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Re: st: STATA graph question: Combining Histograms
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Re: st: STATA graph question: Combining Histograms
From Christoph Engel <[email protected]>
To [email protected]
Subject Re: st: STATA graph question: Combining Histograms
Date Wed, 17 Aug 2011 23:26:40 +0200
If all you want is 10 graphs in one line, the following would do
hist response26, freq, by(group, cols(10))
maybe together with
hist response26, freq, by(group, cols(10)) xsize(10) ysize(2)
If you also want to change the order, you need memory graphs. You could do the following
forvalues i = 1/10 {
hist response26 if group == `i', freq title("group `i'") name(group`i', replace)
and could then arrange them at will, using
graph combine group2 group4 group6
Does that help?
Christoph Engel
Am 17.08.2011 22:15, schrieb Marlis Gonzalez Fernandez:
Excuse my naiveté... have not figured out all the rules yet.
Yes I did:
. histogram response26, freq
and repeated for 10 different responses thus generating 10 different histograms that I would like to combine (basically as they are) side by side so people can visually compare the frequencies between questions. Not worried about labeling at this point.
. graph combine graph1 graph2 graph3
But that is not what I am looking for.
Any thoughts?
M. González
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Nick Cox
Sent: Wednesday, August 17, 2011 3:32 PM
To: [email protected]
Subject: Re: st: STATA graph question: Combining Histograms
We can let -rep78- of the auto data serve as a Likert scale. (The term
Likert honours Rensis Likert.)
Without code examples of what you tried what you want remains a bit
vague. Do you want axis scales? labels on the bars? or what?
Is this what you want?
. sysuse auto
(1978 Automobile Data)
. histogram rep78, freq
. tab rep78
Repair |
Record 1978 | Freq. Percent Cum.
1 | 2 2.90 2.90
2 | 8 11.59 14.49
3 | 30 43.48 57.97
4 | 18 26.09 84.06
5 | 11 15.94 100.00
Total | 69 100.00
. histogram rep78, freq discrete barw(0.8) yaxis(1 2) yla(6.9 "10"
13.8 "20" 20.7 "30" 27.6 "40" , axis(2)) ytitle("Percent", axis(2))
Here I did it by mental arithmetic after noting that 69 is 100% and 6.9 is 10%.
On Wed, Aug 17, 2011 at 8:19 PM, Marlis Gonzalez Fernandez
<[email protected]> wrote:
I am trying to combine frequency histograms for several questions into one graph. Each variable has the answers to questions in a likert scale (always, frequently, infrequently, never, don't know). I want to plot the frequency and percent for each question side by side. Have not found a good way to do this.
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Prof. Dr. Christoph Engel
Max-Planck-Institut zur
Erforschung von Gemeinschaftsgütern
Max Planck Institute for Research on Collective Goods
Kurt-Schumacher-Strasse 10
D 53113 Bonn
Tel. +49/228/91416-10
Fax +49/228/91416-11
e-mail: [email protected]
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* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2011-08/msg00825.html","timestamp":"2024-11-10T14:58:04Z","content_type":"text/html","content_length":"16176","record_id":"<urn:uuid:bcda4953-1605-47d1-a6fd-22b1a602ba35>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00292.warc.gz"} |
%0 Journal Article %A LU Weiyang %A GAO Li %T On the Hybrid Mean Value of the Smarandache Double-Factorial Function %D 2017 %R 10.3969/j.cnki.jdxb.2017.03.002 %J Journal of Jishou University(Natural
Sciences Edition) %P 4-7 %V 38 %N 3 %X
The elementary method and analytic method were performed to study the mean value problem of hybrid function (Sdf(n)-P(n))^β and δ[α](n)(Sdf(n)-P(n))^β involving the Smarandache double-factorial
function Sdf(n) and the largest prime factor divisor function P(n),where δ[α](n) denotes the divisor function.Two sharper asymptotic formulas were proposed.
%U https://zkxb.jsu.edu.cn/EN/10.3969/j.cnki.jdxb.2017.03.002 | {"url":"https://zkxb.jsu.edu.cn/EN/article/getTxtFile.do?fileType=EndNote&id=2564","timestamp":"2024-11-09T23:31:01Z","content_type":"application/x-endnote-refer","content_length":"1234","record_id":"<urn:uuid:f8fba12b-7d54-4111-8e64-3af447b66866>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00867.warc.gz"} |
Direct modeling support
class Gecode::REG
Regular expressions over integer values. More...
class Gecode::Matrix< A >
Matrix-interface for arrays. More...
Linear expressions and relations
Linear float expressions and relations
Set expressions and relations
Boolean expressions
Posting of expressions and relations
Arithmetic functions
Transcendental functions
Trigonometric functions
Channel functions
Aliases for integer constraints
Aliases for set constraints
Support for cost-based optimization | {"url":"https://www.gecode.org/doc/4.4.0/reference/group__TaskModelMiniModel.html","timestamp":"2024-11-07T00:27:06Z","content_type":"text/html","content_length":"6730","record_id":"<urn:uuid:eba84218-042b-4093-8d81-7b2e852be92f>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00353.warc.gz"} |
Improved Performance of FOPI Based Wind Energy Conversion System Using Multilevel Inverter
Improved Performance of FOPI Based Wind
Energy Conversion System Using Multilevel
Disha Tiwari1, S.P. DUBEY2
PG Student [Power Electronics], Department of ELE, RCET, Bhilai, Chhattisgarh, India1
Professor, Department of ELE, RCET, Bhilai, Chhattisgarh, India 2
ABSTRACT:In this paper for improvement in the performance of system and reduction in total harmonic distortion (THD) three-level inverter is used in place of two-level inverter. The system comprises
of permanent magnet synchronous generator (PMSG), Fractional Order PI Controller (FOPI), Neutral-Clamped three-level inverter. The FOPI controller is connected across the inverter which shows better
results when comparing with two-level inverter. The use of three-level inverter provides improved load voltage, output active power, output reactive power and reduction in harmonic distortion whereas
the use of FOPI controller gives smooth and stable output for above system at resistive load.
KEYWORDS: FOPI Controller, PMSG, Neutral -Clamped three- level inverter.
The rapid exhaustion of non-renewable energy sources, air contamination, and ozone harming substance outflow has constrained us to consider some renewable energy sources, as sunlight based, wind,
tidal/wave, biomass and little hydro energy. Among these assets, wind vitality is the most favored and broadly utilized sustainable power source asset because of its favorable circumstances [1]. The
greater part of the new improvements for substantial multi megawatt turbines found in industry depend on synchronous generators using permanent magnet with various arrangements of three phase
windings [2, 3, 4]. The utilization of multilevel inverter and the manufactured strategies add to enhanced execution of the control of wind vitality transformation framework [5, 6]. Multilevel
inverter innovation is being consolidated in different sustainable power source based power generation advancements like wind and sunlight based for higher power and voltage application [7].
Another power change framework is investigated in this paper pointing wind turbines evaluated at the megawatt level. The proposed setup comprises of a medium-voltage, lasting magnet synchronous
generator associated with a minimal effort three phase diode connect rectifier, a dc–dc four-level lift converter as the middle stage, and a four-level diode-braced inverter on the matrix side. The
dc-interface capacitor voltages are adjusted by the lift converter, and in this way the control many-sided quality for the matrix tied inverter is incredibly rearranged. To control the lift converter
and matrix tied inverter, a basic strategy in view of a two-stage display prescient methodology is exhibited [8].
Performance Improvement of Grid Connected DFIG Fed by Three Level Diode Clamped MLI Using Vector Control [22], Giribabu Dynamina (2016) et al. presents a vector control method for doubly fed
induction generator (OFIG) based Wind energy conversion systems (WECS) to independently control the active and reactive power. The performance of OFIG based WECS at the rotor side is poor. To improve
the performance of the system two level inverter (TLI) is replaced with Diode c1amped multilevel inverter (OCMLI) which is commonly used MLI topology due to its merits compared to other. The PI
controllers in the VC of DFIG systems are replaced with AN FIS controllers to improve steady state and transient performance. The proposed OCMLI fed WECS is simulated in MATLAB/SIMULINK environment
where the MLI fed system using ANFIS controller results better performance compared to TLI.
Dual-Boost NPC Converter for a Dual Three-Phase PMSG Wind Energy Conversion System [23], G. Estay (2012) et al. proposes the development of wind energy conversion systems (WECS) is currently focused
in reaching higher power ratings (_10MW). Medium voltage operation is at this power level, particularly at grid side, a desirable feature. Therefore, great attention has been given to high-power
medium-voltage multilevel converters for WECS. Nevertheless, most modern multi-megawatt turbines use low-voltage multiphase permanent magnet synchronous generators (PMSG) operating at 690V, which
requires the use of several converters connected in parallel to reach the multi-megawatt level. This paper presents a low- to medium-voltage converter interface for a dual three-phase PMSG based
WECS. Two full bridge diode rectifiers followed by boost converters are used to control each set of three-phase windings of the dual PMSG. The boost converters elevate the voltage to feed each one of
the capacitors of the dc-link of the medium voltage NPC grid-tied inverter. The generator side converter has high power density and is a cost effective solution compared to traditional back-to-back
solutions. Simulation results are presented to provide a preliminary overview of the performance of the system.
Wind energy obtained from wind turbine is transferred to the permanent magnet synchronous generator. The generated electrical output power is then transferred to the load via back to back, AC-DC-AC
converter. In this paper inverter is controlled by fractional order PI controller.
The regular converter utilized for the vector control is two-level inverter. This two level inverter output voltage
PMSG DIODE [RECTIFIER ]
mutilated and the total harmonic distortion (THD) of the voltage and current is poor. Multilevel inverter output voltage and current is more sinusoidal and the total harmonic distortion (THD) is
The circuit comprises of three half- bridges, which are commonly phase shifted by 120 to produce the three stage voltage waves. The dc supply is normally gotten from a diode connect rectifier.
By utilizing multilevel inverter smooth AC waveform can be created which have low twisting in its output. Thus total harmonic distortion (THD) is diminished in output voltage and current waveforms.
The multilevel inverters are mainly classified as follows:
1. Flying Capacitor 2. Diode Clamped 3. Cascaded H- Bridge.
The most usually utilized multi level topology is the diode clamped inverter, in this the diode is utilized as the clamping gadget to clamp the dc transport voltage which accomplishes load voltage in
step. Hence, the primary idea of utilizing diodes in this MLI is to restrain the voltage weight on exchanging devices. An m-level inverter requires (m-1) voltage sources, 2(m-1) switching devices and
(m-1) (m-2) diodes. As the quantity of voltage levels expands the load voltage quality is enhanced in this way the voltage waveform turns out to be more sinusoidal. The three-level diode-clamped
inverter with one stage leg is represented in Figure 2 . The capacitors CI and C2 are arrangement associated, these capacitors are utilized to part the DC-transport voltage into three-levels where
the neutral point n, is signified as center purpose of capacitors. The three levels of output voltage Van are denoted as E, 0 and - E. To acquire the voltage level E, the switches SI and S2 are to be
exchanged ON. So also, to acquire the voltage level - E and 0.
D1 and D2 these two diodes help in clipping the changing voltage to half of the DC transport voltage. The operation of
inverter utilizing this clipping diodes is clarified as when the two switches SI and S2 are exchanged ON, the voltage crosswise over " a " and '0'is Vdc, i.e., Vao = Vdc.In this case, DI' adjusts the
voltage sharing amongst SI' and S2',by SI' hindering the voltage crosswise over CI and S2' obstructing the voltage crosswise over C2.
The fractional order PI controller has been utilized as a part of different control fields. Fractional calculus is speculation of ordinary calculus. The fractional order PI controller’s transfer
function is given by:
( ) =
Where KP is the proportional constant, Ki is the integral constant and α is the fractional PI order. In the case of
conventional PI controller (α =1).
The fractional order PI controller block contains transfer function. In simulation the transfer function can be written in the form of
b = {0.4, 500}; nb={1,0}; a={1,0}; na={0.8,0}; G = fotf(a,na,b,nb)
Where b is coefficients present in numerator of transfer function, nb is order of s domain, a is the coefficients in numerator and na is the fractional order term. It gives transfer function as
( ) =
(A) For System Using Two-Level Inverter- The result for three-phase inverter using IGBT is represented in the figure 4. The figure 4(a) shows the value of load voltage, the magnitude of voltage is
-100to100V. The result has been taken for single-phase the positive half square wave is achieved when switch S1 is in ON condition and negative
Figure 4(a) Load voltages in square wave mode for three phase two-level inverter
Figure4 (b) shows output active power whose magnitude is approx 350W after 3sec the magnitude of power became stable due to FOPI Controller which gives smooth and stable output.
Figure 4 (b) Output Active Power for two-level inverter
Figure 4(c) shows reactive power whose value is very small which is essential because the load is resistive in nature. After analysing the two results from figure 4(b) and 4 (c) shows that the useful
power is maximum and reactive powere is minimum.
Figure 4(d) shows THD for two level inverter which has been measured from 4.8 sec for 2 cycle. From figure it is clear that the value of THD is 68.53%.
Figure 4 (d) Total Harmonic distortion (THD) for two-level inverter
(B)For System using Three-Level Inverter-The result for three-level inverter is represented in the figure 5(a)
Figure 5(a) Load Voltage for three-level inverter
The result shows the value of load voltage, the magnitude of voltage is -200to200V, -100to100V and 0. Figure 5 (b) shows output active power whose magnitude is approx 350 W.
Figure 5(b) Output Active Power for Three-level inverter
Figure 5(c) Output Reactive Power for Three-Level Inverter
Figure 5(d) shows THD whose value is 46% for 2cycle from 4.8sec.
Figure 5(d) Total Harmonic Distortion for Three-Level inverter
After comparing the THD values for system using two level inverter and three-level inverter it is cleared that there is 22% reduction in harmonic distortion for the system using three-level inverter.
This paper presented the improvement in the value of load voltage and total harmonic distortion, usage of three-level inverter provides following advantages:-
(a) The output voltage level is twice the DC sources. (b) The harmonic distortion is less.
(c) The increasing fundamental value is achieved
[1] Giribabu Dyanamina, Amit Kumar. “Performance improvement of grid connected DFIG fed by three level diode clam ped MLI using
vector control,” IEEE Region 10 Conference(TENCON),pp.560-565,2016.
[2] G. Stay, L. Vattuone, S. Kouro, M.Duran, B.Wu. “Dual-boost NPC converter for a dual three-phase PMSG wind energy conversion system,” IEEE International Conference on Power Electronics, Drives and
Energy Systems (PEDES), pp.1-6, 2012.
[3] X.Xiong, H. Xin. “Research on multiple boost converter based on MW-level wind energy conversion system,” Proceedings of the Eighth International Conference on Electrical Machines and Systems
(ICEMS), vol. 2, pp. 1046–1049, 2005.
[4] J. Birk, B. Andresen, “Parallel-connected converters for optimizing efficiency, reliability and grid harmonics in a wind turbine.”European Conference on Power Electronics and Applications(EPE),
pp. 1-7, 2007.
[5] A. Moualdia, Dj. Boudana, O. Bouchhida. “ Direct Torque Control Based Multi-level Inverter and artificial neural networks of wind energy conversion system,” 8th International Conference on
Modelling Identification and Control (ICMIC), pp.49-54,2016.
[6] M.Pichan, M. Rategar, H.Monfared,M. “Two fuzzy-based direct power control strategies for doubly fed induction generators in wind energy conversion systems,” Elsevier Energy, vol. 6. No.5,
[7] Tremblay, E.Atayde, S, Chandra,A. “Comparative Study of Control Strategies for the Doubly Fed Induction Generator in Wind Energy Conversion System,” IEEE Transactions on Sustainable Energy, vol.
2, No.3,pp. 288-299, 2011.
[8] Sarika Shrivastava, Anurag Tripathi, K.S.Verma. “Reduction in total harmonic distortion by implementing multi-level inverter techonology in grid integrated DFIG,” pp.491-495,2015.
[9] Venkata Yaramasu, and BinWu. “Predictive Control of a Three-Level Boost Converter and an NPC Inverter for High-Power PMSG-Based Medium Voltage Wind Energy Conversion Systems,” IEEE Trans ON POWER
ELECTRONICS, vol. 29, No. 10, pp. 5308-5322,2014. [10] Chunyang Wang, Weicheng Fu, Yaowu Shi. “Tunning fractional order proportional integral differentiation controller for fractional order
system,” 32nd Chinese Control Conference, pp. 552-555,2013.
[11] Mei Li, Keyue Smedley. “One-cycle control of PMSG for wind power generation,”IEEE Power Electronics and Machines in Wind Applications,pp. 1-6,2009
[12] Qiang Wei, Bin Wu. “Analysis and Comparison of current –source-converter- based –medium-voltage PMSG wind energy conversion systems,” IEEE 6th International Symposium on Power Electronics for
Distributed Generation Systems (PEDG), pp. 1-6,2015.
[13] R. Melício, V.M.F. Mendes, J.P.S. Catalão. “Fractional-order control and simulation of wind energy systems with PMSG/full-power converter topology,” Energy Convers and Manage, vol. 51, No. 6,
pp. 1250–1258,2010. | {"url":"https://1library.net/document/q731epky-improved-performance-based-energy-conversion-using-multilevel-inverter.html","timestamp":"2024-11-14T07:34:59Z","content_type":"text/html","content_length":"164375","record_id":"<urn:uuid:fd5ab60b-1446-450d-9233-bcfae84b22b7>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00749.warc.gz"} |
ABZ, Inc. | Design Flow Solutions - Fluid Flow Software
Common Assumptions Made By Other Fluid Flow Programs and The DFS Approach
Many fluid flow analysis programs make questionable or clearly incorrect assumptions when solving problems. Most of these assumptions remain undocumented, or documented where they will not be read by
the average users. When questioned, most of these companies will downplay any of these issues by stating that they just don’t matter. We think you should decide for yourself.
Below are some of the common assumptions that can result in significant errors in calculated results, as well as a simple problem for each assumption that one can use to demonstrate the error. In no
case has ABZ provided the worst case situation in an attempt to bias the results. All of these errors can be worse or better, depending on the specific details of the problem being analyzed.
Assumption: Fluid velocity is constant throughout any system.
This assumption is mandated by the fact that most other programs ignore the velocity terms of Bernoulli’s equation, thus ignoring conservation of energy. One such program even goes so far as to state
that if the user wants to conserve energy, that the velocity component must be manually factored into the system pressures!
The impact of this assumption is that any system which contains tees or crosses (with flow in more than two legs) or reducers or enlargers will be analyzed incorrectly!
How to check:
Draw a pipeline consisting of two different sizes of pipe (with the appropriate reducer or enlarger between the pipes). Make the sizes very different and the lengths of pipe a reasonable length to
make the error larger (such as 36 inch pipe and 1 inch pipe with 10 feet of each size of pipe). Determine the flow rate with a 20 psid with flow in both directions. The flow rates should be very
different because of the change in velocity (increases in one case, decreased in the other).
DFS Approach:
An energy balance is automatically performed across each and every component in the system. The user does not need to perform any separate steps to include the effect of changing velocities.
Assumption: Component resistances do not depend on flow direction.
Most programs do not correctly calculate component resistances unless the flow direction is properly chosen when the problem is specified. Thus, in situations where the flow direction is not known or
changes for different conditions, the component resistances will be calculated incorrectly.
The impact of this assumption is that all flowpaths must be drawn in the correct direction initially (and this direction must be known), and anytime a different configuration of running pumps/open
valves or known pressures and flows is to be analyzed, any flowpaths where flow direction can change may require the user to respecify those flowpaths.
How to check:
Draw a pipeline with a size change (reducer or enlarger). Define flow in one direction. Look at the calculated flow resistance of the size change. Now define flow in the other direction. Look at the
calculated flow resistance of the size change again. The values should be different.
DFS Approach:
Each time flow direction is changed (whether during the analysis of a problem or due to the user specifying known flow information) the resistance of each item is reviewed and recalculated if
Assumption: Tees and Crosses have no flow resistance.
Most programs use tees and crosses to combine and split flow only, and include no resistance for the tee or cross. If desired by the user, such resistance must be added separately. Of course, the
resistance depends on the total flow, the direction of flow for each leg, and the amount of flow in each leg. All of these values must be known prior to determining the resistance of the fitting.
The impact of this assumption is that any problem or flowpath which includes tees will calculate incorrect flow rates or differential pressures unless the user adds the correct resistance for the tee
or cross. This generally requires that the user know the flow rates prior to analyzing the problem.
How to check:
Draw a flow network consisting of a tee with 10 feet of 2 inch pipe connected to each leg. Define the flow at the end of one of the pipes to be zero. Define pressures at the ends of the other two
pipes to specify a differential pressure of 20 psid. The differential pressures across the two pipes should not add up to the total differential pressure (since the tee has resistance) if the program
includes resistance for the tee.
DFS Approach:
The resistance of a tee or cross is determined automatically based on the flow direction and flow rate for each leg of the tee or cross. The user does not need to separately add any additional
resistance nor to know the flow rates prior to adding a tee or a cross.
Assumption: Compressible flow analysis can be performed using the Darcy-Weisbach Equation.
The Darcy-Weisbach equation for liquid flow assumes that the density of the fluid flowing in the system is constant. By definition, properties (such as density) change for a compressible fluid as
pressure changes. For example, a 40 percent change in pressure for air at 200 psia and natural gas at 200 psia results in about a 40 percent change in density. This change results in a calculated
flow that is too high if differential pressure is specified, or a calculated differential pressure that is too high if flow is specified. Further, for any differential pressure greater than 10
percent, these programs require that average fluid properties (calculated from both the inlet and outlet properties) be used throughout. This, of course, requires that the properties at both ends be
known before the problem is solved, or that the problem be solved multiple times (changing the fluid properties each time), until this condition is matched. For a larger network, this would require
that separate fluid properties be calculated and specified at numerous locations throughout the network.
On any single flow path, this assumption can result in a significant error (which is worse for smaller component resistances). In fact, problems with small resistances can even reach sonic conditions
prior to a reaching a pressure change of 40 percent. This assumption, together with a larger flow network, can result in significant errors and even errors in flow direction.
How to check:
Analyze a piece of pipe at a flow rate which results in a pressure drop of 40 percent of the inlet pressure. Compare the inlet and outlet volumetric flow rates. If they are the same, the program is
not performing a correct compressible analysis. Since the density changes from inlet to outlet, the volumetric flow rates must change as well.
DFS Approach:
DFS uses a true compressible analysis, and allows for two heat transfer assumptions (adiabatic and isothermal). Further, DFS is the only program to provide for conservation of energy across tees,
reducers and enlargers, and changes in elevation.
Assumption: Component order within a pipeline has no effect on the calculated results.
Most programs allow the user to enter the number of specific valves and fittings that are contained within a pipeline, as well as the presence of a size change or pump. None allow the user to define
the order of components within the pipeline.
For liquid systems, the order of components is important when looking at pressures along the pipeline, and to ensure that the proper size component has been specified when the pipeline contains a
size change. Ignoring pressures along the pipeline can result in incorrect flow rates being calculated when cavitation occurs (which would not be foreseen if component order is ignored). For
compressible systems, in addition to these two reasons, the calculation may be incorrect if an incorrect component order is assumed (depending on the specific hardware being analyzed) since the flow
velocity changes as the fluid pressure changes, and thus the head loss across each component is different depending on its location in the pipeline.
How to check:
Build two pipelines in series, each with 1 foot of NPS 2, schedule 40 pipe. Add an additional resistance with a K factor of 10 to the first pipeline. Analyze the system with a differential pressure
of 40 percent of the inlet pressure. Note the differential pressure across the pipe without the added resistance. Now remove the additional resistance from the first pipeline and add it to the second
pipeline. Analyze the system again for the same differential pressure. Again note the differential pressure across the pipe without the added resistance. The two noted values should be different. If
they are the same, then the program does not consider component order and all calculated values may be incorrect.
Alternatively, build a pipeline with 100 feet of NPS 1, schedule 40 pipe oriented vertically with two inline globe valves at the top of the pipe. Analyze this pipeline with atmospheric pressure at
both ends. If a flow rate is calculated and no error about cavitation is provided, then the program does not consider component order.
DFS Approach:
DFS allows the user to input components in their correct order, and analyzes systems on a component by component basis. Thus, component order and its effect on the calculations is always considered
Assumption: The user is not interested in differential pressures across individual components and flow rates or velocities within a pipeline.
Most programs require that valves and fittings be specified as contained in a given pipeline, but they do not show calculated values across individual components; rather they provide values for the
entire pipeline as a single item. In some programs, the user may instead choose to add each fitting or valve as a separate item (independent of any given pipeline). This approach, however, quickly
exceeds the capabilities of such programs to display a network with even a normal amount of valves and fittings.
While not generally a calculational issue, the inability to view calculated values across each valve or fitting may make obtaining desired information difficult if not impossible.
How to check:
Build a pipeline with several valves and fittings. Analyze this pipeline for a given flow rate. Attempt to view the differential pressure across each valve and fitting.
DFS Approach:
DFS allows the user to view flow rate and pressure information before, after, and across each component within a pipeline. In addition, available printouts include “big picture” graphical printouts
which illustrate information on a flowpath level, and “detailed” graphical printouts which provide flow and pressure information for each component within every pipeline.
Assumption: Negative absolute pressures are a valid result of a flow calculation.
May programs show calculated results with pressures less than absolute zero. They may provide a warning, but even if such a warning is provided, it is typically difficult to see.
Calculations can provide nonsensical results that the user is not made aware off (either due to the lack of a warning or error message, or due to the lack of a visual flag that such an error exists).
How to check:
Build a system with two pipelines in series. Make the first pipeline NPS 2, schedule 40, 100 feet long, with an increase in elevation of 50 feet (from inlet to outlet). Make the second pipe NPS 2,
schedule 40, 100 feet long, with a decrease in elevation of 100 feet (from inlet to outlet). Specify atmospheric pressure at both the inlet and outlet. The program should indicate that flashing has
occurred (if the program is designed to perform two-phase flow calculations), or should provide a clear indication that an error exists (since the fluid will flash in the middle of the two pipelines
and flow will not be single-phase).
DFS Approach:
For many problems, DFS simply does not consider flows which result in cavitation within liquid systems. Where this is not possible (such as when flow rates have been specified by the user), DFS
provides a clear indication of the resulting error condition.
Assumption: The user always knows whatever flow or pressure information the program is designed to need.
Most programs are designed to accept only one type of flow information, such as mass flow rate (e.g., lbm/hr). This makes the job of the programmer easier at the cost of increased difficulty when
using the program. Further, certain types of known flow information can rarely be specified by the user, such as volumetric flow rate relative to standard conditions (e.g., scfm) and velocity (e.g.,
When information is known by the user in units other than that accepted by the program, the user must convert the known information into whatever the program demands. In the case of a compressible
problem, this is difficult if not impossible (since these conversions generally depend on the fluid state which is not known until the problem has been solved).
How to check:
Add a simple pipeline with any hardware and attempt to specify flow information. Observe what types of information the program allows (most demand mass flow rate; a few also allow volumetric flow
rate). Look specifically for standard volumetric flow rate (scfm) and velocity (fps).
DFS Approach:
DFS allows the user to specify flow information as mass flow rate, volumetric flow rate (relative to flowing or standard conditions), or velocity. This information can be entered in any of the unit
types supported by the program (54 types for flow information alone; more can be added by the user).
Assumption: Flow velocities higher than sonic velocity can be calculated with any geometry fitting.
With the exception of some very specific fittings designed to obtain supersonic flows, the flow rate of all fluids is limited to the velocity of sound of that fluid at that temperature. This limit
can change throughout a system as the fluid temperature changes.
Calculating flows that are physically not possible provides no useful information to the user; rather, unless he somehow figures out that a physical limit has been violated he may unknowingly use the
incorrect flow rates.
How to check:
Add a pipeline with 5 feet of NPS 4, schedule 40 pipe. Define the fluid at the inlet of this pipe to be Air at 60 degrees Fahrenheit and 200 psia. Define the outlet pressure to be 120 psia (a 40
percent drop in pressure). The program should indicate clearly that this pressure drop cannot be obtained due to sonic flow limitations.
DFS Approach:
DFS checks for sonic flow limitations for each and every component. DFS also addresses the limitations in flow rate associated with component reduced flow areas. Further, DFS considers limitations
associated with heat transfer when analyzing compressible isothermal flow. | {"url":"https://abzinc.com/dfs_vs_others.php","timestamp":"2024-11-08T01:50:23Z","content_type":"text/html","content_length":"29574","record_id":"<urn:uuid:ce73d249-cdc7-4871-9882-289585f6bf49>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00102.warc.gz"} |
Mla Paper Spell Out Ordinal Numbers - OrdinalNumbers.com
Mla Paper Spell Out Ordinal Numbers
Mla Paper Spell Out Ordinal Numbers – By using ordinal numbers, it is possible to count infinite sets. You can also use them to generalize ordinal number.
The fundamental concept in math is that of the ordinal. It is a numerical value that represents the place of an item within a list. In general, a number between one and 20 is utilized to indicate the
ordinal number. Although ordinal numbers serve various purposes however, they are typically utilized to represent the sequence that items are placed in a list.
It is possible to display ordinal numbers using numbers or words. Charts, charts, and even charts. They may also be used to explain how pieces of a collection are arranged.
The majority of ordinal numbers are classified into one or one or. Transfinite ordinals will be represented with lowercase Greek letters. The finite ordinals will be represented in Arabic numbers.
According to the axiom that every set well-ordered should have at minimum one ordinal. For instance, the highest possible grade could be given to the first student in the class. The student who
received the highest grade was declared to be the contest’s winner.
Combinational ordinal figures
Multidigit numbers are also known as compound ordinal number. They are generated by multiplying an ordinal by its last number. These numbers are used typically for dating purposes and for ranking.
They don’t have a unique end to the last digit, as do cardinal numbers.
Ordinal numerals are used to indicate the order of elements in the collection. These numbers also serve to denote the names of items in collections. Regular numbers are found in both regular and
suppletive formats.
Regular ordinals are created by prefixing a cardinal number by the suffix -u. The numbers are then typed in the form of a word. A hyphen then added to it. There are numerous suffixes available.
Suppletive ordinals are derived from prefixing words with -u. The suffix can be used to count. It’s also wider than the normal one.
Limit of ordinal importance
Limits on ordinal numbers are ordinal numbers that aren’t zero. Limit ordinal quantities suffer from the disadvantage of not having a maximum element. They are created by joining non-empty set with
no any maximum elements.
Infinite transfinite-recursion definitions use limited ordinal number. Each infinite number of cardinals, as per the von Neumann model can also be considered an ordinal limit.
An ordinal number with the limit is in reality equivalent to the total of all the ordinals beneath it. Limit ordinal figures are enumerated with arithmetic but could also be expressed as an order of
natural numbers.
The ordinal numbers used to arrange the data are employed. They are used to explain an object’s numerical place. These numbers are frequently used in set theory or math. Although they have a similar
structure to natural numbers, they’re not part of the same class.
The von Neumann Model uses a well-ordered and ordered set. It is assumed that fyyfy represents one of the subfunctions g’ of a function that is described as a singular operation. If fy is the only
subfunction (ii) then it must be able to meet the specifications.
The Church-Kleene oral is an limit order in a similar fashion. A limit ordinal is a properly-ordered collection of smaller ordinals. It is an ordinal with a nonzero value.
Ordinal number examples in stories
Ordinal numbers are used frequently to show the structure of entities and objects. They are essential in organising, counting and ranking purposes. They are able to explain the location of the object
as well as the order of their placement.
The ordinal number is usually indicated by the letter “th”. Sometimes however, the letter “nd” is able to be substituted. The titles of books usually contain ordinal numbers.
Even though ordinal number are commonly used in list format they can still be written in words. They may also be referred to as numbers and acronyms. Comparatively speaking, these numbers are more
comprehensible than cardinal ones.
Three different kinds of ordinal numbers are offered. Learn more about them through practicing or games as well as other activities. It is important to learn about them to improve your arithmetic
skills. Try coloring exercises as an easy and enjoyable approach to improving. A simple marking sheet can be used to track your progress.
Gallery of Mla Paper Spell Out Ordinal Numbers
Apa Or Mla Style Quick Answers References 6th Edition 2022 10 31
Mla Spell Out Numbers Numbers Spell Out Or Use Numerals Number
When To Spell Out Numbers Rules For Writing Numbers In APA Chicago
Leave a Comment | {"url":"https://www.ordinalnumbers.com/mla-paper-spell-out-ordinal-numbers/","timestamp":"2024-11-06T07:33:01Z","content_type":"text/html","content_length":"62317","record_id":"<urn:uuid:827df606-232e-4c50-9ff5-3442f9313299>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00109.warc.gz"} |
Sunset Valley [RCT2]
its says you have to open it with something its not an rct2 saved game and i cant even see whats in the file.
Top Posters In This Topic
It's a RCT2 Saved game but when I download it it looks like it's a .jpg. I know it's not... and a lot of other things in the game exchange are doing the same thing for me (even files I didn't
upload). I tried downloading it in Internet Explorer and it worked but the game exchange is really screwed up in Firefox.
I've attached it in this post, so if the game exchange isn't working... try this.
Thanks again!
Edited by coasterbill
Your welcome!
New problem . It says it cant load " AE JUNGlE". Sorry for double post.
I hit the export button but here are all the rides you'll need if they're not exporting.
Let me know if this works.
Have you considered submitting this at NE (New Element)? I think it would do well for its sheer size and great coasters.
Would've been cool if you built a Giovanola Hyper, but ohh well, anyways I downloaded this from the exchange, & it worked just fine!
I thought about it but with Cannonball Run, Desperado, Colossus, Vampire, Cheetah Run and Shangaan being over 200 feet tall I think I have that coaster type more than covered.
RCT2Day... I hadn't thought of it but I take that as a big compliment. It's not a bad idea, I might release it over there.
Still not working It says I'm missing AE-JUNGL
Do you have to have 8 car trainer for this park to work? I have all the rides and stuff in the right place but it gives me a "Violation" error. Does 8 car work on windows XP?
Let me get some stuff straightened out here -
All custom rides featured in this saved game must be manually placed into your ObjData folder or the saved game will not open. Custom rides do not export like objects do.
Edited by A.J.
Still not working It says I'm missing AE-JUNGL
That's included in that last attachment, and it's now included in the download on the exchange. Download that last attachment or re-download the file and put everything in the object data file and it
will work. There are some really neat custom rides in there so you won't regret downloading it.
The file you're talking about sounds like Amazing Earl's Jungle Cruise boats.
how do you put object files into rct2 files?
I'm missing RIDETESJ.
Well, you right click on the RCT2 Icon seen on your desktop, or Start, choose open file location then you should see ObjData folder, then you drag the file into the objdata folder, it works for me!
it dosent say it when i right click on it. i have windows XP BTW.
Well, I have Windows Ultimate 7, and it says!
I'm missing RIDETESJ.
Thanks for your patience everyone. I'm a rookie at uploading parks but I'll get it to work for everyone... and you'll get some cool rides out of it. I've updated the download in the exchange so that
file is included (I thought I got everything... sorry!!!)
it dosent say it when i right click on it. i have windows XP BTW.
Go to "My Computer" in the green start menu. Then click on "C" then "program files" then "infogames interactive" then "Rollercoaster Tycoon 2" then "objdata".
Also I have a problem with the park while playing it, the guests flow down, & the rating goes down quickly something not right!
There's a reason for that.
This park seems to be at it's peak efficiency at about 8,200 - 8,300 guests. Before saving it I brought in some trams so it's a little more crowded than that (It's at full capacity). Until it gets
down to that number very few guests come in but once it does the park runs efficiently and more guests start coming in to balance the few guests that are leaving.
Also, because of the sheer size of the park once you bring in some trams with the trainer, some guests are bound to leave because they all go to the same places and get frustrated when the lines are
all full. Since most guests immediately head to Cheetah Run and Colossus and they're near the back of the park... and when some of them (naturally) try to leave they have a long walk to the entrance
and their presence in the park brings the rating down. This balances out after 30-45 minutes.
Other than that, what do you think of the park? Thanks for playing!
I really like your park, just as I like your Sunset Lake, but, let me ask you, if this park was real, where could this park be really located, I'll take my first guess, Colorado Springs CO.
I'm glad to hear it! I think Sunset Lake might be next to get a big overhaul. I have some other Sunset Parks that will never be on par with the parks I build now but I think with some custom supports
and redesigns of some areas that park could be a lot better. Don't be shocked to see a re-release somewhere down the line.
Regarding the location of this park, I thought about that and I'm really not sure. Hasta Fiesta has a lot of palm trees so it would have to be in a warm climate, and it's very hilly so I'm not really
sure. I decided not to give the parks specific locations so people were free to envision them wherever they wanted.
Edited by coasterbill
You mean redo the Sunset Lake? Well that would be really great!
Loved this. The atmosphere is great in every area. By far my favorite was Vampire Forest. That little area was phenomenal and the coaster was great. The only thing that I have to nit pick would be
some of the coasters. Going over a hill or around a corner doing 70+ is a little extreme, lol. Overall I liked this.
P.S. I blamed you for not getting all my work done today for I was completely enthralled in this park. lol. I think I spent forty minutes on VF alone. Really love that little area.
Thanks! I'm really happy to hear that you enjoyed the park. I see your point with the fast cornering hills (I assume you're referring to Vampire especially). I know Cheetah Run blows over the hill at
30 MPH but with an empty train it goes over at 4-5 MPH so I can't slow it down any more. That annoys me a little bit.
Is anyone still having problems opening the park? | {"url":"https://themeparkreview.com/forum/topic/35956-sunset-valley-rct2/page/7/","timestamp":"2024-11-06T08:32:00Z","content_type":"text/html","content_length":"390937","record_id":"<urn:uuid:4666fac7-52c9-4d7e-b7d1-594585c25959>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00739.warc.gz"} |
How Do I
Enter a Simple Expression?
In Maple, mathematical expressions, such as $\sqrt{{x}^{2}+\mathrm{π}}$, can be entered in a natural,
textbook-like notation. In Maple, this is called 2-D Math notation. The following examples demonstrate the
process of entering expressions.
To begin, create a new document in Document mode. To begin entering a 2-D mathematical expression, ensure that
you are in Math mode. In this mode, the cursor is inside a box with a dotted outline that indicates the
boundaries of the mathematical expression. By default, all new documents created in Document mode start in Math
mode so that you can begin entering mathematical expressions immediately. To switch between Math mode and Text
mode, press F5.
Example: $\frac{2x}{5}+\frac{3x}{7}$
Example: $\sqrt{{x}^{2}+\mathrm{π}}$
Related Topics
See Also
Example: $\frac{2x}{5}+\frac{3x}{7}$
Ensure that you are in Math mode, and follow the steps below to enter the expression $\frac
{2x}{5}+\frac{3x}{7}$. 2-D math
Step Description Illustration complete command
1 Type $2$, then $x$. Expression
2 Press forwards slash (/) to enter
the denominator.
3 Type $5$.
To bring the cursor back onto the
4 base line, press the right arrow
(→) key on your keyboard.
Type (+). There is no need to
5 type a space, as Maple
automatically adjusts the spacing
for you.
6 Repeat steps 1 through 4 to type
If you wish to continue typing
7 non-mathematical text, press F5
to switch to Text mode.
Note on Multiplication: In general, you need to indicate multiplication by using * or a
space. In 2-D math, * appears as a center dot ($\cdot$). In this example, the
multiplication of a number by a name is typed as simply 2, then x: $2x$. However, in other
cases you will need to make sure to use * or a space. Here are a few examples:
$\left(x+1\right)\cdot \left(x-2\right)$
$x\cdot y$
For more information, see Entering 2-D Math: Multiplication.
Example: $\sqrt{{x}^{2}+\mathrm{π}}$
Ensure that you are in Math mode, and follow the steps below to enter the expression $\sqrt
{{x}^{2}+\mathrm{\pi }}$.
Step Description Illustration
Click the $\sqrt{{a}}$ template in the
1 Expression palette. The green placeholder${a}$
is selected automatically.
2 Type $x$ to overwrite the ${a}$.
To place the cursor in the exponent position,
3 press Shift + ^.
Type $2$.
To bring the cursor back onto the base line,
4 press the right arrow [→] key on your
5 Type +. There is no need to type a space, as
Maple automatically adjusts spacing for you.
Type the word pi, and press Esc. The command
completion list appears, listing every
6 available symbol and command whose name begins
with pi. For the Greek symbol for pi, select $
If you wish to continue typing
7 non-mathematical text, press F5 to switch to
Text mode.
Related Topics
The How Do I... topics cover the essentials for doing mathematics in Maple. Learn more
about available tools and features, such as palettes and the context panel.
How Do I... Tools and Features
...Enter a Complex Number? Palettes
...Enter a Function? Context Panel
...Enter a Matrix? Command Completion
...Evaluate an Expression? Equation Labels
...Import Tabular Data? Assistants
...Plot a Function? Maple Help
...Plot a Straight Line? Plotting Guide
...Plot Multiple Functions? Applications
...Solve an Ordinary Differential Equation? Example Worksheets
...Work with Random Generators? Manuals
Refer to Help>Quick Reference for basic Getting Started tips.
In Maple, mathematical expressions, such as $\sqrt{{x}^{2}+\mathrm{π}}$, can be entered in a natural, textbook-like notation. In Maple, this is called 2-D Math notation. The following
examples demonstrate the process of entering expressions.
To begin, create a new document in Document mode. To begin entering a 2-D mathematical expression, ensure that you are in Math mode. In this mode, the cursor is inside a box with a dotted outline
that indicates the boundaries of the mathematical expression. By default, all new documents created in Document mode start in Math mode so that you can begin entering mathematical expressions
immediately. To switch between Math mode and Text mode, press F5.
See Also
Example: $\frac{2x}{5}+\frac{3x}{7}$
Ensure that you are in Math mode, and follow the steps below to enter the expression $\frac
{2x}{5}+\frac{3x}{7}$. 2-D math
Step Description Illustration complete command
1 Type $2$, then $x$. Expression
2 Press forwards slash (/) to enter
the denominator.
3 Type $5$.
To bring the cursor back onto the
4 base line, press the right arrow
(→) key on your keyboard.
Type (+). There is no need to
5 type a space, as Maple
automatically adjusts the spacing
for you.
6 Repeat steps 1 through 4 to type
If you wish to continue typing
7 non-mathematical text, press F5
to switch to Text mode.
Note on Multiplication: In general, you need to indicate multiplication by using * or a
space. In 2-D math, * appears as a center dot ($\cdot$). In this example, the
multiplication of a number by a name is typed as simply 2, then x: $2x$. However, in other
cases you will need to make sure to use * or a space. Here are a few examples:
$\left(x+1\right)\cdot \left(x-2\right)$
$x\cdot y$
For more information, see Entering 2-D Math: Multiplication.
Example: $\sqrt{{x}^{2}+\mathrm{π}}$
Ensure that you are in Math mode, and follow the steps below to enter the expression $\sqrt
{{x}^{2}+\mathrm{\pi }}$.
Step Description Illustration
Click the $\sqrt{{a}}$ template in the
1 Expression palette. The green placeholder${a}$
is selected automatically.
2 Type $x$ to overwrite the ${a}$.
To place the cursor in the exponent position,
3 press Shift + ^.
Type $2$.
To bring the cursor back onto the base line,
4 press the right arrow [→] key on your
5 Type +. There is no need to type a space, as
Maple automatically adjusts spacing for you.
Type the word pi, and press Esc. The command
completion list appears, listing every
6 available symbol and command whose name begins
with pi. For the Greek symbol for pi, select $
If you wish to continue typing
7 non-mathematical text, press F5 to switch to
Text mode.
Related Topics
The How Do I... topics cover the essentials for doing mathematics in Maple. Learn more
about available tools and features, such as palettes and the context panel.
How Do I... Tools and Features
...Enter a Complex Number? Palettes
...Enter a Function? Context Panel
...Enter a Matrix? Command Completion
...Evaluate an Expression? Equation Labels
...Import Tabular Data? Assistants
...Plot a Function? Maple Help
...Plot a Straight Line? Plotting Guide
...Plot Multiple Functions? Applications
...Solve an Ordinary Differential Equation? Example Worksheets
...Work with Random Generators? Manuals
Refer to Help>Quick Reference for basic Getting Started tips.
Example: $\frac{2x}{5}+\frac{3x}{7}$
Ensure that you are in Math mode, and follow the steps below to enter the expression $\frac{2x}{5}+\frac{3x}{7}$.
Step Description Illustration
1 Type $2$, then $x$.
2 Press forwards slash (/) to enter
the denominator.
3 Type $5$.
To bring the cursor back onto the
4 base line, press the right arrow
(→) key on your keyboard.
Type (+). There is no need to
5 type a space, as Maple
automatically adjusts the spacing
for you.
6 Repeat steps 1 through 4 to type
If you wish to continue typing
7 non-mathematical text, press F5
to switch to Text mode.
Note on Multiplication: In general, you need to indicate multiplication by using * or a space. In 2-D math, * appears as a center dot ($\cdot$). In this example, the multiplication of a number by a
name is typed as simply 2, then x: $2x$. However, in other cases you will need to make sure to use * or a space. Here are a few examples:
$\left(x+1\right)\cdot \left(x-2\right)$
$x\cdot y$
For more information, see Entering 2-D Math: Multiplication.
Ensure that you are in Math mode, and follow the steps below to enter the expression $\frac{2x}{5}+\frac{3x}{7}$.
Step Description Illustration
1 Type $2$, then $x$.
2 Press forwards slash (/) to enter
the denominator.
3 Type $5$.
To bring the cursor back onto the
4 base line, press the right arrow
(→) key on your keyboard.
Type (+). There is no need to
5 type a space, as Maple
automatically adjusts the spacing
for you.
6 Repeat steps 1 through 4 to type
If you wish to continue typing
7 non-mathematical text, press F5
to switch to Text mode.
To bring the cursor back onto the base line, press the right arrow (→) key on your keyboard.
Type (+). There is no need to type a space, as Maple automatically adjusts the spacing for you.
If you wish to continue typing non-mathematical text, press F5 to switch to Text mode.
Note on Multiplication: In general, you need to indicate multiplication by using * or a space. In 2-D math, * appears as a center dot ($\cdot$). In this example, the multiplication of a number by a
name is typed as simply 2, then x: $2x$. However, in other cases you will need to make sure to use * or a space. Here are a few examples:
Example: $\sqrt{{x}^{2}+\mathrm{π}}$
Ensure that you are in Math mode, and follow the steps below to enter the expression $\sqrt{{x}^{2}+\mathrm{\pi }}$.
Step Description Illustration
Click the $\sqrt{{a}}$ template in the
1 Expression palette. The green placeholder${a}$
is selected automatically.
2 Type $x$ to overwrite the ${a}$.
To place the cursor in the exponent position,
3 press Shift + ^.
Type $2$.
To bring the cursor back onto the base line,
4 press the right arrow [→] key on your
5 Type +. There is no need to type a space, as
Maple automatically adjusts spacing for you.
Type the word pi, and press Esc. The command
completion list appears, listing every
6 available symbol and command whose name begins
with pi. For the Greek symbol for pi, select $
If you wish to continue typing
7 non-mathematical text, press F5 to switch to
Text mode.
Ensure that you are in Math mode, and follow the steps below to enter the expression $\sqrt{{x}^{2}+\mathrm{\pi }}$.
Step Description Illustration
Click the $\sqrt{{a}}$ template in the
1 Expression palette. The green placeholder${a}$
is selected automatically.
2 Type $x$ to overwrite the ${a}$.
To place the cursor in the exponent position,
3 press Shift + ^.
Type $2$.
To bring the cursor back onto the base line,
4 press the right arrow [→] key on your
5 Type +. There is no need to type a space, as
Maple automatically adjusts spacing for you.
Type the word pi, and press Esc. The command
completion list appears, listing every
6 available symbol and command whose name begins
with pi. For the Greek symbol for pi, select $
If you wish to continue typing
7 non-mathematical text, press F5 to switch to
Text mode.
Click the $\sqrt{{a}}$ template in the Expression palette. The green placeholder${a}$ is selected automatically.
To place the cursor in the exponent position, press Shift + ^.
To bring the cursor back onto the base line, press the right arrow [→] key on your keyboard.
Type +. There is no need to type a space, as Maple automatically adjusts spacing for you.
Type the word pi, and press Esc. The command completion list appears, listing every available symbol and command whose name begins with pi. For the Greek symbol for pi, select $\mathbf{π}$.
Related Topics
The How Do I... topics cover the essentials for doing mathematics in Maple. Learn more about available tools and features, such as palettes and the context panel.
How Do I... Tools and Features
...Enter a Complex Number? Palettes
...Enter a Function? Context Panel
...Enter a Matrix? Command Completion
...Evaluate an Expression? Equation Labels
...Import Tabular Data? Assistants
...Plot a Function? Maple Help
...Plot a Straight Line? Plotting Guide
...Plot Multiple Functions? Applications
...Solve an Ordinary Differential Equation? Example Worksheets
...Work with Random Generators? Manuals
Refer to Help>Quick Reference for basic Getting Started tips.
The How Do I... topics cover the essentials for doing mathematics in Maple. Learn more about available tools and features, such as palettes and the context panel.
How Do I... Tools and Features
...Enter a Complex Number? Palettes
...Enter a Function? Context Panel
...Enter a Matrix? Command Completion
...Evaluate an Expression? Equation Labels
...Import Tabular Data? Assistants
...Plot a Function? Maple Help
...Plot a Straight Line? Plotting Guide
...Plot Multiple Functions? Applications
...Solve an Ordinary Differential Equation? Example Worksheets
...Work with Random Generators? Manuals | {"url":"https://www.maplesoft.com/support/help/maple/view.aspx?path=HowDoI/EnterAnExpression","timestamp":"2024-11-06T18:26:26Z","content_type":"text/html","content_length":"157447","record_id":"<urn:uuid:f3b5492a-e2cc-4b40-9724-4e051aad5605>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00216.warc.gz"} |
Suguru Endo | Innovators Under 35
Among the challenges for quantum computers is how to mitigate calculation errors caused by noise, which refers to the breakdown of the quantum state as a result of slight vibrations and changes in
temperature. Suguru Endo established the world's first practical method of quantum error mitigation and proved that it is a useful countermeasure for noise in NISQ (Noisy Intermediate-Scale Quantum)
At present, NISQ computers, which are being developed by companies like Google and IBM, have an extremely large amount of noise, rendering them useless in the real world unless the impact of this
noise can be mitigated. Since the time that he started his doctoral program at the University of Oxford, Endo has engaged in research on algorithms for NISQ computers, as well as quantum error
mitigation to reduce their calculation errors, and the papers he has written have been cited over 1,400 times.
The concept of quantum error mitigation had been proposed prior to Endo's research. However, it had no practicality due to its limitations, such as only being available for use when there is complete
knowledge beforehand that a specified error would occur. Based on the characteristics of incomplete errors that could actually be measured through experimentation, Endo proposed the world's first
method that would allow quantum error mitigation to function against unspecified errors and significantly expanded the range of application for quantum error mitigation.
Furthermore, Endo is also researching hybrid quantum/classical algorithms, which are believed to be optimal for NISQ computers, in parallel. At the time that these algorithms were first proposed,
their use was limited to tasks such as calculating the ground state of molecules. However, Endo has developed algorithms that enable essential linear algebra operations for property analysis and
machine learning, the design of quantum sensors, and simulation of open quantum systems in order to analyze nanodevices. He greatly expanded the potential of NISQ computers.
Endo's activities go beyond the academic world. He has also thrived in his collaborations with industry institutions, such as through his internship with a venture capital firm that develops quantum
computers and software and his joint research with Mercari, the major online marketplace. Endo's words that he will "bring the spread of quantum computers across society forward by at least 5 years
by conducting research and sharing the results with the world" carry a significant amount of weight. | {"url":"https://www.innovatorsunder35.com/the-list/suguru-endo/","timestamp":"2024-11-08T12:27:39Z","content_type":"text/html","content_length":"54165","record_id":"<urn:uuid:3c8695ba-d9a1-42c3-a84d-5824c3bda05b>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00100.warc.gz"} |
Hyperbolic cosine function
To use this function, choose .
Calculates the hyperbolic cosine of an angle. Hyperbolic trigonometric functions are based on the hyperbola with the equation x^2 – y^2 = 1. These functions differ from those in standard trigonometry
(also called circular trigonometry), whose functions are based on the unit circle with the equation x^2 + y^2 = 1. However, they share many similar identities, such as sinh^2 x + cosh^2 x = 1, where
h represents hyperbolic.
For number, specify the radians or the column of radians.
Column Calculator expression Result
C1 contains -5 COSH (C1) 7.42099485248E+01
Hyperbolic functions have many useful applications in engineering, such as electrical transportation (to calculate length, weight, and stress of cables and conducting wires), superstructure (to
compute elastic curves and deflection of suspension bridges), and aerospace (to determine ideal surface coatings for aircraft). In statistics, the inverse hyperbolic sine is used in the Johnson
transformation to transform the data so it follows a normal distribution. Normality is a necessary assumption for some capability analyses.
For a specified value of x, cosh x = (e^x + e^−x) / 2, where h represents hyperbolic, and e is the constant equal to approximately 2.718.
The inverse of the function is arccosh x (cosh^−1 x). | {"url":"https://support.minitab.com/en-us/minitab/help-and-how-to/calculations-data-generation-and-matrices/calculator/calculator-functions/trigonometry-calculator-functions/hyperbolic-cosine-function/","timestamp":"2024-11-06T02:30:32Z","content_type":"text/html","content_length":"11950","record_id":"<urn:uuid:170595ed-f0ba-4146-8289-40d8f91f2321>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00210.warc.gz"} |
Water Acre Feet Calculator - Easily Estimate Water Storage Needs - Calculator Pack
Water Acre Feet Calculator
Greetings, water resource managers, irrigators, and hydrology enthusiasts! Are you tired of manually calculating acre-feet of water storage or usage? Look no further! With our convenient Water Acre
Feet Calculator, you can quickly and efficiently determine the volume of water in an area in the commonly used unit of acre-feet.
Simply input the dimensions of your area, such as length and width, as well as the depth of water, and our calculator will handle the rest. Whether you're measuring the amount of water in a
reservoir, a field, or a pond, our tool will provide quick and accurate results.
Stop wasting time and effort on manual calculations - let our Water Acre Feet Calculator do the work for you. Try it out today!
Water Acre Feet Calculator
Calculate the volume of water in acre feet based on length, width, and depth of a reservoir or pond.
Water Acre Feet Calculator Results
Length 0
Width 0
Depth 0
Volume (cubic feet) 0
Water Acre Feet 0
Share results with your friends
estimating water acre feet is essential for managing water resources, particularly in agricultural settings. Our water acre feet calculator streamlines this calculation. To understand pond water
volume more comprehensively or assess water requirements, link it with our pond water volume calculator. This pairing offers comprehensive guidance for managing water resources.
How to use this Calcualtor
To use this Water Acre Feet Calculator, follow the simple steps below.
Step 1: Enter Length The first input field in the calculator is for the length of the pond or reservoir in feet. Enter the length of the pond or reservoir in this field. You can use decimal points
for more precise measurements.
Step 2: Enter Width The second input field is for the width of the pond or reservoir in feet. Enter the width of the pond or reservoir in this field. As with the length field, you can use decimal
points for more precise measurements.
Step 3: Enter Depth The third input field is for the depth of the pond or reservoir in feet. Enter the depth of the pond or reservoir in this field. Again, you can use decimal points for more precise
Step 4: Calculate Acre Feet Once you have entered the values for length, width, and depth, click the "Calculate Acre Feet" button. The calculator will then provide you with the volume in cubic feet
and the volume in acre feet.
The volume in acre feet is a measure of the volume of water that would cover an area of one acre to a depth of one foot. It is commonly used in the management and conservation of water resources,
particularly in agriculture and irrigation.
Water Acre Feet Formula:
To calculate the volume of water in acre feet based on the length, width, and depth of a reservoir or pond, you can use the following formula:
Volume (cubic feet) = Length * Width * Depth
Water Acre Feet = Volume (cubic feet) / 43560
The formula for calculating the volume of water in cubic feet is derived from the basic concept of multiplying the three dimensions of length, width, and depth. By multiplying these values together,
we can determine the total volume of the reservoir or pond in cubic feet.
To convert the volume from cubic feet to acre feet, we divide the volume by a conversion factor of 43560. This conversion factor represents the number of square feet in an acre. Dividing the volume
by this factor gives us the volume in terms of acre feet, which is a commonly used unit for measuring large quantities of water.
Let's consider an example where the length of the reservoir is 100 feet, the width is 50 feet, and the depth is 10 feet. We can use the formula to calculate the volume and water acre feet as follows:
Volume (cubic feet) = 100 * 50 * 10 = 50,000 cubic feet
Water Acre Feet = 50,000 / 43560 = 1.15 acre feet
In this example, the volume of water in the reservoir is 50,000 cubic feet, which is equivalent to approximately 1.15 acre feet.
Here is a table showing multiple rows of calculations using the formula for different reservoir dimensions:
Length (ft) Width (ft) Depth (ft) Volume (cubic feet) Water Acre Feet
100 50 10 50,000 1.15
75 60 12 54,000 1.24
120 80 8 76,800 1.76
90 70 15 94,500 2.17
In each row of the table, the values for length, width, and depth are given. Using the formula, the corresponding volume in cubic feet and water acre feet are calculated and presented in the
respective columns.
Please note that these examples are for illustrative purposes only, and the actual values may vary depending on the specific dimensions of the reservoir or pond.
Using this calculator can help you determine the volume of water in a pond or reservoir and how much water you have available for irrigation or other uses. It can also help you monitor the water
level in your pond or reservoir and make informed decisions about water management. | {"url":"https://calculatorpack.com/water-acre-feet-calculator/","timestamp":"2024-11-09T07:05:25Z","content_type":"text/html","content_length":"34814","record_id":"<urn:uuid:14f67c14-8765-4989-bd40-ab1ca9d206a4>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00195.warc.gz"} |
Can I pay for Python assignment help for projects requiring the development of algorithms for analyzing and interpreting data from wearable devices in healthcare? | Pay Someone To Do My Python Assignment
Can I pay for Python assignment help for projects requiring the development of algorithms for analyzing and interpreting data from wearable devices in healthcare? Do you need to read the blog post on
Algorithms, which discusses the benefits of using software to analyze device data, particularly as technology evolves? If not, I would look into the use case for this approach. Unfortunately, the
information I have requested is not currently available in the GitHub repository. So when I looked, as an example, I wanted to understand the data in a wearable computer, not as a physician-managed
collection of data. With my research, I see a small data file that looks like this: % 0$s a dataset (numbers) 2×10, 8, 7 x10 and 4×10, 17 x4 A dataset looks like this, as shown in 10 images each: 4,
4 x8, 16, 39 x34, 100 x34. Deduced the resulting images are annotated with 5, 12 and 26 images when you walk by using the algorithm I presented above. As for the problems with data loss, the 3 image
problem is a very interesting one, because you do not have enough data, you may need to set up another algorithm, which is faster, perform better and has better results than other algorithms that
call it your dataset. These methods generally operate according to the loss function, which measures the loss of the data, or the quantity of data you have processed. I would agree with you, however,
that what I suggest, as outlined in the use case, is not the loss of the data itself. It is a much smaller loss than you can actually measure in a little bit at a time. You will have a much better
chance of getting the data in a better way, regardless of the difficulty of the task. The algorithm I discussed contains an algorithm called Quant+Min-Size, which is essentially a form of
Normalization of size. The algorithm needs moreCan I pay for Python assignment help for projects requiring the development of algorithms for analyzing and interpreting data from wearable devices in
healthcare? I have been investigating and planning for a paper on our work (that I have written for the last 3 months) about the “stray” algorithm developed by Professor John W. Odenkas. I was
wondering if it is possible that a study about computer algorithms for classification of patients and control of them under such conditions would be useful in my research. The goal of my project was
to detect the algorithms used to divide patients into four groups via a distribution of normal and abnormally abnormal activity (positive and negative). These groups would be separated by a
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it is a must take it a shot. If you have my valuable and detailed comments, please notify me shortly.
Is Using A Launchpad Cheating
To be more precise, this is not what I would do for the algorithm that I have presented elsewhere, although I know of articles on a number of others; I have read the paper and heard the references.
This paper was written by Oxford University Press, and was based on a discussion of the algorithms used to divide patients into four groups (in terms of their abnormal behaviour). In doing their
search of the paper I relied mostly on what is now a number of documents from my library: some are articles posted on my blog. There is no evidence that these articles are included in the papers
because of the abstract. I would in most cases prefer to be seen in a journal style paper but where there isn’t this concern, I would argue that this will be a better way to do this and to have the
article published so there are no errors contained. For example, a very interesting number of papers concerning self-therapy programmes are from my international library. It is published in
E-publishers with a minimum of two online editing costs, and is an effective way toCan I pay for Python assignment help for projects requiring the development of algorithms for analyzing and
interpreting data from wearable devices in Home Well, I don’t have the experience to answer this, but I should point out that every time I see people who don’t have experience with an application
(which really depends to a large degree on my knowledge of a lot of software, especially software in terms of architecture and system as well), it makes me think of this: When, for instance, when
they know something new is going on in their home and it has an algorithm for analyzing it, it comes up a lot, including when they have not done this in the past and they do so here. Also, pop over
to this site takes a while to find out what algorithm to use for evaluating health outcomes, like the two things the FDA recently applied in their 2012 United States medical literature. So the reason
my interest is, is that people think it’s really simple, but it looks like it may be extremely complicated. So are you just going to have to spend a little bit more time researching the algorithms,
methods, and applications of algorithms with your application, making sure it’s a coherent science? Today is no more interesting than it has ever been last year. Most of the healthcare data is used
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NCERT Solutions for Class 6 Maths | Cuemath
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
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Class 6 Maths
Class 6 math ensures to strengthen and build the reasoning skills and the basic concepts of students. Class 6 is the upper primary stage where students begin to understand and apply the mathematical
language of symbols and equations. It focuses on implementing a profound knowledge for working with advanced concepts by learning various methods, formulas, and theorems. Therefore, it is necessary
for students to have a detailed understanding of each topic covered in the class 6 maths syllabus.
NCERT Solutions for Class 6 Maths PDF Download
NCERT solutions for class 6 maths comprise a detailed explanation of all the textbook questions available in the NCERT mathematics class 6 coursebook. As they provide a stepwise analysis of all
problems, these solutions are the best cross-reference material. Kids can check their answers as well as get an idea of the correct way to write the solutions to questions. The NCERT solutions for
class 6 maths are prepared by experts in the field of mathematics. They have spent a great amount of time researching the most precise and optimized method to reach the final result.
The NCERT solutions for class 6 maths have several study tips and tricks that help to streamline the learning process. They also use illustrative examples, graphs, and real-life problems that help
kids relate to the topics in a better manner. The well-structured and precise solutions to all the NCERT chapters can be accessed via the links given below:
Class 6 Maths NCERT Solutions Chapter 1 to 14 (PDFs)
NCERT Solutions for Class 6 Maths: Syllabus
The NCERT class 6 maths syllabus comprises topics related to numbers, addition, subtraction, and more. A detailed list of these topics is given below.
