemma, an ordering of HTL means that they saw the HumanProof proof
followed by textbook proof, followed by the Lean proof. The columns are always arranged in the LHT order, so the L column is always the rating that they assigned to the Lean proof. 'Unders.' and
'Conf.' are the understandability and confidence qualities, respectively. Figure 6.4 and Figure 6.5 plot these raw results as bar charts with various groupings. These results are interpreted and
analysed in Section 6.5.
Table 6.3
The full (anonymised) results table of the quantitative data collected from the experiment. Print note: a full, selectable version of this table can be found at https://www.edayers.com/thesis/
Lemma 1: composition of group homomorphisms is a group homomorphism Lemma 2: A ∪ B is open Lemma 3: kernel is normal Lemma 4: A ∩ B is open
№ Education Area ITP experience Question ordering Ordering Unders. Conf. Ordering Unders. Conf. Ordering Unders. Conf. Ordering Unders. Conf.
L H T L H T L H T L H T L H T L H T L H T L H T
1 PhD Statistics None 1324 HTL 4 5 5 5 5 5 HLT 3 2 5 4 4 5 LTH 5 5 5 5 5 5 HLT 3 5 5 5 5 5
2 PhD Physics None 1234 HLT 4 3 4 4 4 5 THL 4 4 3 2 4 4 TLH 5 4 5 5 5 5 THL 3 3 4 4 3 4
3 Undergrad Pure None 1423 THL 3 5 5 4 5 5 HTL 2 5 4 2 5 5 HTL 5 5 4 5 5 5 LTH 3 4 5 2 5 5
4 Post-doc Pure None 1423 THL 5 5 5 5 5 5 HTL 3 5 5 4 5 1 TLH 5 5 4 5 5 5 LHT 3 5 4 4 5 5
5 PhD Pure None 1324 LHT 5 4 5 5 5 5 HTL 4 4 5 5 5 5 HTL 5 4 5 5 5 5 HTL 4 5 4 5 5 5
6 Masters Pure None 1432 LHT 2 4 4 5 5 5 THL 4 4 5 5 5 5 TLH 5 4 5 5 5 5 THL 4 4 5 5 5 5
7 Post Doc Applied None 1234 TLH 4 5 5 5 5 5 LHT 3 5 4 5 5 3 LHT 5 5 5 5 5 5 LHT 3 5 4 4 5 5
8 Undergrad Pure Novice 1324 HLT 3 5 5 3 5 5 HLT 3 3 4 3 3 3 LTH 4 5 4 4 5 5 LTH 3 3 5 4 4 4
9 PhD Physics None 1432 THL 5 5 5 5 5 5 LTH 3 5 5 5 5 4 TLH 5 4 4 5 5 5 LHT 2 4 5 4 5 5
10 PhD Pure None 1243 LHT 5 4 5 5 5 5 TLH 3 4 4 5 4 4 HLT 5 5 5 5 5 5 LTH 4 5 5 5 4 4
11 Masters Pure None 1432 LHT 3 4 4 4 4 4 TLH 2 4 3 3 4 2 LTH 4 3 3 4 4 4 THL 3 5 4 3 4 4
Figure 6.4
Bar chart of the ratings grouped by lemma. That is, each row is a different lemma and for each of the values [1,5] the number of people who assigned that rating are plotted as a bar, with a different
colour for each of the proofs. The columns correspond to the two 'qualities' that the participants rated understandability and confidence.
Figure 6.5
Bar chart of the ratings grouped by proof. The same data is used as in Figure 6.4 but with separate vertical plots for each of the types.
6.5. Quantitative analysis of ratings
6.5.1. Initial observations
On all the lemmas, people on average ranked the understandability of HumanProof and Textbook proofs to be the same. On Lemma 3, people on average ranked the Lean proof as more understandable than the
HumanProof and Textbook proofs. But on Lemmas 2 and 4, they on average ranked understandability of the Lean proof to be less than the HumanProof and Textbook proofs. For participants' confidence in
the correctness of the proof, the participants were generally certain of the correctness of all of the proofs. On Lemma 2, confidence was ranked Textbook < HumanProof < Lean. On Lemma 3 the
confidence was almost always saturated at 5 for all three proofs, meaning that this result had a ceiling effect. On Lemma 4, confidence was ranked Lean < HumanProof < Textbook.
6.5.2. Likelihood of preferences
Now consider the hypotheses outlined in Section 6.1. There are not enough data to use advanced models like ordinal regression, but we can compute some simple likelihood curves for probabilities of
certain propositions of interest for a given participant.
Let's find a likelihood curve for the probability that some participant will rank proof above proof , where and are from the set . Fixing a statistical model with parameters and dataset , define the
likelihood as a function of proportional to the probability of seeing the data given parameters .
Take a pair of proofs and a fixed lemma and quality . Write to be the number of data in the dataset that evaluate to true. Hence from the data we get three numbers , and . In the case of a tie ,
let's make the assumption that the participants 'true preference' is either or , but that them reporting a tie means that the result could go either way. So let's model the proposition that for a
random mathematician being drawn from a Bernoulli distribution with parameter . For example, if the true value for was 0.4 for on Lemma 2, we would expect a new participant to rank Lean below
Textbook 40% of the time and Textbook below Lean 60% of the time. Our goal is to find a likelihood function for given the data. We can write this as
How to implement ? The assumption above tells us to break ties randomly. So that means that if there is one tie, there is a 50% chance of that being evidence for or 50% for . Hence
Plotting the normalised likelihood curves for for each lemma, quality and pair of proofs is shown in Figure 6.8.
Figure 6.8
Normalised likelihood curves for the probability that a random mathematician will rate A less then B for a given lemma and quality. Note that each plot has a different scale, what matters is where
the mass of the distribution is situated. This plot is analysed in Section 6.5.3.
Each curve has been scaled to have an area of . However each of the six plots have a different y-axis scaling to make the shapes as clear as possible. These graphs encapsulate everything that the
data can tell us about each independent . We can interpret them as telling us how to update our prior to a posterior . A curve skewed to the right for means that it is more likely that a participant
will rank above . A curve with most of the mass in the center means that participants could rank and either way.
In the confidence plot for (mid right of Figure 6.8), the horizontal line for "kernel is normal" means that the given data tell us nothing about whether users prefer HumanProof or the Textbook proof.
Consulting the raw data in Table 6.3, we can see that all 11 participants ranked these proofs equally, so the data can't tell us anything about which one they would really prefer if they had to
choose a strict ordering.
6.5.3. Testing the hypotheses
We can interpret the likelihood curves in Figure 6.8 to test the three hypotheses given in Section 6.1. Each of the hypotheses takes the form of a comparison between how the participants ranked pairs
of proof script types. The external validity of the conclusions (i.e., do the findings tell us anything about whether the hypothesis is true in general?) is considered in Section 6.7.
For the first hypothesis, I ask whether HumanProof natural language proofs are rated as more understandable than Lean proofs. The top left plot comparing the ratings of Lean to HumanProof proofs in
Figure 6.8 contains the relevant likelihoods. To convert each of these likelihood functions on in to a posterior distribution , first multiply by a prior since . In the following analysis I choose a
uniform prior .
One way to answer the hypothesis is to take each posterior and compute the area under the curve where . This will be the prior probability that the statement "HumanProof natural language proofs is
preferred" for that particular lemma. The full set of these probabilities are tabulated in Table 6.9.
1. The participants ranked HumanProof natural language proofs as more understandable than Lean proofs. (The top-left plot in Figure 6.8.) Reading this gives a different conclusion for Lemma 3Lemma 3:
the kernel of a group homomorphism is a normal subgroup (Appendix D.3.3). versus Lemmas 2 and 4Lemma 2: If A and B are open then A ∪ B is open (Appendix D.3.2). Lemma 4: If A and B are open then A ∩
B is open (Appendix D.3.4).. For Lemma 3, we can see that users actually found the Lean proof to be more understandable than the natural language proofs , we will see some hints as to why in Section
6.6.2. For Lemmas 2 and 4: and of the mass is above .
2. The participants rank HumanProof natural language proofs as less understandable than proofs from a textbook. (The middle-left plot in Figure 6.8.) For all three lemmas a roughly equal amount of
the area is either side of , suggesting that participants do not have a rating preference of textbook proofs versus HumanProof proofs for understandability. Hence there is no evidence to support
hypothesis 2 from this experiment. As will be discussed in Section 6.7, I suspect that finding evidence for hypothesis 2 requires longer, more advanced proofs.
3. The participants are more confident in the correctness of a Lean proof to the natural language-like proofs. (The top-right and bottom-right plots in Figure 6.8.) For Lemma 3: the straight line
indicates that only one participant gave a different confidence score for the proofs. As such a significant amount of the area lies on either side of , and so the evidence is inconclusive. For Lemma
2: we have an 89% chance that they are more confident in HumanProof over Lean and a 25% chance that they are more confident in Textbook over Lean. So this hypothesis resolves differently depending on
whether it is the HumanProof or Textbook proofs. Finally with Lemma 4: the numbers are 85% and 94%, so the hypothesis resolves negatively; on Lemma 4 we should expect mathematicians to be more
confident in the natural-language-like proofs, even though there is a guarantee of correctness from the Lean proof. The verbal reasons that participants gave when justifying this choice are explored
in Section 6.6.1.
Table 6.9
A table of probabilities for whether a new participant will rank for different pairs of proof scripts. As detailed in Section 6.5.2 and Section 6.5.3, these numbers are found by computing on a
uniform prior . For a fixed , and quality : is the parameter of a Bernoulli distribution modelling the probability that a particular participant will rate for .
Understandability Confidence
Lemma 2 96% 89%
p(Lean < HumanProof) Lemma 3 6% 75%
Lemma 4 99.8% 85%
Lemma 2 50% 11%
p(HumanProof < Textbook) Lemma 3 50% 50%
Lemma 4 62% 75%
Lemma 2 99.7% 25%
p(Lean < Textbook) Lemma 3 3% 75%
Lemma 4 99.95% 94%
So here, we can see that how the hypotheses resolve are dependent on the lemma in question. In order to investigate these differences, let us now turn to the verbal comments and responses that
participants gave during each experiment session.
