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https://git.scc.kit.edu/GPIAG-Software/IFOS3D/-/raw/62c16b9dcfedd7a7b0b47fc4bfaf09a9927a90b9/doc/latex/manual_toy.tex
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\chapter{Toy example - the box model} This chapter describes the inversion of a simple toy example with IFOS3D, which is a subdivided box in a homogeneous full space. This example can be performed as a first synthetic application to test IFOS3D on your system. The necessary input files and scripts are included in the IFOS3D folders and you can follow the following guidelines to perform a successfull inversion. Your results can be compared to those given in this chapter.\\ The 3D inversion is costly even for a simple model. To enable a successfull inversion we therefore recommend the use of a parallel system with a larger number of CPU's as for example found in supercomputing centers. We performed this inversion using 512 CPU's for 7.5 hours. Before starting this toy example, IFOS3D needs to be installed on your system (see chapter~\ref{Gettingstarted}). \section{Inversion setup} \subsection{Model and grid system} \begin{figure}[h!] \begin{center} \includegraphics[width=\textwidth]{fig_toy/toy_real_model} \caption[Toy example - the real model and acquisition geometry]{The toy example - real model $v_p$ (a) and $v_s$ (b), sources and receivers indicated as stars and crosses, respectively}\label{fig:toy_model} \end{center} \end{figure} The true model for the seismic velocities is plotted in figure~\ref{fig:toy_model}. A box divided into four differently-sized parts with different positive and negative velocity variations is placed into a homogeneous full space of $v_p=6200$\,m/s and $v_s=3600$\,m/s. The $v_p$/$v_s$ ratio is not constant. The density model is homogeneous with 2800\,kg/m$^3$ and kept constant during inversion. The elastic model is generated in IFOS3D on the fly (\textbf{READMOD}=0) using the function \textit{src/hh\_toy.c} as defined in \textit{src/Makefile}.\\ The model size is 128\,m$\times$128\,m $\times$147.2\,m in $x$-, $y$- and $z$-direction. We use a grid point distance of \textbf{DX}=\textbf{DY}=\textbf{DZ}=0.8\,m and thus a grid size of \textbf{NX}=160, \textbf{NY}=160 and \textbf{NZ}=184 points. \\ We use spatial FD-operators of \textbf{FDORDER}=4 and Holberg coefficients (\textbf{FDCOEFF=2}). This fullfills the dispersion criterion (see SOFI manual) up to a frequency of 500\,Hz ($v_{s,min}=3450$\,m/s): $$dh <= \frac{v_{s,min}}{nf_{max}}=\frac{v_{s,min}}{8.32f_{max}}$$ For our maximum frequency of 320\,Hz used for inversion this is well sufficient. \\ The total calculation time is \textbf{TIME}=0.06\,s with a time stepping of 5e$^{-5}$s resulting in 1200 tiesteps per forward modeling. The stability criterion is fullfilled ($v_{p,max}=6700$\,m/s): $$dt<=\frac{dh}{0.487v_{p,max}}=2.4e^{-4}$$ The model does not contain a free surface (\textbf{FREE\_SURF}=0) and as boundary condition we use a PML-layer (\textbf{FW}) of 10 grid points width at each model side.\\ \subsection{Sources and receivers} \begin{figure}[h!] \begin{center} \includegraphics[width=0.8\textwidth]{fig_toy/source_toy} \caption[Toy example - source wavelet and spectrum]{The toy example - normalised amplitude and spectrum of sin$^3$-source wavelet }\label{fig:toy_wavelet} \end{center} \end{figure} Sources and receivers are arranged within $x$-$y$-planes, as indicated in Figure~\ref{fig:toy_model}. The sources are listed in \textit{sources/sources\_toy.dat} (\textbf{SOURCE\_FILE}). We use 12 (3$\times$4) sources in 92\,m depth with a central frequency of \textbf{FC}=300\,Hz. The sources are vertical directed point forces (\textbf{QUELLTYP}=4) with sin$^3$-wavelets (\textbf{QUELLART=4}) as source time functions. The corresponding source wavelet can be seen in figure~\ref{fig:toy_wavelet}a. This source wavelet comprises the low frequencies with $40\%$ of the maximum amplitude left at about 400\,Hz (figure~\ref{fig:toy_wavelet}b). \\ The receivers are arranged using a horizontal receiver array (\textbf{READREC}=2, \textbf{REC\_ARRAY}=1) in 24\,m depth. The distance between the receivers is \textbf{DRX}=\textbf{DRY}=10 gridpoints, which gives a total number of 169 receivers. \subsection{Inversion parameters} For the inversion we use homogeneous starting models with $v_s=3600$\,m/s and $v_p=6200$\,m/s. In total 60 iterations are performed (\textbf{ITMIN, ITMAX }= 1, 60) and five frequencies are employed simultanously (\textbf{NFMAX}=5). For preconditioning of gradients a local taper is applied around source and receiver positions (\textbf{DAMPTYPE}=2). We do not use the diagonal Hessian approximation (\textbf{HESS}=0) or the L-BFGS method (\textbf{LBFGS}=0). Out of the total number of 12 shots we use 4 shots for the steplength estimation (\textbf{NSHOTS\_STEP}=4) and start with an initial test steplength of 0.02 (\textbf{TESTSTEP}=0.02). The filenames for gradient and model output are given in the input file. \subsubsection*{The workflow} The workflow file \textit{in\_and\_out} defines the different frequency stages. They are listed in table~\ref{tab:toy_fstage}.\\ \begin{table}[h!] \centering \begin{tabular}{|c|c|c|c|}\hline iteration& frequencies in Hz & $\lambda_{min}(v_p)$ in m &$\lambda_{min}(v_s)$ in m \\ 1-15& 160,170,180,190,200 & 31 & 18\\ 16-30& 200,210,220,230,240 & 26 & 15\\ 31-45& 240,250,260,270,280 & 22 & 13\\ 46-60& 280,290,300,310,320 & 19 & 11\\ \hline \end{tabular} \caption[Transmission example: frequency stages]{Frequency stages used for the transmission geometry box example with minimum wavelengths $\lambda_{min}(v_p)$ and $\lambda_{min}(v_s)$.} \label{tab:toy_fstage} \end{table} We used four frequency stages with five frequencies, which increase from stage to stage. The frequency bands are slightly overlapping. The table also lists the ninimum and maximum wavelengths, which define the resolution of the result. In the first stage lower frequencies enable the reconstruction of rough model structures, whereas higher frequencies and smaller wavelengths result in a higher resolution of smaller model structures. \section{Step by step guideline} \subsection{Step 1 - forward modeling} In a first step, a forward modeling is performed to calculate data of the box model, which can be used as observed data in the inversion. This corresponds to a SOFI3D simulation. The model is created on the fly by the function \textit{src/hh\_toy.c}. The parameters are already set in \textit{in\_and\_out/ifos3D\_toy.inp}. For the forward modeling we use \textbf{METHOD}=0. The seismograms are written in the folder \textit{su\_obs}. For the option \textbf{METHOD}=0 the seismograms are written in one file for each processor containing receiver locations. From these files the data can be directly used as input in the inversion. Note, that \textbf{FILT}=0 and the program calculates the unfiltered seismograms. The box model is saved in the folder \textit{model}.\\ To perform the modeling the following steps are applied: \begin{itemize} \item compile IFOS3D (execute \textit{./compileIFOS3D.sh}) \item adapt processor number in \textit{in\_and\_out/ifos3D\_toy.inp} and your job script (e.g. \textit{startIFOS3D.sh}) \item start IFOS3D (e.g. execute \textit{./startIFOS3D.sh}) \end{itemize} The progress of IFOS3D can be viewed in \textit{in\_and\_out/ifos3D\_toy.out}. \subsection{Step 2 - the FWI} In a second step, the inversion is performed. Before staring the inversion perform the following steps: \begin{itemize} \item set the parameter (kasten=0) in \textit{src/hh\_toy.c} to gain homogeneous starting models \item compile IFOS3D (execute \textit{./compileIFOS3D.sh}) \item change the following parameters in \textit{in\_and\_out/ifos3D\_toy.inp}: \textbf{METHOD}=1, \textbf{FILT}=1, \textbf{SEIS\_FILE} = ./su/cal\_toy \item start IFOS3D (e.g. execute \textit{./startIFOS3D.sh}) \end{itemize} Now the inversion is performed. The proceedings of IFOS3D can be seen in the outputfile \textit{in\_and\_out/ifos3D.out}. An overview how the inversion succeeds, e.g. misfit values and steplengths can be found in the output file \textit{in\_and\_out/ifos3D\_invers.out}. Both files are written during simulation. \section{Results} In this section we look at the output of IFOS3D and show the inversion results. All output is located in the folder \textit{par}. For processing and plotting of data we use Matlab programs, located in the folder \textit{mfiles} and Seismic Unix. \subsection{Seismograms} \subsubsection*{Output} The observed seismograms are calculated in step 1- forward modeling. They are located in the folder \textit{su\_obs} and provide unfiltered particle velocities for each shot and component in the su-format (e.g. \textit{obs\_toy\_vx\_it1.su.shot3}). \\ The inverted seisograms are stored in each iteration for each shot and each component in the folder \textit{su}. They are also in the su-format and named for instance \textit{cal\_toy\_vy\_it30.su.shot2}. Note that these data are filtered below the highest frequency used within the corresponding frequency stage.\\ The su-header in front of each trace contains i.a. information about source and receiver location, time stepping and time samples. It can be viewed using the Seismic Unix function \textit{surange} $<$ FILENAME or the Matlab function \textit{su2matlab}. \subsubsection*{Data plot} For a first look, the seismograms can be plotted with the SU-routine \textit{suxwigb} $<$ FILENAME. Note, that it can be necessary to use the function \textit{segyclean} $<$ FILENAME $>$ FILENAME\_OUT before plotting.\\ For a comparison of obeserved and inverted data, it is necessary to apply a lowpass filter to the observed data, using the maximum frequency of the corresponding stage. Figure~\ref{fig:toy_seismo1} shows a comparison between initial, observed and inverted seismograms for one source and receiver and $x$-, $y$- and $z$-component of the particle velocity. Note, that the observed data was filtered with the maximum frequency of the first iteration for a comparison in a) and with the maximum inversion frequency for a comparison with the final inverted data in b). The plot was produced with the Matlab program \textit{seismo\_trace\_toy.m}. This program uses the binary format as input, which is why the files need to be transformed before use, applying the SU-routine \textit{sustrip} $<$ FILENAME $>$ FILENAME\_OUT. \begin{figure}[h!] \begin{center} \includegraphics[width=\textwidth]{fig_toy/seismo1_toy} \caption[Toy example - observed, initial and inverted seismograms]{Multi-component seismograms exemplarily for source at ($x_s$, $y_s$, $z_s$) = (32\,m, 96\,m, 92\,m) and receiver at ($x_r$, $y_r$, $z_r$) = (56\,m, 40\,m, 24\,m), normalised to one trace a) initial vs. observed data filtered below 200\,Hz and b) observed vs. inverted data filtered below 320\,Hz} \label{fig:toy_seismo1} \end{center} \end{figure} \subsubsection*{Discussion} Figure~\ref{fig:toy_seismo1} shows the comparison of initial and observed data for the low frequencies of the first inversion stage. The waveforms show only small differences and clearly no cycle skipping. Thus the homogeneous starting model is already sufficient for this simple example. \\ The success of the inversion can be seen when comparing the observed and inverted data. The waveforms, including the small oscillations are fitted very well for all components. \subsection{The gradient} Gradients for $v_p$, $v_s$ and $\rho$ are stored in binary format in each iteration in the folder \textit{grad} (e.g. \textit{toy\_grad.vp\_200.00Hz\_it5}). Additionally to the raw gradients named by iteration number, the conjugate gradients are stored with labels of (iteration number +1000), like (e.g. \\ \textit{toy\_grad.vp\_200.00Hz\_it1005}). \\ The gradients can be plotted with the Matlab program \textit{slice\_3D\_toy\_grad.m}. This program plots the 3D grid as two perpendicular slices. The acquisition geometry of the toy example is also included. \\ Here, we show the raw'' gradients and the preconditioned gradients of $v_p$ and $v_s$ normalised to their maximum value for the first iteration (figure~\ref{fig:toy_grad}). In the raw gradients (a,b) the high amplitudes around sources and receivers are clearly visible. Especially the source artefacts are very distinct compared to the small scaled receiver artefacts. By preconditioning these artefacts can be removed for the greater part (c,d) and the main update concentrates on the box area. Due to the smaller wavelengths of the shear wave, the $v_s$-gradient already shows more structure than the $v_p$-gradient. \begin{figure}[h!] \begin{center} \includegraphics[width=\textwidth]{fig_toy/grad_toy} \caption[Toy example - gradient before and after preconditioning]{Normalised gradients of first iteration : a) raw'' gradient $v_p$, b) raw'' gradient $v_s$, c) preconditioned gradient $v_p$ and d) preconditioned gradient $v_s$; sources (stars) and receivers (crosses) are indicated. }\label{fig:toy_grad} \end{center} \end{figure} \subsection{Misfit and steplength proceedings} The misfit and steplength values are stored in \textit{in\_and\_out/ifos3D\_invers.out}. They give insights into the proceedings of the inversion. After storing the values one in a row in textfiles, they can be plotted with the Matlab functions \textit{misfit\_toy.c} and \textit{steplength\_toy.c}. \\ The misfit proceedings are plotted in figure~\ref{fig:toy_misfit}. The values are normalised to the initial misfit. The curve shows a high convergence at the beginning of the inversion, where the rough model structures are included. The curve flattens as the inversion proceeds and only finer changes in the model are performed. Note, that with each new frequency stage data is added to the misfit and the misfit steps to higher values, which is then reduced within this stage.\\ \begin{figure}[h!] \begin{center} \includegraphics[width=0.7\textwidth]{fig_toy/toy_misfit} \caption[Toy example - misfit]{Misfit normalised to initial value shows the convergence }\label{fig:toy_misfit} \end{center} \end{figure} The steplength curve (figure~\ref{fig:toy_steplength}) shows a similar behaviour. At the beginning steplengths of few percent are found. That means the size of the model update is few percent of the average model parameter. These steplengths quickly decrease to less than 0.5\% after few iterations. Most model changes are thus made in the first few iterations whereas only the fine work is performed in the later iterations and at higher frequencies. \begin{figure}[h!] \begin{center} \includegraphics[width=0.7\textwidth]{fig_toy/toy_steplength} \caption[Toy example - misfit]{Steplength proceedings}\label{fig:toy_steplength} \end{center} \end{figure} \subsection{The final models} Model parameters are written into the folder \textit{model} in each iteration for each parameter ($v_p$, $v_s$, $\rho$) in binary format (e.g. \textit{toy.vp\_it22}). They can be plotted similarly to the gradient with the program \textit{slice\_3D\_toy\_model.m}. Note, that gradient and model files share the same structure.\\ The results for the toy example is plotted for two 2D slices. Figure~\ref{fig:toy_result1} shows a horizontal slice through the box area. The three different box areas seen in the real model in a) and c) are well resolved for $v_p$ (b) and $v_s$ (d) after 60 iterations. However the $v_s$-model offers a clearly higher resolution due to the smaller wavelengths compared to the $v_p$-model. In transmission geometry a resolution down to one wavelength is possible and higher frequencies would be necessary to gain a better resolved box in $v_p$. \begin{figure}[h!] \begin{center} \includegraphics[width=\textwidth]{fig_toy/toy_model_result_new} \caption[Toy example - final inverted models, horizontal slice]{Final inverted models (60 iteartions) compared to real models for horizontal slice at $z$=60\,m: a) real model $v_p$, b) inverted model $v_p$, c) real model $v_s$ and d) inverted model $v_s$. The dashed line indicates the absorbing frame; }\label{fig:toy_result1} \end{center} \end{figure} A vertical slice of the models is plotted in figure~\ref{fig:toy_result2}. Overall, the vertical direction is more difficult to reconstruct, as can be ssen in the inverted models of $v_p$ (b) and $v_s$ (d) compared to the real models (a,b). The boundaries of the box are relatively well recovered for both parameters. The inversion of $v_p$ reconstructs the two main areas of the box, however the small high velocity area (dark red) cannot be resolved. This area is indicated in the inverted $v_s$ model, again showing the higher resolution of this parameter. \begin{figure}[h!] \begin{center} \includegraphics[width=\textwidth]{fig_toy/toy_model_result1} \caption[Toy example - final inverted models, vertical slice]{Final inverted models (60 iterations) compared to real models for vertical slice at $x$=56\,m: a) real model $v_p$, b) inverted model $v_p$, c) real model $v_s$ and d) inverted model $v_s$; }\label{fig:toy_result2} \end{center} \end{figure} \section{Outlook} This example is also included by \cite{But15}. Here the applications using the diagonal Hessian approximation and the L-BFGS method are shown. You can also look at the other applications \citep{But13, But15} which include the inversion of a random medium model in different transmission geonetries and a surface geometry model.
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A rhomboid, is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles.
The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ... - Side 2
av Euclides - 1816 - 528 sider
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## The elements of plane geometry; or, The first six books of Euclid, ed. by W ...
Euclides - 1863
...an obtuse ./f xxrx. An acute-angled triangle is that which has three acute ./\ angles. ^-—t XXX. Of four-sided figures, a square is that which has...oblong is that which has all its angles right angles, I but has not all its sides equal. ' L- 1 A rhombus is that which has all its sides equal, but its...
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## Euclid's Elements of geometry, the first four books, by R. Potts. Corrected ...
Robert Potts - 1864
...triangle is that which has three acute angles. XXX. Of quadrilateral or four-sided figures, a square has all its sides equal and all its angles right angles....but has not all its sides equal. XXXII. A rhombus has all its sides equal, but its angles are not right angles. XXXIII. A rhomboid has its opposite sides...
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## The school edition. Euclid's Elements of geometry, the first six books, by R ...
Robert Potts - 1864
...triangle is that which has three acute angles. XXX. Of quadrilateral or four-sided figures, a square has all its sides equal and all its angles right angles....but has not all its sides equal. XXXII. A rhombus has all its sides equal, but its angles are not right angles. XXXIII. A rhomboid has its opposite sides...
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## The college Euclid: comprising the first six and the parts of the eleventh ...
...has an obtuse angle. XXIX. An acute-angled triangle, is that which has three acute angles. B3 XXX. Of four-sided figures, a square is that which has...its sides equal, and all its angles right angles. XXXL An oblong, is that which has all its angles right angles, but has not all its sides equal. XXXIL...
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## An English primer; compiled under the superintendence of E.C. Lowe
Edward Clarke Lowe - 1866
...that which has an obtuse angle. 29. An acute-angled triangle is that which has three acute angles. 30. Of four-sided figures, a square is that which has...its sides equal, and all its angles right angles. 31. An oblong is that which has all its angles right angles, but not all its sides equal. 32. A rhombus...
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## The Elements of Euclid for the Use of Schools and Colleges: Comprising the ...
Euclid, Isaac Todhunter - 1867 - 400 sider
...angle : 29. An acute-angled triangle is that which has three acute angles. Of four-sided figures, 30. A square is that which has all its sides equal, and all its angles right angles : 31. An oblong is that which has all its angles right angles, but not all its sides equal : 32. A...
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## The Elements of Euclid for the Use of Schools and Colleges: Comprising the ...
Euclid, Isaac Todhunter - 1867 - 400 sider
...angle : 29. An acute-angled triangle is that which has three acute angles. Of four-sided figures, 30. A square is that which has all its sides equal, and all its angles right angles : 31. An oblong is that which has all its angles right angles, but not all its sides equal : 32. A...
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## Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with ...
Robert Potts - 1868 - 410 sider
...triangle is that which has three acute angles. XXX. Of quadrilateral or four-sided figures, a square has all its sides equal and all its angles right angles....angles right angles, but has not all its sides equal. XXXIL A rhombus has all its sides equal, but its angles are not right angles. XXXIII. A rhomboid has...
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## Right lines in their places; or, The first principles of drawing and design ...
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## The Educational Speeches of the Hon'ble John Bruce Norton
1870 - 329 sider
...that, especially when the rudiments are mastered. Euclid always to speak of a square as " that figure which has all its sides equal, and all its angles right angles;" or to avoid the term circle by a round about allusion to it as " a plain figure contained by one line...
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https://www.quantstart.com/articles/The-Markov-and-Martingale-Properties
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The Markov and Martingale Properties
In order to formally define the concept of Brownian motion and utilise it as a basis for an asset price model, it is necessary to define the Markov and Martingale properties. These provide an intuition as to how an asset price will behave over time.
The Markov property states that a stochastic process essentially has "no memory". This means that the conditional probability distribution of the future states of the process are independent of any previous state, with the exception of the current state. The Martingale property states that the future expectation of a stochastic process is equal to the current value, given all known information about the prior events.
Both of these properties are extremely important in modeling asset price movements.
## The Markov Property
A sensible way to introduce the Markov property is through a sequence of random variables $Z_i$, which can take one of two values from the set $\{1,-1\}$. This is known as a coin toss. We can calculate the expectations of $Z_i$:
\begin{eqnarray*} \mathbb{E}(Z_i) = 0, \mathbb{E}(Z_i^2) = 1, \mathbb{E}(Z_i Z_k) = 0 \end{eqnarray*}
The key point is that the expectation of $Z_i$ has no dependence on any previous values within the sequence. Let us take the partial sums of our random variables within our coin toss, which we will denote by $S_i$:
\begin{eqnarray*} S_i = \sum^i_{k=1} Z_i \end{eqnarray*}
We can now calculate the expectations of our partial sums, using the linearity of the expectation operator:
\begin{eqnarray*} \mathbb{E}(S_i) = 0, \mathbb{E}(S_i^2)= \mathbb{E}(Z_1^2 + 2 Z_1 Z_2 + ...) = i \end{eqnarray*}
We see that, again, there is no dependence on the expectation of $S_i$ of any previous value within the sequence of partial sums. We can extend this to discuss conditional expectation. Conditional expectation is the expectation of a random variable with respect to some conditional probability distribution. Hence, we can ask that if $i=4$ (i.e. we carry out four coin tosses), what does this mean for the expectation of $S_5$?
\begin{eqnarray*} \mathbb{E}(S_5 | Z_1, Z_2, Z_3 , Z_4) = S_4 \end{eqnarray*}
That is, the expected value of $S_i$ is only dependent upon the previous value $S_{i-1}$, not on any values prior to that. This is known as the Markov Property. Essentially, there is no memory of past events beyond the point our variable is currently at within the sequence. Nearly all financial models discussed in these articles will possess the Markov property.
## The Martingale Property
An additional property that holds for our sequence of partial sums is the Martingale property. It states that the conditional expectation of the sequence of partial sums, $S_i$ is simply the current value:
\begin{eqnarray*} \mathbb{E}(S_i | S_k, k<i) = S_k \end{eqnarray*}
Essentially, the martingale property ensures that in a "fair game", knowledge of the past will be of no use in predicting future winnings.
These properties will be of fundamental importance in regard to defining Brownian motion, which will later be used as a model for an asset price path.
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# Why is a sphere in an $n$-dimensional space called $(n-1)$-sphere?
Why is a sphere in an $n$-dimensional space called $(n-1)$-sphere? Isn't it natural to call a sphere in 3D a 3-sphere, a sphere in 2D (i.e. a circle) a 2-sphere, etc?
• It is the dimension viewed as a manifold May 18 '14 at 11:52
• It takes only two coordinates to specify a point on a sphere $S \subset \mathbb R^3$, hence $S$ is two dimensional. A plane in $\mathbb R^3$ is certainly two dimensional, and a sphere is also. May 18 '14 at 11:56
tl;dr: You are describing the sphere itself, not a particular space in which it is represented.
It should first be made clear the distinction between a sphere and a ball: a sphere consists of the surface only, whereas a ball consists of both the surface and the contained volume.
Now that that's out of the way, then there's the question of how you define a sphere. A sphere in n-dimensional space with radius $r$ centered at $(c_1,c_2,\cdots,c_n)$ is very easily represented by $(x_1-c_1)^2+(x_2-c_2)^2+\cdots+(x_n-c_n)^2=r^2$. However, this is an implicit representation, not an explicit representation. An explicit representation entails a mapping of some other space, and in the case of a sphere in $n$-dimensional space, it happens to be the case that the lowest possible dimension of this other space is $n-1$. As such, spheres in $n$-dimensional space are said to be $n-1$-spheres.
The reason for this nomenclature is that this creates a description specific to the sphere itself and not to a specific environment of the sphere in which it is convenient to describe.
Consider a circle, that is, a $1$-sphere, for instance. Certainly, the equation $x_1^2+x_2^2=r^2$ may come to mind, and there are only two variables in this equation, $x$ and $y$, which might inspire you to call this a $2$-sphere. But a circle can also be represented in $\mathbb{R}^3$, $\mathbb{R}^4$, and so on (although of course, you would need to fix $x_3, x_4$ and so on as well). Why, then, should this circle not also be called a $3$-sphere or $4$-sphere? For that matter, is a $3$-sphere a circle or sphere?
To expand a little - a small piece of a circle is like a line, a small piece of the surface of a sphere is like a plane.
The only way that a line can live in one dimension is if it is straight - a curve can live in many dimensions and still look like a line when you look at a short piece.
Likewise a surface can only live in two dimensions if it is flat. Surfaces exist in many dimensions, but they look like a plane when you take a small enough piece.
If we are creating a coherent useful idea of dimension, it turns out that we want a curve to be one-dimensional and a surface to be two dimensional - but what do we mean by a curve or a surface if we have many-dimensional space?
The idea of a manifold captures this - if our curve is locally like a line, and the local pieces can be coherently glued together to make the whole, it is a one-dimensional manifold. Likewise a surface is a two dimensional manifold if it looks locally like a plane wherever you look and you can glue the local pieces together in a coherent way.
There are some technicalities about precise definitions, to make these intuitive ideas rigorous. But that is why the dimensions come out as they do.
Because sphere in $n$-dimensional space is $(n-1)$-dimensional object.
• So, we include the temporal dimension aswell? May 18 '14 at 11:49
• I do not know what is temporal dimension, but sphere in an $n$-dimensional space is a $(n-1)$-dimensional manifold. This is waht I meant in the answer. May 18 '14 at 11:51
• ball ≠ sphere. The surface of the ball exists in 3D but is is a 2 dimensional object in the sense that you only need 2 real numbers to specify the point on the sphere. May 18 '14 at 12:12
This is because the "$n-1$" sphere is the surface of the $n$ dimensional solid contained in the $n-1$ sphere, I.e it is one dimension lower.
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Jeopardy! Super Storms Written by Seymour Simon
Presentation on theme: "Jeopardy! Super Storms Written by Seymour Simon"— Presentation transcript:
Jeopardy! Super Storms Written by Seymour Simon
This Jeopardy game was created based on the Nonfiction title – “Super Storms”. Many questions have an open ended type question but is based explicitly on the text from Mr. Simon. Feel free to modify this power point to meet your needs.
Hurricanes Blizzards Tornadoes Thunderstorms 100 100 100 100 200 200 200 200 300 300 300 300 400 400 400 400 500 500 500 500
What are thunderstorms?
Heavy rain showers What are thunderstorms?
Hurricanes _____ more people than any other type of storm.
Kill and Injure Hurricanes _____ more people than any other type of storm.
A huge snowstorm What is a blizzard?
Up to 300 mph How fast can tornadoes travel?
What can you see during thunderstorms?
Lightning What can you see during thunderstorms?
What is the eye of the storm like?
Calm What is the eye of the storm like?
How fast are the winds blowing
At least 35 mph How fast are the winds blowing during a blizzard?
Pick up cars, Rip out houses
What kind of damage can a tornado cause?
You can tell how far away lightning is by
Five You can tell how far away lightning is by counting the time between the lightning strikes. For every ___ seconds equals one mile?
Where do hurricanes start?
On the warm ocean Where do hurricanes start?
How much snow comes down during a blizzard?
2 inches in one hour How much snow comes down during a blizzard?
What does a tornado watch mean?
Conditions are likely for a tornado to form What does a tornado watch mean?
During a thunderstorm, how much rain can fall in one minute?
Millions of gallons During a thunderstorm, how much rain can fall in one minute?
Which best describes the winds in a hurricane?
D. 74 – 200 mph Which best describes the winds in a hurricane? A. 10 – 50 mph B – 2000 mph C. 7 – 20 mph mph D mph
How cold is it during a blizzard?
20 degrees or colder How cold is it during a blizzard?
What does a tornado warning mean?
Get somewhere safe fast! One has been spotted. What does a tornado warning mean?
Thunderstorms have lightning, thunder, more rain.
How are thunderstorms different than storms?
Hurricanes are different sizes but how large can they get?
Hundreds of miles Hurricanes are different sizes but how large can they get?
different than a snowstorm?
The fierce winds blow the snow making it hard to see. What makes a blizzard different than a snowstorm?
Where are most US tornados found?
Central United States Where are most US tornados found?
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Jump to content
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# Scale bezier curve along normal
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Hi!
I want to resize a bezier curve along its normals in order to resize a free form shape. I want to resize the object, but not _scale_ it.
What i do right now is the following:
Take the set of control points
Calculate the normal on the tangent points
Add normal * size to the control points
Of course, this doesn't work exactly as it should. The tangents should be changed, but i don't know how. Right now, my shape (that is tesselated later) looks like it is liquified
Any idea?
My control point setup gives me a point, and a tangent in, and tangent out value (defining the other 2 control points). But if i add the normal to these points, they get eased out, as they are handled as tangents and the original point is already translated. I hope you can follow me...
I also tried to tesselate first and move every vertex along its vertex normal, but this gives me intersections pretty soon.
Thank you very much, any help is appreciated and i am happy to share a working solution as soon as i get there...
snow
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The name of the curve you're trying to find is an offset curve. Here's what I found about calculating those with a quick google search:
http://processingjs....erinfo/#offsets
I had to do this for quadratic bezier curves to implement stroking in my vector graphics library, and it was easy enough.
I started by using the same method as you (shifting control points along normals), which gives a good approximation when the curve is flat enough. When it isn't, you get the right distance at the extremities but much less than it at the center. Fortunately it is easy to solve by calculating the distance between the middle point of the original curve and of the approximate offset curve. If this distance is too far from the one you want, you just split the original curve in two and redo all this recursively.
It is probably more complicated for higher order curves, though. Edited by Zlodo
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awesome, i check it out, thanks!
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That worked perfectly!
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# 9 graph the system below and write its solution note
• Test Prep
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Chapter 9 / Exercise 13
Elementary and Intermediate Algebra
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9. Graph the system below and write its solution. Note that you can also answer "No solution" or "Infinitely many" solutions. 10. Use substitution to solve the system.