The topics included in the class 6 chapters have concepts that lay the groundwork for higher-order lessons. If children are not familiar and well-versed with these it could prove to be problematic as
they reach further grades. Thus, to avoid superfluous understanding children should periodically visit the links given above to revise the chapters thoroughly. To effectively prepare for exams the
NCERT class 6 maths book is given below:
>> Download Class 6 Maths NCERT Book
NCERT Solutions Class 6 Maths Chapter 1 - Knowing Our Numbers
• Number comparison
• Place value notation
• Usage of commas and brackets in numbers
• Arithmetic operations on numbers
• Estimation by round off
NCERT Solutions Class 6 Maths Chapter 1: Exercise
NCERT Solutions Class 6 Maths Chapter 2: Whole Numbers
• Predecessor and successor
• The number line
• Properties of whole numbers
• Patterns in whole numbers
NCERT Solutions Class 6 Maths Chapter 2: Important Formulas
• Associative Property for addition: (a + b) + c = a + (b + c)
• Associative Property for multiplication: (a * b) * c = a * (b * c)
• Commutative Property for addition: a + b = b + a
• Commutative Property for multiplication: a * b = b * a
NCERT Solutions Class 6 Maths Chapter 2: Exercise
NCERT Solutions Class 6 Maths Chapter 3: Playing with Numbers
• Factors and multiples
• Prime numbers and Composite numbers
• Test for divisibility of numbers
• Common factors and common multiples
• Prime factorization
• Highest Common Factor (HCF)
• Lowest Common Multiple (LCM)
NCERT Solutions Class 6 Maths Chapter 3: Important Formulas
• Divisibility by 2: A number is divisible by 2 if it has any of the digits 0, 2, 4, 6, or 8 in its one’s place.
• Divisibility by 3: If the sum of the digits is a multiple of 3, then the number is divisible by 3.
• Divisibility by 4: A number with 3 or more digits is divisible by 4 if the number formed by its last two digits (i.e. ones and tens) is divisible by 4.
• Divisibility by 5: A number that has either 0 or 5 in its one’s place is divisible by 5.
NCERT Solutions Class 6 Maths Chapter 3: Exercise
NCERT Solutions Class 6 Maths Chapter 4: Basic Geometrical Ideas
• Points
• Line segment
• Line
• Intersecting lines
• Parallel lines
• Curves
• Polygons
• Angles
• Triangles
• Quadrilaterals
• Circles
NCERT Solutions Class 6 Maths Chapter 4: Exercise
NCERT Solutions Class 6 Maths Chapter 5: Understanding Elementary Shapes
• Measuring line segments
• Angles - right angle and straight angle
• Angles - acute angle, obtuse angle, reflex angle
• Measuring angles
• Perpendicular lines
• Classification of triangles
• Quadrilaterals
• Polygons
• Three-dimensional shapes (3D shapes)
NCERT Solutions Class 6 Maths Chapter 5: Exercise
NCERT Solutions Class 6 Maths Chapter 6 - Integers
• Tag with a sign
• Integers
• Representation of integers on a number line
• Ordering of Integers
• Addition of integers
• Addition of integers on a number line
• Subtraction of Integers with the help of a number line
NCERT Solutions Class 6 Maths Chapter 6 Important Formulas
• When two positive integers are added, we get a positive integer.
• When two negative integers are added, we get a negative integer.
• When one positive and one negative integer are added we subtract them as whole numbers by considering the numbers without their sign and then put the sign of the bigger number with the
subtraction obtained
NCERT Solutions Class 6 Maths Chapter 6 Exercises
NCERT Solutions Class 6 Maths Chapter 7 - Fractions
• Fractions
• Fraction on a number line
• Proper fractions
• Improper fractions, mixed fractions
• Equivalent fractions
• Simplest form of a fraction
• Like fractions
• Comparing fractions, comparing like fractions, comparing unlike fractions
• Addition of fractions, subtraction of fractions
NCERT Solutions Class 6 Maths Chapter 7 Exercises
NCERT Solutions Class 6 Maths Chapter 8 - Decimals
• Tenths
• Hundredths
• Comparing decimals
• Using decimals in money, length, weight
• Addition of numbers with decimals
• Subtraction of decimals
NCERT Solutions Class 6 Maths Chapter 8 Exercises
NCERT Solutions Class 6 Maths Chapter 9 - Data Handling
• Recording data
• Organization of data
• Pictograph
• Interpretation of a pictograph
• Drawing a pictograph
• A bar graph
• Interpretation of a bar graph
• Drawing a bar graph
NCERT Solutions Class 6 Maths Chapter 9: Exercise
NCERT Solutions Class 6 Maths Chapter 10 - Mensuration
• Introduction to the concept of Perimeter
• Perimeter for a rectangle
• Perimeter of regular shapes, perimeter of a square
• Introduction to the concept of area
• Area of a rectangle
• Area of a square
NCERT Solutions Class 6 Maths Chapter 10 Important Formulas
• Perimeter of a rectangle = 2 × (length + breadth)
• Perimeter of a square = 4 × length of a side
• Perimeter of an equilateral triangle = 3 × length of a side
• Area of a rectangle = (length × breadth)
• Area of the square = side × side
NCERT Solutions Class 6 Maths Chapter 10 Exercises
NCERT Solutions Class 6 Maths Chapter 11 - Algebra
• Introduction to algebra
• Matchstick patterns
• The idea of a variable
• Use of variables in common rules, rules from geometry, rules from arithmetic
• Expressions with variables
• Using expressions practically
• Equations
• Solution of an equation
NCERT Solutions Class 6 Maths Chapter 11 Exercises
NCERT Solutions Class 6 Maths Chapter 12 - Ratio and Proportion
• Understanding the concept of ratios
• Proportion as equality of two ratios
• Unitary method
NCERT Solutions Class 6 Maths Chapter 12 Exercises
Important Formulas in NCERT Solutions Class 6 Maths
Class 6 maths covers many important formulas and procedures that are vital to attempting questions. Formulas can be seen as tools that help simplify the process of solving an otherwise complicated
sum. It is advised for children to keep formula charts with handwritten notes to help them understand the underlying concepts as well as memorize the formulas quickly. Additionally, it is recommended
that kids apply these formulas to problems with a gradually increasing level of difficulty to get the most of these lessons. A few of the formulas used in the NCERT solutions for class 6 maths are
given below:
• Area of a square = (side)^2
• Perimeter of a square = 4 * side
• Area of a triangle = ½ * base * height
• Perimeter of a triangle = sum of all sides
• General form of a quadratic equation: ax^2 + bx + c = 0
Number System
• √(ab) = √(a)√(b)
• [√(a) + √(b)][√(a) - √(b)] = a - b
• Associative property: p + (q + r) = (p + q) + r
• Commutative property: p + q = q + p
Importance of NCERT Solutions Class 6 Maths
Class 6 Maths NCERT Solutions are of great importance for CBSE students as they give them an organized way to prepare for any exam. Given below are the benefits of using these comprehensive sources
for learning.
• Time Management - These NCERT Class 6 Maths Solutions set a good pace for students so that they not only complete all chapters well within the given time limit but also have enough buffer to
revise each lesson. During exams, they are very useful as these solutions give a bird’s eye view of all the topics so that kids can glance through them and recall concepts quickly.
• Attempting Exams - If the test papers are attempted in a step-wise manner with proper explanations given for each computation performed it enables kids to get the best possible score. Thus, by
using the NCERT Class 6 Maths Solutions students get an idea of how to efficiently present their answers in an examination to maximize the scope of getting good marks.
• Understanding Concepts - The NCERT Class 6 Maths Solutions have been written in a way to convey tough concepts in the easiest way. Thus, irrespective of the difficulty all kids can quickly
understand all topics and apply them to problems effectively.
FAQs on NCERT Solutions for Class 6 Maths
Do I Need to Practice all Questions Provided in NCERT Solutions Class 6 Maths?
Each question in the NCERT solutions for class six maths has been designed to give an insight into a different aspect of the topic at hand. If kids practice each sum, they will be sure to get a
holistic understanding of the chapter rather than only having superficial knowledge. Thus, children should revise all problems and give special attention to topics they find tough.
Do I Need to Make Notes while Referring to NCERT Solutions Class 6 Maths?
If kids take down notes while going through the NCERT Solutions Class 6 Maths then they have simplified pointers that act as guidelines while revising that topic. Thus, it is imperative for students
to make their own handwritten notes while studying these solutions as it helps them to recall concepts within a fraction of a second as well as improves their understanding of that section.
How are NCERT Solutions Class 6 Maths Promoting Problem Solving in Students?
The NCERT Solutions Class 6 Maths takes the step-by-step approach to solving questions. These solutions first break the big problem into simpler and more manageable chunks then proceed to solve each
part with logical explanations. This method encourages students to inculcate the same techniques to solving any question thus instilling them with a problem-solving mindset.
What are the Best Ways to Learn Concepts Covered in NCERT Solutions Class 6th Maths?
To learn the concepts covered in the NCERT Solutions Class 6th Maths kids first need to go through the complete theory before each exercise. The next step is to memorize all the associated formulas.
Children then need to attempt the solved examples and move on to the exercise questions. Finally, they should consult the NCERT solutions to cross-check their answers and get an idea of the
procedures used to solve these sums.
Where Can I Get Chapter-wise NCERT Solutions for Class 6 Maths?
Students can access the chapter-wise pdfs of the NCERT Solutions for Class 6 Maths via the links mentioned in this article. Each exercise question has been explained in a simplified language. They
also provide several study cues that can help kids understand and practice the sums more efficiently.
Can We Use Different Methods to Solve Problems Apart From Those Mentioned in NCERT Solutions Class 6 Maths?
The beauty of mathematics lies in the flexibility of solving problems using different and creative methods. Kids are encouraged to use techniques apart from those mentioned in the NCERT Solutions
Class 6 Maths to solve problems however, they should be universally accepted. Additionally, when appearing for exams students should make sure to explain all the steps in a detailed manner with the
required logic in order to get a good score.
Why Should I Practice NCERT Solutions Class 6 Maths Regularly?
Practice ensures perfection. The topics in the NCERT Solutions Class 6 Maths make their way in higher classes as well. If students can build a flawless foundational base they will not only excel in
the grade 6 exams but will also be prepared to assimilate more complicated topics that are yet to come. Thus, it is necessary to practice each and every question at least twice so as to build a
robust understanding of all chapters.
Which is the hardest chapter in class 6 maths?
In class six mathematics, there is no chapter that is difficult. However, there are a few topics that are introduced in this class which need a little more time and attention. For example, algebra,
geometry, and mensuration are a few new topics that need time, but these can be mastered once the concepts are clear. Proper planning and regular practice of the related questions from different
resources also helps because this makes the student think from different perspectives.
How to study mental maths for class 6?
Mental maths for class 6 is similar to physical fitness training. It may seem to be painful and difficult initially, but with regular practice, it always improves and helps. Mental maths improves our
memory and can be learnt using the following techniques:
• Try to create games related to mental maths along with other students. This helps in making it interesting and it includes some fun elements as well.
• Set a timer of say, 30 seconds and try to finish the given problem within that time.
• Try to use tricks like adding many numbers at once. Use the associative property for simplifying problems.
• Learn the multiplication tables up to 20. This saves a lot of time in calculation.
• Initially give yourself time, and then reduce the time for each question and then focus on your accuracy.
How can I get full marks in class 6 maths?
Class 6 maths can be excelled with a little bit of effort, sincerity, and consistent practice. Here are a few tips that can help a student score well in 6th-class mathematics:
• Practice regularly: This is the most important technique which is always effective and assures the best results. Make sure that you give one hour to maths every day.
• Make a schedule: Planning the topics day-wise or week-wise helps in proper execution. Make a plan and a list that covers all the topics of maths that you need to cover. Allot 2 topics for one
week giving 1 hour every day from Monday to Friday and 2 hours on the weekends because that is usually a holiday.
• Use visuals and figures: Start solving a problem by at least reading it twice and move on to create a picture or any figure that is helpful in understanding the problem. This helps you think
• Practice other resources: Try to practice questions related to the topic from different textbooks or online resources. This makes you come across a variety of questions and will build up your
How to solve word problems for class 6 maths?
Word problems in class 6 maths can be solved using a few tips given below:
• Read the problem: Read the given problem at least twice to understand what the story is all about.
• Note and plan: Note the information that is given and the information that is required. Think about the method and the arithmetic operation which needs to be used here.
• Solve and verify: Solve the problem with a calm mind avoiding any careless mistakes and verify the solution.
Math worksheets and
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Learning Objectives
By the end of this section, you will be able to:
• Discuss the general characteristics of friction.
• Describe the various types of friction.
• Calculate the magnitude of static and kinetic friction.
Friction is a force that is around us all the time that opposes relative motion between systems in contact but also allows us to move (which you have discovered if you have ever tried to walk on
ice). While a common force, the behavior of friction is actually very complicated and is still not completely understood. We have to rely heavily on observations for whatever understandings we can
gain. However, we can still deal with its more elementary general characteristics and understand the circumstances in which it behaves.
Friction is a force that opposes relative motion between systems in contact.
One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion or attempted motion of the systems relative
to each other. If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on ice.
But when objects are stationary, static friction can act between them; the static friction is usually greater than the kinetic friction between the objects.
Kinetic Friction
If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction.
Imagine, for example, trying to slide a heavy crate across a concrete floor—you may push harder and harder on the crate and not move it at all. This means that the static friction responds to what
you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion it is easier to
keep it in motion than it was to get it started, indicating that the kinetic friction force is less than the static friction force. If you add mass to the crate, say by placing a box on top of it,
you need to push even harder to get it started and also to keep it moving. Furthermore, if you oiled the concrete you would find it to be easier to get the crate started and keep it going (as you
might expect).
Figure 1 is a crude pictorial representation of how friction occurs at the interface between two objects. Close-up inspection of these surfaces shows them to be rough. So when you push to get an
object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. A considerable force can be
resisted by friction with no apparent motion. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. Part of the friction
is due to adhesive forces between the surface molecules of the two objects, which explain the dependence of friction on the nature of the substances. Adhesion varies with substances in contact and is
a complicated aspect of surface physics. Once an object is moving, there are fewer points of contact (fewer molecules adhering), so less force is required to keep the object moving. At small but
nonzero speeds, friction is nearly independent of speed.
Frictional forces, such as f, always oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded
view. In order for the object to move, it must rise to where the peaks can skip along the bottom surface. Thus a force is required just to set the object in motion. Some of the peaks will be broken
off, also requiring a force to maintain motion. Much of the friction is actually due to attractive forces between molecules making up the two objects, so that even perfectly smooth surfaces are not
friction-free. Such adhesive forces also depend on the substances the surfaces are made of, explaining, for example, why rubber-soled shoes slip less than those with leather soles.
The magnitude of the frictional force has two forms: one for static situations (static friction), the other for when there is motion (kinetic friction).
When there is no motion between the objects, the magnitude of static friction f[s] is f[s] ≤ μ[s]N, where μ[s] is the coefficient of static friction and N is the magnitude of the normal force (the
force perpendicular to the surface).
Magnitude of Static Friction
Magnitude of static friction f[s] is f[s] ≤ μ[s]N, where μ[s] is the coefficient of static friction and N is the magnitude of the normal force.
The symbol ≤ means less than or equal to, implying that static friction can have a minimum and a maximum value of μ[s]N. Static friction is a responsive force that increases to be equal and opposite
to whatever force is exerted, up to its maximum limit. Once the applied force exceeds f[s(max)], the object will move. Thus f[s(max)] = μ[s]N.
Once an object is moving, the magnitude of kinetic friction f[k ] is given by f[k] = μ[k]N, where μ[k] is the coefficient of kinetic friction. A system in which f[k] = μ[k]N is described as a system
in which friction behaves simply.
Magnitude of Kinetic Friction
The magnitude of kinetic friction f[k] is given by f[k]=μ[k]N, where μ[k] is the coefficient of kinetic friction.
As seen in Table 1, the coefficients of kinetic friction are less than their static counterparts. That values of μ in Table 1 are stated to only one or, at most, two digits is an indication of the
approximate description of friction given by the above two equations.
Table 1. Coefficients of Static and Kinetic Friction
System Static friction μ[s] Kinetic friction μ[k]
Rubber on dry concrete 1.0 0.7
Rubber on wet concrete 0.7 0.5
Wood on wood 0.5 0.3
Waxed wood on wet snow 0.14 0.1
Metal on wood 0.5 0.3
Steel on steel (dry) 0.6 0.3
Steel on steel (oiled) 0.05 0.03
Teflon on steel 0.04 0.04
Bone lubricated by synovial fluid 0.016 0.015
Shoes on wood 0.9 0.7
Shoes on ice 0.1 0.05
Ice on ice 0.1 0.03
Steel on ice 0.4 0.02
The equations given earlier include the dependence of friction on materials and the normal force. The direction of friction is always opposite that of motion, parallel to the surface between objects,
and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight, W = mg
= (100 kg)(9.80 m/s^2) = 980 N, perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than f[s][(max)] =μ[s]N =
(0.45)(980)N = 440N to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N f[k] = μ[k]N = (0.30)(980N) =
290N would keep it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than they would be without lubrication. Coefficient of friction is a unit less
quantity with a magnitude usually between 0 and 1.0. The coefficient of the friction depends on the two surfaces that are in contact.
Take-Home Experiment
Find a small plastic object (such as a food container) and slide it on a kitchen table by giving it a gentle tap. Now spray water on the table, simulating a light shower of rain. What happens now
when you give the object the same-sized tap? Now add a few drops of (vegetable or olive) oil on the surface of the water and give the same tap. What happens now? This latter situation is particularly
important for drivers to note, especially after a light rain shower. Why?
Many people have experienced the slipperiness of walking on ice. However, many parts of the body, especially the joints, have much smaller coefficients of friction—often three or four times less than
ice. A joint is formed by the ends of two bones, which are connected by thick tissues. The knee joint is formed by the lower leg bone (the tibia) and the thighbone (the femur). The hip is a ball (at
the end of the femur) and socket (part of the pelvis) joint. The ends of the bones in the joint are covered by cartilage, which provides a smooth, almost glassy surface. The joints also produce a
fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic joint can be replaced by an artificial joint (Figure 2). These replacements can be made of metals (stainless steel or
titanium) or plastic (polyethylene), also with very small coefficients of friction.
Other natural lubricants include saliva produced in our mouths to aid in the swallowing process, and the slippery mucus found between organs in the body, allowing them to move freely past each other
during heartbeats, during breathing, and when a person moves. Artificial lubricants are also common in hospitals and doctor’s clinics. For example, when ultrasonic imaging is carried out, the gel
that couples the transducer to the skin also serves to to lubricate the surface between the transducer and the skin—thereby reducing the coefficient of friction between the two surfaces. This allows
the transducer to mover freely over the skin.
Example 1. Skiing Exercise
A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N.
The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force N as f[k] = μ[k]N; thus, the coefficient of kinetic friction can be found if we can find
the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the
component of the skier’s weight perpendicular to the slope. (See the skier and free-body diagram in Figure 3.)
The motion of the skier and friction are parallel to the slope and so it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is
perpendicular (axes shown to left of skier). N (the normal force) is perpendicular to the slope, and f (the friction) is parallel to the slope, but w (the skier’s weight) has components along both
axes, namely w[⊥] and W[//]. N is equal in magnitude to w[⊥], so there is no motion perpendicular to the slope. However, f is less than W[//] in magnitude, so there is acceleration down the slope
(along the x-axis).
That is, N = w[⊥] = w cos 25º = mg cos 25º.
Substituting this into our expression for kinetic friction, we get f[k] = μ[k]mg cos 25º, which can now be solved for the coefficient of kinetic friction μ[k].
Solving for μ[k] gives [latex]\displaystyle\mu_k=\frac{f_k}{N}=\frac{f_k}{w\text{ cos }25^{\circ}}=\frac{f_k}{mg\text{ cos }25^{\circ}}\\[/latex]
Substituting known values on the right-hand side of the equation, [latex]\displaystyle\mu_k=\frac{45.0}{(62\text{ kg})(9.80\text{ m/s}^2)(0.906)}=0.082\\[/latex].
This result is a little smaller than the coefficient listed in Table 5.1 for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary greatly. In
situations like this, where an object of mass m slides down a slope that makes an angle θ with the horizontal, friction is given by f[k] = μ[k] mg cos θ. All objects will slide down a slope with
constant acceleration under these circumstances. Proof of this is left for this chapter’s Problems and Exercises.
Take-Home Experiment
An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As
shown in Example 1, the kinetic friction on a slope f[k] = μ[k] mg cos θ. The component of the weight down the slope is equal to mg sin θ (see the free-body diagram in Figure 3). These forces act in
opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out:
f[k] = Fg[x]
μ[k] mg cos θ = mg sin θ.
Solving for μ[k], we find that
Put a coin on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move. Measure the angle of tilt relative to the
horizontal and find μ[k]. Note that the coin will not start to slide at all until an angle greater than θ is attained, since the coefficient of static friction is larger than the coefficient of
kinetic friction. Discuss how this may affect the value for μ[k] and its uncertainty.
We have discussed that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Furthermore, simple friction is always proportional to the
normal force.
Making Connections: Submicroscopic Explanations of Friction
The simpler aspects of friction dealt with so far are its macroscopic (large-scale) characteristics. Great strides have been made in the atomic-scale explanation of friction during the past several
decades. Researchers are finding that the atomic nature of friction seems to have several fundamental characteristics. These characteristics not only explain some of the simpler aspects of
friction—they also hold the potential for the development of nearly friction-free environments that could save hundreds of billions of dollars in energy which is currently being converted
(unnecessarily) to heat.
Figure 4 illustrates one macroscopic characteristic of friction that is explained by microscopic (small-scale) research. We have noted that friction is proportional to the normal force, but not to
the area in contact, a somewhat counterintuitive notion. When two rough surfaces are in contact, the actual contact area is a tiny fraction of the total area since only high spots touch. When a
greater normal force is exerted, the actual contact area increases, and it is found that the friction is proportional to this area.
But the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is now being determined. In other words, why do surfaces get
warmer when rubbed? Essentially, atoms are linked with one another to form lattices. When surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate—essentially creating sound waves
that penetrate the material. The sound waves diminish with distance and their energy is converted into heat. Chemical reactions that are related to frictional wear can also occur between atoms and
molecules on the surfaces. Figure 5 shows how the tip of a probe drawn across another material is deformed by atomic-scale friction. The force needed to drag the tip can be measured and is found to
be related to shear stress, which will be discussed later in this chapter. The variation in shear stress is remarkable (more than a factor of 10^12 ) and difficult to predict theoretically, but shear
stress is yielding a fundamental understanding of a large-scale phenomenon known since ancient times—friction.
PhET Explorations: Forces and Motion
Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position,
velocity, and acceleration vs. time. Draw a free-body diagram of all the forces (including gravitational and normal forces).
Section Summary
• Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force N pushing the systems together. (A normal
force is always perpendicular to the contact surface between systems.) Friction depends on both of the materials involved. The magnitude of static friction [latex]{f}_{\text{s}}\\[/latex] between
systems stationary relative to one another is given by [latex]{f}_{\text{s}}\le {\mu }_{\text{s}}N\\[/latex], where [latex]{\mu }_{\text{s}}\\[/latex] is the coefficient of static friction, which
depends on both of the materials.
• The kinetic friction force [latex]{f}_{\text{k}}\\[/latex] between systems moving relative to one another is given by [latex]{f}_{\text{k}}={\mu }_{\text{k}}N\\[/latex], where [latex]{\mu }_{\
text{k}}\\[/latex] is the coefficient of kinetic friction, which also depends on both materials.
Conceptual Questions
1. Define normal force. What is its relationship to friction when friction behaves simply?
2. The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to vertical walls and even to ceilings.
3. When you learn to drive, you discover that you need to let up slightly on the brake pedal as you come to a stop or the car will stop with a jerk. Explain this in terms of the relationship between
static and kinetic friction.
4. When you push a piece of chalk across a chalkboard, it sometimes screeches because it rapidly alternates between slipping and sticking to the board. Describe this process in more detail, in
particular explaining how it is related to the fact that kinetic friction is less than static friction. (The same slip-grab process occurs when tires screech on pavement.)
Problems & Exercises
Express your answers to problems in this section to the correct number of significant figures and proper units.
1. A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction
between the two materials, he quickly calculates the normal force. What is it?
2. When rebuilding her car’s engine, a physics major must exert 300 N of force to insert a dry steel piston into a steel cylinder. (a) What is the magnitude of the normal force between the piston
and cylinder? (b) What is the magnitude of the force would she have to exert if the steel parts were oiled?
3. (a) What is the maximum frictional force in the knee joint of a person who supports 66.0 kg of her mass on that knee? (b) During strenuous exercise it is possible to exert forces to the joints
that are easily ten times greater than the weight being supported. What is the maximum force of friction under such conditions? The frictional forces in joints are relatively small in all
circumstances except when the joints deteriorate, such as from injury or arthritis. Increased frictional forces can cause further damage and pain.
4. Suppose you have a 120-kg wooden crate resting on a wood floor. (a) What maximum force can you exert horizontally on the crate without moving it? (b) If you continue to exert this force once the
crate starts to slip, what will the magnitude of its acceleration then be?
5. (a) If half of the weight of a small 1.00 × 10^3 kg utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete? (b) Will
a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (c) Solve both problems assuming the truck has four-wheel drive.
6. A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its rider has a mass of 210 kg. (a) Calculate the
magnitude of the acceleration starting from rest if each dog exerts an average force of 185 N backward on the snow. (b) What is the magnitude of the acceleration once the sled starts to move? (c)
For both situations, calculate the magnitude of the force in the coupling between the dogs and the sled.
7. Consider the 65.0-kg ice skater being pushed by two others shown in Figure 6. (a) Find the direction and magnitude of [latex]{\mathbf{F}}_{\text{tot}}\\[/latex], the total force exerted on her by
the others, given that the magnitudes [latex]{F}_{1}\\[/latex] and [latex]{F}_{2}\\[/latex] are 26.4 N and 18.6 N, respectively; (b) What is her initial acceleration if she is initially
stationary and wearing steel-bladed skates that point in the direction of [latex]{\mathbf{F}}_{\text{tot}}\\[/latex]? (c) What is her acceleration assuming she is already moving in the direction
of [latex]{\mathbf{F}}_{\text{tot}}\\[/latex]? (Remember that friction always acts in the direction opposite that of motion or attempted motion between surfaces in contact.)
8. Show that the acceleration of any object down a frictionless incline that makes an angle θ with the horizontal is a = g sin θ. (Note that this acceleration is independent of mass.)
9. Show that the acceleration of any object down an incline where friction behaves simply (that is, where f[k ]= μ[k]N) is a =g(sin θ − μ[k]cos θ). Note that the acceleration is independent of mass
and reduces to the expression found in the previous problem when friction becomes negligibly small (μ[k]=0).
10. Calculate the deceleration of a snow boarder going up a 5.0º, slope assuming the coefficient of friction for waxed wood on wet snow. The result of question 9 may be useful, but be careful to
consider the fact that the snow boarder is going uphill. Explicitly show how you follow the steps in Problem-Solving Strategies.
11. (a) Calculate the acceleration of a skier heading down a 10.0º slope, assuming the coefficient of friction for waxed wood on wet snow. (b) Find the angle of the slope down which this skier could
coast at a constant velocity. You can neglect air resistance in both parts, and you will find the result of question 9 to be useful. Explicitly show how you follow the steps in the
Problem-Solving Strategies.
12. If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for
steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is [latex]\theta=\tan^{-1}\mu _{\text{s}}\\[/latex]. You may use the result
of the previous problem. Assume that a = 0 and that static friction has reached its maximum value.
13. Calculate the maximum deceleration of a car that is heading down a 6º slope (one that makes an angle of 6º with the horizontal) under the following road conditions. You may assume that the weight
of the car is evenly distributed on all four tires and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the deceleration. (Ignore rolling.)
Calculate for a car: (a) On dry concrete; (b)On wet concrete; (c) On ice, assuming that [latex]{\mu }_{\text{s}}=0.100\\[/latex] , the same as for shoes on ice.
14. Calculate the maximum acceleration of a car that is heading up a 4º slope (one that makes an angle of 4º with the horizontal) under the following road conditions. Assume that only half the weight
of the car is supported by the two drive wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.) (a)
On dry concrete; (b) On wet concrete; (c) On ice, assuming that [latex]\mu _{\text{s}}=0.100\\[/latex], the same as for shoes on ice.
15. Repeat question 14 for a car with four-wheel drive.
16. A freight train consists of two [latex]8\text{.}\text{00}\times {\text{10}}^{5}\text{-kg}\\[/latex] engines and 45 cars with average masses of [latex]5\text{.}\text{50}\times {\text{10}}^{5}\text
{kg}\\[/latex]. (a) What force must each engine exert backward on the track to accelerate the train at a rate of [latex]5\text{.}\text{00}\times {\text{10}}^{-2}m/{s}^{2}\\[/latex] if the force
of friction is [latex]7\text{.}\text{50}\times {\text{10}}^{5}N\\[/latex], assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling
friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the magnitude of the force in the coupling between the 37th and 38th cars (this
is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines?
17. Consider the 52.0-kg mountain climber in Figure 7. (a) Find the tension in the rope and the force that the mountain climber must exert with her feet on the vertical rock face to remain
stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms; (b) What is the minimum coefficient of friction between her shoes and the
18. A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake as shown in Figure 8a. (a) Calculate the minimum force F he must exert to get the block moving; (b) What
is the magnitude of its acceleration once it starts to move, if that force is maintained?
19. Repeat Question 18 with the contestant pulling the block of ice with a rope over his shoulder at the same angle above the horizontal as shown in Figure 8b.
friction: a force that opposes relative motion or attempts at motion between systems in contact
kinetic friction: a force that opposes the motion of two systems that are in contact and moving relative to one another
static friction: a force that opposes the motion of two systems that are in contact and are not moving relative to one another
magnitude of static friction: [latex]{f}_{\text{s}}\le {\mu }_{\text{s}}N\\[/latex] , where [latex]{\mu }_{\text{s}}\\[/latex] is the coefficient of static friction and N is the magnitude of the
normal force
magnitude of kinetic friction: [latex]{f}_{\text{k}}={\mu }_{\text{k}}N\\[/latex], where [latex]{\mu }_{\text{k}}\\[/latex] is the coefficient of kinetic friction
Selected Solutions to Problems & Exercises
1. 5.00 N
4. (a) 588 N; (b) 1.96 m/s^2
6. (a) 3.29 m/s^2; (b) 3.52 m/s^2; (c) 980 N, 945 N
10. 1.83 m/s^2
14. (a) 4.20 m/s^2; (b) 2.74 m/s^2; (c) –0.195 m/s^2
16. (a) 1.03 × 106 N; (b) 3.48 × 105 N
18. (a) 51.0 N; (b) 0.720 m/s^2 | {"url":"https://courses.lumenlearning.com/atd-austincc-physics1/chapter/5-1-friction/","timestamp":"2024-11-07T09:56:42Z","content_type":"text/html","content_length":"83531","record_id":"<urn:uuid:2be65491-5df8-4a18-9997-67d2804073bd>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00574.warc.gz"} |
Exponential Distribution
Exponential Distribution MLE Applet
$X \sim exp(\lambda)$
This applet computes probabilities and percentiles for the exponential distribution: $$X \sim exp(\lambda)$$ It also can plot the likelihood, log-likelihood, asymptotic CI for $\lambda$, and
determine the MLE and observed Fisher information.
• $f(x) = \lambda e^{-\lambda x}$ for $x>0$ (and 0 otherwise)
• $E(X) = 1/\lambda$
• $Var(X) = 1/\lambda^2$
Probability Density Function (pdf) mode
• Enter the rate in the $\lambda$ box.
• Hitting "Tab" or "Enter" on your keyboard will plot the pdf.
To compute a right-tail probability, select $P(X \lt x)$ from the drop-down box, enter a numeric $x$ value, and press "Tab" or "Enter" on your keyboard. The probability $P(X \gt x)$ will appear in
the pink box. Select $P(X \gt x)$ from the drop-down box for a left-tail probability (i.e. the cdf).
Maximum Likelihood (MLE) mode
• Leave the $\lambda$ box empty.
• Enter the one or more data values in the $x$ box (separate multiple $x$ values by commas).
• Hitting "Tab" or "Enter" on your keyboard will plot the likelihood, log-likelihood, and 95% asymptotic CI for $\lambda$. The MLE and observed Fisher information are also displayed.
Note that the green line is drawn at $$max(log(likelihood))-\frac{3.84}{2}$$ The resulting asymptotic 95% CI is shown in green on the horizontal axis. The asymptotic CI is based on large sample
theory; it is shown in this applet for all sample sizes, however. | {"url":"https://homepage.divms.uiowa.edu/~mbognar/applets/exp-like.html","timestamp":"2024-11-02T06:27:22Z","content_type":"text/html","content_length":"7994","record_id":"<urn:uuid:74431df2-4f6b-4381-b88d-77e0fea572c4>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00794.warc.gz"} |
Indices Starters:
This lesson starter presents a number of statements about indices and pupils are asked if they can spot the mistake.
Dr Tim's Indices Challenge
Dr Tim Honeywill has come up with these challenges to test your understanding of indices and algebra.
A self marking exercise on indices (powers or exponents) including evaluating expressions and solving equations.
Indices True False
Arrange the given statements involving indices to show whether they are true or false.
Power Play
Exercises on powers and roots and simplifying index expressions involving numbers of the same base.
Power Shift
Arrange the given numbers as bases and indices in the three-term sum to make the target total.
Standard Form
Test your understanding of standard form (scientific notation) with this self-marking quiz.
Standard Form Algebra
Deeply test your understanding of standard form (scientific notation) by involving a little algebra.
Standard Order
Arrange the numbers given in standard form with the smallest at the top and the largest at the bottom.
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Variational Inverting Network for Statistical Inverse Problems of Partial Differential Equations
Junxiong Jia, Yanni Wu, Peijun Li, Deyu Meng.
Year: 2023, Volume: 24, Issue: 201, Pages: 1−60
To quantify uncertainties in inverse problems of partial differential equations (PDEs), we formulate them into statistical inference problems using Bayes' formula. Recently, well-justified
infinite-dimensional Bayesian analysis methods have been developed to construct dimension-independent algorithms. However, there are three challenges for these infinite-dimensional Bayesian methods:
prior measures usually act as regularizers and are not able to incorporate prior information efficiently; complex noises, such as more practical non-i.i.d. distributed noises, are rarely considered;
and time-consuming forward PDE solvers are needed to estimate posterior statistical quantities. To address these issues, an infinite-dimensional inference framework has been proposed based on the
infinite-dimensional variational inference method and deep generative models. Specifically, by introducing some measure equivalence assumptions, we derive the evidence lower bound in the
infinite-dimensional setting and provide possible parametric strategies that yield a general inference framework called the Variational Inverting Network (VINet). This inference framework can encode
prior and noise information from learning examples. In addition, relying on the power of deep neural networks, the posterior mean and variance can be efficiently and explicitly generated in the
inference stage. In numerical experiments, we design specific network structures that yield a computable VINet from the general inference framework. Numerical examples of linear inverse problems of
an elliptic equation and the Helmholtz equation are presented to illustrate the effectiveness of the proposed inference framework.
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bayesImageS: An R Package for Bayesian Image Segmentation using a Hidden Potts Model
Image segmentation can be viewed as the task of labelling the observed pixels \(\mathbf{y}\) according to a finite set of discrete states \(\mathbf{z} \in \{ 1, \dots, k \}\). The hidden model allows
for spatial correlation between neighbouring labels in the form of a Markov random field. The latent labels follow a Gibbs distribution, which is specified in terms of its conditional probabilities:
\[$$\label{eq:Potts} p(z_i | z_{\setminus i}, \beta) = \frac{\exp\left\{\beta\sum_{i \sim \ell}\delta(z_i,z_\ell)\right\}}{\sum_{j=1}^k \exp\left\{\beta\sum_{i \sim \ell}\delta(j,z_\ell)\right\}}$$\]
where \(\beta\) is the inverse temperature, \(z_{\setminus i}\) represents all of the labels except \(z_i\), \(i \sim \ell\) are the neighbouring pixels of \(i\), and \(\delta(u,v)\) is the Kronecker
delta function. Thus, \(\sum_{i \sim \ell}\delta(z_i,z_\ell)\) is a count of the neighbours that share the same label.
The observation equation links the latent labels to the corresponding pixel values: \[$$\label{eq:obs} p(\mathbf{y} | \mathbf{z}, \boldsymbol\theta) = \prod_{i=1}^n p(y_i | z_i, \theta_{z_i})$$\]
where \(\theta_{j}\) are the parameters that govern the distribution of the pixel values with label \(j\). The hidden Potts model can thus be viewed as a spatially-correlated generalisation of the
finite mixture model . We assume that the pixels with label \(j\) share a common mean \(\mu_j\) corrupted by additive Gaussian noise with variance \(\sigma_j^2\): \[$$\label{eq:obs2} y_i | z_i=j, \
mu_j, \sigma^2_j \;\sim\; \mathcal{N}\left( \mu_j, \sigma^2_j \right)$$\]
The Gibbs distribution is a member of the exponential family and so there is a sufficient statistic for this model, as noted by : \[$$\label{eq:potts_stat} \mathrm{S}(\mathbf{z}) = \sum_{i \sim \ell
\in \mathcal{E}} \delta(z_i,z_\ell)$$\] This statistic represents the total number of like neighbour pairs in the image. The likelihood \(p(\mathbf{y},\mathbf{z} | \boldsymbol\theta, \beta)\) can
therefore be factorised into \(p(\mathbf{y} | \mathbf{z}, \boldsymbol\theta) p(\mathrm{S}(\mathbf{z}) | \beta)\), where the second factor does not depend on the observed data, but only on the
sufficient statistic. The joint posterior is then: \[$$\label{eq:joint_post} p(\boldsymbol\theta, \beta, \mathbf{z} | \mathbf{y}) \propto p(\mathbf{y} | \mathbf{z}, \boldsymbol\theta) \pi(\boldsymbol
\theta) p(\mathrm{S}(\mathbf{z}) | \beta) \pi(\beta)$$\] The conditional distributions \(p(\boldsymbol\theta | \mathbf{z}, \mathbf{y})\) and \(p(z_i | z_{\setminus i}, \beta, y_i, \boldsymbol\theta_
{z_i})\) can be simulated using Gibbs sampling, but \(p(\beta | \mathbf{y}, \mathbf{z}, \boldsymbol\theta)\) involves an intractable normalising constant \(\mathcal{C}(\beta)\): \[\begin{eqnarray} \
label{eq:beta_post} p(\beta \mid \mathbf{y}, \mathbf{z}, \boldsymbol\theta) &=& \frac{p(\mathrm{S}(\mathbf{z}) | \beta) \pi(\beta)}{\int_\beta p(\mathrm{S}(\mathbf{z}) | \beta) \pi(d \beta)} \\ \
label{eq:beta} &\propto& \frac{\exp\left\{ \beta\, \mathrm{S}(\mathbf{z}) \right\}}{\mathcal{C}(\beta)} \pi(\beta) \end{eqnarray}\] The normalising constant is also known as a partition function in
statistical physics. It has computational complexity of \(\mathcal{O}(n k^n)\), since it involves a sum over all possible combinations of the labels \(\mathbf{z} \in \mathcal{Z}\): \[$$\label
{eq:norm} \mathcal{C}(\beta) = \sum_{\mathbf{z} \in \mathcal{Z}} \exp\left\{\beta\, \mathrm{S}(\mathbf{z})\right\}$$\] It is infeasible to calculate this value exactly for nontrivial images, thus
computational approximations are required.
This paper describes the package , which is available from the Comprehensive R Archive Network (CRAN) at https://CRAN.R-project.org/package=bayesImageS. This package implements five major algorithms
for intractable likelihoods in Bayesian image analysis. These methods provide alternative means to simulate parameter values from () without computing the normalising constant. We describe the
algorithms in terms of Markov chain Monte Carlo (MCMC) to enable direct comparison, although we also mention other approaches where applicable, such as particle-based (SMC and PMCMC) methods.
Reference implementations of all of these methods are available from various sources described below, but for the purpose of comparison we have reimplemented the algorithms using .
There are a number of contributed packages available for , for example on CRAN, which provide image segmentation using Potts and other models: , , ,
Algorithms for Intractable Likelihoods
Pseudolikelihood and Composite Likelihood
Pseudolikelihood is the simplest of the methods that we have implemented and also the fastest. showed that the intractable distribution () could be approximated using the product of the conditional
densities given by (): \[$$\label{eq:pseudo} p(\mathrm{S}(\mathbf{z}) | \beta) \approx \prod_{i=1}^n p(z_i | z_{\setminus i}, \beta)$$\] This enables updates for the inverse temperature at iteration
\(t\) to be simulated using a Metropolis-Hastings (M-H) step, with acceptance ratio: \[$$\label{mh:ratio} \rho = \min\left( 1, \frac{p(\mathbf{z}|\beta') \pi(\beta') q(\beta_{t-1} | \beta')}{p(\
mathbf{z}|\beta_{t-1}) \pi(\beta_{t-1}) q(\beta' | \beta_{t-1})} \right)$$\]
The M-H proposal density \(q(\beta'|\beta_{t-1})\) can be any distribution such that \(\int q(\beta'|\beta_{t-1})\, d\beta' = 1\). However, there is a tradeoff between exploring the full state space
and ensuring that the probability of acceptance is sufficiently high. We use the adaptive random walk (RWMH) algorithm of , which automatically tunes the bandwidth of the proposal density to target a
given M-H acceptance rate. When a symmetric proposal density is used, \(q(\beta'|\beta_{t-1}) = q(\beta_{t-1}|\beta')\) and so this term cancels out in the M-H ratio . Likewise, under a uniform prior
for the inverse temperature, \(\pi(\beta') = \pi(\beta_{t-1}) = 1\). The natural logarithm of \(\rho\) is used in practice to improve numerical stability.
Pseudolikelihood is exact when \(\beta=0\) and provides a reasonable approximation for small values of the inverse temperature. However, the approximation error increases rapidly for \(\beta \ge \
beta_{crit}\), as illustrated by Figure . This is due to long-range dependence between the labels, which is inadequately modelled by the local approximation. The implications of this inaccuracy for
posterior inference will be demonstrated in Section .
referred to Equation as point pseudolikelihood, since the conditional distributions are computed for each pixel individually. They suggested that the accuracy could be improved using block
pseudolikelihood. This is where the likelihood is calculated exactly for small blocks of pixels, then is modified to be the product of the blocks: \[$$p(\mathbf{z}|\beta) \approx \prod_{i=1}^{N_B} p
(\mathbf{z}_{B_i} | \mathbf{z}_{\setminus B_i}, \beta) \label{eq:pl_comp}$$\] where \(N_B\) is the number of blocks, \(\mathbf{z}_{B_i}\) are the labels of the pixels in block \(B_i\), and \(\mathbf
{z}_{\setminus B_i}\) are all of the labels except for \(\mathbf{z}_{B_i}\). This is a form of composite likelihood, where the likelihood function is approximated as a product of simplified factors .
compared point pseudolikelihood to composite likelihood with blocks of \(3 \times 3\), \(4 \times 4\), \(5 \times 5\), and \(6 \times 6\) pixels. showed that () outperformed () for the Ising (\(k=2\)
) model with \(\beta < \beta_{crit}\). discuss composite likelihood for the Potts model with \(k > 2\) and have provided an open source implementation in the package .
Evaluating the conditional likelihood in () involves the normalising constant for \(\mathbf{z}_{B_i}\), which is a sum over all of the possible configurations \(\mathcal{Z}_{B_i}\). This is a
limiting factor on the size of blocks that can be used. The brute-force method that was used to compute Figure is too computationally intensive for this purpose. showed that the normalising constant
can be calculated exactly for a cylindrical lattice by computing eigenvalues of a \(k^r \times k^r\) matrix, where \(r\) is the smaller of the number of rows or columns. The value of () for a free
boundary lattice can then be approximated using path sampling. extended this method to larger lattices using a composite likelihood approach.
The reduced dependence approximation (RDA) is another form of composite likelihood. introduced a recursive algorithm to calculate the normalising constant using a lag-\(r\) representation. divided
the image lattice into sub-lattices of size \(r_1 < r\), then approximated the normalising constant of the full lattice using RDA: \[$$\mathcal{C}(\beta) \approx \frac{\mathcal{C}_{r_1 \times n}(\
beta)^{r - r_1 + 1}}{\mathcal{C}_{r_1 - 1 \times n}(\beta)^{r - r_1}} \label{eq:rda}$$\] compared RDA to pseudolikelihood and the exact method of , reporting similar computational cost to
pseudolikelihood but with improved accuracy in estimating \(\beta\).
Source code for RDA is available in the online supplementary material for . | {"url":"https://cran.mirror.garr.it/CRAN/web/packages/bayesImageS/vignettes/Background.html","timestamp":"2024-11-03T01:36:01Z","content_type":"text/html","content_length":"20454","record_id":"<urn:uuid:5547d495-ad71-4bf9-bbf8-06debd1e6552>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00161.warc.gz"} |
Draw From a Multivariate t-Distribution — rmvt
Fast ways to draw from a multivariate t-distribution the scale (covariance) matrix is sparse.
rmvt(n, Sigma, df = 1, delta = rep(0, nrow(Sigma)),
type = c("shifted", "Kshirsagar"), ..., sigma)
rmvt.spam(n, Sigma, df = 1, delta = rep(0, nrow(Sigma)),
type = c("shifted", "Kshirsagar"), ..., sigma)
number of observations.
scale matrix (of class spam).
degrees of freedom.
vector of noncentrality parameters.
type of the noncentral multivariate t distribution.
arguments passed to rmvnorm.spam.
similar to Sigma. Here for portability with mvtnorm::rmvt()
This function is very much like rmvt() from the package mvtnorm. We refer to the help of the afore mentioned.
See also | {"url":"https://www.math.uzh.ch/pages/spam/reference/rmvt.html","timestamp":"2024-11-01T19:25:54Z","content_type":"text/html","content_length":"9816","record_id":"<urn:uuid:2033062e-cda5-4c84-8f57-ac8f083a9cb4>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00101.warc.gz"} |
Re: BIAS in Regression Parameter Estimates
Is there a proc or formula to calculate the magnitude of bias in regression parameter estimates?
4 REPLIES 4
Well, we know that the MSE is equal to the bias squared plus the variance for an estimator. So now it all depends on what you know about the distribution of the estimator. If the distribution has a
known variance, you can calculate the MSE from the estimator's standard error, subtract the population variance, and get the squared bias.
This all requires knowing the population mean and variance for the estimator in question.
Steve Denham
Thank you, Steve and Paige.
Currently, I am working on a project that involves missing data analysis of sample data. I wanted to know if there is a way to measure bias for regression models for different "Missing Data" Deletion
or Imputation methods -- I mean, in Listwise or Pairwise or Mean Substitution methods, I know the estimates are highly biased compared to those in Multiple Imputation or Expectation Maximization, but
is it possible to calculate bias in the regression parameter estimates.
It seems since there is no "correct estimate" for non-missing sample data, the "amount of bias" cannot be calculated.
Please let me know if you have any other thoughts.
If you are talking about Ordinary Least Squares Regression, and you are estimating the correct model, then it is my understanding that the bias is zero.
Paige Miller
Are you asking about Linear or Logistic regression? If Logistic Regression, then there is a very good paper by King and Zeng, "Logistic Regression in Rare Events Data" that gives a formula for bias
to account for missing data. | {"url":"https://communities.sas.com/t5/Statistical-Procedures/BIAS-in-Regression-Parameter-Estimates/m-p/146615","timestamp":"2024-11-10T15:37:27Z","content_type":"text/html","content_length":"207951","record_id":"<urn:uuid:71692b24-4442-4765-beeb-981793329cfe>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00721.warc.gz"} |
The diffusion equation is,
$\large \frac{\partial C}{\partial t}= D\nabla^2C$ .
If a limited source of dopants is deposited in a thin layer with a thickness $w$ at the surface such that the dose $Q$ in dopants per square meter is $Q=\int\limits_0^w dz$, the concentration as a
function of time
$\large C(z,t)=\frac{Q\exp\left(\frac{-z^2}{4Dt}\right)}{\sqrt{4\pi Dt}}$.
The concentration falls at the surface and the total number of dopants remains constant. | {"url":"http://lampz.tugraz.at/~hadley/num/apps/pde/limited_source.php","timestamp":"2024-11-05T23:33:05Z","content_type":"text/html","content_length":"2377","record_id":"<urn:uuid:911a2a45-afa8-45f3-b3c6-d394dee002ff>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00396.warc.gz"} |
Discrete time and continuous time - Wikiwand
In mathematical dynamics, discrete time and continuous time are two alternative frameworks within which variables that evolve over time are modeled.
Discrete sampled signal
Discrete time views values of variables as occurring at distinct, separate "points in time", or equivalently as being unchanged throughout each non-zero region of time ("time period")—that is, time
is viewed as a discrete variable. Thus a non-time variable jumps from one value to another as time moves from one time period to the next. This view of time corresponds to a digital clock that gives
a fixed reading of 10:37 for a while, and then jumps to a new fixed reading of 10:38, etc. In this framework, each variable of interest is measured once at each time period. The number of
measurements between any two time periods is finite. Measurements are typically made at sequential integer values of the variable "time".
A discrete signal or discrete-time signal is a time series consisting of a sequence of quantities.
Unlike a continuous-time signal, a discrete-time signal is not a function of a continuous argument; however, it may have been obtained by sampling from a continuous-time signal. When a discrete-time
signal is obtained by sampling a sequence at uniformly spaced times, it has an associated sampling rate.
Discrete-time signals may have several origins, but can usually be classified into one of two groups:^[1]
• By acquiring values of an analog signal at constant or variable rate. This process is called sampling.^[2]
• By observing an inherently discrete-time process, such as the weekly peak value of a particular economic indicator.
In contrast, continuous time views variables as having a particular value only for an infinitesimally short amount of time. Between any two points in time there are an infinite number of other points
in time. The variable "time" ranges over the entire real number line, or depending on the context, over some subset of it such as the non-negative reals. Thus time is viewed as a continuous variable.
A continuous signal or a continuous-time signal is a varying quantity (a signal) whose domain, which is often time, is a continuum (e.g., a connected interval of the reals). That is, the function's
domain is an uncountable set. The function itself need not to be continuous. To contrast, a discrete-time signal has a countable domain, like the natural numbers.
A signal of continuous amplitude and time is known as a continuous-time signal or an analog signal. This (a signal) will have some value at every instant of time. The electrical signals derived in
proportion with the physical quantities such as temperature, pressure, sound etc. are generally continuous signals. Other examples of continuous signals are sine wave, cosine wave, triangular wave
The signal is defined over a domain, which may or may not be finite, and there is a functional mapping from the domain to the value of the signal. The continuity of the time variable, in connection
with the law of density of real numbers, means that the signal value can be found at any arbitrary point in time.
A typical example of an infinite duration signal is:
${\displaystyle f(t)=\sin(t),\quad t\in \mathbb {R} }$
A finite duration counterpart of the above signal could be:
${\displaystyle f(t)=\sin(t),\quad t\in [-\pi ,\pi ]}$ and ${\displaystyle f(t)=0}$ otherwise.
The value of a finite (or infinite) duration signal may or may not be finite. For example,
${\displaystyle f(t)={\frac {1}{t}},\quad t\in [0,1]}$ and ${\displaystyle f(t)=0}$ otherwise,
is a finite duration signal but it takes an infinite value for ${\displaystyle t=0\,}$.
In many disciplines, the convention is that a continuous signal must always have a finite value, which makes more sense in the case of physical signals.
For some purposes, infinite singularities are acceptable as long as the signal is integrable over any finite interval (for example, the ${\displaystyle t^{-1}}$ signal is not integrable at infinity,
but ${\displaystyle t^{-2}}$ is).
Any analog signal is continuous by nature. Discrete-time signals, used in digital signal processing, can be obtained by sampling and quantization of continuous signals.
Continuous signal may also be defined over an independent variable other than time. Another very common independent variable is space and is particularly useful in image processing, where two space
dimensions are used.
Discrete time is often employed when empirical measurements are involved, because normally it is only possible to measure variables sequentially. For example, while economic activity actually occurs
continuously, there being no moment when the economy is totally in a pause, it is only possible to measure economic activity discretely. For this reason, published data on, for example, gross
domestic product will show a sequence of quarterly values.
When one attempts to empirically explain such variables in terms of other variables and/or their own prior values, one uses time series or regression methods in which variables are indexed with a
subscript indicating the time period in which the observation occurred. For example, y[t] might refer to the value of income observed in unspecified time period t, y[3] to the value of income
observed in the third time period, etc.
Moreover, when a researcher attempts to develop a theory to explain what is observed in discrete time, often the theory itself is expressed in discrete time in order to facilitate the development of
a time series or regression model.
On the other hand, it is often more mathematically tractable to construct theoretical models in continuous time, and often in areas such as physics an exact description requires the use of continuous
time. In a continuous time context, the value of a variable y at an unspecified point in time is denoted as y(t) or, when the meaning is clear, simply as y.
Discrete time
Discrete time makes use of difference equations, also known as recurrence relations. An example, known as the logistic map or logistic equation, is
${\displaystyle x_{t+1}=rx_{t}(1-x_{t}),}$
in which r is a parameter in the range from 2 to 4 inclusive, and x is a variable in the range from 0 to 1 inclusive whose value in period t nonlinearly affects its value in the next period, t+1. For
example, if ${\displaystyle r=4}$ and ${\displaystyle x_{1}=1/3}$, then for t=1 we have ${\displaystyle x_{2}=4(1/3)(2/3)=8/9}$, and for t=2 we have ${\displaystyle x_{3}=4(8/9)(1/9)=32/81}$.
Another example models the adjustment of a price P in response to non-zero excess demand for a product as
${\displaystyle P_{t+1}=P_{t}+\delta \cdot f(P_{t},...)}$
where ${\displaystyle \delta }$ is the positive speed-of-adjustment parameter which is less than or equal to 1, and where ${\displaystyle f}$ is the excess demand function.
Continuous time
Continuous time makes use of differential equations. For example, the adjustment of a price P in response to non-zero excess demand for a product can be modeled in continuous time as
${\displaystyle {\frac {dP}{dt}}=\lambda \cdot f(P,...)}$
where the left side is the first derivative of the price with respect to time (that is, the rate of change of the price), ${\displaystyle \lambda }$ is the speed-of-adjustment parameter which can be
any positive finite number, and ${\displaystyle f}$ is again the excess demand function.
A variable measured in discrete time can be plotted as a step function, in which each time period is given a region on the horizontal axis of the same length as every other time period, and the
measured variable is plotted as a height that stays constant throughout the region of the time period. In this graphical technique, the graph appears as a sequence of horizontal steps. Alternatively,
each time period can be viewed as a detached point in time, usually at an integer value on the horizontal axis, and the measured variable is plotted as a height above that time-axis point. In this
technique, the graph appears as a set of dots.
The values of a variable measured in continuous time are plotted as a continuous function, since the domain of time is considered to be the entire real axis or at least some connected portion of it. | {"url":"https://www.wikiwand.com/en/articles/Continuous-time_signal","timestamp":"2024-11-07T08:50:45Z","content_type":"text/html","content_length":"266977","record_id":"<urn:uuid:3c01ae02-82e5-4052-8578-c3da8110f479>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00617.warc.gz"} |
August 30, #18
2013, 05:19
Member Quote:
xuhe-openfoam Originally Posted by
Join Date: jms
Aug 2013
Hi all!
DaLian，china I have been using OpenFOAM for 2 months, so I am quite new with this software.
I am using it to do a study in 2D of the flow over a NACA0015, at Re=2x10^6 at steady state. Using simpleFoam and the k-omega SST turbulence model.
Posts: 82 I have been doing a sensitivity study of the numerical schemes. Thus, I have been changing the divSchemes. I have tried changing all the entries in there to QUICK/QUICKV, linear,
linearUpwind and upwind. I couldn´t change the entry "div((nuEff*dev(grad(U).T())))" to any of those (I have to keep at as linear, otherwise the programme does not recognize it), why?
Rep Power: I have upload the file fvSchemes used so you can have a look at it.
I have also uplodaded a figure showing the results obtained compared with a reference. They do not look as expected since the closest solution obtained for the lift coefficient
13 calculations is for the upwind numerical scheme, while this one should give the worst results, shouldn´t it?
Thank you for your attention.
I will really appreciate your help.
default none;
div(phi,U) Gauss QUICKV cellLimited Gauss linear 1;
div(phi,k) Gauss QUICK cellLimited Gauss linear 1;
div(phi,omega) Gauss QUICK cellLimited Gauss linear 1;
div((nuEff*dev(grad(U).T()))) Gauss linear cellLimited Gauss linear 1;
The above is your divschemes. Does the number "1" in every schemes indicate the non-orthogonal correction? thank you! | {"url":"https://www.cfd-online.com/Forums/openfoam/85246-fvschemes.html","timestamp":"2024-11-05T17:29:11Z","content_type":"application/xhtml+xml","content_length":"158580","record_id":"<urn:uuid:2b655a07-a629-46f3-93e6-df0918ced963>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00862.warc.gz"} |
Fault-tolerant quantum computing – from theory to practice
We work on various aspects of quantum error correction and fault tolerance. Our recent direction has been on reducing the gap between theoretical quantum error correction and fault tolerance ideas
and their implementation in experiments. With the rapid development of quantum computing devices, we are beginning to have an inkling of what a quantum computer might look like, and the practical
obstacles, to do with noise and scalability, are taking on more concrete shapes. This is thus the right time to re-examine the often generic and abstract theoretical proposals for noise removal, in
the light of recent experiments, for progress towards large-scale, useful quantum devices.
Below, we highlight some recent projects within the group.
Fault-tolerant embedding of circuits via swap gates
HK Ng, with Entropica Labs (arXiv:2406.17044)
Embedding surface-code syndrome extraction circuits onto a heavy-hex lattice using our swap strategy, under depolarizing noise of strength p. peff = p for the abstract circuit; peff = 3.63p for the
embedded circuit, giving a noise deterioration of the factor 3.63.
In near-term quantum devices, qubit connectivity remains limited by architectural constraints. A computational circuit with given connectivity requirements for multi-qubit gates has to be embedded in
physical hardware with fixed connectivity. Long-distance gates have to be done by first routing the information together.
The simplest routing strategy uses swap gates to swap information carried by two unconnected qubits to connected ones. Ideal SWAPs just permute qubits; real SWAPs, however, can cause simultaneous
errors on the qubits involved and spread errors across the circuit. General swap schemes can thus destroy fault-tolerant features carefully designed into the original circuit.
Here, we show that, by a simple restriction of allowed swap moves, we can embed an arbitrary circuit in a fault-tolerant manner.The embedded circuit will be noisier, but we show, in the examples of
surface codes on heavy-hexagonal and hexagonal lattices, that the noise deterioration is not severe.
Our approach is easily incorporated into existing circuit compilation algorithms, and offers an immediate solution to implementing circuits on current hardware in a fault-tolerant manner.
Circuit-level fault tolerance of cat codes
LDH My, S Qin, and HK Ng (arXiv:2406.04157)
Noise regions (below curves) where EC works under circuit-level noise, for fixed 𝜏wait, optimized 𝜏wait (varied wait) and squeezing; N specifies the order of the cat code.
Bosonic codes, which encode quantum information in the infinite Hilbert space of a harmonic oscillator, are viable alternatives to conventional qubit codes. The family of rotationally symmetric
bosonic (RSB) codesis capable of correcting for both photon loss and phase (i.e., rotation) errors, offering robustness against arbitrary physical errors at the base layer of encoding.
We extend the formalism of fault tolerance to RSB codes, and assess the performance of previously proposed teleportation-based error correction (EC) circuits [Grimsmo et al., 2020] for cat codes (a
type of RSB codes), accounting for circuit-level noise, i.e., where every physical component of the circuit can be faulty. We find that the noise threshold is significantly worse than found in
previous more idealised studies. Through our analysis, we identify crucial circuit settings, such as the choice of code order, the optimal waiting time between EC cycles, and the addition of
squeezing to the code states, that improve the noise threshold by an order of magnitude, restoring the noise requirement to a level achievable with near-term quantum hardware.
Bosonic codes in quantum-dot–resonator systems
M Ma, HK Ng, with the group of TS Koh in NTU
Recent advancements in coupling quantum-dots to superconducting (SC) resonatorsenable long-range gates between quantum-dot qubits, and present the intriguing possibility of implementing circuit-QED
ideas, originally for SC qubits and cavities, in quantum-dot–resonator systems. We study how arbitrary bosonic resonator states can be prepared using a double-quantum-dot system as control, and
further investigate how computational operations can be performed on information carried by bosonic codes.
We explore how to prepare arbitrary bosonic resonator states through interaction with a double-quantum-dot system, for different coupling regimes. In the coherent regime, the dual-channel Law-Eberly
protocol creates arbitrary superpositions of Fock states. In the dispersive regime, the qcMAP protocol commonly used in SC contexts allows for preparation of superpositions of coherent states. More
generally, GRAPE optimization gives flexible state preparation through simultaneous resonator and quantum-dot drives.
While we follow well-understood schemes from standard SC c-QED contexts, the main research thrust here is to examine how well the effective Hamiltonians assumed in those schemes describe the exact
dynamics of the double-quantum-dot–resonator system. Our system also offers different tuning knobs than those in SC systems, presenting further opportunities for improved control under noise.
Surface codes: numerical studies
M Myers II, M Lavialle, ET Duong, O Valette, and HK Ng
We study the performance of surface codes, one of the most popular routes to error-corrected quantum devices, under realistic noise scenarios.
The performance of surface codes under depolarizing (Pauli), amplitude-damping, and random non-Clifford and non-unitary noise.
Quasi-probability methods for simulating general noise. The stabilizer formalism allows for efficient classical simulation of error correction circuits with only Clifford gates and Pauli measurements
subjected to Pauli errors. Surface codes, however, can correct general noise, but studies of its experimentally relevant properties have been limited to Pauli noise only because of the stabilizer
restrictions. We make use of quasi-probability methods that circumvent the stabilizer restrictions to study the performance of surface codes under non-Pauli noise. While computational costs grow
rapidly as one steps away from stabilizer noise, relying on the specific features of error correction circuits, we can access noise types never before simulated for surface codes. Unitary noise
(i.e., coherent errors), however, remains difficult and requires novel ideas for feasible simulation.
Logical operations. Another direction of interest is the investigation of fault-tolerance properties of current methods of lattice surgery for implementing logical operations, like the Hadamard and
CNOT gates. The significantly higher simulation resource demands for lattice surgery as well as the question of proper decoding in such contexts form our current research focus.
Noise-adapted fault tolerance
LDH My and HK Ng, with P Mandayam (IIT Madras) and A Jayashankar (TCG CREST)
Standard fault-tolerant (FT) schemes are designed with codes that correct arbitrary errors and assume no knowledge of the physical noise. Noise-adapted FT schemes, tailor-made to deal with the
dominant noise in the device, may have lower resource overheads and less stringent thresholds. Here, we develop a full fault-tolerant quantum computing protocol for amplitude-damping (AD) noise,
using Bacon-Shor codes. We describe a universal set of fault-tolerant encoded gadgets and estimate the noise thresholds below which our scheme leads to more accurate computation. This is the first
example of a full FT scheme adapted to non-Pauli-type noise.