6.6. Qualitative analysis of verbal responses
In this section, I seek answers to the questions listed in Section 6.1. That is, what do the participants interpret the word 'understandable' to mean and what are their reasons for scoring the proofs
as they did?
Schreier's textbook on qualitative content analysis [Sch12[Sch12]Schreier, MargritQualitative content analysis in practice (2012)publisher SAGE Publicationshttps://uk.sagepub.com/en-gb/eur/
qualitative-content-analysis-in-practice/book234633] was used as a reference for determining the methodology here. There are a few different techniques in sociology for qualitative analyses:
'coding', 'discourse analysis' and 'qualitative content analysis'. The basic idea in these is to determine the kinds of responses that would help answer ones research question (coding frame) and then
systematically read the transcripts and categorise responses according to this coding frame.
To analyse the verbal responses, I transcribed the recordings of each session and segmented them into sentences or sets of closely related sentences. The sentences that expressed a reason for a
preference between the proofs or a general comment about the understandability were isolated and categorised. If the same participant effectively repeated a sentence for the same Lemma, then this
would be discarded.
6.6.1. Verbal responses on understandability
In the debriefing phase, participants gave their opinions on what it means for a proof to be understandable. I have coded these responses into four categories.
• Syntax: features pertaining to the typesetting and symbols used to present the proof.
• Level of detail (LoD): features to do with the amount of detail shown or hidden from the reader, for example, whether or not associativity steps are presented in a calculation. Another example is
explicitly vs. implicitly mentioning that A ⊆ A ∪ B.
• Structure: features to do with the wider layout of the arguments and the ordering of steps in the proof. I also include here the structure of exposition. So for example, when picking a value for
introducing an exists statement, is the value fixed beforehand and then shown to be correct or is it converted to a metavariable to be assigned later?
• Signposting: features to do with indicating how the proof is going to progress without actually performing any steps of the proof, for example, explaining the intuition behind a proof or pausing
to restate the goal.
• Other: anything else.
The presence of the 'other' category means that these codings are exhaustive. But are they mutually exclusive? We can answer this by comparing the categories pairwise:
• Syntax/LoD: An overlap here would be a syntactic device which also changes the level of detail. There are a few examples of this which are supported by Lean: implicit argumentsFor example, for
the syntax representing a group product 𝑎 * 𝑏, the carrier group is implicit. and castingIn 0.5 + 1, the 1 is implicitly cast from a natural number to a rational number.. Both casting and
implicit arguments hide unnecessary detail from the user to aid understanding. However since these devices are used by both Lean and natural language proofs they shouldn't be the reason that one
is preferred over the other.
• Syntax/Structure: the larger layout of a proof document could be mediated by syntax such as begin/end blocks, but here a complaint on syntax would be a complaint against the choices of token used
to represent the syntax. If the comment is robust to changing the syntax for another set of tokens, then it is not a comment about the syntax.
• Syntax/Signposting: similarly to Syntax/Structure, signposting has syntax associated with it, but we should expect any issue with signposting to be independent of the syntax used to denote it.
• LoD/Structure: LoD is different to structure in that if a comment is still valid after adding or removing detail (e.g., steps in an equality chain or an explanation for a particular result) then
it is not a comment about LoD.
• LoD/Signposting: There is some overlap between LoD and signposting, because if a signpost is omitted, then this could be considered changing the level of detail. However I distinguish them by
stipulating that signposts are not strictly necessary to complete the proof, whereas the omitted details are necessary but are not mentioned.
• Structure/Signposting: Similarly to LoD/Signposting, signposts can be removed while remaining a valid proof.
The verbal answers to the meaning of understandability are coded and tabulated in Table 6.10.
Table 6.10
Verbal responses prompted by 'What does it mean for a proof to be understandable to you?'. Each phrase in the first column is paraphrased to be a continuation of 'A proof is understandable when...'.
A proof is understandable when ... count category
"it provides intuition or a sketch of the proof" 6 signposting
"it clearly references lemmas and justifications" 4 level of detail
"it emphasises the key difficult steps" 4 signposting
"each step is easy to check" 3 level of detail
"it is aimed at the right audience" 3 other
"it allows you to easily reconstruct the proof without remembering all of the details" 2 structure
"it hides bits that don't matter" (for example, associativity rewrites) 2 level of detail
"it explains the thought process for deriving 'inspired steps' such as choosing ε" 2 structure
"it is clear" 1 syntax, structure
"it is concise" 1 syntax, structure
"it has a balance of words and symbols" 1 syntax
Additionally, 4 participants claimed that just being easy to check does not necessarily mean that a proof is understandable. So the data show that signposting and getting the right level of detail
are the most important aspects of a proof that make it understandable. The syntax and structure of the proof mattered less in the opinion of the participants.
6.6.2. Reasons for rating proofs in terms of understandability
Now I turn to the two qualitative questions asked in Section 6.1:
1. What properties of a proof contribute to, or inhibit, understandability?
2. How does having a computer-verified proof influence a mathematician's confidence in the correctness of the proof?
In order to answer these, we need to find general reasons why respondents ranked certain proofs above others on both the understandability and confidence quality.
The main new category to introduce here is whether the participants' reasons for choosing between the proofs are an intrinsic part of the proof medium or a presentational choice that can be fixed
easily. For example, it may be preferred to skip a number of steps in an equality calculation, but it might be awkward to get this to verify in Lean in a way that might not be true for smaller
Since the numerical results for the metric space lemmas tell a different story to the group theory lemma, I separate the analysis along these lines.
For the group homomorphism question, the Lean proof was usually preferred in terms of understandability to the HumanProof and textbook proofs. The main reasons given are tabulated in Table 6.11. I
write H+ or L- and so on to categorise whether the comment is a negative or positive opinion on the given proof. So T+ means that the quote is categorised as being positive about the textbook proof.
Write = to mean that the statement applies to all of the proofs.
Table 6.11
Table of reasons participants gave for their ranking of the understandability of proofs for Lemma 3.
paraphrased quote count category judgement
1 "I would rather thatf (g * k * g⁻¹) = f g * f k * (f g)⁻¹ be performed in one step" 2 level of detail L-
2 "I would rather thatf (g * k * g⁻¹) = f g * f k * (f g)⁻¹ be performed in two steps" 2 level of detail L+
3 "I like how the HumanProof proof does not need to apply(f g)⁻¹ = f (g⁻¹)" 2 structure H+
4 "I prefer the Lean proof because it explains exactly what is happening on each step with therewrite tactics" 2 level of detail L+
5 "I don't like how none of the proofs explicitly state that provingf (g * k * g⁻¹) = e implies that the kernel is therefore normal" 2 level of detail =
6 Express a lack of difference between the proofs of Lemma 3 2 structure =
7 "The textbook proof is too hard to read because the equality chain is all placed on a single line" 2 syntax, level of detail T-
8 "In the textbook proof, I dislike how applyingf (g * k * g⁻¹) = f g * f k * (f g)⁻¹ together with (f g)⁻¹ = f (g⁻¹) is performed in one step 1 level of detail T-
9 "It is not clear in the equality chain exactly where the kernel property is used in the HumanProof proof" 1 level of detail H-
Rows 2, 4 and 7 in Table 6.11 suggest that the main reason why Lean tended to rank higher in understandability for Lemma 3 is because the syntax of Lean's calc block requires that the proof terms for
each step of the equality chain be included explicitly. Participants generally found these useful for understandability, because it meant that each line in the equality chain had an explicit reason
instead of having to infer which rewrite rule was used by comparing expressions (this is a case where the Lean proof's higher level of detail was actually a good thing rather than getting in the
way). This suggests that future versions of HumanProof should include the option to explicitly include these rewrite rules.
Rows 1 and 2 in Table 6.11 show that there is also generally disagreement on what the most understandable level of detail is in terms of the number of steps that should be omitted in the equality
Now let us turn to the metric space lemmas in Table 6.12.
Table 6.12
Table of reasons participants gave in their ranking of the understandability of proofs for Lemmas 2 and 4.
paraphrase quote count category judgement
1 "The textbook proof of Lemma 2 is too terse" 5 level of detail T-
2 "The textbook proof is what I would see in a lecture but if I was teaching I would use HumanProof" 5 structure, level of detail H+
3 Expressing shock or surprise upon seeing the Lean proof 5 level of detail L-
4 "I like the use of an explicit 'we must show' goal statement in the HumanProof proof" 3 structure H+
5 "The Lean proof includes too much detail" 3 level of detail L-
6 "In Lemma 2, the last paragraph of HumanProof is too wordy / useless" 3 level of detail, structure H-
7 "It is difficult to parse the definition ofis_open" 2 syntax =
8 "I prefer 'x ∈ A whenever dist y x < ε for all x' to '∀ x, dist y x < ε → x ∈ A'" 2 syntax =
9 "In the Lean proof, it is difficult to figure out what each tactic is doing to the goal state" 2 structure, syntax L-
10 "Lean gives too much detail to gain any intuition on the proof" 2 level of detail L-
11 "I prefer HumanProof's justification of choosing ε, generally" 2 structure H+
12 "I prefer HumanProof's justification of choosing ε, but only for the purposes of teaching" 2 structure H+
13 "I prefer the lack of justification of choosing ε" 2 structure T+
14 "It is difficult to parse the definition ofis_open" 2 syntax =
15 "I prefer '∀ x, dist y x < ε → x ∈ A' to 'x ∈ A whenever dist y x < ε for all x'" 1 syntax =
16 "Knowing that the 'similarly' phrase in Textbook proof of Lemma 2 requires intuition" 1 level of detail T
17 "Both HumanProof and textbook proofs of Lemma 4 are the same" 1 structure H=T
Here, most of the criticisms of Lean are on the large level of detail that the proof needs (rows 1, 2, 3, 5, 6, 10, 16 of Table 6.12). Now, to some extent the amount of detail included has been
inflated by my representation of the lemma to ensure that the participants can read through the proof without having an interactive window open, so this might be more of a complaint about how I have
written the proof than an intrinsic problem with Lean.
Another common talking point (rows 12, 13, 14 in Table 6.12) was the way in which HumanProof structured proofs by delaying the choice of ε to when it is clear what the value should be, rather than
the Lean and Textbook proofs which choose ε up-front. There was not agreement among the participants about which structure of proof is better. One participant noted that they preferred proofs where '
ε is pulled out of a hat and checked later'.