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Chapter 9 / Exercise 13
Elementary and Intermediate Algebra
Tussy/Gustafson
Expert Verified
ALEKS Systems and Rational Equations Quiz #1 - 09/05/2014 6:49:58 PM MDT - Copyright © 2014 UC Regents and ALEKS Corporation. P. 7/16 11. Solve for
12. For each system of linear equations shown below, classify the system as "consistent dependent," "consistent independent," or "inconsistent." Then, answer the question about its solutions. L1: L2: This system of equations is: - consistent dependent - consistent independent - inconsistent This means the system has: - a unique solution:
ALEKS Systems and Rational Equations Quiz #1 - 09/05/2014 6:49:58 PM MDT - Copyright © 2014 UC Regents and ALEKS Corporation. P. 8/16 L1: L2: This system of equations is:
ALEKS Systems and Rational Equations Quiz #1 - 09/05/2014 6:49:58 PM MDT - Copyright © 2014 UC Regents and ALEKS Corporation. P. 9/16 - consistent dependent - consistent independent - inconsistent This means the system has: - a unique solution: Solution: - no solution - infinitely many solutions L1: L2: This system of equations is: - consistent dependent - consistent independent - inconsistent This means the system has: - a unique solution: Solution: - no solution - infinitely many solutions
ALEKS Systems and Rational Equations Quiz #1 - 09/05/2014 6:49:58 PM MDT - Copyright © 2014 UC Regents and ALEKS Corporation. P. 10/16 13. Graph the system below and write its solution. Note that you can also answer "No solution" or "Infinitely many" solutions. 14. The sum of two numbers is One number is times as large as the other. What are the numbers? 15. Solve for
ALEKS Systems and Rational Equations Quiz #1 - 09/05/2014 6:49:58 PM MDT - Copyright © 2014 UC Regents and ALEKS Corporation. P. 11/16 Systems and Rational Equations Quiz #1 Answers for class Beginning and Intermediate Algebra Combined / MATH 101 - Fall 2014 – 504 1.
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T-SQL: Converting between Decimal, Binary, and Hexadecimal
By ronen ariely on 14/12/2015 10:38
I see in the forum this question raised all the time, and yet the solution that people get usually based on looping and complex math, while there is a simple and much better solution as shown in this code. In this short blog I will wrote a simple and VERY fast solution to convert Decimal number to Binary number, Decimal number to Hexadecimal number, and vise versa without using any loop. My solution for binary base is based on using BITWISE.
Representing list of values using a single value
By ronen ariely on 29/10/2014 16:18
In some case we have to use single value, in order to represent a list of elements (values). This is a very common needs in hardware and software developing. All the logics that we discus in this article, can be implements regardless of database use. In this article we will focus on using SQL Server to implement our logics. This might be an external demand (applications for example) or even a result of bad database design. Yet, the needs is clear, we want to store multiple elements in one column. Those elements can be properties list, options, security permissions, dates, or any other data.
In this article we will go over several solutions, using different logics. Our challenge is to find a good logic, which give us a one-to-one correspondence, between each available combination of our elements, to a single value which represent this combination.
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• To illustrate further:You personally endorse 1 in TEAM A and 1 in TEAM B. You get 500 for both the direct endorsement.But since the system detects 1 is to 1 you get another 500 pesos.Now those 2 friends of yours had endorsed 10 each, therefore you get 500 x 10 = 5000Now those in TEAM A endorsed 20 and those in TEAM B endorsed 20 also, therefore You get 20 pairs x 500 = 10,000 TSBNow those in TEAM A endorsed 60 and those in TEAM B endorsed 100, thereforeYou only get 60 pairs x 500 = 30,000 TSB.The remaining 40 in TEAM B will not expire but will just wait until there are endorsements in TEAM A.
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## Power leaders complete bomPresentation Transcript
• PPPP
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# 10 topics : how much is 5ml ?
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How Much is 5ml? Understanding the Measurement
# How Much is 5ml? Understanding the Measurement
## What is 5ml?
5ml is a unit of measurement used to measure volume. It is equivalent to 5 cubic centimeters (cc) or 0.17 fluid ounces (fl oz). This measurement is commonly used in various industries, including medicine and cooking.
## How is 5ml Used in Medicine?
In medicine, 5ml is often used to measure liquid medications. This is because it is a small enough amount to accurately measure doses for children and adults alike. For example, a doctor may prescribe a patient to take 5ml of cough syrup every 4 hours.
It is important to note that different medications may have different concentrations, so it is always important to follow the instructions provided by a healthcare professional or pharmacist.
## How is 5ml Used in Cooking?
In cooking, 5ml is often used to measure small amounts of liquid ingredients, such as vanilla extract or lemon juice. It is also commonly used to measure small amounts of spices and seasonings.
When measuring ingredients in cooking, it is important to use precise measurements to ensure the recipe turns out correctly. Using too much or too little of an ingredient can greatly affect the taste and texture of a dish.
## Converting 5ml to Other Units of Measurement
While 5ml is a commonly used measurement, it may be necessary to convert it to other units of measurement depending on the situation. Here are some common conversions:
• 5ml = 5cc
• 5ml = 0.17 fl oz
• 5ml = 1 teaspoon (tsp)
It is important to note that these conversions are approximate and may vary slightly depending on the specific substance being measured.
## Conclusion
5ml is a small but important unit of measurement used in various industries, including medicine and cooking. Understanding how to accurately measure and convert 5ml can help ensure the proper dosages and ingredient amounts are used.
Whether you are measuring liquid medication or adding spices to a recipe, knowing how much 5ml is can make a big difference in the outcome.
You are looking : how much is 5ml
## 10 how much is 5ml for reference
### 1.How much is 5ml liquid medicine in teaspoons? – Quora
• Author: How
• Publish: 27 days ago
• Rating: 2(1047 Rating)
• Highest rating: 5
• Lowest rating: 1
• Descriptions: 5 ml is about 1 teaspoon.
• More : 5 ml is about 1 teaspoon.
• Source : https://www.quora.com/How-much-is-5ml-liquid-medicine-in-teaspoons
### 2.How much is 5 milliliters in tablespoons? convert ml to tbsp
• Author: How
• Publish: 21 days ago
• Rating: 2(1281 Rating)
• Highest rating: 3
• Lowest rating: 1
• Descriptions:
• More :
• Source : https://tapproom.com/recipes/how-much-is-5-milliliters-in-tablespoons
### 3.How to Measure Without Measuring Spoons – Betty Crocker
• Author: How
• Publish: 29 days ago
• Rating: 4(972 Rating)
• Highest rating: 4
• Lowest rating: 2
• Descriptions: A teaspoon is 5ml, so if you have metric measuring items, such as a measuring jug or even a clean medicine cap, you can do a quick measurement that way.
• More : A teaspoon is 5ml, so if you have metric measuring items, such as a measuring jug or even a clean medicine cap, you can do a quick measurement that way.
• Source : https://www.bettycrocker.co.uk/how-to/measuring-tip-5ml-spoon
### 4.How Much Is 5 Milliliters In Tablespoons? 5 Tbsp To ML
• Author: How
• Publish: 24 days ago
• Rating: 4(1128 Rating)
• Highest rating: 3
• Lowest rating: 1
• Descriptions:
• More :
• Source : https://baccocharleston.com/cooking-tips/how-much-is-5-milliliters-in-tablespoons
### 5.Is 5 mL the same as 1 teaspoon? – The Donut Whole
• Author: Is
• Publish: 30 days ago
• Rating: 4(685 Rating)
• Highest rating: 4
• Lowest rating: 3
• Descriptions: So, 5mL is equal to 1 teaspoon. How much is 5 mL of liquid medicine? 5 milliliters (mL) of liquid medicine is equivalent to 1 teaspoon ( …
• More : So, 5mL is equal to 1 teaspoon. How much is 5 mL of liquid medicine? 5 milliliters (mL) of liquid medicine is equivalent to 1 teaspoon ( …
• Source : https://www.thedonutwhole.com/is-5-ml-the-same-as-1-teaspoon/
### 6.Is 5 mL or 10 mL a teaspoon? – The Donut Whole
• Author: Is
• Publish: 18 days ago
• Rating: 3(1805 Rating)
• Highest rating: 5
• Lowest rating: 3
• Descriptions: The measurement of 5ml in teaspoons is equal to one teaspoon. How many mL does a teaspoon hold?
• More : The measurement of 5ml in teaspoons is equal to one teaspoon. How many mL does a teaspoon hold?
• Source : https://www.thedonutwhole.com/is-5-ml-or-10-ml-a-teaspoon/
### 7.5 ml to tsp – How Many Teaspoons is 5ml? – Online Calculator
• Author: 5
• Publish: 28 days ago
• Rating: 4(908 Rating)
• Highest rating: 5
• Lowest rating: 3
• Descriptions: 5ml equals to 1.01 teaspoons or there are 1.01 tsp in 5 milliliters. ML Conversion. Milliliter: 5. US Teaspoon: 1.01442.
• More : 5ml equals to 1.01 teaspoons or there are 1.01 tsp in 5 milliliters. ML Conversion. Milliliter: 5. US Teaspoon: 1.01442.
• Source : https://online-calculator.org/5-ml-to-tsp
### 8.Convert 5 milliliters to tablespoons, ounces, cups, etc
• Author: Convert
• Publish: 18 days ago
• Rating: 5(874 Rating)
• Highest rating: 3
• Lowest rating: 1
• Descriptions: 5 Milliliters = 0.3381 Tablespoons … Volume Calculator Conversions. How much is 5 milliliters? … How big is 5 milliliters? Converting from 5 milliliters to …
• More : 5 Milliliters = 0.3381 Tablespoons … Volume Calculator Conversions. How much is 5 milliliters? … How big is 5 milliliters? Converting from 5 milliliters to …
• Source : https://www.carinsurancedata.org/calculators/volume/milliliters/5
### 9.How Much Is 5 mL? – Yea Big
• Author: How
• Publish: 19 days ago
• Rating: 4(595 Rating)
• Highest rating: 5
• Lowest rating: 3
• Descriptions: 5mL is an important measurement term as its value is equal to 1 teaspoon, tsp. This is a standard measurement for many medicines and other culinary needs …
• More : 5mL is an important measurement term as its value is equal to 1 teaspoon, tsp. This is a standard measurement for many medicines and other culinary needs …
• Source : https://yeabig.com/how-much-is-5-ml/
### 10.Convert 5 ml to cups. How much is 5 milliliter in a cup? | Mltocups.net
• Author: Convert
• Publish: 24 days ago
• Rating: 1(1858 Rating)
• Highest rating: 4
• Lowest rating: 1
• Descriptions:
• More :
• Source : https://mltocups.net/5-ml-to-cups/
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# ncpol2sdpa 1.0
A converter from polynomial optimization problems of noncommutative variables to sparse SDPA input format.
Latest Version: 1.9
## Ncpol2sdpa
Ncpol2sdpa is a set of scripts to convert a noncommutative polynomial optimization problem to a sparse semidefinite programming (SDP) problem that can be processed by the SDPA family of solvers. The optimization problem can be unconstrained or constrained by equalities and inequalities.
The objective is to be able to solve very large scale optimization problems. For example, a convergent series of lower bounds can be obtained for ground state problems with arbitrary Hamiltonians.
The implementation has an intuitive syntax for entering Hamiltonians and it scales for a larger number of noncommutative variables using a sparse representation of the SDP problem.
## Dependencies
The code requires SymPy>=0.7.2 in the Python search path. The code is known to work with Python 2.6.8 and 2.7.5, and also with Pypy 1.8 and 2.0.2. Using Pypy is highly recommended, as execution time is several times faster and memory use is reduced. The code is compatible with Python 3, but using Python 3.3.2 incurs a major decrease in performance; the case is likely to be similar in with other Python 3 versions.
## Usage
The following code replicates the toy example from Pironio, S.; Navascues, M. & Acin, A. Convergent relaxations of polynomial optimization problems with noncommuting variables SIAM Journal on Optimization, SIAM, 2010, 20, 2157-2180.
```from ncpol2sdpa.ncutils import generate_ncvariables
from ncpol2sdpa.sdprelaxation import SdpRelaxation
#Number of Hermitian variables
n_vars = 2
#Order of relaxation
order = 2
#Get Hermitian variables
X = generate_ncvariables(n_vars)
#Define the objective function
obj = X[0] * X[1] + X[1] * X[0]
# Inequality constraints
inequalities = [ -X[1]**2 + X[1] + 0.5 ]
# Equality constraints
equalities = []
#Simple monomial substitutions
monomial_substitution = {}
monomial_substitution[X[0]**2] = X[0]
#Obtain SDP relaxation
sdpRelaxation = SdpRelaxation(X)
sdpRelaxation.get_relaxation(obj, inequalities, equalities,
monomial_substitution, order)
sdpRelaxation.write_to_sdpa('examplenc.dat-s')
```
Further examples are under the examples folder.
## Installation
Follow the standard procedure for installing Python modules:
\$ sudo python_interpreter setup.py install
If you install the module to your CPython library, but you want to use Pypy, please ensure that the PYTHONPATH variable is set up correctly, otherwise Pypy will not find the relevant modules.
## Acknowledgment
This work is supported by the European Commission Seventh Framework Programme under Grant Agreement Number FP7-601138 PERICLES, by the Red Espanola de Supercomputacion grants number FI-2013-1-0008 and FI-2013-3-0004, and by the Swedish National Infrastructure for Computing project number SNIC 2014/2-7.
## More Information
For more information refer to the following manuscript:
http://arxiv.org/abs/1308.6029
File Type Py Version Uploaded on Size
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Ordered set ,filed, definition of =
1. Sep 13, 2009
plsmail2mark
ordered set ,filed, definition of "="
Hi everyone,
an ordered set is a set S that for any x,y in S, there is a order definition "<" so that one and only one of the followings will be true:
x<y, y<x, x=y.
and also if x<y, y<z and x,y,z in S, then x<z.
but there is no such a definition that if x=y,y=z and x,y,z in S, then x=y=z and also a definition like x=x if x is in S.
An filed(F) should fullfill the axioms of addition,mulplification and distribution law.
from the addition axiom we can get the proposition that if x+y=x+z then y=z by following proof:
y=0+y=(-x+x)+y=-x+(x+y)
given above condition x+y=x+z then
y=-x+x+z=0+z=z.
but there is no such a definition or axioms that if x=y,a=b and x,y,z,b are in F, then x+a=y+b.
it really confuse me about "if x=y,y=z then x=z".
could you help me?
thanks.
2. Sep 13, 2009
D H
Staff Emeritus
Re: ordered set ,filed, definition of "="
Where did you get the idea that "there is no such a definition that if x=y,y=z and x,y,z in S, then x=y=z and also a definition like x=x if x is in S"? Equality is reflexive, symmetric, and transitive.
3. Sep 13, 2009
plsmail2mark
Re: ordered set ,filed, definition of "="
Hi DH,
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Understanding Decimals
This is a set of three, one-page problems about the sizes of moons in the solar system. Learners will use decimals to compare the sizes and distances of Saturn's moons to the center of Saturn. Options are presented so that students may learn about... (View More)
Audience: Middle school
Materials Cost: Free
Number Theory and Fractions
This is a set of two, one-page problems about the sizes of moons in the solar system. Learners will use fractions to compare the sizes and distances of Jupiter's moons. Options are presented so that students may learn about the Juno mission through... (View More)
Audience: Middle school
Materials Cost: Free
The Coordinate Plane
This is a set of three, one-page problems about how astronomers use coordinate systems. Learners will plot a constellation on a coordinate plane and/or plot the route of Mars Science Lab (MSL aka Curiosity) on the surface of Mars. Options are... (View More)
Audience: Middle school
Materials Cost: Free
Scale Models and Diagrams
This is a set of three, one-page problems about the scale of objects in images returned by spacecraft. Learners will measure scaled drawings using high-resolution images of the lunar and martian surfaces. Options are presented so that students may... (View More)
Audience: Middle school
Materials Cost: Free
Ratios, Proportions, Similarity
This is a set of four, one-page problems about the size of planets compared to earth. Learners may use ratios to compare planets within our solar system or those outside of our solar system with the earth. Options are presented so that students may... (View More)
Audience: Middle school
Materials Cost: Free
Geometry and Angle Relationships
This is a set of four, one-page problems about the distance craft travel on Mars. Learners will use the Pythagorean Theorem to determine distance between a series of hypothetical exploration sites within Gale Crater on Mars. Options are presented so... (View More)
Audience: Middle school
Materials Cost: Free
Measurement and Geometry
This is a set of three, one-page problems about the size and area of solar panels used to generate power. Learners will calculate area fractions to compare the sizes and distances of Jupiter's moons. Options are presented so that students may learn... (View More)
Audience: Middle school
Materials Cost: Free
Integer Arithmetic
This is a set of three, one-page problems about the size and area of solar panels used to generate power. Learners will will use integer arithmetic to tally the number of hydrogen, oxygen and carbon atoms in a molecule and determine the number of... (View More)
Audience: Middle school
Materials Cost: Free
The Volume of Spheres and Cylinders
This is a set of three, one-page problems about calculating the volume of objects. Learners may calculate the volume of an asteroid, Vesta, or the stacking of satellites inside an atlas V rocket nose cone. Options are presented so that students may... (View More)
Audience: Middle school
Materials Cost: Free
Multi-Step Equations
This is a set of three, one-page problems about mass and power of spacecraft. Learners will use multi-step equations to solve several diverse problems. Options are presented so that students may learn about different types of power systems to... (View More)
Audience: Middle school
Materials Cost: Free
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# C/C++ Archive
## Mathematical Morphology – Opening and Closing, Top Hat, Black Hat, Gradient in OpenCV
We discussed dilation and erosion in previous article where we discussed that one expands the object pixels and the other shrinks them. What if we combine these operations? Fine these interesting details written below. Opening An erosion operation followed by dilation operation with the same structuring element is known as opening operation. An opening removes
## Motion Detection using Frame Differencing and Mean Background Model
Instead of using static background model to find difference in current frame, I am using frame difference technique where current frame is compared or subtracted from previous frame. The difference depends upon speed of moving objects. Making a fair analysis is very difficult. Find the code bellow to find difference in frames. VideoCapture capture("ewap/seq_hotel.avi"); Mat
## Opencv 3 installation, Difference between Frames and Averaging Frames
Opencv 3 installation on windows 10 and configure with visual studio 2012 Step1:Download windows version of opencv from www.opencv.org Step2: Extract into a folder Step3: Configure opencv with Visual studio Set envirenment Vairable variable name: OPENCV3_DIR value: D:\\development\opencv\build\x86\vc11 Set Path: %OPENCV_DIR%\bin Configure visual studio for opencv project C/C++ additional include
## Convert Hex to Dec and Dec to Hex
Convert from Decimal to Hexadecimal WORD DecToHex(WORD wDec) { return (wDec / 1000) * 4096 + ((wDec % 1000) / 100) * 256 + ((wDec % 100) / 10) * 16 + (wDec % 10); } Convert from Hexadecimal to Decimal WORD HexToDec(WORD wHex) { return (wHex / 4096) * 1000 + ((wHex % 4096)
## Reading data from multiple COM serial ports part II
Remain part of reading data from multiple serial ports is as follows. Handler.h file #include "SerialCom.h" class Handler { private: CSerialCom port; BYTE data; DWORD BaudRate; BYTE byteSize; DWORD fparity; BYTE parity; BYTE stopBit; public: Handler(void); void setPortDetail(DWORD BaudRate,BYTE byteSize,DWORD fparity,BYTE parity, BYTE stopBit); void COMListen(); void listenCOM(); ~Handler(void); }; Handler.cpp class #include "Handler.h" #include
## Reading data from multiple COM serial ports part I
I downloaded CSerialCom.cpp class from following link – A Simple Class for Implementing Serial Communication in Win-9X/2000 and modified it as follow to handle multiple communication ports. Here is SerialCom.h header file. #include atlstr.h class CSerialCom { // Construction public: CSerialCom(); // Attributes public: // Operations public: // Implementation public: void ClosePort(); BOOL ReadByte(BYTE &resp);
## Configure MinGW, Eclipse IDE for C/C++ Developers and OpenCV 2.4.10
I spent few days to find suitable tutorial with complete information, how to configure OpenCV 2.4.10 with eclipse. New versions do not give pre-compiled binaries whereas OpenCV has support for JAVA and other languages. But it was developed in C++ and there are many ready made codes available in C/C++. Basically research community is still
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# CRLB for $\theta$ using random vectors $(x,y)$
Let $(x_1,y_1)(x_2,y_2)... (x_n,y_n)$ be independent pairs such that $y_i=\theta x_i+\epsilon_i$ where $x_i$ and $\epsilon_i$ are iid Normal (0,1) for $i=1,2..n$
It was previously computed that
$f(x,y)=\frac{1}{2\pi}e^{\frac{-1}{2}x^2-\frac12(y-\theta x)^2 }$
And that the MLE for $\theta$ is $\theta^*=\frac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx_i^2}$ which is unbiased for $\theta$
I need to know if this estimator achieves the CRLB.
Here is what I have so far:
For any estimator W(M) of $\theta$ using sample $M_i's$ of$M$ which are iid $i=1..n$, the CRLB is given by
$\frac{[\frac{d}{d\theta}E(W(M))]^2}{nE_\theta[(\frac{d}{d\theta}lnf(M|\theta)]^2}$
So the numerator is 1 since $E(W(M))=\theta$.
For the denominator, I worked it out as $n(\theta^2(2-\theta^2))$.
Is this correct? It seems iffy coz it can have negative values.
Another problem is in computing the variance of the estimator.
$Var(\theta^*)=Var(\frac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx_i^2}$). How do I go about this?
The expression of your MLE isn't quite correct. The log-likelihood takes the form,
$$\log(L(\textbf{x},\textbf{y};\theta) = constant - \frac{1}{2}\sum\limits_{i=1}^n(y_i - \theta x_i)^2$$
Taking the derivative and setting equal to zero gives,
$$\frac{d}{d\theta}\log(L(\textbf{x},\textbf{y};\theta) = \sum\limits_{i=1}^n(y_i - \theta x_i)x_i = 0 \Longrightarrow \hat{\theta} = \frac{\sum\limits_{i=1}^nx_iy_i}{\sum\limits_{i=1}^nx_i^2}$$
You can verify this gives a maximum using the second derivative.
• You're right. I made a mistake evaluating the derivative. I will fix it now. Jul 7, 2017 at 4:06
Ok. Here's my solution in determining if the estimator reaches the CRLB.
$Var(\theta^*)=Var(\frac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx_i^2})=Var(\frac{\sum_{i=1}^nx_i(\theta x_i+\epsilon_i)}{\sum_{i=1}^nx_i^2})$
$=Var(\frac{\sum_{i=1}^n(\theta x_i^2+x_i\epsilon_i)}{\sum_{i=1}^nx_i^2})$
$=Var(\frac{\theta\sum_{i=1}^n x_i^2}{\sum_{i=1}^nx_i^2}+\frac{\sum_{i=1}^nx_i\epsilon_i}{\sum_{i=1}^nx_i^2})$
$=Var(\theta+\frac{\sum_{i=1}^nx_i\epsilon_i}{\sum_{i=1}^nx_i^2})=Var(\frac{\sum_{i=1}^nx_i\epsilon_i}{\sum_{i=1}^nx_i^2})$
At this point, I can argue that since $x_i's$ and $\epsilon_i's$ are iid $\sim N(0,1)$ then $Var(\theta^*)$ is independent of $\theta$ which means that it cannot attain the CRLB which is dependent on $\theta$. That is, there will always be a theta that can make the CRLB smaller than $Var(\theta^*)$.
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# Analyzing the NYC Subway Dataset
## Questions
### Overview
This project consists of two parts. In Part 1 of the project, you should have completed the questions in Problem Sets 2, 3, and 4 in the Introduction to Data Science course. This document addresses part 2 of the project. Please use this document as a template and answer the following questions to explain your reasoning and conclusion behind your work in the problem sets. You will attach a document with your answers to these questions as part of your final project submission.
### Section 0. References
Please include a list of references you have used for this project. Please be specific - for example, instead of including a general website such as stackoverflow.com, try to include a specific topic from Stackoverflow that you have found useful.
``````
In [2]:
import pandas as pd
import pandasql as pdsql
import datetime as dt
import numpy as np
import scipy as sc
import scipy.stats
import statsmodels.api as sm
from sklearn.linear_model import SGDRegressor
from ggplot import *
%matplotlib inline
``````
### Section 1. Statistical Test
#### 1.1 Which statistical test did you use to analyze the NYC subway data? Did you use a one-tail or a two-tail P value? What is the null hypothesis? What is your p-critical value?
``````
In [ ]:
``````
### Section 2. Linear Regression
#### 2.1 What approach did you use to compute the coefficients theta and produce prediction for ENTRIESn_hourly in your regression model:
• OLS using Statsmodels or Scikit Learn
• Gradient descent using Scikit Learn
• Or something different?
#### 2.3 Why did you select these features in your model? We are looking for specific reasons that lead you to believe that the selected features will contribute to the predictive power of your model.
Your reasons might be based on intuition. For example, response for fog might be: “I decided to use fog because I thought that when it is very foggy outside people might decide to use the subway more often.” Your reasons might also be based on data exploration and experimentation, for example: “I used feature X because as soon as I included it in my model, it drastically improved my R2 value.”
### Section 3. Visualization
Please include two visualizations that show the relationships between two or more variables in the NYC subway data. Remember to add appropriate titles and axes labels to your plots. Also, please add a short description below each figure commenting on the key insights depicted in the figure.
#### 3.1 One visualization should contain two histograms: one of ENTRIESn_hourly for rainy days and one of ENTRIESn_hourly for non-rainy days.
You can combine the two histograms in a single plot or you can use two separate plots. If you decide to use to two separate plots for the two histograms, please ensure that the x-axis limits for both of the plots are identical. It is much easier to compare the two in that case. For the histograms, you should have intervals representing the volume of ridership (value of ENTRIESn_hourly) on the x-axis and the frequency of occurrence on the y-axis. For example, each interval (along the x-axis), the height of the bar for this interval will represent the number of records (rows in our data) that have ENTRIESn_hourly that falls in this interval. Remember to increase the number of bins in the histogram (by having larger number of bars). The default bin width is not sufficient to capture the variability in the two samples.
#### 3.2 One visualization can be more freeform. You should feel free to implement something that we discussed in class (e.g., scatter plots, line plots) or attempt to implement something more advanced if you'd like. Some suggestions are:
Ridership by time-of-day Ridership by day-of-week
### Section 5. Reflection
#### 5.1 Please discuss potential shortcomings of the methods of your analysis, including:
• Dataset,
• Analysis, such as the linear regression model or statistical test
## Data Analysis and Source Code
### A.1 Import NYC Subway Data
``````
In [3]:
weather_data["hour"] = weather_data["hour"].astype('category')
weather_data["rain"] = (weather_data["rain"]+1).astype('category')
weather_data["fog"] = (weather_data["fog"]+1).astype('category')
``````
``````
Out[3]:
UNIT
DATEn
TIMEn
ENTRIESn
EXITSn
ENTRIESn_hourly
EXITSn_hourly
datetime
hour
day_week
...
pressurei
rain
tempi
wspdi
meanprecipi
meanpressurei
meantempi
meanwspdi
weather_lat
weather_lon
0
R003
05-01-11
00:00:00
4388333
2911002
0
0
2011-05-01 00:00:00
0
6
...
30.22
1
55.9
3.5
0
30.258
55.98
7.86
40.700348
-73.887177
1
R003
05-01-11
04:00:00
4388333
2911002
0
0
2011-05-01 04:00:00
4
6
...
30.25
1
52.0
3.5
0
30.258
55.98
7.86
40.700348
-73.887177
2
R003
05-01-11
12:00:00
4388333
2911002
0
0
2011-05-01 12:00:00
12
6
...
30.28
1
62.1
6.9
0
30.258
55.98
7.86
40.700348
-73.887177
3 rows × 27 columns
``````
### A.2 Exploratory Data Analysis for Features
#### a) Number of Riders over the Day Dependent on Rain
``````
In [24]:
p = ggplot(aes(x = 'rain', y='ENTRIESn_hourly', color = "meantempi"), data=weather_data)
p + geom_point(position = "jitter", alpha = 0.7) + scale_y_continuous(limits = [0,45000]) + \
facet_wrap('hour') + ggtitle("Number of Riders over the Day Dependent on Rain") + theme_bw()
``````
``````
Out[24]:
<ggplot: (43556500)>
``````
#### b) Number of Riders over the Day Dependent on Weekday
``````
In [23]:
p = ggplot(aes(x = 'rain', y='ENTRIESn_hourly', color = "meantempi"), data=weather_data)
p + geom_point(position = "jitter", alpha = 0.7) + scale_y_continuous(limits = [0,45000]) + theme_bw() + \
facet_wrap('day_week', nrow = 4) + ggtitle("Number of Riders over the Week Dependent on Rain")
``````
``````
Out[23]:
<ggplot: (35601289)>
``````
#### c) Number of Riders over the Day Dependent on Fog
``````
In [22]:
p = ggplot(aes(x = 'fog', y='ENTRIESn_hourly', color = "meantempi"), data=weather_data)
p + geom_point(position = "jitter", alpha = 0.7) + scale_y_continuous(limits = [0,45000]) + \
facet_wrap('hour') + ggtitle("Number of Riders over the Day Dependent on Fog") + theme_bw()
``````
``````
Out[22]:
<ggplot: (38316972)>
``````
### A.3 Test of Normal Distribution
``````
In [26]:
p = ggplot(aes(x = 'ENTRIESn_hourly', color = 'rain'), data=weather_data)
p + geom_density(size = 3, alpha = 0.25) + theme_bw() + \
scale_x_continuous(limits = [-1000,5000]) + ggtitle("Number of Riders Dependent on Rain")
``````
``````
Out[26]:
<ggplot: (35249850)>
``````
Based on the plot, the sample of entries does not seem normally distributed. Hence, the Mann-Whitney-Wilcoxon RankSum test (no assumptions about any underlying distributions) is conducted to test if the two samples of the number of entries in the NYC subway on rainy and non rainy days come from the same population:
H0: The distribution of number of entries on rainy days \$F_{rain}(x)\$ is identical with the distribution on non rainy days \$F_{no-rain}(x-a)\$, hence a = 0
H1: The distributions are not the same, a \$\neq\$ 0
``````
In [52]:
no_rain = weather_data["ENTRIESn_hourly"][weather_data["rain"]==1].dropna()
with_rain = weather_data["ENTRIESn_hourly"][weather_data["rain"]==2].dropna()
without_rain_mean = np.mean(no_rain)
with_rain_mean = np.mean(with_rain)
print without_rain_mean
print with_rain_mean
U, p = sc.stats.mannwhitneyu(no_rain, with_rain)
z, pval = sc.stats.ranksums(no_rain, with_rain)
print U, p
print z, pval
``````
``````
0 0
1 0
2 0
3 0
4 0
Name: ENTRIESn_hourly, dtype: float64
16 83
17 24
18 532
19 454
20 247
Name: ENTRIESn_hourly, dtype: float64
1845.53943866
2028.19603547
153635120.5 nan
-4.54541569645 5.48269387142e-06
``````
### A.4 Linear Regression
``````
In [ ]:
``````
``````
In [ ]:
``````
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https://www.codeproject.com/Questions/1107803/How-do-I-sort-a-vector-of-struced-data-with-sortor
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15,919,434 members
1.00/5 (1 vote)
See more:
Hi,
unfortunately im sitting at this problem for 4hours now, and it drives me crazy:
i got an 2 vectors as a members of my class.
the firstvector contains structs < with cstring Name, enum Kind and float transformation-factor>
and the vector with the sorting informations is filled by Enums for "Name"(0), "Kind"(1) and "transformation-factor"(2).