Our published article [PR Research 4, 023034 (2022)] details the protocol for the smallest instance of the 4-qubit code; a manuscript in preparation gives the generalization to higher-distance
Bacon-Shor codes.
Reinforcement learning for context-aware dynamical gate calibration
A Strauss, L Voss, and HK Ng
Model-free reinforcement learning
Quantum control techniques have enabled significant improvements in gate fidelities. However, most methods do not provide dynamical and contextual error robustness, likely important for near-term
devices. Here, we present (1) a gate calibration procedure based on reinforcement learning (RL) to suppress errors arising from a specific circuit context; (2) a concrete use case showing how
contextual and dynamical gate calibrations can successfully increase quantum circuit fidelity.​
Context-aware calibration. Each gate instance carries a unique calibration, adapted to its location in the circuit, i.e., its context.
Resource costs of quantum metrology
YL Len, T Acharya, T Lim, and HK Ng, with A Auffèves from MajuLab
Quantum metrology promises the possibility of beyond-classical precision in estimation problems, through nonclassical features in the probe states. More exotic states often offer greater advantages,
but are more difficult to prepare in practice. Here, we compare the efficiencies of quantum metrology with different probe-state choices by taking into proper account the experimental costs and
constraints in preparing the quantum states. Through such a resource-based figure of merit, we arrive at significantly different conclusions about the quantum advantages of different metrology
protocols, compared to the oversimplified conventional benchmarks based only on counting the number of probe states consumed. | {"url":"https://quantum-nghk.commons.yale-nus.edu.sg/research-2/","timestamp":"2024-11-06T09:24:15Z","content_type":"text/html","content_length":"45251","record_id":"<urn:uuid:b043c3a8-f531-4fee-ad1c-42ad0ec6dcf3>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00636.warc.gz"} |
1. for Each Parabola Below Find
1. For each parabola below find
(i) the point of crossing the y-axis
(ii) the roots of the parabola
(iii) the minimum or maximum turning point
(a) y = x2 – 6x (b) y = x2 + 4x
(c) (d)
y = x2 – 9
y = 8x – x2
(e) y = 2x2 – 32 (f)
y = 25 – x2
(g) y = x2 – 6x + 8 (h) y = x2 + 8x + 15
(i) y = x2 + 2x - 15 (j) y = x2 – 4x – 12
(k) (l) y = 2x2 + 5x – 3
y = -x2 + 10x - 9
2. The diagram shows the parabola
y = x2 + 2x.
(a) Find the coordinates of the point A.
(b) Find the coordinates of B, the minimum
turning point of the parabola.
3. The diagram shows the parabola
y = 12x – 2x2
(a)Find the coordinates of the point A.
(b)Find the coordinates of B, the maximum
turning point of the parabola.
4. The parabola with equation
y = 4 – x2
is shown opposite.
(a) Find the coordinates of A and B,
the roots of the parabola.
(b) Find the coordinates of C.
5. The diagram shows the graph of
y = 3x2 – 27
(a) Find A and B.
(b) Find the coordinates of C,
the minimum turning point.
6. The diagram opposite shows part of the graph of
y = x2 – 8x – 9.
The graph cuts the y-axis at A and the x-axis at B
and C.
(a) Write down the coordinates of A
(b) Find the coordinates of B and C
(c) Calculate the minimum value of x2 – 8x – 9.
7. The diagram shows the parabola
y = x2 – 10x + 16
(a) Write down the coordinates of E
(b) Find the coordinates of F and G
(c) Find the coordinates of H, the minimum
turning point.
8. The parabola with equation
y = x2 – 4x – 5
is shown opposite.
(a) Write down the coordinates of A
(b) Find the coordinates of B and C
(c) Find the coordinates of D, the minimum
turning point.
9. The diagram shows the parabola
y = x2 – 10x – 11
(a) Write down the coordinates of A
(b) Find the coordinates of B and C
(c) Find the minimum value of
y = x2 – 10x – 11.
10. The graph of
y = x2 + 8x + 7
is shown opposite.
(a) Write down the coordinates of A
(b) Find the coordinates of B and C
(c) Find the coordinates of D, the minimum
turning point.
11. The diagram shows the parabola
y = – x2 – 2x + 15
(a) Write down the coordinates of N
(d) Find the coordinates of K and L
(e) Find the coordinates of M, the maximum
turning point.
12. The diagram shows the parabola
y = – x2 + 6x + 7
(a) Write down the coordinates of A
(b) Find the coordinates of B and C
(c) Find the maximum value of
y = – x2 + 6x + 7
13. The graph of
y = x2 – x – 2
is shown opposite.
(a) Write down the coordinates of A
(b) Find the coordinates of B and C
(c) Find the coordinates of D, the minimum
turning point.
14. The graph of
y = x2 + 5x – 6
is shown opposite.
(a) Write down the coordinates of T
(b) Find the coordinates of Q and R
(c) Find the coordinates of P, the minimum
turning point.
15. The diagram opposite shows part of the
graph of
y = 4x2 + 4x – 3.
The graph cuts the y-axis at A and the
x-axis at B and C.
(a) Write down the coordinates of A
(b) Find the coordinates of B and C.
(c) Calculate the minimum value of 4x2 + 4x – 3
16. The diagram opposite shows part of the graph of
y = - 3x2 + 2x + 1.
The graph cuts the y-axis at P and the x-axis
at Q and R.
(a) Write down the coordinates of P.
(b) Find the coordinates of Q and R.
(c) Find the maximum turning point of
the parabola.
17. The diagram opposite shows part of the
graph of y = k(x – a)(x – b).
The graph cuts the y-axis at (0,-6) and the
x-axis at (-1,0) and (3,0).
(a) Write down the values of a and b.
(b) Calculate the value of k.
(c) Find the coordinates of the minimum turning
point of the parabola.
18. The diagram opposite shows part of
the graph of y = k(x – a)(x – b).
The graph cuts the y-axis at (0,-18)
and the x-axis at (-3,0) and (2,0).
(a) Write down the values of a and b.
(b) Calculate the value of k.
(c) Find the minimum value of the parabola.
19. The diagram opposite shows part of
the graph of y = k(x + a)(x + b).
The graph cuts the y-axis at (0,4) and
the x-axis at (-1,0) and (2,0).
(a) Write down the values of a and b.
(b) Find the value of k.
(c) Find the coordinates of the maximum
turning point of the parabola.
20. The diagram opposite shows part of
the graph of y = p(x + a)(x + b).
The graph cuts the y-axis at (0,-16)
and the x-axis at (-4,0) and (1,0).
(a) Write down the values of a and b.
(b) Find the value of p.
(c) Find the minimum value of y. | {"url":"https://docsbay.net/doc/493019/1-for-each-parabola-below-find","timestamp":"2024-11-11T07:57:49Z","content_type":"text/html","content_length":"17283","record_id":"<urn:uuid:1f69fa92-63ca-41b1-964d-7802c7a27db7>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00593.warc.gz"} |
Difference potentials method based on LOD splitting technique for nonlinear convection–diffusion equations with interfaces
[1] J. Albright, Y. Epshteyn, M. Medvinsky, and Q. Xia, High-order numerical schemes based on difference potentials for 2D elliptic problems with material interfaces, Appl. Numer. Math., 111 (2017),
pp. 64–91.
[2] A. R. Appadu and H. H. Gidey, Time-splitting procedures for the numerical solution of the 2d advectiondiffusion equation, Mathematical Problems in Engineering, 2013 (2013), p. 634657.
[3] D. S. Britt, S. V. Tsynkov, and E. Turkel, A high-order numerical method for the Helmholtz equation with nonstandard boundary conditions, SIAM J. Sci. Comput., 35 (2013), pp. A2255–A2292.
[4] S. Britt, S. Petropavlovsky, S. Tsynkov, and E. Turkel, Computation of singular solutions to the Helmholtz equation with high order accuracy, Appl. Numer. Math., 93 (2015), pp. 215–241.
[5] Y. Epshteyn, Upwind-difference potentials method for Patlak-Keller-Segel chemotaxis model, J. Sci. Comput., 53 (2012), pp. 689–713.
[6] Y. Epshteyn and M. Medvinsky, On the solution of the elliptic interface problems by difference potentials method, in Spectral and high order methods for partial differential equations—ICOSAHOM
2014, vol. 106 of Lect. Notes Comput. Sci. Eng., Springer, Cham, 2015, pp. 197–205.
[7] S. Huang and Y. Liu, A fast multipole boundary element method for solving the thin plate bending problem, Eng. Anal. Bound. Elem., 37 (2013), pp. 967–976.
[8] E. Lee and D. Kim, Stability analysis of the implicit finite difference schemes for nonlinear Schr¨odinger equation, AIMS Math., 7 (2022), pp. 16349–16365.
[9] J. Liu and Z. Zheng, IIM-based ADI finite difference scheme for nonlinear convection-diffusion equations with interfaces, Appl. Math. Model., 37 (2013), pp. 1196–1207.
[10] M. Medvinsky, S. Tsynkov, and E. Turkel, The method of difference potentials for the Helmholtz equation using compact high order schemes, J. Sci. Comput., 53 (2012), pp. 150–193.
[11] , High order numerical simulation of the transmission and scattering of waves using the method of difference potentials, J. Comput. Phys., 243 (2013), pp. 305–322.
[12] , Solving the Helmholtz equation for general smooth geometry using simple grids, Wave Motion, 62 (2016), pp. 75–97.
[13] A. A. Reznik, Approximation of surface potentials of elliptic operators by difference potentials, Dokl. Akad. Nauk SSSR, 263 (1982), pp. 1318–1321.
[14] V. S. Ryaben’kii, Method of difference potentials and its applications, vol. 30 of Springer Series in Computational Mathematics, Springer-Verlag, Berlin, 2002. Translated from the 2001 Russian
original by Nikolai K. Kulman.
[15] V. S. Ryaben’kii, V. I. Turchaninov, and E. Y. Epshte ` ˘ın, An algorithm composition scheme for problems in composite domains based on the method of difference potentials, Zh. Vychisl. Mat.
Mat. Fiz., 46 (2006), pp. 1853–1870.
[16] Q. Sheng, The legacy of ADI and LOD methods and an operator splitting algorithm for solving highly oscillatory wave problems, in Modern mathematical methods and high performance computing in
science and technology, vol. 171 of Springer Proc. Math. Stat., Springer, Singapore, 2016, pp. 215–230.
[17] G. D. Smith, Numerical solution of partial differential equations, Oxford Applied Mathematics and Computing Science Series, The Clarendon Press, Oxford University Press, New York, third ed.,
1985. Finite difference methods.
[18] D. A. Voss and A. Q. M. Khaliq, Parallel LOD methods for second order time dependent PDEs, Comput. Math. Appl., 30 (1995), pp. 25–35.
[19] R. F. Warming and B. J. Hyett, The modified equation approach to the stability and accuracy analysis of finite-difference methods, J. Comput. Phys., 14 (1974), pp. 159–179.
[20] H. Zhu, H. Shu, and M. Ding, Numerical solutions of two-dimensional Burgers’ equations by discrete Adomian decomposition method, Comput. Math. Appl., 60 (2010), pp. 840–848. | {"url":"https://ajmc.aut.ac.ir/article_5105.html","timestamp":"2024-11-10T14:10:54Z","content_type":"text/html","content_length":"50061","record_id":"<urn:uuid:85270e59-2af0-412c-86b5-4dec92d3b445>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00408.warc.gz"} |
calculate number of IP addresses in a subnet
Reading Time: 3 minutes
In this blog post I will take you through working out how many IPv4 IP addresses there are in a subnet using the built in Windows calculator.
Let’s take an example of a /24 mask, and I want to know how many IP addresses I can use,
192.168.1.0/24 and a mask of 255.255.255.0
If you haven not already read my post on CIDR notation simplified, I would recommend you have a read.
Add up the 1’s, that’s a total of 24 bits, that’s where the /24 comes from. See my post CIDR notation simplified if you wish to dive deeper.
How do I calculate how many available IP addresses I can assign from the above. We’ll use the Windows calculator,
Click on your start menu and search for calculator,
Switch from standard to scientific
Image showing standard calculator in Windows Operating System
Image showing the option to switch to Scientific mode
In the example above we had a subnet mask of /24. The sum to calculate the number of IP addresses available would be,
A total of 32 if we total all the bits together, we minus the bits which are turned on, 24, so that’s 32 – 24 leaving us with 8 bits (The last box to the right below).
The sum is below, but let’s input this into the calculator.
1. 32 minus 24 = 8
2. we now calculate 2 to the power of 8
3. To do this, clear the calculator and type 2
4. Next, click the Xy button as shown in the screenshot below
5. Type 8 and click the = button
That gives us 256 IP addresses in a /24 subnet mask. We take away two as we don’t use .0 (Network address) and 255 (Broadcast address) = 254 IP’s that we can use
Try the sum with another example,
What do you get if you calculate /16, /27 and /32 using the method above. What numbers appear on your calculator.
Let me know in the comments section below
/16 = ? post the answer below
/27 = ? post the answer below
/23 = ? post the answer below
That’s it. I hope you found this post useful. | {"url":"https://cloudbuild.co.uk/tag/calculate-number-of-ip-addresses-in-a-subnet/","timestamp":"2024-11-06T01:25:31Z","content_type":"text/html","content_length":"133873","record_id":"<urn:uuid:002fe147-3e7d-4b86-ad99-dedabac9c935>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00053.warc.gz"} |
sing Rails & MySQL
With an explosive growth of the internet, data need to be localized more than ever. Many DBMS already include some GIS features but support is often incomplete. InnoDB’s MySQL engine currently
supports some spacial data types but not indexes on it. Building an efficient GIS system from scratch is quite easy, let’s see how to do it using Ruby on Rails & MySQL.
The distances between two points on a sphere from their longitudes and latitudes can be calulated using the Haversine formula (half-versed-sine). Because the earth is not a perfect sphere but an
oblate spheroid, this formula gives an approximated distance with a margin of error generally below 0.3% (the earth radius varies from 6356.752 km at the poles to 6378.137 km at the equator). For a
greater accuracy, the Vincenty’s formula could be used.
Because we are using high-precision floating point numbers, we are able to use the simpler Spherical law of cosines formula : d = acos(sin(φ1).sin(φ2) + cos(φ1).cos(φ2).cos(Δλ)).R where φ is
latitude, λ is longitude.
The first thing to do is to implement the Spherical law of cosines formula in a MySQL function to be able to use it easily into queries.
The earth radius can be chosen depending on where you want to make those calculations. Miles can also be used instead of kilometers.
1 DELIMITER //
3 DROP FUNCTION IF EXISTS distance//
4 CREATE FUNCTION distance(lat1 DOUBLE, lng1 DOUBLE, lat2 DOUBLE, lng2 DOUBLE) RETURNS DOUBLE
5 LANGUAGE SQL
6 DETERMINISTIC
7 COMMENT 'Calculate distance in km between two points on earth'
8 RETURN ACOS(SIN(RADIANS(lat1)) * SIN(RADIANS(lat2)) + COS(RADIANS(lat1)) * COS(RADIANS(lat2))
9 * COS(RADIANS(lng1 - lng2))) * 6371;//
So if we consider that we have a table point_of_interests including 2 fields lat (latitude) and lng (longitude), a way to find all POIs in a 50 km radius around a given point 48.852842, 2.350333
could be :
1 SELECT * FROM PointOfInterests
2 WHERE distance(48.852842, 2.350333, lat, lng) < 50;
The big performance issue with this query is that it does a whole table scan to calculate distance between the point and all POIs. It can be ok if you have few POIs but in most cases it sucks hard.
These boundaries are illustrated on the schema and are located at 4 differents bearings and obviously at a radius distance.
• θ 0 : Maximum latitude
• θ 90 : Maximum longitude
• θ 180 : Minimum latitude
• θ 270 : Minimum longitude
Using the Haversine formula, we are able to find geographic coordinates of a point with another point, a bearing and the distance between the two points.
Latitude φ2 = asin(sin(φ1)*cos(d/R) + cos(φ1)*sin(d/R)*cos(θ))
Longitude λ2 = λ1 + atan2(sin(θ)*sin(d/R)*cos(φ1), cos(d/R)−sin(φ1)*sin(φ2))
where φ is latitude, λ is longitude, θ is the bearing (in radians, clockwise from north), d is the distance, R is the earth’s radius (d/R is the angular distance, in radians).
To be able to add the distance search feature to any Active Record models, the search code can be included into a Concern.
1 module Localizable
2 extend ActiveSupport::Concern
4 included do
5 scope :near, -> lat, lng, radius {
6 d =-> b { destination_point(lat, lng, b, radius) }
7 where(["lat BETWEEN ? AND ? AND lng BETWEEN ? AND ?", d[180][:lat], d[0][:lat], d[270][:lng], d[90][:lng]])
8 .where(["COALESCE(distance(?, ?, lat, lng), 0) < ?", lat, lng, radius])
9 }
10 end
12 module ClassMethods
14 # Return destination point given distance and bearing from start point
15 def destination_point(lat, lng, initial_bearing, distance)
16 d2r =-> x { x * Math::PI / 180 }
17 r2d =-> x { x * 180 / Math::PI }
18 angular_distance = distance / 6371.0
19 lat1, lng1, bearing = d2r.(lat), d2r.(lng), d2r.(initial_bearing)
20 lat2 = Math.asin(Math.sin(lat1) * Math.cos(angular_distance) + Math.cos(lat1) * Math.sin(angular_distance) * Math.cos(bearing))
21 lng2 = lng1 + Math.atan2(Math.sin(bearing) * Math.sin(angular_distance) * Math.cos(lat1), Math.cos(angular_distance) - Math.sin(lat1) * Math.sin(lat2))
22 { :lat => r2d.(lat2).round(7), :lng => r2d.(lng2).round(7) }
23 end
25 end
26 end
Just include the Localizable concern into each models you need to (off course the corresponding table must have lat & lng fields).
1 class PointOfInterest < ActiveRecord::Base
2 include Localizable
5 end
You’re done !
1 PointOfInterest.near(48.852842, 2.350333, 50).count
2 (23.4ms) SELECT COUNT(*) FROM `point_of_interests` WHERE (lat BETWEEN 48.4031812 AND 49.3025028 AND lng BETWEEN 1.6669714 AND 3.0336946) AND (COALESCE(distance(48.852842, 2.350333, lat, lng), 0) < 50)
3 => 969 | {"url":"http://kochka.org/2013/08/07/geo-distance-search-from-scratch-using-rails-and-mysql/","timestamp":"2024-11-12T22:28:16Z","content_type":"text/html","content_length":"28159","record_id":"<urn:uuid:af96e8e3-c704-4ca7-a11c-5e0f4fb9a845>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00250.warc.gz"} |
hard to find yankee candles
A function is invertible if and only if it is a bijection. This implies a2 = b2 by the de nition of f. Thus a= bor a= b. Then \(f\) is injective if and only if the restriction \(f^{-1}|_{\range(f)}\)
is a function. }\) Thus \(b = f(a) = y\text{,}\) so \(f^{-1}\) is injective. f: X → Y Function f is one-one if every element has a unique image, i.e. Then for a few hundred more years, mathematicians
search for a formula to the quintic equation satisfying these same properties. I have to prove two statements. Now suppose \(a \in A\) and let \(b = f(a)\text{. Moreover, if \(f : A \to B\) is
bijective, then \(\range(f) = B\text{,}\) and so the inverse relation \(f^{-1} : B \to A\) is a function itself. The crux of the proof is the following lemma about subsets of the natural numbers. 1.
The inverse of a permutation is a permutation. Wikidot.com Terms of Service - what you can, what you should not etc. Bijective functions are also called one-to-one, onto functions. An injection may
also be called a one-to-one (or 1–1) function; some people consider this less formal than "injection''. To prove that a function is injective, we start by: “fix any with ” Then (using algebraic
manipulation etc) we show that . }\) That means \(g(f(x)) = g(f(y))\text{. Check out how this page has evolved in the past. }\) That is, for every \(b \in B\) there is some \(a \in A\) for which \(f
(a) = b\text{.}\). \newcommand{\amp}{&} iii)Function f is bijective i f 1(fbg) has exactly one element for all b 2B . when f(x 1 ) = f(x 2 ) ⇒ x 1 = x 2 Otherwise the function is many-one. If \(f,g\)
are bijective then \(g \circ f\) is also bijective by what we have already proven. for every y in Y there is a unique x in X with y = f ( x ). Suppose \(f : A \to B\) is bijective, then the inverse
function \(f^{-1} : B \to A\) is also bijective. The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office
of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. \newcommand{\lt}{<} If m>n, then there is no injective function from N m to N
n. Proof. Let \(b_1,\ldots,b_n\) be a (combinatorial) permutation of the elements of \(A\text{. Injections and surjections are `alike but different,' much as intersection and union are `alike but
different.' Proof: We must (⇒) prove that if f is injective then it has a left inverse, and also (⇐) that if fhas a left inverse, then it is injective. A function f:A→B is injective or one-to-one
function if for every b∈B, there exists at most one a∈A such that f(s)=t. }\) Since \(f\) is injective, \(x = y\text{. Then \(f(a_1),\ldots,f(a_n)\) is some ordering of the elements of \(A\text{,}\)
i.e. Consider the following function that maps N to Z: f(n) = (n 2 if n is even (n+1) 2 if n is odd Lemma. Groups were invented (or discovered, depending on your metamathematical philosophy) by
Évariste Galois, a French mathematician who died in a duel (over a girl) at the age of 20 on 31 May, 1832, during the height of the French revolution. De nition 67. A function f: X→Y is: (a)
Injective if for all x1,x2 ∈X, f(x1) = f(x2) implies x1 = x2. View and manage file attachments for this page. A function is said to be bijective or bijection, if a function f: A → B satisfies both
the injective (one-to-one function) and surjective function (onto function) properties. Functions can be injections (one-to-one functions), surjections (onto functions) or bijections (both one-to-one
and onto). So, what is the difference between a combinatorial permutation and a function permutation? The function \(f\) is called injective (or one-to-one) if it maps distinct elements of \(A\) to
distinct elements of \(B.\)In other words, for every element \(y\) in the codomain \(B\) there exists at most one preimage in the domain \(A:\) A function f: R !R on real line is a special function.
That is, let \(f: A \to B\) and \(g: B \to C\text{.}\). (c) Bijective if it is injective and surjective. General Wikidot.com documentation and help section. See pages that link to and include this
page. Problem 2. \DeclareMathOperator{\dom}{dom} In the following proofs, unless stated otherwise, f will denote a function from A to B and g will denote a function from B to A. I will also assume
that A and B are non-empty; some of these claims are false when either A or B is empty (for example, a function from ∅→B cannot have an inverse, because there are no functions from B→∅). The function
\(f\) that we opened this section with is bijective. (b) Surjective if for all y∈Y, there is an x∈X such that f(x) = y. How to check if function is one-one - Method 1 In this method, we check for
each and every element manually if it has unique image Example 4.3.4 If A ⊆ B, then the inclusion map from A to B is injective. A group is just a set of things (in this case, permutations) together
with a binary operation (in this case, composition of functions) that satisfy a few properties: Chances are, you have never heard of a group, but they are a fundamental tool in modern mathematics,
and they are the foundation of modern algebra. . Definition. Let \(A\) be a nonempty set. Prove there exists a bijection between the natural numbers and the integers De nition. Notice that nothing in
this list is repeated (because \(f\) is injective) and every element of \(A\) is listed (because \(f\) is surjective). This is what breaks it's surjectiveness. Shopping. It should be noted that Niels
Henrik Abel also proved that the quintic is unsolvable, and his solution appeared earlier than that of Galois, although Abel did not generalize his result to all higher degree polynomials. Tap to
unmute. Informally, an injection has each output mapped to by at most one input, a surjection includes the entire possible range in the output, and a bijection has both conditions be true. Let a;b2N
be such that f(a) = f(b). However, we also need to go the other way. (A counterexample means a speci c example Something does not work as expected? Proof. Now if I wanted to make this a surjective
and an injective function, I would delete that mapping and I … A function \(f : A \to B\) is said to be injective (or one-to-one, or 1-1) if for any \(x,y \in A\text{,}\) \(f(x) = f(y)\) implies \(x
= y\text{. ii)Function f is surjective i f 1(fbg) has at least one element for all b 2B . Example 7.2.4. Click here to edit contents of this page. injective. A function \(f : A \to B\) is said to be
bijective (or one-to-one and onto) if it is both injective and surjective. Galois invented groups in order to solve this problem. In this case the statement is: "The sum of injective functions is
injective." A function f: A → B is: 1. injective (or one-to-one) if for all a, a′ ∈ A, a ≠ a′ implies f(a) ≠ f(a ′); 2. surjective (or onto B) if for every b ∈ B there is an a ∈ A with f(a) = b; 3.
bijective if f is both injective and surjective. Because f is injective and surjective, it is bijective. Well, let's see that they aren't that different after all. If $f_{\big|N_k}$ is injective
function for all $k\in\mathbb{N}$, then $f$ is injective function(one to one) and second if $f[N_k]=N_k$ for all $k\in\mathbb{N}$, then $f$ is identity function. Discussion In Example 2.3.1 we prove
a function is injective, or one-to-one. We use the definition of injectivity, namely that if f(x) = f(y), then x = y. Let \(f : A \to B\) be a function from the domain \(A\) to the codomain \(B.\). \
DeclareMathOperator{\perm}{perm} Change the name (also URL address, possibly the category) of the page. Lemma 1. Suppose \(f,g\) are injective and suppose \((g \circ f)(x) = (g \circ f)(y)\text{. As
we established earlier, if \(f : A \to B\) is injective, then the restriction of the inverse relation \(f^{-1}|_{\range(f)} : \range(f) \to A\) is a function. There is another similar formula for
quartic equations, but the cubic and the quartic forumlae were not discovered until the middle of the second millenia A.D.! Proof: Composition of Injective Functions is Injective | Functions and
Relations. }\) Therefore \(z = g(f(x)) = (g \circ f)(x)\) and so \(z \in \range(g \circ f)\text{. }\) Thus \(b = f(a) = y\text{,}\) so \(f^{-1}\) is injective. If it isn't, provide a counterexample.
Galois invented groups in order to solve, or rather, not to solve an interesting open problem. Proof. }\) Since \(g\) is injective, \(f(x) = f(y)\text{. the binary operation is associate (we already
proved this about function composition), applying the binary operation to two things in the set keeps you in the set (, there is an identity for the binary operation, i.e., an element such that
applying the operation with something else leaves that thing unchanged (, every element has an inverse for the binary operation, i.e., an element such that applying the operation to an element and
its inverse yeilds the identity (. If a function is defined by an even power, it’s not injective. =⇒ : Theorem 1.9 shows that if f has a two-sided inverse, it is both surjective and injective …
Although, instead of finding a formula, he proved that no such formula exists for the quintic, or indeed for any higher degree polynomial. If it passes the vertical line test it is a function; If it
also passes the horizontal line test it is an injective function; Formal Definitions. It is clear, however, that Galois did not know of Abel's solution, and the idea of a group was revolutionary. \
quad (f \circ i)(x) = f(i(x)) = f(x) , \quad (i \circ f)(x) = i(f(x)) = f(x) , Unless otherwise stated, the content of this page is licensed under. Basically, it says that the permutations of a set \
(A\) form a mathematical structure called a group. Suppose \(f,g\) are surjective and suppose \(z \in C\text{. }\) Since \(g\) is surjective, there exists some \(y \in B\) with \(g(y) = z\text{. Here
is the symbolic proof of equivalence: Notice that we now have two different instances of the word permutation, doesn't that seem confusing? There is another way to characterize injectivity which is
useful for doing proofs. }\) Since any element of \(A\) is only listed once in the list \(b_1,\ldots,b_n\text{,}\) then \(f\) is injective. Proving a function is injective. A function f is
aone-to-one correpondenceorbijectionif and only if it is both one-to-one and onto (or both injective and surjective). View/set parent page (used for creating breadcrumbs and structured layout).
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License Click here to toggle editing of individual sections of the page (if possible).
\), Injective, surjective and bijective functions, Test corrections, due Tuesday, 02/27/2018, If \(f,g\) are injective, then so is \(g \circ f\text{. Well, two things: one is the way we think about
it, but here each viewpoint provides some perspective on the other. The identity map \(I_A\) is a permutation. All Injective Functions From ℝ → ℝ Are Of The Type Of Function F. If You Think That It
Is True, Prove It. Claim: fis injective if and only if it has a left inverse. Definition4.2.8. If it is, prove your result. "If y and x are injective, then z(n) = y(n) + x(n) is also injective." A
function \(f: A \rightarrow B\) is bijective if it is both injective and surjective. Note that $f_{\big|N_k}$ is restricted domain of function and $f[N_k]=N_k$ is image of function. This formula was
known even to the Greeks, although they dismissed the complex solutions. }\) Thus \(A = \range(f^{-1})\) and so \(f^{-1}\) is surjective. Example 1.3. Therefore, d will be (c-2)/5. Stated in concise
mathematical notation, a function f: X → Y is bijective if and only if it satisfies the condition. Theidentity function i A on the set Ais de ned by: i A: A!A; i A(x) = x: Example 102. Append content
without editing the whole page source. }\), If \(f,g\) are bijective, then so is \(g \circ f\text{.}\). View wiki source for this page without editing. The composition of permutations is a
permutation. This is another example of duality. }\), If \(f\) is a permutation, then \(f \circ f^{-1} = I_A = f^{-1} \circ f\text{. Let \(A\) be a nonempty set. Let, c = 5x+2. De nition 68. However,
the other difference is perhaps much more interesting: combinatorial permutations can only be applied to finite sets, while function permutations can apply even to infinite sets! If you want to
discuss contents of this page - this is the easiest way to do it. This shows 8a8b[f(a) = f(b) !a= b], which shows fis injective. Copy link. You should prove this to yourself as an exercise. Share.
Injective but not surjective function. \renewcommand{\emptyset}{\varnothing} Well, no, because I have f of 5 and f of 4 both mapped to d. So this is what breaks its one-to-one-ness or its
injectiveness. 2. The graph of $i$ is given below: If we instead consider a finite set, say $B = \{ 1, 2, 3, 4, 5 \}$ then the identity function $i : B \to B$ is the function given by $i(1) = 1$, $i
(2) = 2$, $i(3) = 3$, $i(4) = 4$, and $i(5) = 5$. \DeclareMathOperator{\range}{rng} This function is injective i any horizontal line intersects at at most one point, surjective i any }\) Thus \(g \
circ f\) is surjective. When we say that no such formula exists, we mean there is no formula involving only the coefficients and the operations mentioned; there are other ways to find roots of higher
degree polynomials. Let X and Y be sets. Watch headings for an "edit" link when available. This means a function f is injective if a1≠a2 implies f(a1)≠f(a2). Called one-to-one, onto functions ), then
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Simplify The Following Expression. (19x + 4y) + (49x + 32y) (2024)
Answer 1
Answer: 68x + 36y
Step-by-step explanation:
Combine Like Terms:
= 19x + 4y + 49x + 32y
= (19x + 49x) + (4y + 32y)
= 68x + 36y
Hope this helped you!
Answer 2
68x + 36y
Step-by-step explanation:
You can compeletely get rid of the parenthesis because it is addition through and thorough.
You have 19x + 4y + 49x + 32y
You can combine like terms to get
19x +49x which is 68x
4y + 32y which is 36y
You now have 68 + 36y which cannot be simplified anymore because the variables are different.
Hope this helps!
Related Questions
An architect's scale drawing of a new school is 8.4 inches
long. The scale used in the drawing is 2 inches = 8 feet.
What is the actual length, in feet, of the school?
Select one:
Step-by-step explanation:
What is the slope of the line through (2, -2) and (9,3)?
The slope I got using a calculator was 0.71 or 5/7. The slope can be calculated through the slope formula.
Use the slope formula y1-y2/x1-x1 and you will get the answer is 5/7!
ratios that are equivalent to 40:28
Step-by-step explanation:
Using eight eights and addition only, can you make 1000?
Step-by-step explanation:
Step-by-step explanation:
A grocery store sells a bag of 6 oranges for $3.90. What is the unit cost?
Pls Answer and I shall give u a- Thank You - Vielen Dank
it would be 0.64
Step-by-step explanation:
Create a list of steps, in order, that will solve the following equation. 3(x+1)^2=108
Simplify the following expression (87x + 63y) - (6x + 32y)
Answer: 81x + 31y
Step-by-step explanation:
Explain the difference between parallel and perpendicular lines in terms of their slopes. What form of equation of a line is easiest to determine slope?
Parallel lines have the same slope, while perpendicular lines have completely opposite slopes. So a line has a slope of 2, a parallel line would have a slope of 2, and a perpendicular line would have
a slope of -1/2. The easiest form to identify the slope is standard form because it shows the slope within the equation with out having to solve the equation or convert it into another form.
plz help
Solve the system by Substitution
A(-6, 3)
B(4, -7)
C(6, -7)
D(-4, 7)
Step-by-step explanation:
Write 0.09 as a fraction.
A) 1/11
B) 1/7
C) 1/99
D) 1/9
Step-by-step explanation:
The answer is a
Step-by-step explanation:
Its D because its literly 1/11
Which function has a domain of all real numbers?
A. y=-x½+5 B. y=(2x)⅓-7 C. y=(x+2)¼ D. y=-2(3x)⅙?
Step-by-step explanation:
B. Because the others have exponents whose denominators are even so the results of , for example x^1/2, if x is negative is not defined.
For example:
For A, when x is -4, (-4)^1/2 is not real.
For B, when x = -4, (2*-4)^ 1/3 = -8^(1/3) = = -2.
Step-by-step explanation:
For all Plato users
Will give brainiest and thx for the help
C would be the correct answer I hope you have an amazing day!
katy had 162 pieces of candy. everyday she gave 6 pieces to her little brother. which function can be used to find how many candles remain x days since katy began giving candy to her brother?
6x = 162 or 162 = 6x
Step-by-step explanation:
x = days, since how many days is unknown, we substitute it as x
hurryyy!! Otero translated the phrase "three less than a number" into the expression 3 - X. Which best describes the accuracy of
Otero's expression?
It is accurate. The phrase can be translated as "three" = 3, "less than" = subtraction, and "a number" = X, SO 3- X is
the correct expression.
It is inaccurate. Three is being subtracted from a number, so X-3 is what he should have written.
It is inaccurate. Three is being compared to a number, so 3 It is inaccurate. Three is being added to a number, so 3+ x is what he should have written.
A. It is accurate. The phrase can be translated as “three” = 3, “less than” = subtraction, and “a number” = x, so 3-x is the correct expression.
Step-by-step explanation:
See below
Step-by-step explanation:
On Edge 2020 it is B.
It is inaccurate. Three is being subtracted from a number, so x minus 3 is what he should have written.
Answer:1144.2 for question 5
Step-by-step explanation:
what is 4\5 equal ?\20 round by tens
Step-by-step explanation:
In the diagram, AB= DC and AC= DB. Why does DBC = ACB?
Because they are congruent sides
AB = DC Given
AC = DB Given
BC = BC Reflexive property
△ACB = △DBC SSS
m∠DBC = m∠ACB CPCTC
Step-by-step explanation:
Reflexive property basically means that something is congruent to itself
SSS means that if all 3 sides of 2 triangles are congruent, the triangles are congruent
CPCTC means C.orresponding P.arts of C.ongruent T.riangles are C.ongruent
meaning because the triangles are congruent, all the corresponding sides and angles are also congruent
Deku and reaper also what is 2-4=?
the answer is -2
Step-by-step explanation: The answer is from google
Step-by-step explanation:
indeed brother
Use the Pythagorean Theorem to solve for x.
The other person provided a great answer but i do it a little different....
[tex]\left[\begin{array}{ccc}A^2=C^2-B^2\\A^2=10^2-6^2\\A^2=100-36\\A^2=64\\\sqrt{64}=8 \end{array}\right][/tex]
All i did was use the formula A^2=C^2-B^2. I Plugged in the numbers to the formula and got A^2=10^2-6^2. Next, I found what 10^2 equals by doing 10*10 which equals 100. Then, i found out what 6^2
equals by doing 6*6 which equals 36. Next, what i did was subtracted 100 from 36 and got 64. Finally i found the square root of the number 64 which equal 8 since 8x8=64.
The measure of the unknown variable {x] is 8 inches.
What is Pythagoras theorem?
The Pythagorean theorem or Pythagoras' theorem is a fundamental relation in Euclidean geometry between the three sides of a right triangle. Mathematically, it can be written as -
(hypotenuse)² = (base)² + (perpendicular)²
Given is a right angle triangle.
Using the Pythagoras theorem, we can write -
(hypotenuse)² = (base)² + (perpendicular)²
(10)² = (base)² + (6)²
(base)² = (10)² - (6)²
(base)² = (10 - 6)(10 + 6)
(base)² = 4 x 16
(base)² = 64 inches
(base) = 8 inches
x = 8 inches
Therefore, the measure of the unknown variable {x] is 8 inches.
To solve more questions on Pythagoras theorem, visit the link below -
Step-by-step explanation:
Since 12/4=3, 20/4=x, which is 5
5 = x
Step-by-step explanation:
12. 3
__ = __
20 ?
12 ÷ 4 = 4
20 ÷ 4 = 5
PLEASE HELP!!!!
When you decrease -1 by 4 the result is -5.
Please select the best answer from the choices provided
Step-by-step explanation:
The answer is false because negatives work backwards
What is the simplified form of the expression?
7c²d-7c+4d-10, you first do 5c²d+2c²d, then you do -4c-3c, after that you do +3d + d and finally you do -3-7.
I hope I helped you
7c ^2 d-7c+4d-10
hope this helps
Straws are sold in packs and boxes.
there 15 straws in each pack.
there are 48 straws in each box.
Tricia buys p packs of straw and boxes of straw.
write down an expression in terms of p and b, for the total number of straws bought by tricia.
15p + 48b = total amount of straws?
help me with this math please
She spent two times as long on the phone then she did cleaning the kitchen.
Step-by-step explanation:
1/4 times 2 is 1/2.
In total she spent 45 minutes on the phone and in the kitchen cleaning
Step-by-step explanation:
1/4 of an hour is 15 minutes and 1/2 of an hour is 30 min. You add them together to get 45 minutes and that how much time she has spent in total on the phone and in the kitchen.
Write an equation in slope- intercept for of each line
The cost of a gym membership is $25 per month. There is an initial fee of $40
y = 25x + 40
Step-by-step explanation:
Using the formula, y = mx + b we can find out what this equation would be.
m = the rate of change/slope
b = the y- intercept
25 per month is telling us that this is the rate of change.
Since the $40 is only paid once, you can tell that this is the y intercept.
The equation will look like:
y = 25x + 40
On a graph it would look like:
Let the total cost of a gym membership be y, with the number of months as x.
The total cost would be 40 dollars, the initial fee, plus 25 times the number of months (x).
y = 40 + 25x
|6| + |-2|
show your work
Answer: This just adding so it would be 8
6+2=8 6, 7, 8
Step-by-step explanation:
Step-by-step explanation:(6) + (-2) the plus would cancel out the -
im trying to find the answer for 7f + 15 - 2g (please make it clear of what the answer is)
22-2g I think this is the answer
Which of the following represents a
fraction that converts to a repeating
decimal with a value less than one?
A) 1/2
B) 4/3
C) 9/10
D) 479
Step-by-step explanation:
Please help ASAP In the diagram, the radius of the inner circle is 4 meters. The area of the shaded region is 481 square meters. Find the radius of the outer circle.
The answer is 8!
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Perseids & Aurora
Moderated by: Fred Miranda
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Archive 2016 · Perseids & Aurora
Hardcore p.1 #1 · p.1 #1 · Perseids & Aurora
Upload & Sell: On
Was out taking some photos of the perseid meteor shower. Was lucky enough to also get a small aurora show.
First image is a panorama of 8 shots encompassing about 180 degree view facing north.
Second image is a composite of 4 images to show the perseid meteors with the aurora.
Thanks for looking,
I usually don't post links to my 500px but the panorama really does look much better larger so here is a link to a larger version.
Edited on Aug 13, 2016 at 01:37 PM · View previous versions
Aug 13, 2016 at 07:53 AM
douter p.1 #2 · p.1 #2 · Perseids & Aurora
Upload & Sell: Off Crazy wonderful Corey! I went out Thursday night, but did not see any meteors, even though that was to be the highpoint of the shower. These are both very fine.
Aug 13, 2016 at 09:30 AM
CosmicCruiser p.1 #3 · p.1 #3 · Perseids & Aurora
Upload & Sell: On both are great. love the meteors!
Aug 13, 2016 at 09:44 AM
srkbar p.1 #4 · p.1 #4 · Perseids & Aurora
Upload & Sell: Off Awesome, love them both but the second one is special
Thanks for sharing.
Aug 13, 2016 at 10:42 AM
Jim Dockery p.1 #5 · p.1 #5 · Perseids & Aurora
Upload & Sell: Off Both are great. I sure struck out on the perseid a couple days ago. Set my alarm for 11:30 am instead of pm!
Aug 13, 2016 at 12:33 PM
Hardcore p.1 #6 · p.1 #6 · Perseids & Aurora
Upload & Sell: On douter wrote:
Crazy wonderful Corey! I went out Thursday night, but did not see any meteors, even though that was to be the highpoint of the shower. These are both very fine.
Thanks Douglas! Ya, the "storm" of meteors wasn't much more than a heavy trickle imo.
CosmicCruiser wrote:
both are great. love the meteors!
srkbar wrote:
Awesome, love them both but the second one is special
Thanks for sharing.
Thank you!
Jim Dockery wrote:
Both are great. I sure struck out on the perseid a couple days ago. Set my alarm for 11:30 am instead of pm!
Thanks Jim! Well at least you got a good nights sleep!
Aug 13, 2016 at 03:50 PM
Dave Dillemuth p.1 #7 · p.1 #7 · Perseids & Aurora
Upload & Sell: Off Awesome! Love them both.
Aug 13, 2016 at 04:22 PM
mitchel674 p.1 #8 · p.1 #8 · Perseids & Aurora
Upload & Sell: Off Fantastic. That second one is superb!
Aug 13, 2016 at 05:44 PM
DaleBerlin p.1 #9 · p.1 #9 · Perseids & Aurora
Upload & Sell: Off Incredible images, you da man, very well done. 92 on 500px lol, what a joke, it's a popularity contest over there.
Aug 13, 2016 at 08:22 PM
bill s p.1 #10 · p.1 #10 · Perseids & Aurora
Upload & Sell: Off Both are my favs. Great technique. You know what you are doing, I could definitely learn from you!
Aug 14, 2016 at 02:06 AM
Brad Williams p.1 #11 · p.1 #11 · Perseids & Aurora
Upload & Sell: Off Beautiful photos Corey! The meteors are a nice addition to the Auroras!
Aug 15, 2016 at 09:34 AM
Hardcore p.1 #12 · p.1 #12 · Perseids & Aurora
Upload & Sell: On Dave Dillemuth wrote:
Awesome! Love them both.
Thanks very much Dave!
mitchel674 wrote:
Fantastic. That second one is superb!
Thanks Mitchel!
DaleBerlin wrote:
Incredible images, you da man, very well done. 92 on 500px lol, what a joke, it's a popularity contest over there.
Ya, not sure why I bother sometimes with 500px. I did sell some images off it so it does pay a bit.
bill s wrote:
Both are my favs. Great technique. You know what you are doing, I could definitely learn from you!
Thanks Bill!
Brad Williams wrote:
Beautiful photos Corey! The meteors are a nice addition to the Auroras!
Thanks Brad!
Aug 15, 2016 at 06:29 PM
Fred Miranda p.1 #13 · p.1 #13 · Perseids & Aurora
Upload & Sell: On Great set!
The second is more dynamic and my personal favorite.
I really like the silhouette on the first though. Perhaps the very top right branches are not even needed. I don't know, but it's worth a try to see how it looks without
All the best,
Aug 15, 2016 at 07:45 PM
psharvic p.1 #14 · p.1 #14 · Perseids & Aurora
Upload & Sell: On Nice work on these, Corey.
Aug 15, 2016 at 07:53 PM
Rohanban p.1 #15 · p.1 #15 · Perseids & Aurora
Upload & Sell: Off #2! is pretty cool especially with the meteors. I like the aurora better in #1.
Aug 17, 2016 at 11:33 AM
m.sommers00 p.1 #16 · p.1 #16 · Perseids & Aurora
Upload & Sell: Off Excellent as usual!
Aug 17, 2016 at 11:52 AM
Hardcore p.1 #17 · p.1 #17 · Perseids & Aurora
Upload & Sell: On Fred Miranda wrote:
Great set!
The second is more dynamic and my personal favorite.
I really like the silhouette on the first though. Perhaps the very top right branches are not even needed. I don't know, but it's worth a try to see how it looks without
All the best,
Thanks Fred! I thought about removing the branches on the top right but I think I'd prefer to leave them in. Thanks for the suggestion though!
psharvic wrote:
Nice work on these, Corey.
Rohanban wrote:
#2! is pretty cool especially with the meteors. I like the aurora better in #1.
m.sommers00 wrote:
Excellent as usual!
Thanks Mathew!
Aug 18, 2016 at 09:07 AM
IndyFab p.1 #18 · p.1 #18 · Perseids & Aurora
Upload & Sell: Off Have to love the Aurora, my pick is #2
Aug 19, 2016 at 04:38 PM
dswiger p.1 #19 · p.1 #19 · Perseids & Aurora
Upload & Sell: Off Very nice.
I agree w/Fred. The 2nd has just the right balance/dynamics of Aurora vs meteors.
Aug 19, 2016 at 05:42 PM
7.5 Ire p.1 #20 · p.1 #20 · Perseids & Aurora
Upload & Sell: Off Very Nice, Corey! Thanks for sharing.
Aug 20, 2016 at 11:52 AM | {"url":"https://www.fredmiranda.com/forum/topic/1445279/","timestamp":"2024-11-02T11:21:21Z","content_type":"application/xhtml+xml","content_length":"96937","record_id":"<urn:uuid:90f56cb9-bbe4-43c6-a26b-81efbaf3cd95>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00631.warc.gz"} |
In a 100 m race, Bipin, Chandan, Danny and Feroz won the first four prizes, not necessarily in the same order. Each of these four sprinters hails from a different city among Delhi, Mumbai, Kolkata and Chennai. Who got the first prize?
In a 100 m race, Bipin, Chandan, Danny and Feroz won the first four prizes, not necessarily in the same order. Each of these four sprinters hails from a different city among Delhi, Mumbai, Kolkata
and Chennai. Who got the first prize?
A. Bipin is not from Kolkata and didn't get the first prize. The sprinter from Chennai is the winner. Danny is from Mumbai.
B. Feroz is either from Delhi or from Chennai.
- one statement is sufficient and the other statement is not sufficient to answer the question.
- either statement taken alone is sufficient to answer the question.
- the two statements together are sufficient but neither statement alone is sufficient to answer the question.
- even both statements together are not sufficient to answer the question CORRECT ANSWER : the two statements together are sufficient but neither statement alone is sufficient to answer the question.
Discussion Board
wrong answer
Let''s analyze both statements separately to see if they are individually sufficient to determine the winner of the first prize in the 100 m race:
Statement A:
1. Bipin is not from Kolkata and didn''t get the first prize.
2. The sprinter from Chennai is the winner.
3. Danny is from Mumbai.
From statement A, we know that the winner is from Chennai (point 2). However, we don''t have enough information about the other sprinters'' cities to uniquely determine the winner. We know Danny is
from Mumbai, but we don''t know the cities of Chandan and Feroz.
Statement B:
1. Feroz is either from Delhi or from Chennai.
From statement B, we only have information about Feroz''s potential cities. We know he can be from Delhi or Chennai, but we don''t know anything about the other sprinters or the winner.
When we combine both statements:
- The winner is from Chennai (from statement A).
- Feroz is either from Delhi or Chennai (from statement B).
Unfortunately, even when combining both statements, we still don''t have enough information about the cities of the other sprinters to uniquely determine the winner. We know Feroz is not from Chennai
(as the winner is from Chennai), but we don''t know if Feroz is from Delhi or Chandan''s city, and we have no information about Chandan''s city.
Therefore, both statements, when taken individually or together, are not sufficient to give a definite answer to the question of who won the first prize in the 100 m race.
ashok jangid 07-29-2023 02:57 PM
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Singular Point: Regular and Irregular Examples
Singular Point in Differential Equations
A differential equation of the form y′′ + p(x)y′ + q(x)y = 0 has a singular point at x[0] if either of the following limits do not exist [1]:
What this means for second order differential equations is that an initial value problem will not have a unique solution. Alternatively, it may not have any solution, or its solution or derivatives
might be discontinuous. For linear homogeneous differential equations, a singular point happens when at least one coefficient is either undefined (i.e. discontinuous) or multivalued at that point
Regular and Irregular Singular Point
Singular points can be regular or irregular. Regular singular points are well-behaved, defined in terms of ratios of the differential equation’s polynomial coefficients Q(x)/P(x) and R(x)/P(x) [3].
Irregular singular points exhibit bizarre behavior and cannot easily be pinned down or defined, other than to say that if a point isn’t regular, then it is irregular. To put this more concretely, a
regular singular point can be defined as follows: It is where a singularity of P(x) is no worse than 1 / (x – x[0]) and the singularity of Q(x) is no worse than 1 / (x – x[0])^2. In other words, it’s
where both of the following limits exist:
Otherwise, the point is an irregular singular point.
Examples of Regular and Irregular Singular Points
1. Example of a regular single point x[0] [4]:
Here, p[2](x) is singular but xp[0](x) = -1 is analytic* is x[0] = 0 (and for all x).
2. Example of an irregular single point x[0]:
Here, x[0] = -1 is an irregular singular point because (x + 1)p[1](x) is singular at x = -1.
Example question: Are the singular points of (x^3 − 3x^2)y′′ + y′ + 2y = 0 regular or irregular?
Step 1: Find the singular points. As every coefficient is a polynomial, the singular points (0 and 3) are the roots of the leading coefficient, x^3 – 3x^2.
Step 2: Find the limits of each point.
x = 0 is an irregular singular point because the limit is undefined:
x = 0 is a regular singular point because both limits exist:
*Most functions you come across are “analytic.” All polynomial functions, rational functions, exponential functions, logarithmic functions, and trigonometric functions are analytic away from their
Singular Point in Complex Analysis
A singular point, also called a singularity, is a point where a complex function isn’t analytic. In other words, it’s an obstacle to analytic continuation where the function can’t be expressed as an
infinite series of powers of z. Singular points can be classified as regular points or irregular points (also called essential singularities).
A singular point may be an isolated point, or a point on the curve (e.g. a cusp). If there aren’t any other singular points in the neighborhood of z, the point is called an isolated singularity. In
some cases, you might be able to assign a value to the discontinuity to fill in the “gap”. If that’s the case, the point is called a removable singularity.
[1] Binegar, B. Lecture 19: Regular Singular Points and Generalized Power Series. Retrieved August 11, 2021 from: https://math.okstate.edu/people/binegar/4233/4233-l19.pdf
[2] Dobrushkin, V. MATHEMATICA TUTORIAL for the First Course. Part V: Singular and ordinary points. Retrieved August 11, 2021 from: https://www.cfm.brown.edu/people/dobrush/am33/Mathematica/ch5/
[3] 9.2. Classifying Singular Points as Regular or Irregular. Retrieved August 11, 2021 from: https://www.oreilly.com/library/view/differential-equations-workbook/9780470472019/
[4] Bertherton, C. Regular and Irregular Singular Points of ODEs. Retrieved August 11, 2021 from: https://atmos.washington.edu/~breth/classes/AM568/lect/lect15.pdf
Comments? Need to post a correction? Please Contact Us. | {"url":"https://www.statisticshowto.com/singular-point/","timestamp":"2024-11-04T05:03:37Z","content_type":"text/html","content_length":"72261","record_id":"<urn:uuid:c30bb73d-b97d-414e-8356-d9f66d1b2f3d>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00030.warc.gz"} |
Prediction of turbulent jets and plumes in flowing ambients
A differential approach for the analysis of the behavior of turbulent, axisymmetric buoyant jets and plumes discharged into a crossflowing ambient is presented. The thin shear layer form of the
governing partial differential equations for the conservation of mass, momentum and energy, derived in a curvilinear, orthogonal coordinate system is solved numerically by a fully implicit finite
difference technique. A 'lumped' analysis is proposed to simplify the momentum equation in the transverse direction. The turbulent shear stress and heat flux terms appearing in the governing
equations are evaluated through turbulent viscosity and conductivity models which contain parameters related to buoyancy and streamline curvature.
NASA STI/Recon Technical Report N
Pub Date:
August 1978
□ Jet Flow;
□ Plumes;
□ Turbulent Flow;
□ Finite Difference Theory;
□ Numerical Integration;
□ Partial Differential Equations;
□ Three Dimensional Flow;
□ Fluid Mechanics and Heat Transfer | {"url":"https://ui.adsabs.harvard.edu/abs/1978STIN...7915282H/abstract","timestamp":"2024-11-04T15:37:58Z","content_type":"text/html","content_length":"34765","record_id":"<urn:uuid:cf68aaad-1e4e-489b-bac3-6f06884b17b6>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00800.warc.gz"} |
What is the best method to determine an improvement curve slope in contracting? - Answers
How do you determine a slope from a graph?
Slope = (vertical change)/(horizontal change), commonly referred to as rise/run. If the graph is a straight line, then you can count squares or measure how much change in vertical, over a specified
change in horizontal. If it is a curve, then you need to have a tangent line (a line that touches the curve at a specific point and has the same slope as the line), then you can determine the slope
of that line using the method described, above. | {"url":"https://math.answers.com/math-and-arithmetic/What_is_the_best_method_to_determine_an_improvement_curve_slope_in_contracting","timestamp":"2024-11-04T02:34:44Z","content_type":"text/html","content_length":"163713","record_id":"<urn:uuid:ed14c411-9cf6-45ff-9481-7e6aa2997551>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00259.warc.gz"} |
Estimating weight of oven: does water count?
I know how much concrete, concrete block, rebar, brick, mortar, hardibacker (concrete board), and stucco went into each stage of construction (I forgot to note the number of bags of concrete for the
hearth but I can approximately extrapolate it from the foundation which is geometrically similar). This means I can make a reasonable estimate of the weight of the entire structure.
But for a weight-estimate, I'm not sure if I'm supposed to count all the water that went into the concrete, mortar, and stucco. I understand that concrete doesn't "dry out", but rather that it
"sets", but ultimately, what does that mean in terms of water retention or loss -- is the water actually retained forever, or does it work its way entirely out of the cement such that years later,
the total weight is only that of the cement and not of the water that was mixed into it? | {"url":"https://community.fornobravo.com/forum/pizza-oven-design-and-installation/tools-tips-and-techniques/16608-estimating-weight-of-oven-does-water-count","timestamp":"2024-11-03T03:43:00Z","content_type":"application/xhtml+xml","content_length":"105668","record_id":"<urn:uuid:cd4a6d98-d959-48a2-8acc-ca67a7a3a918>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00387.warc.gz"} |
Selecting Procedures for Calculating Derivatives - Knowunity
Selecting Procedures for Calculating Derivatives: AP Calculus AB/BC Study Guide
Welcome to the land of derivatives, where slopes reign supreme and every function has a tale to tell! 🎢 Whether you're dealing with composite, implicit, or inverse functions, mastering the art of
derivatives is like unlocking the secret levels of a video game. This study guide will help you navigate the maze and come out victorious. 🏆
Understanding Derivative Rules
Over the past units, you've collected quite the arsenal of derivative rules like your very own toolbox. These tools range from the Power Rule, which is your basic screwdriver, to the Chain Rule,
which is more like your multi-tool gadget. Let’s break down how to appropriately use these tools to tackle any derivative problem that comes your way.
Remember that knowing when to apply the right rule is key! It’s like knowing when to use the Force in a Star Wars duel. 🌟
Common Derivative Procedures
Quotient Rule: Used when you’re dealing with a function divided by another function. It’s like the Cookie-Cutter of calculus—cuts through the complexities of fractions to give you a neat result.
Product Rule: This one’s for when functions are getting cozy and multiplying with each other. Think of it as the "Friendship Rule"—you're dealing with products of two functions, ensuring each part
gets its proper derivative credit.
Chain Rule: Ah, the Chain Rule—complex yet satisfying, much like solving a Rubik's Cube. Perfect for composite functions, where one function is nested inside another like a calculus Matryoshka doll.
Implicit Differentiation: Used for equations where y hangs out with x, and you need to find dy/dx. Think of it as detective work—solving for the hidden y’s.
Practice Problems and Solutions
Let’s dive into some derivative problems! Just like training at Hogwarts, practice makes perfect.
Question 1: Differentiating [ f(x) = \frac{\cos(x^3)}{5x} ]
Answer: B) Quotient rule, then chain rule
Explanation: The function [ f(x) = \frac{\cos(x^3)}{5x} ] involves a quotient, so we begin with the Quotient Rule. Then, we notice that (\cos(x^3)) is a composite function, requiring the Chain Rule.
It’s like a two-layer cake—tackle each layer with the right tools.
Question 2: Differentiating [ g(x) = 4x \cos(x) \sin(x) ]
Answer: D) Product rule, then product rule again
Explanation: This fancy function [ g(x) = 4x \cos(x) \sin(x) ] is a product of three separate functions. It’s time to roll out the Product Rule twice—each application peels back another layer, like
an onion. 🌰
Question 3: What is the derivative of [ h(x) = (3x^3 - 15x)(2x - x^7) ]?
[ h(x) = (3x^3 - 15x)(2x - x^7) ]
Using the Product Rule:
[ h'(x) = (9x^2 - 15)(2x - x^7) + (3x^3 - 15x)(2 - 7x^6) ]
Simplified, the answer is:
[ h'(x) = -30x^9 + 120x^7 + 24x^3 - 60x ]
Like mixing a perfect potion, each ingredient and step must be followed precisely for the magic to work!
Question 4: What is the derivative of [ f(x) = 6e^{x^3 + 4} ]?
[ f(x) = 6e^{x^3 + 4} ]
First, apply the Chain Rule:
[ f'(x) = 6e^{x^3 + 4} \cdot 3x^2 ]
Which simplifies to:
[ f'(x) = 18e^{x^3 + 4}x^2 ]
It’s a chain reaction! One leads to the next, resulting in a smooth derivative.
Key Terms to Review
1. Quotient Rule: For derivatives of two divided functions, like a divorce lawyer—splitting responsibilities.
2. Product Rule: Handles the derivatives of products, ensuring both functions play nice.
3. Chain Rule: Derivatives for composite functions, the ultimate nesting tool.
4. Implicit Differentiation: When ( y ) is entangled with ( x ); solving for (\frac{dy}{dx}) is like untangling earbuds.
You’ve got the rules, now it’s time to play the game. Knowing which procedure to select when differentiating functions is key to mastering calculus. With practice, these rules will become second
nature, like knowing how to navigate your favorite theme park without a map. 🎢
Now go forth and tackle those derivatives like a calculus ninja! 🥷 You got this! 🍀 | {"url":"https://knowunity.com/subjects/study-guide/selecting-procedures-for-calculating-derivatives","timestamp":"2024-11-13T01:29:54Z","content_type":"text/html","content_length":"269848","record_id":"<urn:uuid:b2e3aae3-89d4-4700-9787-1dfabb55730b>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00165.warc.gz"} |
seminars - Intensive Lecture on truncated Hankel operators
In this lecture we introduce truncated Hankel operators, which has been introduced by Caixing Gu. We focus the basic theory and difference between truncated Hankel operators and general Hankel
operators. Moreover we review the recent developments and open problems in the theory of truncated Hankel operators.
- Basic theory of truncated Hankel operators
- Difference between truncated Hankel operators and general Hankel operators.
- Recent developments on truncated Hankel operators | {"url":"http://www.math.snu.ac.kr/board/index.php?mid=seminars&l=en&sort_index=speaker&order_type=desc&page=52&document_srl=732116","timestamp":"2024-11-06T17:59:48Z","content_type":"text/html","content_length":"46895","record_id":"<urn:uuid:d89327c6-37ff-4b55-9694-6d1e2acbce46>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00162.warc.gz"} |
Jacobian-Based Topology Optimization Method Using an Improved Stiffness Evaluation
A Jacobian-based topology optimization method is recently proposed for compliant parallel mechanisms (CPMs), in which the CPMs' Jacobian matrix and characteristic stiffness are optimized
simultaneously to achieve kinematic and stiffness requirement, respectively. Lately, it is found that the characteristic stiffness fails to ensure a valid topology result in some particular cases. To
solve this problem, an improved stiffness evaluation based on the definition of stiffness is adopted in this paper. This new stiffness evaluation is verified and compared with the characteristic
stiffness by using several design examples. In addition, several typical benchmark problems (e.g., displacement inverter, amplifier, and redirector) are solved by using the Jacobian-based topology
optimization method to show its general applicability.
Compliant mechanisms are elastic structures that can transmit force or motion from input to output. Due to the combined characteristics of mechanism and structure, the analysis and design of
compliant mechanisms are more challenging than that of rigid-body mechanisms. There are two main synthesis approaches for compliant mechanisms, i.e., rigid-body replacement approach [1–3] and
topology optimization approach [4–7].
The rigid-body replacement approach has a wide design scope including all kinds of degree-of-freedom (DOF) compliant mechanisms. This approach synthesizes most of compliant parallel mechanisms
(CPMs), whose synthesis method is the focus of this paper, by replacing the kinematic joints of existing rigid-body mechanisms with flexure hinges. Thus, the compliant mechanisms designed by this
approach rely on the topologies of rigid-body mechanisms. While this approach is successful in designing multi-DOF CPMs for precision applications, it is limited by the fact that a compliant
mechanism may still be unable to fully reproduce the motion of its rigid-body counterpart even using rigorous analysis and optimization techniques [8]. Moreover, this approach cannot select the best
topology for a specific problem currently, which is quite important for developing mechanisms with high performance.
The topology optimization approach regards the synthesis of compliant mechanism as finding out the optimal material distribution within a given design domain, by maximizing the motion or force
transmission between specific input and output ports. Due to this advantage, the topology optimization approach has been successfully applied to the synthesis problems of multiple input and output
ports [5,9–13], multiple materials or physics [9–11], three-dimensional simple compliant mechanisms [14–16], etc. The compliant mechanisms designed by this approach possess structural type of
topology, i.e., no flexure hinges, which is different from the flexure hinge-based mechanisms obtained by the rigid-body replacement approach.
To introduce the idea of topology optimization into the design scope of the rigid-body replacement approach, our previous works [17–19] tried to synthesize the flexure-based compliant mechanisms with
simple motion based on the idea of topology optimization. Lum et al. [20–22] presented a hybrid topological and structural optimization method. This method first synthesizes the compliant joints with
the optimal stiffness characteristics by topology optimization. The resulting compliant joints are then assembled into a CPM based on existing rigid-body mechanism topology.
Recently, we proposed a Jacobian-based topology optimization method [23] for the optimal synthesis of planar CPMs. Traditional topology optimization methods realize multiple outputs by predefining
specific output displacements at output ports. The premise of this realization is that the position and direction of output ports are known. However, the output motion of multi-DOF CPMs is unknown.