Row 1 of Table 6.12 shows that participants generally struggled with the textbook proof of Lemma 2 (Appendix D.3.2). This might be because the original proof was stated in terms of open ε-balls which
I replaced with a ∀∃ expression, and this change unfairly marred the understandability of the textbook proof. This was done to ensure that the same definitions were used across all versions of the
I also provide below in Table 6.13 some general remarks on the proofs that weren't specific to a particular lemma.
Table 6.13
Table of general reasons participants gave in their ranking of the understandability of proofs that apply to all of the Lemmas.
Paraphrase quote count category judgement
1 "I am finding Lean difficult to read mostly because I am not used to the syntax rather than because the underlying structure is hard to follow" 6 syntax L-
2 "The Lean proof makes sense even though I don't know the tactic syntax well" 5 syntax L+
3 "I am getting used to reading Lean", "This is much easier now that I am familiar with the syntax" 4 syntax L+
4 "Lean really focusses the mind on what you have to prove" 2 level of detail L+
5 "In HumanProof and textbook proofs I find reading the Lean-like expressions to be difficult" 1 syntax =
The trend in Table 6.13 is syntax. Many participants were keen to state that they found the Lean tactic syntax difficult to read, but they also stated that this shouldn't be a reason to discount a
proof as being difficult to understand because it is just a matter of not being used to the syntax rather than an intrinsic problem with the proof. When asked to give a reason why the Lean proof was
found to be less understandable for Lemmas 2 and 4, the reason was usually about level of detail and syntax rather than about the structure of the proof. The take-home message here is that while
newcomers are likely to find the new syntax of Lean strange, they are willing to learn it and do not see it as a problem with formal provers.
6.6.3. Verbal responses for confidence
When the participants gave a rating of their confidence in the correctness of the proofs, I reminded them that the Lean proofs had been verified by Lean's kernel. Looking at the numerical results in
the right column of Figure 6.8 and the results of Section 6.5.3, we can see that participants are more likely to be confident in the HumanProof proofs than the Lean proofs, and more likely to be
confident in the Textbook proofs over the Lean proofs (with the exception of Lemma 2). As discussed, the signal is not very strong but this suggests that knowing that a proof is formally verified
does not necessarily make mathematicians more confident that the proof is correct.
Three participants volunteered that they were less certain of Lean because they don't know how it works and it might have bugs. Meanwhile one participant, after ranking their confidence in Lean
lower, stated
"I can't tell if the reason I said that I am less confident was just an irrational suspicion or something else... I can't figure out what kind of mistake... in my mind there might exist some kind
of mistake where it would logically be fine but somehow the proof doesn't work, but I don't know if that's a real thing or not."
This suggests a counterintuitive perspective; convincing mathematicians to have more confidence in formal methods such as provers is not a problem of verifying that a given proof is correct. Instead
they need to be able to see why a proof is true.
6.6.4. Summary of verbal responses
To wrap up, in this section we have explored the evidence that the verbal responses of participants provide for the research questions laid out in Section 6.1.
What properties of a proof contribute to, or inhibit, the understandability of a proof? The most commonly given property is on signposting a proof by providing an intuition or sketch of the proof,
followed by getting the level of detail right; skipping over parts of the proof that are less important while remaining easy to check. Syntactic clarity was less important.
Does having a formal guarantee of correctness increase a mathematician's confidence in the correctness of the proof? The answer seems to be no, but perhaps this will change as the mathematicians have
more experience with an interactive theorem prover.
6.7. Threats to validity and limitations
Here, I list some of the ways in which the experiment could be invalidated. There are two kinds of threats: internal and external. An internal threat is one which causes the experiment to not measure
what it says it is measuring. An external threat is about generalisation; do the results of the experiment extend to broader claims about the whole system?
Below I list the threats to validity that I have identified for this study.
Confounding – Some other aspect of how the proofs are presented might be causing participants to be biased towards one or the other (e.g., if one uses typeset mathematics vs monospace code, or
dependent on the choice of variable names). Because of this I have changed the human-written natural language proof scripts to use Lean notation for mathematical expressions instead of . Another
defence against this threat is to simply ask the participants about why they chose to rate the proofs as they did. However, it is possible that this bias is subconscious and therefore would not be
picked up in the verbal responses ("I don't know why I prefer this one"...).
Selection bias – Participants are not drawn randomly from the population but are drawn from people who answer an advert for a study. This could cause a bias towards users who are more interested in
ITP and ATPInteractive Theorem Proving and Automated Theorem Proving, see Section 2.1.2.. I defend against this with the question in the debrief phase asking what their prior experience is with ITP
and ATP. Additionally, it is not mentioned that the HumanProof-generated proof is generated by a machine, so being biased towards ITP would cause the bias to be towards the Lean proof rather than
towards the generated proof.
Maturation/ learning effects – During the course of the experiment, the participants may become used to performing the rankings and reading the two Lean-based proofs. The training phase and throwing
away the results of the first task should help remove any effects from the participants being unfamiliar with the material. The lemmas and proofs are presented in a randomised order so at least any
learning effect bias is distributed evenly over the results.
'Cached' scoring – Similar to maturation, participants may remember their scores to previous rounds and use that as a shortcut to thinking about the new problem presented to them. This is partly the
purpose of asking the participants to explain why they ranked as they do, which should cause them to think about their rating more than relying on cached ratings.
Experimenter bias – I have a bias towards the HumanProof system because I built it. This could cause me to inadvertently behave differently when the participant is interacting with the HumanProof
proof script. In order to minimise this, I tried to keep to the experimenter script as much as possible. I have also introduced bias in how I have selected the example questions. The textbook proofs
were chosen from existing mathematical textbooks, but I perhaps introduced bias when I augmented them to bring them in line with the other two proofs. To some extent the Lean proofs are not
representative of a Lean proof that one would find 'in the wild'. I couldn't use an existing mathlibMathlib is Lean's mathematical library. proof of the same theorem because Lean proofs in mathlib
can be difficult to read without the Lean goal-state window visible and without first being familiar with the definitions and conventions which are highly specific to Lean. Although this bias may
cause participants perceptions to be changed, an impression I got during the sessions was that the participants were generally sympathetic to the Lean proofs, possibly because they assumed that I was
primarily interested in the Lean proof scripts as opposed to the HumanProof proof which was not declared to be computer generated.
Generalisation - The lemmas that have been selected are all elementary propositions from the Cambridge undergraduate mathematics course. This was necessary to make sure that the proofs could be
understood independently of the definitions and lemmas that are required to write down proofs of later results and to ensure that the participants could review the proofs quickly enough to complete
each session in under an hour. However this choice introduces a threat to external validity, in that it is not clear whether any results found should generalise to more advanced lemmas and proofs.
More experiments would need to be done to confirm whether the results and responses generalise to a more diverse range of problems and participants. Do we have a strong prior reason to expect that
participants would answer differently on more advanced lemmas and proofs? My personal prior is that they would rank Lean even worse, most notably because the result would rely on some unfamiliar
lemmas and concepts such as filters and type coercions. I also suspect that HumanProof would perform worse on more advanced proofs, since the proofs would be longer and hence a more sophisticated
natural language generation system would be needed to generate the results.
Ceiling effects - Some of the ratings have ceiling effects, where the ratings data mostly occurs around the extrema of the scale. This can be seen to occur for the understandability and confidence
scores for Lemma 3.
Sample size - There were only 11 participants in the study. This low sample size manifests in the results as the fat likelihood curves in Figure 6.8. However, the study was designed to be fairly
informal, with the quantitative test's purpose mainly to spark discussion, so it is not clear to me that any additional insight would be gained from increasing the study size.
6.8. Conclusions and future work
In this chapter I have evaluated the natural language write-up component of HumanProof by asking mathematicians to compare it against Lean proofs and proofs derived from textbooks. The results give
different answers for equational reasoning proofs as opposed to the more structural natural language proofs and Lean proofs. This suggests that we should come to two conclusions from this experiment.
In the case of structural proofs, users generally prefer HumanProof and textbook proofs to Lean proofs for understandability. The participants reported that to some extent it was because they found
the natural language proofs more familiar, but also because the Lean proofs were considered too detailed.
However the qualitative, verbal results showed (Section 6.6.2) that participants were usually sympathetic towards the Lean structure, only wishing that the proofs be a little more terse, provide
signposts and hide unnecessary details.
In the wider sense, we can use the results from this study to see that mathematicians are willing to move to different syntax, but that the high-level structure and level of detail of formalised
proofs needs to be improved in order for them to adopt ITP. The study confirms that producing natural language proofs at a more human-like level of detail is useful for mathematicians. It also
provides some surprising complaints about how proofs are typically laid out in textbooks where a more formal and detailed approach would actually be preferred, as we saw for the group homomorphisms
question, where the Textbook version of the proof was considered to be too terse.
Within the wider scope of the thesis and the research questions in Section 1.2, the study presented in this chapter seeks to determine whether software can produce formalized, understandable proofs.
The study shows that HumanProof system developed in Chapter 3 and Chapter 4 can help with understandability (as shown in Section 6.5 for the case of Lemmas 2 and 4).
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Laminar to turbulent transition in terms of information theory
In the present study we investigate turbulence using information theory. We propose a methodology to study mutual information, entropy and complexity utilizing data from direct numerical simulations.
These quantities are defined in their continuous form and estimated using the Kozachenko-Leonenko and Kraskov-Stogbauer-Grassberger estimators, for entropy and mutual information, respectively. To
validate our tools and methodology, we compare the entropy of three direct numerical simulations of forced isotropic turbulence at different Reynolds numbers (Re lambda = 433, Re lambda = 648 y Re
lambda = 1300) with results previously reported in the literature. The main body of the study is the analysis of the transition to turbulence in a transitional boundary layer, our results show that
the mutual information between the velocity field and temporal velocity increments reaches a maximum close to the transition to turbulence followed by a decrease during the transition, while at the
same time there is a change in entropy due to both, the change in variance of the velocity field and the deformation of the probability density function. Finally, we examine the vertical profiles of
the information-theoretic quantities in a turbulent channel flow (Re tau = 5200). Here we study spatial velocity increments and observe an increase of the entropy due to the deformation of the
probability density function of the velocity field, as it becomes more Gaussian away from the wall. (c) 2023 Elsevier B.V. All rights reserved.