The Position of the Enum in the Vector tells us the Sort Order.
My Goal is to display the first vector data in a data grid, sorted by the users order.
Now i have problems to write a working functor, which shall tell the std::sort how to sort.
What I have tried:
The sorting functions should get the parameters and should stop the sortion as soon theres is a first matching pair, what is NOT l. == r.,
right now i not even can get the struct for the functor down in the .header,
also i have problems with the call of the functions in the .cpp, cause every step depends on the sorting-vector.
Who can help me and give me an example?
Posted
Updated 21-Jun-16 4:37am
[no name] 20-Jun-16 11:30am
How is about to let your original data like it is and introduce a third vector managing "only" the sort indices?
Sergey Alexandrovich Kryukov 20-Jun-16 11:43am
Not clear. What you want to achieve cannot be called "sorting", it it should stop before the end of sorting. No matter what it is, you should strictly formulate it. "l. == r." is not clear; you never introduced l or r, and this is not C++ code.
Probably you need to provide the code where you define vectors and the element types and clearly formulate the problem.
What have you tried so far?
—SA
Patrice T 20-Jun-16 12:21pm
Show a sample of your data hexibiting the situation.
## Solution 1
To achieve your own sort order you must overwrite the sorting algorithm, by overwriting the comparison operator "<" which is repeatly called in a sort call.
For a deeper inside read and understand the great article STL Sort Comparison Function.
Tip: make a small sample list and debug it to understand what is happening.
Dennis1990 5-Aug-16 8:49am
thank you KarstenK! your great article made it in my Literature List of my Thesis.
helped me a lot :)
## Solution 2
i used the STL Sort Comparision Funciton and could use the
Member < Operator
like it was described in the Article
next time ill use a functor :)
v2
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, 08.11.2019 13:31 nanda22
# Is the diameter of a circle with center p. the coordinates of the endpoints of the diameter are (1, 3) and (6, 6). what are the coordinates of point p? a(2,6) b(2.2, 3) c(2.5, 3.5) d(4.5, 5) e(3.4, 4.5)
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# Structure properties of Q-anti fuzzy left H-ideals in a hemi rings.
Introduction
Ideals of hemi rings play a central role in the structure theory and are very useful for many purposes. However, they do not in general coincide with the usual ring ideals. Many results in rings apparently have no analogues in hemi rings using only ideals. Henriksen defined in [4] a more restricted class of ideals in semi rings, which is called the class of k-ideals, with the property that if the semi ring R is a ring then a complex in R is a k-ideal if and only if it is a ring ideal. Another more restricted, but very important class of ideals, called h-ideals, has been given and investigated by Izuka [5] and La Torre [7] other important results connected with fuzzy ideals in hemi rings were obtained in [6].The concept of Q-fuzzy subgroups can be obtained in [11] [12] [13] [14]. In this paper, we introduced the notion of Q-anti fuzzy left h-ideals in terms of hemi rings and investigate their properties.
Preliminaries
In this section, we review some elementary aspects that are necessary for this paper.
Definition 2.1: An algebra (R, +, *) is said to be a semi ring if it satisfies the following conditions
(R, +) is a semi group
(R, *) is a semi group
a.(b + c) = a.b + a.c and (b + c).a = b.a + c.a [for all] a, b, c[member of] R.
Definition 2.2: A semi ring (R, +, *) is called a hemi ring
[H.sub.1]: + is commutative and
[H.sub.2]: there exists an element 0[right arrow]R such that 0 is the identity of (R, +) and the zero element of (R, *) i.e, 0.a = a.0 = 0 [for all] a[member of]R. A subset I of a semi ring R is called a left ideal of R if I is closed under addition and RI [subset or equal to] I. A left ideal of R is called a left K-ideal of R if y, z [member of] I and x[member of] R, x + y = z implies x[member of] I.
A left h ideal of a hemi ring R is defined to be a left ideal A of R such that (x + a + z = b + z [right arrow] x[member of] A, [for all](x, z[member of]R), ([for all]a, b[member of] A)).
Right h-ideals are defined similarity.
Definition 2.3: A mapping f: [R.sub.1] [right arrow] [R.sub.2] is said to be hemi ring homomorphism of [R.sub.1] is to [R.sub.2] if f(x + y) = f(x) + f(y) and f(xy) = f(x).f(y) for all x, y [member of] R.
Definition 2.4: A mapping [m,u]: X [right arrow] [0,1] where X is an arbitrary non-empty set is called a fuzzy set is X. For any fuzzy set [mu] is X and any X [member of] [0,1] we defined the set L([mu]: [alpha]) = {x[member of] X/[mu](x) [subset or equal to] [alpha]} which is called lower level cut of [mu].
Definition 2.5: Let Q and G be a set and a group respectively. A mapping [mu]: G x Q [right arrow] [0,1] is called a Q-fuzzy set.
Definition 2.6: A fuzzy subset is of a semi ring R is said to be Q-fuzzy left h- ideal of R if [mu](x + y, q) [greater than or equal to] min{[mu](x, q), [mu](y, q)} [for all]x, y[member of] R, q[member of]Q [mu](xy, q) [greater than or equal to] [mu](y,q) [for all]x, y[member of] R, q[member of] Q
Note that if [mu] is a Q-fuzzy left h-ideal if a hemi ring R, then [mu](0,q) [greater than or equal to] [mu](x,q) [for all] x[member of] R.
Definition 2.7: A fuzzy subset [mu] of a hemi ring R is said to be an Q-anti fuzzy left h-ideal of R if 1. [mu](x + y, q) [less than or equal to] max{[mu](x, q), [mu](y, q)} [for all]x, y[member of] R, q[member of] Q
2. [mu](xy, q) [less than or equal to] [mu](y, q) [for all] x, y[member of] R, q[member of] Q
3. x + a + z = b + z [right arrow] [mu](x, q) [less than or equal to] max{[mu](a, q), [mu](b, q)}
Example: Let R = {0, 1, 2, 3, 4} be a hemi ring with zero multiplication and addition defined by the following table
``` 0 1 2 3 4
0 0 1 2 3 4
1 1 1 4 4 4
2 2 4 4 4 4
3 3 4 4 4 4
4 4 4 4 4 4
```
We define a fuzzy set [mu]: R [right arrow] [0,1] by letting [mu](0) = [t.sub.1] and [mu](x) = [t.sub.2] [for all]x [not equal to] 0, [t.sub.1] [subset or equal to] [t.sub.2]. By routine computations, we can also easily check that [mu] is an anti fuzzy left h-ideal of hemi ring R.
Properties of Q-anti fuzzy left h-ideals
Proposition 3.1 let 'R' be a hemi ring and '[mu]' be a Q-fuzzy set in R. Then [mu] is an Q-anti fuzzy left h-ideal in R if and only if [[mu].sup.c] is a Q-fuzzy left h- ideal in R.
Proof: Let [mu] be an Q-anti fuzzy left h-ideal in R. For x, y [member of] R, We have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Let x, z, a, b [member of] R be such that x + a + z = b + z
Then
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Hence [[mu].sup.c] is a Q-fuzzy left h-ideal of R.
Conversly, [[mu].sup.c] is a Q-fuzzy left h-ideal of R.
For x, y [member of] R, q[member of] Q, we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Let x, z, a, b [member of] R be such that x + a + z = b + z,
Then [mu](x, q) = 1 - [[mu].sup.c](x, q)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Hence [mu] is a Q-anti fuzzy left h-ideal of R.
Proposition 3.2 let '[mu]' be Q-anti fuzzy left h-ideal in a hemi ring R such that L([mu]: [alpha]) is a left h-ideal of R for each [alpha][member of] [I.sub.m]([mu]), [alpha][member of] [0,1].Then [mu] is an Q-anti fuzzy left h-ideal in R.
Proof: Let x, y [member of] R, q [member of] Q be such that [mu](x, q) = [[alpha].sub.1], [mu](y, q) = [[alpha].sub.2] then x + y [member of] L([mu]: [alpha]) without loss of generality, we may assume that [[alpha].sub.1] > [[alpha].sub.2]. It follows that L([mu]; [[alpha].sub.2]) [subset or equal to] L([mu]; [[alpha].sub.1]) so that x [member of] L([mu]; [[alpha].sub.1]) and y [member of] L([mu]; [[alpha].sub.2]). Since L([mu]; [[alpha].sub.1]) is a left is a left h-ideal of R. We have x + y [member of] L([mu]; [[alpha].sub.1]). Thus
[mu](x + y, q) [less than or equal to] [[alpha].sub.1] = max {[mu](x, q), [mu](y, q)} [mu](xy, q) [less than or equal to] [[alpha].sub.1] = [mu](y, q)
Let x, z, a, b [member of] R be such that x + a + z = b + z, then
[mu](x, q) [less than or equal to] [[alpha].sub.1] = max {[mu](a, q), [mu](b, q)}.
This shows that [mu] is an Q-anti fuzzy left h-ideal in R.
Corollorry 3.3: Let [mu] be Q-anti fuzzy left h-ideal in R then [mu] is an Q- anti fuzzy left h-ideal in R if and only if L([mu]; [alpha]) is a left h-ideal in R for every [alpha][member of] [[mu],(0,q),1] with [alpha][member of] [0,1].
Proposition 3.4: let '[mu]' be Q-anti fuzzy set in a hemi ring R then two lower level subsets L([mu]; [t.sub.1]) and L([mu]; [t.sub.2]), ([t.sub.1] < [t.sub.2]) are equal iff there is no x[member of] R such that [t.sub.1] [less than or equal to] [mu](x, q) [less than or equal to] [t.sub.2].
Proof: From definition of L([mu]; [alpha]) it follows that L([mu]; t) = [[mu].sup.-1]([[mu](0,q);t]) for t[member of] [0,1]. Let [t.sub.1], [t.sub.2][member of] [0,1] be such that [t.sub.1] < [t.sub.2] then L([mu]; [t.sub.1]) = L([mu]; [t.sub.2])
[??] [[mu].sup.-1]([[mu](0,q); [t.sub.1]]) = [[mu].sup.-1]([[mu](0,q); [t.sub.2]])
[??] [[mu].sup.-1]([t.sub.1], [t.sub.2]) = [phi]
[??] There is no x [member of] R such that [t.sub.1] [less than or equal to] [mu](x, q) [less than or equal to] [t.sub.2].
This completes the proof.
Definition 3.5: A left h-ideal A of hemi ring R is said to be characteristic iff f(A) = A, for all f [member of] Aut (R), Where Aut (R) is the set of all automorphisms of R. Q-anti fuzzy left h-ideal [mu] of hemi ring R is said to be Q-anti fuzzy characteristic if [[mu].sup.f](x, q) = [mu](x, q) for all and f [member of] Aut (R).
Lemma3.6: Let [mu] be an Q-anti fuzzy left h-ideal of a hemi ring R and let x [member of] R then [mu](x, q) = s iff x [member of] L([mu]; s) and x [not member of] L([mu]; t) for all s > t.
Proof: Straight forward.
Proposition 3.7: let '[mu]' be an Q-anti fuzzy left h-ideal of a hemi ring R then each level left h-ideal of [mu] is characteristic iff [mu] is an Q-anti fuzzy Characteristic.
Proof: Suppose that [mu] is an Q-anti fuzzy Characteristic and let S [member of] [I.sub.m]([mu]), f [member of] Aut (R) and x [member of] L([mu]; s) then
[[mu].sup.f] (x, q) = [mu](x, q) [less than or equal to] s [??] [mu](f(x, q)) [greater than or equal to] s [??] f(x, q) [member of] L([mu]; s)
Thus
f(L([mu]; s)) and y [member of] R such that f(y, q) = (x, q)
then
[mu](y, q) = [[mu].sup.f](y, q) = [mu](f(y, q)) = (x, q) [less than or equal to] s [??] y [member of] L([mu]: s) so that (x, q) = f(y, q) [member of] L([mu]; s).
Consequently, L([mu]; s) [subset or equal to] f (L([mu]; s)).
Hence f (L([mu]; s)) = L([mu]; s)
i.e. L([mu]; s) is characteristic. Conversely, suppose that each level h-ideal of [mu] is characteristic and let x [member of] R, f [member of] Aut (R) and [mu](x, q) = s. then by virtue Lemma, x[member of] L([mu]; s) and x [not member of] L([mu]; t) for all s > t. It follows from the assumption that
f(x, q) [member of] f (L([mu]; s))= L([mu]) so that [[mu].sup.f](x, q) = [mu](f(x, q)) [less than or equal to] s.
Let t = [[mu].sup.f](x, q) and assume that s > t. then f(x, q) [member of] L([mu]; t)= f(L([mu]; t)) which implies from the injectivity of f that x[member of] L([mu]: t), a contradiction.
Hence [[mu].sup.f](x, q) = [mu](f(x, q)) = s = [mu](x, q) showing that [mu] is an Q-anti fuzzy characteristic.
Proposition 3.8: Let f: [R.sub.1] [right arrow] [R.sub.2] be an epimorphism of hemi rings. If V is an Q-anti fuzzy left h-ideal of [R.sub.2] and [mu] is the pre-image of V under f then [mu] is an Q-anti fuzzy left h-ideal of [R.sub.1].
Proof: For any x, y [member of] [R.sub.1] and q[member of] Q, we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Let x, z, a, b [member of] R be such that x + a + z = b + z, then we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Hence [mu] is an Q-anti fuzzy left h-ideal of [R.sub.1].
Definition 3.9: Let [R.sub.1] and [R.sub.2] be two hemi rings and f be a function of [R.sub.1] into [R.sub.2]. If [mu] is a Q-anti fuzzy in [R.sub.2] then the Pre-image of [mu] under f then [mu] is the Q-anti fuzzy in [R.sub.1] defined by [f.sup.-1]([mu])(x, q) = [mu](f(x, q)) [for all] x[member of] [R.sub.1], q[member of] Q.
Proposition 3.10: Let f: [R.sub.1] [right arrow] [R.sub.2] be an onto homomorphism of hemi rings. If [mu] is an Q-antifuzzy left h-ideal of [R.sub.2] then [f.sup.-1] ([mu]) is an Q-anti fuzzy left h-ideal of [R.sub.1].
Proof: Let [x.sub.1], [x.sub.2] [member of] [R.sub.1], then we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Let x, z, a, b [member of] [R.sub.1] be such that x + a + z = b + z, then we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Hence [f.sup.-1]([mu]) is an Q-anti fuzzy left h-ideal of [R.sub.1].
Definition 3.11: Let [R.sub.1] and [R.sub.2] be any sets and let f: [R.sub.1] [right arrow] [R.sub.2] be any function. A Q-fuzzy subset [mu] of [R.sub.1] is called f-invariant if f(x) = f(y) implies [mu](x, q) = [mu](y, q)) [for all] x, y[member of] R, q[member of] Q.
Proposition 3.12: Let f: [R.sub.1] [right arrow] [R.sub.2] be an epimorphism of hemi rings. Let [mu] be an f-invariant Q-anti fuzzy left h-ideal of [R.sub.1], then f([mu]) is an Q-anti fuzzy left h- ideal of [R.sub.2].
Proof: Let x, y[member of] [R.sub.2] then there exists a, b [member of] [R.sub.1], such that f(a) = x and f(b) = y then x + y = f (a + b) and xy = f(ab). Since [mu] is invariant,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Let x, z, a, b [member of] [R.sub.2] be such that x + a + z = b + z, then there exist [bar.x], [bar.z], [bar.a], [bar.b] such that f([bar.x]) = x, f([bar.y]) = y, f([bar.a]) = a and f([bar.b] = b. Since [mu] is f- invariant,
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Hence f([mu]) is an Q-anti fuzzy left h-ideal of [R.sub.2].
Definition 3.13: A Q-fuzzy left h-ideal [mu] of a hemi ring R is said to be normal if there exist x [member of] R such that [mu](x, q) = 1. Note that if [mu] is a normal Q-anti fuzzy left h-ideal of [R.sub.1] then [mu](0, q) = 1 and hence [mu] is normal if and only if [mu](0, q) = 1.
Proposition 3.14: Let [mu] be an Q-anti fuzzy left h-ideal of a hemi ring R. Let [[mu].sup.+] be a Q-fuzzy set in R defined by [[mu].sup.+] (x, q) = [mu](x, q) = 1 - [mu](0, q) for all x[member of] R then [[mu].sup.+] is a normal Q-anti fuzzy left h-ideal of R which contains [mu].
Proof: For any x, y[member of] R, we have [[mu].sup.+](x, q) = [mu](0, q) + 1 - [mu](0,q) = 1
And [[mu].sup.+] (x + y, q) = [mu](x + y, q) + 1 - [mu](0,q)
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
This Shows that [[mu].sup.+] is an q-anti fuzzy left H-ideal of R. Let a, b, x, z, [member of] R be such that x + a + z = b + z, then
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Hence [[mu].sup.+] is a normal Q-anti fuzzy left h-ideal of hemi ring of R. Clearly [mu] [less than or equal to] [[mu].sup.+].
Definition 3.15: Let N(R) denote the set of all normal Q-anti fuzzy left h- ideals of R. Note that N(R) is a Poset under the set inclusion. A Q-fuzzy set [mu] in a hemi ring R is called a Maximal Q-anti fuzzy left h-ideal of R if it is non-constant and [[mu].sup.+] is a maximal element of (N(R), [subset or equal to]).
Proposition 3.16: Let [mu] [member of]N(R) be non-constant such that it is a maximal element of (N(R), [subset or equal to]) then it takes only two values {0,1}.
Proof: Since [mu] is normal, [mu](0,q) = 1. We claim that [mu](x, q) = 0. If not, then there exists [x.sub.0] [member of] R such that 0 [less than or equal to] [mu]([x.sub.0], q) < 1. Define on R a Q-fuzzy set V by putting V(x, q) = 1/2 {[mu](x, q) + [x.sub.0], q)} for each x [member of]R then clearly V is well defined and for all x, y [member of] R. we have
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Thus V is an Q-anti fuzzy left h-ideal of R.
Let a, b, x, z, [member of] R be such that x + a + z = b + z, then
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
Hence V is an Q-anti fuzzy left h-ideal of R. By theorem 3.16, [V.sup.+] is a maximal Q-anti fuzzy left h-ideal of R. Note that
[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]
and [V.sup.+]([x.sub.0], q) [less than or equal to] 1 = [V.sup.+](0,q).
Hence [V.sup.+] is a non-constant and [mu] is not a maximal non-constant and [mu] is not a maximal element of N(R). This is a contradiction.
Conclusion
Y.B.Jun[6] introduced the concept on fuzzy h-ideals in hemi rings and [14] investigated the idea of anti fuzzy left h-ideals in hemi rings. In this paper, we established the some structure properties of Q-anti fuzzy left h-ideals in hemi rings. One can obtain similar results, by using the intuitionistic anti fuzzy ideals in a hemi rings.
Acknowledgements
The authors are highly greatful to the referees for their valuable comments and suggestions for improving the paper. This research work was supported by J J Educational & Charitable Trust, Tirchirappalli.
References
[1] R.Biswas, Fuzzy subgroups and anti fuzzy subgroups, Fuzzy sets and systems, 44 (1990), 121-124.
[2] S.Ghose, Fuzzy k-ideals of semi rings, Fuzzy sets and systems, 95(1998), 103-108.
[3] T.K.Dutta and B.K.Biswas, Fuzzy k-ideals of semi rings, Bull.Calcutta Math.Soc.87(1995), 91-96.
[4] M.Henriksen, Ideals in Semi rings with commutative addition, Am.Math.Soc Notices, 6(1958) 3-21
[5] K.Izuka, On the Jacobson radical of a semi ring, Tohoku, Math. J.,11(2)(1959), 409-421.
[6] Y.B.Jun, M.A.Ozturk and S.Z.Song, On fuzzy h-ideals in hemi rings, info. Scien.162(2004), 211-226.
[7] D.R.LaTorre, On h-ideals and k-ideals in hemi rings, Publ.Math.Debrecen, 12(1965), 219-226.
[8] D.M.Olson, A note on the homomorphism theorem for hemi rings, IJMMS, 1(1978), 439-439.
[9] A.Rosenfeld, Fuzzy groups, J.Math.Anal.Appl.35(1971), 512-517.
[10] A.Solairaju and R.Nagarajan, Q-fuzzy left R-subgroups of near rings with respect to T-norms, Antarctica Journal of mathematics, 5(2)(2008), 59-63.
[11] A.Solairaju and R.Nagarajan, A New Structure and Constructions of Q-Fuzzy groups, Advances in Fuzzy mathematics, 4(1)(2009), 23-29.
[12] A.Solairaju and R.Nagarajan, Lattice valued Q-fuzzy sub modules of near rings with respect to To norms, Advances in Fuzzy mathematics, 4(2)(2009), 137-145.
[13] A.Solairaju and R.Nagarajan, Q-fuzzy subgroups of Beta fuzzy congruence relations on a group, Accepted for publications in International Journal of Computer Science, Network and Security(IJCSNS), 2010.
[14] A.Solairaju and R.Nagarajan, characterization of interval valued anti fuzzy left h-ideals over hemi rings, Advances in Fuzzy Mathematics, 4(2)(2009), 129-136.
[15] L.A.Zadeh, Fuzzy sets, Information control, 8(1965), 338-353.
R. Nagarajan (1) and M. Muruga Ganesan (2)
(1) Assistant Professor, (2) Senior Lecturer
(1,2) Dept. of Mathematics, JJ College of Engineering & Technology Trichirappalli--620 009, India
E-mail: (1) nagalogesh@yahoo.co.in, (2) murugaganesan.m@gmail.com
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What Discount Rate To Use
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Discount Rate Definition - Investopedia
(12 days ago) Discount Rate: The discount rate is the interest rate charged to commercial banks and other depository institutions for loans received from the Federal Reserve's discount window.
https://www.investopedia.com/terms/d/discountrate.asp
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Discount Rate - Definition, Types and Examples, Issues
(11 days ago) When using the WACC as a discount rate, the calculation centers around the use of a company’s beta Beta The beta (β) of an investment security (i.e. a stock) is a measurement of its volatility of returns relative to the entire market. It is used as a measure of risk and is an integral part of the Capital Asset Pricing Model (CAPM).
https://corporatefinanceinstitute.com/resources/knowledge/finance/discount-rate/
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What Is the Discount Rate? | The Motley Fool
(10 days ago) A higher discount rate implies greater uncertainty, the lower the present value of our future cash flow. Calculating what discount rate to use in your discounted cash flow calculation is no easy ...
https://www.fool.com/knowledge-center/discount-rate.aspx
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Discount Rate Formula: Calculating Discount Rate [WACC/APV]
(8 days ago) The definition of a discount rate depends the context, it's either defined as the interest rate used to calculate net present value or the interest rate charged by the Federal Reserve Bank. There are two discount rate formulas you can use to calculate discount rate, WACC (weighted average cost of capital) and APV (adjusted present value).
https://www.profitwell.com/blog/discount-rate-formula
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Discount Rate Formula | How to calculate Discount Rate ...
(10 days ago) Discount Rate = 2 * [(\$10,000 / \$7,600) 1/2*4 – 1] Discount Rate = 6.98%; Therefore, the effective discount rate for David in this case is 6.98%. Discount Rate Formula – Example #3. Let us now take an example with multiple future cash flow to illustrate the concept of a discount rate.
https://www.educba.com/discount-rate-formula/
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How to Select the Appropriate Discount Rate | CREentrepreneur
(9 days ago) = Discount Rate. A Few Examples . . . If you are acquiring an existing stabilized asset with credit tenants then you could use a discount rate of around 7%. If you are analyzing a speculative development, you discount rate should be in the high teens. In general, discount rates in real estate fall between 6-12%.
http://www.creentrepreneur.com/how-to-select-the-appropriate-discount-rate/
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What is a discount rate in cost benefit analysis ...
(10 days ago) Discount rates can vary from 0 to infinity. A discount rate of 0% means that someone is indifferent between having a benefit or cost now vs. any time in the future. A discount rate of 0% implies that future generations are treated exactly the same as current generations.
https://www.freeeconhelp.com/2018/06/what-is-discount-rate-in-cost-benefit.html
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Discounted Cash Flow (DCF) Definition
(9 days ago) You are looking to invest in a project and your company's WACC is 5%, so you will use 5% as your discount rate. The initial investment is \$11 million and the project will last for five years, with ...
https://www.investopedia.com/terms/d/dcf.asp
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Discounted Cash Flow: What Discount Rate To Use? | Seeking ...
(10 days ago) The discount rate is by how much you discount a cash flow in the future. For example, the value of \$1000 one year from now discounted at 10% is \$909.09. Discounted at 15% the value is \$869.57.
https://seekingalpha.com/article/462411-discounted-cash-flow-what-discount-rate-to-use
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What Discount Rate To Use To Determine The Present Value ...
(10 days ago) If you want to get, say, a 10% rate of return on your money, then you should use a discount rate of 10% per year when translating future dollars into present dollars.
https://seekingalpha.com/article/3610726-what-discount-rate-to-use-to-determine-present-value-of-stock
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Discount Rate – Meaning, Importance, Uses And More
(12 days ago) Use of discount rate in the financial tools may appear scientific, but it is not. Several assumptions go in for estimating or calculating this rate. Also, a company or analyst generally uses one discount rate for all future cash flows. This is not as practical as risk profiles and interest rates are constantly changing.
https://efinancemanagement.com/investment-decisions/discount-rate
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What Discount Rate to Use to Determine the Present Value ...
(14 days ago) The discount rates I use vary between 9% and 12%. A company with stellar numbers that will stabilize my portfolio such as 3M Co (MMM) will be discounted at 9% while a company evolving in a highly volatile environment such as Helmerich & Payne (a drilling company in the oil industry) will require a higher return on my investment and will be ...
https://www.dividendmonk.com/what-discount-rate-to-use-to-determine-the-present-value-of-a-stock/
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How to Calculate Discount Rate in a DCF Analysis
(10 days ago) How to Discount the Cash Flows and Use the Discount Rate in Real Life. Finally, we can return to the DCF spreadsheet, link in this number, and use it to discount the company’s Unlevered FCFs to their Present Values using this formula: Present Value of Unlevered FCF in Year N = Unlevered FCF in Year N /((1+Discount_Rate)^N)
https://breakingintowallstreet.com/biws/how-to-calculate-discount-rate/
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Cost of Capital vs. Discount Rate: What's the Difference?
(12 days ago) The discount rate is the interest rate used to calculate the present value of future cash flows from a project or investment. Many companies calculate their WACC and use it as their discount rate ...
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G100 Discount Rates
(10 days ago) A G100 discount rate curve will be regularly produced by Milliman and will be available as a transparent central reference point for the industry to use on an ongoing basis with the first curve available in June 2015.
https://www.milliman.com/G100_Discount_Rates/
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What is the Discount Rate and Why Does It Matter? - SmartAsset
(10 days ago) Businesses considering investments will use the cost of borrowing today to figure out the discount rate, For example, \$200 invested against a 15% interest rate will grow to \$230. Working backwards, \$230 of future value discounted by 15% is worth \$200 today.
https://smartasset.com/investing/discount-rate
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Federal Discount Rate: Definition, Impact, How It Works
(10 days ago) The Federal Reserve discount rate is the rate that the U.S. central bank charges member banks to borrow from its discount window to maintain the bank's cash reserve requirements. On March 16, 2020, the Federal Reserve Board of Governors lowered the rate to 0.25% in response to the COVID-19 coronavirus outbreak.
https://www.thebalance.com/federal-reserve-discount-rate-3305922
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What You Should Know About the Discount Rate
(10 days ago) This is the appropriate discount rate to use for this corporate investor. Any investment that the company makes must at least achieve a 6.80% return in order to satisfy debt and equity investors. Any return greater than 6.80% will create additional value for the shareholders.
https://propertymetrics.com/blog/npv-discount-rate/
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How Do I Calculate a Discount Rate Over Time Using Excel?
(10 days ago) The discount rate is the interest rate used when calculating the net present value (NPV) of an investment. NPV is a core component of corporate budgeting and is a comprehensive way to calculate ...
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How to Determine a Discount Rate - Vintage Value Investing
(12 days ago) So What Discount Rate Should We Use Instead of WACC? There is a lot of debate about what discount rate Buffett uses, and the man himself has given conflicting information on the matter over the years. But I think this exchange from the 2003 shareholder meeting sheds the most light on the issue: Buffett: Charlie and I don’t know our cost of ...
https://www.vintagevalueinvesting.com/how-to-determine-a-discount-rate/
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How to calculate the Discount Rate to use in a Discounted ...
(10 days ago) This discounted cash flow (DCF) analysis requires that the reader supply a discount rate. In the blog post, we suggest using discount values of around 10% for public SaaS companies, and around 15-20% for earlier stage startups, leaning towards a higher value, the more risk there is to the startup being able to execute on it’s plan going forward.
https://www.forentrepreneurs.com/discount-rate-for-dcf/
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Why Do Lease Discount Rates Matter? | Deloitte US
(9 days ago) Accurately estimating lease discount rates can have a significant impact on your company’s lease liabilities and right-of-use (ROU) assets. Under the new standard, every lease with a lease term of more than a year must be recorded on the balance sheet as a right-of-use asset and a corresponding lease liability.
https://www2.deloitte.com/us/en/pages/audit/articles/why-lease-discount-rates-matter.html
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Choosing a Discount Rate - Impact DataSource
(10 days ago) Remember, a low discount rate means the organization is very patient. We believe using a discount rate in the 2% to 3% range may distort the city or county’s decision-making process. Recently, we’ve recommended economic development organizations use a discount rate of 4% to 5%.
https://impactdatasource.com/choosing-a-discount-rate/
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What Discount Rate Should Be Used? | Customer Lifetime Value
(11 days ago) A discount rate of 10% is commonly used, as it is generally around the return that firms make on their other investments. In some organizations, it is known as a “hurdle” rate. This is the minimum level of return that a firm is willing to accept for its investment/expansions as this is what it would make if it reinvested in its own business.
https://www.clv-calculator.com/discount-rates/discount-rate-used/
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Understanding the discount rate l Grant Thornton Insights
(10 days ago) Discount rates are also used to determine lease classification for a lessor and to measure a lessor’s net investment in a lease. Download IFRS 16 - Understanding the discount rate [ 78 kb ] Download this article. Download PDF [78 kb] IFRS 16 - Understanding the discount rate [ 78 kb ]
https://www.grantthornton.global/en/insights/ifrs-16/ifrs-16---understanding-the-discount-rate/
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What is a Discount Rate? - Definition | Meaning | Example
(10 days ago) The discount rate is most often used in computing present and future values of annuities. For example, an investor can use this rate to compute what his investment will be worth in the future. If he puts in \$10,000 today, it will be worth about \$26,000 in 10 years with a 10 percent interest rate.
https://www.myaccountingcourse.com/accounting-dictionary/discount-rate
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How To Select A Discount Rate For A Commercial Real Estate ...