The traditional way of predefining specific output ports cannot be applied to multi-DOF CPMs topology optimization directly. To solve this problem, Jacobian matrix [24,25] is introduced into the
field of topology optimization by the proposed method. The Jacobian matrix describes all the freedoms of CPM's mobile platform in a unified and concise form, and contains the information of CPM's DOF
and direct kinematics simultaneously. By optimizing the Jacobian matrix, one can synthesize a compliant mechanism with desired DOF (mechanism's function) and optimized direct kinematics (mechanism's
performance). In addition to the above kinematic realization, the mechanism's input and output characteristic stiffness [26,27] (C-stiffness for short) are also optimized to achieve enough stiffness
to bear the external loads.
Lately, we found that the C-stiffness fails to ensure a valid topology result in some particular problems. In this paper, an improved stiffness evaluation based on its definition is incorporated into
the problem formulation, and will be compared with the C-stiffness formulation by using several design problems.
The rest of this paper is organized as follows: Section 2 describes the problem formulations of Jacobian-based topology optimization method. Section 3 illustrates the topology analysis of CPMs. The
sensitivity analysis and optimization algorithm are described in Sec. 4. Section 5 gives numerical examples to verify the stiffness formulation. The conclusions are presented in Sec. 6.
Jacobian-Based Topology Optimization Method
Problem Description.
The topology of a CPM is determined by the number, arrangement, and topology structures of its constituent compliant limbs. The proposed method regards the problem of CPMs topology optimization as
finding the best topology of compliant limbs within several given design domains.
A CPM composed of n compliant limbs is shown in Fig. 1. The mechanism's compliant limbs are assumed to be synthesized within n predefined design domains, $Γ={Ω1,Ω2,…,Ωn}$. The corresponding design
variables to these design domains are illustrated by $X={X1,X2,…,Xn}$. The number of compliant limbs, n, is determined by the number of CPM's DOF. The position of input points a[i] ($i=1,2,…,n$)
and output point o is defined by designers.
Properties of Jacobian Matrix.
The kinematics of multi-DOF CPMs is much more complicated than that of the compliant mechanisms designed by current topology optimization methods. Since the motion of CPMs is unknown, i.e., the
output trajectory of the mobile platform is not fixed, we cannot use fixed output loads to define its output motion any more. Considering the application situation, the Jacobian matrix, which is
previously used in analysis of rigid-body mechanisms and flexure-based CPMs, is introduced into topology optimization as an alternative kinematic formulation. As shown in Eq.
, the Jacobian matrix
describes transmission relation between the displacements of input and output freedoms, i.e., forward kinematics of the CPM
where $Uo=[ux,uy,θz]T$ contains the three freedoms' displacement of output point o at the mobile platform, and $Ua=[u1,u2,…,un]T$ is the vector of input displacements at input points a[i]. Since the
compliant mechanisms are usually driven by linear motion actuators, we limit the displacements in U[a] to the translational freedoms of input points.
The element J[ji] in J represents the geometry advantage (GA) between the jth freedom of U[o] and the ith input. If all the elements of the jth row in J are equal or close to zero, all the inputs in
U[a] will not produce displacement in the jth freedom of mobile platform, i.e., this freedom is suppressed. While the other freedoms corresponding to nonzero row vectors are considered as the CPM's
DOF. By maximizing the absolute value of the elements in the nonzero row vectors, the forward kinematics and motion transmission performance of CPM can be optimized. Thus, the Jacobian matrix
contains the information of CPM's function and performance simultaneously. By optimizing the Jacobian matrix, we can synthesize a compliant mechanism with desired DOF to realize the mechanism's
function and optimized direct kinematics to achieve higher performance.
Problem Formulation.
The objective function of CPMs topology optimization used in Ref. [23] is developed on the basis of Chen and Wang's formulation [26,27] that utilizes the C-stiffness. This paper tries to modify the
formulation using a new stiffness evaluation that calculates the stiffness based on its definition.
Differences Between C-Stiffness and Stiffness.
The formulation proposed by Chen and Wang is used for the compliant mechanisms with single input and single output. Take the compliant system shown in Fig.
as an example to illustrate the formulation. For this compliant system, the relationship between forces (
) and displacements (
) at input and output ports can be described by a mechanism stiffness matrix
] shown in the following equation:
The diagonal elements of the mechanism stiffness matrix
] are the C-stiffness [
]. As shown in Eq.
, they incorporate the C-stiffness at input and output ports (
) into the formulation to achieve topology optimization of hinge-free compliant mechanisms
$min−e−(GA-GA*)2︸f k11k22︸S$
where GA is the geometry advantage of the mechanism, and GA* is the desired geometry advantage, f and S represent the kinematic and stiffness requirement, respectively.
According to Eq. (2), the physical meanings of k[11] and k[21] are the forces that should be acted on input and output ports, if u[in]=1 and u[out]=0 are expected. The input C-stiffness k[11]
just describes part of the stiffness relationship between f[in] and u[in], and so does the output C-stiffness k[22].
On the contrary, the stiffness based on its definition can fully describe the force and displacement relationship at one specific freedom. To obtain the input and output stiffness of this simple
compliant system, the input and output compliance are first calculated according to the physical meaning of compliance. Let
=1 and
=0, and solve Eq.
. The resulting input displacement
is the input compliance
=0 and
=1, and solve Eq.
. The resulting output displacement
is the output compliance
The inverse of input and output compliance are the input and output stiffness, respectively
By comparing k[11] and k[22] with k[in] and k[out], respectively, C-stiffness is part of the stiffness. While the C-stiffness has successfully evaluated the mechanism's stiffness property in many
design problems, some problems show that the stiffness is more reliable than C-stiffness. Thus, this paper will modify the formulations of Jacobian-based topology optimization method by replacing the
C-stiffness with stiffness.
Formulations Using New Stiffness Evaluation.
As can be seen in Eq.
, the kinematic requirement
and stiffness requirement
are two conflicting subobjectives. On one hand, the mechanism should be soft enough to deform and deliver motion. On the other hand, it should be stiff enough to transmit forces to the mobile
platform and bear the external force. A general problem formulation for CPMs topology optimization can be written as follows:
$min ζ(X)=−fωS(1−ω)s. t. V(X)≤Vo$
where ω (0<ω<1) is the weight indicating the relative significance of kinematic requirement f. V(X) is the volume fraction of topology candidate, and V[o] is the allowed volume fraction.
The kinematic requirement
has two different forms according to the design problem. The first form of kinematic requirement
is suitable for the CPMs whose kinematics is simple enough to be predefined by designers. This form is to force the Jacobian matrix
of CPM to be close to a desired Jacobian matrix
by minimizing the differences between
, which are the elements of
, respectively. As a result, the desired DOF and kinematic properties of CPM can be expressed in the desired Jacobian matrix
$max f1=e−∑(Jji−Jji*)2j=1,2,…,3 i=1,2,…,n$
The second form of kinematic requirement
is suitable for the CPMs with complex kinematics. This form tries to maximize the motion in desired freedoms
and suppress the rest
evaluates the workspace in the
th freedom, which is the quadratic sum of the elements in corresponding row vector shown in Eq.
is the workspace of a desired freedom, while
is the workspace of a constrained freedom. The natural exponential function forces each
to be close to zero
In our previous work [
], the stiffness requirement
is achieved by maximizing the input and output C-stiffness of the CPM. This paper uses the stiffness calculated by its definition as a new stiffness evaluation instead of the C-stiffness.
Mathematically, the stiffness requirement is formulated as the product of the input stiffness
and output stiffness
$max S=∏i=1nkinikouti$
The calculation of Jacobian matrix J, input and output stiffness will be illustrated in Secs. 3.3 and 3.4, respectively.
Unification of the Units in Rotational and Translational Freedoms.
Since the units in rotational and translational freedoms are different, it is unfair to compare the parameters in rotational and translational freedoms directly during the optimization. Thus, the
characteristic length [
is introduced in this paper. Based on the characteristic length
, we can define equivalent moment
and equivalent rotational displacement
as follows:
For the Jacobian matrix, each element
is the ratio of the
th freedom of
and the
th input. Since all the inputs are translational freedoms, only the rotational displacement
should be transformed into equivalent rotation displacement
. Thus, the elements in the third row vector of
are multiplied by
to obtain equivalent Jacobian matrix
$J̃3i=J3ilc, i=1,2,…,n$
For the input and output stiffness, the output stiffness related to rotational freedom should be transformed into equivalent rotational stiffness. The relationship among the applied moment
, rotational stiffness
, and rotational displacement
is as follows:
Solve the moment
and rotational displacement
in Eq.
, and substitute them into Eq.
The equivalent rotational stiffness
is calculated as follows:
Topology Analysis
This section shows how the Jacobian matrix, and input and output stiffnesses of CPMs can be obtained by using the finite element analysis and matrix methods [31].
Discretization and Parameterization.
One advantage of using multiple design domains is that the compliant limbs of CPM can be discretized, parameterized, and analyzed separately in their local coordinates. A compliant limb in its local
coordinate is shown in Fig.
, and consists of design domain Ω
and part of the mobile platform
. The compliant limb (
) is discretized by using the quadrilateral elements that possess three freedoms (
) at each node, and parameterized by the simplified isotropic material with penalization scheme [
]. The stiffness matrix of the
th compliant limb can be obtained by the following equation:
$Ki(Xi)=∑e=1Ni(xe)ρKe+∑e=1NpiKe0<xmine≤xe≤1, xe∈Xi$
where N[i] is the number of the elements in design domain Ω[i], $Npi$ is the number of the elements in $Ωpi$, K[e] is the element stiffness matrix in the global level, x^e is the material density
(design variable) of each element in Ω[i] with value between the lower limit $xmine$ (void) and 1 (solid), ρ is the penalty factor, the elements in $Ωpi$ are solid.
Stiffness Modeling of CPMs.
For a compliant limb in its local coordinate (as shown in Fig.
), only the input freedom at point
and the three freedoms of endpoint
are considered in its stiffness modeling. First, a compliance matrix
that characterizes the compliance relationship between these four concerned freedoms is calculated according to physical meaning of compliance. Four load cases
$Fj (j=1,2,…,4)$
, in which a unit dummy load is applied to each concerned freedom of
in sequence, are used to calculate corresponding displacements by solving the following equation:
The physical meaning of element
in a compliance matrix
is the displacement of the
th freedom due to a unit load that only acts on the
th freedom. According to this physical meaning, the element of
is obtained by Eq.
. The displacements of the four concerned freedoms in displacement vector
form the
th column in
$Caoi(j,k)=FjTUk, j,k=1,2,…,4$
Then, the compliance of endpoint
is transferred into the coordinate of the output point
by a transformation matrix
, whereas the compliance of the input freedom at
remains in its local coordinate. The inverse of the resulting compliance matrix, i.e., stiffness matrix
, is shown in Eq.
. For more information about the transformation matrix
, please refer to Refs. [
where 1 is to keep the compliance of the input freedom at
remain in its local coordinate, and the notation ⊗ is defined as follows:
Finally, the stiffness model of compliant limbs will be combined into the stiffness model of CPM. The stiffness related to the output point
of all the compliant limbs are superimposed to form stiffness of the output point
at mobile platform, while the stiffness of all the input freedoms at points
) remain in their local coordinates. The transformation is shown in Eq.
. The resulting
is the mechanism stiffness matrix that characterizes the stiffness relationship between the
input freedoms and the three freedoms of output point
Kinematic Analysis of CPMs.
According to the input and output freedoms of the CPM, the mechanism's displacements are partitioned into two sets
, respectively, for displacements of the input and output freedoms. As shown in Eq.
, the mechanism loads are also partitioned into two sets as
, accordingly. This would in effect partition the mechanism stiffness matrix
into the following form:
Assuming that there is no external load applied to the mobile platform of CPMs, i.e.,
by solving the second equation in Eq.
, the relationship between the input displacement
and output displacement
, i.e., the Jacobian matrix
, can be obtained
Input and Output Stiffness of CPMs.
By using the mechanism stiffness matrix
, we can calculate the input and output stiffness of CPMs according to the physical meaning of compliance. As shown in Eq.
+3 unit load vectors (
) are applied to the freedoms of
in sequence to obtain corresponding displacement vectors
$Fini=KmUini, i=1,2,…,nFoutj=KmUoutj, j=1,2,3$
The displacement of the freedom where the unit load is applied is the compliance in this freedom, which can be extracted from the corresponding displacement vector by using the related unit load
vector. The inverse of these compliances are the stiffness in the input and output freedoms of this CPM
$kini=((Fini)TUini)−1, i=1,2,…,nkoutj=((Foutj)TUoutj)−1, j=1,2,3$
Sensitivity Analysis
The sensitivity of the objective function discussed in Sec.
is determined by the sensitivities of
, and
. The sensitivity of
can be extracted from sensitivity of
which is calculated as follows:
are parts of
, the sensitivity of
is determined by the sensitivity of
. According to Eq.
, the sensitivity of
is calculated by Eq.
, while the sensitivity of
can be obtained by the same way
where the sensitivity of
is as follows:
Substitute Eq.
into Eq.
, the sensitivity of
can be turned into the following form:
According to Eqs.
, the sensitivity of
is determined by the sensitivities of
where the sensitivity of the element in
can be calculated as follows:
On the basis of sensitivity analysis, the topology optimization problem is solved by modifying the 99line matlab code proposed by Sigmund [34]. The optimality criteria-based optimizer and filtering
technique [7] of the 99line Matlab code are used to update the design variables of each domain and ensure existence of solutions, respectively. For each numerical example in Sec. 5, the initial
design X is defined by setting the material density of each element to be the value of the allowed volume fraction, i.e., x^e=V[o]. The convergence criterion is the change in design variables,
which is set to 0.005 in this paper. The move limit in the heuristic updating scheme is 0.1. The filter radius r[min] is set to 1.2, i.e., the filter length scale d[min] (d[min]=2r[min]) is 2.4. It
should be pointed out that the volume constraint is active during the whole optimization process. For more detail about the optimality criteria-based optimizer and filtering technique, readers can
refer to Ref. [34].
Numerical Studies
This section will compare several topology optimization results obtained by using the C-stiffness and stiffness formulations, respectively. The artificial material properties for these examples are
described as: Young's modulus is E=1GPa and Poisson's ratio is υ=0.3. The characteristic length l[c] is set to 10mm in this study. All the numerical examples are carried out on a computer with
Intel Core i7 – 6700 (3.40 GHz) CPU, 8.00GB RAM, and MatlabR2009a. Note that the filter length scale d[min]=2.4 is represented by a red bar in each figure showing the final topology.
Design of 2DOF CPMs.
In this section, a 2DOF CPM will be synthesized using two asymmetrical compliant limbs within the design domain shown in Fig. 4. As can be seen, the two compliant limbs have the same size and
boundary conditions. Each compliant limb is discretized by 100×100 finite elements for elastic analysis. The allowable amount of material is 20%.
Solved by Using f[1].
Since the kinematics of 2DOF CPMs is simple enough to predefine its desired Jacobian matrix
, this design problem can be solved using the first form of objective function in Eq.
. The
for this example is given by Eq.
, in which
are the input freedoms. The zero vector in the third row of the
means that the freedom
should be suppressed, i.e., the CPM is expected to have only two translational freedoms (
). Moreover, the two translational freedoms are expected to be decoupled. For example, the input
only induces the translational displacement
with desired geometry advantage GA
and has no impact on the other two freedoms
The GA^* is first set to −3 and ω is set to 0.5. The corresponding topology optimization problem is solved by using the C-stiffness and stiffness formulations, respectively. The optimizations were
run for 200 iterations. The resulting topologies of the two formulations are shown in Fig. 5, which shows that both of C-stiffness and stiffness formulations can obtain valid topologies in this case.
Figures 6 and 7 show the iteration history of objective value, kinematic requirement f, and stiffness requirement S in the optimization process of the two formulations, respectively. It can be seen
that oscillations exist in the iteration curves. The oscillations may be caused by the material distribution at some specific elements. Fortunately, the topologies at later period of iteration are
stable, which can be regarded as the optimal topology. The corresponding J of the two final topologies are listed in the first two rows of Table 1. It should be noted that only the elements in the
first column vector of J are displayed for brevity, since J[11] ≃ J[22], J[21] ≃ J[12], and J[31] ≃ J[32]. One can see that both of C-stiffness and stiffness formulations can force J of CPMs to be
close to J^*, i.e., the kinematics requirement f is realized. In addition, the C-stiffness (Ck[a][1] and Ck[ox]) and stiffness (k[a][1] and k[ox]) of the input and output ports in x-axis of the two
final topologies are given in the first two rows of Table 2. The results show that both of C-stiffness and stiffness formulations realize the stiffness requirement S effectively. The resulting output
stiffness is smaller than the input stiffness to achieve GA, whereas the value of C-stiffness is larger than stiffness for the same mechanism.
Table 1
Formulation GA^* J[11] J[21] J[31]
C-stiffness −3 −2.52 8×10^−4 0.02
Stiffness −3 −2.26 1.1×10^−3 0.02
C-stiffness 2.5 1 0 −3.5×10^−3
Stiffness 2.5 2.27 −4×10^−4 −0.02
Formulation GA^* J[11] J[21] J[31]
C-stiffness −3 −2.52 8×10^−4 0.02
Stiffness −3 −2.26 1.1×10^−3 0.02
C-stiffness 2.5 1 0 −3.5×10^−3
Stiffness 2.5 2.27 −4×10^−4 −0.02
Table 2
Formulation GA^* Ck[a1] Ck[ox] k[a1] k[ox]
C-stiffness −3 38.7 4.3 11.3 1.3
Stiffness −3 51.4 3.1 35.8 2.1
C-stiffness 2.5 101.4 101.4 2×10^−6 2×10^−6
Stiffness 2.5 77.5 6.9 42.1 3.7
Formulation GA^* Ck[a1] Ck[ox] k[a1] k[ox]
C-stiffness −3 38.7 4.3 11.3 1.3
Stiffness −3 51.4 3.1 35.8 2.1
C-stiffness 2.5 101.4 101.4 2×10^−6 2×10^−6
Stiffness 2.5 77.5 6.9 42.1 3.7
However, when GA^* is set to be positive, e.g., GA^*=2.5, it is found that the C-stiffness formulation fails to ensure the stiffness requirement S and results in invalid topologies. Figure 8(a)
shows the final topology obtained by using the C-stiffness in the case of GA^*=2.5. Although the input and output points of the CPM are connected by solid material successfully, there is no
material connection between the compliant limbs and fixed ports. Consequently, the displacements of the input and output ports in one direction are equal, i.e., J[11]=1 (in the third row of Table 1
). As shown in the third row of Table 2, the C-stiffness of the input and output ports in x-axis is 101.4 N/mm, whereas the corresponding stiffness k[a][1] and k[ox] of this invalid topology are
approximate to zero. Obviously, the C-stiffness fails to evaluate the stiffness of mechanism in this case.
Then, this problem is solved by using the stiffness formulation. The final topology is shown in Fig. 8(b), where ω is set to 0.7. One can see that there are valid material connections between the
input, output and fixed ports of the CPM. As shown in the last rows of Tables 1 and 2, the resulting J is close to J^*, e.g., J[11]=2.27. The values of its C-stiffness and stiffness are reasonable.
Thus, the problem of C-stiffness is avoided and a valid final topology can be ensured by the stiffness formulation.
Solved by Using f[2].
To compare the two forms of objective function, the 2DOF CPM design problem is also solved by using
shown in Eq.
. According to this design problem, the kinematic requirement
is to maximize the workspace of the two translational freedoms and minimize the workspace of the rotational freedom, which is formulated as follows:
$max f2=e−J3c∏j=12Jjd$
is combined with the stiffness requirement
based on C-stiffness, the resulting final topology is similar to the topology in Fig.
. The final topology obtained by using
and stiffness based
is shown in Fig.
is set to 0.7. The corresponding Jacobian matrix is described by Eq.
. One can see that the workspace of the two translational freedoms (
) is much larger than that of rotational freedom
. The resulting values of
are positive, i.e.,
cannot control the sign of
Design of 3DOF CPMs.
The second design problem of CPMs is to synthesize a 3DOF CPM with three symmetrically arranged compliant limbs. The positions of the CPM's input, output, and fixed points are shown in Fig. 10. Each
design domain Ω[i] is discretized by 50×80 finite elements for elastic analysis under the same boundary condition. The allowable amount of material is 20%. Since the kinematics of 3DOF CPM is
complex, it is hard to predefine a desired Jacobian matrix for the optimization. The second form of kinematic requirement f[2] shown in Eq. (9) will be used in the objective function. For the planar
3DOF CPMs, no freedom should be suppressed, i.e., the design objective is to maximize the workspace of these three freedoms.
The problem is solved by using the C-stiffness and stiffness formulations, respectively. The optimization process was run for 200 iterations. Figure
gives the final topology of 3DOF CPM obtained by using the C-stiffness formulation and setting
=0.7. Its corresponding Jacobian matrix
is shown in Eq.
. While the final topology obtained by using the stiffness formulation (
=0.9) and its resulting Jacobian matrix
is shown in Fig.
and Eq.
, respectively. One can see that both of the C-stiffness and stiffness formulations are able to achieve valid topology in this example, whereas the topology obtained by the stiffness formulation is
Solving Benchmark Problems.
Although the Jacobian-based topology optimization method is developed for the CPMs, this method is applicable to the typical compliant mechanisms designed by the current topology optimization
methods, e.g., displacement inverter, amplifier, and redirector.
Displacement Inverter and Amplifier.
A design problem of 1DOF compliant mechanisms using single design domain is shown in Fig.
. The top left corner and the bottom left corner of the design domain are fixed. The input point
and output point
are in the middle of the left and right sides, respectively. The whole design domain is discretized using 100×100 finite elements for elastic analysis. The material usage is restricted to 20%. The
Jacobian matrix of the 1DOF compliant mechanism is a 3×1 vector, whose desired form is shown in Eq.
. One can see that the desired output motion of this mechanism is in the direction of
-axis, whereas the freedoms
should be suppressed. Obviously, when only the element
of Jacobian matrix is considered, our objective function is equal to the formulation proposed by Chen and Wang (Eq.
When GA^* is negative, e.g., GA^*=−3, the design problem is to synthesize a displacement inverter. The corresponding topology optimization problem is also solved by using the C-stiffness and
stiffness formulations, respectively. Figure 14(a) shows the resulting topology obtained by using the C-stiffness formulation and setting the weight as ω=0.5. Its corresponding Jacobian matrix is J
=[−2.7, 0, 0]^T, i.e., the optimized GA of displacement inverter is −2.7. The C-stiffness at the input and output ports are 40.9 and 4.7 N/mm, while the stiffness at input and output ports are 6.4
and 0.7 N/mm, respectively. Figure 14(b) shows the resulting topology obtained by using the stiffness formulation and setting ω to 0.5. Its corresponding Jacobian matrix is J=[−2.8, 0, 0]^T. The
C-stiffness at the input and output ports are 57.2 and 2.7 N/mm, while the stiffness at input and output ports are 41.3 and 2.0 N/mm, respectively. One can see that both of the C-stiffness and
stiffness formulations can obtain valid topologies in this case.
When GA^* is positive, e.g., GA^*=3, the design problem is to synthesize a displacement amplifier. It is found that the C-stiffness formulation results in the invalid topology shown in Fig. 15(a).
Without material connected to the fixed ports, the Jacobian matrix of the resulting displacement amplifier is J=[1, 0, 0]^T. Both of the input and output C-stiffness are 116.1 N/mm, whereas both of
the corresponding input and output stiffness are 4.3×10^−7 N/mm. Figure 15(b) shows the resulting topology obtained by using the stiffness formulation and setting ω to 0.6. Its corresponding
Jacobian matrix is J=[2.3, 0, 0]^T. The C-stiffness at the input and output ports are 79.5 and 6.2 N/mm, while the stiffness at input and output ports are 47.8 and 3.8 N/mm, respectively. This case
shows again that the C-stiffness formulation fails to ensure the stiffness requirement.
Solved by Using the Artificial I/O Spring and MSE/SE Formulations.
There are several popular formulations developed for topology optimization of compliant mechanisms. Deepak et al. [
] have made a comparative study of these formulations. The popular artificial I/O spring and mutual strain energy/strain energy (MSE/SE) formulations are adopted here to solve the problem of
displacement amplifier. In the artificial I/O spring formulation, two artificial springs are added to the input and output ports, respectively. The problem is formulated as follow:
$max uouts. t. V(X)≤Vo$
For the problem of displacement amplifier discussed above, it is found that the artificial I/O spring formulation also results in the invalid topology shown in Fig.
, although the stiffness values of the artificial springs have been varied from 10
to 10
N/mm by ten times. On the contrary, the MSE/SE formulation (Eq.
) ensures the valid topology result shown in Fig.
$min −MSE+SEs. t. V(X)≤Vo$
The computational expense of the proposed method is compared with the artificial I/O spring formulation by solving the design problem of displacement inverter in Sec. 5.3.1. It takes the artificial I
/O spring formulation 47.59s and 220 iterations to find out the solution, i.e., 216ms per iteration. The C-stiffness formulation spends 181.5s and 300 iterations to obtain the topology in Fig. 14
(a), i.e., 605ms per iteration. The stiffness formulation spends 209.37s and 300 iterations to obtain the topology in Fig. 14(b), i.e., 698ms per iteration. Obviously, the proposed method is more
expensive than the artificial I/O spring formulation in computation. One reason for this is that the finite element with 12 nodal freedoms increases the computational expense of finite element
analysis. The other reason is that all the freedoms of the output point o are considered by the proposed method, whereas the spring formulation only concerns the freedom u[x].
Displacement Redirector.
This example illustrates the application in the case of single input and two outputs compliant mechanisms. The function of a displacement redirector is sketched in Fig.
. The input port
is at the middle of the left side and causes two output displacement at ports
, respectively. The whole design domain is descretized using 100×100 finite elements for elastic analysis. The material usage is restricted to 20%. In this case, only the freedoms of the output
are considered, so the Jacobian matrix
of the displacement redirector is a 2×1 vector, whose desired form is shown in the following equation:
The corresponding topology optimization problem is solved by using the C-stiffness and stiffness formulations, respectively. The optimization process was run for 300 iterations. Figure 18(a) shows
the resulting topology obtained by using the C-stiffness formulation and setting ω to 0.5. Its corresponding Jacobian matrix is J=[1.7, −1.7]^T. The C-stiffness at the input and output ports are
41.2, 5.9, and 5.9 N/mm, while the stiffness at input and output ports are 10.7, 2.3, and 2.3 N/mm, respectively. Figure 18(b) shows the resulting topology obtained by using the stiffness formulation
and setting ω to 0.5. Its corresponding Jacobian matrix is J=[1.5, −1.5]^T. The C-stiffness at the input and output ports are 58.1, 4.1, and 4.1 N/mm, while the stiffness at input and output ports
are 40.1, 3.3, and 3.3 N/mm, respectively. This case shows that the proposed method is applicable to the compliant mechanisms with multiple output ports, and both of the C-stiffness and stiffness
formulations can obtain valid topologies.
Analysis of Mesh Independency.
This section is devoted to analyzing the mesh independency of the proposed method. The problem of 2DOF CPM, whose parameter settings except discretization are the same as the case in Fig. 5, is used
to illustrate the mesh independency. By using three different element discretizations of 40×40, 200×200 and 300×300, we obtain the corresponding results shown in Fig. 19. The three topologies
on the left side of Fig. 19 are solved by the C-stiffness formulation, whereas the other side contains the results of the stiffness formulation. One can see that the results are almost stable under
mesh refinement or mesh coarsening. In other words, the proposed method is mesh independent, which is ensured by the filtering technique of the 99line Matlab code.
This paper presents a new stiffness evaluation based on the definition of stiffness for the Jacobian-based topology optimization method. The proposed stiffness formulation is compared with the
C-stiffness formulation by using two synthesis problems of CPMs and three traditional benchmark design problems. The results show that both of the two formulations can achieve valid topologies in
most of design cases. In some cases like displacement amplifier, the C-stiffness formulation even the artificial I/O spring formulation cannot obtain valid result, while the stiffness formulation
gives an improved stiffness evaluation. Besides, the Jacobian-based topology optimization method shows a general applicability in multi-DOF CPMs and benchmark design problems.
According to the results, the topologies produced by the proposed method are easy to exhibit hinges, especially when the kinematic requirement is much higher than stiffness requirement. Relatively
speaking, the stiffness formulation has a better performance than the C-stiffness formulation in avoiding hinges, e.g., the two topologies in Fig. 18. The strategies for alleviating these hinges will
be addressed in our future works.
Funding Data
• National Natural Science Foundation of China (Grant Nos. 51275174, 51605166, U1609206, and 51675189).
• Natural Science Foundation of Guangdong Province (Grant No. 2014A030313460).
• Fundamental Research Funds for the Central Universities.
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100X100 Sudoku Printable Printable Template Free | Sudoku Printables
100X100 Sudoku Printable Printable Template Free
100X100 Sudoku Printable Printable Template Free – If you’ve ever had trouble with sudoku, you know that there are numerous types of puzzles available and it can be difficult to choose which ones to
work on. There are many options to solve them. In fact, you’ll discover that solving a printable version can be an excellent method to get started. The guidelines for solving sudoku are the same as
those for other puzzles however the way they are presented differs slightly.
What Does the Word ‘Sudoku’ Mean?
The word ‘Sudoku’ is derived from the Japanese words suji and dokushin, which mean “number” and “unmarried person’, respectively. The objective of the puzzle is to fill every box with numbers, so
that each number between one and nine appears only once on every horizontal line. The word Sudoku is a trademark belonging to the Japanese puzzle maker Nikoli, which originated in Kyoto.
The name Sudoku originates by the Japanese word”shuji wa Dokushin Ni Kagiru, which means ‘numbers must not be separated’. The game is composed of nine 3×3 squares that have nine smaller squares
within. Originally called Number Place, Sudoku was an exercise that stimulated mathematical development. Although the origins of the game aren’t known, Sudoku is known to have roots that go back to
the earliest number puzzles.
Why is Sudoku So Addicting?
If you’ve played Sudoku and you’ve played it before, you’ll be aware of how addictive this game can be. The Sudoku addicted person will be unable to not think about the next puzzle they can solve.
They’re always thinking about their next puzzle, while various aspects of their lives seem to fall by the sidelines. Sudoku is a game that can be addictive however it’s essential that you keep the
addictive nature of the game in check. If you’ve developed a craving for Sudoku Here are some ways to curb your addiction.
One of the best ways to detect that the addict you are to Sudoku is to watch your behaviour. Many people carry magazines and books with them as well as scroll through social media posts. Sudoku
addicts, however, carry books, newspapers, exercise books, and phones wherever they go. They can be found for hours working on puzzles and aren’t able to stop! Some people even discover it is easier
to complete Sudoku puzzles than their regular crosswords. They simply can’t quit.
100×100 Sudoku Printable
What is the Key to Solving a Sudoku Puzzle?
The best way to solve an printable sudoku problem is to practice and experiment using different methods. The best Sudoku puzzle solvers do not follow the same formula for each puzzle. The most
important thing is to practice and experiment with various approaches until you can find one that works for you. After some time, you’ll be able to solve sudoku puzzles without a problem! However,
how do you learn to solve an printable sudoku problem?
To begin, you need to grasp the basics of suduko. It’s a game of logic and deduction that requires you to view the puzzle from many different perspectives to find patterns and then solve it. When you
are solving a suduko puzzle, do not try to figure out the numbers; instead, you should search the grid for clues to recognize patterns. It is also possible to apply this method to rows and squares.
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Coding drops quantum computing error rate
Errors in quantum computing have limited the potential of the emerging technology. Now, however, researchers at Australia’s University of Sydney have demonstrated a new code to catch these bugs.
The promised power of quantum computing lies in the fundamental nature of quantum systems that exist as a mix, or superposition, of all possible states.
A traditional computer processes a series of “bits” that can be either 1 or 0 (or, on or off). The quantum equivalent, called a “qubit”, can exist as both 1 and 0 simultaneously, and can be “solved”
One outcome of this is an exponential growth in computing power. A traditional computer central processing unit is built on 64-bit architecture. The equivalent-size quantum unit would be capable of
representing 18 million trillion states, or calculations, all at the same time.
The challenge with realising the exponential growth in qubit-powered computing is that the quantum states are fragile and prone to collapsing or producing errors when exposed to the electrical
‘noise’ from the world around them. If these bugs could be caught by software it would make the underlying hardware much more useful for calculations.
“This is really the first time that the promised benefit for quantum logic gates from theory has been realised in an actual quantum machine,” says Robin Harper, lead author of a new paper published
in the journal, Physical Review Letters.
Harper and his colleague Steven Flammia implemented their code on one of tech giant IBM’s quantum computers, made available through the corporation’s IBM Q initiative. The result was a reduction in
the error rate by an order of magnitude.
The test was performed on logic gates, the building blocks of any quantum computer, and the equivalent of classical logic gates.
“Current devices tend to be too small, with limited interconnectivity between qubits and are too ‘noisy’ to allow meaningful computations,” Harper says.
“However, they are sufficient to act as test beds for proof of principle concepts, such as detecting and potentially correcting errors using quantum codes.”
Everyday devices have electronics which can operate for decades without error, but a quantum system can experience an error just fractions of a second after booting up.
Improving that length of time is a critical step in the quest to scale up from simple logic gates to larger computing systems.
The team’s code was able to drop error rates on IBM’s systems from 5.8% to 0.60%.
“One way to look at this is through the concept of entropy,” explains Flammia.
“All systems tend to disorder. In conventional computers, systems are refreshed easily, effectively dumping the entropy out of the system, allowing ordered computation.
“In quantum systems, effective reset methods to combat entropy are much harder to engineer. The codes we use are one way to dump this entropy from the system.”
Companies such as IBM, Google, Rigetti and IonQ have started or are about to start allowing researchers to test their theoretical approaches on these small, noisy machines.
“These experiments are the first confirmation that the theoretical ability to detect errors in the operation of logical gates using codes is advantageous in present-day devices, a significant step
towards the goal of building large-scale quantum computers,” Harper says.
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Gadgets, approximation, and linear programming
Trevisan, Luca, Sorkin, Gregory B. ORCID: 0000-0003-4935-7820, Sudan, Madhu and Williamson, David P. (1996) Gadgets, approximation, and linear programming. In: 37th Annual IEEE Symposium on
Foundations of Computer Science, 1996-10-14 - 1996-10-16, VT, United States, USA.
Full text not available from this repository.
The authors present a linear-programming based method for finding “gadgets”, i.e., combinatorial structures reducing constraints of one optimization problem to constraints of another. A key step in
this method is a simple observation which limits the search space to a finite one. Using this new method they present a number of new, computer-constructed gadgets for several different reductions.
This method also answers the question of how to prove the optimality of gadgets-they show how LP duality gives such proofs. The new gadgets improve hardness results for MAX CUT and MAX DICUT, showing
that approximating these problems to within factors of 60/61 and 44/45 respectively is NP-hard (improving upon the previous hardness of 71/72 for both problems). They also use the gadgets to obtain
an improved approximation algorithm for MAX 3SAT which guarantees an approximation ratio of 0.801, This improves upon the previous best bound of 0.7704.
Actions (login required) | {"url":"http://eprints.lse.ac.uk/35886/","timestamp":"2024-11-05T23:49:11Z","content_type":"application/xhtml+xml","content_length":"21858","record_id":"<urn:uuid:bfdd77e9-d247-4ac3-ad7c-c6557feaf51a>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00880.warc.gz"} |
How to Solve Multi-Step Equations with Examples and Answers (Arithmetic) - Knowunity
Multi-Step Equations: A Comprehensive Guide to Solving Complex Algebraic Problems
This guide provides a detailed exploration of multi-step equations, covering various types of solutions, step-by-step solving processes, and practical examples. It's an essential resource for
students learning to tackle more complex algebraic problems.
• Covers three types of solutions: one solution, null set, and identity
• Provides detailed examples of solving multi-step equations
• Includes practice problems with answers for self-assessment | {"url":"https://knowunity.com/knows/arithmetic-multistep-equations-ea1979d1-feab-47d6-92db-d57848319f40?utm_content=taxonomy","timestamp":"2024-11-03T19:31:41Z","content_type":"text/html","content_length":"391198","record_id":"<urn:uuid:9c90607d-1a10-4547-8fcb-4874960be607>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00411.warc.gz"} |
The six circles theorem revisited
The six circles theorem of C. Evelyn, G. Money-Coutts, and J. Tyrrell concerns chains of circles inscribed into a triangle: the first circle is inscribed in the first angle, the second circle is
inscribed in the second angle and tangent to the first circle, the third circle is inscribed in the third angle and tangent to the second circle, and so on, cyclically. The theorem asserts that if
all the circles touch the sides of the triangle, and not their extensions, then the chain is 6-periodic. We show that, in general, the chain is eventually 6-periodic but may have an arbitrarily long
All Science Journal Classification (ASJC) codes
Dive into the research topics of 'The six circles theorem revisited'. Together they form a unique fingerprint. | {"url":"https://pure.psu.edu/en/publications/the-six-circles-theorem-revisited","timestamp":"2024-11-12T03:15:16Z","content_type":"text/html","content_length":"45710","record_id":"<urn:uuid:fc5a2f09-38c2-4ff1-bd40-6538253dba60>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00552.warc.gz"} |
Elementary Math Formulas, Solutions and Explanations
Although formulas are an important component of mathematics study in middle and high school, they're typically not a big part of elementary math. However, your child likely will be introduced to a
handful of formulas - including those for perimeter, area and volume - by the time he or she completes fifth grade.
What Elementary Math Formulas Does My Child Need To Know?
In third and fourth grades, students learn to calculate the perimeter and area of 2-dimensional shapes, including squares, rectangles and triangles, and in fifth grade, they move on to determining
the volume of cubes and rectangular prisms. In some cases, students may be required to memorize the formulas for these calculations; in others, the teacher may provide students with these formulas
for assignments and tests. As you review the sample problems below, keep in mind that many formulas use notations like 'l' for length, 'w' for width, 'h' for height and 'b' for base.
Formulas and Sample Problems
The perimeter (P) of a shape can be calculated by adding together the lengths of all its sides. Although the formula for perimeter differs slightly for some shapes, the concept is always the same.
For instance, the formula for the perimeter of a rectangle is P = 2l + 2w, while the formula for the perimeter of a triangle is P = a + b + c. Have your child put these formulas to practice using the
problems below.
1. The sides of a square are 15 cm long. What is its perimeter?
Because all sides of a square are the same length, the perimeter can be calculated by adding the length of one side four times or by multiplying the length of the side by four. The equation
should look like this: P = 15 + 15 + 15 + 15 = 60 or P = 4(15) = 60. The answer is 60 cm.
2. The base of a triangle is 5 inches, and its two sides are 4 inches long. Find the perimeter.
The perimeter here is calculated by adding the sides together: P = 5 + 4 + 4 = 13 inches.
The formula for calculating the area (A) of rectangles and parallelograms is A = lw, also written as A = bh. The area for a triangle can be found by calculating A = ½bh. Remind your child to label
the answers using square units.
1.Find the area of a rectangle that's two inches long and five inches wide.
For this problem, A = 2 x 5 = 10 square inches.
2. A triangle has a base of four feet and a height of ten feet. Find the area.
The formula should look like this: A = ½(4)(10). Your child should begin by multiplying four by ten, which equals 40. Then, he or she should take half of 40, which is 20. The area of this
triangle is 20 square feet.
The formula used to calculate the volume of rectangular prisms is V = lwh. Another way to think of this formula is multiplying the height of the shape by the area of its base. Remember that all
answers relating to volume should be labeled using cubic units.
1. A cube has a height of ten centimeters. What is its volume?
Because all the sides of a cube area the same, the length and width of this cube are also ten centimeters. Using the formula, plug in the numbers like this: V = 10 x 10 x 10. The answer is 1,000
cubic centimeters.
2. The area of a rectangular prism's base is 42 square feet. Its height is three feet. What is its volume?
Remember that another way to find volume is multiplying the area of the base by the height. In this case, all we need to do is multiply 42 (the area of the base) by the height: 42 x 3 = 126. The
answer is 126 cubic feet.
Other Articles You May Be Interested In
10 Top Math Apps for Elementary School Children
Is your child struggling with math in elementary school? You may be able to get him or her excited about learning math by using an educational app on your iPad, iPhone, Android or other mobile
device. Here are ten apps for elementary school children that may hold the key to math success.
Elementary School Math Test Anxiety
Too much anxiety about tests can be debilitating and result in poor performance. It can interfere with studying, cause difficulty with learning new material and even have a negative effective on
a child's social interactions
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Where to Aim on a Dart Board
I’ve used much of lockdown as an opportunity to teach myself Python, the computer coding language. I started lockdown as an absolute beginner and am now a fan. I’ve found that giving myself projects
and problems to solve has been the most effective way of learning, so I was pleased to be challenged with a new project by some colleagues recently.
A conversation with Rob Eastaway* (maths author and speaker) inspired me to use my newly acquired Python skills to investigate which area of the dartboard it was best for players of different skill
levels to aim at. Rob’s own analysis of this question can be read in his book ‘The Hidden Mathematics of Sport’. He and co-author John Haigh mathematically analysed the numbers on the board to reach
their conclusions.
My dart board project reassuringly reaches roughly the same conclusions, but achieves this by means of computer simulation and repeated trials, rather than with a purely mathematical approach. In
this post I write briefly about how my program works and share my code.
The maths behind the code
My program simulates a player throwing a dart at the dart board. The player has a variable skill level and they are aiming for a target on the board that can be varied.
When a player takes a shot, the computer chooses an x coordinate and a y coordinate for the shot to land at. A normal distribution centred around the x coordinate of the target is used to generate
the x coordinate of the shot, and a second normal distribution is used to generate the y coordinate of the shot. We set the skill level of the player and the target they are aiming at when we decide
the standard deviation and the mean of the normal distribution, respectively. A high standard deviation means a poor player – a player whose shots will deviate greatly from the target they are aiming
Once the computer has generated an x and y coordinate for the shot, we then need the program to calculate the score that we would get for landing a dart in that position. To do this we first need to
convert the cartesian coordinate into polar form because the score is dictated by the distance of the shot radially from the centre, and by its angle from the horizontal. Once we have the shot
coordinate in polar form, we can use a look up system to find the score that shot is worth.
Now that we can simulate a single dart being thrown and return the score that that shot would equate to, we can do this for 10,000 shots and find the average score that those shots achieved.
Repeating this experiment for lots of different targets on the board (the program allows the user to choose these targets in polar form), and for many different skill levels, means we can find for
each skill level, which target gave the player the highest average score.
The next step was to illustrate this information visually. I first drew a dart board in Python to ensure it had the right proportions (most of the images I found on the internet were not proportional
to a professional dart board). Then for each skill level, I plotted a circle on any of the targets that gave an average score in the top 5% of scores for that player. The size of the circles
indicates the average score for that target, relative to the others shown.
The results
As an example, the image below shows where a medium level player should target. The top place to aim at for a player with this specific skill level came out to be triple 7 (it has the biggest
circle), but the other targets shown also gave very good scores that were not much different.
The animation below takes the image created for each skill level and plays them together (with the help of a GIF creating website). The animation starts by showing us where a professional player
should aim and ends by showing us where an awful player should aim.
We learn from the animation that only the very best players should aim for triple 20, followed by ‘good’ players aiming for triple 19. Medium players should aim for generally around the 8/11/16
sectors, and as players get worse they should aim closer and closer to the centre. The worst players should aim dead centre as this gives them the highest chance of getting the dart on the board.
The code also visualises what 10,000 shots at a particular target by a player of a particular skill level, looks like.
The image below shows what it looks like when a professional player takes 10,000 shots aiming at triple 20.
And the image below shows what it looks like when a good, but not excellent, player aims at triple 20. The high likelihood that their shot will land in the low scoring 5 or 1 sectors explains why
only the very best players should aim for triple 20!
For anyone interested to see all the code, I’ve shared it here:
Enjoy your next game of darts with this analysis in mind!
*In fact, it was Ben Sparks who suggested I use Python to investigate the question that Rob raised in the conversation. Thanks Ben!
Join the Conversation
1. Here’s a puzzle I recently investigated with Python: Player A has m coins and player B has n coins. Player A tosses all her coins, then counts the number of heads. Player B does the same. What’s
the probability that player A tossed more heads than B? The solution I came up with is a recurrence relation. Does a closed form exist? I had a lot of fun with the puzzle and hope the same for | {"url":"https://zoelgriffiths.co.uk/index.php/2020/07/09/where-to-aim-on-a-dart-board/","timestamp":"2024-11-08T23:19:24Z","content_type":"text/html","content_length":"48399","record_id":"<urn:uuid:d87d9228-d1dd-4d30-bc0a-9e98fa4512dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00700.warc.gz"} |
3: MATRIX Operations (2024)
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MATLAB serves as a powerful tool to solve matrices. To use matrices as a tool to solve equations or represent data a fundamental understanding of what a matrix is and how to compute arithmetical
operations with it is critical.
What is a Matrix?
A matrix is a rectangular array or grid of values which arranged in rows and columns. Matrices are used to operate on a set of numbers with variations of traditional mathematical operations. Matrices
serve valuable rolls within many engineering and mathematic tasks due to their useful ability to effectively store and organize information. Understanding matrices proves valuable when trying to
solve systems of equations, organizing data collected during experiments, computing mathematical operations on large quantities of numbers, and complicated applications in linear algebra, machine
learning, and optimization.
When describing matrices, we will name them based on the number of rows and columns. For example, the following matrix is a 2×3 matrix as it has two rows and three columns.
And this matrix is a 4×3 matrix:
Matrix Arithmetic
Matrices are an effective way to modify an entire set of numbers in one operation. Simple ways to modify matrices include addition, subtraction, multiplication, and division by a scalar, or
individual number. When completing these operations, complete the calculation with each number in the matrix, as denoted below.
\[\left[\begin{matrix}1&2\\4&3\\\end{matrix}\right]+2=\left[\begin{matrix}1+2&2+2\\4+2&3+2\\\end{matrix}\right]=\left[\begin{matrix}3&4\\6&5\\\end{matrix}\right]\gets Answer\]
\[\left[\begin{matrix}2&-4\\1.5&3\\\end{matrix}\right]\ast3=\left[\begin{matrix}2\ast3&-4\ast3\\1.5\ast3&3\ast3\\\end{matrix}\right]=\left[\begin{matrix}6&-12\\4.5&9\\\end{matrix}\right]\gets Answer
Matrices with the same dimensions (i.e. two 2×2 matrices) can have more mathematical operations completed with them. For example, you can add or subtract matrices with the same dimensions by
completing operations on the values in each corresponding location in a matrix. The following shows a template for adding or subtracting two matrices.
Multiplying matrices is more difficult than adding and subtracting and does not follow the format listed above. The process known as element-wise matrix multiplication is shown below. This process
for multiplying matrices is a fundamental concept of linear algebra and occurs when working with matrices in MATLAB. Be aware of the general form shown below and that it can be extrapolated to
include matrices of different sizes. An alternative method of multiplying two matrices that are the same size is called component-wise multiplication, which would follow the same form as the matrix
addition shown above. The procedure for coding these into MATLAB are shown below.
Vectors and Matrices in MATLAB
Inputting Matrices
It is easy to input matrices into MATLAB scripts. To make a standard matrix in the command window, use the following format with values of a matrix listed with spaces between each value. Use a
semicolon to separate each line of the matrix. To see how this process looks within MATLAB, refer to the examples at the end of this section.
>> [1 2 3;4 5 6;7 8 9]
Which produces
Note that to create an array list each number in a row separated only by spaces. To move down to a new row, use a semicolon. To save time making a large array, a colon can be used to “list” numbers.
For example, 1:5 would create a row containing 1, 2, 3, 4, and 5. For example,
>> [1:3;4:6;7:9]
creates the same matrix as the first example. If you would like to create a matrix that counts by a unit other than one, add a second colon that denotes what numbers will be included. For example,
>> [2:2:10;12:2:20]
will create the following 2 row by 5 column matrix which counts by twos between 2 and 10 in the top row and 12 and 20 in the bottom row
Matrix Operations and Concatenating Matrices
1) Enter the following matrix efficiently into MATLAB.
2) Enter the following matrix efficiently into MATLAB.
3) Use the following matrices in the following parts.
3a) Input the above matrices into MATLAB. Assign each the variable name shown.
Note that by placing semicolons at the end of the line the output is suppressed. As a result, the actual matrices are not printed in the code, which saves space in this instance.
3b) Add matrix
3c) Subtract matrix
3d) Multiply matrix
Efficiently type the following matrices into MATLAB’s command window.
Use these matrices to complete the following computations using MATLAB.
\[a=\left[\begin{matrix}-8&4\\5&12\\\end{matrix}\right];\ \ b=\left[\begin{matrix}3&5\\2&3\\\end{matrix}\right];\ \ c=\left[\begin{matrix}-2&1.5\\12&-4.25\\\end{matrix}\right];\ \ d=\left[\begin | {"url":"https://cozool.online/article/3-matrix-operations","timestamp":"2024-11-13T11:27:54Z","content_type":"text/html","content_length":"129016","record_id":"<urn:uuid:14cf1767-e9a2-4214-bce8-c27f341a5170>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00329.warc.gz"} |
Deep (learning) like Jacques Cousteau – Part 6 – Dot products | R-bloggersDeep (learning) like Jacques Cousteau – Part 6 – Dot products
Deep (learning) like Jacques Cousteau – Part 6 – Dot products
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(TL;DR: Start with two vectors with equal numbers of elements. Multiply them element-wise. Sum the results. This is the dot product.)
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Hmmm…this is a tricky one!
Uhhh…did you know that Kendrick Lamar’s stage name used to be “K.Dot”?
Last time, we learnt how to add vectors. It’s time to learn about dot products!
Today’s topic: dot products
Let’s define two vectors:
Let’s multiply these vectors element-wise. We’ll take the first elements of our vectors and multiply them:
Let’s take the second elements and multiply them:
Now add the element-wise products:
This, my friends, is the dot product of our vectors.
More generally, if we have an arbitrary vector $\boldsymbol{u}$ of $n$ elements and another arbitrary vector $\boldsymbol{v}$ also of $n$ elements, then the dot product $u \cdot v$ is:
The dot product $\boldsymbol{u} \cdot \boldsymbol{v}$ is equivalent to $\boldsymbol{u}^T \boldsymbol{v}$. Let’s come back to this next time when we talk about matrix multiplication.
What is that angular ‘E’ looking thing?
For anyone who doesn’t know how to read the dot product equation, let’s dissect its right-hand side!
$\sum$ is the uppercase form of the Greek letter ‘sigma’. In this context, $\sum$ means ‘sum’. So we know that we’ll need to add some things.
We have $\boldsymbol{u_i}$ and $\boldsymbol{v_i}$. In an earlier post, we learnt that this refers to the $i$th element of some vector. So we can refer to the first element of our vector $\boldsymbol
{u}$ as $u_1$. We notice that $v$ also shares the same subscript $i$. So we know that whenever we refer to the second element in $u$ (i.e. $u_2$), we will be referring to the second element in $v$
(i.e. $v_2$).
We notice that $\boldsymbol{u_i}$ is next to $\boldsymbol{v_i}$. So we’re going to be multiplying elements of our vectors which occur in the same position, $i$.
We see that below our uppercase sigma there is a little $i=1$. We also notice that there is a little $n$ above it. These mean “Let $i = 1$. Keep incrementing $i$ until you reach $n$”.
What is $n$? It’s the number of elements in our vectors!
If we expand the right-hand side, we get:
This looks somewhat similar to the equation from the example earlier:
Easy! These are the mechanics of dot products.
What the hell does this all mean anyway?
For a deeper understanding of dot products (which is unfortunately beyond me right at this moment!) please refer to this video:
The entire series in the playlist is so beautifully done. They are mesmerising!
How can we perform dot products in R?
Let’s define two vectors:
x <- c(1, 2, 3)
y <- c(4, 5, 6)
We can find the dot product of these two vectors using the %*% operator:
x %*% y
## [,1]
## [1,] 32
What does R do if we simply multiply one vector by the other?
x * y
## [1] 4 10 18
This is the element-wise product! If the dot product is simply the sum of the element-wise product, then x %*% y is equivalent to doing this:
sum(x * y)
## [1] 32
In our previous posts, R allowed us to multiply vectors of different lengths. Notice how R doesn’t allow us to calculate the dot product of vectors with different lengths:
x <- c(1, 2)
y <- c(3, 4, 5)
x %*% y
This is the exception that gets raised:
Error in x %*% y : non-conformable arguments
We have learnt the mechanics of calculating dot products. We can now finally move onto matrices. Ooooooh yeeeeeah. | {"url":"https://www.r-bloggers.com/2019/06/deep-learning-like-jacques-cousteau-part-6-dot-products/","timestamp":"2024-11-02T02:16:30Z","content_type":"text/html","content_length":"94271","record_id":"<urn:uuid:d87c08f1-f836-440b-9ad6-66ef7c48daf0>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00118.warc.gz"} |
Basic Algebra - A Simple Introduction to Algebra(video lessons, examples, solutions)
The Basics
The first thing to grasp is that when we have an equation, both sides have exactly the same value.
Let’s start with:
8 = 8
That is an equation. Simple enough? Now we change the equation a little by introducing simple arithmetic operations that you already know:
5 + 3 = 8
8 = 2 × 4
Thus: 5 + 3 = 2 × 4
Easy to follow so far? OK, the next step is something you may done in arithmetic quizzes in grade school:
5 + ☐ = 2 × 4
If you are asked to fill in the box, you can do the simple arithmetic and know that the answer should be 3. Now we are ready for basic algebra. Let’s substitute the box with the letter ‘k’ and we
5 + k = 2 × 4
In the equation above, the letter ‘k’ is known as a variable.
Of course we know that it is 3, so why is it called a variable? Well, that’s the way algebra is - there are just some terms where the meaning is not as straightforward. You may think of it this way -
if you were just given the equation
5 + k = 2 × 4
without any of the earlier discussions, then k would be unknown until you solve the arithmetic. That’s the idea for variables in algebra. Anyway, variables are defined as numbers that can change
value or represent a missing value (an unknown value). Variables are usually represented by letters of the alphabet, and the letters x, y, and z are most commonly used.
Now we have a real basic algebra equation, and the goal is to solve for the variable k - that means to find the value of ‘k’ in the equation. Of course we know from earlier our earlier exercises that
k = 3, but hey, where’s the fun if algebra is just like that?
So, an algebra equation would be given as: 5 + k = 2 × 4 without any of the earlier exercises and you would be asked to solve for the unknown k.
The Fundamental Principle of Equation
Before we go about solving for the variable k, there’s just one simple principle of equations that we need to grasp. Since we know that both sides of the equation are the same, whatever we do on one
side (arithmetically), if we do the same to the other side, and the result is still an equation - that means both sides would still be equal.
For example, we can do any of these:
5 + k - 2 = 2 × 4 - 2
5 + k + 4 = 2 × 4 + 4
(5 + k) × 3 = (2 × 4) × 3
Solving Our First Equation
Now we are ready to tackle our first algebra equation. What we want to do is to isolate the variable k on one side of the equation. Let’s start with the equation:
5 + k = 2 × 4
We can see that on the left side, there’s an extra 5 added to k. So we must get rid of the 5 to isolate k. We can do this be subtracting 5 from the left side. Remember that we must do the same thing
to the right side to maintain equality:
5 + k - 5 = 2 × 4 - 5
Now we are almost done solving our first algebra equation!
Looking at the left side
5 + k - 5, the two 5s (5 and -5) would cancel out, leaving us with:
k = 2 × 4 - 5
So we only need to do the arithmetic on the right side:
k = 2 × 4 - 5
k = 8 - 5
k = 3
Voila! We have solved our first algebra equation! Remember, the goal is to get the variable alone by doing the same thing to each side of the equation.
With this you have a good understanding of basic algebra, and now you should be able to solve other equations like 6 + k = 11 or 11 - m = 7. Otherwise, you may want to re-read this lesson.
Just one more simple thing to finish up. In algebra you would often see something like 6k or 14m used in equations. They just mean 6 × k and 14 × m - just think of it as a mathematician’s shorthand.
You can figure out why they prefer to omit the × sign especially when the letter x is most commonly used as the variable in algebra equations.
If you are comfortable with the basic algebra in this lesson, you are now ready to go to Isolate the Variable (Transposition).
You may also want to practice with some basic algebra worksheets.
How to Solve Basic Equations (first step to understand algebra)?
This video shows students the basic concepts and steps to solve equations in algebra. The linear equations he focuses on are those first introduced in middle school and mastered in high school
1. 2x = 10
2. y - 3 = 12
3. 1/3 x = 5
4. z + 6 = -3
Steps to solve a basic two-step algebra equation
This video walks students through the steps to solve a basic two-step algebra equation. This lesson on equations should be very useful to students in middle and high school math.
1. 2x + 8 = 14
2. -3y - 2 = 10
How to Solve Basic Liner Equations in Algebra?
This video explains the steps involved to solve equations in algebra. Middle school and high school math students will need to understand the steps to solve basic linear equations.
Example: 4(x - 2) + 6x = 14
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. | {"url":"https://www.onlinemathlearning.com/basic-algebra.html","timestamp":"2024-11-05T02:42:27Z","content_type":"text/html","content_length":"43972","record_id":"<urn:uuid:c442aa3b-4c57-447a-a49a-61f78d2ada95>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00344.warc.gz"} |
Development of a multi-GPU solver for atmospheric entry flows with gas-surface interactions
The understanding of atmospheric re-entry is fundamental in the aerospace engineering field. The heat load experienced by a space vehicle while entering in the atmosphere is extreme and its correct
prediction is necessary in the view of an appropriate design of the thermal shield. Technology progression allows to exploit sophisticated facilities able to reproduce the macroscopic features of
entry flows. However, high fidelity experimental reproduction are still hard due to two main reasons, namely the cost of an experiment and the difficulty in reproducing each aspect of the flight
conditions. This led many companies to invest more and more in numerical tool, representing a valid alternative to provide accurate prediction of interesting information, such as heat flux, pressure
distribution or shock stand-off distance. Of course, the development of an efficient numerical tool is not trivial and requires particular attention. Indeed, dealing with hypersonic flows, one must
account for 'real gas' effects, known as non-equilibrium phenomena. By the years, many researchers have been devoting efforts to the development of physical models able to describe the correct
evolution of the challenging conditions encountered during the re-entry. The high velocities of a space vehicle induce the formation of strong shock waves in front of it, across which the temperature
reaches values of the order of 10000 K. It is immediate to understand that these extreme conditions implicate the conversion of the kinetic energy into internal energy, whose total content involves
translational, rotational, vibrational and electronic modes. Also, molecular dissociation occurs due to the particle collisions in the shock layer and, if the temperature is large enough, ionization
occurs. The latter is a relevant aspect of re-entry flows as the presence of electrons in the mixture is responsible for the well-known blackout. For the purpose of heat mitigation, several
strategies are adopted. The employment of ablative material for the Thermal Protection System (TPS) has become very common. Thanks to material degradation, the heat flux on the surface of the vehicle
is reduced, even if this introduces further complexity in the numerical modeling. The material directly interacts with the species in the mixture, leading the the occurrence of gas-surface
interactions (GSI) such as catalysis and ablation. Classical numerical approaches exploit finite-volume method applied in a body-fitted multi-block grids, very common in Computational Fluid Dynamics
(CFD). Nevertheless, when dealing with complex and/or moving geometries, the employment of body-conformal domains can be complicated due to the need of run time remeshing procedures. In this context,
Immersed Boundary Methods (IBM) are suitable for a more versatile numerical solver. Such an approach allows for a unique Cartesian grid generation, that can be refined in the most critical region to
increase the accuracy of the numerical solution. Taking into account all the above mentioned phenomena is a complex task as the numerical model employed must be accurate and cheap at the same time.
Indeed, given the huge computational cost required by these kind of numerical simulations, an affordable strategy must be thought in order to speed-up the calculations. Graphics Processing Units
(GPUs) provide high performances for general purposes in the scientific field. NVIDIA Corporation is still actively working in the development of efficient interfaces between hardware and software.
The most famous one is Compute Unified Device Architecture (CUDA) that allows a very easy interface with basic programming languages such as C/C++ or Fortran. Thanks to GPU programming, very fast
simulations are possible even in the most demanding configurations. All the aforementioned aspects are addressed in this manuscript, which aims at illustrating the main challenges in modeling
hypersonic flows. A comparison of the current tools is presented for interesting aerospace applications, with the hope it can inspire further developments for technology progression.
Development of a multi-GPU solver for atmospheric entry flows with gas-surface interactions / Ninni, Davide. - ELETTRONICO. - (2022). [10.60576/poliba/iris/ninni-davide_phd2022]
Development of a multi-GPU solver for atmospheric entry flows with gas-surface interactions
The understanding of atmospheric re-entry is fundamental in the aerospace engineering field. The heat load experienced by a space vehicle while entering in the atmosphere is extreme and its correct
prediction is necessary in the view of an appropriate design of the thermal shield. Technology progression allows to exploit sophisticated facilities able to reproduce the macroscopic features of
entry flows. However, high fidelity experimental reproduction are still hard due to two main reasons, namely the cost of an experiment and the difficulty in reproducing each aspect of the flight
conditions. This led many companies to invest more and more in numerical tool, representing a valid alternative to provide accurate prediction of interesting information, such as heat flux, pressure
distribution or shock stand-off distance. Of course, the development of an efficient numerical tool is not trivial and requires particular attention. Indeed, dealing with hypersonic flows, one must
account for 'real gas' effects, known as non-equilibrium phenomena. By the years, many researchers have been devoting efforts to the development of physical models able to describe the correct
evolution of the challenging conditions encountered during the re-entry. The high velocities of a space vehicle induce the formation of strong shock waves in front of it, across which the temperature
reaches values of the order of 10000 K. It is immediate to understand that these extreme conditions implicate the conversion of the kinetic energy into internal energy, whose total content involves
translational, rotational, vibrational and electronic modes. Also, molecular dissociation occurs due to the particle collisions in the shock layer and, if the temperature is large enough, ionization
occurs. The latter is a relevant aspect of re-entry flows as the presence of electrons in the mixture is responsible for the well-known blackout. For the purpose of heat mitigation, several
strategies are adopted. The employment of ablative material for the Thermal Protection System (TPS) has become very common. Thanks to material degradation, the heat flux on the surface of the vehicle
is reduced, even if this introduces further complexity in the numerical modeling. The material directly interacts with the species in the mixture, leading the the occurrence of gas-surface
interactions (GSI) such as catalysis and ablation. Classical numerical approaches exploit finite-volume method applied in a body-fitted multi-block grids, very common in Computational Fluid Dynamics
(CFD). Nevertheless, when dealing with complex and/or moving geometries, the employment of body-conformal domains can be complicated due to the need of run time remeshing procedures. In this context,
Immersed Boundary Methods (IBM) are suitable for a more versatile numerical solver. Such an approach allows for a unique Cartesian grid generation, that can be refined in the most critical region to
increase the accuracy of the numerical solution. Taking into account all the above mentioned phenomena is a complex task as the numerical model employed must be accurate and cheap at the same time.
Indeed, given the huge computational cost required by these kind of numerical simulations, an affordable strategy must be thought in order to speed-up the calculations. Graphics Processing Units
(GPUs) provide high performances for general purposes in the scientific field. NVIDIA Corporation is still actively working in the development of efficient interfaces between hardware and software.