Más información
Título según WOS: ID WOS:001086267600001 Not found in local WOS DB
Título según SCOPUS: ID SCOPUS_ID:85172175039 Not found in local SCOPUS DB
Título de la Revista: PHYSICA A-STATISTICAL MECHANICS AND ITS APPLICATIONS
Volumen: 629
Fecha de publicación: 2023
DOI: 10.1016/J.PHYSA.2023.129190
Notas: ISI, SCOPUS - WOS/ISI
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FE Symp
Monday, September 9, 2019
09:00 Opening
Scientific Computing
Chair: Arnd Rösch
09:05 Stefan Turek
09:55 Jan Philipp Thiele
10:20 Coffee Break
Scientific Computing Space Time
Chair: Stefan Turek Chair: Steffen Münzenmaier
10:50 Mathias Anselmann Ulrich Langer
11:15 Aurora Ferrja Julia Hauser
11:40 Roland Herzog Gunar Matthies
12:05 Max Winkler Douglas Ramalho Queiroz Pacheco
12:30 Lunch
Preconditioning A Posteriori
Chair: Ulrich Langer Chair: Gerhard Starke
13:45 Jan Blechta Ullrich Heupel
14:10 Stephan Köhler Andreas Rademacher
14:35 Friederike Röver Kemal Suntay
15:00 Coffee Break
Fractional and Nonlinear PDEs Optimal Control
Chair: Gundolf Haase Chair: Roland Herzog
15:30 Nabi Chegini Livia Betz
15:55 Mohadese Ramezani Arnd Rösch
16:20 Reza Mokhtari Monika Weymuth
16:45 Break
Applications Optimal Control
Chair: Andreas Rademacher Chair: Thomas Apel
17:00 Gonzalo G. de Diego Jens Baumgartner
17:25 Aurora Ferrja Fernando Gaspoz
17:50 Lucas Schöbel-Kröhn Marita Holtmannspötter
18:15 Hendrik Pasing Huidong Yang
19:00 Conference Dinner
Tuesday, September 10, 2019
Computational Mechanics
Chair: Oliver Rheinbach
09:00 Philipp Junker
09:50 Marco Favino
10:15 Coffee Break
Computational Mechanics Numerical Analysis
Chair: Philipp Junker Chair: Gunar Matthies
10:45 Bernhard Kober Philipp Morgenstern
11:10 Marcel Moldenhauer Johannes Riesselmann
11:35 Lisa Julia Nebel Michael Sievers
12:00 Korinna Rosin Timo Sprekeler
12:30 Lunch
14:00 Excursion
19:30 Meeting of the Scientific Committee
Wednesday, September 11, 2019
Least-Squares Finite Element Method
Chair: Sven Beuchler
09:00 Fleurianne Bertrand
09:50 Maximilian Bernkopf
10:15 Coffee Break
Least-Squares Finite Element Method and High Order
Chair: Fleurianne Bertrand
10:45 Solveigh Averweg
11:10 Steffen Münzenmaier
11:35 Tim Haubold
12:00 Gozel Judakova
12:25 Closing
12:30 Lunch
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Ow an increase, just by random variation. The purpose of the | PDGFR inhibitor-pdgfr.com
Ow a rise, just by random variation. The objective of the proposed model-based strategy would be to define where to draw the line to define a considerable enhance, and to adjust for the
multiplicities.Biom J. Author manuscript; readily available in PMC 2014 May 01.Le -Novelo et al.Page3 The Selection ProblemThe proposed strategy to pick peptide/tissue pairs for reporting is
independent in the underlying probability model. It is according to a formalization with the inference problem as a selection difficulty having a specific utility function. The certain probability
model only adjustments the distribution with respect to which we compute posterior expected utilities. The only assumptions that we need in the upcoming discussion are that the model contains
parameters ” 0, 1 that may be interpreted as indicators for rising imply counts of i peptide/tissue pair i across the three stages. Recall that i = p(= 1 | y) denotes the posterior i probabilities.
We also assume that the model involves parameters … that can be i” interpreted as the extent with the enhance, with = I(… 0). We use mE(…y) for the i i i= i| marginal posterior suggests. We currently
introduced d – (1) as a reasonable choice rule to select peptide/tissue pairs in for reporting as preferentially binding. Rule d – be justified as control with the false can discovery price (FDR)
(Newton, 2004) or, alternatively, as an optimal Bayes rule. To define an optimal rule we ought to augment the probability model to a decision issue by introducing a utility function. Let , and y
generically denote all unknown parameters and all observable information. A utility function u(d, , , y) formalizes relative preferences for decision d beneath hypothetical outcomes y and beneath an
assumed truth , . For example, in our application a utility function could beNIH-PA Author Manuscript NIH-PA Author Manuscript NIH-PA Author Manuscript(2)i.e., a linear combination on the quantity of
true good selections di and correct negatives. To get a provided probability model, information and utility function, the optimal Bayes rule is defined because the rule that maximizes u in
expectation more than all not observed variable, and conditional on all observed variables,(3)in It could be shown that the rule d – (1) arises as Bayes rule under many utility functions that trade
off false constructive and false adverse counts, including the utility in (two) and other individuals.Rosin Inhibitor See, as an example, M ler et al.NH125 Autophagy (2007), for any discussion.
Alternatively, d – be derived as FDR control. Recall the posterior expected FDR, can(four)Similarly, the posterior expected false adverse rate (FNR) is usually computed as .PMID:35116795 It can be
simply noticed that the pairs selected by d – report the biggest list for any given worth of posterior anticipated FDR. Characterizing d – the Bayes rule (3) under (two) highlights a crucial
limitation with the rule. because the utility function (2) weights every correct constructive, or equivalently, just about every false damaging, equally. Recall that we assume that the model contains
a parameter … is often interpreted i that as the strength of a correct comparison, i.e., in our application, because the amount of preferential binding with the i-th peptide/tissue pair. A correct
good with tiny … is unlikely to bring about i thatBiom J. Author manuscript; obtainable in PMC 2014 May 01.Le -Novelo et al.Pageany meaningful follow-up experiments is of far less interest for the
investigator than a accurate optimistic with massively significant …Equivalently, a false negative, i.e., missing to report a definitely i. preferentially binding tripep.
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Problem 1 – Hint 3
Let $x$ be the height of the cylinder from the bottom of the vessel; you should obtain an expression for the kinetic energy in the form $T=a\frac {\dot x^2}x$, where $\dot x=\frac {\mathrm dx}{\
mathrm dt}$, and $a$ is a constant. Now there are two ways to proceed. First, you can express $\mathrm dt$ from the energy conservation law $T+K=E_0$ in terms of $\mathrm dx$ and integrate. Second
and mathematically simpler way is to substitute $x=\xi^2$ to obtain a nice equation for $\ddot \xi$.
Please submit the solution to this problem via e-mail to physcs.cup@gmail.com. The next hint for the Problem 1 will be published alongside with the Problem No 2 at 13:00 GMT, 18th December 2022,
together with the next updates of the intermediate results.
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How to Read a Ruler (and Other Simple Tricks).
Introduction: How to Read a Ruler (and Other Simple Tricks).
While it may seem to be a very basic skill, being able to read a ruler is the foundation to just about any project you make by hand or even with a Shopbot! Reading a metric ruler is pretty simple- no
fractions, everything converts nicely in factors of 10, and its pretty straight forward. The English system, however, can be kind of confusing- fractions, units, and symbols. This instructable will
help you understand how to use a "standard" ruler better; specifically being able to read fractions of an inch or "Drawing the Inch" as I call it with my students.
The cool thing about knowing how to label the fractions inside of an inch is that you can use it as a calculator to reduce fractions! Follow along and I'll show you how. This instructable may be a
little hard to follow if you don't read it through all the way to the end before trying it out.
Step 1: Start Your Inches!
Gather supplies. See Picture 1. You need:
Pencil or Pen
That's it!
A few definitions that might help you understand this instructable a little better:
Numerator: The number above the line in a common fraction
Denominator: The number below the line in a common fraction
The easiest way I have found to explain an inch is to draw it out, starting with a mark representing zero and a mark representing 1. These DO NOT HAVE TO BE ACCURATE!!! I find it is a lot easier if
you make your "inch" BIG to give you more room to write in the fractions. SO:
On the paper, Draw a rectangle as shown in Picture 2 and 3. This is going to be our "ruler". Remember to make it big! On the ruler, make a mark, also as shown in Pictures 2, 3, and 4. This will be
out 1 inch mark.
Step 2: Cut & Double- Halves
Now that you have your inch started, we can "Cut and Double". Inside of the inch mark you have drawn, put another mark in the middle, cutting the inch in half as shown in Picture 1. Most people
understand that this is a half an inch, or 1/2, as shown in Picture 2 and 3.
To explain this further, lets talk about the "Cut and Double" for a minute. When we marked the 1/2 inch on our ruler, we CUT the inch in half, and DOUBLED the denominator of the previous fraction.
See Picture 2. For example, we cut the inch in half and made a mark. Then we took the previous fraction- ONE INCH which converted to a fraction with a numerator and denominator is 1/1- and DOUBLED
the denominator. 2 x 1 = 2, so our new fraction will be 1/2. See Pictures 3 and 4. Confused? Maybe a little... Lets do it one more time in the next step and see if we start to catch on.
Step 3: Cut & Double- Quarters
If you understood the last step, the rest is simple. Just repeat it as many times as you want!
CUT again! For each section of your inch, cut it in half as shown in Picture 1. Notice there are TWO marks now instead of one, because we have TWO sections- one on each side of the 1/2 inch mark.
DOUBLE again! The denominator of the last fraction is 2, so 2 x 2 = 4 as shown in Picture 2. That means our new fraction is going to be 1/4 as shown in Picture 3.
Here's where it gets a little different. There are TWO unlabeled marks on the ruler. We are going to count them off in quarters- the first mark being 1/4, the second mark would be 2/4, but it's
already labeled as 1/2, and the 3rd mark will be labeled as 3/4. Count by 1/4'ths- 1/4, 2/4, 3/4, 4/4. See Picture 4.