(9 days ago) The discount rate is often defined as the opportunity cost of making a particular investment instead of making an alternative investment of an almost identical nature. Opportunity cost is a slippery concept, though, because we need to subjectively conclude what that “alternative investment of an almost identical nature” is.
https://www.getrefm.com/how-to-select-a-discount-rate-for-a-commercial-real-estate-investment/
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Discount Rates for Value Investors | Old School Value
(10 days ago) So What Is A Good Discount Rate? I like to use a post-tax discount rate of 7-12%. Like Buffett, I have a minimum return rate that I want and that happens to be between 7-12% in today’s world of low interest rates and dependent on the type of company. In the example above using SIRI, I used 7% and 9% to show the difference it can make.
https://www.oldschoolvalue.com/investing-strategy/explaining-discount-rates/
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US Discount Rate - YCharts
(1 year ago) US Discount Rate is at 0.25%, compared to 0.25% the previous market day and 2.75% last year. This is lower than the long term average of 2.05%.
https://ycharts.com/indicators/us_discount_rate
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Discount Rate in NPV | WACC vs Risk-Adjusted Rate
(10 days ago) Where β a is the project beta we should use, β e, D/E and t are the equity beta, debt-to-equity ratio, and tax rate of the pure play company.. The project beta should be plugged into the CAPM equation to get the appropriate discount rate. The discount rate determined using this approach will be higher or lower than the weighted average cost of capital.
https://xplaind.com/595242/discount-rate
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The Role of the Discount Rate in CLV | Customer Lifetime Value
(10 days ago) What does a discount rate do? Discount rate converts future cash flows (that is revenue/profits) into today’s money for the firm. For example, if you put \$100 into a bank account today that have 10% interest, then in 12 months’ time you would have \$110 in the bank.
https://www.clv-calculator.com/discount-rates/role-discount-rate-clv/
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On Investors and Discount Rates | FiniCulture
(14 days ago) It is not unreasonable to state that discount rates for valuing startups can be as high as 90% and depend on the maturity of said startup. I have come up with the following graph to illustrate the discount rates one can use based on the stage of a startup. Obviously some will argue for different start and end points discount rates.
https://finiculture.com/on-investors-and-discount-rates/
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Bond Discount
(10 days ago) The bond discount rate is, therefore, \$41.31/\$1,000 = 4.13%. Bonds trade at a discount to par value for a number of reasons. Bonds on the secondary market with fixed coupons will trade at ...
https://www.investopedia.com/terms/b/bond-discount.asp
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The Discounted Cash Flow Approach to Business Valuation
(9 days ago) The use of a terminal growth rate may seem sloppy or conservative, but in valuing a small business with an appropriately high discount rate, the value of cash flows 11-plus years out is going to be worth very little today.
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Discount Rate vs Interest Rate | Top 7 Differences (with ...
(9 days ago) Discount Rate vs Interest Rate Key Differences. The followings are the key differences between Discount Rate vs Interest Rate: The use of discount rate is complex as compared to the interest rate as the discount rate is used in discounted cash flow analysis for calculating the present value of future cash flows over the period of time whereas the interest rate is generally charged by the ...
https://www.wallstreetmojo.com/discount-rate-vs-interest-rate/
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Discount Rate vs Interest Rate | 7 Best Difference (with ...
(10 days ago) Discount Rate : Meaning: An interest rate is an amount charged by a lender to a borrower for the use of assets. Discount Rate is the interest rate that the Federal Reserve Banks charges to the depository institutions and to commercial banks on its overnight loans. Charged on:
https://www.educba.com/discount-rate-vs-interest-rate/
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Discount Rate - ReadyRatios Financial Analysis
(12 days ago) The discount rate, sometimes also referred as the annual effective discount rate, can be defined as the annual interest divided by the capital plus that interest. This rate is lower than rate of interest. Moreover, it corresponds to the use of value after a year in the form of a nominal value less a discount.
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What is Discount Rate? - Definition from Divestopedia
(9 days ago) The discount rate of a business is closely tied to the riskiness of the operation and a company's ability to access capital. Although discount rates for any company can vary significantly, it is important for business owners to understand that, in general, discount rates will fall within the following ranges:
https://www.divestopedia.com/definition/5039/discount-rate
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Discount Rates For Social Security Or Pension Decisions
(10 days ago) For instance, in the example above, Jeremy’s decision to use a 10% discount rate was based on having a (comparable risk) alternative investment that already had an expected return of 10%, which effectively became the “hurdle rate” that the new investment had to clear (and in this case, couldn’t surpass).
https://www.kitces.com/blog/net-present-value-discount-rate-formula-retirement-plan-pension-lump-sum-or-social-security-breakeven/
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Difference Between Cap Rate and Discount Rate
(8 days ago) The discount rate, on the other hand, is the investor’s required rate of return. The discount rate is used to discount future cash flows back to the present to determine value and account’s for all years in the holding period, not just a single year like the cap rate. If a property’s cash flows are expected to increase or decrease over ...
https://propertymetrics.com/blog/difference-between-cap-rate-and-discount-rate/
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Discounted cash flow - Wikipedia
(9 days ago) In finance, discounted cash flow (DCF) analysis is a method of valuing a security, project, company, or asset using the concepts of the time value of money.Discounted cash flow analysis is widely used in investment finance, real estate development, corporate financial management and patent valuation.It was used in industry as early as the 1700s or 1800s, widely discussed in financial economics ...
https://en.wikipedia.org/wiki/Discounted_cash_flow
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Lease Accounting: Simplifying the discount rate debacle
(11 days ago) Discount rates are one of the new data points that need to be captured when implementing the new lease accounting standard.The guidance for non-public companies is more lenient than that for public companies, but it will require significant consideration as the adoption date quickly approaches.
https://www.lbmc.com/blog/lease-accounting-simplifying-the-discount-rate/
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Discount rates vs. capitalization rates - FiskeCo.com ...
(10 days ago) Some clients mistakenly use the terms “discount rate” and “capitalization rate” (cap rate) interchangeably. But they are two different concepts. It’s important to understand how these terms differ to prevent erroneous conclusions and courtroom blunders. Dueling techniques Both rates come into play when an appraiser applies the income approach to valuation. The underlying theory is
https://fiskeco.com/discount-rates-vs-capitalization-rates/
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A Refresher on Net Present Value
(10 days ago) If shareholders expect a 12% return, that is the discount rate the company will use to calculate NPV. If the firm pays 4% interest on its debt, then it may use that figure as the discount rate ...
https://hbr.org/2014/11/a-refresher-on-net-present-value
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3.4 Choosing the Discount Rate - Issues in Capital ...
(12 days ago) With the estimated cost of capital of 6.5%, we can argue that this company should use 6.5%, as the discount rate whenever it analyzes new projects by using NPV or IRR methods. The company would need to make sure that the new project has the expected rate of return of at least 6.5%.
https://www.coursera.org/lecture/capital-budgeting/3-4-choosing-the-discount-rate-zntHE
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How to Calculate the Discount Factor or Discount Rate ...
(9 days ago) Discount rates, also known as discount factors, are a critical component of the time value of money. Investors can use discount rates to translate the value of future investment returns into today's dollars. If your investment provides you dividends or interest proceeds over time, you will need to calculate multiple discount rates.
https://www.sapling.com/6516198/calculate-discount-factor
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https://hoogle.haskell.org/?hoogle=%28a%20-%3E%20b%29%20-%3E%20%5Ba%5D%20-%3E%20%5Bb%5D%20-package%3Abrittany%20-package%3Autility-ht%20-package%3Arelude%20-package%3Aghc-lib-parser%20-package%3Agraphviz
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# :: (a -> b) -> [a] -> [b] -package:brittany -package:utility-ht -package:relude -package:ghc-lib-parser -package:graphviz
map f xs is the list obtained by applying f to each element of xs, i.e.,
```map f [x1, x2, ..., xn] == [f x1, f x2, ..., f xn]
map f [x1, x2, ...] == [f x1, f x2, ...]
```
```>>> map (+1) [1, 2, 3]
[2,3,4]
```
map f xs is the list obtained by applying f to each element of xs, i.e.,
```map f [x1, x2, ..., xn] == [f x1, f x2, ..., f xn]
map f [x1, x2, ...] == [f x1, f x2, ...]
```
```>>> map (+1) [1, 2, 3]
```
map f xs is the list obtained by applying f to each element of xs, i.e.,
```map f [x1, x2, ..., xn] == [f x1, f x2, ..., f xn]
map f [x1, x2, ...] == [f x1, f x2, ...]
```
O(n). map f xs is the list obtained by applying f to each element of xs, i.e.,
```map f [x1, x2, ..., xn] == [f x1, f x2, ..., f xn]
map f [x1, x2, ...] == [f x1, f x2, ...]
```
```>>> map (+1) [1, 2, 3]
```
Using ApplicativeDo: 'fmap f as' can be understood as the do expression
```do a <- as
pure (f a)
```
with an inferred Functor constraint.
An infix synonym for fmap. The name of this operator is an allusion to \$. Note the similarities between their types:
```(\$) :: (a -> b) -> a -> b
(<\$>) :: Functor f => (a -> b) -> f a -> f b
```
Whereas \$ is function application, <\$> is function application lifted over a Functor.
#### Examples
Convert from a Maybe Int to a Maybe String using show:
```>>> show <\$> Nothing
Nothing
>>> show <\$> Just 3
Just "3"
```
Convert from an Either Int Int to an Either Int String using show:
```>>> show <\$> Left 17
Left 17
>>> show <\$> Right 17
Right "17"
```
Double each element of a list:
```>>> (*2) <\$> [1,2,3]
[2,4,6]
```
Apply even to the second element of a pair:
```>>> even <\$> (2,2)
(2,True)
```
An infix synonym for fmap. The name of this operator is an allusion to \$. Note the similarities between their types:
```(\$) :: (a -> b) -> a -> b
(<\$>) :: Functor f => (a -> b) -> f a -> f b
```
Whereas \$ is function application, <\$> is function application lifted over a Functor.
#### Examples
Convert from a Maybe Int to a Maybe String using show:
```>>> show <\$> Nothing
Nothing
>>> show <\$> Just 3
Just "3"
```
Convert from an Either Int Int to an Either Int String using show:
```>>> show <\$> Left 17
Left 17
>>> show <\$> Right 17
Right "17"
```
Double each element of a list:
```>>> (*2) <\$> [1,2,3]
[2,4,6]
```
Apply even to the second element of a pair:
```>>> even <\$> (2,2)
(2,True)
```
```map = fmap
```
map generalized to Functor.
```>>> map not (Just True)
Just False
>>> map not [True,False,True,True]
[False,True,False,False]
```
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You are on page 1of 2
# Name
:
Class:
Find the pattern.
1. 5, 4.7, 4.4, 4.1, 3.8, 3.5, 3.2,
2. 5.8, 6.2, 6.6, 7, 7.4, 7.8, 8.2,
3. 5.8, 6.2, 6.6, 7, 7.4, 7.8, 8.2,
4. 8, 7.4, 6.8, 6.2, 5.6, 5, 4.4,
5. 2.7, 3.4, 4.1, 4.8, 5.5, 6.2, 6.9,
6. 2.7, 3.4, 4.1, 4.8, 5.5, 6.2, 6.9,
7. 7, 6.7, 6.4, 6.1, 5.8, 5.5, 5.2,
8. 0, 0.9, 1.8, 2.7, 3.6, 4.5, 5.4,
9. 0, 0.9, 1.8, 2.7, 3.6, 4.5, 5.4,
10. 9.3, 8.7, 8.1, 7.5, 6.9, 6.3, 5.7,
www.mathgames4children.com
-1-
mphowehnzembayie@gmail.com
Name:
Class:
Find the pattern.
1. 5, 4.7, 4.4, 4.1, 3.8, 3.5, 3.2, 2.9, 2.6 (- 0.3)
2. 5.8, 6.2, 6.6, 7, 7.4, 7.8, 8.2, 8.6, 9 (+ 0.4)
3. 5.8, 6.2, 6.6, 7, 7.4, 7.8, 8.2, 8.6, 9 (+ 0.4)
4. 8, 7.4, 6.8, 6.2, 5.6, 5, 4.4, 3.8, 3.2 (- 0.6)
5. 2.7, 3.4, 4.1, 4.8, 5.5, 6.2, 6.9, 7.6, 8.3 (+ 0.7)
6. 2.7, 3.4, 4.1, 4.8, 5.5, 6.2, 6.9, 7.6, 8.3 (+ 0.7)
7. 7, 6.7, 6.4, 6.1, 5.8, 5.5, 5.2, 4.9, 4.6 (- 0.3)
8. 0, 0.9, 1.8, 2.7, 3.6, 4.5, 5.4, 6.3, 7.2 (+ 0.9)
9. 0, 0.9, 1.8, 2.7, 3.6, 4.5, 5.4, 6.3, 7.2 (+ 0.9)
10. 9.3, 8.7, 8.1, 7.5, 6.9, 6.3, 5.7, 5.1, 4.5 (- 0.6)
www.mathgames4children.com
-1-
mphowehnzembayie@gmail.com
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https://www.physicsforums.com/threads/harness-energy-of-earth-magnetic-field.668209/
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# Harness energy of Earth magnetic field?
• Stanley514
In summary, someone estimated that if cables will be extended from pole to pole and contacts placed on equator we could obtain 100.000 V, as Earth rotates in its own magnetic field.
Stanley514
Somebody estimated that if cables will be extended from pole to pole and contacts placed on equator we could obtain 100.000 V, as Earth rotates in its own magnetic field.
Ok. Did you have a question?
Is it true or not?
Who is "somebody" and do you have a reference?
Voltage isn't energy.
Question more.
Heard of any such effects on N-S railroad tracks, like Miami to Quebec?
Or power lines?
Ocean water is a conductor and very nearly spans the poles.
Crackpottery is easy to test.
jim hardy said:
Question more.
Heard of any such effects on N-S railroad tracks, like Miami to Quebec?
Or power lines?
Ocean water is a conductor and very nearly spans the poles.
Crackpottery is easy to test.
The only possibility would have to involve a ring round a meridian N-S-N. That would produce the Faraday Paradox Induction Effect. There isn't enough Area involved with any other loop moving through the Earth's field. Perhaps someone could calculate the emf and then the possible power output for us?
The thing that people forget when dreaming about electricity from Earth's magnetic field is that the field is stationary looking to it in overall , ofcourse it changes a little and drifts but the changes are so little that on a large scale it just won't do , as to induce current in a wire you either need a time varying magnetic field or you have to move the coil, just putting a coil in a stationary field won't do.
So you would need a bearing type cable around Earth and you would need to spin it and then you would get current induced , the problem again is spinning something requires energy, so in the end reality is harsh.
@sophiecentaur again writes faster than me:D
I guess I got the idea about the faraday paradox you mentioned , the whole system would then function as a simple homopolar engine, like a single wire from a AA battery + to the minus terminal?
The problem Is Earth itselft isn't a battery which gives voltage at certain terminals nor a magnet in a way that soil would conduct so I have a little bit of problem understanding your paradox case scenario.?
Last edited:
The Faraday paradox involves generating an emf even when magnet and loop are rotating together. Isn't that what would happen with a loop around the Earth n-s-n?
I guess so but doesn't the ends of the wire have to be connected to the magnet so that current could flow.Otherwise it would be like spinning a magnet and wire around it both at the same rpm but not physically connected and that wouldn't induce current in the wire as there would be no closed path for the electrons to move.And the field seen by the moving coil would be static just like you would place a magnet next to a wire sitting on a table.
So I understand that they have to be connected physically then the question is Earth as we know are not all made of magnetic minerals and ore , how does one connect the N-S-N wire to the ground for the electrons to flow?
Dig until reach mantle but that would melt the cable and again no luck.
@sophiecentaur so how it is actually , could you please see my last post, as I believe this thread just got forgotten a little bit, but not 100% finished.
:)
Crazymechanic said:
@sophiecentaur so how it is actually , could you please see my last post, as I believe this thread just got forgotten a little bit, but not 100% finished.
:)
Sorry. Mind elsewhere.
The magnet doesn't need to be part of the circuit - why did you think that was necessary? Was it because of some of the demos on YouTube? All you need is a loop or loops of wire and a changing magnetic flux. Some magnets are not even of conductive material but can be used in generators.
Actually, I was trying to think up a suitable geometry for the loop of wire and I found that most feasible layouts seem to involve induced voltages almost cancelling out. The problem is that the Earth's field is very weak and needs a massive loop area to obtain any more than a 'just detectable' amount of power.
Well I thought of it as the Faraday disc,in his case the rotating magnet was electrically connected to the wire via rolling brushes or other mechanism.
But If you don't do that I think the current induced in the wire will be even less than in the wire to Earth scenario as basically the Earth spins and the wire does that together with Earth and the Earth magnetic field speaking generally doesn't change that much at all as much as I know. Basically the wire would see pretty much static magnetic field am i right? Hence very small almoust laughable current would be induced?
IFFF you could drop a wire down through the Earth, to connect north and south poles, and then ran a semicircular wire on the surface, the rotation of that loop, once a day, would generate a significant emf. The Faraday paradox is explained by the fact that the Earth's magnet, spinning on its axis produces a stationary field - not an intuitive thing at all and hard to cope with (for me, anyway - I just have to accept it). Those demos on YouTube show this happening.
Ok now I understand more what you meant.yes I have seen them too, at the first moment i double checked the information as I don't have much trust in all those "random guy in a garage" videos on youtube but yes this seems to be so.
I think the important part here is that you have to have one wire a loop around Earth N-S-N and then another one straight through Earth and then connect the loop with the straight one via some brushes or so and in between a capacitor, now that reminds me the Farday thing quite clearly yes.
I think the problem arises about when one would want to put the straight wire through earth. :D
You woulds see those demos in School too (I did when I was a lad) but Health and Safety forbids the use of mercury! So far, they can't control what you do in your garage (at least what you do with mercury).
hi, friends, i have heard about something called as electrodynamic tethers. These are long conducting wires usually of aluminium which are used to change the orbits of spacecraft s. I took this subject for my class seminar. These electrodynamic tethers actually uses Earth magnetic field. They have also been used to produce higher electric current for experiments.
## 1. How can we harness energy from the Earth's magnetic field?
To harness energy from Earth's magnetic field, we can use a device called a magnetohydrodynamic generator. This device converts the kinetic energy of ionized gases in the Earth's atmosphere, which are affected by the magnetic field, into electrical energy.
## 2. What is the potential of harnessing energy from the Earth's magnetic field?
The potential of harnessing energy from Earth's magnetic field is vast. The Earth's magnetic field is estimated to contain about 10^25 joules of energy, which is equivalent to the total energy consumption of the world for 200,000 years.
## 3. How does harnessing energy from the Earth's magnetic field impact the environment?
Harnessing energy from Earth's magnetic field is a clean and renewable energy source, so it has minimal impact on the environment. Unlike fossil fuels, it does not produce greenhouse gases or other pollutants, making it a more sustainable option.
## 4. Can we use the Earth's magnetic field to power our homes and cities?
While the Earth's magnetic field has a vast amount of energy, the technology to harness it is still in its early stages. Currently, it is not feasible to use the Earth's magnetic field to power homes and cities on its own. However, it can be used as a supplemental energy source alongside other renewable energy sources.
## 5. Are there any challenges to harnessing energy from the Earth's magnetic field?
One of the main challenges of harnessing energy from Earth's magnetic field is the low efficiency of current magnetohydrodynamic generators. Additionally, the technology is still in the development phase, so there are still many obstacles to overcome to make it a viable energy source on a larger scale.
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https://calculatorguru.net/converter/kilopound-square-inch-to-ton-force-short-sq-foot/
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# Kilopound/Square Inch to Ton-force (short)/sq. foot converter| ksi to tonf/ft^2 conversion
Are you struggling with converting Kilopound/Square Inch to Ton-force (short)/sq. foot? Don’t worry! Our online “Kilopound/Square Inch to Ton-force (short)/sq. foot Converter” is here to simplify the conversion process for you.
Here’s how it works: simply input the value in Kilopound/Square Inch. The converter instantly gives you the value in Ton-force (short)/sq. foot. No more manual calculations or headaches – it’s all about smooth and effortless conversions!
Think of this Kilopound/Square Inch (ksi) to Ton-force (short)/sq. foot (tonf/ft^2) converter as your best friend who helps you to do the conversion between these pressure units. Say goodbye to calculating manually over how many Ton-force (short)/sq. foot are in a certain number of Kilopound/Square Inch – this converter does it all for you automatically!
## What are Kilopound/Square Inch and Ton-force (short)/sq. foot?
In simple words, Kilopound/Square Inch and Ton-force (short)/sq. foot are units of pressure used to measure how much force is applied over a certain area. It’s like measuring how tightly the air is pushing on something.
The short form for Kilopound/Square Inch is “ksi” and the short form for Ton-force (short)/sq. foot is “tonf/ft^2”.
In everyday life, we use pressure units like Kilopound/Square Inch and Ton-force (short)/sq. foot to measure how much things are getting squeezed or pushed. It helps us with tasks like checking tire pressure or understanding the force in different situations.
## How to convert from Kilopound/Square Inch to Ton-force (short)/sq. foot?
If you want to convert between these two units, you can do it manually too. To convert from Kilopound/Square Inch to Ton-force (short)/sq. foot just use the given formula:
``tonf/ft^2 = Value in ksi * 72``
here are some examples of conversion,
• 2 ksi = 2 * 72 = 144 tonf/ft^2
• 5 ksi = 5 * 72 = 360 tonf/ft^2
• 10 ksi = 10 * 72 = 720 tonf/ft^2
### Kilopound/Square Inch to Ton-force (short)/sq. foot converter: conclusion
Here we have learn what are the pressure units Kilopound/Square Inch (ksi) and Ton-force (short)/sq. foot (tonf/ft^2)? How to convert from Kilopound/Square Inch to Ton-force (short)/sq. foot manually and also we have created an online tool for conversion between these units.
Kilopound/Square Inch to Ton-force (short)/sq. foot converter” or simply ksi to tonf/ft^2 converter is a valuable tool for simplifying pressure unit conversions. By using this tool you don’t have to do manual calculations for conversion which saves you time.
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https://mapleprimes.com/users/nepomukk/replies
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15 Reputation
2 years, 277 days
Perfect!...
@Rouben Rostamian
Perfect. Many many thanks. This has brought me a lot further. I would still have a little question ...
How can I get the expression:
`sin(k1*(c*t-x))`
change to these:
`-sin(k1*(x-c*t)) `
Thanks again!
fg
New problem description...
Thank you for the quick response and sorry for the unclear description.
I try to describe my problem again.
The approach function
`u := a(x)*sin(k1*(-c*t+x))`
is used in the following pde:
`pde := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x));`
That is, I have simply formed the appropriate derivatives. The goal later is to determine the variable a(x). I would like to control this "hand calculation". With "pdsolve" I have the problem to integrate the approach function "u" correctly with. "pdetest" is a good suggestion as a last step for comparison. But how do I get here?
Many many thanks again!
Way to solution...
Many thanks for the help!
We now come to the following solution:
`-c^2*(d^2*a*sin(k1*(-c*t+x))/dx^2+2*d*a*k1*cos(k1*(-c*t+x))/dx)`
which intermediate steps would I have to insert this in maple yet? Since I have to check several approach functions, I want to "automate" the control as much as possible.
Thanks a lot!
Frank
Started successfully...
Hello,
Thank you for the help and tips. I'll get on quite well with that.
kf1, A1 etc. are constant variables (wavenumber, amplitudes ...)
In my case, the sulution assumption ends with
u
(3). The equation could, however, be extended at desired, higher non-linearities.
How exactly do I deal with the constant variables in my case?
Do I value these in advance?
Thanks a lot!
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https://au.mathworks.com/help/ecoder/ug/simplify-multiply-operations-in-array-indexing.html
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## Simplify Multiply Operations in Array Indexing
The generated code might have multiply operations when indexing an element of an array. You can select the optimization parameter Simplify array indexing to replace multiply operations in the array index with a temporary variable. This optimization can improve execution speed by reducing the number of times the multiply operation executes.
### Example Model
If you have the following model:
The Constant blocks have the following Constant value:
• `Const1`: `reshape(1:30,[1 5 3 2])`
• `Const2`: `reshape(1:20,[1 5 2 2])`
• `Const3`: `reshape(1:90,[1 5 9 2])`
The Concatenate block parameter Mode is set to `Multidimensional array`. The Constant blocks Sample time parameter is set to `–1`.
### Generate Code
Building the model with the Simplify array indexing parameter turned off generates the following code:
```int32_T i; int32_T i_0; int32_T i_1; for (i = 0; i < 2; i++) { for (i_1 = 0; i_1 < 3; i_1++) { for (i_0 = 0; i_0 < 5; i_0++) { ex_arrayindex_Y.Out[(i_0 + 5 * i_1) + 70 * i] = ex_arrayindex_P.Constant1_Value[(5 * i_1 + i_0) + 15 * i]; } } } for (i = 0; i < 2; i++) { for (i_1 = 0; i_1 < 2; i_1++) { for (i_0 = 0; i_0 < 5; i_0++) { ex_arrayindex_Y.Out[(i_0 + 5 * (i_1 + 3)) + 70 * i] = ex_arrayindex_P.Constant2_Value[(5 * i_1 + i_0) + 10 * i]; } } } for (i = 0; i < 2; i++) { for (i_1 = 0; i_1 < 9; i_1++) { for (i_0 = 0; i_0 < 5; i_0++) { ex_arrayindex_Y.Out[(i_0 + 5 * (i_1 + 5)) + 70 * i] = ex_arrayindex_P.Constant3_Value[(5 * i_1 + i_0) + 45 * i]; } } }```
### Generate Code with Optimization
Open the Configuration Parameters dialog box and select the Simplify array indexing parameter. Build the model again. In the generated code, `[(i_0 + tmp_1) + tmp]` replaces a multiply operation in the array index, `[(i_0 + 5 * i_1) + 70 * i]`. The generated code is now:
```int32_T i; int32_T i_0; int32_T i_1; int32_T tmp; int32_T tmp_0; int32_T tmp_1; tmp = 0; tmp_0 = 0; for (i = 0; i < 2; i++) { tmp_1 = 0; for (i_1 = 0; i_1 < 3; i_1++) { for (i_0 = 0; i_0 < 5; i_0++) { ex_arrayindex_Y.Out[(i_0 + tmp_1) + tmp] = ex_arrayindex_P.Constant1_Value[(i_0 + tmp_1) + tmp_0]; } tmp_1 += 5; } tmp += 70; tmp_0 += 15; } tmp = 0; tmp_0 = 0; for (i = 0; i < 2; i++) { tmp_1 = 0; for (i_1 = 0; i_1 < 2; i_1++) { for (i_0 = 0; i_0 < 5; i_0++) { ex_arrayindex_Y.Out[((i_0 + tmp_1) + tmp) + 15] = ex_arrayindex_P.Constant2_Value[(i_0 + tmp_1) + tmp_0]; } tmp_1 += 5; } tmp += 70; tmp_0 += 10; } tmp = 0; tmp_0 = 0; for (i = 0; i < 2; i++) { tmp_1 = 0; for (i_1 = 0; i_1 < 9; i_1++) { for (i_0 = 0; i_0 < 5; i_0++) { ex_arrayindex_Y.Out[((i_0 + tmp_1) + tmp) + 25] = ex_arrayindex_P.Constant3_Value[(i_0 + tmp_1) + tmp_0]; } tmp_1 += 5; } tmp += 70; tmp_0 += 45; }```
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https://polypad.amplify.com/hu/lesson/land-area-and-water-area-in-the-usa
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A Google hitelesítés sikertelen. Kérlek, próbáld újra később!
# Land Area and Water Area in the USA
## Overview and Objective
In this lesson, students will research, graph, and analyze the land area and water area for various states in the United States of America. Students will explore three types of column or row charts and analyze the benefits of each chart type. Working through the lesson will also incorporate thinking about ratios and percent.
## Warm-Up
Welcome students into the lesson by asking them to predict the following:
• The total number of square miles in the United States (including land and water).
• The percentage of the total square miles in the United States that is water.
• The state with the largest percentage of the total area that is water.
• The state with the smallest percentage of the total area that is water.
After a short discussion in which students share their predictions, reveal to the students that 7.8% of the United States area is water. According to data provided by the United States Census, the United States has 3,532,316 square miles of land and 277,209 square miles of water.
## Main Activity
In small groups or individually, invite students to research the land area and water area of 5 states of their choosing. Depending upon your preference, you may or may not want to ensure that all 50 states are selected. Share this link to the US Census website with students to find the data. They will need to select "Explore Your State" on the bottom left and then select a state. The land area and water area for each state is shown in the first paragraph. As students find the data for their various states, they can enter the data into a table on a blank Polypad canvas. Numbers in the table in Polypad cannot include commas.
After gathering the data, invite students to create a column chart or row chart of the data. Students should explore the three types of charts as shown in the video below. This video shows 5 states as an example. Students should research states of their own.
After students have explored the different chart types, invite them to respond to the following prompts. Students can record their answers in text boxes on their Polypad.
• What do you notice about the chart? What does the chart make you wonder?
• Which version of the chart (Grouped, Stacked, or Percentage) represents the data best?
• Which version of the chart (Grouped, Stacked, or Percentage) represents the data worst?
• Change the chart type to a line chart. Does this represent the data well? Why or why not?
## Closure
Gather for a class discussion after students have finished. One option would be for groups to report out the states that seems to have the smallest and largest percentage of water area. As students report out, you can add the data to a table on a new Polypad and create a percentage column chart. The chart will update as you add data to the table. Encourage students to find examples of states with both higher and lower percentages shown on the class chart.
If some states look close on the chart, you could calculate the specific percentage for each state to get a more detailed comparison. This will also be a good time to revisit students' predictions from earlier in the lesson and compare those with the class data. If time allows, you could assign any states not on the class list to specific students so your class data is complete. You could also add the data for the United States and compare specific states to the overall breakdown for the entire country.
## Extension
Interested students could explore the ratio of land area to water area in various other countries. Students could also explore other data available on the Census site for each state and see if there any any correlations between the percentage of water area and and other statistical measures.
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http://www.risc.jku.at/people/schreine/papers/swcomm99/index_3.html
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Go backward to 2 The System ModelGo up to TopGo forward to 4 Correctness of the Algorithm
# 3 The Termination Detection Algorithm
We will extend the System interface to
SystemTn in N(`act` in Zn->B, `chan` in Zn x Zn->Seq(MSG), `term` in B)
such that `term` initialized by
Init(...) :<=>
/\
• ...
• ~`term`
signals termination of the system.
Basic Idea The derivation of the termination detection algorithm is based on the following idea (see Figure 2): If Process 0 wants to detect termination, it sends a signal to process n-1. If an inactive process i > 0 receives the signal, it forwards the signal to process i-1. An active process keeps the signal until it becomes inactive. The signal therefore represents a "token" that circulates through the ring of processes.