The most famous one is Compute Unified Device Architecture (CUDA) that allows a very easy interface with basic programming languages such as C/C++ or Fortran. Thanks to GPU programming, very fast
simulations are possible even in the most demanding configurations. All the aforementioned aspects are addressed in this manuscript, which aims at illustrating the main challenges in modeling
hypersonic flows. A comparison of the current tools is presented for interesting aerospace applications, with the hope it can inspire further developments for technology progression.
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Unit Circle Calculator
Unit Circle Calculator:
This unit circle calculator helps to calculate trigonometric values (sine, cosine, and tangent) for any angle on the unit circle. Simply enter the angle in degrees, radians, or pie (π) radians and
the calculator will determine:
• The coordinates (sine and cosine) of the point on the unit circle corresponding to the given angle
• Tangent values
What Is A Unit Circle?
“A unit circle is a circle on a cartesian plane having a radius equal to 1, centered at the origin (0,0)”
By marking the endpoint (terminal point) of a rotation from the positive x-axis (1, 0) on the unit circle, we can calculate the values of the cosine, sine, and tangent for a specific angle θ. The
unit circle is very helpful when you have to work with core trigonometry functions and need to find the angle measurements.
While the equation of the unit circle can be derived using the basic circle equation and the Pythagorean theorem, Let's see how: The equation of a circle, having center (x1, y1) and radius r can be
written as:
\(\ (x-\ x_{1})^{2}\ \ +\ \ (y-\ y_{1})^{2}=\ r^{2}\)
• (x, y) are the coordinates of the unit circle point
According to the Pythagorean theorem, the radius of the unit circle is always 1 so the equation will be transformed as:
\(\ (x-\ 0)^{2}\ \ +\ \ (y-\ 0)^{2}=\ r^{2}\) \(\ (x-\ 0)^{2}\ \ +\ \ (y-\ 0)^{2}=\ 1\) \(\ x^{2}\ \ +\ \ y^{2}=\ 1\)
\(\cos^{2}\theta\ \ +\ \sin^{2}\theta =\ 1\)
Trigonometry Ratios on Unit Circle and Derivation:
In trigonometry, the unit circle is used to visualize and derive trigonometric functions and their relationships. Let's see how the unit circle helps us derive the trigonometric ratios!
Unit Circle: Sine and Cosine
Sine and cosine relationship with unit circle:
• Sine is the y-coordinate
• Cosine is the x-coordinate
• Let's take a look at the point A in the above image
• Xx and Yy are the coordinates of the point P. As it’s a unit circle the radius is equal to 1
• After projecting the radius on the x and y coordinates, you will get the right triangle
• The horizontal side(adjacent to the unit circle angle) has a length equal to the x-coordinate
• The vertical side of this triangle (opposite to the angle) has a length equal to the y coordinate
• The hypotenuse (the diagonal side that creates the right angle) is the radius, which remains equal to 1
\(\ sin\ \alpha =\frac{ Opposite }{ Hypotenuse }=\frac{y}{1}=\ y\)
\(\ cos\ \alpha =\frac{ Adjacent }{ Hypotenuse } =\frac{x}{1}=\ x\)
As we have discussed above the equation of the unit circle that comes from the Pythagorean theorem is as follows:
\(\ x^{2}\ \ +\ \ y^{2}=\ 1\) \(\ cos^{2}\theta\ \ +\ \ sin^{2}\theta=\ 1\)
Unit Circle & Tangent:
According to the definition of tangent, it's the ratio of opposite sides and adjacent sides to an angle in the right triangle.
\(\tan\alpha =\frac{opposite}{adjacent}\)
\(\tan\alpha =\frac{y}{x}\)
The tangent of an angle (α) is calculated using the sine (y) and cosine (x) values from the unit circle or a right triangle using the following formula:
\(\tan\alpha =\frac{\ sine\ (y)}{cose\ (x)}\)
\(\tan\alpha =\frac{\ sine\ (y)}{cose\ (x)}=\frac{y}{x}\) (when x = 0, the tangent is undefined)
How To Find Trig Ratios From an Angle on The Unit Circle?
• Determine the given point on the unit circle. The coordinates of the unit circle are \(\ (sin\theta,\ cos\theta)\), where \(\theta\) is the angle between the positive x-axis and the line that
connects the point with the origin
• Find the \(\ cos\theta\) as it’s the x-coordinate
• Find the \(\ sin\theta\) as it’s the y-coordinate
• Get the value of tangent by dividing the \(\ cos\theta\) by \(\ sin\theta\), \(\ tan\theta=\frac{\ cos\ \theta}{sin\ \theta}\)
The unit circle coordinates calculator eliminates the need for stepwise trig function (coordinates) calculation from an angle.
Solved Examples For Trig Functions (Unit Circle):
#1: Find the trig functions if the angle of the unit circle is 60°.
Mark Point P on the unit circle where the angle is formed
Find coordinate on unit circle:
sin(60°) = y-coordinate of point P = \(\frac{\sqrt{3}}{2} = 0.86602…\) (positive because it is on the upper half of the circle)
cos(60°) = x-coordinate of point P = \(\frac{1}{2} =0.5\) (positive because it lies on the right half of the circle)
Calculate Tangent (tan(60°)):
\(\ tan(60°) =\frac{sin(60°)}{cos(60°)} = \frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})} =\sqrt{3} = 1.7320…\)
#2: Find trigonometric ratios for an angle of \(\frac{π}{3}\) radians on the unit circle.
Mark Point P on the unit circle as done in the previous example
Find the Coordinates: X-coordinate \(\cos(\frac{π}{3}) = \frac{1}{2}=\ 0.5\) (positive because it is present on the right half of the circle)
Y-coordinate (sin(π/3)) =\(\frac{\sqrt{3}}{2} =\ 0.86602 \) (positive because it is available on the upper half of the circle)
Calculate Tangent \(\ (tan(\frac{π}{3})\):
\(\ tan(\frac{π}{3}) =\frac{sin(\frac{π}{3})}{cos(\frac{π}{3})} = \frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})} =\sqrt{3} = 1.7320…\)
#3: Calculate the unit circle trig values for an angle of π radians.
π radians corresponds to 180 degrees, so for the unit circle radians the coordinates of point P will be:
X-coordinate (cos(π)) = P = -1 (negative because it is present on the left half of the circle)
Y-coordinate (sin(π)) = P = 0 (because it lies on the x-axis)
Tangent (tan(π)):
tan(π) = undefined
Also, you can use the unit circle trig calculator to find the exact values of sine, cosine, and tangent for any angle in radians, including π radians.
Unit Circle Chart with Radians and Degrees:
This chart shows a circle with angles marked in degrees or radians. With the help of this unit circle chart, you can easily find the sine (y-coordinate), cosine (x-coordinate), and tangent values on
the circle's edge.
A unit circle goes from 0 to 360 degrees (0 to 2 π radian). So whenever you get an angle bigger than 360 degrees, you should keep subtracting 360 until it reaches the normal value from 0 to 360
The given table can be considered for calculating the coordinates (x, y) of the unit circle from the value of angle.
Angle (Degrees) Angle (Radians) Unit Circle Coordinates
30° \(\frac{\pi}{6}\) (\(\frac{\sqrt{3}}{2}\),\(\frac{1}{2}\))
45° \(\frac{\pi}{4}\) (\(\frac{\sqrt{2}}{2}\),\(\frac{\sqrt{2}}{2}\))
60° \(\frac{\pi}{3}\) (\(\frac{1}{2}\),\(\frac{\sqrt{3}}{2}\))
90° \(\frac{\pi}{2}\) (0, 1)
120° \(\frac{2\pi}{3}\) (\(\frac{-1}{2}\),\(\frac{\sqrt{3}}{2}\))
135° \(\frac{3\pi}{4}\) (\(\frac{-{\sqrt{2}}}{2}\),\(\frac{\sqrt{2}}{2}\))
150° \(\frac{5\pi}{6}\) (\(\frac{-{\sqrt{3}}}{2}\),\(\frac{1}{2}\))
180° π (-1, 0)
210° \(\frac{7\pi}{6}\) (\(\frac{-{\sqrt{3}}}{2}\),\(\frac{-1}{2}\))
225° \(\frac{5\pi}{4}\) (\(\frac{-{\sqrt{2}}}{2}\),\(\frac{-{\sqrt{2}}}{2}\))
270° \(\frac{3\pi}{2}\) (0, -1)
300° \(\frac{5\pi}{3}\) (\(\frac{1}{2}\),\(\frac{-{\sqrt{3}}}{2}\))
315° \(\frac{7\pi}{4}\) (\(\frac{\sqrt{2}}{2}\),\(\frac{-{\sqrt{2}}}{2}\))
330° \(\frac{11\pi}{6}\) (\(\frac{\sqrt{3}}{2}\),\(\frac{-1}{2}\))
360° 2π (1, 0)
Skip memorizing all the values! Find trigonometric ratios using the unit circle calculator.
What Are The Applications of Unit Circle In Real Life?
The unit circle has a wide range of applications in real-world scenarios, including:
• Understanding Periodic Phenomena
• Solving Engineering Problems
• Computer Graphics and Animation
• Navigation and Positioning
• Signal Processing and Data Analysis
What Are The Positive Angles on The Unit Circle?
Positive angles on the unit circle are measured starting from the positive x-axis and rotating counterclockwise around the origin. They lie between the positive x-axis and the terminal side.
How Do Special Right Triangles Create The Unit Circle?
The special right triangles (30-60-90 and 45-45-90) serve as a roadmap for the unit circle. By scaling them to fit inside (hypotenuse = 1 unit), their side lengths directly give sine (y) and cosine
(x) values for key angles (30°, 45°, 60°) on the circle. Using trigonometry, this foundation helps assign sine and cosine values to other angles.
A source of Wikipedia: All you need to know about the unit circle
From the source of Khan Academy: Unit: Trigonometric functions, intro to radians & much more!
From the source of clarku: tangent to the circle | {"url":"https://calculator-online.net/unit-circle-calculator/","timestamp":"2024-11-13T15:42:03Z","content_type":"text/html","content_length":"69727","record_id":"<urn:uuid:73a3381f-f2da-4fdf-8a09-59470ab778f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00683.warc.gz"} |
Anfangswerte für einen späteren Schwarzes-Loch-Kollaps von kugelsymmetrischen relativistischen Flüssigkeiten - Existenzsätze und Numerik
Translated title
Initial values for a later black hole collaps of spherically relativistic fluids - existence theorems and numeric
Document type
PhD thesis (dissertation)
Müller zum Hagen, Henning
Granting institution
Helmut-Schmidt-Universität / Universität der Bundeswehr Hamburg
Part of the university bibliography
Method of Diagonalization
Proof of Existence
Initial boundary value problems for quasilinear, partial differential equations of first order $\partial_t u+{\bf A}(u)\cdot \partial_x u={\bf b}(u)$ in two unknowns of hyperbolic type are
considered. An astrophysically interesting and challenging (1) example hereof is a spherically symmetric perfect fluid space time, whose later collapse can be achieved by suitable choice of inital
values. By that a timely global statement 'occurence of an event horizon from innocuous initial data' is won from a timely local existence theorem. The initial data must hold physical conditions
which prevent a smooth solution in the star boundary (2). The proof stays clearly arranged since the system is brought into diagonal form. The diagonalisation method is used on the one hand, on the
above mentioned Einstein equations and on the other hand, on the equations of one-dimensional gas dynamics (written in comoving and after transformation also in Eulerian coordinates) A new hybride
algorithm is numerically testet for a single equation with creasing inital data. footnotes (1) Apart from the non-linearity of the equations system, which is not given in divergent form, the
coordinates are concatenated with the underlying geometry, where the components of the metric are unknowns of the partial differential system. (2) where a vacuum space time is connected and the mass
energy is positive (surface of a fluid) (3) new in that method is that it works also if the matrix ${\bf A$ is not invertible, because some eigenvalues disappear identically, provided that the system
is written in such a way that one found so called constraints for all trivially propagated unknowns
Not applicable (or unknown) | {"url":"https://openhsu.ub.hsu-hh.de/handle/10.24405/587","timestamp":"2024-11-14T17:22:39Z","content_type":"text/html","content_length":"715797","record_id":"<urn:uuid:29b97db9-3e4a-493a-9358-a8b2b185ce31>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00210.warc.gz"} |
Backtesting in Excel and R | R-bloggersBacktesting in Excel and R
Backtesting in Excel and R
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This post is the introduction to a series that will illustrate how to backtest the same strategy in Excel and R. The impetus for this series started with
this tweet
Jared Woodard
Condor Options
. After
Soren Macbeth
introduced us, Jared suggested backtesting a simple
strategy in Excel and R.
The three-post series will show you:
Since I know next to nothing about testing strategies in Excel, I will be writing posts 1 and 3. Jared was kind enough to create the Excel framework for post 2, but did not have time to devote to a
full post. Thankfully,
Damian Roskill
has agreed to write post 2 using Jared’s Excel file.
Hopefully this will be a useful example for those of you who currently use Excel but would like to explore how to use R. | {"url":"https://www.r-bloggers.com/2011/02/backtesting-in-excel-and-r/","timestamp":"2024-11-09T17:46:44Z","content_type":"text/html","content_length":"83069","record_id":"<urn:uuid:ad1f9283-24f2-402c-b956-3085164ab1a8>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00372.warc.gz"} |
Must-Read: Remember our rules!
• Rule 1: Paul Krugman is right.
• Rule 2: If you think Paul Krugman is wrong, consult Rule 1:
Paul Krugman: Some Misleading Geometry on Corporate Taxes (Wonkish): "There’s a fairly simple geometric way to see where the optimistic view that cutting corporate taxes is great for wages comes
...[Then] it becomes a lot easier to ask “What’s wrong with this picture?” (Answer: a lot).... Envision a small open economy with a fixed labor force (because labor supply isn’t of the essence
here) that can import from or export capital to the rest of the world.... Saving... also isn’t of the essence: the stock of capital, we’ll assume, changes only through capital inflows or
outflows.... Let’s... assume... factors of production are paid their marginal products. Then we can represent the economy with Figure 1, which has the stock of capital [per worker] on the
horizontal axis and the rate of return on capital on the vertical axis. The curve MPK is the marginal product of capital, diminishing in the quantity of capital because of the fixed labor force.
The area under MPK – the integral of the marginal products of successive units of capital – is the economy’s real GDP, its total output....
The economy faces a given world rate of return r*. However, the government imposes a profits tax at a rate t, so that to achieve a post-tax return r* domestic capital must earn r*/(1-t)....
That... determines the size of the domestic capital stock.... In the initial equilibrium real output is a+b+d. Of this, d is the after-tax return to capital, b is profit taxes, and a – the rest –
is wages.
Now imagine eliminating the profits tax (we can also do a small cut, but that’s harder and this is already sufficiently wonky). In equilibrium, the capital stock rises by ∆K, and... [GNP] to
a+b+c+d... (e is returns to foreign capital).... Profit taxes disappear: that’s a revenue loss of b. But wages rise to a+b+c, a gain of b+c....
What’s wrong with this picture?.... Four reasons.... First, a lot of what we tax with the corporate profits tax is... monopoly profits and other kinds of rents. There is no reason to believe that
these rents would be bid down by capital inflows.... Second, capital mobility is far from perfect. Third, the US isn’t a small open economy.... Finally... what we’re showing here is long-run
equilibrium... [after] capital inflows take place as the counterpart of trade deficits, which in turn have to be created by a temporarily overvalued real exchange rate. And the kind of adjustment
we’re talking about here would require moving a lot of capital, meaning very big trade deficits, meaning a strongly overvalued dollar, which would itself be a deterrent to capital inflows. So
we’re talking about a slow process.... Long-run analysis is a very poor guide to the incidence of corporate taxes in any politically or policy-relevant time horizon...
Paul is, of course, 100% correct.
There is a footnote with respect to Mankiw to be written here...
Suppose that you cut the corporate tax rate in this model from its initial level t by an amount Δt. The reduction in revenue collected is then:
${k}\left(\frac{r}{1-t}\right){t} - {k}{\left(\frac{r}{1-t+Δt}\right)}{\left(t-Δt\right)}$
By contrast, Mankiw miscalculates the "static" reduction in revenue as simply:
That is where the factor 1/(1-t) that puzzles him—"dw/dx = 1/(1 - t). I must confess that I am amazed at how simply this turns out. In particular, I do not have much intuition for why, for example,
the answer does not depend on the production function"—comes from.
And do note that Alan Auerbach is right when he writes:
this result... is a combination of (1) the standard result that in a small open economy labor bears 100% of a small capital income tax...
and wrong when he writes:
the burden of a tax increase exceeds revenue collection due to the first-order deadweight loss...
The burden of a tax increase exceeds revenue collection due to the first-order deadweight loss, but Mankiw is claiming that even for an infinitesimal change in the tax rate—for which the deadweight
loss term is infinitesimal relative to the distribution term—the ratio of revenue lost to wages gained is 1/(1-t). And that is simply a miscalculation of what the revenue loss is. | {"url":"https://www.bradford-delong.com/2017/10/must-read-remember-our-rules-rule-1-paul-krugman-is-right-rule-2-if-you-think-paul-krugman-is-wrong-consult-ru.html","timestamp":"2024-11-11T09:31:25Z","content_type":"text/html","content_length":"36343","record_id":"<urn:uuid:039bb8d2-182e-460f-b912-c1cd37ff445a>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00054.warc.gz"} |
Present Value of a Series of Cash Flow
The present value of a series of cash flow is used to obtain the initial/current worth of a series of cash flow to be invested for some years at a prevalent rate of interest. In other words, it is
the amount required to be invested today at a certain rate of interest to meet future uniform cash flow needs. E.g. Money to be invested today to meet rent payment liabilities for the next 5 years.
Uniform series and Non-Uniform series of payments
• The word series means something one after the other in a pattern, like weekly episodes of a daily soap, 5 volumes of a book, etc.
• If the episodes of a daily soap are telecasted weekly, then it is known as uniform series and the duration between two episodes is the same.
• An Annuity is a uniform series of payments made at equal intervals. Examples of Annuity include recurring deposit account, loan repayment in fixed EMIs, etc.
• If the amount paid is variable at intervals or different amounts is paid at the same intervals, then it is called as uneven cash flow instead of Annuity.
Example. 1:
• Sanjay invests Rs 25000 every year in mutual funds and Rakesh pays ₹15000 every month as loan EMI are examples of the annuity.
Types of Annuities (Annuity Due, Ordinary Annuity and Perpetuity)
• Annuity Due: In this Annuity, the amount is paid at the beginning of each period.
• Ordinary Annuity: In this Annuity, the amount is paid at the end of each period.
• Perpetuity: This is an Annuity for the Infinite Period.
Example. 2:
• Ruchi pays her LIC premium of Rs 10000 at the beginning of each year is an Annuity Due
• Nisha deposits Rs 500 in her recurring deposit account at the end of every month is an Ordinary Annuity
• Dividend paid by a company on its share for a lifetime, hence it’s a Perpetuity
• In case of an Annuity Due, as the installment is paid at the beginning of the year, hence all installments are considered for interest payments.
• In the case of Ordinary Annuity, the installment is paid at the end of the year, hence the last installment is not considered for interest calculation.
Examples and Formula for Present Value of Cash Flow
Example 3:
• Priyanka wants Rs 10000 after two years and goes to the bank to invest in an FD that offers interest at the rate of 6%. Let us help Priyanka in finding out how much she should invest. In other
words, it is the Present Value of Rs 10000 receivable after 2 years.
□ Solution: The formula for calculating the present value of cash is PV= FV (1 + R) ^-T
□ where FV is the maturity value to be obtained in T years at R% interest
□ Hence, PV = 10000 (1 + 0.06) ^-2 = Rs 8899.96
• The Present Value of an ordinary annuity and annuity due is calculated by adding present values of individual installments as shown below:
Example 4:
• Consider an ordinary annuity of Rs 10000 for 5 years at the rate of 10% PA compounded annually.
│ Year Completed │0│ 1 │ 2 │ 3 │ 4 │ 5 │
│Instalment Due │0│Rs 10000 │Rs 10000 │Rs 10000 │Rs 10000 │Rs 10000 │
│Number of years after which the installment is due to be paid (T) │0│1 │2 │3 │4 │5 │
│Calculation of Amount (Using PV of single cash flow) │0│Rs 9090.90│Rs 8264.46│Rs 7513.15│Rs 6830.13│Rs 6209.21│
• Here, the Present Value will be some of the present values of all payments made, i.e. PV = Rs 37907.85
• Instead of following this lengthy process, there is a direct formula to calculate the present value of a series of cash flow
Present Value of Ordinary Annuity:
• PV= A {[1 - (1 + r)^-N] /r}
• where A is the amount paid per year for a period of N years and r is the rate of interest.
Example 5:
• Consider an annuity due to Rs 10000 for 5 years at the rate of 10% PA compounded annually.
│ Year Completed │ 0 │ 1 │ 2 │ 3 │ 4 │5│
│Instalment Due │Rs 10000 │Rs 10000 │Rs 10000 │Rs 10000 │Rs 10000 │0│
│Number of years after which the installment is due to be paid (T) │0 │1 │2 │3 │4 │5│
│Calculation of Amount (Using PV of single cash flow) │Rs 10000.00│Rs 9090.90│Rs 8264.46│Rs 7513.14│Rs 6830.13│0│
• Here, the Present Value will be some of the present values of all payments made, i.e. PV = Rs 41698.63
□ Instead of following this lengthy process, there is a direct formula to calculate the present value of a series of cash flows.
Present Value of Ordinary Annuity:
• PV= A {[1 - (1 + r)^-N] /r} (1+r)
• where A is the amount paid per year for a period of N years and r is the rate of interest.
Example 6:
Amit has rented his house at Rs 50000 per year to Sunil for 5 years. The current rate of interest is 5%. Amit wants to know that what is the Present Value of the total rent he will receive from Sunil
in 5 years and Sunil is also concerned and wants to know that how much amount should he invest today so that he is able to pay rent for 5 years.
• Solution: Both Amit and Sunil are talking about one and the same thing
• Rent is always paid in advance, so it is a case of Annuity Due.
• A = 50000, r = 5% and N = 5 years.
• PV = 50000 {[1-(1 + 0.05) ^-5 ] /0.05 } (1+0.05) = Rs 227297.53
• Hence, if Sunil invests Rs 227297.53 at an interest rate of 5% for 5 years, he will be able to pay all his rent for the next 5 years.
• Also, Amit can now understand that the present value of the total rent receivable in the next 5 years is Rs 227297.53
Important Points
• In case of uneven cash flows, there cannot be any standard formula as the installment amount keeps changing, therefore it is advised to calculate the future value of uneven cash flows by adding
future values of individual installments.
• The present value of a perpetuity is A/R, where A is the periodic payment to be received forever and the expected rate of interest is r%. This is very important in calculations like share price.
Example 7:
• M/S ABC pays an annual dividend of Rs 10 indefinitely, ten what should be the per-share price of M/S ABC to attract the investors assuming no capital growth in share price and rate of return
expected is 5%.
□ Solution: PV of all the dividends received is = A / r =10/.1 = Rs 100,
□ If M/S ABC will keep share price below Rs 100 then the investor will be attracted as it will be profitable because the net present value of dividends will be more than the investment.
□ If the company quotes share price more than Rs 100 then the investors will not get attracted to the scheme. | {"url":"https://www.testpanda.com/2020/04/present-value-of-series-of-cash-flow.html","timestamp":"2024-11-09T16:09:14Z","content_type":"application/xhtml+xml","content_length":"109676","record_id":"<urn:uuid:3cc8d7d7-a119-4c4e-9c7f-85ddc4168c13>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00490.warc.gz"} |
Brown Bear Population Trends from Demographic and Monitoring-based Estimators
How to translate text using browser tools
1 November 2007 Brown Bear Population Trends from Demographic and Monitoring-based Estimators
Jedediah F. Brodie, Michael L. Gibeau
A primary goal of monitoring wildlife populations is the estimation of population growth rate, λ. Two common methods by which biologists estimate λ are demographic studies of marked individuals,
which tend to be expensive and labor-intensive, and estimators derived from time series of population indices. We compare grizzly bear (Ursus arctos) population growth rates in the Banff ecosystem
(Alberta, Canada) from a published demographic study to estimates from concurrent monitoring of an index of population size, the number of females with cubs-of-the-year (F[cub]). We estimated
population trends by transforming the index into 2 population estimators (bias-corrected Chao and summation), and used each to estimate λ. The 95% confidence intervals of λ̂ from the 2
monitoring-based estimators overlapped the point estimate of the demographic study. Precision of the bias-corrected Chao estimator was very low (95% CI of λ = 0.572–1.679); its application to the
time-series used here is essentially fruitless. Precision of the summation estimator (95% CI of λ = 0.847–1.137) and the demographic study (0.99–1.09) were higher, but the CI of the former at least
could be artificially narrow. Because all estimates were close to 1.00, the long-term fate of this population may depend critically on subtle changes in growth rate and on environmental
stochasticity. Given that long-term demographic studies are not feasible in this system, population monitoring may be a worthwhile way to assess population dynamics. However, given the low power of
many monitoring techniques to detect trends and the low precision of the F[cub] estimators in particular, long time-series and explicit measures to remove sampling variance should be employed to
increase trend estimate precision.
Research has consistently indicated that 3 dominant factors drive extinction risk: population size, average population growth rates, and temporal variation in population growth rates (Fagan et al.
2001, Inchausti and Halley 2003, Reed and Hobbs 2004). The minimum criterion for a population to persist is a geometric mean annual population growth rate ≥1.0, meaning that births exceed or
numerically balance deaths. However, this criterion alone is not sufficient for population persistence; small or highly variable populations can go extinct despite relatively high population growth
rates (Dennis et al. 1991, Mangel and Tier 1994). Therefore, population size and variance in growth rate are both key parameters for understanding population status.
The true abundance and growth rate of vertebrate populations can seldom, if ever, be ascertained by direct enumeration. Instead, wildlife biologists have several options. Individuals in the
population can be marked and their fates followed over time, allowing demographic rates (e.g., age-specific survivorship or fecundity) to be calculated (Leslie 1945, Wakkinen and Kasworm 2004, Kovach
et al. 2006). Dominant eigenvalues of the resulting transition matrices then provide a measure of λ (Caswell 2000). If large sample sizes are available over relatively long periods, such studies can
estimate trends precisely and accurately. However, because they are labor intensive and expensive, demographic studies of marked individuals are seldom conducted over enough years to estimate
variance in λ over time (Fieberg and Ellner 2000). In addition, small sample sizes obtained in many vertebrate population studies can seriously affect performance of demographic estimators (McKelvey
and Pearson 2001).
Population size, λ, and variance in λ can also be estimated from long-term monitoring of population size estimates. These repeated estimates can be derived from, for example, mark–recapture studies (
Pradel 1996, Kendall et al. 1997) or index-based density estimates (Knight et al. 1995, Keating et al. 2002). Indices are measurable quantities that are assumed to be proportional to actual
population size. If the functional relationship between the index and true population size is known, the index can be turned into an estimator of the actual population size. An advantage of
index-derived estimators is that they are typically easier to measure than actual population sizes and, therefore, make monitoring over large scales in space and time feasible. A disadvantage is that
the relationship between index values and true population size is seldom known. Use of population parameters derived from indices can be especially problematic if underlying assumptions are not met (
Thompson 2003).
Problems resulting from extrapolation from small samples are particularly acute in studies of large, free-ranging carnivores. These species usually occur at very low densities, are difficult to
locate and count, and yet are often of extremely high conservation or management concern (Soulé and Terborgh 1999). In principle, demographic studies of marked (e.g., radiocollared) individuals may
provide the best estimates of population growth rates, yet these types of studies face several limitations, described above. Sampling techniques using genetic tagging can provide population estimates
non-invasively (Bellemain et al. 2005, Solberg et al. 2006), but remain expensive and labor-intensive and are also, therefore, usually short in duration.
Demographic and monitoring-based estimators each have advantages and disadvantages, and we make no claim here that either method is necessarily preferable. Although many studies have assessed the
efficacies of index-based estimators (e.g., Don 1984, Hallett et al. 1991, Calvert and Robertson 2002), few have compared index-based estimators with demographic estimators to assess similarity in
calculated growth rates. This is surprising because both of these methods for monitoring populations are common. It would be useful for biologists using one monitoring method to have a sense of how
their results would differ if using an alternative method. Here, we compare grizzly bear (Ursus arctos) population growth rates calculated from monitoring-based estimators with that from a previously
published demographic study in Banff National Park and Kananaskis Country, Alberta, Canada (hereafter, “Banff–Kananaskis”). We generated estimates of λ using yearly counts of unduplicated female
bears with cubs-of-the-year (hereafter F[cub]) after conversion by 2 common formulae to estimators of population size (Knight et al. 1995, Keating et al. 2002). We compared growth rates generated
from these analyses with estimates derived from a concurrent demographic study by Herrero (2005; also see Garshelis et al. 2005a) to ascertain: (1) how well population growth rates generated from the
index-based estimators matched those of the demographic estimator, and (2) which of the F[cub] estimators generated growth rate estimates closest to that of the demographic estimator.
Materials and Methods
Study Area
The Bow River watershed of southwestern Alberta constituted the core of the study area. This area is 11,400 km^2 of mountainous terrain 50–180 km west of Calgary and was the focus of an intensive
grizzly bear research program during 1994–2004 (Herrero 2005). The area includes roughly 50% of Banff National Park (BNP) and all adjacent Alberta Provincial land known as Kananaskis Country. Neither
jurisdiction permitted grizzly bear hunting, although bears were exposed to hunting outside the Bow River Watershed. Differing agencies oversee preservation, industrial tourism, recreation, forestry,
oil and gas extraction, mining, and stock grazing. Native councils, towns and municipalities, commercial developers, and residential owners all manage lands.
Field Methods
Reproductive status of female grizzly bears was determined as part of a larger ongoing research effort (Herrero 2005) that maintained 10–15 radiocollared females out of a low density population of
approximately 100 bears. Observations of F[cub] were by research staff as well as sighting records from provincial and federal agencies. Both agencies employ a bear monitoring system where the public
is encouraged to report grizzly bear sightings. We did not include in the F[cub] calculations sightings where the females were located for demographic monitoring; the F[cub] sighting records only
included bears (collared or not) seen without specifically looking for them or by radiotracking by biologists or the general public. Agency and research staff followed up on all reports of family
groups for data verification.
We followed procedures and criteria similar to Knight et al. (1995; also see Interagency Conservation Strategy Team 2003) to determine whether sequential sightings belonged to the same family group
or different groups. The combination of range size, physical barriers, and population density made distinguishing individual females with litters relatively straightforward. The mean standard
diameter for annual ranges of radiocollared females with cubs of the year was 13 km (n = 27, SD = 5.2 km, Gibeau unpublished data). Given the similarity in range size to Knight et al. (1995), we
simply adopted their criteria of 30 km separation in judging whether 2 females with litters of the same size were distinct. Based on evidence from radiotelemetry (Herrero 2005) and genetics (Proctor
2005), we considered the Trans Canada Highway, which bisects the area, a barrier to female bears with cubs. In 6 of the 12 year monitoring period, there were overlapping litters of the same size. In
all 6 cases this involved only 2 family groups.
We made decisions whether family groups were distinct following the rule set outlined by Knight et al. (1995:246), “Once a female with a specific number of cubs was sighted in an area, no other
female with the same number of cubs in that same area was regarded as distinct unless 2 family groups were seen by the same observer on the same day, or by 2 observers at different locations but
similar times, or 1 or both of the females were radio-marked. Because of possible cub mortality, no female with fewer cubs was considered distinct in that area unless she was seen on the same day as
the first female or unless both were radio-marked.” Cubs were classified from their size and, if known, the reproductive status of the female from the previous year. The maximum number of cubs
observed was considered the litter size, although cubs lost very early in the season would not have been recorded.
Population Estimators
Raw counts of F[cub] alone are not good metrics of population size in a given year because a different fraction of bears breed in each year (Eberhardt and Knight 1996) and because individual F[cub]
vary in their detectability (Mattson 1997). A simple method of using F[cub] to establish a lower bound for population size is to sum the observations across the mean interbirth interval; this is a
commonly used estimator in the continental US as part of the recovery plan under the Endangered Species Act (16 US Code 1531–1544; Knight et al. 1995), where F[cub] are summed over 3 years, the mean
interbirth interval (in Yellowstone National Park, USA; see Knight and Eberhardt 1985, Eberhardt and Knight 1996). However, Keating et al. (2002) pointed out that this method biased the trend
estimation by using minimum counts rather than actual population estimates, and furthermore that the method does not permit calculation of valid confidence limits. They supported other estimator
functions based on recording the number of sightings of each female over the course of the year (analogous to building a “capture history” in a mark–recapture study). These asymptotic estimates of
population size are less affected by variation in detectability (Boyce et al. 2001). Sighting history can then be used with various models to estimate the total number of females in the population.
Keating et al. (2002) used Monte Carlo simulations to test a number of non-parametric F[cub] estimator models and determined that the Chao (Chao 1984, 1989) and second-order sample coverage (Chao and
Lee 1992) estimators were the best in terms of robustness to variation in number of unique females, overall sample size, and coefficient of variation. Further simulations suggested that the
bias-corrected Chao estimator (Chao 1989, referred to as “Chao2” in Cherry et al. 2007) should be used for management applications because it is less likely to be biased high than the sample-coverage
estimator (Cherry et al. 2007). It should be noted that the bias of the Chao estimator increases as detectability becomes more heterogeneous and sample size decreases (K. Keating, US Geological
Survey, personal communication, Bozeman, Montana, USA, 2006), but that the magnitude of this bias is much lower than the potential bias incurred by assuming a female population size equal to the
number of raw F[cub] observed.
We tested the F[cub] summation estimator (Knight et al. 1995) and the bias-corrected Chao estimator (Chao 1989, Keating et al. 2002) using data collected in Banff–Kananaskis from 1993–2004,
concurrent with the demographic study of Garshelis et al. 2005a. For each estimator, we converted raw estimates of F[cub] into indices of female population size.
The F[cub] summation estimator (Knight et al. 1995) was:
where m̂
is the estimated number of unique
in year
. This is essentially
Eberhardt and Knight's (1996)
method, as exemplified by
Morris and Doak (2002)
, except that we used a 4 instead of a 3-year summation because Banff–Kananaskis grizzlies have 4-year mean interbirth intervals (
Herrero 2005
The second estimator we tested was the bias-corrected Chao estimator (Chao 1989, Keating et al. 2002):
are the numbers of unique
seen once and twice, respectively, in year
For both estimators, annual log growth rates were calculated as:
where n̂
was either n̂
or n̂
. The exponent of the mean of these rates provides an estimate of λ (
Dennis et al. 1991
Morris and Doak 2002
). We regressed the ln λ̂
array against an array of ones (the square root of 1-year time intervals between censuses) with the intercept forced through zero (
Dennis et al. 1991
); the slope of this regression was μ̂. The 95% confidence limits for λ̂ were then: where
(μ̂) is the standard error of the regression slope and
is the critical value of the 2-tailed Student's
distribution with a significance level α = 0.05 and degrees of freedom equal to the number of transitions in the time series minus 1 (
Morris and Doak 2002
). We note that the confidence interval for the Chao estimator is based on variance in annual counts, whereas the variance for the summation estimator is reduced by the autocorrelation inherent in
the method.
Finally, we used simulations in MATLAB (version 7.0.4; The Math Works Inc., Natick, Massachusetts, USA) to estimate the sensitivity of the indices to variation in the number of unique females
observed (e.g., caused by observation error such as missing observations that would underestimate population size in a year as well as misidentifications that could overestimate population size). For
each index-based estimator (Chao and summation), we created 10,000 simulated time-series with 12 years each (1993–2004). For each year we randomly selected a population size from a Poisson
distribution with a mean equal to the observed m[t] for that year. We randomly drew the ratios of f[1-simulated] and f[2-simulated ]to m[simulated] in each year from the range of ratios observed in
the data. We then estimated λ for each simulated time-series using equation 3. The degree to which random variation in the annual population estimates affected measures of population growth rate was
assessed as the difference between the mean simulated λ estimates and the observed λ.
Eleven years of field monitoring recorded year-to-year fluctuations in records of F[cub] (Fig. 1, Table 1). Because human activity was relatively common in both jurisdictions, sightings from all
regions within the study area (n = 407) were easily obtained. Annual female abundance estimates from the Chao estimator were very close to actual counts of F[cub]. When applied to the interval over
which the demographic study took place (1993–2004), the λ point estimate from the Chao estimator was 0.980, that of the summation estimator was 0.981. The λ̂ of the matrix model was 1.04. The
confidence interval for the Chao estimator was 0.572–1.679; that of the summation estimator was 0.847–1.137 (Fig. 2), closer to the precision of the matrix model (95% CI = 0.99–1.09).
Table 1
Number of unduplicated females with cubs of the year (m), total number (n) of sightings of m grizzly bears, number of m bears seen i times (fi), and number of females estimated from summation (n̂sum)
and Chao (n̂Chao) monitoring-based estimators.
Sensitivity analysis showed that random fluctuation in bear sightings caused the trend estimates of both monitoring-based estimators to deviate little from zero. Mean differences between simulated
and observed λ were close to zero for both the Chao (Δλ = 0.0217, SD = 0.0676) and the summation (Δλ = -0.0002, SD = 0.0340) estimator (Fig. 3).
The bias-corrected Chao estimator inflates raw counts of known F[cub] seen only once, but reduces this inflation by females seen twice. In this study most F[cub] were seen more than twice per year (
Table 1), implying that relatively few escaped detection; thus, the Chao-estimated populations are largely the same as actual annual F[cub] counts. This may be a relatively common scenario for
relatively small populations in well-studied areas. We also note that n/m ratios (Table 1) differed widely across years. This variability is due to differences in visibility of individual bears, not
differences in search effort. In certain years highly habituated animals or those whose home ranges made them especially visible (e.g., from park roads) were observed almost daily due to their
proximity to people.
We cannot determine which of the 3 methods (2 monitoring-based estimators and the demographic model) best approximates the true population growth rate. As presented here, the results of the Chao
estimator-based analysis are unusable due to their extremely low precision. The wide confidence intervals are due to high inter-annual variance in population estimates which, again, may be fairly
common for small populations. For example, a random fluctuation in 2 bear sightings per year has much more effect on abundance estimates in a small population (40% change from n = 5) than in a
larger one (4% change from n = 50). Importantly, the performance of any estimator varies with population size (Keating et al. 2002). Thus, precision should increase positively with population size,
introducing a potential source of bias in the resulting estimates of λ. No study has yet explored the implications of such bias for management (K. Keating, personal communication, 2006). Moreover
trend analysis using this estimator cannot account for variation in the proportion of females that breed each year. The λ̂ point estimate from the demographic study indicates 4% annual growth. This
implies a 40% increase in abundance over the duration of the study, but no such increase was noticed. Exclusion of stochasticity and density dependence from their model, as well as imprecision in
age-of-senescence estimates, could have biased their λ̂ upwards (Garshelis et al. 2005b).
Though the summation estimator appears to give more precise trend estimates than the Chao estimator, those of the former may be artificially narrow. Inter-annual variance in population estimates is
necessarily reduced in a running sum, and to our knowledge no methods account for this in the calculation of confidence limits. Furthermore, given that the female bear population in Banff–Kananaskis
is relatively small and that the summation method relies on summing F[cub] across the average interbirth interval, this method risks random bias due to over-counting or under-counting bears whose
interbirth interval differs from four years (D. Garshelis, University of Minnesota, Grand Rapids, Minnesota, USA, personal communication, 2005).
Sensitivity analysis shows that the bias of both monitoring-based estimators was relatively unaffected by random variation in annual counts. Nevertheless, the variance in the differences between
simulated and observed λ̂ was relatively high, implying that the trend predictions of both estimators will be importantly affected by, for example, observation error. While the variance of the
summation estimator was lower than that of the Chao estimator, this may again be an artifact for the reasons discussed above.
Only through continued F[cub]-based monitoring will we learn the minimum time-series length necessary for reasonably precise trend estimates; though if trends are not monotonic during this time, we
will likely have very little power to detect thresholds or inflection points. Indeed, trend estimation from monitoring data is often bedeviled by low power (see Doak 1995).
For F[cub]-based monitoring to be useful, even over longer time-series, we strongly recommend methods to remove sampling variance and other forms of observation error before estimating confidence
intervals. Observation error refers to inaccuracies in population size and trend estimation, part of which (sampling variation) comes from the extrapolation of subsets of the population up to the
entire population (Morris and Doak 2002). Some facets of observation error can be reduced by careful attention to detail and accuracy during data collection; there are also techniques to reduce
sampling variation during the annual censuses (Morris and Doak 2002). Also, recently developed statistical models allow researchers to separate sampling variance from environmental stochasticity
using relatively short (15–20 year) time-series (de Valpine and Hastings 2002, Lindley 2003, Holmes 2004). For example Lindley (2003) shows that a time-series of F[cub] counts in Yellowstone can be
converted to state space (where the population process and observation process are modeled separately) and, through the application of a Kalman filter (Harvey 1989), likelihood functions can be
generated to partition total variance into process and sampling components. It may be possible to apply these methods after only a few more years of F[cub]-based monitoring in the Banff–Kananaskis
Demographic and monitoring-based estimators remain 2 of the most common methods by which biologists monitor populations, yet few studies compare results generated from the 2 methods. Thus, it is
difficult for biologists using one monitoring method to assess how their results would change using an alternative method. In Banff National Park and Kananaskis Country, biologists have conducted a
decade-long demographic study of radiocollared grizzly bears that was terminated due to funding and political pressure. It was not feasible to continue the demographic study over the long term, but
it may be possible that an intensive F[cub]-based monitoring program would eventually be able to confidently estimate population trends and the magnitude of process variation.
A decade of research was funded and guided by the many partners contributing to the Eastern Slopes Grizzly Bear Project. E. Crone gave advice on the analysis presented here; the manuscript also
benefited greatly from comments by R. Harris, K. Keating, A. Ordiz, P. McLoughlin, C. Schwartz, D. Garshelis, M. Kauffman, J. Maron, and G. White. These analyses were partly funded by Parks Canada.
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[Solved] Find the equation of the parabola whose: focus is (2,... | Filo
Find the equation of the parabola whose:
focus is (2, 3) and the directrix x − 4y + 3 = 0.
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Let P (x, y) be any point on the parabola whose focus is S (2, 3) and the directrix is x − 4y + 3 = 0.
Draw PM perpendicular to x − 4y + 3 = 0.
Then, we have:
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Find the equation of the parabola whose:
Question Text focus is (2, 3) and the directrix x − 4y + 3 = 0.
Updated On Dec 30, 2022
Topic Conic Sections
Subject Mathematics
Class Class 11
Answer Type Text solution:1 Video solution: 1
Upvotes 155
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Math Worksheets - Homeschooling In Indiana
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117. A numerical tool for obtaining wave eigenvalues in non-uniform solar waveguides
Author: Samuel Skirvin, Viktor Fedun & Gary Verth from the University of Sheffield.
<< previous nugget — next nugget >>
The modern ground- and space-based instruments (DST, SST, DKIST, SDO, Hinode, Solar Orbiter) provide solar physicists with ample observations of solar plasma processes, i.e. magnetic bright points,
spicules, plasma flows, structure of magnetic fields etc. at different temporal and spatial scales. However, direct measurements of important plasma properties such as e.g. magnetic field strength in
the corona, using traditional observational techniques is incredibly difficult. Fortunately, magnetohydrodynamic (MHD) waves which permeate almost all structures observed in the solar atmosphere can
be used as a proxy to determine the properties of the plasma, through a tool known as solar magnetoseismology. Therefore, advanced theoretical modelling becomes essential to explain the ever
increasing quality of observational results and provide more accurate information about MHD wave propagation and solar atmospheric plasma properties.
MHD waves in a spatially non-uniform plasma
When it comes to an analytical description of wave properties in a solar plasma, the traditional technique of solving the linearised MHD equations for small perturbations is usually adopted. This
method ultimately obtains a dispersion relation which relates the frequency of the wave and its wavenumber along with known characteristic properties of the background plasma. Pioneering work [1]
first applied this in a solar context for a uniform magnetic slab model and provided an analytical description of magnetoacoustic kink and sausage modes split into two physical categories namely
surface and body waves. In a non uniform magnetic waveguide however (see Figure 1) the governing differential equations develop coefficients which are spatially varying, along the coordinate of
inhomogeneity. As a result, the governing equations now have no known closed form analytical solution and consequently no dispersion relation can be obtained, therefore, a numerical approach must be
Figure 1. Example cartoon of a solar waveguide modelled as a non-uniform magnetic slab. The interior region in this work is allowed to be spatially inhomogeneous, the magnetic field is assumed
vertical and constant but of different magnitude inside and outside the slab.
Numerical approach
We present a numerical approach based on the shooting method and bisection method to obtain the eigenvalues for a magnetic slab with an arbitrary non-uniform background plasma and/or plasma flow [2].
Real frequencies are obtained such that information about trapped modes of the system can be analysed, complex frequencies such as those in the leaky or continuum regimes are left for future work.
The initial wave phase space is used as the domain to find eigenvalues that provide exact solutions to satisfy the relevant boundary conditions of the waveguide, namely the continuity of perturbation
of radial displacement and total pressure. Additional information from the definitions of the sausage mode and the kink mode are utilised to obtain the relevant (anti-)symmetric eigenfunctions. The
governing equations describing these properties are derived and solved numerically as no analytical solution exists without making simplifying assumptions about the model. The values of wave
frequency and wavenumber that satisfy both boundary conditions simultaneously will be classified as a solution and used in further analyses of the wave modes.
This is an extremely powerful numerical tool as, provided the initial equilibrium is stable, a wave analysis of any non-uniform or non-linear plasma can be investigated without the need for a
dispersion relation. It should also be noted that this numerical approach is not limited to a purely planar geometry, a cylindrical or spherical geometry would only modify the mathematical vector
operators used in the initial analytical description – the physics of the numerical tool still remains the same.
Figure 2. Radial spatial profiles of background plasma density considered in this work for a coronal slab. The width of the Gaussian profiles decreases with colour such that black line represents the
uniform case (W=1e5) and the red curve denotes the extreme non-uniform case (W=0.9). The blue curve models a sinc(x) profile which has been observed in magnetic bright points (MBPs).
Non-uniform plasma density in a coronal slab
We investigate the properties of magnetoacoustic waves in a coronal slab with a non-uniform background plasma flow modelled with the profiles in Figure 2. A sinc(x) profile models the spatial
distribution seen in intensity images of magnetic bright points [3]. The width of the Gaussian profiles is determined by a parameter W, where a smaller W indicates a more inhomogeneous profile. The
numerical algorithm obtains the eigenvalues plotted on the dispersion diagram (Figure 3) for each case which allows the resulting eigenfunctions for total pressure and horizontal perturbation of
velocity to be calculated.
Figure 3. Dispersion diagrams for all non-uniform cases shown in Figure 2. Blue curves represent the kink mode whereas the red curves show the sausage mode. (a) Denotes the uniform dispersion diagram
of a coronal slab, (b-d) an increasingly narrower Gaussian, and (e) the sinc(x) profile. The shaded bands denote non-uniform regions where the characteristic speeds have varying frequencies. Orange
region shows the Alfven continuum, blue region the cusp continuum, green region shows the non-uniform band of sound speed.
In Figure 4 the eigenfunctions for the different cases of non-uniform equilibria are shown for the slow body sausage and kink modes. It can be seen by comparing the uniform and extreme non-uniform
cases that additional nodes and points of inflexion are present for the non-uniform equilibrium case which may cause difficulty when interpreting the wave modes of observed/simulated perturbed
velocity fields in non-uniform waveguides.
Figure 4. Resulting eigenfunctions for the slow body sausage mode (left) and slow body kink mode (right) for the non-uniform cases shown in Figure 2 where the colour scheme is consistent.
• A numerical approach has been developed to obtain the eigenvalues for magnetoacoustic waves in an arbitrary symmetrically non-uniform magnetic slab.
• The algorithm is heavily tested against numerous well known analytical results and successfully obtains the correct eigenvalues for both the sausage and kink modes in the long and short
wavelength limits.
• Investigations of non-uniform plasma and background flow modelled as a series of Gaussian profiles reveal that slow body modes are more affected by the non-uniform equilibria. For investigation,
analysis and discussion into the non-uniform background flow case, refer to [2].
• Additional nodes and points of inflexion appear in the resulting eigenfunctions which may be of interest to observers when interpreting observational results of MHD waves in the highly structured
non-uniform solar atmosphere.
• Future work investigating similar cases to those considered in this work but in the case of a cylindrical model can be found in [4].
• [4] Skirvin et al. (2021b), MNRAS, submitted | {"url":"https://www.uksolphys.org/uksp-nugget/117-a-numerical-tool-for-obtaining-wave-eigenvalues-in-non-uniform-solar-waveguides/","timestamp":"2024-11-11T03:29:41Z","content_type":"text/html","content_length":"41936","record_id":"<urn:uuid:750ab1c4-dfbb-436e-b2af-80878475b082>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00558.warc.gz"} |
M* Path Algorithm
M-Star Path Algorithm
The M-Star (M*) Path Algorithm is a Patent Protected Algorithm that is aimed at finding a route between two points in a graph or grid without exploration of the entire domain.
In its basic form, the algorithm makes use of a matrix that defines the grid (usually in 2D but can equally be used in 3D or even n-D) through a parametric representation of all the nodes and
The path between the two points is a property of the matrix and as such only a single segment (i.e. the next node in the path) can be computed -or the entire path- depending on the situation at hand.
Since the matrix is dependent only on the grid being used, it can be pre-calculated and stored without requiring an update when the two points are modified.
Further use of the Algorithm includes it use in Obstacle Avoidance (where 'No-Go' regions may be dynamically added or removed without re-computation of the grid) and in Social Grids (where the
shortest distance is less of a concern rather than whether two points are connected at all).
In general, the M* Path Algorithm (once the grid has been defined) achieves a time complexity as low as O(S*C) (where S is the number of steps that it takes to connect the points and C is the average
number of connections at each of those steps). Determination of the 'next' step from a current location towards a destination has thus a complexity of O(C).
The purpose of this site is to provide information to interested parties.
Currently, the algorithm has been implemented into development libraries and the focus is on finding a partner to support further work or an interested party to license/purchase the algorithm for
further commercial development and implementation.
At this time, only invited users may access the details and overview. If you are interested in this algorithm for commercial use, please use the Contact page to get in touch and request further
Note: this site is best viewed in FireFox 45 or Internet Explorer 11
Already received an invitation? | {"url":"http://mstarpath.com/Default.aspx?ReturnUrl=%2f","timestamp":"2024-11-03T18:16:15Z","content_type":"application/xhtml+xml","content_length":"10259","record_id":"<urn:uuid:74c739cb-d0ec-43f2-970c-1814c2846d27>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00564.warc.gz"} |
Statistical inference issues
Statistics is an infamously difficult field of mathematics (or, as some pure mathematician will say, of "pretend mathematics") in that many of its conclusions seem very counter-intuitive at first.
Human brains naturally operate on deduction: we hold a model of the world in our brain and apply this model to estimate something, to choose between two actions, et cetera. Statistics, on the other
hand, is inductive and inferential: we look at the data and try to understand how it came about. Unfortunately, induction is much harder than deduction: deduction always leads to a definitive
conclusion, while induction, in principle, may have us consider infinity of possibilities before we land on something that works.
Here is an example illustrating the difference.
Deduction: "What is 71 * 83?" The answer is 5893, and a strong middle school student should be able to do the math by hand and obtain the answer.
Induction: "What are the two prime factors that, when multiplied, give us 5893?" The answer is 71 and 83, and even professional number theorists will be stumped by this question.
In both cases we are talking about the same equation: 71 * 83 = 5893. However, in the first case we calculate the product of two known numbers, and in the second we look for two numbers the product
of which equals the known number. In the first case we look for the answer; in the second case we look for the question giving the answer.
As a result, statistics is frequently misused by people leading them to obtain bogus conclusions. I highly recommend Darrell Huff's "How to Lie with Statistics" book illustrating just how common this
is. Virtually all statistical claims you run into in newspapers or popular science articles are, at best, misleading, and at worst, flat-out wrong.
Here I would like to talk about the particular error people make: ignoring prior probabilities when choosing between multiple hypotheses. According to the Bayes Theorem (which is a hard mathematical
result that can be rigorously derived), probabilities of different models being true have to be adjusted by the probabilities of their conditions being true. People often forget about this adjustment
and, therefore, pick the wrong model.
Here is an illustration.
Suppose you have a bag of 1000 coins, and 1 of those coins is two-headed while 999 other coins are fair. You pull out 1 coin from the bag, flip it 6 times and get 6 heads. What is more likely, that
you pulled out a two-headed coin, or a fair coin?
Wrong reasoning: "If this was a two-headed coin, the probability of getting 6 heads would be 100%. Otherwise, it would be less than 2%. I should bet on it being a two-headed coin."
Correct reasoning: "There are two possibilities: I pulled out a two-headed coin, then got 6 heads - and I pulled out a one-headed coin, then got 6 heads. The probability of the first outcome is 0.1%,
and the probability of the second outcome is a little less than 2%. I should bet on the coin being fair."
What happened here? In the wrong reasoning one forgot to take into account the fact that the vast majority of the coins are fair, and that fact changes the calculation entirely. In the correct
reasoning, one realized that, even though with a fair coin the probability of getting 6 heads in a row is low, the odds of pulling out a double-headed coin in the first place are far lower than that
This error is not just made in mathematical calculations, but it comes up all over the place: when choosing between different medical treatments, when deciding between different business decisions...
and when choosing between multiple historical hypotheses.
Let us talk about "Jesus rising from the grave". According to our beloved Doctor William Lane Craig (from his first appearance on Alex O'Connor's podcast), between two explanations - naturalistic and
religious - one should pick the one that best explains the evidence. He argues that the naturalistic explanation struggles with explaining the allegedly found Jesus' empty grave and a bunch of eye
witness accounts records of which are spread across centuries - while religious explanation does not have this issue. Hence we should go with the religious one.
Believe it or not, but he makes the same error as in the coin example. He simply asks, "Given A is true, what is the probability of the observed outcome? And what is it given B is true? The latter is
bigger than the former, so we should choose B over A". He does not take into account the fact that A and B might have the same probabilities of being true to begin with.
But what are those probabilities? What does it even mean to talk about prior probability that naturalism is true? Well, in Bayesian statistics we look at the evidence of similar things happening and
ask ourselves, "How often historically did A happen and how often did B happen?" Now, even the most prominent Christian advocates such as Dr Craig admit that, aside from Jesus', there are no known
cases of human resurrection. Furthermore, the claim that Jesus was resurrected is exactly the hypothesis we test, so we cannot use it as evidence of itself... Therefore, here our prior probability
has to be incredibly low: since there are no other known cases of human resurrection despite tens of billions of humans having died throughout the recorded history, there are few reasons to believe
that it is even possible.
On the other hand, history is littered with examples of erroneous claims, false witness testimonies, misunderstood writings, straight out fantasies and so on. The prior probability of this being just
one more such story is not 100%, of course, but it is very high - since it has happened to be the case in every single instance ever recorded.
So now we have two hypotheses: naturalism, and religionism. The naturalistic hypothesis struggles a little with explaining a couple of historical facts, but given other observations, there is a very
high chance that it still holds and the facts can be explained eventually. On the other hand, religious hypothesis explains everything perfectly well if it holds - but we have absolutely zero data
supporting it even being possible in principle, let alone actually occurring here. In the first case we have an explanation that has always worked well and just struggles a little here; in the second
case we have an explanation that has never worked at all, but if it worked, it would explain everything. Which one is better?
Instead of answering the question, I will suggest yet another analogy, a food for thought.
For the past 100 days my friend has been meditating every morning, until yesterday. Yesterday she skipped her meditation, and in the evening got a raise at her job. Consider two competing hypotheses:
1) Meditating exactly for 100 days and then stopping always causes one to get a raise.
2) It is a coincidence.
Which is a better explanation of what happened?
Post Argument Now Debate Details + | {"url":"https://debateisland.com/discussion/9941/statistical-inference-issues","timestamp":"2024-11-09T07:16:43Z","content_type":"text/html","content_length":"59796","record_id":"<urn:uuid:fbeb77eb-ca7a-4db9-a6d6-f14c5aacbfe8>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00025.warc.gz"} |
Comments on Aggregated Intelligence: BODMAS or PEMDAS: It is elementary!As a math teacher, I feel sad reading this.