Step 4: Cut & Double- Eighths
CUT again! For each section of your inch, cut it in half as shown in Picture 1. Notice there are FOUR marks now instead of 2, because we have FOUR sections- one on each side of the 1/4 inch marks.
DOUBLE again! The denominator of the last fraction is 4, so 2 x 4 = 8 as shown in Picture 2. That means our new fraction is going to be 1/8 as shown in Picture 3.
Fill in the missing fractions by counting each mark as an eighth- 1/8, 2/8, 3/8, 4/8, 5/8, 6/8, 7/8, and 8/8. You will ONLY LABEL THE FRACTIONS WITH AN ODD NUMERATOR! See Picture 4.
Step 5: Cut & Double- Sixteenths
CUT again! For each section of your inch, cut it in half as shown in Picture 1. Notice there are EIGHT marks now instead of 4, because we have EIGHT sections- one on each side of the 1/8 inch marks.
DOUBLE again! The denominator of the last fraction is 8, so 2 x 8 = 16 as shown in Picture 2. That means our new fraction is going to be 1/16 as shown in Picture 3.
Fill in the missing fractions by counting each mark as an eighth- 1/16, 2/16, 3/16, 4/16, 5/16, 6/16, 7/16, 8/16, 9/16, 10/16, 11/16, 12/16, 13/16, 14/16, 15/16, and 16/16. You will ONLY LABEL THE
FRACTIONS WITH AN ODD NUMERATOR! See Picture 4.
Step 6: Tips, Tricks, and Continuing On.
You've Drawn your Inch! Here's what your completed Inch should look like- see
Pictures 1 and 4
. Now lets show you a couple of patterns and give you some tips and tricks!
Trick 1
Take a minute and look at the fractions. Do you see any patterns? There are two that stand out that can come in handy to check your work to make sure you drew your inch correctly...
1. Look at picture one again. What do you notice about ALL of the numerators?! THEY ARE ALL ODD. If you have an even number as a numerator, it needs to be reduced or you haven't got it in the right
2. Look at the last fraction in each set as shown in
Picture 2
. Notice that in each fraction, the numerator is ONE LESS than the denominator!
Trick 2
You can use your completed inch as a calculator for reducing fractions. If you were to write ALL of the fractions down every time you did a set, your Inch would look like
Picture 3
. Each mark on the ruler that ends up with multiple fractions can be reduced to the top most fraction in the set!
Trick 3
Continuing on! You can continue the Cut and Double forever! Each time just split the last section in half and double the denominator of the last fraction. After 16ths, you would have 32nds, 64ths,
128ths, 256ths, 512ths, 1024ths, 2048ths, and on and on and on... But it does get a little hard to draw. :)
I'm sure there are a lot more cool things you can do with a ruler, and I'd love to hear about them! In the meantime, the best way to get good at this is to practice practice practice! After doing
this about 20 times, my students can just about do it in their sleep- and their ability to read a ruler shows in their metals projects.
Once you have your Inch down, you can try some of my other projects. Making a Perfect Paper Cube is great practice at reading and using a ruler! See it here:
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ACM Other ConferencesA Polynomial Kernel for Multicut in Trees
The {\sc Multicut In Trees} problem consists in deciding, given a tree, a set of requests (i.e. paths in the tree) and an integer $k$, whether there exists a set of $k$ edges cutting all the
requests. This problem was shown to be FPT by Guo and Niedermeyer (2005). They also provided an exponential kernel. They asked whether this problem has a polynomial kernel. This question was also
raised by Fellows (2006).
We show that {\sc Multicut In Trees} has a polynomial kernel.
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On Fermi Calculations
Big Falling Fermi Confetti
Take a piece of standard paper. Rip it up, roughly, into an 8 by 8 grid, which will net you 64 pieces of over-sized confetti. Hold them in your cupped hands at chest height, and let them fall to the
What can you learn from doing this? Well, for one thing, it can help you estimate the strength of Trinity, the first atom bomb test explosion.
No, really, it can.
But I’ve left out two things. You have to have been at the observing site at the time when Trinity was actually detonated. And you have to be Enrico Fermi.
Fermi was born in 1901 in Rome. His was the trajectory common for physics geniuses. Graduated top of his high school, took first place in his university entrance exam, was acknowledged to be
“unteachable” by his physics prof (not that he was dumb; there was just nothing new to teach him, as he already knew it all), and so on. When 19 he started work on x-ray crystallography, and in his
third year was already publishing papers. By the age of 24 he was appointed professor at Sapienza University in Rome. He won the Nobel Prize in Physics at the age of 37.
Brilliant in both theoretical and practical physics, Fermi was one of the early experimenters in nuclear fission, and designed Chicago Pile-1, the first reactor to achieve a self-sustaining chain
reaction. He was not slow to see the dangers of this technology, and advised the US military on its potential uses. Eventually he worked on the atom bomb at Los Alamos, and observed the Trinity Test
on 16 July, 1945.
It was here he did his famous paper test: by pacing off how far the paper confetti was blown by the blast wave, guessing the weight of each piece of paper, and doing a few back-of-the-envelope
calculations, he estimated the yield as 10 kilotons of TNT. The currently accepted value is between 18 and 19 kilotons.
That in fact is astonishingly accurate, given what he had to work with, which was basically nothing. Don’t forget, this was the first-ever blast, and this was the age when the word “computer” meant a
human being who did calculations as a member of a team, with pencil and paper or a mechanical adder. (Although it must be said there were a very few electronic computers at the time, and the Los
Alamos project did use them.)
Two Wrongs Can Make A Right
Throughout his career, Fermi was noted for his ability to produce reasonable guesstimates in situations in which he had little or no solid data. LIke the confetti test. His basic method was to break
a problem down into many constituent parts, and estimate values for each. By adopting this strategy, a thinker benefits in two ways:
First, the component problems frequently can be estimated with more accuracy than just a single shot at the larger whole. If you have no idea of what the Trinity yield might be, you can guess the
weight and surface area of your confetti, pace off the distance, know how far you are from the blast, and so on.
Second, in the absence of any systematic bias, over- and under-estimates of all components tend to cancel each other out. In curious contradiction to the adage “two wrongs don’t make a right”
(stolen, I would admit, from the world of home spun ethics, not physics), you can home in on the correct value (with a bit of luck).
Fermi became so good at this technique, that today it bears his name. We call them Fermi Calculations or Fermi Problems.
Let’s take an example.
VWs Sold In Germany
A while back I was in Frankfurt, visiting a very good friend, and somehow I happened to mention Fermi Calculations. She hadn’t heard of them before, I explained, and she immediately wanted to have a
go. So we set ourselves the problem of trying to calculate how many Volkswagens are sold every year in Germany. We were in my VW Tiguan at the the time, so had no decent access to the Internet. Thus,
we were forced to guess every step of the way.
There are many ways of approaching such a task; we tried a top-down strategy, starting with the overall population of Germany. As I cannot remember exactly how our guesses went, I can only supply a
rough reconstruction of our steps, which appears below. Real answers, since researched where possible via Wolfram|Alpha, Wikipedia and other web sites, follow our estimates, in brackets:
1. How many people live in Germany? We guessed 70 million. (81.6 million is the current 2014 estimate, so we were significantly low on that one.)
2. What is the ratio of cars per person? We broke this down a little bit. We reckoned most single people own a car. Maybe close to 50% of families own one car, the other near 50% own two, and only a
negligible number of families own more than two cars. Perhaps we should have attempted to estimate the ratio of single people to families, but in the end we simply guessed one car for every two
people, which would be about 35 million cars in Germany, based on our 70 million population estimate.
3. But of course this leaves out a lot of business vehicles. VW don’t actually do fully-fledged trucks, but they do vans and pickups. There are far fewer of these than cars, and we didn’t see any
easy way to break this estimate down any further. So we threw in 5 million more vans and pickups, to arrive at a total number of 40 million vehicles. (True values: cars 46.6 million, vans and
pickups 2.6, for a total of 49.2 million vehicles. So we were low on cars, high on vans, and low on the overall total).
4. Assuming the total number of vehicles in Germany stays roughly the same year in, year out, we decided to have a wild stab at guessing the average vehicle life-expectancy. I recall we guessed this
was 10 years (which seems high to me now, but I do believe that was the figure we used). So in order to replace those dying vehicles, 4 million new cars need to be sold each year. (According to
Statista, the number of vehicle sales in Germany in 2015 was 3.5 million http://www.statista.com/statistics/265951/vehicle-sales-in-germany/. So we were pretty close, but a bit high.)
5. Guessing the proportion of VWs to other brands in Germany was not an easy task. The country would appear to be a nation of nationalistic car-buyers, with German cars far outweighing anything
else. In some place comparatively rich like Frankfurt (where we were), you see a lot of higher-end Mercedes, BMWs and Audis. In less-wealthy areas, cheaper cars predominate, and VWs really come
into their own. So we guessed VWs made up one fifth of the total car population, nationwide. So, in order to keep VW’s market share roughly stable over time, they have to sell one-fifth of our 4
million new cars estimate, or 800,000 vehicles per annum. And that was our final guess.
6. The web site Best Selling Cars (http://www.best-selling-cars.com/germany/2014-full-year-germany-best-selling-car-manufactures-brands/) states that the Volkswagen brand had 656,494 vehicle
registrations in 2014.
So we got pretty close. As I stated earlier, I cannot remember exactly each estimate we came up with in every step along the way (the exercise occurred over a year ago), but this does present a fair
To be clear, I am not trying to claim we are brilliant estimators; quite the opposite. What is interesting about the exercise is it reveals how breaking the problem down genuinely does make it
tractable. Without the breakdown, I would have had no idea how many cars VW sold in a year. Any estimate would have been a complete stab in the dark. On the other hand, I had a certain amount of
confidence that each of our individual steps were at least reasonable. Add up a bunch of reasonables, and, with a bit of luck benefit from under- and over-estimates canceling each other out, you end
up at a new reasonable.