We model this circulation by introducing a variable
`sig` in Zn->B
initialized by
Init(...) :<=>
/\
• ...
• forall i in Zn: ~`sig`i
and by two actions
• Starti(...) :<=>
/\
• i = 0
• `sig``'` = `sig`[i-n1 |-> true]
• Forwardi(...) :<=>
/\
• `sig``'` = if i != 0 then `sig`[i |-> false, i-n1 |-> true] else `sig`[i |-> false]
where a-nb denotes the difference modulo n, i.e., 0-n1 = n-1).
(Modeling token circulation by an array of boolean signals reflects the behavior of the distributed algorithm closer than modeling it by a single integer position that "automatically" guarantees token unicity.)
Figure 2: Token Circulation
However, above specification allows Process 0 to submit a new token before it has received the token it has previously submitted. We therefore introduce another variable
`run` in B
that records whether there is a token in the system and then disables Start:
• Starti(...) :<=>
/\
• ...
• ~`run`/\ `run``'`
• Forwardi(...) :<=>
/\
• ...
• `run``'` = if i != 0 then `run` else false
However, when Process 0 finally receives the token, it can only deduce that each process has been inactive at some time in the past. Since a process mave have received a message after forwarding the token, it may have become active again. Only if the algorithm could maintain the invariant
forall k in Zn: `sig`k =>
forall j in Zn: j > k => ~`act`j,
Process 0 could deduce from the receipt of the token (i.e., `sig`0) and its own inactivity (i.e., `~``act`0) that all processes are currently inactive.
#### Message Counters
Even if Process 0 receives the token under above invariant, there may be still messages pending in the network that can cause processes to become active again. We therefore need some mean to keep track of the number of messages pending in the network. Since a process only knows about the messages this it itself sends or receives, this can be only achieved by introducing a distributed counter
`cnt` in Zn->Z
such that every process sending a message increases its counter and every process receiving a message decreases its counter (see Figure 3):
• Sendi, j(...) :<=>
/\
• ...
• `cnt``'` = `cnt`[i |-> `cnt`i+1],
/\
• ...
• `cnt``'` = `cnt`[j |-> `cnt`j-1]
Figure 3: Message Counters
Then clearly the total sum of the distributed counter equals the number of the messages in the network, i.e., the system maintains the invariant
sumi in Z_n`cnt`i = sumi in Z_nsumj in Z_nlen(`chan`i, j)
where len(`ch`) := such l in N: exists m0, ..., ml-1: `ch` = < m0, ..., ml-1 > .
#### Token Value
How can Process 0 learn about this sum? Apparently this is only possible, if the circulating token collects this information and delivers it to this process. We therefore introduce
`tval` in Z
modeling the value carried by the token. Process 0 initializes the value to 0 when submitting the token; each forwarding process adds the value of its counter (see Figure 4):
• Starti(...) :<=>
/\
• ...
• `tval``'` = 0
• Forwardi(...) :<=>
/\
• ...
• `tval``'` = `tval`+`cnt`i
Figure 4: Token Value
If the algorithm could maintain the invariant
forall k in Zn: `sig`k =>
/\
• forall j in Zn: j > k => ~`act`j
• `tval` = sumj in Z_n, j > k`cnt`j
then Process 0 could on receipt of the token deduce from ~`act`0 that all processes are inactive and from `cnt`0+`tval` = 0 that no message is in the network, i.e., that the system has terminated.
Unfortunately, the algorithm cannot maintain this invariant. We therefore have to disjoin the conclusion part (the "core") of the invariant with some weakening conditions such that the algorithm is able to maintain the weaker invariant but Process 0 can still conclude termination from the information stored locally and received by the token.
#### Awaking a Process
Figure 5: Invalidating the Core Invariant
Let us investigate how above invariant can be falsified by an action (i.e., the invariant holds in the state before the action but does not any more hold in the state after the action). Since all processes that have already forwarded the token are inactive, the only possibility is that such a process receives a message and becomes active again (see Figure 5). This however means that there must be still pending messages in the network. Since the invariant still holds before invalidation, a weaker version of the variant is
forall k in Zn: `sig`k =>
\/
• /\
• forall j in Zn: j > k => ~`act`j
• `tval` = sumj in Z_n, j > k`cnt`j
• 0 < `tval`+sumj in Z_n, j <= k`cnt`j
If Process 0 receives the token (i.e., `sig`0 holds), it can still conclude termination from ~`act`0 and `cnt`0+`tval` = 0.
#### Process Marks
Is it still possible for an action to falsify the weaker invariant? Clearly it can be invalidated, if a process that has not yet forwarded the token receives a message, which causes the summation term in above condition to drop by one (see Figure 6). A simple way to record whether such an incident has taken place is to associate to each process a marker that is set on message receipt. We therefore introduce
`mark` in Zn->B
which is initialized to false in every position and which is set to true when a message is received:
• Init(...) :<=>
/\
• ...
• forall i in Zn: ~`mark`i
/\
• ...
• `mark``'` = `mark`[j |-> true],
Then we can weaken the invariant further to
forall k in Zn: `sig`k =>
\/
• /\
• forall j in Zn: j > k => ~`act`j
• `tval` = sumj in Z_n, j > k`cnt`j
• 0 < `tval`+sumj in Z_n, j <= k`cnt`j
• exists j in Zn: j <= k/\ `mark`j
If Process 0 receives the token, i.e., `sig`0 holds, it can still conclude termination from ~`act`0, `cnt`0+`tval` = 0, and ~`mark`0.
Figure 6: Process Markers
#### Token Marker
Clearly this weakened invariant is still falsified when the marked process with the least index forwards the token (see Figure 7). Therefore the token itself has to record the information when it passes a marked process. We introduce a token marker
`tmark` in B
which is initialized by Process 0 and forwarded by each process as
• Starti(...) :<=>
/\
• ...
• ~`tmark``'`
• Forwardi(...) :<=>
/\
• ...
• `tmark``'` = `tmark`\/ `mark`i
Then we can weaken the invariant further by
forall k in Zn: `sig`k =>
\/
• /\
• forall j in Zn: j > k => ~`act`j
• `tval` = sumj in Z_n, j > k`cnt`j
• 0 < `tval`+sumj in Z_n, j <= k`cnt`j
• exists j in Zn: j <= k/\ `mark`j
• `tmark`
This invariant cannot be falsified any more by the activity of any process.
Process 0 can then conclude termination from `sig`0, ~`act`0, `cnt`0+`tval` = 0, `cnt`0+`tval` = 0, and ~`tmark`. By definition of `tmark``'` and `tval``'`, this can be stated as
Forwardi(...) :<=>
/\
• ...
• `term``'` <=> if i != 0 then `term` else (~`tmark``'`/\ `tval``'` = 0).
Figure 7: Token Marker
#### Resetting Process Markers
If Process 0 cannot yet conclude termination after a termination detection round, it may later start another round. Without resetting the process markers, however, such an attempt is doomed to fail. Fortunately, after it has marked the token, the process marker has fulfilled its task and can be reset:
Forwardi(...) :<=>
/\
• ...
• `mark``'` = `mark`[i |-> false]
Since this action takes place simultaneously with token transmission (i.e., ~`sig``'`i holds), it cannot falsify the invariant.
If the system terminates during a termination detection round, Process 0 may not yet conclude termination at the end of this round. However, after this run, no process is active, there are no messages in the network any more, and the sum of the message counters equals 0. Still some processes may be marked; the next termination detection round may thus fail, too. Anyway, after this run all processes are unmarked, and the next termination detection round will succeed.
#### Algorithm Specification
The complete algorithm is compiled in Specification 2. The existential quantifier
var `v` in T: ...
introduces a local program variable `v` in a formula like (exists v in T: ...) introduces a local mathematical variable v. The crucial difference between both kinds is that the "variable" `v` may have different values in different states of a program behavior while the "constant" v has always the same value in all states.
Specification 2: The Termination Detection Algorithm
SystemTn in N(`act` in Zn->B, `chan` in Zn x Zn->Seq(MSG), `term` in B) :<=>
(a system that may detect its termination)
let
• Init(in `act`, in `chan`, in `term`, in `run`, in `cnt`, in `sig`, in `mark`) :<=>
/\
• ...(as in the basic system)
• ~`term`/\ ~`run`
• forall i in Zn: `cnt`i = 0/\ ~`sig`i/\ ~`mark`i,
• Sendi, j(in `a`, io `chan`, io `cnt`) :<=>
/\
• ...(as in the basic system)
• `cnt``'` = `cnt`[i |-> `cnt`i+1],
• Receivei, j(io `act`, io `chan`, io `cnt`, io `mark`) :<=>
/\
• ...(as in the basic system)
• `cnt``'` = `cnt`[j |-> `cnt`j-1]
• `mark``'` = `mark`[j |-> true],
• Inactivatei(io `act`) :<=> ...(as in the basic system)
• Starti(io `run`, io `sig`, io `mark`, io `tval`, out `tmark`) :<=>
(Process 0 starts termination detection)
/\
• i = 0/\ ~`run`
• `run``'`
• `sig``'` = `sig`[i-n1 |-> true]
• `mark``'` = `mark`[i |-> false]
• `tval``'` = 0
• ~`tmark``'`,
• Forwardi(in `a`, in `c`, io `sig`, io `mark`, io `tval`, io `tmark`, io `term`) :<=> (process i gets token, forwards it, or detects termination)
/\
• ~`a`/\ `sig`i
• `run``'` = if i != 0 then `run` else false
• `sig``'` = if i != 0 then `sig`[i |-> false, i-n1 |-> true] else `sig`[i |-> false]
• `mark``'` = `mark`[i |-> false]
• `tval``'` = `tval`+`c`
• `tmark``'` = `tmark`\/ `mark`i
• `term``'` <=> if i != 0 then `term` else (~`tmark``'`/\ `tval``'` = 0),
• Actionn, i(io `act`, io `chan`, io `term`, io `run`, io `cnt`, io `sig`, io `mark`, io `tval`, io `tmark`) :<=>
\/
• exists j in Zn: Sendi, j(in `act`i, io `chan`, io `cnt`)
• exists j in Zn: Receivei, j(io `act`, io `chan`, io `cnt`, io `mark`)
• Inactivatei(io `act`)
• Starti(io `run`, io `sig`, io `mark`, out `tval`, out `tmark`)
• Forwardi(in `act`i, in `cnt`i, io `sig`, io `mark`, io `tval`, io `tmark`, io `term`):
var
(internal variables of the algorithm)
• `run` in B, `cnt` in Zn->Z, `sig` in Zn->B, `mark` in Zn->B,
• `tval` in Z, `tmark` in B:
/\
• Init(in `act`, in `chan`, in `term`, in `run`, in `cnt`, in `sig`, in `mark`)
• [][exists i in Zn: Actionn, i(io `act`, io `chan`, io `term`, io `run`, io `cnt`, io `sig`, io `mark`, io `tval`, io `tmark`)]
Maintainer: Wolfgang Schreiner
Last Modification: August 20, 1998
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https://cpep.org/mathematics/1311368-what-makes-this-statement-true-7-2.html
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7 November, 12:02
# What makes this statement true? 7^-2
+2
1. 7 November, 12:33
0
2. 7 November, 12:57
0
The negative power means 1/the number to the given power
This means that 7^-2 can be rewritten as 1 / (7^2)
7^2 = 7*7
Therefore:
1 / (7^2) = 1 / (7*7)
7*7 = 49
Therefore:
1 / (7*7) = 1/49
Based on the above calculations:
7^-2 is equal to 1/49
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https://arduino.stackexchange.com/questions/92936/how-much-time-does-it-take-to-propagate-the-control-signal-through-all-leds-in-t
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# How much time does it take to propagate the control signal through all LEDs in the ws2812b strip?
In my next small Arduino project, I wanted to use some addressable rgb led strips. I made some small research and choose to use a strip with ws2812b. Generally, I know how this works but I am not 100% sure if what I think is correct. So please correct me if I am wrong.
For each pixel in the strip, we need to send 24 bits to control it. If we have 10 pixels in a strip then we need to send 240 bits. Then first pixel reads all 240 bits, sets itself color, and lastly propagates other 216 bits to the next pixel. And so on to the last pixel.
I am wondering, how much time does it take to update state of the last pixel in strip?
If sending 24 bit takes something about 30us (1.25us for 1 bit) then is that mean that sending signals to the last pixel in a strip with 10 pixels would take 300us + 270us + 240us + ... + 30us or is it rather just 300us?
• it works like a row of buckets ... data fills the first bucket ... when the bucket is full then everything gets passed to the next bucket ... so, after reset, the first 24 bits go to the first LED ... the rest of the bits pass through... same thing happens at the second LED Apr 23, 2023 at 0:03
I have a page about the WS2812 chip along with a suggested way of writing code simply for it. It also describes the timing.
The documented timing for a 0-bit is 1150 ns and for a 1-bit is 1300 ns.
Thus, if we assume all bits are on, then 24 bits would take 24 * 1300 ns which is 31.2 µs.
Since the bits are propagated along the strip, then it would take 10 * 31.2 µs (312 µs) to send all 10 colours.
What I mean by "propagated" is that as each bit arrives at each chip it also forwards it onward to the next chip, much like a shift register. Thus the more bits you send the further the original ones are "pushed" towards the end of the strip.
As the documentation describes, each chip "reshapes" the incoming bit, so that any jitter or noise in the arriving bit is cleaned up as it is forwarded so that the individual bits do not degrade as they are forwarded. Think of Chinese Whispers - because of the reshaping by each chip the information is not corrupted as it is passed on.
Then you send 50 µs of 0V to latch the new colours to the entire strip.
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| 2.890625
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CC-MAIN-2024-22
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latest
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en
| 0.948629
|
https://www.spudd64.com/what-is-planar-density/
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What is planar density?
What is planar density?
Planar density is the fraction of total crystallographic plane area that is occupied by atoms. Linear and planar densities are one- and two-dimensional analogs of the atomic packing factor.
What is the planar density of 111 plane in FCC?
For (111): From the sketch, we can determine that the area of the (111) plane is (v2a./2) (va/V2) = 0.866a.. There are (3) (1/2) + (3) (1/6) = 2 atoms in this area. planar density = 2 points 0.866(3.5167 x 10-8 cm)?
How do you do planar density?
Calculate planar density with the formula: PD = Number of atoms centered on a given plane / Area of the plane.
What is the difference between planar and linear?
It does not, of course, apply to planar structures, in which the apparent lineations should all be in the one plane, whereas those of a linear structure should conform to a double-cone-shaped body with the true lineation as its axis. Planar structures.
What is a family of directions?
The concept of a family of directions. A set of directions related by symmetry operations of the lattice or the crystal is called a family of directions. A family is a symmetry related set. A family of directions is represented (Miller Index notation) as: .
What are the similarities and dissimilarity between fcc and hcp structure?
Dear Sohail, FCC or the Face centres cubic and Hexagonal close packed structures both have a packing factor of 0.74, consists of closely packed planes of atoms and have a coordination number of 12. That is where the similarities end.
What is the most dense plane in FCC?
Structure Close packed planes Close packed directions
Face-centered cubic (FCC) {111} <110>
Hexagonal close-packed (HCP) Basal planes: (0001), (0002); Prismatic planes: one of the three {10-10} planes; Pyramidal planes: one of the six {10-11} <100>, <110> (three-axis notation) or <11-20> (four-axis notation)
What are linear features?
A Linear Feature can denote a single or multiple Themes along a line. By changing the ‘From’ and ‘To’ Data Fields along the length of the Linear Feature, different Themes can be allocated to a single line along different parts of the length.
What is a linear structure?
A linear or chronological structure is where the story is told in the order it happens. With a chronological or linear structure, the reader finds out what happens in the ‘correct’ order – this can lead the reader through events clearly.
What is the surface density of a planar plane?
The length of one of the sides of the plane is a*sqrt(2). Hence the surface density is 1/(a*sqrt(2)) atoms per unit area. The [111] plane is a plane that touches the three far corners of the unit cell and it looks like a triangle.
Which is the formula for the density of an object?
This is done by dividing the object’s mass by its volume. The formula for density is: Density = Mass / Volume. This equation can also be written: ρ=m/V. In the formula, ρ is the symbol for density. Scientists measure density in kilograms per cubic metre (kg/m3).
Which is the best definition of the word planar?
Definition of planar 1 : of, relating to, or lying in a plane 2 : two-dimensional in quality Other Words from planar Example Sentences Learn More about planar
What do you need to know about density?
Lesson Summary. Density measures the mass of an object or substance compared to its volume. The equation we use to find density is: density = mass / volume. If an object is heavy and compact, it has a high density. If an object is light and takes up a lot of space, it has a low density.
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# Calculate thermal resistance of heatsink required
1. Jan 9, 2016
### pbonesteak
1. The problem statement, all variables and given/known data
Q4) Calculate the thermal resistance of the heat sink required for the regulator of Q3 above given the information below [from the datasheet] and the data in the table.
Q3 FIGURE 2 shows an adjustable voltage regulator using the LT1083*. The LT1083 develops a 1.25V reference voltage between the output and the adjust terminal. By placing a resistor R1 between these two terminals, a constant current is caused to flow through R1 and down through R2 to set the overall output voltage. If VIN = 18 V, determine the range over which the output voltage can be varied.
2. Relevant equations
My formula
PD=Vin-Vout*Iout
θ Jct = T(jmax) - TA / PD
3. The attempt at a solution
PDmax = 18V+10% - 15V*1A = 4.8W
Control Section θ Jct = 125-75/4.8 = 10.4 °C/W
Power section θ Jct = 150-75/4.8 = 15.6 °C/W
The data sheet given is here
This is as far as I get. Theres so much info on the LT1083 datasheet I just need steering in a direction.
Any suggesgions much appreciated
Regards
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2. Jan 9, 2016
### pbonesteak
Just noticed the max 'Iout' is actually 7.5A not 1A
3. Jan 9, 2016
### Staff: Mentor
There's a discussion of Thermal Considerations in the APPLICATIONS INFORMATION section of the datasheet. It includes a calculated example that may help your understanding.
You'll need to pin down your operating conditions a bit better than just assuming a maximum input voltage and maximum device current; The device would not survive under such conditions. An actual maximum load current would be good. Look for the maximum power dissipation curve. It will tell you what the maximum case temperature can be for a given power dissipation.
4. Jan 9, 2016
### pbonesteak
So if i use a Vin as 18v use the adjustable resistor and set an output of 12v and am drawing 5amps load on a device that is connected on the output.
Using the formula Pd=Vin-Vout*Iout
Pd = 18v-12v*5=30W
For the control calculation: JuncT = T(amb)+Pd(θHsink+θJct-cas+θcas-Hsink)
I would get 75°+30W(1°C/W+0.6°C/W+0.2°C/W) = 105*1.8
JuncT=189° > 125° I would have 64°C that would need a heat sink to dissipate
I could do the same for the Power side
5. Apr 2, 2016
### StripesUK
Okay so here is what I have so far....
Control:
$ΔT=Tjmax-Ta=125-75=50°C$
$PD=Ilm(Vin-Vout)=1(18-15)=3W$
$θ=\frac{ΔT}{PD}=\frac{50}{3}=16.667°C/W$
This value is well above the value of 1.6°C/W stated on the datasheet so it will require a heat sink.
$θhs=θ-θjc-θchs=16.667-0.6-0.2=15.867°C/W$
I have done the same for the power section, but I feel I'm not on the right track due the thermal resistance values for the heat sinks being so drastically high. Especially if I'm right in thinking they then have to be added together? The question gives the maximum load current as 1A is this the correct value to use in the PD calculation? I've also seen the mention of pinning down operating conditions in a previous post, but should a heatsink not be able to handle the maximum values of the regulator?
6. Apr 16, 2016
### gazp1988
this is what i have.
PD= (Vin-Vout) x (Iout) = (18 - 15) x 1 = 3w
TJ = TA + PD (θHS + θCHS + θJC)
TJ= 75 + 3 (θHS + 0.2 + 0.6)
to work out the heat sink value
PD = ΔT/(θJC + θHS) =
3 = 50/(0.6 + θHS) =
θHS = 50/(0.6 + 3) = 13.89 C/W
TJ= 75 + 3(13.89 + 0.2 + 0.6) = 119.07°C <125°C
this shows that the calculated Tj is within the range of the maximum junc temperature.
am i on the right track here.
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# 95
TLDR; though you can't be 100% certain of anything, a lot of the people who go around talking about how you can't be 100% certain of anything would be surprised at how often you can be 99.99% certain. Indeed, we're often justified in assigning odds ratios well in excess of a million to one to certain claims. Realizing this is important for avoiding certain rookie Bayesian's mistakes, as well as for thinking about existential risk.
53 is prime. I'm very confident of this. 99.99% confident, at the very least. How can I be so confident? Because of the following argument:
If a number is composite, it must have a prime factor no greater than its square root. Because 53 is less than 64, sqrt(53) is less than 8. So, to find out if 53 is prime or not, we only need to check if it can be divided by primes less than 8 (i.e. 2, 3, 5, and 7). 53's last digit is odd, so it's not divisible by 2. 53's last digit is neither 0 nor 5, so it's not divisible by 5. The nearest multiples of 3 are 51 (=17x3) and 54, so 53 is not divisible by 3. The nearest multiples of 7 are 49 (=7^2) and 56, so 53 is not divisible by 7. Therefore, 53 is prime.
(My confidence in this argument is helped by the fact that I was good at math in high school. Your confidence in your math abilities may vary.)
I mention this because in his post Infinite Certainty, Eliezer writes:
Suppose you say that you're 99.99% confident that 2 + 2 = 4. Then you have just asserted that you could make 10,000 independent statements, in which you repose equal confidence, and be wrong, on average, around once. Maybe for 2 + 2 = 4 this extraordinary degree of confidence would be possible: "2 + 2 = 4" extremely simple, and mathematical as well as empirical, and widely believed socially (not with passionate affirmation but just quietly taken for granted). So maybe you really could get up to 99.99% confidence on this one.
I don't think you could get up to 99.99% confidence for assertions like "53 is a prime number". Yes, it seems likely, but by the time you tried to set up protocols that would let you assert 10,000 independent statements of this sort—that is, not just a set of statements about prime numbers, but a new protocol each time—you would fail more than once. Peter de Blanc has an amusing anecdote on this point, which he is welcome to retell in the comments.
I think this argument that you can't be 99.99% certain that 53 is prime is fallacious. Stuart Armstrong explains why in the comments:
If you say 99.9999% confidence, you're implying that you could make one million equally fraught statements, one after the other, and be wrong, on average, about once.
Excellent post overall, but that part seems weakest - we suffer from an unavailability problem, in that we can't just think up random statements with those properties. When I said I agreed 99.9999% with "P(P is never equal to 1)" it doesn't mean that I feel I could produce such a list - just that I have a very high belief that such a list could exist.
In other words, it's true that:
• If a well-calibrated person claims to be 99.99% certain of 10,000 independent statements, on average one of those statements should be false.
• If a well-calibrated person claims to be 99.99% certain of one statement, they should be able to produce 9,999 other independent statements of equal certainty and be wrong on average once.
If it's not clear why this doesn't follow consider the anecdote Eliezer references in the quote above, which runs as follows: A gets B to agree that if 7 is not prime, B will give A $100. B then makes the same agreement for 11, 13, 17, 19, and 23. Then A asks about 27. B refuses. What about 29? Sure. 31? Yes. 33? No. 37? Yes. 39? No. 41? Yes. 43? Yes. 47? Yes. 49? No. 51? Yes. And suddenly B is$100 poorer.
Now, B claimed to be 100% sure about 7 being prime, which I don't agree with. But that's not what lost him his $100. What lost him his$100 is that, as the game went on, he got careless. If he'd taken the time to ask himself, "am I really as sure about 51 as I am about 7?" he'd probably have realized the answer was "no." He probably didn't check he primality of 51 as carefully as I checked the primality of 53 at the beginning of this post. (From the provided chat transcript, sleep deprivation may have also had something to do with it.)
If you tried to make 10,000 statements with 99.99% certainty, sooner or later you would get careless. Heck, before I started writing this post, I tried typing up a list of statements I was sure of, and it wasn't long before I'd typed 1 + 0 = 10 (I'd meant to type 1 + 9 = 10. Oops.) But the fact that, as the exercise went on, you'd start including statements that weren't really as certain as the first statement doesn't mean you couldn't be justified in being 99.99% certain of that first statement.
I almost feel like I should apologize for nitpicking this, because I agree with the main point of the "Infinite Certainty" post, that you should never assign a proposition probability 1. Assigning a proposition a probability of 1 implies that no evidence could ever convince you otherwise, and I agree that that's bad. But I think it's important to say that you're often justified in putting a lot of 9s after the decimal point in your probability assignments, for a few reasons.
One reason is arguments in the style of Eliezer's "10,000 independent statements" argument lead to inconsistencies. From another post of Eliezer's:
I would be substantially more alarmed about a lottery device with a well-defined chance of 1 in 1,000,000 of destroying the world, than I am about the Large Hadron Collider being switched on.
On the other hand, if you asked me whether I could make one million statements of authority equal to "The Large Hadron Collider will not destroy the world", and be wrong, on average, around once, then I would have to say no.
What should I do about this inconsistency? I'm not sure, but I'm certainly not going to wave a magic wand to make it go away. That's like finding an inconsistency in a pair of maps you own, and quickly scribbling some alterations to make sure they're consistent.
I would also, by the way, be substantially more worried about a lottery device with a 1 in 1,000,000,000 chance of destroying the world, than a device which destroyed the world if the Judeo-Christian God existed. But I would not suppose that I could make one billion statements, one after the other, fully independent and equally fraught as "There is no God", and be wrong on average around once.
Okay, so that's just Eliezer. But in a way, it's just a sophisticated version of a mistake a lot of novice students of probability make. Many people, when you tell them they can never be 100% certain of anything, respond switching to saying 99% or 99.9% whenever they previously would have said 100%.
In a sense they have the right idea—there are lots of situations where, while the appropriate probability is not 0, it's still negligible. But 1% or even 0.1% isn't negligible enough in many contexts. Generally, you should not be in the habit of doing things that have a 0.1% chance of killing you. Do so on a daily basis, and on average you will be dead in less than three years. Conversely, if you mistakenly assign a 0.1% chance that you will die each time you leave the house, you may never leave the house.
Furthermore, the ways this can trip people up aren't just hypothetical. Christian apologist William Lane Craig claims the skeptical slogan "extraordinary claims require extraordinary evidence" is contradicted by probability theory, because it actually wouldn't take all that much evidence to convince us that, for example, "the numbers chosen in last night's lottery were 4, 2, 9, 7, 8 and 3." The correct response to this argument is to say that the prior probability of a miracle occurring is orders of magnitude smaller than mere one in a million odds.
I suspect many novice students of probability will be uncomfortable with that response. They shouldn't be, though. After all, if you tried to convince the average Christian of Joseph Smith's story with the golden plates, they'd require much more evidence than they'd need to be convinced that last night's lottery numbers were 4, 2, 9, 7, 8 and 3. That suggests their prior for Mormonism is much less than one in a million.
This also matters a lot for thinking about futurism and existential risk. If someone is in the habit of using "99%" as shorthand for "basically 100%," they will have trouble grasping the thought "I am 99% certain this futuristic scenario will not happen, but the stakes are high enough that I need to take the 1% chance into account in my decision making." Actually, I suspect that problems in this vicinity explain much of the problems ordinary people (read: including average scientists) have thinking about existential risk.
I agree with what Eliezer has said about being ware of picking numbers out of thin air and trying to do math with them. (Or if you are going to pick numbers out of thin air, at least be ready to abandon your numbers at the drop of a hat.) Such advice goes double for dealing with very small probabilities, which humans seem to be especially bad at thinking about.
But it's worth trying to internalize a sense that there are several very different categories of improbable claims, along the lines of:
• Things that have a probability of something like 1%. These are things you really don't want to bet your life on if you can help it.
• Things that have a probability of something like one in a million. Includes many common ways to die that don't involve doing anything most people would regard as especially risky. For example, these stats suggest the odds of a 100 mile car trip killing you are somewhere on the order of one in a million.
• Things whose probability is truly negligible outside alternate universes where your evidence is radically different than what it actually is. For example, the risk of the Earth being overrun by demons.
Furthermore, it's worth trying to learn to think coherently about which claims belong in which category. That includes not being afraid to assign claims to the third category when necessary.
# 95
Mentioned in
New Comment
Christian apologist William Lane Craig claims the skeptical slogan "extraordinary claims require extraordinary evidence" is contradicted by probability theory, because it actually wouldn't take all that much evidence to convince us that, for example, "the numbers chosen in last night's lottery were 4, 2, 9, 7, 8 and 3." The correct response to this argument is to say that the prior probability of a miracle occurring is orders of magnitude smaller than mere one in a million odds.
This only talks about the probability of the evidence given the truth of the hypothesis, but ignores the probability of the evidence given its falsity. For a variety of reasons, fake claims of miracles are far more common than fake TV announcements of the lottery numbers, which drastically reduces the likelihood ratio you get from the miracle claim relative to the lotto announcement.
The specific miracle also has lower prior probability (miracles are possible+this specific miracle's details), but that's not the only issue.
8Will_Sawin10y
Even if true announcments are just 9 times more likely than false announcements, then a true announcment should raise your confidence that the lottery numbers were 4 2 9 7 9 3 to 90%. This is because the probability P (429783 announced | 429783 is the number) is just the probability of a true announcement, but the probability P( 429783 announced | 429783 is not the number) is the probability of a false announcement, divided by a million. A false announcer would have little reason to fake the number 429793. This already completely annihilates the prior probability.
1ChrisHallquist10y
Actually, I'd consider it fairly important. It's one reason the probabilities ought to get very small very fast, but if you're reluctant to assign less than one in a million odds...
I think it's important to grasp the general principle under which a person telling you that this week's winning lotto numbers are some particular sequence is stronger evidence than their telling you a miracle took place. It offers a greater odds ratio, because they're much less likely to convey a particular lottery number in the event of it not being the winning one than they are to convey a miracle story in the event that no miracle occurred (even people who believe in miracles should be able to accept that miracles have a very high false positive rate if they believe that miracles only occur within their own religion.)
9SilentCal10y
To illustrate: Suppose you're checking the lottery results online, and you see that you won, and you're on your laptop at the house of a friend who knows what lottery numbers you buy and who has used his wi-fi to play pranks on guests in past. Suddenly the evidence doesn't fare so well against that million-to-one prior. This reminds me of reading about the Miracle of the Sun (http://en.wikipedia.org/wiki/Miracle_of_the_Sun) in The God Delusion and in a theist's response. I found Dawkins fairly unpersuasive; the many agreeing testimonials weren't enough to overcome the enormous prior improbability, but they were still disconcertingly strong evidence. The theists' response cleared this up by giving historical background that Dawkins omitted. Apparently, the miracle was predicted in advance by three children and had become a focal point in the tensions between the devout and the secular. Suddenly, it was not at all surprising that the gathered crowd witnessed a miracle. So I'd agree that miracles often have probability of under one in a million, but it's also vitally important to understand the effect of motivation on the likelihood of the evidence. If I thought every testimony to every reported miracle was based on unbiased reporting of fact, I'd have to conclude that many of them happened (caused by aliens messing with us or something).