Both ...Result = p*q/s+r*t/y*t; //where p=9, q=8, r=7, s=6...100% agree with this statement:
As others have st...Still a lot of people literally follow PEMDAS or B...Rao
48 - 48+3
48- 51
The - is just a sign to be ...Lol u didn't go left to right, that's ur p...As others have stated, you have worked through the...I don't use a specific procedure, the way I ge...The rule is simple - when you have operations of t...PEMDAS and BODMAS are not rules of math. They are...Why not! It can be the part of denominator. Bodmas...Using PEMDAS for multiplication and division requi...M and D have the same rank as well as A and S are ...You did pemdas wrong. Multiply & divide left ...PEMDAS and BODMAS ..... reverse the order of multi...1+2/1 or 1/1+2 . 😅 Actually, we can have a different answer in the si...A late comment ... PEMDAS and BODMAS are just m...Everyone is missing the whole point of brackets. ...6÷2(3) is not equal to 6÷2x3
In first case the sim...Maam u have to add all the positive numbers togeth...Maam u have to add all the positive numbers togeth...Hi,
PEMDAS = Parenthesis, Exponent, Multiplication...In pemdas you are supposed to multiply and divide ...The BODMAS calculator at https://play.google.com/s...
tag:blogger.com,1999:blog-7656888.post8680405049934213934..comments2024-10-26T01:47:42.386-06:00Raj Raohttp://www.blogger.com/profile/
02773005045522319448noreply@blogger.comBlogger71125tag:blogger.com,1999:blog-7656888.post-56751641845580314152018-11-27T21:09:37.139-07:002018-11-27T21:09:37.139-07:00As a math teacher, I feel sad
reading this.<br /><br />Both BODMAS and PEMDAS are used to help children learn mathematics. As you grow, you can ignore these acronyms.<br /><br />I will try to show how to use these acronyms using
the problems given in this forum.<br /><br />8÷2(2+0)<br /><br />BODMAS:<br />=8÷2(2)<br />=4(2)<br />=8<br /><br />PEMDAS:<br />=8÷2(2)<br />=16÷2 (note that 8÷2 can be seen as a fraction, 8÷2 = 8/
2. When we multiply a fraction with a number, we just multiply the numerator so 8/2 x 2 = 16/2)<br />=8<br /><br />Rule 3:<br />=8(1+0)<br />=8(1)<br />=8<br /><br />Rule 4:<br />=(16+0)÷2<br />=16÷2
<br />=8<br /><br />Rule 5:<br />=4(2+0)<br />=4(2)<br />=8<br /><br />Rule 6:<br />=4(2+0)<br />=(8+0)<br />=8<br /><br />How many rules do you need to learn?<br />If you are still bounded to these
acronym, then you are not supposed to go to secondary school yet. Master your basic on primary school first.Lhttps://www.blogger.com/profile/
14639620140517958961noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-65863324054498799352018-10-21T03:29:58.386-06:002018-10-21T03:29:58.386-06:00Result = p*q/s+r*t/y*t; //where p=9, q=8, r=
7, s=6, t=5, y=5<br /><br />= 9*8/6+7*5/5*5<br />= 72/6+35/5*5<br />= 12+7*5<br />= 12+35 = 47<br />Using BODMAS, I get 47<br /><br />= 9*8/6+7*5/5*5<br />= 72/6+35/25<br />= 12+1.4 = 13.4<br />Using
PEMDAS, I get 13.4Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7656888.post-64066475961610388402018-06-08T15:12:51.518-06:002018-06-08T15:12:51.518-06:00100% agree with this statement:<br />
<br />As others have stated, you have worked through the problem incorrectly using PEMDAS. Yes, the parentheses will be solved first (2), however then you complete the equation left to right if there
are no exponents, doing either division or multiplication (whichever comes first, left to right). Once all multiplication and division is completed left to right, you complete any remaining addition/
subtraction left to right.<br /><br />You do PEMDAS and BODMAS the same - in the direction the equation is written. They'll both give you the correct answer.<br /><br />Like I said before try
Excel and you will get the correct answer.GLallyhttps://www.blogger.com/profile/
10144188975292080730noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-41941485718116417692018-06-08T15:08:53.491-06:002018-06-08T15:08:53.491-06:00Still a lot of people literally follow
PEMDAS or BODMAS. When it comes to group of Multiplication and Divide in an equation, you should solve left to right. Same goes for Addition and Subtraction. Therefore:<br /><br />4-6/3+5*3<br />=
4-2+15<br />= 2+15<br />= 17 Correct<br /><br />If you have a doubt put the same equation in MS Excel and get your answer. Also read this article with examples.<br /><br />https://www.mathsisfun.com/
10144188975292080730noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-91579620933528777152018-05-22T14:40:38.582-06:002018-05-22T14:40:38.582-06:00Rao<br />48 - 48+3<br />48- 51<br /><br />
<br />The - is just a sign to be used after addition in Bodmas.<br />Addition before subtractionAnonymoushttps://www.blogger.com/profile/
13363696386727168344noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-41300008915725857782017-10-01T12:18:36.524-06:002017-10-01T12:18:36.524-06:00Lol u didn't go left to right, that'
s ur problem. Funny thing is you were so matter of fact about your answer and didn't even bother to follow the steps. PEMDAS and BODMAS are literally the same but I like pemdas cause I think it
sounds betterAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-7656888.post-17438320663175283792017-09-19T00:45:50.783-06:002017-09-19T00:45:50.783-06:00As others have stated, you have worked
through the problem incorrectly using PEMDAS. Yes, the parentheses will be solved first (2), however then you complete the equation left to right if there are no exponents, doing either division or
multiplication (whichever comes first, left to right). Once all multiplication and division is completed left to right, you complete any remaining addition/subtraction left to right.<br /><br />You
do PEMDAS and BODMAS the same - in the direction the equation is written. They'll both give you the correct answer.mortalsolacehttps://www.blogger.com/profile/
18336912885516733936noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-27945008392283210522017-09-06T05:03:24.178-06:002017-09-06T05:03:24.178-06:00I don't use a specific procedure, the
way I get it right according to a calculator is that you do brackets first, and then rewrite the equation with the answer of the brackets in place for example 3+8÷2(2+0) would be 3+8÷2x2 as I
answered the brackets. Then you would answer all multiplication problems from left to right which would make the above equation equate to 3+8. Then you proceed to complete any addition or subtraction
from left to right again and you will come to your answer of 11, check it on a calculator.Anonymoushttps://www.blogger.com/profile/
13860254437134401828noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-53551410762937627512017-08-19T15:57:12.590-06:002017-08-19T15:57:12.590-06:00The rule is simple - when you have
operations of the same level do them starting from the left. So it doesn't matter division or multiplications - they are of the same level of importance so just do one on the left first, then the
next which becomes left. Example: 6/2/1*2/2*4*5/2 => 6/2 => 3/1*2/2*4*5/2 => 3/1 => 3*2/2*4*5/2 => 3*2 => 6/2*4*5/2 => 6/2 => 3*4*5/2 => 12 => 12*5/2 => 12*5 => 60
/2 = 30. This is how it works and either BODMAS(BEDMAS) and PEMDAS will work fine.Anonymoushttps://www.blogger.com/profile/
02688789383630898518noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-8864953072805285422017-08-16T18:29:02.472-06:002017-08-16T18:29:02.472-06:00PEMDAS and BODMAS are not rules of math. They
are both just acronyms used to try to apply the SAME order of operations rules. They CANNOT and DO NOT provide different results. They also mean exactly the same thing.<br />Parentheses or Brackets
<br />Exponents or Orders<br />Multiplication and Division or Division and Multiplication<br />Addition and Subtraction<br /><br />MD means exactly the same thing as DMAnonymoushttps://
www.blogger.com/profile/08972644168705988560noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-63820633465187600562017-08-15T03:40:08.439-06:002017-08-15T03:40:08.439-06:00Why not! It can be
the part of denominator. Bodmas or bedmas is the correct way.. According to you it is actually (6/2)(1+2).. so actually you are not following pemdas..you are just adding parenthesis and seperating
07676281656176386463noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-22690625399339482122017-08-15T02:22:03.285-06:002017-08-15T02:22:03.285-06:00Using PEMDAS for multiplication and division
requires a rule::You have to start solving from left to right... <br /><br />On the other hand using BODMAS does not require any rule.. Just do the division before you do the multiplication.. Whether
left to right or right to left or middle to anywhere as long as division is done before multiplication.. Eg:<br /><br /> 6/2*3 =9 <br /><br />6*2/3 using PEMDAS: (6*2)/3=4<br />6*2/3 using BODMAS: 6*
(2/3)=4<br /><br /><br />Just make sure to start from left to right while using PEMDAS... on the other hand, while using BODMAS, do not worry, you will get the correct solution. Anonymoushttps://
www.blogger.com/profile/11881572900260163504noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-57500738015296086622017-07-23T18:15:14.895-06:002017-07-23T18:15:14.895-06:00M and D have the
same rank as well as A and S are on the same rank.<br />Whoever is at the left side will be the commanding officer. So, as the officer in command will get the priority to lead.<br />Anonymoushttps://
www.blogger.com/profile/00234271459604291647noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-61350170371165217552017-07-21T08:33:29.088-06:002017-07-21T08:33:29.088-06:00You did pemdas
wrong. Multiply & divide left to right you will also get 8.Anonymoushttps://www.blogger.com/profile/
05658248134377006718noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-90124990560228150262017-07-19T13:30:50.089-06:002017-07-19T13:30:50.089-06:00PEMDAS and BODMAS ..... reverse the order of
multiplication vs division ..... but in REALITY they are EQUAL ...not actually prioritized because the order is opposite for dm and md. So the ONLY way to get the same answer is to solve the equation
left to right and treat multiplication and division as equal. The correct way to write the acronym cannot be linear.<br />tblounthttps://www.blogger.com/profile/
06918516185618471984noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-12308551319907432472017-06-17T20:29:43.989-06:002017-06-17T20:29:43.989-06:001+2/1 or 1/1+2 . 😅 Anonymoushttps://
www.blogger.com/profile/09293363258342416530noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-39589139959001470752017-06-12T14:05:02.319-06:002017-06-12T14:05:02.319-06:00Actually, we can
have a different answer in the simple math problem of 8÷2×2.<br /><br />Using PEMDAS will give you an answer of 2, but using BODMAS will give you 8.<br /><br />
Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7656888.post-73663059215237635662017-06-04T11:27:24.799-06:002017-06-04T11:27:24.799-06:00A late comment ... PEMDAS and BODMAS are just mnemonics
help children learn precedence of operations. They have the same effect!<br /><br />Division and multiplication are equivalent operations, since you can replace them in an equation, as is often done
when solving problems:<br /><br />6÷2 is the same as 6x(1/2)<br /><br />Addition and subtraction have a similar relationship:<br /><br />6 - 2 is the same as 6 + (-2) <br /><br />So when dealing with
multiplication and division (or addition and subtraction) neither takes precedence over the other, and you perform the calculation from left to right.<br /><br />Part of the problem (confusion?) is
from the typesetting of the content<br /><br />6÷2x(2+1) can be written (as on a piece of paper) as:<br /><br /> 6 <br />--- x (2+1) = 9<br /> 2<br /><br />6÷2x(2+1) is NOT<br /><br /> 6<br />
--------<br /> 2(2+1)<br /><br /><br />The issue stems from how we visualise the problem. Have a look at https://en.wikipedia.org/wiki/Order_of_operationsrhphttps://www.blogger.com/profile/
09321590669006883483noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-88813850372430885052017-04-04T15:31:00.463-06:002017-04-04T15:31:00.463-06:00Everyone is missing the whole point of
brackets. Division and multiplication are equal. And addition and subtraction are equal. When writing an equation you write it from left to right and you insert brackets if anything needs to be done
in a different order than left to right. Keeping in mind that div/multi comes before add/Sub.<br /><br />6÷2x(2+1) would be written like this:<br /><br />6÷2(2+1)<br />6÷2(3)<br />6÷2x3<br />3x3<br
/>9<br /><br /><br />The wrong way people do it is by forgetting multi and division are equals and you do whichever of them is farther left first unless there's a bracket.<br /><br />6÷(2×(2+1))
is the equation you're solving when you're doing 6÷2(2+1) the wrong way. 6÷2(3) the 3 is in brackets does not mean you multiply it first . The brackets at that point are moot and it's
just a multiplication. 6÷2×3 . Now left to right because there are no more brackets and all the equations are even. You could convert it all from a mix of division to only containing multiplication
(to 6 × 1/2 × 3 ) and you'll still get 9. But that's for advanced mathematics.<br /><br /><br />6÷(2x3)<br />6÷6<br />1<br /><br />Or<br /><br />(1/6)(2×3)<br />(1/6)(6)<br />1<br /><br />
This is actually correct 100% guaranteed<br />Anonymoushttps://www.blogger.com/profile/
16020220159590264310noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-68454333398550314582017-02-23T09:43:44.378-07:002017-02-23T09:43:44.378-07:006÷2(3) is not equal to 6÷2x3<br />In first
case the simplification is 6÷6=1 and in the next case it is 3x3=9FIRSANhttps://www.blogger.com/profile/
01069151080169491097noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-746619659389907952017-02-08T20:32:39.785-07:002017-02-08T20:32:39.785-07:00Maam u have to add all the positive numbers
together and then the negative numbers together and then substract.u cannot add -48 and +3 and even if u do that u donot get +51!!u get-45!!Foodies' Paradisehttps://www.blogger.com/profile/
13269342023070692649noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-61411598603262789892017-02-08T20:29:52.194-07:002017-02-08T20:29:52.194-07:00Maam u have to add all the positive numbers
together and then the negative numbers together and then substract.u cannot add -48 and +3 and even if u do that u donot get +51!!u get-45!!Foodies' Paradisehttps://www.blogger.com/profile/
13269342023070692649noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-23175258156545304612017-01-14T02:35:35.760-07:002017-01-14T02:35:35.760-07:00Hi,<br />PEMDAS = Parenthesis, Exponent,
Multiplication, Division, Addition and Substraction;<br />BOMDAS (Brackets, order, multiplication/division, addition/subtraction);<br /><br />are not the unique expressions. They incite us to think
that a priority exists between multiplications and divisions (same for additions and substractions). They confuses, they are not the unique names:<br /><br />PEMDAS = PEMDSA = PEDMAS = PEDMSA<br />
BOMDAS = BOMDSA = BODMAS = BODMSA<br /><br />A unique name should not be used to indicate the priority of the operations.<br /><br />Best regards.VERDAD VERDADERAhttps://www.blogger.com/profile/
06450092998338648973noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-8922252125589350922017-01-08T00:29:28.736-07:002017-01-08T00:29:28.736-07:00In pemdas you are supposed to multiply and
divide left to right so for your equation 6÷3×2=4 6÷3=2 2×2=4. Do you see where you get it they both give the same answer but the more everyone trys to complicate it the more confusion we creat. By
just simply doing it in two steps its easier. It does not matter if you have division fist or multiply first in the equation as long as you go from left to right. <br />If you just follow pemdas with
out remembering that when you reach md and as you do that from left to right if you have add before subtract you add first if you have divide before multiply you divide first. Everyone is trying to
make ot so hard yet its so easyAnonymoushttps://www.blogger.com/profile/
03246247829199016276noreply@blogger.comtag:blogger.com,1999:blog-7656888.post-11682135070511424802017-01-07T07:13:30.537-07:002017-01-07T07:13:30.537-07:00The BODMAS calculator at https://
play.google.com/store/apps/details?id=com.mbradley.sumtree.view will provide interactive checks on results.<br /><br />Its a nice way to look at the BODMAS/PEDMAS questions.<br /><br />thanks<br /> | {"url":"https://blog.aggregatedintelligence.com/feeds/8680405049934213934/comments/default","timestamp":"2024-11-07T14:01:34Z","content_type":"application/atom+xml","content_length":"54146","record_id":"<urn:uuid:e1d2a0a8-af01-4c19-841f-b3c6e948558c>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00177.warc.gz"} |
Michael Harris : L-functions and the local Langlands correspondence
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Michael Harris : L-functions and the local Langlands correspondence
Henniart derived the following theorem from his numerical local Langlands correspondence: If $F$ is a non-archimedean local field and if $\pi$ is an irreducible representation of $GL(n,F)$, then,
after a finite series of cyclic base changes, the image of $\pi$ contains a fixed vector under an Iwahori subgroup. This result was indispensable in all demonstrations of the local correspondence.
Scholze gave a different proof, based on the analysis of nearby cycles in the cohomology of the Lubin-Tate tower (and this result also appears, in a somewhat different form, in proofs based on the
global correspondence for function fields). An analogous theorem should be valid for every reductive group, but the known proofs only work for GL(n). I will sketch a different proof, based on
properties of L-functions and assuming the existence of cyclic base change, that also applies to classical groups; I will also explain how the analogous result for a general reductive group is
related to the local parametrization of Genestier-Lafforgue.
0 Comments
Comments Disabled For This Video | {"url":"https://www4.math.duke.edu/media/watch_video.php?v=465eaca56d99c2d0c2a8376027cc8d7f","timestamp":"2024-11-11T10:08:28Z","content_type":"text/html","content_length":"48261","record_id":"<urn:uuid:91bf0254-21f1-468e-8221-dfe6df98474e>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00316.warc.gz"} |
Unscramble UGLN
How Many Words are in UGLN Unscramble?
By unscrambling letters ugln, our Word Unscrambler aka Scrabble Word Finder easily found 7 playable words in virtually every word scramble game!
Letter / Tile Values for UGLN
Below are the values for each of the letters/tiles in Scrabble. The letters in ugln combine for a total of 5 points (not including bonus squares)
What do the Letters ugln Unscrambled Mean?
The unscrambled words with the most letters from UGLN word or letters are below along with the definitions.
• lung (n.) - An organ for aerial respiration; -- commonly in the plural. | {"url":"https://www.scrabblewordfind.com/unscramble-ugln","timestamp":"2024-11-05T22:43:30Z","content_type":"text/html","content_length":"37160","record_id":"<urn:uuid:3252730a-c879-4d99-ba41-ddee2aa452c5>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00201.warc.gz"} |
Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to
the wavelength. The higher the photon's frequency, the higher its energy. Equivalently, the longer the photon's wavelength, the lower its energy.
Photon energy can be expressed using any energy unit. Among the units commonly used to denote photon energy are the electronvolt (eV) and the joule (as well as its multiples, such as the microjoule).
As one joule equals 6.24×10^18 eV, the larger units may be more useful in denoting the energy of photons with higher frequency and higher energy, such as gamma rays, as opposed to lower energy
photons as in the optical and radio frequency regions of the electromagnetic spectrum.
Photon energy is directly proportional to frequency.^[1] ${\displaystyle E=hf}$ where
This equation is known as the Planck relation.
Additionally, using equation f = c/λ, ${\displaystyle E={\frac {hc}{\lambda }}}$ where
The photon energy at 1 Hz is equal to 6.62607015×10^−34 J, which is equal to 4.135667697×10^−15 eV.
Photon energy is often measured in electronvolts. One electronvolt (eV) is exactly 1.602176634×10^−19 J^[3] or, using the atto prefix, 0.1602176634 aJ, in the SI system. To find the photon energy in
electronvolt using the wavelength in micrometres, the equation is approximately
${\displaystyle E{\text{ (eV)}}={\frac {1.2398}{\lambda {\text{ (μm)}}}}}$
since ${\displaystyle hc/e}$ = 1.239841984...×10^−6 eV⋅m^[4] where h is the Planck constant, c is the speed of light, and e is the elementary charge.
The photon energy of near infrared radiation at 1 μm wavelength is approximately 1.2398 eV.
An FM radio station transmitting at 100 MHz emits photons with an energy of about 4.1357×10^−7 eV. This minuscule amount of energy is approximately 8×10^−13 times the electron's mass (via mass–energy
Very-high-energy gamma rays have photon energies of 100 GeV to over 1 PeV (10^11 to 10^15 electronvolts) or 16 nJ to 160 μJ.^[5] This corresponds to frequencies of 2.42×10^25 Hz to 2.42×10^29 Hz.
During photosynthesis, specific chlorophyll molecules absorb red-light photons at a wavelength of 700 nm in the photosystem I, corresponding to an energy of each photon of ≈ 2 eV ≈ 3×10^−19 J ≈ 75 k
[B]T, where k[B]T denotes the thermal energy. A minimum of 48 photons is needed for the synthesis of a single glucose molecule from CO[2] and water (chemical potential difference 5×10^−18 J) with a
maximal energy conversion efficiency of 35%.
See also | {"url":"https://www.knowpia.com/knowpedia/Photon_energy","timestamp":"2024-11-14T04:00:49Z","content_type":"text/html","content_length":"86473","record_id":"<urn:uuid:eff30837-557f-40ef-8490-89ede44260aa>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00898.warc.gz"} |
Multiplying Decimals By Decimals
Multiplying Decimals By Decimals
Price: 300 points or $3 USD
Subjects: math,mathElementary,mathMiddleSchool
Grades: 5,6
Description: In this multiplying with decimals Boom card deck, students practice their multi-digit multiplication and decimal multiplication all in one. Equations include three numbers by two
numbers, three by three, four by two, and four by three. Decimal place values go beyond the thousandths place, as students practice solving each equation and placing the decimal point in the proper
place in the product. Students will calculate the product and type the response into the rectangle. This engaging and challenging Boom card deck will allow you to check for understanding and mastery
of multiplying with decimals. | {"url":"https://wow.boomlearning.com/deck/suAj7v9ND3T5D9wtc","timestamp":"2024-11-10T14:57:18Z","content_type":"text/html","content_length":"2409","record_id":"<urn:uuid:9041d3e5-60a9-48db-996e-60bf811e589d>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00332.warc.gz"} |
Yr8 math papers
yr8 math papers
Related topics:
Home multiply polynomials made easy | "trigonometry, ninth edition" pdf | mixed numbers to decimal converter | beginner fractions | trigonometry facts and
Exponents and Radicals trivias | solve equations with multiple variables and mapple | polynomial-programs in java language | intro to algebra 098 book pearson | poems on
Division of Radicals mathematical subjects | algebra 2 program
Exponents and Radicals
RADICALS & RATIONAL EXPONENTS
Radicals and Rational Exponents Author Message
Radical Equations
Solving Radical Equations NejhdLimks Posted: Sunday 31st of Dec 12:09
Roots and Radicals Hi math gurus! I am about two weeks through the semester, and getting a bit worried about my course work. I just don’t seem to
RADICAL EQUATION understand the stuff I am learning, especially things to do with yr8 math papers. Could somebody out there please enlighten me
Simplifying Radical Expressions with adding numerators, quadratic formula and distance of points. I can’t afford to pay for a tutor, but if anyone knows about
Radical Expressions other ways of mastering topics like angle suplements or graphing circles effectively, please drop me a line Thanks heaps
Solving Radical Equations Registered:
Solving Radical Equations 18.08.2005
Exponents and Radicals From: Bronx, NY
Exponents and Radicals
Roots;Rational Exponents;Radical
Solving and graphing radical oc_rana Posted: Tuesday 02nd of Jan 08:20
equations There are several topics comprising the general category of yr8 math papers, such as monomials, binomial formula and subtracting
Solving Radical Equations exponents. I have conversed with some people who gave up on the pricey options for help as well . However , do not despair because
Radicals and Rational Exponents I obtained an alternative resolution that is low-priced , easy to use and to a greater extent more practical than I would have
exponential_and_radical_properties ever supposed . After trials with demonstrative math computer software programs and nearly surrendering, I picked up Algebrator.
Roots, Radicals, and Root Registered: This software system has precisely furnished results to each mathematics problem I have brought to the software . But just as
Functions 08.03.2007 large , Algebrator as well allows all of the short-lived steps needed to deduce the ultimate solution . Although a user might use
Multiplication of Radicals From: the program just to dispose of class assignments , I dubiousness about someone should be granted permission to apply it for
Solving Radical Equations egypt,alexandria quizzes.
Radical Expressions and Equations
Equations Containing Radicals and
Complex Numbers Gools Posted: Tuesday 02nd of Jan 11:38
Square Roots and Radicals I have tried out quite a lot of software. I would without any doubt say that Algebrator has assisted me to come to grips with my
Solving Radical Equations in One difficulties on difference of squares, distance of points and conversion of units. All I did was to simply key in the problem. The
Variable Algebraically answer showed up almost right away showing all the steps to the answer . It was quite straightforward to follow. I have relied on
Polynomials and Radicals this for my algebra classes to figure out Remedial Algebra and Basic Math. I would highly advice you to try out Algebrator.
Roots,Radicals,and Fractional Registered:
Exponents 01.12.2002
Adding, Subtracting, and From: UK
Multiplying Radical Expressions
Square Formula and Powers with
Simplifying Radicals CHS` Posted: Thursday 04th of Jan 11:09
Exponents and Radicals Practice I would suggest using Algebrator. It not only assists you with your math problems, but also provides all the necessary steps in
Solving Radical Equations detail so that you can improve the understanding of the subject.
Solving Radical Equations
Solving Radical Equations
Lecture-Radical Expressions Registered:
Radical Functions 04.07.2001
From: Victoria City,
Hong Kong Island,
Hong Kong
InjoyMERVANO! Posted: Friday 05th of Jan 07:46
Great! I think that’s what I need . Can you tell me where to buy it?
From: Germany
Vild Posted: Saturday 06th of Jan 11:33
I am sorry; I forgot to give the link in the previous post. You can find the tool here https://mathradical.com/
From: Sacramento, CA
yr8 math papers
Related topics:
Home multiply polynomials made easy | "trigonometry, ninth edition" pdf | mixed numbers to decimal converter | beginner fractions | trigonometry facts and
Exponents and Radicals trivias | solve equations with multiple variables and mapple | polynomial-programs in java language | intro to algebra 098 book pearson | poems on
Division of Radicals mathematical subjects | algebra 2 program
Exponents and Radicals
RADICALS & RATIONAL EXPONENTS
Radicals and Rational Exponents Author Message
Radical Equations
Solving Radical Equations NejhdLimks Posted: Sunday 31st of Dec 12:09
Roots and Radicals Hi math gurus! I am about two weeks through the semester, and getting a bit worried about my course work. I just don’t seem to
RADICAL EQUATION understand the stuff I am learning, especially things to do with yr8 math papers. Could somebody out there please enlighten me
Simplifying Radical Expressions with adding numerators, quadratic formula and distance of points. I can’t afford to pay for a tutor, but if anyone knows about
Radical Expressions other ways of mastering topics like angle suplements or graphing circles effectively, please drop me a line Thanks heaps
Solving Radical Equations Registered:
Solving Radical Equations 18.08.2005
Exponents and Radicals From: Bronx, NY
Exponents and Radicals
Roots;Rational Exponents;Radical
Solving and graphing radical oc_rana Posted: Tuesday 02nd of Jan 08:20
equations There are several topics comprising the general category of yr8 math papers, such as monomials, binomial formula and subtracting
Solving Radical Equations exponents. I have conversed with some people who gave up on the pricey options for help as well . However , do not despair because
Radicals and Rational Exponents I obtained an alternative resolution that is low-priced , easy to use and to a greater extent more practical than I would have
exponential_and_radical_properties ever supposed . After trials with demonstrative math computer software programs and nearly surrendering, I picked up Algebrator.
Roots, Radicals, and Root Registered: This software system has precisely furnished results to each mathematics problem I have brought to the software . But just as
Functions 08.03.2007 large , Algebrator as well allows all of the short-lived steps needed to deduce the ultimate solution . Although a user might use
Multiplication of Radicals From: the program just to dispose of class assignments , I dubiousness about someone should be granted permission to apply it for
Solving Radical Equations egypt,alexandria quizzes.
Radical Expressions and Equations
Equations Containing Radicals and
Complex Numbers Gools Posted: Tuesday 02nd of Jan 11:38
Square Roots and Radicals I have tried out quite a lot of software. I would without any doubt say that Algebrator has assisted me to come to grips with my
Solving Radical Equations in One difficulties on difference of squares, distance of points and conversion of units. All I did was to simply key in the problem. The
Variable Algebraically answer showed up almost right away showing all the steps to the answer . It was quite straightforward to follow. I have relied on
Polynomials and Radicals this for my algebra classes to figure out Remedial Algebra and Basic Math. I would highly advice you to try out Algebrator.
Roots,Radicals,and Fractional Registered:
Exponents 01.12.2002
Adding, Subtracting, and From: UK
Multiplying Radical Expressions
Square Formula and Powers with
Simplifying Radicals CHS` Posted: Thursday 04th of Jan 11:09
Exponents and Radicals Practice I would suggest using Algebrator. It not only assists you with your math problems, but also provides all the necessary steps in
Solving Radical Equations detail so that you can improve the understanding of the subject.
Solving Radical Equations
Solving Radical Equations
Lecture-Radical Expressions Registered:
Radical Functions 04.07.2001
From: Victoria City,
Hong Kong Island,
Hong Kong
InjoyMERVANO! Posted: Friday 05th of Jan 07:46
Great! I think that’s what I need . Can you tell me where to buy it?
From: Germany
Vild Posted: Saturday 06th of Jan 11:33
I am sorry; I forgot to give the link in the previous post. You can find the tool here https://mathradical.com/
From: Sacramento, CA
Exponents and Radicals
Division of Radicals
Exponents and Radicals
RADICALS & RATIONAL EXPONENTS
Radicals and Rational Exponents
Radical Equations
Solving Radical Equations
Roots and Radicals
RADICAL EQUATION
Simplifying Radical Expressions
Radical Expressions
Solving Radical Equations
Solving Radical Equations
Exponents and Radicals
Exponents and Radicals
Roots;Rational Exponents;Radical
Solving and graphing radical
Solving Radical Equations
Radicals and Rational Exponents
Roots, Radicals, and Root
Multiplication of Radicals
Solving Radical Equations
Radical Expressions and Equations
Equations Containing Radicals and
Complex Numbers
Square Roots and Radicals
Solving Radical Equations in One
Variable Algebraically
Polynomials and Radicals
Roots,Radicals,and Fractional
Adding, Subtracting, and
Multiplying Radical Expressions
Square Formula and Powers with
Simplifying Radicals
Exponents and Radicals Practice
Solving Radical Equations
Solving Radical Equations
Solving Radical Equations
Lecture-Radical Expressions
Radical Functions
yr8 math papers
Related topics:
multiply polynomials made easy | "trigonometry, ninth edition" pdf | mixed numbers to decimal converter | beginner fractions | trigonometry facts and trivias | solve equations with multiple
variables and mapple | polynomial-programs in java language | intro to algebra 098 book pearson | poems on mathematical subjects | algebra 2 program
Author Message
NejhdLimks Posted: Sunday 31st of Dec 12:09
Hi math gurus! I am about two weeks through the semester, and getting a bit worried about my course work. I just don’t seem to understand the stuff I am learning,
especially things to do with yr8 math papers. Could somebody out there please enlighten me with adding numerators, quadratic formula and distance of points. I can’t
afford to pay for a tutor, but if anyone knows about other ways of mastering topics like angle suplements or graphing circles effectively, please drop me a line Thanks
From: Bronx, NY
oc_rana Posted: Tuesday 02nd of Jan 08:20
There are several topics comprising the general category of yr8 math papers, such as monomials, binomial formula and subtracting exponents. I have conversed with some
people who gave up on the pricey options for help as well . However , do not despair because I obtained an alternative resolution that is low-priced , easy to use and
to a greater extent more practical than I would have ever supposed . After trials with demonstrative math computer software programs and nearly surrendering, I picked
up Algebrator. This software system has precisely furnished results to each mathematics problem I have brought to the software . But just as large , Algebrator as well
Registered: allows all of the short-lived steps needed to deduce the ultimate solution . Although a user might use the program just to dispose of class assignments , I dubiousness
08.03.2007 about someone should be granted permission to apply it for quizzes.
Gools Posted: Tuesday 02nd of Jan 11:38
I have tried out quite a lot of software. I would without any doubt say that Algebrator has assisted me to come to grips with my difficulties on difference of squares,
distance of points and conversion of units. All I did was to simply key in the problem. The answer showed up almost right away showing all the steps to the answer . It
was quite straightforward to follow. I have relied on this for my algebra classes to figure out Remedial Algebra and Basic Math. I would highly advice you to try out
From: UK
CHS` Posted: Thursday 04th of Jan 11:09
I would suggest using Algebrator. It not only assists you with your math problems, but also provides all the necessary steps in detail so that you can improve the
understanding of the subject.
From: Victoria City,
Hong Kong Island,
Hong Kong
InjoyMERVANO! Posted: Friday 05th of Jan 07:46
Great! I think that’s what I need . Can you tell me where to buy it?
From: Germany
Vild Posted: Saturday 06th of Jan 11:33
I am sorry; I forgot to give the link in the previous post. You can find the tool here https://mathradical.com/radical-expressions-and-equations.html.
From: Sacramento, CA
Author Message
NejhdLimks Posted: Sunday 31st of Dec 12:09
Hi math gurus! I am about two weeks through the semester, and getting a bit worried about my course work. I just don’t seem to understand the stuff I am learning, especially
things to do with yr8 math papers. Could somebody out there please enlighten me with adding numerators, quadratic formula and distance of points. I can’t afford to pay for a
tutor, but if anyone knows about other ways of mastering topics like angle suplements or graphing circles effectively, please drop me a line Thanks heaps
From: Bronx, NY
oc_rana Posted: Tuesday 02nd of Jan 08:20
There are several topics comprising the general category of yr8 math papers, such as monomials, binomial formula and subtracting exponents. I have conversed with some people who
gave up on the pricey options for help as well . However , do not despair because I obtained an alternative resolution that is low-priced , easy to use and to a greater extent
more practical than I would have ever supposed . After trials with demonstrative math computer software programs and nearly surrendering, I picked up Algebrator. This software
system has precisely furnished results to each mathematics problem I have brought to the software . But just as large , Algebrator as well allows all of the short-lived steps
Registered: needed to deduce the ultimate solution . Although a user might use the program just to dispose of class assignments , I dubiousness about someone should be granted permission to
08.03.2007 apply it for quizzes.
Gools Posted: Tuesday 02nd of Jan 11:38
I have tried out quite a lot of software. I would without any doubt say that Algebrator has assisted me to come to grips with my difficulties on difference of squares, distance
of points and conversion of units. All I did was to simply key in the problem. The answer showed up almost right away showing all the steps to the answer . It was quite
straightforward to follow. I have relied on this for my algebra classes to figure out Remedial Algebra and Basic Math. I would highly advice you to try out Algebrator.
From: UK
CHS` Posted: Thursday 04th of Jan 11:09
I would suggest using Algebrator. It not only assists you with your math problems, but also provides all the necessary steps in detail so that you can improve the understanding
of the subject.
From: Victoria City,
Hong Kong Island,
Hong Kong
InjoyMERVANO! Posted: Friday 05th of Jan 07:46
Great! I think that’s what I need . Can you tell me where to buy it?
From: Germany
Vild Posted: Saturday 06th of Jan 11:33
I am sorry; I forgot to give the link in the previous post. You can find the tool here https://mathradical.com/radical-expressions-and-equations.html.
From: Sacramento, CA
Posted: Sunday 31st of Dec 12:09
Hi math gurus! I am about two weeks through the semester, and getting a bit worried about my course work. I just don’t seem to understand the stuff I am learning, especially things to do with yr8
math papers. Could somebody out there please enlighten me with adding numerators, quadratic formula and distance of points. I can’t afford to pay for a tutor, but if anyone knows about other ways of
mastering topics like angle suplements or graphing circles effectively, please drop me a line Thanks heaps
Posted: Tuesday 02nd of Jan 08:20
There are several topics comprising the general category of yr8 math papers, such as monomials, binomial formula and subtracting exponents. I have conversed with some people who gave up on the pricey
options for help as well . However , do not despair because I obtained an alternative resolution that is low-priced , easy to use and to a greater extent more practical than I would have ever
supposed . After trials with demonstrative math computer software programs and nearly surrendering, I picked up Algebrator. This software system has precisely furnished results to each mathematics
problem I have brought to the software . But just as large , Algebrator as well allows all of the short-lived steps needed to deduce the ultimate solution . Although a user might use the program just
to dispose of class assignments , I dubiousness about someone should be granted permission to apply it for quizzes.
Posted: Tuesday 02nd of Jan 11:38
I have tried out quite a lot of software. I would without any doubt say that Algebrator has assisted me to come to grips with my difficulties on difference of squares, distance of points and
conversion of units. All I did was to simply key in the problem. The answer showed up almost right away showing all the steps to the answer . It was quite straightforward to follow. I have relied on
this for my algebra classes to figure out Remedial Algebra and Basic Math. I would highly advice you to try out Algebrator.
Posted: Thursday 04th of Jan 11:09
I would suggest using Algebrator. It not only assists you with your math problems, but also provides all the necessary steps in detail so that you can improve the understanding of the subject.
Posted: Friday 05th of Jan 07:46
Great! I think that’s what I need . Can you tell me where to buy it?
Posted: Saturday 06th of Jan 11:33
I am sorry; I forgot to give the link in the previous post. You can find the tool here https://mathradical.com/radical-expressions-and-equations.html. | {"url":"https://mathradical.com/calculator-with-radical/rational-equations/yr8-math-papers.html","timestamp":"2024-11-14T20:58:37Z","content_type":"text/html","content_length":"99878","record_id":"<urn:uuid:c208f634-ba44-4cc0-b013-dd8d2dc9980c>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00803.warc.gz"} |
How to Calculate the Frequency Factor in Chemical Kinetics
If you've ever wondered how engineers calculate the strength of concrete they create for their projects or how chemists and physicists measure the electrical conductivity of materials, much of it
comes down to how fast chemical reactions occur.
Figuring out how fast a reaction happens means looking at the reaction kinematics. The Arrhenius equation lets you do such a thing. The equation involves the natural logarithm function and accounts
for the rate of collision between particles in the reaction.
Arrhenius Equation Calculations
In one version of the Arrhenius equation, you can calculate the rate of a first-order chemical reaction. First-order chemical reactions are ones in which the rate of reactions depends only on the
concentration of one reactant. The equation is:
Where K is the reaction rate constant, the energy of activation is E[a] (in joules), R is the reaction constant (8.314 J/mol K), T is the temperature in Kelvin and A is the frequency
factor. To calculate the frequency factor A (which is sometimes called Z), you need to know the other variables K, E[a], and T.
The activation energy is the energy that the reactant molecules of a reaction must possess in order for a reaction to occur, and it's independent of temperature and other factors. This means that,
for a specific reaction, you should have a specific activation energy, typically given in joules per mole.
The activation energy is often used with catalysts, which are enzymes that speed up the process of reactions. The R in the Arrhenius equation is the same gas constant used in the ideal gas law PV
= nRT for pressure P, volume V, number of moles n, and temperature T.
The Arrhenius equations describes many reactions in chemistry such as forms of radioactive decay and biological enzyme-based reactions. You can determine the half-life (the time required for the
reactant's concentration to drop by half) of these first-order reactions as ln (2) / K for the reaction constant K. Alternatively, you can take the natural logarithm of both sides to change the
Arrhenius equation into ln (K) = ln (A) − E[a]/RT. This lets you calculate the activation energy and temperature more easily.
Frequency Factor
The frequency factor is used to describe the rate of molecular collisions that occur in the chemical reaction. You can use it to measure the frequency of the molecular collisions that have the proper
orientation between particles and appropriate temperature so that the reaction can occur.
The frequency factor is generally obtained experimentally to make sure the quantities of a chemical reaction (temperature, activation energy and rate constant) fit the form of the Arrhenius equation.
The frequency factor is temperature-dependent, and, because the natural logarithm of the rate constant K is only linear over a short range in temperature changes, it's difficult to extrapolate the
frequency factor over a broad range of temperatures.
Arrhenius Equation Example
As an example, consider the following reaction with rate constant K as 5.4 × 10 ^−4 M ^−1s ^−1 at 326 °C and, at 410 ^ °C, the rate constant was found to be 2.8 × 10 ^−2 M ^−1s ^−1. Calculate the
activation energy E[a] and frequency factor A.
You can use the following equation for two different temperatures T and rate constants K to solve for activation energy E[a].
\ln\bigg(\frac{K_2}{K_1}\bigg) = -\frac{E_a}{R}\bigg(\frac{1}{T_2} - \frac{1}{T_1}\bigg)
Then, you can plug the numbers in and solve for E[a]. Make sure to convert the temperature from Celsius to Kelvin by adding 273 to it.
\ln\bigg(\frac{5.4 ×10^{-4} \;\text{M}^{-1}\text{s}^{-1}}{2.8 ×10^{-2}\; \text{M}^{-1}\text{s}^{-1}}\bigg) = -\frac{E_a}{R}\bigg(\frac{1}{599 \;\text{K}} - \frac{1}{683 \;\text{K}}\bigg)
\begin{aligned} E_a&= 1.92 × 10^4 \;\text{K} × 8.314 \;\text{J/K mol} \\ &= 1.60× 10^5 \;\text{J/mol} \end{aligned}
You can use either temperature's rate constant to determine the frequency factor A. Plugging in the values, you can calculate A.
k = Ae^{-E_a/RT}
5.4 × 10^{-4} \;\text{M}^{-1}\text{s}^{-1} =A e^{-\frac{1.60 × 10^5 \;\text{J/mol}}{8.314 \;\text{J/K mol} ×599 \;\text{K}}} \\ A = 4.73 × 10^{10} \;\text{M}^{-1}\text{s}^{-1}
• If you do not know the rate constant offhand, you may need to determine the value experimentally. In this case, the frequency factor can be found by graphing the relationship between the rate
constant and the temperature.
About the Author
S. Hussain Ather is a Master's student in Science Communications the University of California, Santa Cruz. After studying physics and philosophy as an undergraduate at Indiana University-Bloomington,
he worked as a scientist at the National Institutes of Health for two years. He primarily performs research in and write about neuroscience and philosophy, however, his interests span ethics, policy,
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How do you write a linear function equation that passes through points (-2,5) and (-4,7)? | HIX Tutor
How do you write a linear function equation that passes through points (-2,5) and (-4,7)?
Answer 1
The equation of a line in #color(blue)"point-slope form"# is.
#color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))# where m represents the slope and # (x_1,y_1)" a point on the line"#
To calculate m, use the #color(blue)"gradient formula"#
#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(2/2)|)))# where # (x_1,y_1),(x_2,y_2)" are 2 points on the line"#
The 2 points here are (-2 ,5) and (-4 ,7)
Use either of the 2 given points for # (x_1,y_1)#
substitute m = - 1 and (-2 ,5) into the equation.
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Answer 2
To write a linear function equation that passes through two points (-2,5) and (-4,7), you can use the point-slope form of a linear equation.
First, calculate the slope (m) using the formula: m = (y2 - y1) / (x2 - x1)
Using the points (-2,5) and (-4,7): m = (7 - 5) / (-4 - (-2)) = 2 / (-2) = -1
Now, you have the slope (m). Next, choose one of the points (let's say (-2,5)) and plug it into the point-slope form equation: y - y1 = m(x - x1)
Using (-2,5): y - 5 = -1(x - (-2))
Now, simplify: y - 5 = -1(x + 2)
Distribute -1: y - 5 = -x - 2
Add 5 to both sides: y = -x - 2 + 5
Simplify: y = -x + 3
So, the linear function equation that passes through the points (-2,5) and (-4,7) is y = -x + 3.
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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
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Rationalize the Denominator Calculator
The rationalize the denominator calculator is used to rationalize the denominator for a given input. So rationalizing denominators can be made easy by using our calculator.
What is a Rationalization of Numbers?
“The process of removing imaginary numbers or radicles (square root, cube root) from the denominator of a fraction is known as rationalization”.
In other words, we say that it is a process of multiplying a surd with another surd to get a rational number. The surd that is used to multiply is called a rationalized factor.
Standard Form of Rationalization:
As we know fractions have a denominator and numerator therefore we rationalize the fractions to express them in the standard form with the help of rationalize denominator calculator.
Suppose a fractional term is 1/(n-√m), So the rationalized form of the fractional term is written as:
• [1/( n - √m )] × [( n + √m ) / ( n + √m )]
• Rationalized Form = [( n + √m ) / ( n2 - m )]
How to Rationalize The Denominator?
The rationalization of the denominator is the process of eliminating the radicles from fractional terms by multiplying the both numerator and denominator by the conjugate of the denominator
The rule for how do you rationalize the denominator and simplify is the topic for the whole article. In order to get rid of radicles from the denominator and seek rationalizing denominators take the
help of our rationalize the denominator calculator and attach to the below points:
1. Radicle/Radicle || (a * n√b) / (x * k√y)
Multiply the denominator and the numerator by the radicle which will get rid of the radicle from the denominator. Note that when you multiply the denominator and numerator by the exact same thing the
fractions will be equivalent.
• If the denominator is in the form of √a then multiply the numerator by the denominator √a.
• If the denominator is in the form of n√a^m where m
ᵏ√(b^ᵏ⁻¹) / ᵏ√(b^ᵏ⁻¹)
a * ᵏ√b * ᵏ√(b^ᵏ⁻¹)
= a * ᵏ√(b^ᵏ)
= a * b
2. Sum/Radicle || (a * n√b + c * m√d) / (x * k√y)
This rule is also similar to the above discussed but it has two summands in numerator. In this calculation, rationalizing the denominator calculator takes into service the k√(y^k-1) for both
Here we again multiply;
k√(y^k-1) / k√(y^k-1) and obtain x * y.
3. Radicle/Sum || (a * √b) / (x * √y + z * √u)
In this case, the summands of both denominators need to be rationalized. For the simplifications of rationalizing the denominators, the [a^2 - b^2 = (a + b)(a - b)] formula is used which gives two
squares by eliminating the square root.
So here we multiply our expression;
(x * √y - z * √u) / (x * √y - z * √u)
(x^2 - y^2)
4. Sum/Sum || (a * √b + c * √d) / (x * √y + z * √u)
This point is similar to rule three which has already been discussed multiplying both quantities.
(x * √y - z * √u) / (x * √y - z * √u).
Practical Examples:
Rationalize the Denominator with 1 Term:
Suppose a term that is available for how do you rationalize the denominator and simplify
1 √ 3
We can multiply both the numerator and denominator by √ 3
To rewrite the expression to have a rational denominator: 1 √ 3 = 1 √ 3 × √ 3 √ 3 = 1 × √ 3 √ 3 × √ 3 = √ 3 3
We now have two different forms of the same number: 1 √ 3 = √ 3 3
Rationalize the Denominator with Multiple Terms:
Suppose another but a little complex term in order to clarify the concepts
4 * √64 / 3 * 3√27
First, we simplify 64 and 27 and after move forward.
4 * √4^2 * 4 / 3 * 3√3^2 * 3
4 * 8 / 3 * 3
Working of Rationalize the Denominator Calculator:
In order to avoid square roots and rationalize the denominators use our best tool which is designed with user-friendliness in mind.
What to Do?
• First, select a simple or advanced method
• Select the expression from the drop-down menu
• Insert the values for the numerator and denominator
• Tap on the “Calculate” icon
What to Get?
• Rationalization of denominator
• Complete the calculation in steps
Wikipedia: Rationalisation (mathematics), Rationalisation of a monomial square root and cube root, Dealing with more square roots.
Lumen Learning: Rationalize Denominators, Example. | {"url":"https://www.calculatored.com/rationalize-the-denominator-calculator","timestamp":"2024-11-10T18:55:21Z","content_type":"text/html","content_length":"59609","record_id":"<urn:uuid:a30090d3-93c4-452a-88cf-04fa6bb24403>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00436.warc.gz"} |
Create Transformed, N-Dimensional Polygons with Covariance Matrix - DataScienceCentral.com
The covariance matrix has many interesting properties, and it can be found in mixture models, component analysis, Kalman filters, and more. Developing an intuition for how the covariance matrix
operates is useful in understanding its practical implications. This article will focus on a few important properties, associated proofs, and then some interesting practical applications, i.e.,
extracting transformed polygons from a Gaussian mixture’s covariance matrix.
I have often found that research papers do not specify the matrices’ shapes when writing formulas. I have included this and other essential information to help data scientists code their own
Sub-Covariance Matrices
The covariance matrix can be decomposed into multiple unique (2×2) covariance matrices. The number of unique sub-covariance matrices is equal to the number of elements in the lower half of the
matrix, excluding the main diagonal. A (DxD) covariance matrices will have D*(D+1)/2 -D unique sub-covariance matrices. For example, a three dimensional covariance matrix is shown in equation (0).
It can be seen that each element in the covariance matrix is represented by the covariance between each (i,j) dimension pair. Equation (1), shows the decomposition of a (DxD) into multiple (2×2)
covariance matrices. For the (3×3) dimensional case, there will be 3*4/2–3, or 3, unique sub-covariance matrices.
Note that generating random sub-covariance matrices might not result in a valid covariance matrix. The covariance matrix must be positive semi-definite and the variance for each dimension the
sub-covariance matrix must the same as the variance across the diagonal of the covariance matrix.
Positive Semi-Definite Property
One of the covariance matrix’s properties is that it must be a positive semi-definite matrix. What positive definite means and why the covariance matrix is always positive semi-definite merits a
separate article. In short, a matrix, M, is positive semi-definite if the operation shown in equation (2) results in a values which are greater than or equal to zero.
M is a real valued DxD matrix and z is an Dx1 vector. Note: the result of these operations result in a 1×1 matrix.
A covariance matrix, M, can be constructed from the data with the following operation, where the M = E[(x-mu).T*(x-mu)]. Inserting M into equation (2) lead to equation (3). It can be seen that any
matrix that can be written in the form M.T*M is positive semi-definite. This full proof can be found here.
Note that the covariance matrix does not always describe the covariation between a dataset’s dimensions. For example, the covariance matrix can be used to describe the shape of a multivariate normal
cluster, for Gaussian mixture models.
Geometric Implications
Another way to think about the covariance matrix is geometrically. Essentially, the covariance matrix represents the direction and scale for how the data is spread. To understand this perspective, it
will be necessary to understand eigenvalues and eigenvectors.
Equation (4) shows the definition of an eigenvector and its associated eigenvalue. The next statement is important in understanding eigenvectors and eigenvalues. Z is an eigenvector of M if the
matrix multiplication M*z results in the same vector, z, scaled by some value, lambda. In other words, we can think of the matrix M as a transformation matrix that does not change the direction of z,
or z is a basis vector of matrix M.
Lambda is the eigenvalue (1×1) scalar, z is the eigenvector (Dx1) matrix, and M is the (DxD) covariance matrix. A positive semi-definite (DxD) covariance matrix will have D eigenvalue and (DxD)
eigenvectors. The first eigenvector is always in the direction of highest spread of data, all eigenvectors are orthogonal to each other, and all eigenvectors are normalized, i.e. they have values
between 0 and 1. Equation (5) shows the vectorized relationship between the covariance matrix, eigenvectors, and eigenvalues.
S is the (DxD) diagonal scaling matrix, where the diagonal values correspond to the eigenvalue and which represent the variance of each eigenvector. R is the (DxD) rotation matrix that represents the
direction of each eigenvalue.
The eigenvector and eigenvalue matrices are represented, in the equations above, for a unique (i,j) sub-covariance matrix. The sub-covariance matrix’s eigenvectors, shown in equation (6), across each
columns has one parameter, theta, that controls the amount of rotation between each (i,j) dimension pair. The covariance matrix’s eigenvalues are across the diagonal elements of equation (7) and
represent the variance of each dimension. It has D parameters that control the scale of each eigenvector.
The Covariance Matrix Transformation
A (2×2) covariance matrix can transform a (2×1) vector by applying the associated scale and rotation matrix. The scale matrix must be applied before the rotation matrix as shown in equation (8).
The vectorized covariance matrix transformation for a (Nx2) matrix, X, is shown in equation (9). The matrix, X, must centered at (0,0) in order for the vector to be rotated around the origin
properly. If this matrix X is not centered, the data points will not be rotated around the origin.
An example of the covariance transformation on an (Nx2) matrix is shown in the Figure 1. More information on how to generate this plot can be found here.
Please see this link to see how these properties can be used to draw Gaussian mixture contours and create non-Gaussian, polygon, mixture models.
Originally posted by Rohan Kotwani | {"url":"https://www.datasciencecentral.com/create-transformed-polygons-using-the-covariance-matrix/","timestamp":"2024-11-08T23:44:13Z","content_type":"text/html","content_length":"162798","record_id":"<urn:uuid:63cd0d3f-571c-48aa-aeba-ebaee5dd75cd>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00822.warc.gz"} |
How to Create Monthly Averages for Headcount?
Hi All,
I'm in the process of building an attrition model in Domo in the form:
Total Terms / Average headcount
Terms is easy, that's just a count of everyone with a term status for that time period. Headcount is where I'm getting stuck. My headcount numbers are currently aggregated based on headcount at the
end of each month. I've just been using this number so far, but I'm about 1/10% off in terms of accuracy.
In order to get average headcount, it should be:
((Current Month's headcount + Last Month's headcount) / 2)
I could write this manually for each month for the last decade, but that would take a lot of time and require I update the code every month, which I would prefer not to do. Is there a way to do this
in Beast Mode with a couple of lines instead of a line for every month?
• What columns are currently in your dataset? I'm having trouble picturing what your data looks like. Do you have one row per month right now? If you can explain a bit more what your data looks
like, it will be easier to suggest a solution.
**Check out my
Domo Tips & Tricks Videos
**Make sure to
**Please mark as accepted the ones who solved your issue.
• I'd recommend using a custom date dimension with offsets so you can easily have this month and last month together based on your current month. I've done a more detailed writeup on it here: A
more flexible way to do Period over Period comparisons You may need to tweak it if your data is on a monthly cadence instead of daily.
Then you could use a beast mode like:
(SUM(CASE WHEN `Period Type` = 'Current' THEN `Headcount` ELSE 0 END) + SUM(CASE WHEN `Period Type` = 'Last Month' THEN `Headcount` ELSE 0 END)) / 2
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**Did this solve your problem? Accept it as a solution!**
• @MarkSnodgrass @GrantSmith
My table has a row for every employee for every month with all of their statuses and a count of total employees and total terminations. An employee who has been here 12 months has 12 rows of
data. Example below:
• I would suggest creating an ETL that groups the data by month/year and sums the employee count and term count. You can then add a rank and window tile and use the lag function to get the total
employee count for the previous month as a column right next to the current month. Next, add a formula tile to add your current and previous month and divide by two to get your average headcount.
This should get you all the datapoints you are looking for.
**Check out my
Domo Tips & Tricks Videos
**Make sure to
**Please mark as accepted the ones who solved your issue.
• I appreciate the suggestion, but I would prefer doing it at the BeastMode level if possible. There are about a dozen factors I need to be able to slice this data by and aggregating it at the ETL
level will remove that functionality.
• @nshively Understood. I'm sure @GrantSmith will come up with a very elaborate beast mode for you.
You can whatever you are potentially slicing by into your group by and rank and window tile and then save the final formula tile work that calculates the average headcount for your beast mode. I
have done this multiple times and it allows me to maintain the card flexibility while avoiding complex beast modes that are hard to troubleshoot.
**Check out my
Domo Tips & Tricks Videos
**Make sure to
**Please mark as accepted the ones who solved your issue.
• I'm not certain what all information you need displayed on your card but you could utilize the LAG window function assuming you don't have any missing months:
Average Head Count (Beast Mode)
( SUM(SUM(`EmployeeCount`)) OVER (PARTITION BY `EOMDate`) + LAG(SUM(`EmployeeCount`), 1) OVER (ORDER BY `EOMDate`)) / 2
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**Did this solve your problem? Accept it as a solution!**
• Thanks @GrantSmith
I feel like that should work, but Domo is not recognizing 'OVER' or 'LAG()' as functions in Beast Mode?
• You'll need to talk with your CSM to get Window Functions turned on as it's a feature switch they need to do.
**Was this post helpful? Click Agree or Like below**
**Did this solve your problem? Accept it as a solution!**
• Hi @GrantSmith,
Thanks for the help so far, and my team is going to chat with the CSM about the window function. I did find a temporary solution for building out monthly attrition. It's not perfect, but it's the
closest I think I can get without the Lag function:
SUM(`TermCount`)/ (((SUM(`ActiveCount`))+(Sum(`ActiveCount`) + (Sum(`TermCount`)-SUM(`NewHireCount`))))/2)
Basically, it's finding the average based on the combination of current headcount + last month's headcount (using current headcount +/- this month's terms and new hires).
However, due to how my tables are built, this only works for Monthly attrition. I'm trying to build yearly attrition, which should be easier because I can just divide by total months, however, I
only have 10 months data in 2021. So I'm trying to build a rule to use different formulas based on the year, but it's not working at all. Here's where I'm at:
When YEAR(2021) Then SUM(`TermCount`)/ ((SUM(`ActiveCount`))/10)
When YEAR(2020) Then SUM(`TermCount`)/ ((SUM(`ActiveCount`))/12)
Else 0
I tried building out the Case statement as a different beast mode calculation to = 10 or 12 (based on the year), but when I tried to use this as a variable in my statement it just copied the
whole thing over instead of using it as a measure.
• Nevermind, I realized it was a simple solution since I have the EOMDate:
SUM(`TermCount`)/ ((SUM(`ActiveCount`))/COUNT(DISTINCT `EOM_DateKey`))
It just divides the total count of active employees per month per year by the number of months of data for that year.
• I'm glad you got it figured out! I was going to suggest programmatically calculating the number of months to divide by depending on the current year and month (untested - back of napkin):
When YEAR(`EOMDate`) = YEAR(CURRENT_DATE) Then SUM(`TermCount`)/ ((SUM(`ActiveCount`))/MONTH(CURRENT_DATE))
When YEAR(`EOMDate`) = YEAR(CURRENT_DATE) - 1 Then SUM(`TermCount`)/ ((SUM(`ActiveCount`))/12)
Else 0
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If ABC is a right triangle and angle A=45°, what degrees are B and C? | Socratic
If ABC is a right triangle and angle A=45°, what degrees are B and C?
1 Answer
As the sum of internal angles of a triangle is $180 \mathrm{de} g r e e$
If angle A is $45 \mathrm{de} g r e e$ & angle B is $45 \mathrm{de} g r e e$ and angle c is $90 \mathrm{de} g r e e$
Angle A + Angle B + Angle C=180 degree
Hope it helps!
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SQL Arithmetic | SQL Tutorial
In addition to querying raw data with SQL, you can also use math expressions to transform column values. Here's an example:
Let's dive into each arithmetic operator, one by one.
SQL Arithmetic Operators
SQL Addition +
The operator adds two numbers.
SQL Subtraction -
The operator subtracts one column value from another.
SQL Multiplication *
The operator multiplies two numbers.
Note: this is exactly the same as ... so don't get confused by 's two uses!!
SQL Division /
The operator divides the first column value by the number in the 2nd column.
Always ensure you're not dividing by zero, as it will cause an error. We dive into more nuances and issues with SQL division in a later tutorial.
SQL Modulus %
The operator, also known as the modulus or remainder function, returns the remainder of a division operation.
In the above example, the DBMS returns 5 because 23 divide by 6 is 3, with a remainder of 5.
Odd and Even with Modulus %
The modulus () operator is often used to find odd and even values, like in this hard Google SQL Interview Question.
While solving the entire Google problem is too tricky right now because it needs ranking window functions, let's look at a small snippet from the full solution which finds odd values:
The above SQL query looks at the and looks at the remainder when divided by 2.
Odd numbers like 1, 3, 5, when divided by 2, have a remainder of 1, which is why gets us all odd numbered measurements.
SQL Exponentiation (^)
The operator, also known as the power operator, raises a number to the power of another number.
The above returns 100.
SQL Arithmetic Operator Summary
Here's a summary table that summarizes how the arithmetic operators in SQL work:
Operator Description Example Result
+ Addition 15 + 5 20
- Subtraction 15 - 5 10
* Multiplication 15 * 5 75
/ Division 15 / 5 3
% Modulus (Remainder of Division) 14 % 5 4
^ Exponentiation (Not standard in all DBMS) 15 ^ 2 225
- (as a prefix) Negation -15 -15
SQL Arithmetic Order of Operations
Just like in standard arithmetic, SQL follows the order of operations of PEMDAS:
• P: Parentheses first
• E: Exponents (ie. )
• MD: Multiplication and Division (left-to-right)
• AS: Addition and Subtraction (left-to-right)
Here's some SQL examples of PEMDAS:
SQL Statement Result Explanation
SELECT 3 + 7 * 2; 17 Multiplication comes before addition.
SELECT (3 + 7) * 2; 20 Parentheses means addition happens first.
SELECT 10 / 2 + 3 * 4; 17 10/2 = 5, 3*4=12, so 5 + 12 = 17.
SELECT (10 / 2) + (3 * 4); 17 Same as above, but more explicit with parens!
To make your code more readable and less confusing, feel free to use parenthesis to make your SQL math formulas more explicit.
SQL Arithmetic Practice Exercises
Let's practice combining the arithmetic operators in this lesson to compute some interesting metrics by analzying CVS Pharmacy, JP Morgan, and FAANG stock datasets.
Practice SQL Subtraction: CVS Pharmacy Interview Question
Here's a real SQL interview question asked by CVS Health for a healthcare analytics job: Write a query to find the top 3 most profitable medicines sold, and how much profit they made.
Your output should look like this:
drug total_profit
Humira 81515652.55
Keytruda 11622022.02
Dupixent 11217052.34
Hint #1: Total Profit = Total Sales - Cost of Goods Sold
Hint #2: To find the top 3 drugs, just do a
Practice SQL Arithmetic: JPMorgan Chase SQL Interview Question
In this JPMorgan Data Analyst interview question, imagine that you're on the credit card marketing analytics team at Chase.
You're preparing to launch a new credit card, and to gain some insights, you're analyzing how many credit cards were issued each month. Write a query that outputs the name of each credit card and the
difference in the number of issued cards between the month with the highest issuance cards and the lowest issuance.
Arrange the results based on the largest disparity like as follows:
card_name difference
Chase Sapphire Reserve 30000
Chase Freedom Flex 15000
Hint: You'll want to use the and aggregate functions!
FAANG Stocks That Had 'Big-Mover Months'
A "big-mover month" is when a stock closes up or down by greater than 10% compared to the price it opened at. For example, when COVID hit and e-commerce became the new normal, Amazon stock in April
2020 had a big-mover month because the price shot up from $96.65 per share at open to $123.70 at close β a 28% increase!
ticker date open close percent_change
AMZN 04/01/2020 00:00:00 96.65 123.70 28.0
NFLX 04/01/2022 00:00:00 376.80 190.36 -49.5
Netflix stock had a big-mover month in April 2022 in the reverse direction. That month, Netflix reported that the company lost 200k subscribers in Q1, and expected to lose another two million subs in
Q2. In Apr'22, Netflix stock opened that month at $376.80 per share, but closed at $190.36, representing a 49.5% loss β yikes!
Display the stocks which had "big-mover months", and how many of those months they had. Order your results from the stocks with the most, to least, "big-mover months".
What's Next: MATH Functions
In the last SQL exercise about big-mover months, in order to figure out if the difference between the open & close price was greater than 10% of the opening price, we used this clunky SQL snippet:
We wrote it this way, because could be a positive or negative number, depending on whether the stock closed at 10% higher OR 10% lower compared to the start of the month. We needed the to account for
both cases.
Now, some of you might be thinking...
damn I really wish SQL had an absolute value function rn
If that's your thinking, good job β that's exactly what we need!
We'll cover the function, along with some other powerful SQL commands in the next tutorial on mathematical functions in SQL! | {"url":"https://datalemur.com/sql-tutorial/sql-arithmetic","timestamp":"2024-11-08T22:14:00Z","content_type":"text/html","content_length":"117307","record_id":"<urn:uuid:499b40cd-e71e-45e4-b8ba-e8da0598fd15>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00362.warc.gz"} |
American Mathematical Society
Comment on A.-P. Calderón’s paper: “On an inverse boundary value problem” [in Seminar on Numerical Analysis and its Applications to Continuum Physics (Rio de Janeiro, 1980), 65–73, Soc. Brasil. Mat.,
Rio de Janeiro, 1980; MR0590275 (81k:35160)]
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by David Isaacson and Eli L. Isaacson PDF
Math. Comp. 52 (1989), 553-559 Request permission
Calderón determined a method to approximate the conductivity $\sigma$ of a conducting body in ${R^n}$ (for $n \geq 2$) based on measurements of boundary data. The approximation is good in the ${L_\
infty }$ norm provided that the conductivity is a small perturbation from a constant. We calculate the approximation exactly for the case of homogeneous concentric conducting disks in ${R^2}$ with
different conductivities. Here, the difference in the conductivities is the perturbation. We show that the approximation yields precise information about the spatial variation of $\sigma$, even when
the perturbation is large. This ability to distinguish spatial regions with different conductivities is important for clinical monitoring applications. References
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• Robert V. Kohn and Michael Vogelius, Relaxation of a variational method for impedance computed tomography, Comm. Pure Appl. Math. 40 (1987), no. 6, 745–777. MR 910952, DOI 10.1002/cpa.3160400605
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Maryland, 1988.
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J. G. Webster & W. J. Tompkins, "Comparing reconstruction algorithms for electrical impedance tomography," IEEE Trans. Biomed. Engrg., v. BME-34, 1987, pp. 843-851.
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• © Copyright 1989 American Mathematical Society
• Journal: Math. Comp. 52 (1989), 553-559
• MSC: Primary 35R30; Secondary 35K60
• DOI: https://doi.org/10.1090/S0025-5718-1989-0962208-X
• MathSciNet review: 962208 | {"url":"https://www.ams.org/journals/mcom/1989-52-186/S0025-5718-1989-0962208-X/?active=current","timestamp":"2024-11-14T05:52:52Z","content_type":"text/html","content_length":"62879","record_id":"<urn:uuid:d3cfd859-76dd-4a85-b3ac-5ab39d9e9695>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00298.warc.gz"} |
Han, J.-H. Chang, H.-J. Jee, K.-K. Oh, K.H. 2024-01-20T21:30:47Z 2024-01-20T21:30:47Z 2021-09-02 2009-06 1598-9623 https://pubs.kist.re.kr/handle/201004/132468 The deformation behavior within the
deformation zone of a workpiece during equal channel angular pressing (ECAP) was investigated using the finite element method. The effects of die geometry on the variations of normal and shear
deformations were studied with a deformation rate tensor (D). The zero dilatation line, at which the normal components (D11 and D 22) of the deformation rate tensor (D) are zero, in the die coincided
with the line of intersection of the two die channels irrespective of die geometry such as curvature angle (χ) and oblique angle (φ), while the maximum shear line, at which the shear components (D12
and D 21) of the deformation rate tensor (D) have maximum value, is dependant on the die geometry. ？ KIM and Springe. English KOREAN INST METALS MATERIALS Effects of die geometry on variation of the
deformation rate in equal channel angular pressing Article 10.1007/s12540-009-0439-3 1 Metals and Materials International, v.15, no.3, pp.439 - 445 Metals and Materials International 15 3 439 445
scie scopus kci ART001354220 000267786600013 2-s2.0-75149191001 Materials Science, Multidisciplinary Metallurgy & Metallurgical Engineering Materials Science Metallurgy & Metallurgical Engineering
Article TEXTURE EVOLUTION MICROSTRUCTURAL DEVELOPMENT EXTRUSION STRIP ALUMINUM SHEAR HISTORY ALLOY Curvature angle Deformation rate Equal channel angular pressing (ECAP) Maximum shear deformation
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Bayesian Portfolio Analysis: Analyzing the Global Investment Market
From The Developing Economist VOL. 2 NO. 1
Bayesian Portfolio Analysis: Analyzing the Global Investment Market
The Developing Economist
2015, Vol. 2 No. 1 | pg. 1/1
The goal of portfolio optimization is to determine the ideal allocation of assets to a given set of possible investments. Many optimization models use classical statistical methods, which do not
fully account for estimation risk in historical returns or the stochastic nature of future returns. By using a fully Bayesian analysis, however, I am able to account for these aspects and incorporate
a complete information set as a basis for the investment decision. I use Bayesian methods to combine different estimators into a succinct portfolio optimization model that takes into account an
investor's utility function. I will test the model using monthly return data on stock indices from Australia, Canada, France, Germany, Japan, the U.K. and the U.S.
I. Introduction
Portfolio optimization is one of the fastest growing areas of research in financial econometrics, and only recently has computing power reached a level where analysis on numerous assets is even
possible. There are a number of portfolio optimization models used in financial econometrics and many of them build on aspects of previously defined models. The model I will be building uses Bayesian
statistical methods to combine insights from Markowitz, BL and Zhou. Each of these papers use techniques from the previous one to specify and create a novel modeling technique.
Bayesian statistics specify a few types of functions that are necessary to complete an analysis, the prior distribution, the likelihood function, and the posterior distribution. A prior distribution
defines how one expects a variable to be distributed before viewing the data. Prior distributions can be of different weights in the posterior distribution depending on how confident one is in their
prior. A likelihood function describes the observed data in the study. Finally, the posterior distribution describes the final result, which is the combination of the prior distribution with the
likelihood function. This is done by using Bayes theorem^2, which multiplies the prior times the posterior and divides by the normalizing constant, which conditions that the probability density
function (PDF) of the posterior sums to 1. Bayesian analysis is an ideal method to use in a portfolio optimization problem because it accounts for the estimation risk in the data. The returns of the
assets form a distribution centered on the mean returns, but we are not sure that this mean is necessarily the true mean. Therefore it is necessary to model the returns as a distribution to account
for the inherent uncertainty in the mean, and this is exactly what Bayesian analysis does.
Zhou incorporates all of the necessary Bayesian components in his model; the market equilibrium and the investor?s views act as a joint prior and the historical data defines the likelihood function.
This strengthens the model by making it mostly consistent with Bayesian principles, but some aspects are still not statistically sound. In particular, I disagree with the fact that Zhou uses the
historical covariance matrix, Σ, in each stage of the analysis (prior and likelihood). The true covariance matrix is never observable to an investor, meaning there is inherent uncertainty in modeling
Σ, which must be accounted for in the model. Zhou underestimates this uncertainty by using the historical covariance matrix to initially estimate the matrix, and by re-updating the matrix with the
historical data again in the likelihood stage. This method puts too much confidence in the historical matrix by re-updating the prior with the same historical matrix. I plan to account for this
uncertainty by incorporating an inverse-Wishart prior distribution on the Black-Litterman prior estimate, which will model Σ as a distribution and not a point estimate. The inverse-Wishart prior will
use the Black-Litterman covariance matrix as a starting point, but the investor can now model the matrix as a distribution and adjust confidence in the starting point with a tuning parameter. This is
a calculation that must be incorporated to make the model statistically sound, and it also serves as a starting point for more extensive analysis of the covariance matrix.