There’s More Than One Way to Skin An Estimate
In fact, before we were able to check our first calculation via the Internet, we decided to attack the problem in a completely different way. Our first approach was clearly top-down, starting with
the population of Germany. We tried bottom-up: My friend lives in Bad Homburg, which boasts one VW dealer, albeit fairly large. I had recently bought the car we were driving around in, in
Albertville, France, from a dealer about the same size. I had chanced then to ask him how many cars he sold in a year, and he told me. We used that as an average number of cars sold per normal-sized
Although it is true the French don’t have nearly the proportion of German cars on their roads (they too, are nationalistic in their buying decisions), we decided a dealership count would adjust for
this difference all by itself: Germany would have more German car dealers than France. So we then guessed the number of towns in Germany of similar in size to Bad Homburg. We thought for a while
about how you could adjust this to arrive at an overall national VW dealership count by taking into consideration the many-fewer-but-larger towns, and the many-more-but-smaller towns in the country.
By multiplying our estimate of cars sold per dealership by dealerships per town by towns in the country (adjusting as we went), we eventually arrived at a figure that happened to be pretty close to
our first estimate of 800,000.
I have to admit it is very possible, even probable, that we were unconsciously adjusting our intermediate calculations so we arrived at an agreement with our first estimate. Daniel Kahneman would
call it a kind of constructive confirmation bias. Having said that, at the time we did our second calculation, we hadn’t yet confirmed the first. So at least, if we were unconsciously cheating, we
were favouring a figure we didn’t yet know was fairly accurate.
But (assuming the current reader cares to give us the benefit of the doubt), it is gratifying that both methods got us pretty close to the true answer. Once again, a bunch of small reasonables can
add up to one large reasonable.
Yeah, But It Is Easier to Look It Up….
Of course in this web-enabled age, it is much easier — not to mention more accurate — simply to look things up. Why take the stairs when there’s an escalator?
Curiously, the answer to this question is pretty much the same for Fermi Calculations as it is for getting around in an airport or shopping mall: it is good mental exercise, it is challenging and
educational to break a problem down, and it is fun.
Another point, no doubt already evident to the current reader, is that the whole process is immensely satisfying when you get it even near right. We were both beaming when we looked up the true
answer and discovered we’d come pretty close.
And finally: we do all have the odd, occasional (hopefully non-nuclear) Trinity moments, when a question arises that simply can’t be answered by Wolfram|Alpha or a Google search. Who ya gonna call
when even the Web doesn’t know? Your Fermi calculation skills. Because,
There are more things in heaven and earth, Horatio,
Than are dreamt of in your Internet.
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The sum of three numbers in AP is 12 and the sum of their cubes is 288. Find the numbers.
Hint: Take the second term of AP as a and common difference as d. Write three terms of AP and form equations relating terms of AP based on the data given in the question. Solve those equations to
find the value of variables a and d. Substitute the value of variables to get the terms of AP.
Complete step-by-step answer:
We have three numbers in AP such that the sum of the numbers is 12 and the sum of cubes of the numbers is 288. We have to calculate the three numbers.
Let’s assume that the second term of the AP is a and the common difference of the terms is d.
Thus, we can write the other two terms as \[a-d\] and \[a+d\].
We know that the sum of these three numbers is 12. Thus, we have \[a-d+a+a+d=12\].
& \Rightarrow 3a=12 \\
& \Rightarrow a=4 \\
We have to now find the value of common difference d.
We know that the sum of cubes of the numbers is 288. Thus, we have \[{{\left( a-d
\right)}^{3}}+{{a}^{3}}+{{\left( a+d \right)}^{3}}=288\].
We know that \[{{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}\] and
\[{{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}\].
Thus, we have
Simplifying the above expression, we have \[3{{a}^{3}}+6a{{d}^{2}}=288\].
Further simplifying the equation, we have \[{{a}^{3}}+2a{{d}^{2}}=96\].
Substituting the value \[a=4\] in the above equation, we have \[{{\left( 4 \right)}^{3}}+2\left(
4 \right){{d}^{2}}=96\].
& \Rightarrow 64+8{{d}^{2}}=96 \\
& \Rightarrow 8{{d}^{2}}=32 \\
& \Rightarrow {{d}^{2}}=4 \\
& \Rightarrow d=\pm 2 \\
Substituting the value \[a=4,d=\pm 2\], we have the terms of our AP as 6,4,2 or 2,4,6.
Hence, the terms of AP are 2,4,6 or 6,4,2, which is option (a).
Note: Arithmetic Progression is a sequence of numbers such that the difference between any two consecutive terms is a constant. One need not worry about getting two values of common difference and
first term as they simply represent an increasing AP and a decreasing AP. We should also be careful while expanding the cubic power of equations.
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Rick's Ramblings
WORDLE for TYROS
Tyro is a bit of crosswordese that means beginner or novice. Writing this reminds me of my first WORDLE in which I failed to guess TACIT in six tries. A tweet related to this puzzle which found its
way into Rex Parker’s NYTimes Xword blog said something like the following: The answer reminds me of why I don’t do crosswords they are done by old people writing old words into the grid.
Turning to the main subject, as most of you probably know in WORDLE you get six tries to guess a five-letter word. On each turn you guess a five-letter word, a rule which prevents you from guessing
say AEIOU to find out what vowels are present. If a letter is in the correct location it shows green. If it is in the puzzle but not in the right place then it is white. If it is not in the answer
it is gray. (Colors may vary) A copy of a computer key board on the screen allows you to enter you guesses and shows the status of each letter you have guessed.
As I start to give my advice I must admit I am still a novice but that never stopped TRUMP from pontificating on how to be president. In thinking about how to play WORDLE it is useful to know how
frequently letters are used in the English language.
When Samuel Morse wanted to figure this out in the 1800s, he looked at the frequency of letters in sets of printers type which he found to be (numbers in thousands) E (12), T (9), A, E, I, O, S (8),
H (6.4), R (6.2), D(4.4), L (4), U (3.4), C,M (3), etc. With computers and electronic dictionaries at our disposal we have a more precise idea (numbers are percentages).
E: 11.16 A: 8.50 R: 7.58 I: 7.55 O: 7.16 41.95
T: 6.95 N: 6.65 S: 5.74 L: 5.49 C: 4.54 + 29.73 = 71.68
U: 3.63 D: 3.38 P: 3.17 M: 3.01 H: 3.00 + 16.19 = 87.87
G: 2.47 B: 2.07 F: 1.81 Y: 1.78 W: 1.29 9.42
K: 1.102 V: 1.007 X: 0.290 Z: 0.272 J,Q: 0.196 2.93
Here the numbers in the last column are the sum of the numbers on the row and we have made 26 divisible by 5 by putting J and Q which have the same frequency to 3 significant figures into the same
entry. This table become somewhat irrelevant once you visit
to find the letter frequencies in five letter words.
A: 10.5 E: 10.0 R: 7.2 O: 6.6 I: 6.1 40.4
S: 5.6 T: 5.6 L: 5.6 N: 5.2 U: 4.4 + 26.4 = 66.8
Y: 3.6 C: 3.6 D: 3.3 H: 3.1 M: 3.1 + 16.7 = 83.5
P: 3.0 B: 2.7 G: 2.6 K: 2.1 W: 1.6 12.0
F: 1.6 V: 1.1 Z: 0.6 X,J: 0.4 Q: 0.2 4.3
Here E has fallen from the #1 spot. However, with the exception of Y climbing from 19^th to 11^th and P dropping from 13^th to 16^th it doesn’t seriously change the rankings, so I am not going to
change my blog post due to this late breaking information.
The next thing to decide about WORDLE is what is your definition of success. I think of the game as being like a par-5 in golf. To take the analogy to a ridiculous extreme you can think of the game
as par-5 in a tournament which uses the modified Stableford scoring system (like the Barracuda Open played at a course next to Lake Tahoe). Double bogey or worse (= not solving the puzzle) is -3,
bogey (six guesses) -1, par (five) 0, birdie (four) 2, eagle (three) 5, and double eagle (two) 8 points.
I am not one who is good at brilliant guesses, so my personal metric is to maximize the probability of solving the puzzle. Hence I follow the approach of Zach Johnson who won the 2007 Masters by
“laying up” on each par five. Most of these holes are reachable in two (for the pros) but 13 and 15 have water nearby so trying to hit the green in two and putting your ball in th water can lead to a
bogey or worse. Zach hit his second shots to within 80-100 yards of the green so he could use his wedge to hit the ball close and make old school birdie.
My implementation of his strategy is to start with TRAIL, NODES, and CHUMP which covers all five traditional vowels and has 15 most frequent letters. The expected number of letters in the word this
uncovers is (to use the five letter word frequencies) is 0.835 x 5 = 4.175 if all five letters in the word are different. (Recall from elementary probability that if X[i] is the indicator of the
event that the letter appear among the first 15 in frequency then E(X[1] + … + X[5]) = 5EX[1] Dividing by 5 shows that the expected number of letters in the right position is 0.835 (assuming again
all letters are different), so on the average we expect a green and three yellos..
Of course the answer can have repeated letters and can be chosen by the puzzle creator to be unusual, e.g., EPOXY or FORAY which were recent answers. (It is now April 8). In several cases my first
three guesses have produced only 2 letters in the word, which makes the birdie putt very difficult. Even when one has four letters, as in _OUND, possibilities are bound, found, mound, pound, round,
sound, wound, even though some of these are eliminated if they are in the first 15 guessed.
If there are three (or more) possibilities for the one unknown letter, then it can be sensible to use a turn to see which of these are possible in order to get the answer in two more guesses rather
than three. Or you can be like Tiger one year at Augusta and “go for it all.” give your birdie putt on the 15^th hole a good hard rap and watch it roll off the green into the creek. Fortunately for
him, the rules of golf allowed him to play his next shot from the previous position.
These rules I have described are just to give you a start at finding a better strategy. You should choose your own three words not only to feel good about having done it yourself, but because the
order of the letters can influence the probability of success. Of course you can also choose only to guess two (or only one) and then make your guess based on the result. When I get several letters
on the first two guesses, I have often substituted another word for CHUMP to get to the solution faster but I have often regretted that. On the otherhand sometimes when I play CHUMP I am disappointed
to get no new positive information about what is in the word
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Get base10 mantissa and exponent of real (eg Float64)
What would be the recommended way of getting the base 10 mantissa and exponent of a floating point number?