1ActionItem10y
Craig is just purposely conflating the likelihood of a particular result and the likelihood of given the declaration of a result by the lottery officials, that result being true. If you and I are flipping coins for a million dollars, it's going to take a lot of convincing evidence that I lost the coin flip before I pay up. You just cannot flip the coin in another room where I can't even see, and then expect me to pay up because, well, the probability of heads is 50% and I shouldn't be so surprised to learn that I lost. Therefore, the actual likelihood of a particular set of lottery numbers is totally irrelevant in this discussion. In any case, the only kind of "evidence" that we have been presented for miracles has always been of the form "person X says Y happened', which has been known as hearsay and dealt with without even bothering with probability theory.
1wizzwizz44y
A second, detailed reading might make it seem like this comment's has an error. However, the reasoning is sound; "you said the coin was heads" doesn't distinguish very well between "the coin was heads" and "the coin was tails but you lied about the bet", so doesn't provide much evidence. Likewise, the dismissing of hearsay appears to be an error, but remember that humans have finite computational power. If you take into account (at least) the hypothesis that somebody's trying to deceive you about reality, you effectively end up dismissing the evidence anyway – but then you need to keep track of an extra hypothesis for the rest of your life to avoid scatterings of hearsay consistently nudging up your probability estimate when that's not really founded. (This is assuming that it's cheap to manufacture hearsay; expensive-to-manufacture hearsay shouldn't be dismissed so lightly.)
Generally, you should not be in the habit of doing things that have a 0.1% chance of killing you. Do so on a daily basis, and on average you will be dead in less than three years
Indeed!
It's even worse than that might suggest: 0.999^(3*365.25) = 0.334, so after three years you are almost exactly twice as likely to be dead than alive.
To get 50%, you only need 693 days, or about 1.9 years. Conversely, you need a surprising length of time (6500 days, about 17.8 years) to reduce your survival chances to 0.001.
The field of high-availability computing seems conceptually related. This is often considered in terms of the number of nines - so 'five nines' is 99.999% availability, or <5.3 min downtime a year. It often surprises people that a system can be unavailable for the duration of an entire working day and still hit 99.9% availability over the year. The 'nines' sort-of works conceptually in some situations (e.g. a site that makes money from selling things can't make money for as long as it's unavailable). But it's not so helpful in situations where the cost of an interruption per se is huge, and the length of downtime - if it's over a certain threshold - matters much less than whe...
[-]philh10y130
Christian apologist William Lane Craig claims the skeptical slogan "extraordinary claims require extraordinary evidence" is contradicted by probability theory, because it actually wouldn't take all that much evidence to convince us that, for example, "the numbers chosen in last night's lottery were 4, 2, 9, 7, 8 and 3." The correct response to this argument is to say that the prior probability of a miracle occurring is orders of magnitude smaller than mere one in a million odds.
I'm not sure that response works. Flip a fair coin two hundred times, tell me the results, then show me the video and I'll almost certainly believe you. But if the results were H^200, I won't; I'll assume you were wrong or lying about the coin being fair, or something.
H^200 isn't any less likely than any other sequence of two hundred coin flips, but it's still one of the most extraordinary. Extraordinariness just doesn't feel like it's a mere question of prior probability.
H^200 isn't any less likely under the assumption that the coin is fair, and the person reporting the coin is honest. But! H^200—being a particularly simple sequence—is massively more likely than most other sequences under the alternative assumption that the reporter is a liar, or that the coin is biased.
So being told that the outcome was H^200 is at least a lot of evidence that there's something funny going on, for that reason.
2shminux10y
This has nothing to do with simplicity. Any other apriori selected sequence, such as first 200 binary digits of pi, would be just as unlikely. It seems like it is related to simplicity because "non-simple" sequences are usually described in an aggregate way, such as "100 heads and 100 tails" and in fact include a lot of individual sequences, resulting in an aggregate probability much higher than 1/2^200.
This has nothing to do with simplicity. Any other apriori selected sequence, such as first 200 binary digits of pi, would be just as unlikely.
Yes, under the hypothesis that the coin is fair and has been flipped fairly all sequences are equally unlikely. But under the hypothesis that someone is lying to us or has been messing with the coin simple sequences are more likely. So (via Bayes) if we hear of a simple sequence we will think it's more likely to have be artificially created than if we hear of a complicated one.
3TheOtherDave10y
Well, what's most interestingly improbable here is the prediction of a 200-coin sequence, not the sequence itself. I suspect what's going on with such "extraordinary" sequences is a kind of hindsight bias... the sequence seems so simple and easy to understand that, upon revealing it, we feel like "we knew it all along." That is, we feel like we could have predicted it... and since such a prediction is extraordinarily unlikely, we feel like something extraordinarily unlikely just happened.
0[anonymous]10y
And, that, my friend, is how an algorithm feels from inside. What else could extraordinariness possibly be? It might also help to read "Probability Is Subjectively Objective".
1philh10y
You seem to be suggesting that if I actually flipped a coin 200 times, the actual result would be just as extraordinary as the hypothetical result H^200, having an equal prior probability. I'm not sure why you'd be suggesting that, so maybe we have crossed wires? For one thing, it might be that an extraordinary event is one which causes us to make large updates to the probabilities assigned to various hypotheses. H^200 makes hypotheses like "double headed coin" go from "barely worth considering" to "really quite plausible", so is extraordinary. (I think this idea has a bunch of fiddly little bits that I'm not particularly interested in hammering out. But the idea that the extraordinariness of an event is purely a function of its prior probability, just seems plain wrong.)
0[anonymous]10y
I agree that I seemed to suggest that; I indeed disagree that some arbitrary (expected to come with attached justifications, but those justifications are not present) sequence would be just as "extraordinary". This is where simplicity comes in -- the only reason a sequence of 200 bits would be interesting to humans would be if it were simple -- if it had some special property that allowed us to describe it without listing out the results of all 200 flips. Most sequences of 200 flips won't have this property, which makes the sequences that do extraordinary. So I'd consider T^200, (HT)^100, and (TH)^100 extraordinary sequences, but not 001001011101111001101011001111100101001110001011101000100011100 000111000101110001011100011010101001000011011001011011010110101 011100011000000101011001000100011000010100001100110000110010101 1. However, if I were to take out a coin and flip it now, and get those results, I could say "that sequence I posted on Less Wrong", and thus it would be extraordinary. So I agree that extraordinariness has nothing to do with the prior probability of a particular sequence of flips, but rather the fact that such a sequence of flips belongs to a privileged reference class (sequences of flips you can easily describe without listing all 200 flips), and getting a sequence from that reference class is an event with a low prior probability. The combination of being in that particular reference class and the fact that such an event (being in that reference class, not the individual sequence itself) is unlikely together might provide a sense of extraordinariness.
0TheOtherDave10y
I've suggested elsewhere that this sense of extraordinariness when faced with a result like H^200 is the result of a kind of hindsight bias. Roughly speaking, the idea is that certain notions seem simple to our brains... easier to access and understand and express and so forth. When such an notion is suggested to us, we frequently come to believe that "we knew it all along." A H^200 string of coin-flips is just such a notion; it seems simpler than a (HHTTTHHTHHT)^20 string, for example. So when faced with H^200 we have a stronger sense of having predicted it, or at least that we would have been able to predict it if we'd thought to. But, of course, predicting the result of 200 coin flips is extremely unlikely, and we know that. So when faced with H^200 we have a much stronger sense of having experienced something extremely unlikely (aka extraordinary) than when faced with a more "random-seeming" string.
0christopherj10y
Getting 200 heads only in coinflipping is just as likely as any other result. However, it is of incredibly low entropy -- you should not expect to see a pattern of that sort (less bits to describe than listing the results). It's also impossibly unlikely as a result from a fair coin, compared to as a result of fraud or an unfair coin.
0zslastman10y
Isn't it also the case that you are, in that case, receiving extraordinary evidence? If people were as unreliable about lottery numbers as they are about religion you would in fact remain pretty skeptical about the actual number.
Your list actually doesn't go far enough. There is a fourth, and scarier category. Things which would, if possibly render probability useless as a model. "The chance that probabilities don't apply to anything." is in the fourth category. I would also place anything that violates such basic things as the consistency of physics, or the existence of the external world.
For really small probabilities, we have to take into account some sources of error that just aren't meaningful in more normal odds.
For instance, if I shuffle and draw one card fro...
0NancyLebovitz10y
Would 53 not being prime break mathematics?
1Furslid10y
It would more likely be user error. I believe 53 is prime. If it isn't then either mathematics is broken or I have messed up in my reasoning. It is much more likely that I made an error or accepted a bad argument. 53 not being prime while having no integer factors other than 1 and itself would break mathematics.
0somervta10y
LNC, not the law of identity, I think.
0Furslid10y
First, I really like you pointing out the frequent 99% cop out and your partitioning of low-probability events into meaningful categories.
Second, I am not sure that your example with 53 being prime is convincing. It would be more interesting to ask "what unlikely event would break your confidence in 53 being prime?" and estimate the probability of such an event.
-4Pentashagon10y
The discovery of modular arithmetic and finite fields?
1shminux10y
Presumably ChrisHallquist is already familiar with finite fields.
This is one of the great reasons to do your math with odds rather than probabilities. (Well, this plus the fact that Bayes' Theorem is especially elegant when formulated in the form of odds ratios.)
There is no reason, save the historical one, that the default mode of thinking is in probabilities (as opposed to odds.) The math works just the same, but for probabilities that are even slightly extreme (even a fair amount less extreme than what is being talked about here), our intuitions about them break down. On the other hand, our intuitions when doing calc...
I'm not sure of the value of odds as opposed to probabilities for extreme values. Million-to-one odds is virtually the same thing as a 1/1,000,000 probability. Log odds, on the other hand, seem like they might have some potential for helping people think clearly about the issues.
I'd also note that probabilities are more useful for doing expected value calculations.
6NancyLebovitz10y
Anyone know whether gambling being expressed as odds is cross-cultural?
[-]scav10y60
I've never been completely happy with the "I could make 1M similar statements and be wrong once" test. It seems, I dunno, kind of a frequentist way of thinking about the probability that I'm wrong. I can't imagine making a million statements and have no way of knowing what it's like to feel confidence about a statement to an accuracy of one part per million.
Other ways to think of tiny probabilities:
(1) If probability theory tells me there's a 1 in a billion chance of X happening, then P(X) is somewhere between 1 in a billion and P(I calculated wr...
6Randaly10y
There are numerous studies that show that our brain's natural way of thinking out probabilities is in terms of frequencies, and that people show less bias when presented with frequencies than when they are presented with percentages.
1Lumifer10y
Thinking about which probabilities? Probability is a complex concept. The probability in the sentence "the probability of getting more than 60 heads in 100 fair coin tosses" is a very different beast from the probability in the sentence "the probability of rain tomorrow". There is a reason that both the frequentist and the Bayesian approaches exist.
3gjm10y
You can calculate wrong in a way that overestimates the probability, even if the probability you estimate is small. For some highly improbable events, if you calculate a probability of 10^-9 your best estimate of the probability might be smaller than that.
3scav10y
True. I suppose I was unconsciously thinking (now there's a phrase to fear!) about improbable dangerous events, where it is much more important not to underestimate P(X). If I get it wrong such that P(X) is truly only one in a trillion, then I am never going to know the difference and it's not a big deal, but if P(X) is truly on the order of P(I suck at maths) then I am in serious trouble ;) Especially given the recent evidence you have just provided for that hypothesis.
-1Armok_GoB10y
T-t-t-the Ultimate Insult, aimed at... oh my... /me faints
[-]fela10y40
it actually wouldn't take all that much evidence to convince us that, for example, "the numbers chosen in last night's lottery were 4, 2, 9, 7, 8 and 3." The correct response to this argument is to say that the prior probability of a miracle occurring is orders of magnitude smaller than mere one in a million odds.
That doesn't seem right. If somebody tries to convince me that the result of a fair 5 number lottery is 1, 2, 3, 4, 5 I would have a much harder time believing it, but not because the probability is less then one in a million. I think...
In cases like this where we want to drive the probability that something is true as high as possible, you are always left with an incomputable bit.
The bit that can't be computed is - am I sane? The fundamental problem is that there are (we presume) two kinds of people, sane people, and mad people who only think that they are sane. Those mad ones of course come up with mad arguments which show that their sanity is just fine. They may even have supporters who tell them they are perfectly normal - or even hallucinatory ones. How can I show which category I am...
[-][anonymous]10y40
Things that have a probability of something like one in a million. Includes many common ways to die that don't involve doing anything most people would regard as especially risky. For example, these stats suggest the odds of a 100 mile car trip killing you are somewhere on the order of one in a million.
I am not entirely sure about this, since I have made a similar mistake in the past, but If I am applying my relatively recent learning of this correctly, I think technically it suggests that if 1 million random people drive 100 miles, one of them will pro...
This fits with what I've read, though I'd point out that while we get our share of anti-drunk driving and now anti-texting-while-driving messages, most people don't seem to think driving in the rain, driving when they're a bit tired, or being a bit over the speed limit are particularly dangerous activities.
(Also, even if you're an exceptionally careful driver, you can still be killed by someone else's carelessness.)
0NancyLebovitz10y
I don't think most people believe that driving when they're very tired is especially dangerous.
0Gunnar_Zarncke10y
Lower maybe. But they are still in the order of 1:10^6. The border between the categories 1:100, 1:10^6 and 1:10^10 is - well - no border but continuous. The categorization into three rough areas expresses insufficient experience with all the shades in between. I don't mean that offensively. Dealing with risks appears to be normally done by the subconscious. Lifting it into the conscious is sensible but just assigning three categories will not do. Neither will do assigning words to more differentiated categories like in Lojban ( http://lesswrong.com/lw/9jv/thinking_bayesianically_with_lojban/ ). Real insight comes from training. Training with a suitable didactic strategy. One strategy obviously being to read the sequences as that forces you to consider lots of different more or less unlikely situations. What I am missing is a structured way to decompose these odds. 1:10^6 for a car accident in a 100 mile drive is arbitrary in so far as you can decompose it into either a 10 meter drive (say out of the parking lot) which then immediately moves the risk formally into the latter category. Or alternatively dying in a car accident in your life time which moves it into the first category. So why is it that a 100 mile drive was chosen?
Yeah, that's interesting.
I agree with Eliezer's post, but I think that's a good nitpick. Even if I can't be that certain about 10,000 statements consecutively because I get tired, I think it's plausible that there's 10,000 statements simple arithmetic statements which if I understand, check of my own knowledge, and remember seeing in a list on wikipedia, (which is what I did for 53), that, I've only ever been wrong once on. I find it hard to judge the exact amount, but I definitely remember thinking "I thought that was prime but I didn't really check ...
Of course, it's hard to be much more certain. I don't know what the chance is that (eg) mathematicians change the definition of prime -- that's pretty unlikely, but similar things have happened before that I thought I was certain of. But rarely.
If mathematicians changed the definition of "prime," I wouldn't consider previous beliefs about prime numbers to be wrong, it's just a change in convention. Mathematicians have disagreed about whether 1 was prime in the past, but that wasn't settled through proving a theorem about 1's primality, the way normal questions of mathematical truth are. Rather, it was realized that the convention that 1 is not prime was more useful, so that's what was adopted. But that didn't render the mathematicians who considered 1 prime wrong (at least, not wrong about whether 1 was prime, maybe wrong about the relative usefulness of the two conventions.)
4JackV10y
I emphatically agree with that, and I apologise for choosing a less-than-perfect example. But when I'm thinking of "ways in which an obviously true statement can be wrong", I think one of the prominent ways is "having a different definition than the person you're talking to, but both assuming your definition is universal". That doesn't matter if you're always careful to delineate between "this statement is true according to my internal definition" and "this statement is true according to commonly accepted definitions", but if you're 99.99% sure your definition is certain, it's easy NOT to specify (eg. in the first sentence of the post)
I agree that you can be 99.99% (or more) certain that 53 is prime but I don't think you can be that confident based only on the arguement you gave.
If a number is composite, it must have a prime factor no greater than its square root. Because 53 is less than 64, sqrt(53) is less than 8. So, to find out if 53 is prime or not, we only need to check if it can be divided by primes less than 8 (i.e. 2, 3, 5, and 7). 53's last digit is odd, so it's not divisible by 2. 53's last digit is neither 0 nor 5, so it's not divisible by 5. The nearest multiples of 3 are
...
1ChrisHallquist10y
I wouldn't call that argument my only reason, but it's my best shot at expressing my main reason in words. Funny story: when I was typing this post, I almost typed, "If a number is not prime, it must have a prime factor greater than its square root." But that's wrong, counterexamples include pi, i, and integers less than 2. Not that I was confused about that, my real reasoning was partly nonverbal and included things like "I'm restricting myself to the domain of integers greater than 1" as unstated assumptions. And I didn't actually have to spell out for myself the reasoning why 2 and 5 aren't factors of 53; that's the sort of thing I'm used to just seeing at a glance. This left me fearing that someone would point out some other minor error in the argument in spite of the arguments' being essentially correct, and I'd have to respond, "Well, I said I was 99.99% sure 53 was prime, I never claimed to be 99.99% sure of that particular argument."
2ChrisHallquist10y
Thank you. I'd seen "Advancing Certainty" before I wrote my post, but the comments are really good too.
I should perhaps include within the text a more direct link to Peter de Blanc's anecdote here:
http://www.spaceandgames.com/?p=27
I won't say "Thus I refute" but it is certainly a cautionary tale.
It seems to me to be mostly a cautionary tale about the dangers of taking a long series of bets when you're tired.
8Vulture10y
Definitely agreed. It's basically a variation on the old (very old) "Get a distracted or otherwise impaired person to agree to a bunch of obviously true statements, and then slip in a false one to trip them up" trick. I can't see that it has any relevance to the philosophical issue at hand.
2TheOtherDave10y
Yeah. When I try to do the "can I make a hundred statements yadda yadda" test I typically think in terms of one statement a day for a hundred days. Or more often, "if I make a statement in this class every day, how long do I expect it to take before I get one wrong?"
1A1987dM10y
Not quite, as SquallMage had correctly answered that 27, 33, 39 and 49 were not prime.
0Caspian10y
I believe that was part of the mistake, answering whether or not the numbers were prime, when the original question, last repeated several minutes earlier, was whether or not to accept a deal.
0Vulture10y
The point is, it's fundamentally the same trick, and is just that: a trick.
-1MugaSofer10y
Except it's not the same trick. What you describe relies on the mark getting into the rhythm of replying "yes" to every question; the actual example described has the mark checking each number, but making a mistake eventually, because the odds they will make a mistake is not zero.
5ialdabaoth10y
I think we need to be very careful what system we're actually describing. If someone asks me, "are you 99.99% sure 3 is prime? what about 5? what about 7? what about 9? what about 11?", my model does not actually consider these to be separate-and-independent facts, each with its own assigned prior. My mind "chunks" them together into a single meta-question: "am I 99.99% sure that, if asked X questions of the nature 'is {N} prime', my answer will conform to their own?" This question itself depends on many sub-systems, each with its own probability: * P(1). How likely is my prime number detection heuristic to return a false positive? * P(2). How prone is my ability to utilize my prime number detection heuristic to error? * P(3). How lossy is the channel by which I communicate the results of my utilization of my prime number detection heuristic? * P(4). How likely is it that the apparently-communicated question 'is {N} prime?' actually refers to a different thing that I mean when I utilize my prime number detection heuristic? So the meta-question "am I 99.99% sure that, if asked X questions of the nature 'is {N} prime', my answer will conform to their own?" is at LEAST bounded by ([1 - P(1)] [1 - P(2)] [1 - P(3)] * [1 - P(4)] ) < 0.999 ^ X. Why I feel this is important: When first asked a series of "Is {N} prime?" questions, my mind will immediately recognize the meta-question represented by P(1). It will NOT intuitively consider P(2), P(3) or P(4) relevant to the final bounds, so it will compute those bounds as ([1 - P(1)] 1 1 * 1) < 0.999 ^ X. Then, later, when P(2) turns out to be non-zero due to mental fatigue, I will explain away my failure as "I was tired" without REALLY recognizing that the original failure was in recognizing P(2) as a confounding input in the first place. (I.e.: in my personal case, especially if I was tired, I feel that I'd be likely to ACTUALLY use the heuristic "does my sense of memory recognition ping off when I see these numbers
3A1987dM10y
Sure, P(I'm mistaken about whether 53 is prime) is non-negligible (I've had far worse brain farts myself). But P(I'm mistaken about whether 53 is prime|I'm not sleep-deprived, I haven't answered a dozen similar questions in the last five minutes, and I've spent more than ten seconds thinking about this) is several orders of magnitude smaller. And P(I'm mistaken about whether 53 is prime|[the above], and I'm looking at a printed list of prime numbers and at the output of factor 53) is almost at the blue tentacle level.
I think part of what's troubling you about the test is thus: The claim, X has a probability of 10^-30 despite a prior of 50% is roughly equivalent to saying "I have information whose net result is 100 bits of information that X is false" That is certainly a difficult feat, but not really that hard if you put some effort into it (especially when you chose X). The proposed test to verify such a claim, ie making 10^30 similar statements and being wrong only once, would not only be impossible in your lifetime, but would be equivalent to saying "...
I don't think your comment with the lottery is a good example. If there was a lottery last night, then it was going to be some combination of random numbers, with no combination more or less likely then any others. If you come up and tell me "the winning lottery combination last night was X", the odds of you being correct are pretty high; there's really nothing unlikely in that scenario at all. Taking a look at some random number in the real world and thinking about the probability of it is meaningless, since you could be sitting there ha...
0gjm10y
I don't like this analysis much; in particular "the odds of you being correct are pretty high; there's really nothing unlikely in that scenario at all" seems unclear. Here's what I consider a better one. (It's similar to what fela says in a comment from last month but with more detail.) There are multiple different mechanisms that would lead to you saying "Last night's lottery numbers were [whatever]". You might have just made them up at random. You might have tried to determine them by magic, prayer, ESP, etc., and not checked against reality. You might have done one of those things but actually overheard the numbers without noticing, and been influenced by that. You might have read a report of what they were. Etc. It seems likely (though it might be hard to check) that most of the time when someone reports a set of lottery numbers from the past it's because they looked up what they were. In that case they're probably right, and (if wrong) probably close to right. So when someone tells me the numbers were 4, 2, 9, 7, 8, 3, there's an excellent chance that those really were the numbers. If they tell me the numbers were 1, 2, 3, 4, 5, 6, some quite different mechanisms become more probable -- they might be making them up for fun, have been misled by a hoax, etc. In this scenario I'd still reckon a posterior probability well over 10^-6 for the numbers actually being 1,2,3,4,5,6, but probably not over 10^-1 until I'd got some more evidence. Similarly, if the person were clutching a lottery ticket with the numbers 4, 2, 9, 7, 8, 3 then I would be less inclined to believe that those really were the numbers -- again, because of the increased probability that the person in question is lying, is self-deceiving, etc. (Especially if they had something to gain from convincing me that they held a winning lottery ticket.) Now, make it tomorrow's lottery instead of tonight's. What's different? The most important difference is that the formerly most likely mechanism (they rea
I feel that the sentence
Suppose you say that you're 99.99% confident that 2 + 2 = 4. Then you have just asserted that you could make 10,000 independent statements, in which you repose equal confidence, and be wrong, on average, around once.
is a little questionable to begin with. What exactly is an "independent" statement in this context? The only way to produce a statement about whether 2 + 2 = 4 holds, is to write a proof that it holds (or doesn't hold). But in a meaningful mathematical system you can't have two independent proofs for the same statement. Two proofs for the same thing are either both right or both wrong, or they aren't proofs in the first place.
[Retracted fully]
[This comment is no longer endorsed by its author]Reply
I always thought this must be the case from plain observation of thinking; much thinking is "logical", and pure logic is not a suitable model with significant uncertainty. There must be many situations where you're 9.999+ certain in order to make logical thinking useful.
If humans are bad at mental arithmetic, but good at, say, not dying - doesn't that suggest that, as a practical matter, humans should try to rephrase mathematical questions into questions about danger?
E.g. Imagine stepping into a field crisscrossed by dangerous laser beams in a prime-numbers manner to get something valuable. I think someone who had a realistic fear of the laser beams, and a realistic understanding of the benefit of that valuable thing would slow down and/or stop stepping out into suspicious spots.
Quantifying is ONE technique, and it's bee...
7Swimmer963 (Miranda Dixon-Luinenburg) 10y
I don't think this would help at all. Humans have some built-in systems to respond to danger that is shaped like a tiger or a snake or other learned stimuli, like when I see a patient go into a lethal arrhythmia on the heart monitor. This programmed response to danger pumps you full of adrenaline and makes you very motivated to run very fast, or work very hard at some skill that you've practiced over and over. Elite athletes perform better under the pressure of competition; beginners perform worse. An elite mathematician might do math faster if they felt they were in danger, but an elite mathematician is probably motivated to do mental arithmetic in the first place. I place around 95% confidence that generic bad-at-mental-arithmetic human would perform worse if they felt they were in danger than if they were in a safe classroom environment. If a patient is in cardiac arrest, I'm incredibly motivated to do something about it, but I don't trust my brain with even the simplest mental arithmetic. (Which is irritating, actually). This doesn't address the reward part of your situation, the "something valuable" at the end of the road. Without the danger, or with some mild thrill-adding danger, this might be a workable idea.
8fubarobfusco10y
Can you rule out the selection effect? (Which would be that people who don't happen to perform better under the pressure of competition, don't become elite athletes.)
1Ericf4y
#necro, but yes. Longitudinal observation shows that the same people perform worse under pressure as amateurs, and better under pressure at the end.
This is false modesty. This is assuming the virtue of doubt when none ought exist. Mathematics is one of the few (if not the only) worthwhile thing(s) we have in life that is entirely a priori. We can genuinely achieve 100% certainty. Anything less is to suggest the impossible, or to redefine the world in a way that has no meaning or usefulness.
I could say that I'm not really sure 2+2=4, but it would not make me more intelligent for the doubt, but more foolish. I could say that I'm not sure that 5 is really prime, but it would hinge on redefining '5' ...
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# Geometry Forum
Geometry Help Forum: Euclidean, basic geometric proofs, volumes, perimeters
1. ### Existence of inversion
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May 19th 2010, 02:27 PM
2. ### Problem of Apollonius
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3. ### Locuses
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4. ### I'm looking for a geometry book?
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5. ### Nonagon
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6. ### Proof on Interior Points + Pool Table
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7. ### Fructum
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8. ### circle/ellipse question
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9. ### Find a point on the line when given the equation
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10. ### Circle
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11. ### Angle bisector of paired lines
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12. ### A few analytic geometry questions...
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May 17th 2010, 06:29 PM
13. ### circle problems need help
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14. ### Equation of circle and diameter of hyperbola.
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15. ### Rectangle hyperbola
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May 16th 2010, 06:13 PM
16. ### find the locus
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May 16th 2010, 10:48 AM
17. ### Sketching Repricocals and Cubics Help
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May 14th 2010, 09:02 PM
18. ### diameter of a circle
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May 14th 2010, 08:54 PM
19. ### rearranging equasions
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20. ### L is Tangent to C?
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May 14th 2010, 10:21 AM
21. ### Find coordinates of 3rd vertex(x,y,z) of triangle(3D)
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22. ### Triangle HAT
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May 13th 2010, 08:18 PM
23. ### Concentric Circles
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May 13th 2010, 08:04 PM
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### cosx= .4583?
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# Power required to climb a 20-m-tall building in 55s
• ChetBarkley
#### ChetBarkley
Thread moved from the technical forums to the schoolwork forums
Summary:: A 90 kg firefighter needs to climb the stairs of a 20-m-tall building while carrying 40kg of gear. How much power does he need to reach the top of the building in 55s.
So first the total mass of our system is 130 kg. Using this mass, I found the potential energy the firefighter would have at the top of the building. Using U = mgy, I got 25480 J then knowing that power is just Jules over time I divided this by 55s and got 463 Watts. Is this correct?
Looks good to me. That assumes perfect efficiency, but that's OK. In real life, much of the work done by the firefighter will go to thermal energy.
ChetBarkley
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Revision as of 02:46, 5 November 2012 (edit)Stevenj (Talk | contribs)← Previous diff Revision as of 02:49, 5 November 2012 (edit)Stevenj (Talk | contribs) (→Usage)Next diff → Line 32: Line 32: This function Faddeeva::w(z, relerr) computes ''w''(''z'') to a desired [[w:Approximation error|relative error]] relerr. This function Faddeeva::w(z, relerr) computes ''w''(''z'') to a desired [[w:Approximation error|relative error]] relerr. - Omitting the relerr argument, or passing relerr=0 (or any relerr less than machine precision ε≈10−16), corresponds to requesting [[w:Machine epsilon|machine precision]], and in practice a relative error < 10−13 is usually achieved. Specifying a larger value of relerr may improve performance (at the expense of accuracy). + Omitting the relerr argument, or passing relerr=0 (or any relerr less than machine precision ε≈10−16), corresponds to requesting [[w:Machine epsilon|machine precision]], and in practice a relative error < 10−13 is usually achieved. Specifying a larger value of relerr may improve performance for some ''z'' (at the expense of accuracy). Similarly, the erf, erfc, erfcx, erfi, and Dawson functions are computed by calling: Similarly, the erf, erfc, erfcx, erfi, and Dawson functions are computed by calling:
## Revision as of 02:49, 5 November 2012
Steven G. Johnson has written free/open-source C++ code (with wrappers for other languages) to compute the various error functions of arbitrary complex arguments. In particular, we provide:
• The Faddeeva function $w(z) = e^{-z^2} \mathrm{erfc}(-iz) \!$, where erfc is the complementary error function.
• The error function $\mathrm{erf}(z) \!$
• The complementary error function $\mathrm{erfc}(z) = 1 - \mathrm{erf}(z) \!$
• The scaled complementary error function $\mathrm{erfcx}(z) = e^{z^2} \mathrm{erfc}(z) \!$
• The imaginary error function $\mathrm{erfi}(z) = -i \mathrm{erf}(iz) \!$
• The Dawson function $\mathrm{Dawson}(z) = \frac{\sqrt{\pi}}{2} e^{-z^2} \mathrm{erfi}(z)$
Given the Faddeeva function w(z), one can also easily compute Voigt functions and similar related functions as well. In benchmarks of our code, we find that it is comparable to or faster than most competing software for these functions in the complex plane (but we also have special-case optimizations for purely real or imaginary arguments), and we find that the accuracy is typically at at least 13 significant digits.
Because all of the algorithms are based on algorithms for the Faddeeva function, we call this the Faddeeva Package.