The empirical analysis in Zhou is based on equity index returns from Australia, Canada, France, Germany, Japan, the United Kingdom and the United States. My dataset is comprised of the total return
indices for the same countries, but the data spans through 2013 instead of 2007 like in Zhou. This is a similar dataset to that chosen by BL, which was used in order to analyze different
international trading strategies based on equities, bonds and currencies.
The goal of this paper is to extend the Bayesian model created by Zhou by relaxing his strict assumption on the modeling of the covariance matrix by incorporating the inverse-Wishart prior extension.
This will in turn create a statistically sound and flexible model, usable by any type of investor. I will then test the models by using an iterative out-of-sample modeling procedure.
In section II, I further describe the literature on the topic and show how it influenced my analysis. In section III I will describe the baseline models and the inverse-Wishart prior extension. In
Section IV I will summarize the dataset and provide descriptive statistics. In section V I will describe how the models are implemented and tested. In Section VI I will describe the results and
compare the models, and in Section VII I will offer conclusions and possible extensions to my model.
II. Literature Review
Harry Markowitz established one of the first frameworks for portfolio optimization in 1952. In his paper, Portfolio Selection, Markowitz solves for the portfolio weights that maximize a portfolio?s
return while minimizing the volatility, by maximizing a specified expected utility function for the investor. The utility function is conditional on the historical mean and variance of the data,
which is why it is often referred to as a mean-variance analysis. These variables are the only inputs, so the model tends to be extremely sensitive to small changes in either of them. The model also
assumes historical returns on their own predict future returns, which is something known to be untrue in financial econometrics.
These difficulties with the mean-variance model do not render it useless. In fact, the model can perform quite well when there are better predictors for the expected returns and covariance matrix
(rather than just historical values). The model by BL extends the mean-variance framework by creating an estimation strategy that incorporates an investor?s views on the assets in question with an
equilibrium model of asset performance. Many investors make decisions about their portfolio based on how they expect the market to perform, so it is intuitive to incorporate these views into the
Investor views in the Black-Litterman model can either be absolute or relative. Absolute views specify the expected return for an individual security; for example, an investor may think that the S&P
500 will return 2% next month. Relative views specify the relationship between assets; for example, an investor may think that the London Stock Exchange will have a return 2% higher than the Toronto
Stock Exchange next month. BL specify the same assumptions and use a similar model to Markowitz to describe the market equilibrium, and they then incorporate the investor?s views through Bayesian
updating. This returns a vector of expected returns that is similar to the market equilibrium but adjusted for the investor?s views. Only assets that the investor has a view on will deviate from the
equilibrium weight. Finally, BL use the same mean-variance utility function as Markowitz to calculate the optimal portfolio weights based off of the updated expected returns.
Zhou takes this framework one step further by also incorporating historical returns into the analysis because the equilibrium market weights are subject to error that the historical data can help
fix. The market equilibrium values are based on the validity of the capital asset pricing model (CAPM)^3, which is not always supported by historical data. This does not render the equilibrium
returns useless; they simply must be supplemented by historical data in order to make the model more robust. The combination of the equilibrium pricing model and the investor?s views with the data
strengthens the model by combining different means of prediction. As an extension, it would be useful to research the benefit of including a more complex data modeling mechanism that incorporates
more than just the historical mean returns. A return forecasting model could be of great use here, though it would greatly increase the complexity of the model.
Zhou uses a very complete description of the market by incorporating all three of these elements, but there is one other aspect of the model that he neglects; his theoretical framework does not
account for uncertainty in the covariance matrix. By neglecting this aspect, he implies that the next period's covariance matrix is only described by the fixed historical covariance matrix. This is
in line with the problems that arise in Markowitz, and is also not sound in a Bayesian statistical sense because he is using a data generated covariance matrix in the prior, which is then updated by
the same data. I will therefore put an inverse-Wishart prior distribution on the Black-Litterman estimate of Σ before updating the prior with the data. The primary Bayesian updating stage, where the
equilibrium estimate is updated by the investor views will remain consistent. This way Σ is modeled as a distribution in the final Bayesian updating stage which will allow the prior to have a more
profound effect.
Investment Strategies
Though the Black-Litterman model is quantitatively based it is extremely flexible, unlike many other models, due to the input of subjective views by the investor. These views are directly specified
and can come from any source, whether that is a hunch, the Wall Street Journal, or maybe even an entirely different quantitative model. I will present a momentum based view strategy, but this is only
one of countless different strategies that could be incorporated, whether they are quantitatively based or not. The results of this paper will be heavily dependent on the view specification, which is
based on the nature of the model. The goal of this paper is not to have a perfect empirical analysis, but instead to present a flexible, statistically sound and customizable model for an investor
regardless of their level of expertise.
The investor's views can be independent over time or follow a specific investment strategy. In the analysis I use a function based on the recent price movement of the indices, a momentum strategy, to
specify the views. The conventional wisdom of many investors is that individual prices and their movements have nothing to say about the asset's value, but when the correct time frame is analyzed,
generally the previous 6-12 months, statistically significant returns can be achieved (Momentum). In the last 5 years alone, over 150 papers have been published investigating the significance of
momentum investment strategies (Momentum). Foreign indices are not an exception, as it has been shown that indices with positive momentum perform better than those with negative momentum (AQR).
The basis of momentum strategies lies in the empirical failure of the efficient market hypothesis, which states that all possible information about an asset is immediately priced into the asset once
the information becomes available. This tends to fail because some investors get the information earlier or respond to it in different manners, so there is an inherent asymmetric incorporation of
information that creates shortterm price trends (momentum) that can be exposed. This phenomenon can be further explored in Momentum.
Though momentum investing is gaining in popularity, there are countless other investment strategies in use today. Value and growth investing are both examples, and view functions incorporating these
strategies are an interesting topic of further research.
III. Theoretical Framework
As mentioned in the literature review, Markowitz specifies a mean-variance utility function with respect to the portfolio asset weight vector, w. The investor's goal is to maximize the expected
return while minimizing the volatility and he does so by maximizing the utility function
where R[T] is the current period?s return, R[T+1] is the future period?s return, γ is the investor?s risk aversion coefficient, μ is the sample return vector and Σ is the sample covariance matrix.
This is referred to as a two moment utility function since it incorporates the distribution's first two moments, the mean and variance. The first order condition of this utility function, with
respect to w, solves to
which can be used to solve for the optimal portfolio weights given the historical data.
BL first specify their model by determining the expected market equilibrium returns. To do so, they solve for μ in (2) by plugging in the sample covariance matrix and the market equilibrium weights.
The sample covariance matrix comes from the data and the market equilibrium weights are simply the percentage that each country's market capitalization makes up of the total portfolio market
In equilibrium, if we assume that the CAPM holds and that all investors have the same risk aversion and views on the market, the demand for any asset will be equal to the available supply. The supply
of an asset is simply its market capitalization, or the amount of dollars available of the asset in the market. In equilibrium when supply equals demand, we know that the weights of each asset in the
optimal portfolio will be equal to the supply, or the market capitalization of each asset. Σ is simply the historical covariance matrix, so we therefore know both w and Σ in (2), meaning we can solve
for μ^e, the equilibrium expected excess returns.
It is also assumed that the true expected excess return, μ, is normally distributed with mean μ^e and covariance matrix [τ]Σ. This can be written as
where μ^e is the market equilibrium returns, τ is a scalar indicating the confidence of how the true expected returns are modeled by the market equilibrium, and Σ is the fixed sample covariance
matrix. It is common practice to use a small value of tau since one would guess that long-term equilibrium returns are less volatile than historical returns.
We must also incorporate the investor?s views, which can be modeled by
where P is a K × N matrix that specifies K views on the N assets, and Ω is the covariance matrix explaining the degree of confidence that the investor has in his views. Ω is one of the harder
variables to specify in the model, but [?] provide a method that also helps with the specification of τ. Ω is a diagonal matrix since it is assumed that views are independent of one another, meaning
all covariance (non-diagonal) elements of the matrix are zero. Each diagonal element of Ω can be thought of as the variance of the error term, which can be specified as P[i]ΣP'0[i], where P[i] is an
individual row (view) from the K × N view specifying matrix, and Σ is again the historical covariance matrix. Again, I do not agree with this overemphasis on the historical covariance matrix, but I
include it here for simplicity of explaining the intuition of the model.
Intuition calibrate the confidence of each view by shrinking each view's error team by multiplying it by τ. This makes τ independent of the posterior analysis because it is now incorporated in the
same manner in the two stages of the model. If it is drastically increased, so too are be the error terms of Ω, but the estimated return vector, shown in (5) is not changed because there is be an
identical effect on Σ.
We can combine these two models by Bayesian updating, which leaves us with the Black-Litterman mean and variance
The Black-Litterman posterior covariance matrix is simply [(τΣ[h])^−1+P'Ω^−1P]^−1. The extra addition of Σ occurs because the investor must account for the added uncertainty of making a future
prediction. This final distribution is referred to as the posterior predictive distribution and is derived through Bayesian updating. There is an added uncertainty in making a prediction of an
unknown, future value, and to account for this the addition of Σ is necessary.
It is assumed that both the market equilibrium and the investor?s views follow a multivariate normal distribution, so it is known that the posterior predictive distribution is also multivariate
normal due to conjugacy. In order to find the optimal portfolio weights μ[BL] and Σ[BL] are simply plugged into (2).
Once the Black-Litterman results are specified we have the joint prior for the Bayesian extension. We combine this prior with the normal likelihood function describing the data^4, and based off of
Bayesian updating logic we obtain the posterior predictive mean, μ[bayes] and covariance matrix, Σ[bayes],
where Σ is the historical covariance matrix, μ[h] are the historical means of the asset returns, Δ = (τΣ)^−1+P'Ω^−1P]^−1 is the covariance matrix of the Black-Litterman estimate, and T is the sample
size of the data, which is the weight prescribed to the sample data. The larger the sample size chosen, the larger the weight the data has in the results. It is common practice to let T = n, unless
we do not have a high level of confidence in the data and want T <n. The number of returns is specified independently from the data because only the sample mean and covariance matrix are used in the
analysis, not the individual returns. This is ideal because it allows the investor to set the confidence in the data without the sample size doing it automatically. Historical return data is often
lengthy, but that does not necessarily mean a high degree of confidence should be prescribed to it.
Analogous to the Black-Litterman model, the posterior estimate of Σ in Zhou is [(Δ^−1 +(Σ/T )^−1]^−1. The addition of Σ to the posterior in calculating Σ[bayes] is necessary to account for the added
uncertainty of the posterior predictive distribution. The theory behind this is identical to that in the Black-Litterman model.
It is known that both the prior and likelihood follow a multivariate normal distribution, so due to conjugacy the same is true of the posterior predictive distribution. The posterior mean is a
weighted average of the Black-Litterman returns and the historical means of the asset returns. As the sample size increases, so does the weight of the historical returns in the posterior mean. In the
limit if T = ∞, then the portfolio weights are identical to the mean-variance weights, and if T = 0 then the weights are identical to the Black-Litterman weights.
As it stands, the Zhou model uses the sample covariance matrix in the prior generating stage, even though in a fully Bayesian analysis a full incorporation of historical data is not supposed to occur
outside of the likelihood function. This means the data is used to generate the prior views, and then further update the views by again incorporating the data through the likelihood function.
To account for the uncertainty of modeling Σ under the historical covariance matrix in each stage, I will impose an inverse-Wishart prior on the Black-Litterman covariance matrix. Under this method,
the historical covariance matrix will still be used in both Bayesian updating stages, but I can now better account for the potential problems of doing so through the inverse-Wishart prior.
The inverse-Wishart prior changes only the specification of Σ , not μ, and is specified by W[−1](Ψ, v.0) where Ψ is the prior mean of the covariance matrix, and v.0 is the degrees of freedom of the
distribution. The larger the degrees of freedom, the more confidence the investor has in Ψ as an estimate of Σ. In this case, Ψ = Σ[BL], and v.0 can be thought of as the number of "observations" that
went into the prior5.
The prior is then updated by the likelihood function, the historical estimate of Σ. μ[BL] is also updated by the historical data, but the analysis does not change the specification of μ since the
prior is only put on the Σ[BL]. The posterior distribution of Σ is also an inverse-Wishart distribution due to the conjugate Bayesian update and is defined as W[−1]((Ψ+S[μ]), (v.0+T)), where S[μ] is
the historical data generated sum of squares matrix, and T the number of observations that were used to form the likelihood. T is specified in the same manner as in the Zhou model; it is up to the
investor to set confidence in the data through T as it does not necessarily need to be the actual number of observations.
I use the mean of the posterior inverse-Wishart distribution to define the posterior covariance matrix of the extension. The mean of the posterior is defined as E[Σ|μ, y[1], ..., y[n]] = 1/v.0+T−n−1
(Ψ+S[μ]), where y[1], ..., y[n] is the observed data, and n is the number of potential assets in the portfolio. This posterior matrix is then added to the historical covariance matrix in order to get
the posterior predictive value, σ[ext]. The specification of μ is not affected under this model so μ[ext] = μ[bayes]
IV. Data
Monthly dollar returns from 1970-2013, for the countries in question^6 were obtained from Global Financial Data, and I used that raw data to calculate the n = 528 monthly percent returns. The
analysis is based on excess returns, so assuming the investor is from the U.S. I use the 3-month U.S. Treasury bill return as the risk-free rate.
Data must also be incorporated to describe the market equilibrium state of the portfolio. I collected this data from Global Financial Data and am using the market capitalizations of the entire stock
markets in each country from January, 1980 to December, 2013. Given the rolling window used in my analysis, January, 1980 is the first month where market equilibrium data is needed
Table 1 presents descriptive statistics for the seven country indices I am analyzing. The mean annualized monthly excess returns are all close to seven percent and the standard deviations are all
close to 20 percent. The standard deviation for the U.S. is much smaller than the other countries, which makes sense because safer investments generally have less volatility in returns. All countries
exhibit relatively low skewness, and most countries have a kurtosis that is not much larger than the normal distributions kurtosis of 3. The U.K. deviates the most from the normality assumption given
it has the largest absolute value of skewness and a kurtosis that is almost two times as large as the next largest kurtosis. I am not particularly concerned by these values, however, because the
dataset is large and the countries do not drastically differ from a normal distribution. The U.K. is the most concerning, but a very large kurtosis is less problematic than a very large skewness and
the skewness is greatly influenced by one particularly large observation that occurred in January of 1975, during a recession. Though the observation is an outlier, it seems to have occurred under
legitimate circumstances so I include it in the analysis.
Table 1: Analysis of Country Index Returns
V. Model Implementation
Rolling Window
A predictive model is best tested under repeated conditions when it uses a subset of the data as "in-sample" data to predict the "out-of sample" returns. This simulates how a model would be
implemented in a real investment setting since there is obviously no data incorporated in the model for the future prediction period. If I were to include the observations I was also trying to
predict, I would artificially be increasing the predictive power of the model by predicting inputs.
I am using a 10 year rolling window as the in sample data to predict the following month. I begin with the first 10 years of the dataset, January, 1970 December, 1980, to predict returns and optimal
asset weights for the following month, January 1981. I then slide the window over one month and use February, 1970 January, 1981 to predict returns and optimal asset allocations for February, 1981.
The dataset spans through 2013, giving me 528 individual returns. I therefore calculate 408 expected returns and optimal weights.
It is quite easy to assess performance once each set of optimal weights is calculated since there is data on each realized return. For each iteration I calculate the realized return for the entire
portfolio by multiplying each individual index's weight by its corresponding realized return. I do not have any investment constraints in the model so I also need to account for the amount invested
in, or borrowed from, the risk-free rate. One minus the sum of the portfolio weights is the amount invested in (or borrowed from, if negative) the risk-free rate.
Momentum Based views
In order to be able to run the model in an updating fashion, I must to create a function that will iteratively specify the investor's views, and I will do so using a momentum based investment
strategy. I have created a function that uses both a primary absolute strategy and a secondary relative strategy that is explained below.
The primary strategy estimates absolute views based on the mean and variance of the previous twelve months, since this is the known window for Momentum. This is a loose adaption of our momentum
strategy that specifies that stocks that have performed well in the past twelve months will continue to do so in the following month. By taking the mean I can account for the fact that at many times,
the indices have no momentum, in which case I expect the mean to be close to zero. For this strategy, since I am only specifying absolute views, the P matrix is an identity matrix with a dimension
equal to the number of assets in question. The Omega matrix is again calculated using the method specified by Intuition.
The secondary strategy, which is appended to both of these primary strategies, if the conditions hold, attempts to find indices that are gaining momentum quickly in the short term. To do this I look
at the last 4 months of the returns to see if they are consistently increasing or decreasing. If the index is increasing over the four months, it is given a positive weight, and if it is decreasing
over the four months it is given a negative weight. I use a four-month increasing scheme to catch the indices under momentum before they hit the standard six-month cutoff. The weights are determined
by a method similar to the market capitalization weighting method used by Idzorek. The over-performing assets are weighted by the ratio of the individual market capitalization to the total
over-performing market capitalization, and the same goes for under-performing assets. This puts more weight on large indices, which is intuitive because there is likely more potential for realized
returns in this case. The expected return of this view is a market capitalization weighted mean of each of the indices that have the specified momentum.
This is a fairly strict strategy, which is why I refer to it as secondary. For each iteration, sometimes there are no underperforming or over performing assets under the specifications. In this case,
only the primary strategy is used. If assets do appear to have momentum given the definition, then it is appended to the P matrix along with the primary strategy.
VI. Results
The results of the four models are presented below in Table 2. It must be considered that the results are heavily dependent on the dataset and the view specifying function, two aspects of the model
that are not necessarily generalizable to an investor. Further empirical analysis of the models is therefore necessary to determine which is best under the varying conditions of the current
investment market.
Table 2: Portfolio Optimization Results
The Markowitz model performs the worst of the models, both in terms of volatility and returns. A high volatility implies that the returns for each iteration are not consistent, which is a known
feature of the Markowitz model. The results also imply that given the dataset, the historical mean and covariance do not do a great job on their own as data inputs in the portfolio optimization
problem. This is consistent with the original hypothesis that further data inputs are necessary in conjunction with a more robust modeling procedure to improve the overall model.
The Black-Litterman model outperforms the Zhou model in both returns and volatility, meaning that in this analysis the incorporation of the historical data is not optimal. However, this does not
render the Zhou model useless since repeated empirical analysis is necessary to determine the actual effects of the historical data. In Zhou only one iteration of the model is run as brief example,
so there is currently no sufficient literature on whether the historical data is an optimal addition. A robust model testing procedure could be employed by running a rolling-window model testing
procedure on many datasets, and then running t-tests on the set of returns and volatilities specified under each dataset to find if one model outperforms the other.
The inverse-Wishart prior performs significantly better than in volatility than all the other models, and is only beaten by the Black-Litterman model in returns. This is in line with the hypothesis
that the inverse-Wishart prior will better specify the covariance matrix which will in turn lead to safer investment positions. Low volatility portfolios generally do not have high returns, and given
that the volatility of the extension is so much lower than the Black-Litterman volatility, it is not surprising that the return is also lower.
VII. Discussion
In exploring the results of the extended Zhou model it is clear that fully Bayesian models are able to outperform models that use loosely Bayesian methods. The inverse-Wishart extension outperforms
the Zhou model in portfolio volatility by accounting for the uncertainty of modeling Σ and by allowing the investor to further specify confidence in the Black-Litterman and historical estimates. The
parameters are straightforward and determined by the investor's confidence in each data input, which makes the model relatively simple and usable by any type of investor.
The Black-Litterman model, which is used as a joint prior in extended model, allows the investor to incorporate any sort of views on the market. The views can be determined in a oneoff nature views
or by a complex iterative function specifying a specific investment strategy. The former would likely be employed by an amateur, independent investor while the latter by a professional or investment
team. The data updating stage has similar flexibility in that the historical means, or a more complex data modeling mechanism, can be employed depending on the quantitative skills of the investor.
The incorporation of a predictive model is a topic of further research that could significantly increase the profitability of the Bayesian model, though it would also greatly increase the complexity.
Asset return predictions models can also be incorporated in a much simpler manner through the use of absolute views.
The inverse-Wishart prior is used to model the uncertainty of predicting the next period's covariance matrix, which is not fully accounted for in the original Zhou model. This method works well
empirically in this analysis, but further empirical testing is necessary to see if it consistently out-performs the Zhou model.
A further extension that could account for the problems in modeling Σ is through use a different estimate of Σ in the equilibrium stage, rather than just the historical covariance. When many assets
are being analyzed, the historical covariance matrix does estimate Σ well, so using another method of prediction could be very useful. Factor and stochastic volatility models could both provide
another robust estimate of Σ in the equilibrium stage.
Another possible extension that is possible under the inverseWishart prior is to fully model the posterior predictive distribution, rather than simply using the mean value of the posterior
inverse-Wishart distribution as the posterior estimate. The posterior predictive distribution under the inverseWishart prior is t-distributed, which may also be useful the since financial data is
known to have fatter tails than the normal distribution. This would greatly increase the complexity of the model, however, since the expected utility would need to be maximized with respect to the
posterior t-distribution, and this can only be done through complex integration.
The results presented in this paper give an idea of how the models perform under repeated conditions through the use of the rolling window. However, each iteration of the rolling window is very
similar to the previous one since all but one data point is identical. In order to confidently determine if one model outperforms another, it is necessary to do an empirical analysis on multiple
As exemplified above, an investor can use many different strategies to specify the views, expected returns, and expected covariance matrix incorporated in the model. The method of combining these
estimates is also quite important as seen by the optimal performance of the extended model, which used the same data inputs but incorporated an inverse-Wishart prior. By using Bayesian strategies to
combine these different methods of prediction with the market equilibrium returns, the investor has a straightforward quantitative model that can help improve investment success. Almost all investors
base their decisions off how they view the assets in the market, and by using this model, or variations of it, they can greatly improve their chance of profitability by using robust methods of
1. Asness, C., Liew, J., and Stevens, R. (1997). Parallels between the cross-sectional predictability of stock and country returns. Journal of Portfolio Management, 23(3):79–87.
2. Berger, A., Ronen, I., and Moskowitz, T. (2009). The case for momentum investing. Black, F. and Litterman, R. (1992). Global portfolio optimization. Financial Analysts Journal, 48(5):28.
3. Black, F. and Litterman, R. (1992). Global portfolio optimization. Financial Analysts Journal, 48(5):28.
4. He, G. and Litterman, R. (1999). The intuition behind black-litterman model portfolios. Investment Management Research, Goldman Sachs and Co.Hoff, P. (2009). A First Course in Bayesian
Statistical Methods. Springer, New York, 2009 edition.
5. Idzorek, T. (2005). A step-by-step guide to the blacklitterman model.
6. Markowitz, H. (1952). Portfolio selection. The Journal of Finance, 7(1):77.
7. Zhou, G. (2009). Beyond black-litterman: Letting the data speak. The Journal of Portfolio Management, 36(1):36.
1. I am an undergraduate senior at Duke University double majoring in Economics and Statistics. I would like to thank both Scott Schmidler and Andrew Patton for serving as my advisors on this
thesis. I would also like to thank my parents, Sandra Eller and Greg Roeder, for their love, guidance and support throughout my life.
2. P(Θ|Y) = P(Y|Θ)P(Θ)/ƒP(Y|Θ)P(Θ)dΘ
3. For more information regarding the choice of the market equilibrium model, see BL(1992)
4. Zhou (2009) makes the same assumptions on returns as BL, that they are i.i.d.
5. Though no actual historical data observations were used in forming the prior, this interpretation keeps the model consistent given how the Bayesian updating process is conducted
6. Australia, Canada, France, Germany, Japan, the U.K. and the U.S.
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This module contains procedures and generic interfaces for evaluating the mathematical division and multiplication operators acting on integer, complex, or real values.
This module contains procedures and generic interfaces for evaluating the mathematical division and multiplication operators acting on integer, complex, or real values.
The procedures of this module offer a handy and flexible way of membership checks.
See also
Final Remarks ⛓
If you believe this algorithm or its documentation can be improved, we appreciate your contribution and help to edit this page's documentation and source file on GitHub.
For details on the naming abbreviations, see this page.
For details on the naming conventions, see this page.
This software is distributed under the MIT license with additional terms outlined below.
1. If you use any parts or concepts from this library to any extent, please acknowledge the usage by citing the relevant publications of the ParaMonte library.
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This software is available to the public under a highly permissive license.
Help us justify its continued development and maintenance by acknowledging its benefit to society, distributing it, and contributing to it.
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What is meant by the terms ‘odds-on’ and ‘outsider’? | Horse Racing Guide
What is meant by the terms ‘odds-on’ and ‘outsider’?
Unsurprising, ‘odds-on’ is the opposite of ‘odds-against’. In either case, odds represent the implied probability, or percentage chance, of a particular outcome occurring. For example, if you place a
bet at 2/1, or ‘two to one against’, you have a 33.33% chance of winning and a 66.67% chance of losing. In other words, that particular outcome is twice as likely not to happen as happen, which is
reflected by the fact that you win £2 for each £1 staked, if your bet is successful. However, if you place a bet at 1/2, or ‘two to one on’, the reverse is true, so you win just £1 for each £2
staked, if your bet is successful. Put simply, any selection that has an implied winning chance of better than 50%, which is reflected by odds shorter than 1/1, or even money, is odds-on. An odds-on
selection is deemed more likely to win than lose, sometimes significantly so, such that the profit on a winning bet is always less than the stake.
By contrast, an ‘outsider’ is about as far removed as possible from odds-on betting but, beyond that, a clear, unambiguous and objective definition of exactly what constitutes an outsider is
difficult. Also known as a ‘longshot’, an outsider has, at least on paper, little chance of winning a race and is, consequently offered at relatively long odds when compared with some or all of its
rivals. However, ‘relatively’ is the operative word here, because there is no threshold beyond which a horse becomes an outsider. In a three-runner race, for example, the outsider of the trio might
be offered at odds of, say, 9/4, while in a forty-runner race, such as the Grand National, outsiders at 66/1, 100/1 or even longer odds are commonplace. | {"url":"https://racingguide.co.uk/what-is-meant-by-the-terms-odds-on-and-outsider/","timestamp":"2024-11-06T17:55:53Z","content_type":"text/html","content_length":"33442","record_id":"<urn:uuid:47f77299-6b34-48ce-a732-3bef1c356334>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00326.warc.gz"} |
Making sense of the canonical anti-commutation relations for Dirac spinors
4438 views
When doing scalar QFT one typically imposes the famous 'canonical commutation relations' on the field and canonical momentum: $$[\phi(\vec x),\pi(\vec y)]=i\delta^3 (\vec x-\vec y)$$ at equal times
($x^0=y^0$). It is easy (though tedious) to check that this implies a commutation relation for the creation/annihilation operators $$[a(\vec k),a^\dagger(\vec k')]=(2\pi)^32\omega\delta^3(\vec k-\vec
k') $$
When considering the Dirac (spinor) field, it is usual (see e.g. page 107 of Tong's notes or Peskin & Schroeder's book) to proceed analogously (replacing commutators with anticommutators, of course).
We postulate $$\{\Psi(\vec x),\Psi^\dagger(\vec y)\}=i\delta^3(\vec x -\vec y)$$ and, from them, derive the usual relations for the creation/annihilation operators.
I'd always accepted this and believed the calculations presented in the above-mentioned sources, but I suddenly find myself in doubt: Do these relations even make any sense for the Dirac field? Since
$\Psi$ is a 4-component spinor, I don't really see how one can possibly make sense out of the above equation: Isn't $\Psi\Psi^\dagger$ a $4\times 4$ matrix, while $\Psi^\dagger\Psi$ is a number?! Do
we have to to the computation (spinor-)component by component? If this is the case, then I think I see some difficulties (in the usual computations one needs an identity which depends on the
4-spinors actually being 4-spinors). Are these avoided somehow? A detailed explanation would be much appreciated.
As a follow-up, consider the following: One usually encounters terms like this in the calculation: $$u^\dagger \dots a a^\dagger\dots u- u\dots a^\dagger a \dots u^\dagger $$ Even if one accepts that
an equation like $\{\Psi,\Psi^\dagger\}$ makes sense, most sources simply 'pull the $u,\ u^\dagger$ out of the commutators' to get (anti)commutators of only the creation/annihilation operators. How
is this justified?
EDIT: I have just realized that the correct commutation relation perhaps substitutes $\Psi^\dagger$ with $\bar \Psi$ (this may circumvent any issue that arises in a componentwise calculation). Please
feel free to use either in an answer.
This post imported from StackExchange Physics at 2014-12-06 00:44 (UTC), posted by SE-user Danu
Most voted comments show all comments
@glance I think you should be given the honors, and just include a reference to P&K substantiate your comment. Agreed that 3.89 is inappropriate.
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user joshphysics
@glance deriving the CCR for the creation/annihilation operators of the Dirac field would be more than satisfactory.
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Danu
@joshphysics Of course, 3.89 is not the calculation for $\{\cdot,\cdot\}$ but it's the only explicit calculation, so that's why I was looking there ;)
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Danu
Pretty anticlimactic, duh.
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Daniel
@Daniel lol, yeah... I guess that's how these things are bound to end.
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Danu
Most recent comments show all comments
@joshphysics damnit, I was looking at 3.89 and didn't see any components, and freaked out. I guess that settles it...
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Danu
Somebody should post that as an answer! (FWIW I came to the same conclusion from Tong's notes)
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user David Z
One usually starts from the CCR for the creation/annihilation operators and derives from there the commutation rules for the fields. However, one can start from either (see for example here about
this). Suppose we want then to start from the equal-time anticommutation rules for a Dirac field $\psi_\alpha(x)$: $$ \tag{1} \{ \psi_\alpha(\textbf{x}), \psi_\beta^\dagger(\textbf{y}) \} = \delta_{\
alpha \beta} \delta^3(\textbf{x}-\textbf{y}),$$ where $\psi_\alpha(x)$ has an expansion of the form $$ \tag{2} \psi_\alpha(x) = \int \frac{d^3 p} {(2\pi)^3 2E_\textbf{p}} \sum_s\left\{ c_s(p) [u_s
(p)]_\alpha e^{-ipx} + d_s^\dagger(p) [v_s(p)]_\alpha e^{ipx} \right\}$$ or more concisely $$ \psi(x) = \int d\tilde{p} \left( c_p u_p e^{-ipx} + d_p^\dagger v_p e^{ipx} \right), $$
and we want to derive the CCR for the creation/annihilation operators: $$ \tag{3} \{ a_s(p), a_{s'}^\dagger(q) \} = (2\pi)^3 (2 E_p) \delta_{s s'}\delta^3(\textbf{p}-\textbf{q}).$$ To do this, we
want to express $a_s(p)$ in terms of $\psi(x)$. We have: $$ \tag{4} a_s(\textbf{k}) = i \bar{u}_s(\textbf{k}) \int d^3 x \left[ e^{ikx} \partial_0 \psi(x) - \psi(x) \partial_0 e^{ikx} \right]\\ = i \
bar{u}_s(\textbf{k}) \int d^3 x \,\, e^{ikx} \overset{\leftrightarrow}{\partial_0} \psi(x) $$ $$ \tag{5} a_s^\dagger (\textbf{k}) = -i \bar{u}_s(\textbf{k}) \int d^3 x \left[ e^{-ikx} \partial_0 \psi
(x) - \psi(x) \partial_0 e^{-ikx} \right] \\ =-i \bar{u}_s(\textbf{k}) \int d^3 x \,\, e^{-ikx} \overset{\leftrightarrow}{\partial_0} \psi(x) $$ which you can verify by pulling the expansion (2) into
(4) and (5). Note that these hold for any $x_0$ on the RHS.
Now you just have to insert in the anticommutator on the LHS of (3) these expressions and use (1) (I can expand a little on this calculation if you need it).
most sources simply 'pull the $u, u^\dagger$ out of the commutators' to get (anti)commutators of only the creation/annihilation operators. How is this justified?
There is a big difference between a polarization spinor $u$ and a creation/destruction operator $c,c^\dagger$.
For fixed polarization $s$ and momentum $\textbf{p}$, $u_s(\textbf{p})$ is a four-component spinor, meaning that $u_s(\textbf{p})_\alpha \in \mathbb{C}$ for each $\alpha=1,2,3,4$. Conversely, for
fixed polarization $s$ and momentum $\textbf{p}$, $c_s(\textbf{p})$ is an operator in the Fock space. Not just a number, which makes meaningful wondering about (anti)commutators.
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user glance
By "double arrow" do you mean $\Longleftrightarrow$ or two right arrows ontop of each other?
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Kyle Kanos
I mean the arrow which points on both sides.. the symbol used to indicate a derivative on the right minus a derivative on the left: in $ a \bar{\partial}_\mu b \equiv a \partial_\mu b - (\partial_\mu
a) b$ the symbol that would normally be used instead of the bar in $\bar{\partial}$
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user glance
\overset{\leftrightarrow}{\partial} $\overset{\leftrightarrow}{\partial}$
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Robin Ekman
A single lined arrow that points both ways is \leftrightarrow: $\leftrightarrow$. You can put that over by using \overset{up}{down}: $\overset{\leftrightarrow}{\partial_\mu}$.
This post imported from StackExchange Physics at 2014-12-06 00:45 (UTC), posted by SE-user Kyle Kanos | {"url":"https://www.physicsoverflow.org/25233/making-sense-canonical-commutation-relations-dirac-spinors","timestamp":"2024-11-09T20:13:36Z","content_type":"text/html","content_length":"168150","record_id":"<urn:uuid:e71299c1-e679-4383-b999-622489e11aba>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00636.warc.gz"} |
Part B: The Median and the Three-Number Summary (35 Minutes) - Annenberg Learner
Private: Learning Math: Data Analysis, Statistics, and Probability
Min, Max and the Five-Number Summary Part B: The Median and the Three-Number Summary (35 Minutes)
In This Part: The Median
Another useful summary measure for a collection of data is the median. As you learned in Session 2, the median is the middle data value in an ordered list. Here’s one way to find the median of our
ordered noodles.
First, place your 11 noodles in order from shortest to longest on a new piece of paper or cardboard. Your arrangement should look something like this:
Next, remove two noodles at a time, one from each end, and put them to the side:
Continue this process until only one noodle remains. This noodle is the median. Label it “Med”:
Notice that the median divides the set of 11 noodles into two groups of equal size — the five noodles shorter than the median and the five noodles longer than the median. Another way to say this is
that there are just as many noodles before the median as there are after the median.
Problem B1
If you could see only the median noodle, what would you know about the other noodles?
Problem B2
If you could see only the median noodle, describe some information you would not know about the other noodles.
In This Part: The Three-Noodle Summary
Now remove all the noodles except Min, Med, and Max.
We’ll call this display the “Three-Noodle Summary.
Problem B3
If you could see Min, Med, and Max, what would you know about the other noodles? Be specific about how this compares to Problem A3 (where you only knew Min and Max) and Problem B1 (where you only
knew Med).
Problem B4
Describe some information you still wouldn’t know about the other noodles from the Three-Noodle Summary.
In This Part: The Three-Number Summary
Now let’s convert the Three-Noodle Summary to the Three-Number Summary. If they’re not already there, place the three noodles — Min, Med, and Max — in order on the horizontal axis.
Next add a vertical number line, and mark the lengths of the three noodles. (Left)
Remove the noodles, and you’re left with the Three-Number Summary. (Right)
Problem B5
If we call the length of the fourth noodle N4, how does N4 compare to Min, Med, and Max? What wouldn’t you know about N4 if you only knew Min, Med, and Max?
In This Part: Even Data Sets
In the previous example, it wasn’t hard to find the median because there were 11 noodles — an odd number. For an odd number of noodles, the median is the noodle in the middle. But how do we find the
median for an even number of noodles?
Add a 12th noodle, with a different length from the other 11 noodles, to the original collection. Arrange the noodles in order from shortest to longest.
Problem B6
Using the method of removing pairs of noodles (the longest and the shortest), try to determine the median noodle length. What happens?
This time, there won’t be one remaining noodle in the middle — there will be two! If you remove this middle pair, you’ll have no noodles left.
Therefore, you’ll need to draw a line midway between the two remaining noodles to play the role of the median. The length of this line should be halfway between the lengths of the two middle noodles:
Move the middle pair aside, and you can see your new median:
Notice that this median still divides the set of noodles into two groups of the same size — the six noodles shorter than the median and the six noodles longer than the median:
The major difference is that, this time, the median is not one of the original noodles; it was computed to divide the set into two equal parts.
Note: It is a common mistake to include this median in your data set when you’ve added it in this way. This median, however, is not part of your data set.
In this video segment, participants discuss the process of finding the median of a data set with an even number of values (in this case n = 20). Watch this video segment to review the process you
used in Problem B6 or if you would like further explanation.
Note: The data set used by the onscreen participants is different from the one provided above.
Problem B7
If you could see only the median of a set of 12, what would you know about the other noodles?
You can convert the Three-Noodle Summary for these 12 noodles to the Three-Number Summary in the same way you did it for the set of 11 noodles:
Add a vertical number line, and mark the lengths of the three noodles:
Remove the noodles, and you’re left with the Three-Number Summary:
In This Part: Review
As we have seen with the noodle examples, the median divides ordered numeric data into two groups, each with the same number of data values.
If you only know the Three-Number Summary (Min, Med, and Max) for a set of data, you can still glean quite a bit of information about the data. You know that all the data values are between Min and
Max, and you know that Med divides the data into two groups of equal size. One group contains data values to the left of Med, and the other group contains data values to the right of Med. You also
know that the group of values to the left of the median must be lower than (or equal to) the median in value, and that the group of values to the right of the median must be greater than (or equal
to) the median in value.
Problem B1
You would know that there must be exactly five noodles shorter than the median noodle and five noodles longer than the median noodle.
Problem B2
You would not know the actual values of any of the other noodles: The five shorter noodles could be extremely short, the five longer noodles could be many feet long, they could all be fairly close in
size to the median, etc. You would also not know or be able to estimate the maximum or minimum length of the other noodles.
Problem B3
You would know that all of the noodles are between Min and Max, and you can divide the noodles into two equal groups: five that are shorter than Med (including Min) and five that are longer than Med
(including Max). This information gives you two specific intervals that contain an equal number of noodles, and all of the noodles are contained in these intervals. This is different from Problem A3,
where you knew nothing about the size of the noodles between Min and Max, and from Problem B1, where you knew nothing about the upper and lower boundaries of your data set.
Problem B4
You still wouldn’t know the lengths of the noodles in the two intervals between Min and Med, or between Med and Max. These noodles could be very close to Med, very close to the extreme values, evenly
spread within the intervals, or something else entirely. There is no way to know without more information.
Problem B5
You would know that N4 must be larger than Min, smaller than Med, and smaller than Max. This is true because N6 is the median, and N4 must be smaller than N6. You still wouldn’t know N4’s actual
value or whether N4 was closer to Min or to Med. (A common mistake is to claim that N4 must be closer to Med than it is to Min. This is not necessarily true, since the values of N2 through N5 can be
anywhere in the interval between Min and Med; for example, they could all be very close to Min.) | {"url":"https://www.learner.org/series/learning-math-data-analysis-statistics-and-probability/the-five-number-summary/the-median-and-the-three-number-summary-35-minutes-the-median/","timestamp":"2024-11-11T13:13:49Z","content_type":"text/html","content_length":"121367","record_id":"<urn:uuid:1ec782dd-387b-422d-b4dd-b3490a2cac74>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00454.warc.gz"} |
Benford's Law of Amazon Rankings
Late last year, Matthew Beckler was nice enough to make a sales rank tracker for How to Teach Physics to Your Dog. Changes in the Amazon page format made it stop working a while ago, though, and now
Amazon reports roughly equivalent data via its AuthorCentral feature, with the added bonus of BookScan sales figures. So I've got a new source for my book sales related cat-vacuuming.
Still, there's this great big data file sitting there with thousands of hourly sales rank numbers, and I thought to myself "I ought to be able to do something else amusing with this..." And then
Corky at the Virtuosi did a post about Benford's Law, and I said "Ah-ha!"
Benford's Law, if you're not familiar with it, says that in a large assortment of numbers generated by some process, you expect the first non-zero digits of all the numbers to be distributed in a
logarithmic fashion. About 30% of the first digits should be "1," and only about 4% of the first digits should be "9." This goes against the naive expectation that the numbers ought to be evenly
distributed, and is actually used by "forensic accountants" to catch people who are cooking their books-- someone who is making up numbers to fill a phony set of books is fairly likely to pick
numbers that don't follow a Benford's Law distribution.
So, I've got 6,818 hourly values of the Amazon sales rank for my book, spanning almost three orders of magnitude. How do those digits match up with Benford's Law? Well:
That's... pretty good, really. The blue diamonds are the actual frequency of the digit, the red squares are the prediction of Benford's Law. There's a slight shortage of 1's and a surplus of 5's and
6's, but all the actual frequencies are within about 5% of the expected values. The most basic assumption about the statistics of this sort of data set would lead you to expect an uncertainty of
about 1% (that is, 1 over the square root of 6818), but that's pretty crude.
What does this tell us? Not a whole lot, really. if Amazon is somehow fudging their sales rank data (which I have no reason to suspect them of doing), they're clever enough not to get caught by this
really crude analysis of one book's figures.
Making this graph has, however, given me a way to put off some tedious and annoying work for another hour or so, so let's hear it for Benford's Law!
More like this
A few months ago-- just before the paperback release of How to Teach Physics to Your Dog-- Amazon started providing not only their Sales Rank data, but also sales data from Nielsen BookScan. Of
course, the BookScan data is very limited, giving you only four weeks, and the Sales Rank data, while…
Head down to Box Office Mojo and pull up the list of the top grossing films of the year thus far. Seven of the top ten have a dollar gross beginning with the number 1. Okay, that's not too weird. Big
films tend to pull down somewhere between $100-200 million, while only the real monsters have high…
When I saw the data generated by the sales rank tracker Matthew Beckler was kind enough to put together, I joked that I hoped to someday need a logarithmic scale to display the sales rank history of
How to Teach Physics to Your Dog. Thanks to links from Boing Boing, John Scalzi, and Kevin Drum, I…
I'm trying not to obsessively check and re-check the Dog Physics Sales Rank Tracker, with limited success. One thing that jumped out at me from the recent data, though, is the big gap between the
book and Kindle rankings over the weekend. The book sales rank dropped (indicating increased sales,…
Nice illustration of the law!
Did you know that the eponymous Frank Benford after whom the law was named was a research physicist at GE who lived just down the road from Union College (on Rugby Road)?
Benford was inspired to investigate the law empirically after noticing patterns in the dirtiness of the logarithm pages. (This was back in the days when scientists spent a good chunk of time looking
up the logs of their data in order to speed up their calculations.) If he (or the prior discoverer of the law, Newcomb) had had access to graphing calculators, who knows if or when anyone would have
Is there any similar law for second and subsequent digits? I can't see why there would be, but then I'm no mathematician, and I would have guessed that the first digits would be random.
Neat. Benford wins again!
A very interesting post, and it got me thinking...
I have a script for my own book, and a similarly sized dataset, so I decided to run my own numbers. I get a result very similar to your own (1=24%, 2=15%...9=5%), but it's not entirely obvious why. I
don't mean that it's not obvious why Benford's law has the form that it does. What I mean is that it's not clear that sales ranks should obey it. The distribution of ranks is most definitely not
scale invariant, for example. A certain rate of sales corresponds (more or less) to a certain sale, and at high sales rates, the variance will be relatively small.
And clearly this distribution can't be a general law for a fixed population. Suppose there were exactly a million books. At any given instant, exactly 1/9th have each leading digit (Technically, "1"
has 111,112, but that's just quibbling). From a frequentist perspective, Benford's law simply can't drop out.
Of course, since the number of books isn't 1 million, and isn't fixed, this isn't a perfect line of reasoning, but I must confess, I'm still puzzled why amazon ranks should follow Benford's law.
I have a script for my own book, and a similarly sized dataset, so I decided to run my own numbers. I get a result very similar to your own (1=24%, 2=15%...9=5%), but it's not entirely obvious why. I
don't mean that it's not obvious why Benford's law has the form that it does. What I mean is that it's not clear that sales ranks should obey it. The distribution of ranks is most definitely not
scale invariant, for example. A certain rate of sales corresponds (more or less) to a certain sale, and at high sales rates, the variance will be relatively small.
The addition of BookScan data to the stuff reported by Amazon will be an interesting check on the correspondence between Amazon sales rank and total sales. It'll be a while yet before I have enough
data on that to say anything meaningful, though.
I agree that the full set of sales ranks can't possibly follow Benford's Law, since all the ranks need to be filled (in principle, anyway-- I'm not sure how they handle ties). Any one randomly-chosen
book, though, will follow a trajectory through the ranks that is essentially random. That will presumably result in something close to a Benford's Law distribution for that one book's sales history.
At least, that's the justification I came up with when I started thinking about doing this. It may be that a more careful analysis would show a different distribution, and that the deficit of 1's
that we see is a real effect, but figuring that out is beyond me. Especially with classes starting tomorrow.
To me it seems obvious that if you have a random sample set, starting at zero (OK that is a bit unlikely, but most sets probably do), the chance if the first significant digit being 1 increases as
the first sig fig of the maximum size of the sample set drops toward 1. Once it reaches there, the chance diminishes again until you reach 9, and it starts again.
At any point, you will never have a situation that the chance of the first sig fig is less than 1/9, so that is the minimum. The maximum would be at, say 1 to 199, where the chances are (1/9 + 1)/2
(roughly). This is a range of 11% to 56%. If I simplistically assume an average of these, I get 33%, which is pretty close to what Benford says it is, at 30.1%.
I see comment (elsewhere) that state: âEveryone knows that our number system uses the digits 1 through 9 and that the odds of randomly obtaining any one of them as the first significant digit in a
number is 1/9. â
And that appears immediately false to me.
Benford's law seemed obvious to me as a perfectly natural thing to occur after about 10 seconds thinking about it. Am I missing something?
@Jerome, thanks for convincing me I'm perhaps not crazy. I've been having trouble seeing why this isn't an obvious consequence of the fact that a "random" number occuring in the real world is drawn
from a finite (i.e., less than so-and-so) set of numbers. | {"url":"https://scienceblogs.com/principles/2010/12/31/benfords-law-of-amazon-ranking","timestamp":"2024-11-07T06:58:12Z","content_type":"text/html","content_length":"54803","record_id":"<urn:uuid:5c8fbe0b-ed40-4fb2-8a1a-c7760908ba5d>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00880.warc.gz"} |
The Mega Mystery of our Creator’s Handiwork
Viewpoint of Mark Cadwallader, Creation Moments Board Chairman
In Ephesians 5:32 the Bible talks about a “mega mystery” being foreshadowed by Adam and Eve as well as by husbands and wives, “the two shall become one flesh” (Genesis 2:24 and Ephesians 5:31). It is
not just a “mystery”, according to the Bible, but a “great mystery” – a “mega” mystery in the Greek. With this in mind, let’s consider the following.
. It occurs in many created forms in nature that we consider beautiful. It is, for example, the basic ratio in the dimensions of the human hand and the human face. And it occurs in mathematics in a
uniquely elegant and mysterious manner. This is the Golden Section, ϕ or Phi, also known as the Golden Ratio or the “Divine Proportion” as it was first called by Leonardo Da Vinci. It is the
proportion defined by the illustration to the right for a line of any length.
Stay with me here for a minute to consider the truly remarkable connection between a mysterious signature feature of creation and the “mega mystery” of the Bible!
^2 is 2.6180339887…keeping again the exact same infinite decimal digits and being exactly the sum of the two normalized lengths within the Ratio, 1 + 1.6180339887… ! Starting with squares that are
extended into “Golden Rectangles”, you end up making a number series with a peculiar unique sequence to the ratioed lengths of their sides (diagram at right). Arcs drawn to connect the corners of
these squares within the Golden Rectangles form what are called a Golden Spiral, also very common to nature and to natural beauty!
Phi, the Divine Proportion, is thus connected to a peculiar mathematical sequence of numbers, a number series which is used to construct Golden Spirals from Golden Rectangles in the manner described
above and depicted at the right.
Interestingly, the same series of numbers is also formed by adding two whole numbers to become the next. Beginning from the origin, this gives the series 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,
144,…continuing to infinity where two adjacent numbers add to become the next. It’s called the Fibonacci series after the Italian mathematician Leonardo Fibonacci.
When you take the ratio of any two adjacent numbers within the series, you find their ratio converges to the Golden Ratio, Phi. For example, 21/13 = 1.61538…, and 144/89 = 1.61797… getting ever
closer by more and more significant figures to a more and more precise value of the Divine Proportion the further out in the series you go. Why should that be the case? And the interesting thing is
that the numbers in this series, just like with the Golden Ratio and Golden Spirals, also appear with high frequency in nature. For example, look at your hand. Not only is its length-to-width ratio
roughly a Golden Ratio, but you have 5 fingers, 3 knuckles on each finger and 2 knuckles on your thumb. And the ratio of hand bone lengths from fingertips to each knuckle all the way to the wrist are
in the case of each finger 2 to 3 to 5 to 8 – all numbers in this series of the Golden Ratio.
The Golden Ratio, this Proportion that to great artists and thinkers like Leonardo Da Vinci appears to have Divine status, is intimately connected with a number series that is formed by two numbers
adding together to become one more – starting from the origin! This is quite intriguing from a biblical point of view of the “mega mystery” we mentioned at the beginning – “the two shall become one”,
connected by a mathematical definition of wonder and beauty. It’s as if the Creator has connected His Creation with multiple applications made beautiful by a Proportion that is at the heart of
something even more mysterious regarding the beauty of intimate human fellowship and His Own connection to us! “For this cause shall a man leave his father and mother, and shall be joined unto his
wife, and they two shall become one flesh. This is a great mystery, but I speak concerning Christ and the church” (Ephesians 5:31-32) . The first sentence quoted above is essentially a quotation of
Genesis 2:24 with the creation of Eve out of Adam’s side, “they shall become one flesh”.
As we look back on the revelation of Scripture, Adam and Eve were indeed a reflection of Christ and the Church from the beginning. God’s creation of Eve out of Adam when Adam was in “a deep sleep” is
a picture of God’s formation of the Church out of the death of Christ. In fact, Scripture describes Christ as the second or last Adam (Romans 5:12-19 and 1 Corinthians 15:45). The difference is that
Adam’s DNA, his seed in the flesh, drives us to sin – through Adam we inherit the sin nature. But in Christ we inherit a life-giving spirit, a spiritual “seed that cannot sin” residing in us who are
“born of God” (1 John 3:9).
So here we have a betrothal, a marriage in the making, between the risen Jesus Christ and his redeemed Church, the community of individual believers who each carry within us not only the seed of Adam
– and are, therefore, human beings – but the spiritual seed of Jesus Christ! In effect, His “DNA” is in us, and it makes us a part of His body as DNA in each cell of our body makes it part of our
body. Our essence, seed, is in every part of our body. And His essence, seed, is in every part of His body! This is intimate communion with our Creator. And it is a marvelous, beautiful and wonderful
© 2018 Creation Moments All rights reserved. | {"url":"https://creationmoments.com/newsletter/the-mega-mystery-of-our-creators-handiwork/?print=print","timestamp":"2024-11-02T08:53:58Z","content_type":"text/html","content_length":"8256","record_id":"<urn:uuid:781f3d3f-a829-41c9-a60f-23c89dbf99cf>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00712.warc.gz"} |
Introduction to Mathematical Philosophy - Free Computer, Programming, Mathematics, Technical Books, Lecture Notes and Tutorials
Introduction to Mathematical Philosophy
• Title Introduction to Mathematical Philosophy
• Author(s) Bertrand Russell
• Publisher: George Allen and Unwin, UK (1919); Martino Fine Books (2017 Reprint); eBook (Public Domain)
• License(s): Public Domain Certification, CC BY-SA 3.0 US
• Hardcover/Paperback 220 pages
• eBook HTML, PDF, ePub, Kindle (Mobi), TeX, etc.
• Language: English
• ISBN-10/ASIN: 1684221447
• ISBN-13: 978-1684221448
• Share This:
Book Description
Introduction to Mathematical Philosophy has been a seminal work for more than nine decades. It gives the general background necessary for any serious discussion on the foundational crisis of
mathematics in the beginning of the twentieth century.
Requiring neither prior knowledge of mathematics nor aptitude for mathematical symbolism, the book serves as essential reading for anyone interested in the intersection of mathematics and logic and
in the development of analytic philosophy in the twentieth century. Russell offers to his readers a penetrating discussion on certain issues of mathematical logic that embodies the dawn of modern
analytic philosophy.
About the Authors Reviews, Rating, and Recommendations: Related Book Categories: Read and Download Links:Similar Books:
Book Categories
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Resources and Links | {"url":"https://freecomputerbooks.com/Introduction-to-Mathematical-Philosophy.html","timestamp":"2024-11-09T05:58:50Z","content_type":"application/xhtml+xml","content_length":"32976","record_id":"<urn:uuid:74d4d804-df17-4a81-b6bc-d3c84b4dbc8e>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00179.warc.gz"} |
ACM Other ConferencesSeparating Functional Computation from Relations
The logical foundation of arithmetic generally starts with a
quantificational logic over relations. Of course, one often wishes to have a formal treatment of functions within this setting. Both
Hilbert and Church added choice operators (such as the epsilon
operator) to logic in order to coerce relations that happen to encode functions into actual functions. Others have extended the term language with confluent term rewriting in order to encode
functional computation as rewriting to a normal form. We take a different approach that does not extend the underlying logic with either choice principles or with an equality theory. Instead, we use
the familiar two-phase construction of focused proofs and capture functional computation entirely within one of these phases. As a result, our logic remains purely relational even when it is
computing functions. | {"url":"https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.CSL.2017.23/metadata/acm-xml","timestamp":"2024-11-08T22:20:13Z","content_type":"application/xml","content_length":"4463","record_id":"<urn:uuid:c732329b-10b4-4705-babd-c1d21e392d47>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00308.warc.gz"} |
Bass Length-to-Weight Calculator
To calculate the weight of a bass:
\[ W = L^3 \times g \]
• \(W\) is the weight of the bass (lbs)
• \(L\) is the length of the bass (inches)
• \(g\) is the growth coefficient, typically 0.000011
Bass Length-to-Weight Definition
Bass Length to Weight refers to a calculation used by anglers to estimate the weight of a bass fish based on its length. This is often used in catch-and-release fishing where the fish is not actually
weighed to minimize harm.
Example Calculation
Let's assume the following value:
• Bass Length (\(L\)) = 20 inches
Using the formula:
\[ W = 20^3 \times 0.000011 = 0.88 \text{ lbs} \]
The estimated weight of the bass is 0.88 lbs.
BIT1024 Calculator© - All Rights Reserved 2024 | {"url":"https://waycalculator.com/tool/Bass-Length-to-Weight-Calculator.php","timestamp":"2024-11-10T18:09:24Z","content_type":"text/html","content_length":"5796","record_id":"<urn:uuid:e16737e1-031a-41db-861a-6e435186988a>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00732.warc.gz"} |
Lesson 16
Two Related Quantities, Part 1
Let’s use equations and graphs to describe relationships with ratios.
16.1: Which One Would You Choose?
Which one would you choose? Be prepared to explain your reasoning.
• A 5-pound jug of honey for $15.35
• Three 1.5-pound jars of honey for $13.05
16.2: Painting the Set
Lin needs to mix a specific shade of orange paint for the set of the school play. The color uses 3 parts yellow for every 2 parts red.
1. Complete the table to show different combinations of red and yellow paint that will make the shade of orange Lin needs.
│cups of red paint \((r)\)│cups of yellow paint \((y)\) │total cups of paint \((t)\)│
│2 │3 │ │
│6 │ │ │
│ │ │20 │
│ │18 │ │
│14 │ │ │
│16 │ │ │
│ │ │50 │
│ │42 │ │
2. Lin notices that the number of cups of red paint is always \(\frac25\) of the total number of cups. She writes the equation \(r=\frac25 t\) to describe the relationship. Which is the independent
variable? Which is the dependent variable? Explain how you know.
3. Write an equation that describes the relationship between \(r\) and \(y\) where \(y\) is the independent variable.
4. Write an equation that describes the relationship between \(y\) and \(r\) where \(r\) is the independent variable.
5. Use the points in the table to create two graphs that show the relationship between \(r\) and \(y\). Match each relationship to one of the equations you wrote.
A fruit stand sells apples, peaches, and tomatoes. Today, they sold 4 apples for every 5 peaches. They sold 2 peaches for every 3 tomatoes. They sold 132 pieces of fruit in total. How many of each
fruit did they sell?
Equations are very useful for describing sets of equivalent ratios. Here is an example.
A pie recipe calls for 3 green apples for every 5 red apples. We can create a table to show some equivalent ratios.
We can see from the table that \(r\) is always \(\frac53\) as large as \(g\) and that \(g\) is always \(\frac35\) as large as \(r\).
│green apples (\(g\)) │red apples (\(r\)) │
│3 │5 │
│6 │10 │
│9 │15 │
│12 │20 │
We can write equations to describe the relationship between \(g\) and \(r\).
• When we know the number of green apples and want to find the number of red apples, we can write:
\(\displaystyle r=\frac53g\)
In this equation, if \(g\) changes, \(r\) is affected by the change, so we refer to \(g\) as the independent variable and \(r\) as the dependent variable.
We can use this equation with any value of \(g\) to find \(r\). If 270 green apples are used, then \(\frac53 \boldcdot (270)\) or 450 red apples are used.
• When we know the number of red apples and want to find the number of green apples, we can write:
\(\displaystyle g=\frac35r\)
In this equation, if \(r\) changes, \(g\) is affected by the change, so we refer to \(r\) as the independent variable and \(g\) as the dependent variable.
We can use this equation with any value of \(r\) to find \(g\). If 275 red apples are used, then \(\frac35 \boldcdot (275)\) or 165 green apples are used.
We can also graph the two equations we wrote to get a visual picture of the relationship between the two quantities.
• dependent variable
The dependent variable is the result of a calculation.
For example, a boat travels at a constant speed of 25 miles per hour. The equation \(d=25t\) describes the relationship between the boat's distance and time. The dependent variable is the
distance traveled, because \(d\) is the result of multiplying 25 by \(t\).
• independent variable
The independent variable is used to calculate the value of another variable.
For example, a boat travels at a constant speed of 25 miles per hour. The equation \(d=25t\) describes the relationship between the boat's distance and time. The independent variable is time,
because \(t\) is multiplied by 25 to get \(d\). | {"url":"https://curriculum.illustrativemathematics.org/MS/students/1/6/16/index.html","timestamp":"2024-11-14T12:16:40Z","content_type":"text/html","content_length":"99840","record_id":"<urn:uuid:c5aa84c1-c789-4f47-856c-ecb480717608>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00342.warc.gz"} |
Homology (mathematics)
In mathematics, homology is a powerful tool that allows us to associate a sequence of algebraic objects with other mathematical objects such as topological spaces, manifolds, groups, Lie algebras,
Galois theory, and algebraic geometry. Homology groups were originally defined in algebraic topology, but similar constructions are available in a wide variety of other contexts.
The original motivation for defining homology groups was to distinguish two shapes by examining their holes. For instance, a circle is not a disk because the circle has a hole through it while the
disk is solid, and the ordinary sphere is not a circle because the sphere encloses a two-dimensional hole while the circle encloses a one-dimensional hole. However, defining a hole and categorizing
different kinds of holes in a rigorous mathematical way was not straightforward.
Homology provides a rigorous mathematical method for defining and categorizing holes in a manifold. Loosely speaking, a cycle is a closed submanifold, a boundary is a cycle that is also the boundary
of a submanifold, and a homology class represents a hole, which is an equivalence class of cycles modulo boundaries. A homology class is thus represented by a cycle that is not the boundary of any
submanifold. This cycle represents a hole, namely a hypothetical manifold whose boundary would be that cycle, but which is "not there".
There are many different homology theories, and a particular type of mathematical object may have one or more associated homology theories. When the underlying object has a geometric interpretation
as topological spaces do, the n-th homology group represents behavior in dimension n. Most homology groups or modules can be formulated as derived functors on appropriate abelian categories,
measuring the failure of a functor to be exact. From this abstract perspective, homology groups are determined by objects of a derived category.
In summary, homology provides a rigorous mathematical method for defining and categorizing holes in a manifold. Homology groups are powerful tools that allow us to associate algebraic objects with
other mathematical objects and analyze their behavior in higher dimensions. With homology, we can distinguish and categorize different kinds of holes in a rigorous mathematical way, providing a
deeper understanding of the structure and behavior of mathematical objects.
Homology theory, a branch of algebraic topology, originated from the work of Euler, Riemann, and Betti, who developed the concept of homology numbers to classify manifolds according to their cycles,
which are closed loops or submanifolds that cannot be continuously deformed into each other. Cycles can be thought of as cuts that can be glued back together or zippers that can be fastened and
unfastened. They are classified by dimension, with a line on a surface representing a 1-cycle, while a surface cut through a three-dimensional manifold is a 2-cycle.
On a sphere, all cycles can be continuously transformed into each other and belong to the same homology class, making them homologous to zero. In contrast, cycles on other surfaces such as the torus,
a surface with a hole, cannot be continuously deformed into each other. The torus has cycles that cannot be shrunk to a point, while the projective plane has both joins twisted.
To analyze and classify manifolds, homology theory involves cutting a manifold along a cycle homologous to zero, which separates the manifold into two or more components. For instance, cutting the
sphere along a cycle homologous to zero produces two hemispheres. In contrast, cutting the torus along cycles that are not homologous to zero produces a strip that can be opened out and flattened
into a square, which can be twisted to create four distinct surfaces, including the Klein bottle, which is a torus with a twist in it.
While cycles on the torus cannot be shrunk to a point, the cycle on the Klein bottle that goes around the twist can be shrunk to a point. However, following the other cycle forwards and then
backwards reverses left and right, making the Klein bottle a non-orientable surface. Similarly, the Möbius strip is a twisted surface that can be created by cutting along an equidistant point on one
side of the cycle that goes around the twist.
In conclusion, homology theory provides a powerful tool for analyzing and classifying manifolds based on their cycles, which are like zippers, cuts, and gluing back together. By examining the cycles
on a surface, mathematicians can learn a great deal about the surface's topological properties, such as its genus, orientability, and connectedness.
Informal examples
Topology is the study of geometric objects and the relationships between them that remain unchanged under continuous deformation. Topological invariants are fundamental tools that help to distinguish
different topological spaces from one another. Homology is a set of such topological invariants. It is a branch of algebraic topology that associates algebraic objects with topological spaces to
study their properties. Homology groups are the essential algebraic structures used to represent homology, where each homology group represents a different homology invariant.