Is there a package that does this? Or should I just reparse from printed representation, as in
using Printf
function get_mantissa_exponent(x)
str = @sprintf "%.15e" x # specialize 15 based on type
_mantissa, exponent = split(str, 'e')
(; mantissa = filter(!isequal('.'), _mantissa),
exponent = parse(Int, exponent))
julia> get_mantissa_exponent(3.17)
(mantissa = "3170000000000000", exponent = 0)
1 Like
It depends on what you mean by “the” base-10 mantissa and exponent. I assume you are mainly interested in binary floating-point types like Float64 and Float32.
The one that is shown in the default printed representation of a Float64 is only an approximation — it is the shortest decimal representation that rounds to the same Float64 value. If this is what
you want, I’m not aware of a documented API other than parsing the printed representation. If you are willing to use an undocumented API, there is mantissa, exponent = Base.Ryu.reduce_shortest(x).
Of course, there is an exact decimal representation too, but it typically requires a large number of digits.
5 Likes
Thanks! Sorry for not being clear, yes, I am after the printed approximation.
Is this the result you expected? Based on the title, I would have expected mantissa 317 and exponent -2 (or an equivalent alternative like 3170 and -3) because 317 * 10^-2 = 3.17.
Coincidentally, I work on JuliaMath/Decimals.jl these days. You could use that package to parse a number and get the mantissa and exponent from the resulting Decimal:
using Decimals
x = parse(Decimal, "3.17")
mantissa = x.c # 317
exponent = x.q # -2
The package has some bugs (parsing has been fixed recently in a pending PR), but I am trying to improve it and it might already be sufficient for your case now.
2 Likes
Technically you are right, but I am after a string which I can just cut off at some point and insert a decimal dot, so I am flexible.
The point of the exercise is for printing, into LaTeX. That is, I want to generate output like raw"$4.92 \cdot 10^{-3}$" and raw"$22.71 \cdot 10^{6}$" (engineering notation).
How about a simplistic regex?
julia> replace(string((10rand())^10), r"(\d+.\d+)[ef]\+?(-?\d+)"=>s"$\1 \\cdot 10^{\2}$") |> println
$6.71626435914464 \cdot 10^{6}$
5 Likes
Linking related thread.
This works for positive floating point number. Just check if number is negative and manually insert a negative sign yourself.
julia> a = log(10, Float64(Ď€) )
julia> println("$(10.0^(a-trunc(a))) * 10^$(Int64(trunc(a)))")
3.1415926535897927 * 10^0
julia> a = log(10, 123.456*Float64(Ď€) )
julia> println("$(10.0^(a-trunc(a))) * 10^$(Int64(trunc(a)))")
3.8784846264158133 * 10^2
julia> 123.456*Float64(Ď€)
Doing something with a numeric log is highly likely to have catastrophic rounding problems around powers of 10.
In my view, the best answer in this case is to use standard tooling (like @sprintf) and fix it up as strings — either as a split like Tamas first suggested or with a regex. Both are really fine.
1 Like
I don’t think the original poster cares about rounding errors. He said
“Thanks! Sorry for not being clear, yes, I am after the printed approximation.”
I take that to mean approximation in base 10 (not approximation in base 2)
The problem I was worried about isn’t small errors. I suspect that it’s possible the exponent and mantissa might round different ways. Around powers of 10 that could easily lead to a factor of 10
error — something that I suspect everyone would care about.
I might be wrong — and I think you might be doing it in a way that’s safe — but regardless it’s worrisome.
2 Likes
10 posts were split to a new topic: Typst (alternative to LaTeX)
You misunderstand. The “printed approximation” is how floating point numbers are printed; it is understood to be an approximation in the sense the trailing digits that solely come from base 2 to
base 10 conversion are omitted, but it is accurate in the sense that it would be parsed back to the same base 2 number. This is what the module Ryu implements currently in base Julia, and handling it
correctly is equivalent to reimplementing that. If you are interested in the complexity of the topic, see the papers cited here.
I agree, but it would be great to expose the API of Ryu in the long run. It is one of those niche algorithms where a lot of thought went into speed and correctness and are very useful when you need
3 Likes
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Stabilization with discounted optimal control
Title data
Gaitsgory, Vladimir ; Grüne, Lars ; Thatcher, Neil:
Stabilization with discounted optimal control.
In: Systems & Control Letters. Vol. 82 (2015) . - pp. 91-98.
ISSN 1872-7956
DOI: https://doi.org/10.1016/j.sysconle.2015.05.010
This is the latest version of this item.
Project information
Project title: Project's official title
Project's id
Marie-Curie Initial Training Network "Sensitivity Analysis for Deterministic Controller Design" (SADCO)
ARC Discovery Grant
ARC Discovery Grant
Abstract in another language
We provide a condition under which infinite horizon discounted optimal control problems can be used in order to obtain stabilizing controls for nonlinear systems. The paper gives a mathematical
analysis of the problem as well as an illustration by a numerical example.
Further data
Available Versions of this Item
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Obsolete Keywords and Deprecated Features
Deprecated Features
This keyword prevents the FMM facility from being used even when it would improve performance. It was required in some circumstances when running in parallel on a cluster or LAN with Linda in
Gaussian 03. The associated problems have been fixed, and it is no longer needed.
Specifies a coupled cluster calculation using double substitutions and evaluation of the contribution of single and triple excitations through fourth order using the CCD wavefunction. It is
superseded by CCSD(T).
CBS-Q, CBS-Lq
Request the CBS-Q [Ochterski96] and CBS-q [Petersson91a] methods (i.e., Lq for “little q”). These are superceded by CBS-QB3.
Uses the original parametrization [Montgomery99] of CBS-QB3. It is obsolete and is included for backward compatibility only.
Requests the original parametrization [Ochterski96] of CBS-4. It is obsolete and is included for backward compatibility only.
Indicates that the geometry specification is in Cartesian coordinates. Cartesian coordinates can be included in molecule specifications without any special options being necessary.
Use the Gaussian 94 redundant internal coordinate generator.
Uses the minimal setup for Opt=Big. It may not be used for periodic boundary calculations.
Applies only to SCF=Conventional. Raff requests that the Raffenetti format [Raffenetti73] for the two-electron integrals be used. NoRaff demands that the regular integral format be used. It also
suppresses the use of Raffenetti integrals during direct CPHF. This affects conventional SCF and both conventional and direct frequency calculations.
Use the weighting scheme of Becke for numerical integration.
Use an existing integral file. Both the integral file and checkpoint file must have been preserved from a previous calculation. Only allowed for single point calculations and Polar=Restart.
Forces the integral derivative file to be written in HF frequency calculations. Useful only in debugging new derivative code.
LST, LSTCyc
Requests that an initial guess for a transition structure be generated using Linear Synchronous Transit [Halgren77]. The LST procedure locates a maximum along a path connecting two structures and
thus provides a guess for the transition structure connecting them.
Note that an LST calculation does not actually locate a proper transition state. The LST method has been superseded by Opt=QST2.
The Massage keyword requests that the molecule specification and basis set data be modified after it is generated. This keyword is deprecated in favor of ExtraBasis, Charge, Counterpoise and other
Requests an optimization using a pseudo-Newton-Raphson method with a fixed Hessian and numerical differentiation of energies to produce gradients. This option requires that the Hessian be read in via
ReadFC or RCFC. It can be used to locate transition structures and higher saddle points. Requires the molecule be specified as a Z-matrix. The default for energy-only methods is Opt=(EnOnly,EF).
Requests the Fletcher-Powell optimization algorithm [Fletcher63], which does not require analytic gradients. The maximum number of variables allowed for a Fletcher-Powell optimization is 30. Requires
the molecule be specified as a Z-matrix.
Requests a gradient optimization, using the default method unless another option is specified. This is the default whenever analytic gradients are available and is invalid otherwise.
Requests that the MNDO (or AM1, if possible) force constants be computed and used to start the (presumably ab initio) optimization. We recommend performing a PM6 Freq calculation followed by Opt=RCFC
instead of this option.
Specifies the Murtaugh-Sargent optimization algorithm [Murtaugh70]. The Murtaugh-Sargent optimization method is an obsolete alternative, and is retained in Gaussian only for backwards compatibility.
The maximum number of variables allowed for a Murtaugh-Sargent optimization is 50. Requires the molecule be specified as a Z-matrix.
Requests that a unit matrix be used instead of the usual valence force field guess for the Hessian.
Specifies the use of the modified GDIIS algorithm [Csaszar84, Farkas95, Farkas99]. The default GEDIIS algorithm is always better.
Requests the optimization to be done using the fast equation solving methods [Farkas98] for the coordinate transformations and the Newton-Raphson or RFO step. This method avoids the matrix
diagonalizations. Consequently, the eigenvector following methods (Opt=TS) cannot be used in conjunction with it. This option is unreliable and not recommended.
This requests output of an integral file in one variant of the format originated for the PolyAtom integrals program. The format produced by default is that used by the Caltech MQM programs, but the
code in Link 9999 is easily modified to produce other variations on the same theme.
Write an MO coefficient file in Caltech (Tran2P5) format. This is only of interest to users of the Caltech programs.
Requested the loose SCF convergence criteria appropriate for single points; equivalent to SCF=(Conver=4,VarInt,NoFinal,Direct). SinglePoint is a synonym for Sleazy. It is never recommended for
production quality calculations.
Reduced cutoffs even further; uses Int=CoarseGrid and single-point integral accuracy during iterations, followed by a single iteration with the usual single point grid (MediumGrid). Not recommended
for production quality calculations.
Uses the polarizable dielectric model [Miertus81, Miertus82, Cossi96], which corresponds to the Gaussian 98 SCRF=PCM option except for some minor implementation details [Cossi02]. This model is no
longer recommended for general use. The default SCRF method is IEFPCM.
Force numerical SCRF rather than analytic. This keyword is required for multipole orders beyond Dipole. This option implies the use of spherical cavities, which are not recommended. No gradients are
available for this option.
The options Dipole, Quadrupole, Octopole, and Hexadecapole specify the order of multipole to use in the SCRF calculation. All but Dipole require that the Numer option be specified as well.
Begin the SCRF=Numer calculation with a previously computed reaction field read from the input stream, immediately after the line specifying the dielectric constant and radius (three free-format
Used to specify the location of the .SCR scratch file.
Retain symmetry restrictions. NoSymm relaxes symmetry restrictions and is the default.