## Usage
To use the code, include the Faddeeva.hh header file:
#include "Faddeeva.hh"
and compile and link the Faddeeva.cc source code. You can then call various functions. For example:
extern std::complex<double> Faddeeva::w(std::complex<double> z, double relerr=0);
This function Faddeeva::w(z, relerr) computes w(z) to a desired relative error relerr.
Omitting the relerr argument, or passing relerr=0 (or any relerr less than machine precision ε≈10−16), corresponds to requesting machine precision, and in practice a relative error < 10−13 is usually achieved. Specifying a larger value of relerr may improve performance for some z (at the expense of accuracy).
Similarly, the erf, erfc, erfcx, erfi, and Dawson functions are computed by calling:
extern std::complex<double> Faddeeva::erf(std::complex<double> z, double relerr=0);
extern std::complex<double> Faddeeva::erfc(std::complex<double> z, double relerr=0);
extern std::complex<double> Faddeeva::erfcx(std::complex<double> z, double relerr=0);
extern std::complex<double> Faddeeva::erfi(std::complex<double> z, double relerr=0);
extern std::complex<double> Faddeeva::Dawson(std::complex<double> z, double relerr=0);
Since these functions are purely real for real arguments z=x, we provide the following specialized interfaces for convenience (and a slight performance gain, although the complex functions above automatically execute specialized code for purely real arguments):
extern double Faddeeva::erf(double x);
(These functions always compute to maximum accuracy, usually near machine precision.)
It is also sometimes useful to compute Im[w(x)] for real x, since $\mathrm{Im}[w(x)] = e^{-x^2} \mathrm{erfi}(x)$ in that case (like the Dawson function but without the √π/2 factor). [Note that Re[w(x)] is simply exp(−x2) for real x.] Im[w(x)] can be computed efficiently to nearly machine precision by calling:
extern double Faddeeva::w_im(double x);
## Wrappers: Matlab, GNU Octave, and Python
Wrappers are available for this function in other languages.
• Matlab (also available here): We provide source code for compiled Matlab plugins (MEX files) to interface all of the error functions above from Matlab.
• The provided functions are called Faddeeva_w, Faddeeva_erf, Faddeeva_erfc, Faddeeva_erfi, Faddeeva_erfcx, and Faddeeva_Dawson, equivalent to the C++ functions above. All have usage of the form w = Faddeeva_w(z) [or w = Faddeeva_w(z, relerr) to pass the optional relative error], to compute the function value from an array or matrix z of complex (or real) inputs.
• For convenience, a script to compile all of the plugins using the mex command in Matlab is included. Assuming you have a C++ compiler installed (and have run mex -setup to tell Matlab to use it), you can simply run the Faddeeva_build.m script in Matlab to compile all of the Faddeeva functions.
• Install the resulting *.mex* files, along with the *.m help files, into your Matlab path
• GNU Octave: Similar to Matlab, above, we provide source code for compiled GNU Octave plugins (.oct files) for all of the error functions above.
• The provided functions are called Faddeeva_w, Faddeeva_erf, Faddeeva_erfc, Faddeeva_erfi, Faddeeva_erfcx, and Faddeeva_Dawson, with usage identical to the Matlab plugins above.
• A Makefile is included. Assuming you have a C++ compiler and the mkoctfile command installed (mkoctfile comes with Octave, possibly in an octave-devel or similarly named package in GNU/Linux distributions), you can simply run make to compile the plugins, and sudo make install to install them system-wide (assuming you have system administrator privileges); otherwise put the compiled .oct files somewhere in your octave path.
• Python: Our code is used to provide scipy.special.erf, scipy.special.wofz, and the other error functions in SciPy starting in version 0.12.0 (see here).
## Algorithms
Our implementation uses a combination of different algorithms, mostly centering around computing the Faddeeva function w(z).
To compute the Faddeeva function for sufficiently large |z|, we use a continued-fraction expansion for w(z) similar to those described in
Unlike those papers, however, we switch to a completely different algorithm for smaller |z| or for z close to the real axis:
(I initially used this algorithm for all z, but the continued-fraction expansion turned out to be faster for larger |z|. On the other hand, Algorithm 916 is competitive or faster for smaller |z|, and appears to be significantly more accurate than the Poppe & Wijers code in some regions, e.g. in the vicinity of |z|=1 [although comparison with other compilers suggests that this may be a problem specific to gfortran]. Algorithm 916 also has better relative accuracy in Re[z] for some regions near the real-z axis. You can switch back to using Algorithm 916 for all z by changing USE_CONTINUED_FRACTION to 0 in the code.)
Note that this is SGJ's independent re-implementation of these algorithms, based on the descriptions in the papers only. In particular, we did not refer to the authors' Fortran or Matlab implementations (respectively), which are under restrictive "semifree" ACM copyright terms and are therefore unusable in free/open-source software.
Algorithm 916 requires an external complementary error function erfc(x) function for real arguments x to be supplied as a subroutine. More precisely, it requires the scaled function erfcx(x) = ex2erfc(x). Here, we use an erfcx routine written by SGJ that uses a combination of two algorithms: a continued-fraction expansion for large x and a lookup table of Chebyshev polynomials for small x. (I initially used an erfcx function derived from the DERFC routine in SLATEC, modified by SGJ to compute erfcx instead of erfc, but the new erfcx routine is much faster, and also seems to be faster than the calerf rational-Chebyshev code by W. J. Cody.)
Similarly, we also implement special-case code for real z, where the imaginary part of w is Dawson's integral. Similar to erfcx, this is also computed by a continued-fraction expansion for large |x|, a lookup table of Chebyshev polynomials for smaller |x|, and finally a Taylor expansion for very small |x|. (This seems to be faster than the dawsn function in the Cephes library, and is substantially faster than the gsl_sf_dawson function in the GNU Scientific Library.)
The other error functions can be computed in terms of w(z). The basic equations are:
$\mathrm{erfcx}(z) = e^{z^2} \mathrm{erfc}(z) = w(iz)$ (scaled complementary error function)
$\mathrm{erfc}(z) = e^{-z^2} w(iz) = \begin{cases} e^{-z^2} w(iz) & \mathrm{Re}\,z \geq 0 \\ 2 - e^{-z^2} w(-iz)) & \mathrm{Re}\,z < 0 \end{cases}$ (complementary error function)
$\mathrm{erf}(z) = 1 - \mathrm{erfc}(z) = \begin{cases} 1 - e^{-z^2} w(iz) & \mathrm{Re}\,z \geq 0 \\ e^{-z^2} w(-iz) - 1 & \mathrm{Re}\,z < 0 \end{cases}$ (error function)
$\mathrm{erfi}(z) = -i\mathrm{erf}(iz) \!$; for real x, $\mathrm{erfi}(x) = e^{x^2} \mathrm{Im}[w(x)] = \frac{\mathrm{Im}[w(x)]}{\mathrm{Re}[w(x)]}$ (imaginary error function)
$\mathrm{Dawson}(z) = \frac{\sqrt{\pi}}{2} e^{-z^2} \mathrm{erfi}(z) = \frac{i\sqrt{\pi}}{2} \begin{cases} e^{-z^2} - w(z) & \mathrm{Re}\,z \geq 0 \\ w(-z) - e^{-z^2} & \mathrm{Re}\,z < 0 \end{cases}$; for real x, $F(x) = \frac{\sqrt{\pi}}{2}\mathrm{Im}[w(x)]$ (Dawson function)
Note that we sometimes employ different equations for positive and negative Re(z) in order to avoid numerical problems arising from multiplying exponentially large and small quantities. For erfi and the Dawson function, there are simplifications that occur for real x as noted. In some cases, however, there are additional complications that require our implementation to go beyond these simple formulas. For erf, large cancellation errors occur in these formulas near |z|=0 where w(z) is nearly 1, as well as near the imaginary axis for Re[erf], and in these regimes we switch to a Taylor expansion. Similarly, for the Dawson function we switch to a Taylor expansion near the origin or near the real axis. (Similar problems occur for erfi, but our erfi implementation simply calls our erf code.)
## Test program
To test the code, a small test program is included at the end of Faddeeva.cc which tests w(z) against several known results (from Wolfram Alpha) and prints the relative errors obtained. To compile the test program, #define TEST_FADDEEVA in the file (or compile with -DTEST_FADDEEVA on Unix) and compile Faddeeva.cc. The resulting program prints SUCCESS at the end of its output if the errors were acceptable.
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# Ch 32: SAT Math: Geometry and Measurement
Make sure you're ready for geometry questions on the SAT with 2 hours of engaging video lessons covering geometry concepts on the SAT. Test your skills with lesson quizzes and a chapter exam to make sure you're ready for this section of the SAT.
## SAT Math: Geometry and Measurement - Chapter Summary
This portion of the math SAT covers a wide variety of concepts. Watch our short video lessons to review the following topics:
• Types of angles and lines
• Perimeter and area of triangles and rectangles
• Area and circumference of circles
• Properties of shapes
• Similar triangles
• Pythagorean Theorem
• Volumes of basic shapes
• Properties of 3-D objects
• Parts and applications of graphs
• Transformations
• Visualizing geometry problems
In this chapter, you'll work with the graphing formulas for distance and midpoint, in addition to the point-slope formula for finding the equation of a line. You'll also explore the graphing topics of undefined slope, zero slope, absolute value, dilation, reflections, center and radius. After working your way through each lesson, take the quick quiz to make sure that you're staying on target.
### SAT Math Objectives
Your SAT math section will have 58 questions. Your test booklet will contain the formulas needed to complete the questions. The geometry and measurement portion composes 25%-30% of the test questions. You'll be asked to demonstrate your understanding of the following concepts:
• Geometric visualization
• Circles: area and circumference
• Polygons: area and perimeter
• Volume: boxes, cubes and cylinders
• Slope
• Transformations
• Similarity
• Coordinate geometry
• Properties of parallel and perpendicular lines
• Pythagorean Theorem
• Special properties of equilateral, isosceles and right triangles
Our lessons assist your review of these tested objectives and provide practice problems for working with angles and triangles, in addition to applying the Pythagorean Theorem. You'll want to use a calculator on some of the problems, so make sure that you've checked the list of approved models on the SAT student site. Unanswered questions and incorrect choices don't count against you.
Final Exam
Chapter Exam
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Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.
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# Key PostA guide to mortgage repayment calculations
#### twofor1
##### Frequent Poster
So My query really only further confuses the issue- If you have missed 10 payments of €1500 (monthly repayment) - say €100,000 MortgageLoan- of which 1000 is capital and 500 is interest- thereby creating an arrears balance of €15000 . Loan outstanding at start of year was €100,000-
is the total debt now €115000 or €105,000?
My understanding is;
If you owed €100K on 01st January and you missed 10 repayments of €1,500, then on the 31st October you would be in arrears of a bit more than €15K, as the outstanding balance is slightly higher each day, there would be slightly more interest applied each day.
Your total debt would be roughly €104,245. (Assuming an interest rate of 5%) as Interest is calculated on the outstanding balance on a daily basis.
#### Brendan Burgess
##### Founder
I seem to recall arrears being capitalised onto the mortgage, this had the effect of increasing the Capital Balance by the amount of the Arrears !! ...
I would currently argue that the portion of Arrears that affects the repayment term or amount should only be the interest portion- indeed the interest portion of the missed payment.
An example for one month where the interest rate is 12% per annum is probably the easiest way to explain.
For simplicity, let's say that the repayment due is €3,000 per month
|Total balance|scheduled balance | arrears balance
1 January| €100,000|€100,000
Interest charge|€1,000
31 January| €101,000|€98,000|€3,000
If you had paid your repayment on schedule,the total balance at 31 January would be €98,000
As you did not pay it, you owe €101,000
If they capitalise the arrears (capital and interest), the total balance remains the same, and the scheduled balance rises to €101,000.
#### Belcamp70
##### New Member
I took out a mortgage last year and when reviewing my end of year statement and comparing it with the amortization tools mentioned above, it looks like I was over charged in the amount of interest paid (by over a grand).
My question is, what my best approach to getting this resolved?
The above was posted on your site a long time ago and no one answered
My question is there a company who will examine mortgage account and tell us if overpayed or not.
#### Brendan Burgess
##### Founder
comparing it with the amortization tools mentioned above, it looks like I was over charged in the amount of interest paid (by over a grand).
I don't see how you can lead to that conclusion from looking at amortization tools.
Just calculate the interest each month and forget about amortisation.
You do not need to pay someone to do that. It would cost far more than €1,000 for someone competent to do it.
Brendan
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Title: 利用二次渦旋粘滯紊流模式之非結構性網格計算Unstructured Grid Calculation Using Quadratic Eddy- Viscosity Models Authors: 張明介Ming-Jie Chang崔燕勇Yeng-Yung Tsui機械工程學系 Keywords: 非結構性網格;二次渦旋粘滯紊流模式;紊流;Unstructured Grid;Quadratic Eddy- Viscosity Models;Turbulent Issue Date: 2000 Abstract: 在本論文中引用SIMPLE壓力修正法，用有限容積法來離散統御方程式﹐並採用非交錯式之非結構性網格。 紊流模式採用Speziale與Rubinstein的二次非線性紊流模式求解(1)完全發展流(2)背階流場(3)軸對稱管道流場﹐並將結果與線性k-ε模式和實驗值比較。由結果可以發現對於平均速度、剪應力與紊流動能﹐二次模式與線性k-ε模式有相近的結果﹐而二次模式對於紊流正向應力與接觸長度﹐能得到比線性模式更好的結果。與實驗值相比﹐兩種模式對於迴流區附近的平均速度與接觸長度預測的結果都比較差。In this thesis, SIMPLE algorithm were applied to numerical simulation with Unstructured grid. Numerical simulations were applied to several turbulent flows, including:(1)Fully developed channel flow﹐(2)flow through a back-step, (3)flow through a expansion pipe with various expansion angle, using (I)standard linear k-εmodel(II)second order nonlinear models, Including Speziale’s model and Rubinstein’s model. The predicted mean and turbulent results were compared with experimental results. Predictions of mean velocity, shear stress, and turbulent kinetic energy by Linear model and Nonlinear models are comparable. Nonlinear models predict better results in turbulence intensity and reattachment length. Compared with experimental results, all model performing poorly in the separated regions. URI: http://140.113.39.130/cdrfb3/record/nctu/#NT890489052http://hdl.handle.net/11536/67550 Appears in Collections: Thesis
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Blog
Which financial ratios do you need for the real estate deal analysis?
The topic of analyzing the real estate deal seems daunting to many people. And indeed, there many ratios that could be used and many of them are providing you with different information and insights. That's why I would like to share with you how I have grouped the ratios that I often use.
I differentiate 3 groups of ratios for the real estate deals. The First Group is the group with ratios that don't make any difference whether the cash inflow is from the current period or any other period in the far future. The amounts are taken with their nominal value and the time value of money is not taken into consideration. And the second characteristics of these ratios is that don't account for any risk of the particular cash inflow. Is it coming from a very risky project or from the very safe one – the ratios don't reflect this.
The ratios from this group are the Gross Operating Income, The Net Operating Income, the Cap Rate, the Return and the Rent Multiplier.
These ratios are simple deviation of two numbers and they are very useful for quick analysis of the transaction. They will help you to make quick comparison of your potential deal with the alternative ones and to compare the performance of your potential investment with the overall performance of the market. In this respect the effectiveness of the first group ratios should not be underestimated.
The Second Group of ratios deliver actually everything that the first one is missing. They will provide you with accurate estimation of the present value of your cash flows and they will take adequate care of the risk of your project. These ratios are dealing with dynamic cash flows consisting of numerous periods.
On top of this the main ratio among them – the Net Present Value will deliver just a number as an end result. This number is dimensionless, meaning that it is easy to compare with other project that have different magnitude, risk and duration. Despite of this you will get a fair comparison that you might use for your analysis.
The main ratios of this group are the Present Value (PV), the Net Present Value (NPV) and the IRR. As mentioned these ratios are more sophisticated compared with the ratios of the first group and can't be normally calculated on the back of the envelop.
They require the usage of a financial calculator or computer. Some of them like the IRR will give you a percentage as a result indicating the probable profitability of the potential project and others like the Net Present Value will deliver an absolute number that will give actually pretty good idea how much richer you will become if you undertake the analyzed project.
I have to confess that I am a big advocate of the Third Group of financial ratios. This group includes ratios that reflect the level of debt in the transaction. Because the level of debt depends on the investor's decision and preferences, I often say that the ratio of the third group measure actually the performance of the investor and not the performance of the asset itself. The main idea of these ratios is to deliver accurate estimate about the ability of the project to maintain the chosen level of debt (DSCR), about the Return of the investor's equity in the transaction (ROE) and about the cost of capital in an existing portfolio with certain risk profile and profitability characteristics (WACC). All these parameters depend on the level of debt in the transaction and must be accounted for when the investor chose the level of debt in the project.
The ratios included in this group as it become already clear are the Return on Equity (ROE), the Weighted Average Cost of Capital (WACC) and the Debt Service Cover Ratio (DSCR). The latter is used very often by the banks to estimate if the cash flow of the proposed project will be sufficient to cover the service of the debt and the other costs and covenants related to the debt maintenance.
If you have already a portfolio of real estate units and you have debt in this portfolio the calculation of the WACC will become crucial for you, because the WACC could be/should be used as a discount rate in the NPV calculation for future potential projects. You must know this number.
The Return on Equity is the main reason why you need debt in the transaction. It boosts dramatically the investor's return and makes the real estate investment so attractive that it is almost not comparable with anything else on the market. There are however some prerequisites that must be taken into consideration otherwise the debt could be very harmful to the investor's equity.
I describe in detail how exactly these ratios are calculated and used in the praxis in my online course hosted on Udemy called How to use the top 11 fin real estate ratio in the best way.
I wish you successful investing!
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Calculations in Chemistry
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# Calculations in Chemistry - PowerPoint PPT Presentation
Calculations in Chemistry. To calculate the number of moles in a solid we use the following Mole Triangle. To calculate the number of moles in a solution we use the following Mole Triangle. g. n. c. v. n. gfm. n = number of moles c = concentatration (moles/litre)
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Calculations in Chemistry
• To calculate the number of moles in asolid we use the following Mole Triangle
To calculate the number of moles in a solution we use the following Mole Triangle
g
n
c
v
n
gfm
n = number of moles
c = concentatration (moles/litre)
v= volume in litres
g = Mass in Grams
n= Number of moles
gfm=gram formula mass
Examples using the Mole Triangle
From previous slide : If we cover up the entity we require we see that
n = g/gfm
• Calculate the no. of moles present in 0.4g of Na OH
gfm of NaOH= 23+16+1= 40g
n=0.4/40 =0.01moles
therefore
Calculate the mass of 0.05 moles of Mg(Cl)2
Again from previous slide we see that if we cover up the letter we want that we get
g= n × gfm
gfm = 24+2(35.5) =95g
therefore
0.05×95=4.75g
Calculations contd.
• Calculate the no. of moles present in 50cm3 of 0.05 molar HCl .
From previous triangle we that if we cover the letter we want that
n =c×v/1000
n= 0.05×50/1000= 0.00005moles
therefore
Calculate the concentration if we have 0.1 moles dissolved in 100cm3 of water
From previous triangle we see that c= n/v in litres
C = 0.1/100/1000= 0.01 moles/litre
therefore
Empirical or Simplest formula
Example: A sample of a substance was found to contain 0.12g of Magnesium and 0.19g of Fluorine. Find the simplest Formula.
• Rules
• 1. Write down all the symbols present.
• 2. Calculate the no of moles of each element present.
• 3. Compare ratios(get the smallest number of moles and divide it into all the others.
• 4.Write down the formula.
It doesn’t matter if the original sample is in grams or percentages
Mg and F
n =g/gfm
0.12/24 =
0.005moles
0.19/19 =
0.01moles
0.005 : 0.01
1 : 2
Mg(F)2
Neutralisation Calculations
• One way of Neutralising an Acid is to add an Alkali(for other methods see reactions of acids section).ie.
Acid + Alkali
Salt + water
To do neutralisation calculations we use the following formula. H+ × CA× VA = OH- × CB × VB
acid
alkali
H+ = no. of H+ ions in acid
OH- = no. of OH- ions in alkali
CA =Concentration of acid
CB= Concentration of alkali
VA = volume of acid
VB =Volume of alkali
HCl = 1
H2SO4 = 2
H3PO4 = 3
NaOH = 1
Ba(OH)2 = 2
Al(OH)3 = 3
Neutralisation Calculations contd.
• Example: What volume of 0.1M HCl is required to neutralise 100cm3 of 0.5M NaOH.
H+ ×CA ×VA = OH- × CB × VB
We require to find
1 × 0.1 × VA = 1 × 0.5 × 100
VA = 50/0.1
= 500cm3
Example:What concentration of 50cm3 KOH is used to neutralise 100cm3 of 0.05M H2SO4
H+ × CA × VA = OH- ×CB ×VB
We require to find
2 × 0.05 × 100 = 1 × CB × 50
CB = 10/50
= 0.2M
Calculations from Equations
• Example: What mass of Hydrogen gas is produced when 0.12g of Magnesium is added to excess Hydrochloric Acid.
We first require to write down the balanced equation
2
Mg(s) + HCl(aq)
Mg(Cl)2(aq) + H2(g)
To balance the equation we add a 2 in front of the HCl
From This we can see that:
1M
1M + 1M
24g
2g
1g
2/24 =0.08
therefore
0.12g
0.08×0.12
= 0.0096g of H2
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# Euclid’s Axioms
Euclid’s book Elements is the earliest known systematic discussion on geometry. In it he laid the foundations and rules to what is now known as Euclidean geometry. In Elements, Euclid assumed a small set of axioms, from which theorems could be deduced. So what are these axioms?
#### If you draw a line segment across two straight lines and it creates two angles on the same side which add to less than two right angles, then those two straight lines intersect.
This axiom is equivalent to saying that the angles in a triangle add up to 180 degrees.
Sources: 1 | 2
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Matrices & Systems of Linear Equations
Matrices & Systems of Linear Equations
Matrices & Systems of Linear Equations
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Presentation Transcript
1. Matrices & Systems of Linear Equations
2. Special Matrices
3. Special Matrices
4. Equality of Matrices Two matrices are said to be equal if they have the same size and their corresponding entries are equal
5. Equality of Matrices Use the given equality to find x, y and z
6. Matrix Addition and SubtractionExample (1)
7. Matrix Addition and SubtractionExample (2)
8. Multiplication of a Matrix by a Scalar
9. Matrix Multiplication(n by m) Matrix X (m by k) MatrixThe number of columns of the matrix on the left= number of rows of the matrix on the right The result is a (n by k) Matrix
10. Matrix Multiplication3x3 X 3x3
11. Matrix Multiplication1x3 X 3x3→ 1x3
12. Example (1)
13. Example (2)(1X3) X (3X3) → 1X3
14. Example (3)(3X1) X (1X2) → 3X2
15. Example (4)
16. Transpose of Matrix
17. Properties of the Transpose
18. Matrix ReductionDefinitions (1) 1. Zero Row:A row consisting entirely of zeros 2. Nonzero Row:A row having at least one nonzero entry 3. Leading Entry of a row:The first nonzero entry of a row.
19. Matrix ReductionDefinitions (2) Reduced Matrix: A matrix satisfying the following: 1. All zero rows, if any, are at the bottom of the matrix 2. The leading entry of a row is 1 3. All other entries in the column in which the leading entry is located are zeros. 4. A leading entry in a row is to the right of a leading entry in any row above it.
20. Examples of Reduced Matrices
21. Examples matrices that are not reduced
22. Elementary Row Operations 1. Interchanging two rows 2. Replacing a row by a nonzero multiple of itself 3. Replacing a row by the sum of that row and a nonzero multiple of another row.
23. Interchanging Rows
24. Replacing a row by a nonzero multiple of itself
25. Replacing a row by the sum of that row and a nonzero multiple of another row
26. Augmented Matrix Representing a System of linear Equations
27. Solving a System of Linear Equations by Reducing its Augmented Matrix Using Row Operations
28. Solution
29. Solution of the System
30. The Idea behind the Reduction Method
31. Interchanging the First & the Second Row
32. Multiplying the first Equation by 1/3
33. Adding to the Third Equation 12 times the Second Equation
34. Dividing the Third Equation by 40
35. Adding to the First Equation 7 times the third Equation
36. Systems with infinitely many Solutions
37. Systems with infinitely many Solutions
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Paul's Online Notes
Home / Calculus II / Series & Sequences / Binomial Series
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### Section 4-18 : Binomial Series
In this final section of this chapter we are going to look at another series representation for a function. Before we do this let’s first recall the following theorem.
#### Binomial Theorem
If $$n$$ is any positive integer then,
\begin{align*}{\left( {a + b} \right)^n} & = \sum\limits_{i = 0}^n {n \choose i} {a^{n - i}}\,{b^i} \,\\ & = {a^n} + n{a^{n - 1}}b + \frac{{n\left( {n - 1} \right)}}{{2!}}{a^{n - 2}}{b^2} + \cdots + na{b^{n - 1}} + {b^n}\end{align*}
where,
\begin{align*}{n \choose i} & = \frac{{n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - i + 1} \right)}}{{i!}}\hspace{0.25in}i = 1,2,3, \ldots n\\ {n \choose 0} & = 1\end{align*}
This is useful for expanding $${\left( {a + b} \right)^n}$$ for large $$n$$ when straight forward multiplication wouldn’t be easy to do. Let’s take a quick look at an example.
Example 1 Use the Binomial Theorem to expand $${\left( {2x - 3} \right)^4}$$
Show Solution
There really isn’t much to do other than plugging into the theorem.
\begin{align*}{\left( {2x - 3} \right)^4} & = \sum\limits_{i = 0}^4 { {4 \choose i} \,{{\left( {2x} \right)}^{4 - i}}\,{{\left( { - 3} \right)}^i}} \\ & = {4 \choose 0}{\left( {2x} \right)^4} + {4 \choose 1}{\left( {2x} \right)^3}\left( { - 3} \right) + {4 \choose 2}{\left( {2x} \right)^2}{\left( { - 3} \right)^2} + {4 \choose 3}\left( {2x} \right){\left( { - 3} \right)^3} + {4 \choose 4}{\left( { - 3} \right)^4}\\ & = {\left( {2x} \right)^4} + 4{\left( {2x} \right)^3}\left( { - 3} \right) + \frac{{4\left( 3 \right)}}{2}{\left( {2x} \right)^2}{\left( { - 3} \right)^2} + 4\left( {2x} \right){\left( { - 3} \right)^3} + {\left( { - 3} \right)^4}\\ & = 16{x^4} - 96{x^3} + 216{x^2} - 216x + 81\end{align*}
Now, the Binomial Theorem required that $$n$$ be a positive integer. There is an extension to this however that allows for any number at all.
#### Binomial Series
If $$k$$ is any number and $$\left| x \right| < 1$$ then,
\begin{align*}{\left( {1 + x} \right)^k} & = \sum\limits_{n = 0}^\infty { {k \choose n} {x^n}} \,\\ & = 1 + kx + \frac{{k\left( {k - 1} \right)}}{{2!}}{x^2} + \frac{{k\left( {k - 1} \right)\left( {k - 2} \right)}}{{3!}}{x^3} + \cdots \end{align*}
where,
\begin{align*}{k \choose n} & = \frac{{k\left( {k - 1} \right)\left( {k - 2} \right) \cdots \left( {k - n + 1} \right)}}{{n!}}\hspace{0.25in}n = 1,2,3, \ldots \\ {k \choose 0} & = 1\end{align*}
So, similar to the binomial theorem except that it’s an infinite series and we must have $$\left| x \right| < 1$$ in order to get convergence.
Let’s check out an example of this.
Example 2 Write down the first four terms in the binomial series for $$\sqrt {9 - x}$$
Show Solution
So, in this case $$k = \frac{1}{2}$$ and we’ll need to rewrite the term a little to put it into the form required.
$\sqrt {9 - x} = 3{\left( {1 - \frac{x}{9}} \right)^{\frac{1}{2}}} = 3{\left( {1 + \left( { - \frac{x}{9}} \right)} \right)^{\frac{1}{2}}}$
The first four terms in the binomial series is then,
\begin{align*}\sqrt {9 - x} & = 3{\left( {1 + \left( { - \frac{x}{9}} \right)} \right)^{\frac{1}{2}}}\\ & = 3\sum\limits_{n = 0}^\infty { {\frac{1}{2} \choose n} {{\left( { - \frac{x}{9}} \right)}^n}} \,\\ & = 3\left[ {1 + \left( {\frac{1}{2}} \right)\left( { - \frac{x}{9}} \right) + \frac{{\frac{1}{2}\left( { - \frac{1}{2}} \right)}}{2}{{\left( { - \frac{x}{9}} \right)}^2} + \frac{{\frac{1}{2}\left( { - \frac{1}{2}} \right)\left( { - \frac{3}{2}} \right)}}{6}{{\left( { - \frac{x}{9}} \right)}^3} + \cdots } \right]\\ & = 3 - \frac{x}{6} - \frac{{{x^2}}}{{216}} - \frac{{{x^3}}}{{3888}} - \cdots \end{align*}
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## Lesson 1: Lines, Ellipses and Boxes
##### 6:52 AM, Friday February 23rd 2024
My first attempt at learning fundamentals after 10 years of private classes (:
2 users agree
##### 3:03 PM, Wednesday February 28th 2024
Hello! Congratulations on completing lesson 1! I'll be taking a look at your homework :D
You've done a great job! It shows that you have confidence when drawing lines. I don't see errors in this work in general.
Sometimes you are repeating lines over each other to correct the line. You should not repeat it, but should hold the line as if it were correct and move on.
Rotated boxes! This is a difficult exercise, so don't stress too much. There are problems with perspective here, mistakes are to be expected, but you will have a clearer idea of what to do in the box challenge. I recommend revisiting this exercise after you get a better handle on the boxes through the box challenge to see how much you've improved.
Go to the box challenge!
This community member feels the lesson should be marked as complete, and 2 others agree. The student has earned their completion badge for this lesson and should feel confident in moving onto the next lesson.
##### 6:41 PM, Wednesday February 28th 2024
Hello, thank you so much for your feedback :D
I will try the rotated boxes exercise again after the 250 box challenge as suggested ^ - ^
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When starting on a project, I'll often open it up and start dragging reference images off the internet onto the board. When I'm done, I'll save out a '.pur' file, which embeds all the images. They can get pretty big, but are way more convenient than hauling around folders full of separate images.
Did I mention you can get it for free? The developer allows you to pay whatever amount you want for it. They recommend \$5, but they'll allow you to take it for nothing. Really though, with software this versatile and polished, you really should throw them a few bucks if you pick it up. It's more than worth it.
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# Graph planar drawing, with each edge's length is known
Assuming I have a graph $G$, with edges $E$ and vertices $V$, and the length of each edge is known, but the coordinates of vertices are not.
Further assume that this is a graph that can be embedded on a 2D plane, what is the algorithm that can create a "nice" drawing with no edge intersection?
Obviously there are many ways the graph can be drawn, and the definition of nice is a bit fuzzy (for practical purpose I would define it as something that is pleasing to the human eyes), but I would only need a good enough algorithm for this.