The homology of a topological space 'X' is the set of topological invariants represented by its homology groups, which are given as H_0(X), H_1(X), H_2(X), and so on. Here, the k-th homology group
H_k(X) describes, informally, the number of holes in X with a k-dimensional boundary. For example, a 0-dimensional-boundary hole is merely a gap between two components, and H_0(X) describes the
path-connected components of X.
To understand this concept better, let us consider a few informal examples. A one-dimensional sphere S^1 is a circle. It has only one connected component and one one-dimensional-boundary hole, but no
higher-dimensional holes. The corresponding homology groups are H_k(S^1) = Z for k = 0, 1 and {0} otherwise, where Z is the group of integers and {0} is the trivial group. The group H_1(S^1) = Z
represents a finitely-generated abelian group with a single generator representing the one-dimensional hole contained in the circle.
A two-dimensional sphere S^2 has a single connected component, no one-dimensional-boundary holes, one two-dimensional-boundary hole, and no higher-dimensional holes. The corresponding homology groups
are H_k(S^2) = Z for k = 0, 2 and {0} otherwise. Similarly, for an n-dimensional sphere S^n, the homology groups are H_k(S^n) = Z for k = 0, n and {0} otherwise.
Now, let us consider a two-dimensional ball B^2, which is a solid disc. It has only one path-connected component, but unlike the circle, has no higher-dimensional holes. The corresponding homology
groups are all trivial except for H_0(B^2) = Z. In general, for an n-dimensional ball B^n, the homology groups are H_k(B^n) = Z for k = 0 and {0} otherwise.
The torus T is defined as a product of two circles T = S^1 × S^1. The torus has a single path-connected component, two independent one-dimensional-boundary holes, and one two-dimensional-boundary
hole. The corresponding homology groups are H_k(T) = Z for k = 0, 1, 2, and {0} otherwise.
In conclusion, homology is an essential tool for understanding topological spaces. Homology groups provide algebraic objects that help in studying topological spaces' properties and understanding how
they differ from each other. By associating algebraic objects with topological spaces, homology helps mathematicians make inferences about the properties of these spaces, making it an invaluable tool
for topology.
Construction of homology groups
Homology groups provide an important tool for studying topological spaces. By associating algebraic structures to these spaces, homology groups offer a way to classify and compare spaces by
understanding their topological properties. In this article, we will explore the construction of homology groups and the mathematics behind them.
To construct homology groups, we begin with a topological space X and define a chain complex C(X), which encodes information about X. A chain complex is a sequence of abelian groups or modules
connected by homomorphisms, called boundary operators. Specifically, C(X) consists of groups Cn for each integer n, along with boundary operators δn : Cn → Cn−1. The composition of any two
consecutive boundary operators is trivial, meaning δn−1δn = 0.
The boundary of a boundary is trivial, meaning the image of the boundary operator δn+1 is contained in the kernel of δn. The kernel of δn is the set of cycles Zn(X), while the image of δn+1 is the
set of boundaries Bn(X). Elements of Zn(X) are called cycles, and elements of Bn(X) are called boundaries.
Homology groups are defined as quotients of these groups, where Hn(X) = Zn(X)/Bn(X). Homology classes are equivalence classes over cycles, where two cycles are in the same homology class if they
differ by a boundary. This means that two cycles in the same homology class are homologous.
The homology groups of X measure how far the chain complex associated with X is from being exact. An exact sequence is a sequence where the image of the (n+1)th map is always equal to the kernel of
the nth map. When the chain complex is exact, the homology groups are trivial. In contrast, when the chain complex is not exact, the homology groups can reveal important information about the
topology of the space.
The reduced homology groups of a chain complex C(X) are defined as homologies of the augmented chain complex. This augmented chain complex adds an additional group to C(X) that maps onto the
integers. Specifically, Cn(X) = 0 for n < 0, and C0(X) is augmented with an additional group isomorphic to Z, denoted by Z̃. The boundary operator ε : C0(X) → Z is defined as ε(σ) = 1 for any
generator σ in C0(X)̃. This augmented complex allows us to capture the homology groups of non-compact spaces.
In conclusion, homology groups offer a powerful tool for understanding the topology of spaces. By associating algebraic structures to spaces, we can compare and classify them based on their
topological properties. The construction of homology groups involves defining chain complexes and computing quotients of these groups to reveal important topological information.
Homology vs. homotopy
Homology and homotopy may sound like heavy-duty mathematical terms, but they are both fascinating concepts that can help us understand the structure of spaces. Both homology and homotopy groups help
us determine the number of "holes" in a topological space, but they use different approaches to achieve this.
To understand the relationship between homology and homotopy, let's take a look at the first homotopy group and the first homology group of a topological space 'X'. The first homotopy group <math>\
pi_1(X)</math> is the group of directed loops starting and ending at a predetermined point. Essentially, it is the set of all possible ways you can travel around 'X' and end up where you started. If
we think of 'X' as a figure eight, for example, <math>\pi_1(X)</math> would be the set of all possible paths you could take around the figure eight and end up back at the center.
Now, let's consider the first homology group <math>H_1(X)</math> of 'X'. This group represents the cuts that can be made in a surface without breaking it into separate pieces. For instance, if we cut
the figure eight in one place, we might end up with two separate circles. However, if we cut the figure eight in a different place, we might end up with two circles that are linked together. <math>
H_1(X)</math> captures this information by measuring how many different ways we can cut 'X' without breaking it up.
So, what is the connection between homology and homotopy? It turns out that the first homology group <math>H_1(X)</math> is the abelianization of the first homotopy group <math>\pi_1(X)</math>. This
means that if we take <math>\pi_1(X)</math> and "force" it to become commutative, we get <math>H_1(X)</math>. In other words, <math>H_1(X)</math> is like a "commutative alternative" to <math>\pi_1(X)
</math>. This relationship is an example of the Hurewicz theorem, which relates homotopy groups to homology groups.
But while the connection between the first homotopy and homology groups is straightforward, the relationship between higher homotopy and homology groups can be much more complicated. The higher
homotopy groups are abelian and related to homology groups by the Hurewicz theorem, but they can be vastly more difficult to understand. For example, the homotopy groups of spheres are notoriously
difficult to compute, even for the simplest cases.
In summary, homology and homotopy are both powerful tools that help us understand the structure of topological spaces. Homology groups measure the number of cuts we can make in a surface without
breaking it up, while homotopy groups measure the number of different ways we can travel around a space and end up back where we started. The first homology group is the commutative alternative to
the first homotopy group, but the relationship between higher homotopy and homology groups can be much more complex.
Types of homology
Homology is an important mathematical concept that arises in various branches of mathematics. It can be thought of as a way to study the shape of mathematical objects by associating algebraic
structures to them. In particular, homology groups are a way to measure the number of "holes" or "loops" in a given space.
There are different types of homology theory that arise from functors mapping from various categories of mathematical objects to the category of chain complexes. In each case, the composition of the
functor from objects to chain complexes and the functor from chain complexes to homology groups defines the overall homology functor for the theory.
One of the most well-known types of homology theory is simplicial homology, which arises in algebraic topology. The simplicial homology of a simplicial complex X is defined by associating a chain
group C_n to each dimension n, where C_n is the free abelian group or module whose generators are the n-dimensional oriented simplexes of X. The orientation is captured by ordering the complex's
vertices and expressing an oriented simplex as an n-tuple of its vertices listed in increasing order. The boundary mapping from C_n to C_n−1 sends each simplex to a formal sum, and the dimension of
the n-th homology of X turns out to be the number of "holes" in X at dimension n.
Another type of homology theory is singular homology, which can be defined for any topological space X. A chain complex for X is defined by taking C_n to be the free abelian group or module whose
generators are all continuous maps from n-dimensional simplices into X. The homomorphisms arise from the boundary maps of simplexes.
Group homology is another type of homology theory that arises in abstract algebra. Here, one uses homology to define derived functors, such as the Tor functors. By applying a covariant additive
functor F to a sequence of free modules and homomorphisms, one obtains a chain complex whose homology depends only on F and some module X.
Overall, homology is a powerful tool for studying the properties of mathematical objects and spaces. The different types of homology theory each provide a different perspective on these objects and
can be used to answer different types of questions.
Homology functors
Mathematics can be a tricky subject, with a variety of abstract concepts that can be difficult to grasp. One such concept is homology, which is a fundamental tool in topology. Topology is the study
of properties that are preserved by continuous transformations, and homology helps us understand how different shapes are related to each other.
Homology is closely related to chain complexes, which are sequences of groups that are connected by homomorphisms. These homomorphisms preserve the structure of the groups, and they allow us to study
how different groups are related to each other. A morphism from one chain complex to another is a sequence of homomorphisms that connects the two complexes, and it satisfies a special condition known
as the "commutativity" condition.
Homology is a covariant functor that maps chain complexes to the category of abelian groups or modules. The nth homology group, denoted by Hn, is a group that captures the topology of the nth level
of the chain complex. The elements of Hn represent "holes" in the nth level of the chain complex that cannot be filled by the elements of the chain complex itself.
For example, imagine a chain complex that represents a circle. The first level of the chain complex would be a group with one element, representing the starting point of the circle. The second level
would be a group with one element, representing the endpoint of the circle. The homomorphism that connects these two groups represents the transformation of the circle from its starting point to its
endpoint. The homology group of the first level, H1, would be a group with one element, representing the hole in the center of the circle that cannot be filled by the elements of the chain complex
Homology is a powerful tool for understanding the topology of chain complexes, and it has many applications in mathematics and science. For example, homology can be used to study the topology of
surfaces, such as the surface of a sphere or a torus. Homology can also be used to study the topology of networks, such as the internet or social networks.
The concept of homology is closely related to cohomology, which is a contravariant functor that maps chain complexes to the category of abelian groups or modules. In cohomology, the chain complexes
depend on the object X in a contravariant manner, meaning that any morphism X to Y induces a morphism from the chain complex of X to the chain complex of Y. The cohomology groups, denoted by Hn, form
contravariant functors from the category that X belongs to into the category of abelian groups or modules.
In conclusion, homology is an essential concept in topology that allows us to understand the topology of chain complexes. Homology groups capture the holes in the chain complex that cannot be filled
by the elements of the chain complex itself, and they have many applications in mathematics and science. Co homology is a related concept that maps chain complexes in a contravariant manner, and it
also has many important applications. With the help of homology and cohomology, we can explore the rich and complex topology of the mathematical world.
Homology theory is a fundamental concept in algebraic topology that has numerous applications in a wide range of fields. It provides a powerful tool to study the topological properties of objects by
associating a sequence of abelian groups (or modules) called homology groups to them. These homology groups capture the topological structure of the object at different levels, and their properties
are essential in many areas of mathematics, including algebraic geometry, differential geometry, and number theory.
One of the key properties of homology theory is the Euler characteristic, which is a numerical invariant that measures the topological complexity of an object. The Euler characteristic can be
computed on the level of chain complexes or homology groups, providing two ways to compute this important invariant. If the chain complex satisfies certain finiteness conditions, the Euler
characteristic can be expressed as a sum of ranks of the finitely generated abelian groups or finite-dimensional vector spaces in the complex. This is a powerful tool for computing the Euler
characteristic, especially in algebraic topology.
Another essential property of homology theory is its relationship with short exact sequences of chain complexes. Given a short exact sequence of chain complexes, one can construct a long exact
sequence of homology groups, connecting the homology groups of the three chain complexes involved. This long exact sequence plays a crucial role in the study of algebraic topology, as it provides a
tool to relate the homology groups of different objects and to compute the homology groups of more complicated objects from those of simpler ones.
The connecting homomorphisms that appear in the long exact sequence are provided by the zig-zag lemma, a fundamental result in homological algebra that has numerous applications in algebraic
topology. This lemma can be used in various ways to aid in calculating homology groups, such as the theories of relative homology and Mayer-Vietoris sequences.
In conclusion, homology theory is a powerful tool that plays a crucial role in the study of topological spaces and their properties. The Euler characteristic and the long exact sequence associated
with short exact sequences of chain complexes are two essential properties of homology theory that have numerous applications in a wide range of fields. The study of homology theory and its
properties continues to be an active area of research in mathematics, with new applications and generalizations appearing constantly.
In mathematics, Homology is a tool for studying the topological properties of spaces. It is a branch of algebraic topology that deals with algebraic invariants associated with topological spaces. The
concept of homology is used to measure the number of holes and voids in a space or the connectivity between the various components of the space. The algebraic structure of Homology provides a useful
method for describing the shape of a space and detecting subtle changes in its topology.
Applications in Pure Mathematics Homology has been used to prove several essential theorems in pure mathematics. One of the most famous is the Brouwer fixed point theorem. This theorem states that if
a continuous map is made from a ball to itself, then there will always be at least one point in the ball that is fixed by the map. In other words, the map has a fixed point that does not move. This
theorem has important applications in many fields, such as game theory and economics.
Another significant theorem that was proved using homology is the Invariance of domain theorem. This theorem states that if you have an open set in n-dimensional Euclidean space, and you have a
continuous map that is injective on that set, then the image of the set under the map is also an open set. The map is also a homeomorphism between the two sets. This theorem has applications in
geometry, topology, and differential equations.
The Hairy Ball theorem is another famous result that has been proved using homology. It states that there is no way to comb a hairy ball without creating a cowlick, in other words, there will always
be a point on the ball where the hair is sticking straight up. The theorem has applications in many fields, including robotics and computer graphics.
The Borsuk-Ulam theorem is yet another important result that has been proved using homology. This theorem states that if you have a continuous function that maps an n-sphere into Euclidean n-space,
then there will always be a pair of antipodal points that are mapped to the same point. This theorem has applications in fields such as game theory and economics, and it is also important in physics
and chemistry.
Invariance of dimension is a fundamental theorem in topology that has been proved using homology. It states that if two open sets in Euclidean space are homeomorphic, then they must have the same
dimension. This theorem has important applications in differential equations and geometry.
Applications in Science and Engineering Homology also has applications in science and engineering. In topological data analysis, for example, data sets are treated as point clouds that sample a
manifold or algebraic variety embedded in Euclidean space. By linking the nearest neighbor points in the cloud into a triangulation, a simplicial approximation of the manifold can be created, and its
simplicial homology can be calculated. Finding techniques to robustly calculate homology using various triangulation strategies over multiple length scales is the topic of persistent homology.
In sensor networks, homology can be used to understand the global context of a set of local measurements and communication paths. Homology can be used to evaluate holes in coverage, which can be
essential in many applications, including emergency response and disaster management.
In dynamical systems theory in physics, homology is used to study the interplay between the invariant manifold of a dynamical system and its topological invariants. Morse theory relates the dynamics
of a gradient flow on a manifold to its homology, while Floer homology extends this to infinite-dimensional manifolds. The KAM theorem established that periodic orbits can follow complex
trajectories, which can form braids that can be investigated using Floer homology.
In finite element methods, homology can be used to solve boundary
In the world of mathematics, homology is like the alphabet of topology. It is the foundation upon which more complex concepts are built. It's the building blocks of geometric shapes and their
properties, and helps us understand the fundamental structure of complex objects. As such, it has become an important field of study in its own right, with various software packages developed for
computing homology groups of finite cell complexes.
One such software is Linbox, a C++ library that performs fast matrix operations including the Smith normal form. It interfaces with both Gap and Maple, making it a powerful tool for mathematicians
and computer scientists alike. In addition to Linbox, there are other software packages such as Chomp, CAPD::Redhom, and Perseus, all written in C++. These packages implement pre-processing
algorithms based on simple-homotopy equivalence and discrete Morse theory to perform homology-preserving reductions of the input cell complexes before resorting to matrix algebra.
But homology isn't just limited to C++ libraries. Kenzo, written in Lisp, allows for the generation of presentations of homotopy groups of finite simplicial complexes. This software is not only
useful for computing homology groups but also for understanding the deeper relationships between topology and algebra.
Another software, Gmsh, includes a homology solver for finite element meshes, which generates Cohomology bases directly usable by finite element software. This allows for the computation of homology
groups in engineering applications, providing a crucial tool for understanding the structural properties of complex objects.
In summary, homology and its software tools are the backbone of topology and help us understand the fundamental structure of complex objects. Whether you're a mathematician or a computer scientist,
these tools are an essential part of your toolkit. From the fast matrix operations of Linbox to the deep insights of Kenzo, homology software is the key to unlocking the secrets of topology. So the
next time you're exploring the properties of a geometric shape or developing new engineering applications, remember that behind every complex problem, there is a simple homology group waiting to be | {"url":"https://acearchive.org/homology-mathematics","timestamp":"2024-11-04T02:09:11Z","content_type":"text/html","content_length":"95671","record_id":"<urn:uuid:e41f72fe-384c-4480-b2e6-c6b1697388cb>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00577.warc.gz"} |
Quota Borda system - electowikiQuota Borda systemQuota Borda system
The Quota Borda System (QBS) is a PR electoral system for use in multi-member constituencies. It is based around determining solid coalitions and electing candidates from them using the Borda count.
It was devised by Michael Dummett and published in his 1984 book Voting Procedures.^[1]
According to Schulze,^[2] the way the Quota Borda system is described is somewhat convoluted, but can be boiled down to these points:
• The election procedure proceeds in rounds, called "stages", and terminates when all s seats have been elected.
• First, determine the Borda scores of every candidate in the election.
• Then, in the kth stage, starting from k=1, and ending at k = number of candidates:
□ For each solid coalition of k candidates, with support exceeding at least one Droop quota:
☆ Let q be the maximum number of Droop quotas its support exceeds, and let x be the number of candidates in that coalition who have been elected at a prior stage.
☆ If q > x, elect the candidate in that solid coalition with the highest Borda count that hasn't been elected at a prior stage.^[fn 1]
• If the process ends without every seat having been filled, fill the remaining seats with the unelected candidates with the highest Borda scores.
It's only necessary to elect one candidate at a time from each coalition because either there is only one unelected candidate left in the coalition, or some of them must have been elected in earlier
stages. For example, if the coalition {A, B} exceeds two Droop quotas in the second stage, then either A or B exceeds one Droop quota in the first stage. Thus either A or B must have been elected in
the first stage, so it's only necessary to elect the other one in the second stage.
The procedure can be generalized to base methods other than Borda by using that base method's order of finish instead of Borda's.
Choosing candidates to be elected from solid coalition supported by Droop quotas ensures that the method passes proportionality for solid coalitions. Using a single-winner Borda count to decide which
candidate in each coalition is to be elected will, in the absence of strategic voting, tend to elect the candidate closest to the median voter.
Thus all of QBS's proportionality comes from its PSC compliance; subject to this proportionality constraint, it will elect the most centrist candidates possible.
For instance, if voters and candidates are distributed on a standard normal distribution, every sufficiently large solid coalition supported by voters to the left of the center will elect the
rightmost candidate within the coalition's "slice" of the normal distribution. In the same way, every solid coalition supported by voters to the right of center will elect the leftmost candidate in
that coalition.
The purpose of this behavior is to reduce polarization and sectarianism while still remaining broadly proportional.
In addition, the method can be made more consensus-based at the expense of proportionality by stopping the count at an earlier stage than the number of candidates. Doing so will fill the remaining
seats with the Borda winners who haven't yet been elected.
Nicolaus Tideman argued that QBS can deny representation to minority groups that support irrelevant alternatives.^[3] Three-seat example:
1: c1, c2, c3, e1, e2, e3, e4, d1
1: c2, c3, c1, e1, e2, e3, e4, d1
1: c3, c1, d1, c2, e1, e2, e3, e4
6: e1, e2, e3, e4, c1, c2, c3, d1
4: e1, e2, e3, e4, c1, c2, c3, d1
5: e1, e2, e3, e4, c1, c2, c3, d1
6: e1, e2, e3, e4, c1, c2, c3, d1
7: e1, e2, e3, e4, c1, c2, c3, d1
8: e1, e2, e3, e4, c1, c2, c3, d1
9: e1, e2, e3, e4, c1, c2, c3, d1
Had d1 not run, there would've been a Droop solid coalition for c1-3, guaranteeing one of them a seat. But instead, e1-3 all win. This is an example where the Expanding Approvals Rule and STV, two
other common PSC-compliant methods, would elect one of c1-3.
Imagine a six-seater constituency in a plural society of three dominant groups, where the three groups are roughly 30:30:30. (There were many such constituencies in pre-war Bosnia.) Success in a QBS
depends on a good number of top preferences and/or a good Modified Borda Count MBC score; see below.
Lest their members/supporters split the vote, the matrix vote – like RCV (PR-STV) – prompts all parties to nominate only as many candidates as they think might get elected. At the same time, the MBC
element of a QBS encourages the voters to submit a full ballot. Accordingly, in a 6-seater 30:30:30 constituency (in Bosnia), each faction could expect to win 2 seats; at the same time, those parties
which do not fall into one of the country’s three ethno-religious categories (like Bosnia’s Social Democrats) might also hope for some success.
Now in many countries, not least those democracies which make decisions in binary votes, societies tend to divide into two: left- or right-wing, socialist or capitalist, and so on. Likewise, in many
societies already divided, each ethno-religious grouping tends itself to divide into two, to have a more radical and a more moderate party; (this was true both in Northern Ireland and in Bosnia).
Accordingly, in a 30:30:30 constituency, each of the two main parties in each ethno-religious grouping, might like to nominate 2 candidates; but no grouping would want to nominate more than 4.
Meanwhile, others like the Social Democrats might also have a good chance. So that’s 14 candidates already, but not too many more.
Come the vote, every voter would be encouraged, by the MBC element, to cast a full ballot of 6 preferences. In this way, QBS entices voters to cross the gender gap, the religious divide and even the
sectarian chasm; the methodology is ideally suited to plural societies, and especially conflict zones.
In a six-seater constituency, the analysis proceeds as follows, counting:
(a) all the candidates’ 1^st preferences;
(b) all the candidate pairs’^[1] 1^st and 2^nd preferences; and
(c) all their MBC scores.
At each stage, if there are still candidates to be elected, the count proceeds to the next stage. In the analysis:
Part I
(i) all candidates with a quota of 1^st preferences are deemed elected;
(ii) all pairs of candidates with two quotas of 1^st/2^nd preferences are elected;
then, in
Part II, in which any candidates elected in Part I, in stages (i) or (ii), are no longer counted,
(iii) candidates with the highest MBC scores are elected.
QBS has only one count, albeit of three different types of totals: (a), (b) and (c); next, in the analysis, three different stages.
1. ↑ Consider, for example, the situation (which existed in Northern Ireland) where a father stood alongside his son. If x people vote 1^st/2^nd dad/son, while y people vote 1^st/2^nd son/dad, and
if x + y > 2 quotas, then this dad/son pair is said to have two quotas.
Further reading | {"url":"https://electowiki.org/wiki/Quota_Borda_system","timestamp":"2024-11-02T05:00:14Z","content_type":"text/html","content_length":"59361","record_id":"<urn:uuid:d91c26af-f58f-4d11-81d2-52568c76b8cd>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00045.warc.gz"} |
Is Your Line Chart Lying?
Charts are great at revealing trends, patterns and relationships in your data. We love how they enable us to glean information in the blink of an eye, but beware of the line chart because it can lie.
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Charting Disparate Data
Let’s take the line chart below that plots the trend of revenue and profit over time. It looks like revenue is growing faster than profit; after all revenue shows a steady incline whereas profit is
quite flat.
Present a chart like this and you’ll be inundated with questions from the CEO who is irate that either COGS or overheads, or both have increased, and all the extra revenue they’ve worked so hard to
generate appears to be eaten up by increased costs.
To be fair, it’s not the line chart’s fault. The first line chart is plotting the data correctly. It isn’t a bug in the chart. It’s a case of relativity; a 10% increase on $1m is always going to be a
bigger number than a 10% increase on $100k, and when you plot disparate data that continues to grow over time, even at the same rate, the gap is going to widen.
The Truth Chart 1 – Log Scale
However, if we look at the data plotted on a log scale we can see that revenue and profit are actually growing at similar rates.
Phew, the production and operations managers have been spared, but now the CEO is after you for producing a misleading chart!
Using a Log scale is one option to prevent line chart lies, but some people find these a bit confusing to interpret so let’s look at another Truth Chart.
The Truth Chart 2 – Secondary Axis
The secondary axis is a popular choice, however among data visualisation gurus they are discouraged because they are difficult to quickly see which axis is for which line.
The other problem they pose is since they use different scales they can result in an unclear message. Let’s look at what happens when you add a secondary axis:
In this chart the lines for revenue and profit do follow a similar upward trend but we can’t compare the actual growth rate of one against the other.
The uneducated chart reader may even make the incorrect assumption that profit is performing better than revenue since the profit line ends at a higher point, irrespective of the scales.
We humans make split second assumptions based on what we see, often without even realising. We need to be mindful of this and try to present data in a way that is quick to interpret, but also
consider any misinformation they might inadvertently convey.
Perhaps a compromise would be to add some labels with the start and end dollar values of the revenue and profit and percentage growth.
The Truth Chart 3 – Panel Charts
Panel charts, which in their basic form are simply two separate charts, enables you to clearly see the growth in each series and the separate axis scales enable quicker interpretation than the
secondary axis charts. However, there is still a large gap between the lines which makes it difficult to compare them. This could be aided by adding labels for percentage growth.
The Truth Chart 4 – Index Numbers
The chart above plots the change in revenue and profit figures over time relative to the starting position in 2011. These values are known as index numbers (more on that in a moment).
It tells a completely different story to the original line chart. Here we can clearly see that the growth rate of revenue and profit have followed a fairly similar pattern, and by 2022 they were
almost the same.
The labels on the lines in this chart could be used in any of the charts above to aid interpretation. Download the file to see how they're done.
Calculating the Index
We index numbers by saying that in 2011 (our base period) both revenue and profit were 100, or 100% and from there we calculate the change for each year since 2011.
It’s tricky to explain in words alone so let’s take a look at this table which shows the indexed results in columns D and E (remember 1 = 100%):
In cell D5 we can see that the indexed revenue is 114, which is the same as saying that the revenue in 2012 is 14% higher than it was in 2011. Likewise the profit in 2012 is 7% less than it was in
Each subsequent year is also compared to 2011, our base year, to come up with the indexed value for that year.
The formula in D4 is =B4/$B$4 which is then copied down the column so that D5 contains =B5/$B$4 and so on.
1. Jef
Well written analysis. Thanks for sharing.
□ Mynda Treacy
Thanks, Jef 🙂
2. Jon Peltier
It’s not just line charts; any chart can be used in misleading ways. (Intentionally or not.)
In some cases it is useful to plot the ratio of the two numbers, for example, profit as a percentage of revenue.
□ Mynda Treacy
Yep, and yep. Cheers, Jon. I agree. In fact profit as a percentage of revenue would be the most sensible data to plot in this scenario.
I was originally planning on using a different example for this post as I stumbled upon this issue where different figures were used; it was something like the growth of US Debt vs Imports
from China, but for the life of me I couldn’t find it again.
☆ Stefano B.
Awesome job Mynda and great suggestion Jon!! Chart adjusted to help reader better interpret the more valuable data (profit as a percentage of revenue). Thoughts?
○ Mynda Treacy
Hi Stefano,
Thanks for sharing your chart. I recommend showing the ratio increase in its own chart as bars or a line. The labels are helpful, but they’re slow to read and they clutter the chart a
bit too much for my liking. How about this variation:
3. Geraldine Tatters
Really like the index charts; real alternative to changing series from/to primary and secondary axis
I will use this in my day-to-day reporting
□ Mynda Treacy
Great 🙂 Glad you liked it Geraldine. I definitely recommend indexing over the secondary axis option as you can see in my examples in reply to Asif.
4. Sanjiv Daman
Hi Mynda,
This is awesome. Now we can use excel to add real value instead of just using charts for the sake of showing numbers.
□ Mynda Treacy
Cheers, Sanjiv 🙂
5. Asif
Thanks Mynda!
By Index approach, one can see the trend of both the revenue & the profit in above chart…right but a secondary axis option can be adopted to see the real values.
Now what if we have three variables plotted on Y axis with different variation in values…here index will solve our problem if we want trend but how we can get 3rd axis on a graph?
□ Mynda Treacy
Hi Asif,
The secondary axis is a popular choice, however among data visualisation gurus they are discouraged because they are difficult to quickly see which axis is for which line.
The other problem they pose is since they use different scales they can result in an unclear message. Let’s look at what happens when you add a secondary axis:
In this chart the lines for revenue and profit do follow a similar upward trend but we can’t compare the actual growth rate of one against the other.
The uneducated chart reader may even make the incorrect assumption that profit is performing better than revenue since the profit line ends at a higher point, irrespective of the scales.
We humans make split second assumptions based on what we see, often without even realising. We need to be mindful of this and try to present data in a way that is quick to interpret, but also
consider any misinformation they might inadvertently convey.
Perhaps a comprimise would be to add some labels with the start and end dollar values of the revenue and profit, although this may then confuse interpretation too:
☆ roberto mensa
I think that a clear division of the panels can solve the problem of the secondary axis … something like this:
(only the first hint that I found)
○ Mynda Treacy
Thanks for sharing the link, Roberto. I also like a panel chart and the vertical ones are ideal for this scenario.
If you didn’t want to jump through all those hoops to create it you could always create two separate charts and align them, hiding one of the horizontal axes etc. Not as slick as
Jon’s panel chart but maybe good for quickly cobbling together.
☆ Jon Acampora
Great example of indexing! I like the chart above that shows the revenue in the labels. You could also add labels at the end of each line that state the percentage growth. For example, to
the right of the orange line the label could say, “Revenue grew 42% since 2001”.
You could also accomplish this by formatting the numbers as percentages in the y-axis and adding an axis label that describes it. The reader might be confused by what 1.2 or 1.4 means.
Thanks for sharing!
○ Mynda Treacy
Great idea, Jon.
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A Potential Zero Sum Game
An example of a game which is at the same time zero-sum and potential.
A potential zero-sum game
Consider the zero-sum $2\times 2$ game with payoff bimatrix given by
\[u_1 = \begin{pmatrix} -1,1 & 2,-2 \\ 3,-3 & 6,-6. \end{pmatrix}\]
It’s easy to check that the game is potential with potential function
\[\phi = \begin{pmatrix} 3 & 0 \\ 7 & 4 \end{pmatrix},\]
and that $(\text{down}, \text{left})$ with outcome $(3, -3)$ is strict Nash. Since the game has non-trivial unilateral deviations it is not strategically equivalent to the zero-game, thus showing
that the spaces of zero-sum and potential games intersect non-trivially, even after quotienting away strategical equivalence.
The image below shows the response graph of the game and some trajectories of replicator dynamics converging to the strict Nash equilibrium. | {"url":"https://davidelegacci.github.io/2024/10/11/a-potential-zero-sum-game.html","timestamp":"2024-11-02T17:51:30Z","content_type":"text/html","content_length":"5200","record_id":"<urn:uuid:a73718a1-a5ce-460c-9810-241069cbea86>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00382.warc.gz"} |
[tlaplus] Experience report: teaching TLA+ proofs to ChatGPT
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[tlaplus] Experience report: teaching TLA+ proofs to ChatGPT
Disclaimer: in this post I'll use the words "knows", "understands", "learns", "reads" etc. although that isn't quite accurate for what's happening here, but it works well enough.
I spent 1.5 hours trying to teach ChatGPT4 how to write TLA+ proofs this morning. If you fork out $20/month you can access ChatGPT4 with quite a large memory/token window, so you can get it to
customize its understanding on top of its base-level understanding it presumably got from reading all the TLA+ specs on github. Thanks to Jesse Jiryu Davis for suggesting this teaching approach as
opposed to just using pre-existing understanding. Report:
• The base-level understanding of TLA+ and the model checker is quite good. For example, I was able to paste in a spec and ask ChatGPT4 to derive a type invariant. It came up with a pretty good
one, maybe not restricting variable values as much as it could have but acceptable. Not bad for something that takes humans a few days in a TLA+ workshop to figure out how to do reliably!
• It understands the high-level approach of how to write proofs of the form Spec => []Invariant. It will output a basic skeleton of this proof, more or less accurately.
• It really struggles a lot with actual proof language syntax (relatable tbh lol), especially when to use terminal vs. non-terminal proofs, when to use SUFFICES/ASSUME/PROVE and how that affects
subsequent proof steps, when to use CASE etc.
• It struggles a lot with correct comma usage in BY/DEF terminal proofs, often having too many commas or too few commas between elements. In retrospect I should have focused less on fixing this
very minor issue, burned up too much token/memory window on it when I should have focused on other things.
• It does quite well on proofs with 3 or fewer levels. If you zoom in on a specific subproof it can usually handle it, figuring out which definitions it needs to expand in BY/DEF terminal proofs
and such. But then if you ask it to print out the entire proof it will struggle to synthesize all the subproofs.
• You do need to keep an eye on your token/memory window. If you start out by giving it the spec and then asking it to write a proof and correct it as you go, the spec will eventually fall outside
the window and it can only rely on parroting/modifying the proof itself rather than reasoning about the proof in relation to the spec.
All in all, it shows great promise for basic TLA+ but will probably wait another six months before trying it out on proofs again. Ideally the experience would be that ChatGPT version X could teach me
TLA+ proofs, rather than vice-versa.
Really what's happening is that it feeds the entire preceding conversation that fits inside the token/memory window into the inference engine for every token, which is how it seems to learn as you
explain things to it. If you switch to a fresh chat instance it will have "forgotten" everything you explained. To ensure I would be able to easily rehydrate its understanding I asked it to summarize
what it learned so it could easily re-acquire that knowledge; here's a prompt you can paste into your own instance of ChatGPT4 if you want to try it:
Here is a list of rules about TLA+ proofs that will augment your current understanding; you generated this list after a long conversation, with the intention that a fresh version of yourself
could read this list and get the essence of the conversation:
- Use SUFFICES ASSUME ... PROVE ... OBVIOUS to make assumptions available to all subsequent proof steps in that level or subproofs of those steps.
- Terminal proofs can either be the keyword OBVIOUS or a BY ... DEF ... construct.
- In a terminal proof, BY ... DEF ... means the prover expects the elements before DEF to be proof steps or theorems and the elements after DEF to be definitions which must be expanded so the
prover can look at their structure to prove the step correct.
- Theorems should be located between BY and DEF, not after DEF.
- Never put an extra or unnecessary comma in a BY ... DEF terminal proof.
- CASE statements can be used to consider different conditions within the proof.
- All non-terminal proofs that consist of sequences of steps must have a QED step as the last one.
- When using the Proof of Temporal Logic (PTL) theorem, it should be placed between BY and DEF in a proof step.
Andrew Helwer
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KDTree< Dim >
Finds the closest point to the parameter point in the KDTree.
This function takes a reference to a template parameter Point and returns the Point closest to it in the tree. We are defining closest here to be the minimum Euclidean distance between elements.
Again, if there are ties (this time in distance), they must be decided using Point::operator<(). Recall that an HSLAPixel is defined by three components: hue, saturation, and luminance.
The findNearestNeighbor() search is done in two steps: a search to find the smallest hyperrectangle that contains the target element, and then a back traversal to see if any other hyperrectangle
could contain a closer point, which may be a point with smaller distance or a point with equal distance, but a "smaller" point (as defined by operator< in the point class). In the first step, you
must recursively traverse down the tree, at each level choosing the subtree which represents the region containing the search element (another place to save some duplicate code?). When you reach the
lowest bounding hyperrectangle, then the corresponding node is effectively the "current best" neighbor.
However, it may be the case that a better match exists outside of the containing hyperrectangle. At then end of first step of the search, we start traversing back up the kdtree to the parent node.
The current best distance defines a radius which contains the nearest neighbor. During the back-traversal (i.e., stepping out of the recursive calls), you must first check if the distance to the
parent node is less than the current radius. If so, then that distance now defines the radius, and we replace the "current best" match. Next, it is necessary to check to see if the current splitting
plane's distance from search node is within the current radius. If so, then the opposite subtree could contain a closer node, and must also be searched recursively.
During the back-traversal, it is important to only check the subtrees that are within the current radius, or else the efficiency of the kdtree is lost. If the distance from the search node to the
splitting plane is greater than the current radius, then there cannot possibly be a better nearest neighbor in the subtree, so the subtree can be skipped entirely.
You can assume that findNearestNeighbor will only be called on a valid kd-tree.
See Also
Here is a reference we found quite useful in writing our kd-tree: Andrew Moore's KD-Tree Tutorial.
There is an example in the MP5 instruction.
This function is required for MP 5.1.
query The point we wish to find the closest neighbor to in the tree.
The closest point to a in the KDTree.
Implement this function! | {"url":"https://courses.grainger.illinois.edu/cs225/sp2018/doxygen/mp5/classKDTree.html","timestamp":"2024-11-03T10:56:38Z","content_type":"application/xhtml+xml","content_length":"41999","record_id":"<urn:uuid:7bb869fe-2b4d-4252-a1de-03ab229d1a61>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00040.warc.gz"} |
Multiplication One Step Equations Worksheet
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Multiplication One Step Equations Worksheet
Multiplication One Step Equations Worksheet -
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Create a worksheet Solve simple equations involving multiplication and division Solving One Step Equations Multiplication Division Solve simple equations involving multiplication and division The
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I I2 o0b1B2 O 8K 6u Ztka 1 vSLoVfFt Mw0a yr 1er YL1LwC8 j D LA 8l 9l u CrwiKgIhJtos o dr ceUsVenr ivreuds O T 6M Aa8dgeI UwOilt Phn vI YncfliPn4i ntkeR TAvl g6e 5b 0rnaJ Q1B w Worksheet by Kuta
Software LLC Infinite Algebra 1 Name One Step Equations Date Period Solve each equation 1 26
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Solve One Step Equations With Smaller Values Old
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Solve One Step Equation Multiplication And Division One step equations Algebra equations
One Step Equations Multiplication And Division Worksheet Kuta Times Tables Worksheets
Solving One Step Equations Multiplication And Division Worksheet Answer Key Gannuman
Mrs White s 6th Grade Math Blog November 2013
One Step Equations Multiplication and Division Worksheets Tutoring Hour
Multiplication and Division Shapes Students in 8th grade have to apply the property of the shape to set up the one step equation involving multiplication and division Solve and find the value of the
variable Grab our free printable one step equation worksheets on multiplication and division to solve equations involving integers fractions
One Step Equations DadsWorksheets
The one step equations worksheets on this page include problems with integers and fractions for a variety of math operations These basic algebra worksheets are appropriate practice for 6th grade 7th
grade and 8th grade students Example 3 Integers Multiplication 4x 5 Step 1 Start by simplifying the equation by moving the 4 to
Multiplication and Division Shapes Students in 8th grade have to apply the property of the shape to set up the one step equation involving multiplication and division Solve and find the value of the
variable Grab our free printable one step equation worksheets on multiplication and division to solve equations involving integers fractions
The one step equations worksheets on this page include problems with integers and fractions for a variety of math operations These basic algebra worksheets are appropriate practice for 6th grade 7th
grade and 8th grade students Example 3 Integers Multiplication 4x 5 Step 1 Start by simplifying the equation by moving the 4 to
One Step Equations Multiplication And Division Worksheet Kuta Times Tables Worksheets
One Step Equations Multiplication And Division Worksheet Free Printable
Solving One Step Equations Multiplication And Division Worksheet Answer Key Gannuman
Mrs White s 6th Grade Math Blog November 2013
28 One Step Equations Multiplication And Division Worksheet Answers Free Worksheet Spreadsheet
Solving One Step Multiplication And Division Equations Worksheet Tessshebaylo
Solving One Step Multiplication And Division Equations Worksheet Tessshebaylo
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Constant technique is essential. Regular sessions, ideally a few times a week, can yield significant enhancement.
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Motivating constant technique, offering aid, and creating a positive understanding setting are useful actions. | {"url":"https://crown-darts.com/en/multiplication-one-step-equations-worksheet.html","timestamp":"2024-11-12T07:09:03Z","content_type":"text/html","content_length":"29760","record_id":"<urn:uuid:d1ad7360-0c8d-43a9-a33b-22a14c5bd1b0>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00632.warc.gz"} |
Proceedings of the 2008 Summer Program
Article Newspaper/Magazine
Proceedings of the 2008 Summer Program
• Simulation of flows in shock-tube facilities by means of a detailed chemical mechanism for nitrogen excitation and dissociation
By A. Bourdon, M. Panesi, A. Brandis, T. E. Magin, G. Chaban, W. Huo, R. Jaffe and D. W. Schwenke
• Modeling of reactive plasmas for atmospheric entry flows based on kinetic theory
B. Graille, T. E. Magin and M. Massot
• Aspects of advanced catalysis modeling for hypersonic flows
J. Thoemel, O. Chazot and P. Barbante
• Chemical nonequilibrium effects in the wake of a boundary-layer sized object in hypersonic flows
M. Birrer, C. Stemmer, G. Groskopf and M. Kloker
• Bi-global secondary stability theory for high-speed boundary-layer flows
G. Groskopf, M. J. Kloker and O. Marxen
• Boundary layer transition in high-speed flow
P. A. Durbin, J. W. Joo and O. Marxen
• Modeling differential diffusion in non-premixed combustion: soot transport in the mixture fraction coordinate
J. C. Hewson, D. O. Lignell and A. R. Kerstein
• Coupling tabulated chemistry with large-eddy simulation of turbulent reactive flows
R. Vicquelin, B. Fiorina, N. Darabiha, D. Veynante, V. Moureau and L. Vervisch
• LES of two-phase reacting flows
M. Sanjose, T. Lederlin, L. Gicquel, B. Cuenot, H. Pitsch, N. García-Rosa, R. Lecourt and T. Poinsot
• Turbulent combustion of polydisperse evaporating sprays with droplet crossing: Eulerian modeling and validation in the infinite Knudsen limit
S. de Chaisemartin, L. Fréret, D. Kah, F. Laurent, R. O. Fox, J. Reveillon and M. Massot
• Turbulent combustion of polydisperse evaporating sprays with droplet crossing: Eulerian modeling of collisions at finite Knudsen and validation
L. Fréret, F. Laurent, S. de Chaisemartin, D. Kah, R. O. Fox, P. Vedula, J. Reveillon, O. Thomine and M. Massot
• CFD-based mapping of the thermo-acoustic stability of a laminar premix burner
R. Kaess, W. Polifke, T. Poinsot, N. Noiray, D. Durox, T. Schuller and S. Candel | {"url":"https://ctr.stanford.edu/publications/summer-program-proceedings/proceedings-2008-summer-program","timestamp":"2024-11-08T16:57:59Z","content_type":"text/html","content_length":"36603","record_id":"<urn:uuid:aef6f3d5-ecb8-426b-95fd-3a78ab56a2cc>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00123.warc.gz"} |
Hydraulic Crane | SUGA Co., Ltd.
Shellook hydraulic CraneThis is the hydraulic crane which has been developed especially for vessels.
There are three types of Shellooks (300, 500,500A)which can be used for small vessels, etc.
Shellook Hydraulic Cranes are sold on our own design and fabrication.
Shellook 300 is smaller than other products and is a bestseller model for small vessels with high durability due to the use of stainless parts and corrosion-resistant zinc plating.
This is the hydraulic crane which has been developed especially for vessels.
The structure is very rigid in response to the load caused by the influence of waves.
For more corrosion-resistance, molten zinc plating is used for a main body and the stainless rod or double hard chrome plating processing is being done.
This model has three types; 300kg and 500kg hoisting ability, Automation and Semi-Automation of a boom. And it is widely adopted for small vessels
as well as ones for cultured scallops and oysters.
Construction of Shellook
SUGA’s Marine Products (✔️)
Shellook Specification
Shellook 300 Shellook 500 Shellook 500A
Shellook Outer dimension
I Main calculation formula of hydraulic pressure device
1. Hydraulic pump
□ 1) axis input of a pump
□ 2) Oil power of the pump
Lp= P･Q/60＝η･Ls・10-2
□ 3) Total efficiency of the pump
□ 4) Capacity efficiency of the pump
□ 5) Efficiency of the motor
ηe= Ls/ Le
2. Hydraulic motor
□ 1) Theoretical Displacement volume of Hydraulic motor
□ 2) Output power of Hydraulic motor
□ 3) Input power of Hydraulic motor
□ 4) Capacity efficiency of Hydraulic motor
ηv= (Dth･N/Q)･10-1
□ 5) Torque efficiency of Hydraulic motor
ηt= (2π･T/P･Dth )･102
□ 6) Total efficiency of Hydraulic motor
η=ηv･ηt･102= (Ls/ Lm)･102=(2π･T･N/P･Q)･10-1
3. Cylinder
□ 1) Necessary pressure to a cylinder
P1=(1/ A1)･(F/ηc+P2･A2･102) ･10-2
□ 2) Necessary flow rate for a cylinder
Q= A1･v･10-1+QL
□ 3) Driving force of a cylinder
Acceleration ability
Static wear resistance
F2= μs･m･g
Dynamic wear resistance
F3= μd･m･g
The explanations of symbols
Ls: input of pump shaft, output power of the motor, the output power of the motor (kW)
Lp: oil power of the pump (kW)
Le: input power of the motor (kW)
Lm: input power of the motor (kW)
P: Discharge pressure of pump, differential-pressure of the input/output port of a motor (MPa)
P1: necessary pressure to cylinder (MPa)
P2: pressure of the cylinder inflow (MPa)
Q: discharge amount at the time of discharge pressure P,
inflow oil amount to the motor, necessary flow rate of the cylinder (large/min)
Qth: theoretical discharge amount (l/min)
Qo: discharge amount at the time of discharge pressure P≒0, (l/min)
QL: Leak inside the cylinder (l/min)
T: shaft torque (N・m)
N: number of rotations (min-1)
η: Total efficiency of the pump, total efficiency of the motor (the %)
η v: capacity efficiency of pump, capacity efficiency of the motor (the %)
η t: torque efficiency of pump, torque efficiency of the motor (the %)
η e: efficiency of motor (the %)
η c: driving force efficiency of cylinder (0.9-0.95)
Dth: theoretical displacement volume of the motor (cm3/rev)
A1: inlet side pressure receiving area of cylinder (cm2)
A2: outlet side pressure receiving area of cylinder(cm2)
F: cylinder driving force (N)
F1: cylinder acceleration force (N)
F2: static wear resistance (N)
F3: dynamic wear resistance (N)
v: The speed of the cylinder (m/min)
v1: The speed after acceleration (m/s)
m: The mass of the load (KG)
α: acceleration (m/s2)
t: acceleration time (s)
μ s: static friction coefficient
μ d: dynamic friction coefficient
g: acceleration of the gravity (m/s2)
Please contact us for repair and modification of our systems. | {"url":"https://agus.co.jp/en/?p=667","timestamp":"2024-11-06T08:16:17Z","content_type":"text/html","content_length":"41477","record_id":"<urn:uuid:18d0ce65-e44b-4a9e-9627-2694f3850217>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00020.warc.gz"} |
Section: LAPACK (3)
Updated: Tue Nov 14 2017
Page Index
subroutine cptcon (N, D, E, ANORM, RCOND, RWORK, INFO)
Function/Subroutine Documentation
subroutine cptcon (integer N, real, dimension( * ) D, complex, dimension( * ) E, real ANORM, real RCOND, real, dimension( * ) RWORK, integer INFO)
CPTCON computes the reciprocal of the condition number (in the
1-norm) of a complex Hermitian positive definite tridiagonal matrix
using the factorization A = L*D*L**H or A = U**H*D*U computed by
Norm(inv(A)) is computed by a direct method, and the reciprocal of
the condition number is computed as
RCOND = 1 / (ANORM * norm(inv(A))).
N is INTEGER
The order of the matrix A. N >= 0.
D is REAL array, dimension (N)
The n diagonal elements of the diagonal matrix D from the
factorization of A, as computed by CPTTRF.
E is COMPLEX array, dimension (N-1)
The (n-1) off-diagonal elements of the unit bidiagonal factor
U or L from the factorization of A, as computed by CPTTRF.
ANORM is REAL
The 1-norm of the original matrix A.
RCOND is REAL
The reciprocal of the condition number of the matrix A,
computed as RCOND = 1/(ANORM * AINVNM), where AINVNM is the
1-norm of inv(A) computed in this routine.
RWORK is REAL array, dimension (N)
INFO is INTEGER
= 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.
December 2016
Further Details:
The method used is described in Nicholas J. Higham, "Efficient
Algorithms for Computing the Condition Number of a Tridiagonal
Matrix", SIAM J. Sci. Stat. Comput., Vol. 7, No. 1, January 1986.
Definition at line 121 of file cptcon.f.
Generated automatically by Doxygen for LAPACK from the source code. | {"url":"https://man.linuxreviews.org/man3/cptcon.f.3.html","timestamp":"2024-11-11T23:01:12Z","content_type":"text/html","content_length":"5860","record_id":"<urn:uuid:97cd74f8-d401-4b01-8528-05b3159f0bec>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00321.warc.gz"} |
Neuroimaging Based Diagnosis of Alzheimer's Disease Using Privacy-Preserving Machine Learning (Part 2)
Neuroimaging Based Diagnosis of Alzheimer’s Disease Using Privacy-Preserving Machine Learning (Part 2)
February 21, 2022
Authors: Prof. Bogdan Draganski (Director, LREN), Dr. Ferath Kherif (Vice Director, LREN), Claudio Calvaruso (Technical Account Manager, Inpher) and Dr. Dimitar Jetchev (CTO, Inpher)
Editors Note: This is the second in a series of articles that LREN – CHUV and Inpher are jointly publishing to show how privacy-preserving machine learning allows for neuroimaging-based diagnosis of
Alzheimer’s disease that is readily applicable in a federated network. Read part 1 here
In part 1 of this blog post, we shared how the latest advancements in MRI imaging and Statistical Parametric Mapping (SPM) can benefit from the recent developments in Privacy-Enhancing Technologies
(PETs), Privacy-Preserving Machine Learning (PPML) and Privacy-Preserving Federated Learning (PPFL) to address challenges in understanding the mechanisms (from a researcher’s perspective) in the
early diagnosis of Alzheimer’s Disease (from a clinical perspective):
• Researchers’ perspective: Identifying relevant biological variables including regional changes in brain volume associated with the pathologies of AD and other dementias to understand the
mechanisms underlying Alzheimer’s disease
• Clinicians’ perspective: individual diagnostics estimating accurate risk score and propensity of neuromodulatory treatment for a given patient to support better diagnosis
In part 2 of the blog post, we describe two specific simple models, a linear and a logistic regression model, addressing each of the two aspects above and the workflows needed for the training across
data coming from multiple private data sources (hospitals, private radiology labs or research institutes) using a combination of SPM tools for local processing and Inpher’s XOR Secret Computing©
Platform for PPML.
Training Privacy-Preserving Linear Regression Models
Linear regression is a basic supervised learning method for modeling the relationship between a scalar response and one or more explanatory variables (features) where one wants to estimate the
coefficient of the slope that best fits the input training data.
In our scenarios, the individuals’ brain MRI scans are first mapped at the voxel level to the Jacobian determinant template of the SPM tools (described in part 1 of this post) to quantify temporal
grey matter changes. This preprocessing is local to each private data source. One then uses generalized linear models and hypothesis testing across a set of healthy controls (HC) and Alzheimer
disease subjects (AD) to distinguish if these temporal expansions (or contractions) are due to Alzheimer’s disease or regular temporal changes due to aging.
More precisely, it is defined as matrix Y (the dependent variable matrix) of size N x 20,000 where N is the total number of patients across all data sources and the 20,000 columns correspond to the
20,000 voxels of the images. Each row of this matrix corresponds to the voxel values of the SPM-preprocessed Jacobian template for a given patient (HC or AD patient) and each column corresponds to a
particular voxel in the standard stereotaxic space.
The goal is to compute a linear model for each column expressing the voxel value as a linear function of certain feature values. The features correspond to the columns of a feature matrix referred to
as a design matrix. The rows of the design matrix correspond to the subjects (both HC and AD). Figure 1 represents a typical design matrix (where N_samples, in this case, correspond to the total
number of patients). The first column (consisting of 1’s) is optional and may represent the intercept in the regression model. The second (binary) column is the characteristic vector of the healthy
controls – the binary entry corresponding to a sample (or patient) is 1 if the patient is in HC and 0 otherwise. Similarly, the third one is the characteristic vector of the clinically labeled
Alzheimer patients – the binary entry corresponding to a patient is 0 if the patient is in HC and 1 if the patient has been diagnosed with AD. We can also represent features such as age, gender, race
and other phenotyping features as illustrated in Figure 2.
Figure 1: Example Feature Matrix (Product Matrix)
Figure 2: Fitting a Linear Regression Model per Voxel
For each voxel of the image (column of the matrix Y), the voxel value y is expressed in terms of the above design matrix X as
y= X .(β[0], β[HC], β[AD], β[AGE], β[GENDER], β[RACE])+ e.
Therefore, about 20,000 linear regression parameters are obtained, one for each voxel. Here, β[0] is the intercept, β[HC], β[AD], β[AGE], β[GENDER], β[RACE ]are the coefficients corresponding to HC,
AD, age, gender, and race, respectively. The null hypothesis is β[HC] = β[AD]. Rejecting the null hypothesis (rejecting values less than or equal) means that the particular voxel is functionally
related to Alzheimer’s (see Figure 3)
Figure 3: Linear Hypothesis Test (LHT) to identify voxels where HC>AD
Privacy-Preserving Training of Diagnosis Models for Alzheimer’s Disease
In the second scenario (clinicians’ setting), the goal is to build a tool for early and accurate diagnosis of Alzheimer’s Disease. Here, the role of the voxel matrix described in the previous section
changes – it now becomes the feature matrix itself, or the independent variables matrix used to predict the risk score for a patient. In other words, the input features are the voxel values of an SPM
preprocessed (via the Jacobian determinant template) MRI image of a new subject together with other data for that subject, such as clinical phenotyping data biomarkers. The output is an estimate of
the risk of Alzheimer’s disease for the given subject.
In order to obtain a meaningful and unbiased prediction model, the local training datasets available in a single hospital alone are insufficient. First and foremost, a single hospital is unlikely to
provide enough subjects for the model training. Second, even if one were able to aggregate data from several large hospitals, the training data might still be biased since most of the MRI scans
generated at these locations are not scans of healthy controls, hence the need to include more diverse data sources such as private clinics and radiology labs. Finally, in most cases, patients’ MRI
scans and age features do not follow identical distributions across the distinct private data sources.
To overcome this challenge, Inpher’s XOR Platform is leveraged, particularly XOR MPC (Multi-Party Computation) for PPML and XOR Federated Learning with MPC for secure aggregation for PPFL (for better
scalability and model complexities). This enables the training of highly accurate and precise models by using all the available data in a privacy-preserving way. In addition, the PPFL approach could
enable training highly accurate complex deep learning models as well.
Logistic Regression Models
The most basic technique to address this problem is logistic regression, though it is not the only one. There has been progress using more advanced supervised learning methods such as Support Vector
Machines (SVMs) or deep neural network models among others. Logistic regression is a supervised learning method that models the probability of a binary class prediction using a logistic function such
as the sigmoid function. In our particular problem, the larger the probability value, the higher the risk of Alzheimer’s disease is for the patient. As in the researchers’ setting, the training data
contains samples from both HC and AD patients. The feature matrices have multiple columns corresponding to each of the voxel values for each sample as well as the other features (e.g., age, gender,
race). As the total number of voxels per preprocessed MRI image is 20,000, the model has at least 20,000 features and is therefore very complex to train in real-time. Since not all voxels are
relevant to Alzheimer’s disease, one can use various dimensionality reduction techniques to reduce the model complexity.
Figure 4: Estimating a Patient’s AD Risk score using Logistic Regression Model
Dimensionality Reduction Methods
There are three major dimensionality reduction techniques that one can use:
1. Principal Component Analysis (PCA)
2. Prior spatial knowledge from an Atlas
3. Prior knowledge from the linear regression model in the previous section
The first approach is based on singular value decomposition (SVD) and needs a privacy-preserving algorithm. With the Inpher XOR Platform, data scientists are able to run such computations using
secret-shared data.
The second approach relies on prior spatial knowledge from an Atlas indicating what areas of the brain might be associated with AD. Examples of such Atlases are the Neuromorphometrics atlas released
by “MICCAI 2012 Grand Challenge and Workshop on Multi-Atlas Labeling” (www.neuromorphometrics.com).
The third approach uses the functional areas (voxels) identified to be statistically relevant to an Alzheimer’s Disease patient via the method in the previous section. It can be viewed as prior
knowledge. If one chooses this approach, it is important to ensure independent datasets are used for the supervised training of the two models in order to avoid model biases.
For the purpose of our simple model, here the first approach is chosen, that is, privacy-preserving PCA with Inpher’s XOR Platform.
Workflow for Training the Alzheimer’s Disease Logistic Regression Model
Figure 5 below explains the entire workflow for training the Alzheimer’s Disease logistic regression model. The MRI images for both the HC and AD cohorts are preprocessed with SPM locally at each
hospital, private radiology lab or research institute. The output of the preprocessing are the voxel matrices X1, X2 and X3 (with the additional extra features). These matrices are large (have 20,000
columns) and they need to be vertically stacked to assemble the training dataset.
One can first ingest the three matrices into the XOR Platform by letting each party secret share them among the three parties. Those can be then used with the XOR PCA functionality to obtain
dimension-reduced feature matrix (a matrix with about 200 columns) secret shared among the three compute nodes. Lastly, these secret shares, together with the secret shared labels, are used to run
the XOR logistic regression function to compute the final model.
The output model can be kept “secret” to securely compute the class predictions and the positive class probability vector using a new MRI scan. Alternatively, it can be revealed to the data analyst
or to any party that should have access to it and that should perform the predictions in plaintext, e.g., by instantiating a scikit-learn logistic regression classifier with the secretly trained
The final quality assessment of the prediction on the test dataset can be done without revealing the predicted probabilities. This can be accomplished using the privacy-preserving: Confusion Matrix,
Precision/Recall/F1 Score, and AUC functions available in the XOR Platform.
Figure 4: SPM with XOR MPC Workflow for Training Diagnosis Model
In cases where the goal is to derive the probability of more than one predictor at a time, then multilevel logistic regression is ideal. However, while this method could allow clinicians to draw a
better conclusion, the model’s quality, accuracy and reproducibility are highly dependent on the size and representativeness of the sample data.
Inpher developed a federated approach that incorporated partial observations captured across multiple computing parties and built a high-level inference model based on multilevel Bayesian modeling to
overcome these shortcomings. Intrinsically, the Bayesian Model comparisons scheme allows the comparison of the accuracy of specific models according to real-world data. Although the model was
implemented with “privacy-by-design” principles, where only model parameters are exchanged between the computing parties, there were no explicit methods for protecting these parameters, such as
In order to identify Alzheimer’s disease effectively, it is necessary to have an unbiased prediction model. Unfortunately, if the training uses data available only at the hospital, the model can
skew, since most of the samples are AD patients. Hence, having access to diverse data sources such as private clinics, radiology labs, and even pharmaceutical companies while safeguarding patient
privacy is critical to building great predictive models.
This blog describes how AI researchers and clinicians could leverage the sensitive data from multiple parties to build linear and logistic regression models using MPC in the XOR Platform. The next
blog will provide a step-by-step python notebook walk-through example in order to demonstrate the real-world application for building these models with sensitive healthcare data.
Get started with privacy-preserving machine learning and experience what it’s like to build models by securely leveraging sensitive data from various sources. | {"url":"https://inpher.io/blog/neuroimaging-based-diagnosis-of-alzheimers-disease-using-privacy-preserving-machine-learning-part-2/","timestamp":"2024-11-12T13:38:54Z","content_type":"text/html","content_length":"202457","record_id":"<urn:uuid:993a8829-a033-4fcb-a159-065dcff2d561>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00600.warc.gz"} |
Help with simplify and expression
From: Richmond, VA
Registered: 2022-03-29
Posts: 50
Re: Help with simplify and expression
Registered: 2024-10-09
Posts: 1
Help with simplify and expression
Hello everyone. I am sitting a GCSE exam (higher tier) as a mature student and I am doing a lot of practice at the moment. I am stuck with an expression, I would be so grateful if somebody could
explain me how to solve it in simple term please. This is the expression:
8X² x √2X½
( __________ )³
I assume by "solve" you mean "simplify," as it is not an equation.
I find two things confusing in your expression:
1. Is the small "x" before the radical a "multiply" ? If so, you can delete it. Pretty sloppy notation.
2. Is the "1/2" under the radical supposed to be the X's exponent?
Last edited by Phrzby Phil (2024-10-10 03:11:33)
World Peace Thru Frisbee
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,868
Re: Help with simplify and expression
Hi Fra1990 & Phrzby Phil;
If I've interpreted post #1's expression correctly, then...
Sorry, but simplifying that is beyond what I've learnt in maths...
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Registered: 2010-06-20
Posts: 10,610
Re: Help with simplify and expression
hi Fra1990
Welcome to the forum.
You can simplify powers by using the rules for indices. https://www.mathsisfun.com/algebra/exponent-laws.html
phrontister has tidied up your maths notation to make the expression clearer. But is that what you meant?
Another interpretation of what you've posted is:
Please post again, making it clear which is right.
When you've got x and times in an expression you can use a . for the times to avoid confusion.
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob | {"url":"https://mathisfunforum.com/viewtopic.php?pid=443374","timestamp":"2024-11-11T17:28:54Z","content_type":"application/xhtml+xml","content_length":"12447","record_id":"<urn:uuid:30e72028-8126-42bc-9ca4-17953a4c8e62>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00882.warc.gz"} |
Why do we have to use "combinations of n things taken x at a time" when we calculate binomial probabilities? | Socratic
Why do we have to use "combinations of n things taken x at a time" when we calculate binomial probabilities?
1 Answer
See below on my thoughts:
The general form for a binomial probability is:
The question is Why do we need that first term, the combination term?
Let's work an example and then it'll come clear.
Let's look at the binomial probability of flipping a coin 3 times. Let's set getting heads to be $p$ and of not getting heads #~p# (both #=1/2)#.
When we go through the summation process, the 4 terms of the summation will equal 1 (in essence, we are finding all the possible outcomes and so the probability of all the outcomes summed up is 1):
${\sum}_{k = 0}^{3} = \textcolor{red}{{C}_{3 , 0} {\left(\frac{1}{2}\right)}^{0} \left({\left(\frac{1}{2}\right)}^{3}\right)} + \textcolor{b l u e}{{C}_{3 , 1} {\left(\frac{1}{2}\right)}^{1} \left({\
left(\frac{1}{2}\right)}^{2}\right)} + {C}_{3 , 2} {\left(\frac{1}{2}\right)}^{2} \left({\left(\frac{1}{2}\right)}^{1}\right) + {C}_{3 , 3} {\left(\frac{1}{2}\right)}^{3} \left({\left(\frac{1}{2}\
So let's talk about the red term and the blue term.
The red term describes the results of getting 3 tails. There is only 1 way for that to be achieved, and so we have a combination that equals 1.
Note that the last term, the one describing getting all heads, also has a combination that equals 1 because again there is only one way to achieve it.
The blue term describes the results of getting 2 tails and 1 head. There are 3 ways that can happen: TTH, THT, HTT. And so we have a combination that equals 3.
Note that the third term describes getting 1 tails and 2 heads and again there are 3 ways to achieve that and so the combination equals 3.
In fact, in any binomial distribution, we have to find the probability of a single kind of event, such as the probability of achieving 2 heads and 1 tails, and then multiplying it by the number of
ways it can be achieved. Since we don't care about the order in which the results are achieved, we use a combination formula (and not, say, a permutation formula).
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