Forces the old-fashioned process of the 2PDM in post-SCF gradients (sorted in L1111 and then processed in L702 and L703). This is slow, but it reduces memory requirements. This option cannot be used
for frozen core calculations.
Causes the 2PDM to be generated, used, and discarded by L1111 in post-SCF gradient calculations.
Requests that the original transformation method based on externally stored integrals be used.
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K Anderson - MATLAB Central
K Anderson
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Sum of series II
What is the sum of the following sequence: Σ(2k-1)^2 for k=1...n for different n?
2 months ago
Area of an equilateral triangle
Calculate the area of an equilateral triangle of side x. <<https://i.imgur.com/jlZDHhq.png>> Image courtesy of <http://up...
2 months ago
Flip the vector from right to left
Flip the vector from right to left. Examples x=[1:5], then y=[5 4 3 2 1] x=[1 4 6], then y=[6 4 1]; Request not to use d...
2 months ago
Find max
Find the maximum value of a given vector or matrix.
2 months ago
Select every other element of a vector
Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...
2 months ago
Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
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Answered How do I skip items in a legend?
If you plot multiple lines with the same plot command like this h(1,:) = plot(rand(4,11),'r') hold on h(2,:) = plot(rand(4,11...
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Volume of Cylinder Formula, Unit, and Questions
Volume of Cylinder
A cylinder is a three-dimensional geometric shape that consists of two parallel circular bases connected by a curved surface. The volume of a cylinder refers to the amount of space enclosed by its
surfaces. It is measured in cubic units, such as cubic centimeters (cm³) or cubic meters (m³).
Volume of Cylinder Formula
The formula for calculating the volume of a cylinder is:
V = πr²h
In this formula:
• V represents the volume of the cylinder.
• π (pi) is a mathematical constant approximately equal to 3.14159. It is used to calculate the area of a circle.
• r represents the radius of the circular base of the cylinder. The radius is the distance from the center of the base to its outer edge.
• h represents the height of the cylinder. It is the distance between the two bases, or the length of the curved surface.
Cylinder Volume Formula
The Volume of cylinder defines the holding capacity of a cylinder or in other words, the quantity of materials that can carry or hold by a cylinder, known as volume of cylinder. The volume of
cylinder is calculated by the formula V= πr² h. In this article, we will learn about cylinders, the Volume of a cylinder, and the total surface area of a cylinder.
To calculate the volume of a cylinder, you need to know the values of the radius and height. Simply substitute those values into the formula, square the radius, multiply by pi, and then multiply by
the height. The resulting value will give you the volume of the cylinder.
It’s important to ensure that the radius and height are measured in the same units (e.g., centimeters, meters) to obtain the volume in the corresponding cubic units.
What is a cylinder volume?
In geometry, a cylinder is a three-dimensional object that consists of two parallel bases linked by a curved surface. Although a cylinder is a 3D object, it has Total surface area and volume.
Some examples of cylinders are- Pipes, batteries, Barrels, etc.
Properties of a cylinder volume Formula
• Radius of the cylinder- A cylinder consists of two parts- two parallel bases and the curved surface which linked the bases. Both the parallel bases have an equal radius (denoted by r ). A cylinder
can be solid or hollow, In that case, there are two types of radius called- inner radius (r) and outer radius (R).
• Height of cylinder – The height of the cylinder suggests the distance between two parallel bases of the cylinder.
Properties of a cylinder
Volume of Cylinder Formula in Maths
The volume of the cylinder defines the holding capacity of a cylinder. The volume of a cylinder is nothing but the product value of the Area of the base (A) and the height of the cylinder(h)
The volume of cylinder = Area of base × Height
For any solid cylinder with a base radius ‘r’ and height of the cylinder is ‘h’
Volume of a solid cylinder
The volume of the cylinder is V= πr² h cube units. [π is a constant value, equals 3.14 or 22/7]
Volume of Hollow Cylinder Formula
A hollow cylinder means a cylinder that is empty from the inside. If we see a hollow cylinder from the top, it looks like an annular ring. The hollow cylinder has two types of radius called- inner
radius (r) and outer radius (R).as it is bounded by two concentric circles.
For any hollow Cylinder with height ‘h’, external Radius ‘R’ and internal Radius ‘r’. so. Thickness = (R-r) .
Volume of a hollow cylinder
The volume of the hollow cylinder is V= π(R²-r²)h cubic units. [π is a constant value, equals 3.14 or 22/7]
Formula of Volume of Cylinder Unit
The formula for the volume of the cylinder is V= πr² h cube units.
So, generally, we measure the volume of cylinders in cubic units like cubic centimeters (cm³), cubic meters (m³), cubic feet (ft³), and so on.
Volume of a Cylinder in liters
We use cylinder-shaped objects or barrels to carry mostly liquids .so, it is very necessary to find the volume of the cylinder in liters. We calculate the volume of the cylinder by cubic cm of cm³.
we must convert cubic cm or cm³ into liters or vice-versa by applying the following conversion formula-
1 Litre = 1000 cubic cm or cm³
For example, a cylinder can hold 15 liters of petrol. Find the volume of the cylinder in cm³?
We know that 1 Litre = 1000 cubic cm or cm³
Now, 15 litres = 15000 cubic cm or cm³
Hence the volume of the cylinder is 15000 cubic cm or cm³.
Surface area of Cylinder Formula
The formula used for determining the surface area Of a solid Cylinder is,
A = Area of two bases+ Curved surface area
A = 2πr² + 2πrh square units.
A= 2 πr (h + r) square units.
The formula for the surface area Of a hollow Cylinder is,
A= (Area of both bases + curved surface area)
A= 2 π(R² + r²) + 2 πh( R + r) square units.
Volume of cylinder Formula Questions with answers
Example 1: The height of a cylinder is10 cm and the radius of one circular base is 7 cm. Find its volume.
a) 1540 cm³ b) 1600 cm³ c) 490 cm ³ d) None of these .
Answer- The Formula for the volume of a cylinder is, V = πr² h cubic cm
In the given problem, height is 10cm and radius is 7 cm
Volume = π (7)² × 10 cm³
Or, V = 22/7 × 49 × 10 cm³
Or, V = 1540 cm³
Hence, Option ( a ) is correct.
Example 2: The volume of a cylinder is 440 cm³ and the height is 35 cm. Find the radius of the cylinder of a base.
a) 3 cm b) 2 cm c) 4 cm d) None of these
Answer – Let’s assume the radius of the cylinder is – r cm.
In the given cylinder Volume is 440 cm³ and the height is 35 cm.
The Formula for the volume of a cylinder is, V = πr² h cubic cm.
or, 440 = (22/7) x r² x 35
or, r² = (440 x 7)/(22 x 35) = 3080/770 = 4
or, r = 4 cm
Option ( c) is correct.
Example 3: Total surface area of a cylinder is 660 cm² calculate the height of the cylinder of radius 5 cm.
a) 10 cm b) 16 cm c) 15 cm d) None of these
Answer- Let’s suppose the height of the cylinder is -h cm.
The formula used for determining the total surface area of a cylinder is,
A= 2 πr (h + r) square units.
or, 660= 2 x 22/7 x (5 + h)
or, 660×7 = 220 x (5 + h)
or, (5+h) = 21
or h = 16 cm
So, option (b) is correct.
Example 4: A pipe is 21 cm long. The inside and outside radiuses are 6 cm and 4 cm respectively. What is the volume of the water required to fill the pipe completely?
a) 1300 cm³ b) 3234 cm³ c) 1320 cm³ d) None of the above
Answer – In the given Information,
Inner radius – r = 4 cm
Outer radius – R = 6 cm
Height – h = 21 cm
The volume of the hollow cylinder is, V= π(R²-r²)h cubic units
or. V= π × (6² – 4² ) ×21
or.V= 22/7 ×(36- 16 ) ×21
or.V= (22× 20 ×21) /7
or.V = 1320 cm³
So, option ( C ) is correct.
Volume of Cylinder Formula- Exercise Questions
1. The height of a cylinder is12 cm and the radius of one circular base is 3 cm. Calculate the volume of the cylinder.
2. The volume of the cylinder is 12000 cm³. How many liters of water the cylinder can hold?
3. The total surface area of a cylinder is 165 m². The height of the cylinder is 15 m. Find out the radius of the base of the cylinder.
4. The height of a cylinder is 6 cm and the volume of the cylinder is 180 cm³. Calculate the area of the base of the cylinder?
What is the Volume of Cylinder in Hindi- सिलेंडर का आयतन
सिलेंडर का आयतन एक सिलेंडर की धारण क्षमता को परिभाषित करता है या दूसरे शब्दों में, सामग्री की मात्रा जो एक सिलेंडर द्वारा ले जा सकती है या धारण कर सकती है, जिसे सिलेंडर की मात्रा के रूप में जाना जाता है। बेलन का आयतन सूत्र V= r² h द्वारा परिकलित किया जाता है। इस लेख में हम बेलन,
बेलन का आयतन और बेलन के कुल पृष्ठीय क्षेत्रफल के बारे में जानेंगे।
सिलेंडर का आयतन: सिलेंडर क्या है?
ज्यामिति में, एक सिलेंडर एक त्रि-आयामी वस्तु है जिसमें एक घुमावदार सतह से जुड़े दो समानांतर आधार होते हैं। . हालांकि एक सिलेंडर एक 3D वस्तु है, इसमें कुल सतह क्षेत्र और आयतन होता है।
सिलेंडर के कुछ उदाहरण हैं- पाइप, बैटरी, बैरल आदि।
सिलेंडर का आयतन: एक सिलेंडर के गुण
• बेलन की त्रिज्या- एक बेलन में दो भाग होते हैं- दो समानांतर आधार और वक्र सतह जो आधारों को जोड़ती है। दोनों समान्तर आधारों की त्रिज्या समान होती है (r द्वारा निरूपित)। एक बेलन ठोस या खोखला हो सकता है, उस स्थिति में दो प्रकार की त्रिज्याएँ होती हैं जिन्हें आंतरिक त्रिज्या (r) और बाहरी त्रिज्या
(R) कहा जाता है।
• बेलन की ऊँचाई – बेलन की ऊँचाई बेलन के दो समानांतर आधारों के बीच की दूरी को दर्शाती है।
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