Note there could be many graphs with its embedding and I frankly don't care the algorithm gives me which, as long as it is pleasant looking enough and it has all the above constraint satisfied.
The motivation for this question is this: imagine that you have to design a circuit layout, the path that connects between two electronic components is $E$, and the vertex of each component is $V$, as a part of your engineering design you know that the distance between each component is pre-specified. As a designer, you would have to arrange the components on the board so that it looks nice. How can you design the circuit layout?
• @Graviton:The question is essentially not about embeddings but about drawings. So it's desirable to change the title of your question. Further, I guess, it should be specified whether planar embedding of graph is given and if it's not then whether the problem of finding planar embeddings of the considered class is in P. Mar 28, 2011 at 10:23
• @oleksandr, question updated. Also, I explicitly mention that the graph can be embedded. Mar 28, 2011 at 10:27
• @Graviton: As I understood you have possibly one graph with its embedding and you care only about aesthetic drawing of only this one, doesn't you? Mar 28, 2011 at 10:41
• @Oleksandr, nope. There could be many graphs with its embedding and I frankly don't care the algorithm gives me which, as long as it is pleasant looking enough, Mar 28, 2011 at 11:15
• @Graviton: I mean if you consider arbitrary graphs then how you could find desired drawing with the given edge lengths if even for restricted classes of graphs finding Planar Embedding with Specified Edge Lengths is NP-hard (see scholar.google.com/…). Mar 28, 2011 at 11:49
There is considerable work on the problem of finding planar embeddings with specified edge lengths. Survey and some results can be found in $[1]$.
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Induced EMF: Find Flux for Rectangular Loop in z-y Plane
• robert25pl
In summary, the conversation discusses a rectangular loop in the z-y plane with a moving rod in the x-direction and a fixed loop. The equation for induced emf is given by integrating the magnetic field with respect to path length. The lower limit on the x integral should be 1, not 0.
robert25pl
Can someone check my equation for flux. Thanks
A rectangular loop in the z-y plane is situated at t = 0 at the points (x=1,z=0), (x=1,z=5), (x=4,z=5), and (x=4,z=0). The rod of the loop with end points (x=4,z=0), and (x=4,z=5) is moving in the x-direction with a velocity
$$v=5\vec{i}$$m/s while the rest of the loop remains fixed. Find induced emf in the loop for all t
B = (10/x) cos100t j
$$\psi=\int_{s}B\cdot\,ds=\int_{xo=1}^{xo=4} \int_{z=0}^{5}\frac{10}{x}cos100t\vec{j}\cdot\, dx\,dz\vec{j}$$
Last edited:
robert25pl said:
Can someone check my equation for flux. Thanks
A rectangular loop in the z-y plane is situated at t = 0 at the points (x=1,z=0), (x=1,z=5), (x=4,z=5), and (x=4,z=0). The rod of the loop with end points (x=4,z=0), and (x=4,z=5) is moving in the x-direction with a velocity
$$v=5\vec{i}$$m/s while the rest of the loop remains fixed. Find induced emf in the loop for all t
B = (10/x) cos100t j
$$\psi=\int_{s}B\cdot\,ds=\int_{x=0}^{xo+5t} \int_{z=0}^{5}\frac{10}{x}cos100t\vec{j}\cdot\, dx\,dz\vec{j}$$
Looks OK to me except the lower limit on the x integral should be 1, not zero, with xo = 4 in the upper limit. I assume you can simplify and integrate this.
Yes I can do that and then find emf.
Last edited:
$$\psi=\int_{s}B\cdot\,ds=\int_{x=1}^{4+5t} \int_{z=0}^{5}\frac{10}{x}cos100t\vec{j}\cdot\, dx\,dz\vec{j}$$
Last edited:
1. What is induced EMF?
Induced EMF (electromotive force) is the voltage or potential difference that is generated in a conductor due to a change in magnetic flux through the conductor.
2. How is EMF induced in a rectangular loop in the z-y plane?
EMF is induced in a rectangular loop in the z-y plane when there is a changing magnetic field passing through the loop. This can be caused by either a varying magnetic field or by moving the loop through a stationary magnetic field.
3. What is the formula for calculating induced EMF in a rectangular loop in the z-y plane?
The formula for calculating induced EMF in a rectangular loop in the z-y plane is: E = -N * (dΦ/dt), where E is the induced EMF, N is the number of turns in the loop, and dΦ/dt is the rate of change of magnetic flux through the loop.
4. How does the orientation of the rectangular loop affect the induced EMF?
The orientation of the rectangular loop does not affect the induced EMF. As long as the loop is in the z-y plane, the induced EMF will be the same regardless of its orientation.
5. What is the significance of finding the flux in a rectangular loop in the z-y plane?
Finding the flux in a rectangular loop in the z-y plane allows us to determine the amount of induced EMF in the loop, which can then be used to calculate the current and other electrical properties of the loop. This is important in understanding and analyzing the behavior of electromagnetic systems.
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# How do you solve 2(3a+2)=-8 using the distributive property?
##### 1 Answer
Aug 31, 2016
The final answer will be $a = - 2$
#### Explanation:
First off, what you want to do is use your distributive property to simplify the parentheses.
Take the 2 from the outside, and multiply it times the $3 a$ to get $6 a$ because $3 \cdot 2 = 6$.
Then separately multiply that same 2 on the outside times the $+ 2$ on the inside: $2 \cdot \left(+ 2\right) = 4$.
So then your simplified equation will read $6 a + 4 = - 8$.
After this, you have to still get the $6 a$ by itself, so you subtract $4$ from both sides of the equation.
$4 - 4 = 0$ on the left side, so that 4 disappears. You also have to subtract 4 from the other side too. $- 8 - \left(4\right) = - 12$.
Then you have $6 a = - 12$. Divide both sides by $6$ to get $a = - 2$. There is your final answer using distributive property.
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# Scarab: Program Examples
## (Pandiagonal) Latin Square
• This example is for Latin Square used in CSP Solver Comptitions (CSC2009).
• Compared to a usual Latin Square (Latin Square in Wikipeida), this one has additional constraints for pandiagonal lines.
import jp.kobe_u.scarab._, dsl._
var n: Int = 5
for (i <- 0 until n; j <- 0 until n) int('x (i, j), 1, n)
for (i <- 0 until n) {
add(alldiff((0 until n).map(j => 'x (i, j))))
add(alldiff((0 until n).map(j => 'x (j, i))))
add(alldiff((0 until n).map(j => 'x (j, (i + j) % n))))
add(alldiff((0 until n).map(j => 'x (j, ((n - 1 + i - j)) % n))))
}
if (find)
for (i <- 0 until n)
println((0 until n).map { j => solution.intMap('x (i, j)) }.mkString(" "))
• (Lines 1 to 3) import Scarab classes
• (Line 6) declare integer variables
• (Lines 8 to 11) add alldifferenct constraints for each row, column, and diagonal
• (Line 12) print found solution if it exists
## Square Packing
• Square Packing is a two dimensinal packing problem.
• Its goal is to pack $$n$$ squares each of whose sizes are ranged from 1 to $$n$$ into a given (larger sized) container square.
• The sequences of minimum sized containers for $$n=1,2,3,...,$$ is knwon as A005842 of the on-line encyclopedia of integer sequences.
• Non-overrapping constraint is used to model this problem, which are used in several literature.
• Kim Marriott, Peter J. Stuckey, Vincent Tam, Weiqing He. Removing Node Overlapping in Graph Layout Using Constrained Optimization. Constraints, 8(2): 143–171, 2003.
• Takehide Soh, Katsumi Inoue, Naoyuki Tamura, Mutsunori Banbara, Hidetomo Nabeshima. A SAT-based Method for Solving the Two-dimensional Strip Packing Problem. Fundamenta Informaticae, 102(3–4): 467–487, IOS Press, 2010.
• diffn in Global Constraint Catalog.
import jp.kobe_u.scarab._, dsl._
val n = 15; val s =36
for (i <- 1 to n) { int('x(i),0,s-i) ; int('y(i),0,s-i) }
for (i <- 1 to n; j <- i+1 to n)
add(('x(i)+i <='x(j)) || ('x(j)+j<='x(i)) || ('y(i)+i<='y(j)) || ('y(j)+j<='y(i)))
if (find) println(solution.intMap)
• (Lines 1 to 3) import Scarab classes
• (Line 7) declare integer variables
• (Lines 8 and 9) add non-overlaping constraints
• (Line 11) print found solution if it exists
## Langford Pairing
### Model 1
import jp.kobe_u.scarab._, dsl._
val n = 4
for (i <- 1 to 2*n) int('x(i),1,n)
for (i <- 1 to n)
add(Or(for (j <- 1 to 2*n-i-1) yield And(('x(j) === 'x(j+i+1)), ('x(j) === i))))
if(find) println(solution)
• (Line 7) declare integer variables representing each of $$2n$$ positions has which number.
### Model 2 (with position variable)
import jp.kobe_u.scarab._, dsl._
val n = 4
for (i <- 1 to n) {
int('l(i),1,2*n-i-1)
int('r(i),1,2*n)
}
for (i <- 1 to n) add('l(i) === 'r(i)-i-1)
if(find) println(solution)
• (Lines 7 to 10) declare integer variables representing each pairs of $$n$$ numbers are placed to which positions.
## Graph Coloring
• Graph Coloring (see also Graph Coloring in Wikipedia) is a problem to find a coloring for all nodes of a given graph such that neighbors are colored differently.
• You can find its instances in URL.
import jp.kobe_u.scarab._, dsl._
val nodes = Seq(1,2,3,4,5)
val edges = Seq((1,2),(1,5),(2,3),(2,4),(3,4),(4,5))
var maxColor = 4;
int('color,1,maxColor)
for (i <- nodes) int('n(i),1,maxColor)
for (i <- nodes) add('n(i) <= 'color)
for ((i,j) <- edges) add('n(i) !== 'n(j))
while (find('color <= maxColor)) {
println(solution)
maxColor -= 1
}
• (Lines 5 to 7) declare graph structure.
• (Lines 9 and 10) declare integer variables representing available colors and each node of the given graph.
• (Lines 11 and 12) declare constraints that limit available colors and adjacent nodes have different color.
• (Lines 14 to 17) minimizing number of colors.
## Magic Square
• Magic Square (see also Magic Square in Wikipedia) is a problem to place $$1$$ to $$n^2$$ numbers into $$n \times n$$ matrix so that sum of each row, sum of each column, sum of each diagonal must be equal to $$\frac{n(n^2+1)}{2}$$.
import jp.kobe_u.scarab._, dsl._
val xs = for (i <- 1 to 3; j <- 1 to 3) yield csp.int('x(i,j), 1, 9)
for (i <- 1 to 3)
add(Sum((1 to 3).map(j => 'x(i,j))) === 15)
for (j <- 1 to 3)
add(Sum((1 to 3).map(i => 'x(i,j))) === 15)
add(Sum((1 to 3).map(i => 'x(i,i))) === 15)
add(Sum((1 to 3).map(i => 'x(i,4-i))) === 15)
if (find) println(solution)
• (Lines 1 to 3) import Scarab classes
• (Line 5) declare integer variables and puts them to xs
• (Line 6) declare alldiff for the variables
• (Lines 8 and 11) add constraints such that the sum for each row and column become 15
• (Line 12 and 13) add constraints such that the sum for each main diagonal become 15
• (Line 15) print found solution if it exists
## Alphametic Problem SAT + IS + FUN = TRUE
• Alphametic Problem (see also Verbal arithmetic in Wikipedia) is one kind of puzzle which represent numbers by alphabets.
• Goal is to find hidden numbers represented in alphabets by using relations between given words.
• The following gives an instance SAT + IS + FUN = TRUE (by Prof. Daniel Le Berre) which is originally from an instance CP + IS + FUN = TRUE used in a tutorial of or-tools.
• SAT + IS + FUN = TRUE is understood as $$S*100 + A*10 + T + I*10 + S + F*100 + U*10 + N = T*1000 + R*100 + U*10 + E$$.
import jp.kobe_u.scarab._, dsl._
val base = 10
for (v <- Seq('s,'i,'f,'t)) yield int(v,1,base-1) // S, I, F and T are not zero
for (v <- Seq('a,'u,'n,'r,'e)) yield int(v,0,base-1) // others can be zero
for (v <- Seq('c1,'c2,'c3)) yield int(v,0,2) // carries
add('t + 's + 'n === 'e + 'c1*base)
add('a + 'i + 'u + 'c1 === 'u + 'c2*base)
add('s + 'f + 'c2 === 'r + 'c3*base)
if (find) println(solution.intMap)
• (Lines 11 to 14) constraint model considering each digit and carry, which takes around 1 second;)
## Open-shop Scheduling
• Open-shop scheduling is a scheduling problem.
• The following example uses an instance gp03-01’’ given by the paper:
• (DOI) Guéret, C., & Prins, C. (1999). A new lower bound for the open-shop problem. Annals of Operations Research, 92, 165–183.
• The following model is given by the paper:
• (DOI) Naoyuki Tamura, Akiko Taga, Satoshi Kitagawa, Mutsunori Banbara. Compiling finite linear CSP into SAT. Constraints, 14:254–272, 2009.
import jp.kobe_u.scarab._, dsl._
use(new Sat4j("glucose"))
val pt = Seq(
Seq(661, 6, 333),
Seq(168, 489, 343),
Seq(171, 505, 324))
val n = pt.size
val lb = pt.map(_.sum).max
var ub = (0 until n).map(k => (0 until n).map(i => pt(i)((i + k) % n)).max).sum
int('makespan, lb, ub)
for (i <- 0 until n; j <- 0 until n) {
int('s(i,j), 0, ub)
}
for (i <- 0 until n) {
for (j <- 0 until n; l <- j+1 until n)
add('s(i,j) + pt(i)(j) <= 's(i,l) ||
's(i,l) + pt(i)(l) <= 's(i,j))
}
for (j <- 0 until n) {
for (i <- 0 until n; k <- i+1 until n)
add('s(i,j) + pt(i)(j) <= 's(k,j) ||
's(k,j) + pt(k)(j) <= 's(i,j))
}
while (find('makespan <= ub)) {
println(solution)
val end = (for(i <- 0 until n; j <- 0 until n)
yield solution.intMap('s(i,j))+pt(i)(j)).max
ub = end - 1
println(ub)
}
• (Lines 1 to 3) import Scarab classes
• (Lines 7 to 10) declare an instance
• (Lines 12 to 14) compute size, lower and upper bounds of the instance
• (Line 16) declares an integer variable representing current makespan
• (Lines 18 to 21) forces all operations are ended before makespan
• (Lines 22 to 26) forces for operations in the same job do not overlap each other
• (Lines 27 to 31) forces for operations sharing same resource do not overlap each other
• (Lines 33 to 38) coumputes optimum solution
## Colored N Queen
import jp.kobe_u.scarab._, dsl._
val n = args(0).toInt
val c = n
use(new Sat4j("glucose"))
for (i <- 1 to n; color <- 1 to c)
int('q(i,color), 1, c)
for (color <- 1 to c) {
}
for (i <- 1 to n)
if (find) {
for (color <- 1 to c) {
for (row <- 1 to n) {
var seq: Seq[Int] = Seq.empty
for (column <- 1 to n)
if (encoder.decode('q(row,color)) == column)
seq = seq :+ color
else
seq = seq :+ 0
println(seq.mkString(" "))
}
println("-----------------")
}
}
• (Lines 1 to 3) import Scarab classes
• (Lines 5 to 6) size is given from command line
• (Lines 8) declares the use of Sat4j of Glucose setting.
• (Lines 10 to 11) declares integer variables representing queens
• (Lines 13 to 17) representing N-Queen constraints for each color
• (Lines 19 to 20) forces that Queens of each color do no overlap
• (Lines 22 to 35) compute solutions and show the obtained placement
Created: 2019-05-17 金 12:35
Validate
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Home » MATH » Find the probability that a number selected at random from 41…
# Find the probability that a number selected at random from 41…
Posted by: 536 views
A. 1/8
B. 2/15
C. 3/16
D. 7/8
A
## DETAILS…
The that a number selected at random from 41 to 56 is a multiple of 9 is = 1/8
from 41 to 56 = 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, and 56.
Total number occurrence = 16
Numbers that are multiples of 9 (between 41 and 56) = 45, and 54.
Total number of occurrence of the numbers that are multiples of 9 is = 2
RELATED => The factor that least affects food shortages in sub-Saharan Africa is...
Thus, the probability that a number selected at random from 41 to 56 is a multiple of 9 is = 2/16 = 1/8.
### Now for the right answer to the above question:
1. Option A is correct.
2. Option B is not correct.
3. C is incorrect.
4. D is not the correct answer.
## KEY-POINTS…
• Probability = number of required occurrence /number of possible occurrence.
If you love our answers, you can simply join our community and also provide answers like this, fellow learners like you will appreciate it.
/ culled from 2021 -UTME MATHEMATICS question 25 /
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Value of levered company
is a real pain in azz right ?? but i love you guys and when i graduate i give a chocho too all of youuu in our financial sense is like paying all earnings as dividends no money left to invest in future but trust me i\ll do it !! So, Flag enterprises is expected to have a free cash flow in the coming year \$8 million and this free cash flow is expected to grow at an annual rate of 3% , thereafter in the foreseeable future. Flag has an equity cost of capital of 13% , a (pre-tax) cost of debt 7% and it is in the 35% tax bracket , IF flag maitains .05 debt-to-equity ratio then the value of flag as a levered company is ???
sunny… you post some crazy questions.
supersunny, now’s a good time to share with us the source of the questions you’ve been posting today. 1) While equity valuation using FCF is calculated similarly to DDM, this is LII material. 2) Similarly, the pure-play levered/unlevered beta question you posted earlier is LII material Moreover, it would be helpful if this question specified whether the FCF was Free Cash Flow to Equity (FCFE) for Free Cash Flow to the Firm (FCFF), as the former is discounted at Ke and the latter is discounted at WACC. My understanding is that the D/E is irrelevant to equity valuation using discounted FCF.
u really got me sweat hiredguns1 by showing me this LII thingy its my 3rd course ever in finance … i wonder thats why alot of students fail in our school…but if in case i fail iwill suicide…but this is the same question as iit was given in one of our previous exam
I am not too sure if my answer is correct, but applying what I have learnt in school…Weird, why is Capital Structure in L1? Did thte syllabus change or something!!! Definition of symbols: Rsu= unlevered cost of equity. VL=Value of levered firm Vu=value of unlevered firm First: When there is growth in the firm, the value of the unlevered firm is given by: FCF/(Rsu-g)=\$8million/(0.13-0.03)=\$80 million But then, there is a slight problem, because there is no indication on the amount of debt used by the company. I am not sure if u can acty use the D/E ratio (my guess is that it is a red herring), but from the text I was using, they gave an actual number, say \$5million and that really helped. Anyways, just as an illustration, let’s just assume that the company issused \$5m worth of bonds. Going by the MM model, VL=VU+TD=80m+0.35(5million)=\$81.75million But, there is also this growth model that u can use, and VL=VU+[(Cost of debt*TD)/(Rsu-g)] =80m+[(0.07*0.35*5million)/(0.1)]=\$81.225million. It’ll be really helpful if u posted more dettails, like what was the answer and explanation for it…Cos as a matter of fact, there are 2 ways to get abt doing this question from what I understand. MM basically assumes 0 growth and theoretically speaking, the value of the company is supposed to be larger under the growth model, but it seems that because the growth rate here is too low, the value of the tax shield in the MM model is larger thtan the value of the tax shield under the growth mdoel. Here is my 0.02 on this. It does look a bit weird.
that i am sure that we are discounting it by WACC and the Wacc i get is .08775 and the growth rate is 3% when i discount them i dont get the right answer which is \$111 million
I was assuming a debt amount of \$5million. I am not too sure what exactly is the amount of debt in the company. I am not sure why shld it be discounted by WACC as what was mentioned earlier, if it is FCFE, it should not be discounted at Wacc but at the cost of equity alone
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The MathNet Korea
Information Center for Mathematical Science
### 논문검색
Information Center for Mathematical Science
#### 논문검색
SIAM Journal on Computing
( Vol. 28 NO.4 / (1999))
Linear Time Algorithms for Dominating Pairs in Asteroidal Triple-Free Graphs
Pages. 1284-1297
Abstract An independent set of three vertices is called an \$asteroidal ; triple\$ if between each pair in the triple there exists a path that avoids the neighborhood of the third. A graph is asteroidal triple- free (AT-free) if it contains no asteroidal triple. The motivation for this investigation is provided, in part, by the fact that AT-free graphs offer a common generalization of interval, permutation, trapezoid, and cocomparability graphs. indent Previously, the authors have given an existential proof of the fact that every connected AT-free graph contains a dominating pair, that is, a pair of vertices such that every path joining them is a dominating set in the graph. The main contribution of this paper is a constructive proof of the existence of dominating pairs in connected AT-free graphs. The resulting simple algorithm, based on the well-known lexicographic breadth-first search, can be implemented to run in time linear in the size of the input, whereas the best algorithm previously known for this problem has complexity \$O(|V|^3)\$ for input graph \$G=(V,E)\$. In addition, we indicate how our algorithm can be extended to find, in time linear in the size of the input, all dominating pairs in a connected AT-free graph with diameter greater than 3. A remarkable feature of the extended algorithm is that, even though there may be \$O(|V|^2)\$ dominating pairs, the algorithm can compute and represent them in linear time. 1. Introduction 2. Background 3. Lexicographic breadth-fist search 4. The dominating pair algorithm 5. Computing dominated sets 6. Computing all dominating pairs 7. Conclusions lgorithms, dominating pairs, asteroidal triple-free graphs, lexicographic breadth-first search 05C85, 68R10
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https://byjus.com/questions/how-do-you-expand-x-1-3-using-binomial-expansion/
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# How do you expand (x-1)3 using binomial expansion?
A binomial expression is an expression containing two terms joined by either addition or subtraction sign. For instance, (x + y) and (2 – x) are examples of binomial expressions.
## Binomial Theorem
The Binomial Theorem states the algebraic expansion of exponents of a binomial, which means it is possible to expand a polynomial (a + b)n into the multiple terms.
$(a+b)^{n} =\sum_{k=0}^{n}\begin{pmatrix} n\\ k \end{pmatrix}a^{n-k}b^{k}$
### Binomial formula to expand (a+b)3
Binomial formula for (a+b)33C 0 a 3 b 0 +3C 1 a 2 b 1 +3C2a 1 b 2 +3C 3 a 0 b 3 + ……..
### Expand (x-1)3
So using the above formula we will expand (x-1)3
Here, a=x and b=-1.
3C 0 x 3 +3C 1 x 2×(-1) 1 +3C 2 x 1 ×(-1) 2 +3C 3 ×(-1) 3
We know that
3C 0 =3C 3 =1
3C 1 =3C 2 =3
On substituting above values in the above equation we get
x 3 -3x 2 +3x – 1
(x-1)3= x 3 -3x 2 +3x – 1
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# How To Find The Exterior Angle Of A Nonagon?
Answer: The measure of an exterior angle of a regular nonagon is 40 degree. Let’s understand the solution. Explanation: By using the formula (n – 2) × 180 / n, we can find the measure of an interior angle of a regular polygon. What is the sum of the exterior angle of a triangle?
## Properties of Polygons
### What is the exterior angle of nonagon?
40 degreeAnswer: The measure of an exterior angle of a regular nonagon is 40 degree.
### How do you find the measure of an exterior angle of a nonagon?
Answer: heyyy, Therefore to find the measure of one exterior angle of any regular (all angles are congruent) polygon, divide 360 by the number of angles. In the case of a regular nonagon, the measure of one exterior angle is 40 degrees: 360 / 9 = 40 degrees.
### How do you find the external angle?
The formula for calculating the size of an exterior angle is: exterior angle of a polygon = 360 ÷ number of sides.
### Can you find the exterior angles of any irregular nonagon?
You cannot find the exterior angles of any irregular nonagon. If it is a regular one, then you can. A regular nonagon has 9 sides and so 9 exterior angles all of which are equal to 360/9 = 40 degrees. The interior angle = 180 - 40 = 140 degrees.
### What is the sum of the angles of a nonagon?
The sum of the measures of the exterior angles of a nonagon is 360°. A nonagon is a 9-sided polygon, however, the number of sides of a polygon... See full answer below. Become a Study.com member to unlock this answer! The sum of the measures of the exterior angles of a nonagon is 360°.
### What is the interior angle of a nonagon polygon?
Nonagon is one of the types of polygonal shapes. The nonagon is said to be a 9 sided polygon. It is also called Enneagon. The polygon sides are always straight and it is said to be a two dimensional shape. The interior angle of the regular polygon is 140 degree.
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# Is there a winning strategy for this game?
Andy and Bob play a game using a long straight row of squares, alternating turns. When it’s Andy’s turn, he writes an A in one of the blank squares. When Bob takes a turn, he writes a B in some blank square. (Once a letter is written in a square, neither player can use that square again.) A player wins the game when his initial is written in 4 equally-spaced squares. For example, suppose the following board is the result of several turns: (below)
_ _ B B _ A B A _ _ _ A B _ _ A
^
|
Andy can win by writing A in the indicated square. (Four A’s with spacing 2) Bob can win by writing B in that same square. (Four B’s with spacing 3)
• If Andy goes first, find a strategy Andy can use that guarantees that he wins. How many moves must Andy make to get 4 in a row, no matter what moves Bob makes? (Can Andy always win in just 4 moves?) Justify your answer.
• How many squares are needed in the game board to allow Andy’s strategy to work?
-
Welcome to math.SE: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. – Zev Chonoles Apr 19 '12 at 1:20
@ZevChonoles: (+1) I think this sort of proactive communication with new users is helpful. I realize a moderator can't and shouldn't be expected to do this in every instance. But, I wanted you to know that at least one regular user is appreciative of such efforts. – cardinal Apr 19 '12 at 2:14
It appears that the original question about a game has been completely overwritten with a question about putting spheres in spheres. – Greg Martin Apr 20 '12 at 0:40
@Greg, I rolled it back and flagged it for moderator attention. – Gerry Myerson Apr 20 '12 at 0:43
@Blue Eyes, if you want to ask another question, use "ask question" on top; do not edit this question. – sdcvvc Apr 20 '12 at 1:17
## 1 Answer
Andy wins in 4 moves. Denote Andy's first move as position $0$. Without loss of generality, Bob places his move at $n$, then Andy plays at $2(n+1)$. If Bob does not play at $n+1$, $-(n+1)$ or $3(n+1)$, Andy plays at $n+1$ and wins in the next move by either playing at $-(n+1)$ or $3(n+1)$. If Bob did play one of the mentioned moves, then Andy wins by playing at $4(n+1)$ followed by either $6(n+1)$ or $-2(n+1)$.
Minimal board size looks slightly more annoying to figure out at first glance; considering that Andy's second move could happen at the opposite side of Bob's first, it looks like Bob can only prevent a spacing of $1$ and $2$, so that would mean a board size of $21$ would suffice, but I don't see a quick way to prove that without looking at several different cases, and it's entirely possible I overlooked something with that.
-
I think that with 25 board you can guarantee a win in 4 moves, and under it is not possible (but maybe you can win with more moves). If A play in x, and B in x+2, A should play his next move in x+6 or x-6, so at least 6*4+1 space are requested. – carlop Apr 19 '12 at 15:34
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# 41.21 kg to lbs - 41.21 kilograms to pounds
Do you need to know how much is 41.21 kg equal to lbs and how to convert 41.21 kg to lbs? Here you go. You will find in this article everything you need to make kilogram to pound conversion - theoretical and practical too. It is also needed/We also want to point out that whole this article is dedicated to one number of kilograms - that is one kilogram. So if you need to know more about 41.21 kg to pound conversion - keep reading.
Before we get to the practice - this is 41.21 kg how much lbs conversion - we will tell you a little bit of theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 41.21 kg to lbs? 41.21 kilograms it is equal 90.8524981702 pounds, so 41.21 kg is equal 90.8524981702 lbs.
## 41.21 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, formally known as International System of Units (in abbreviated form SI).
At times the kilogram can be written as kilogramme. The symbol of this unit is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. This definition was simply but hard to use.
Then, in 1889 the kilogram was described by the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was substituted by a new definition.
Today the definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It is also divided to 100 decagrams and 1000 grams.
## 41.21 kilogram to pounds
You know some information about kilogram, so now we can move on to the pound. The pound is also a unit of mass. It is needed to underline that there are more than one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. To be honest, this unit is used also in another systems. The symbol of this unit is lb or “.
The international avoirdupois pound has no descriptive definition. It is defined as exactly 0.45359237 kilograms. One avoirdupois pound is divided into 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 41.21 kg?
41.21 kilogram is equal to 90.8524981702 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 41.21 kg in lbs
The most theoretical section is already behind us. In this part we are going to tell you how much is 41.21 kg to lbs. Now you learned that 41.21 kg = x lbs. So it is high time to know the answer. Just look:
41.21 kilogram = 90.8524981702 pounds.
That is an exact result of how much 41.21 kg to pound. You may also round off this result. After rounding off your outcome is exactly: 41.21 kg = 90.662 lbs.
You learned 41.21 kg is how many lbs, so look how many kg 41.21 lbs: 41.21 pound = 0.45359237 kilograms.
Of course, this time you may also round off the result. After it your result is as following: 41.21 lb = 0.45 kgs.
We also want to show you 41.21 kg to how many pounds and 41.21 pound how many kg outcomes in tables. Look:
We are going to begin with a chart for how much is 41.21 kg equal to pound.
### 41.21 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
41.21 90.8524981702 90.6620
Now look at a chart for how many kilograms 41.21 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
41.21 0.45359237 0.45
Now you know how many 41.21 kg to lbs and how many kilograms 41.21 pound, so it is time to go to the 41.21 kg to lbs formula.
### 41.21 kg to pounds
To convert 41.21 kg to us lbs a formula is needed. We are going to show you two formulas. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 90.8524981702 outcome in pounds
The first formula will give you the most exact result. In some cases even the smallest difference could be significant. So if you need a correct result - this formula will be the best for you/option to know how many pounds are equivalent to 41.21 kilogram.
So let’s go to the second version of a formula, which also enables calculations to learn how much 41.21 kilogram in pounds.
The second formula is down below, see:
Number of kilograms * 2.2 = the result in pounds
As you can see, the second version is simpler. It can be better choice if you need to make a conversion of 41.21 kilogram to pounds in easy way, for instance, during shopping. Just remember that your outcome will be not so correct.
Now we want to learn you how to use these two versions of a formula in practice. But before we are going to make a conversion of 41.21 kg to lbs we are going to show you easier way to know 41.21 kg to how many lbs totally effortless.
### 41.21 kg to lbs converter
An easier way to check what is 41.21 kilogram equal to in pounds is to use 41.21 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. It is based on longer formula which we showed you in the previous part of this article. Due to 41.21 kg pound calculator you can easily convert 41.21 kg to lbs. You only have to enter amount of kilograms which you need to convert and click ‘convert’ button. The result will be shown in a second.
So let’s try to calculate 41.21 kg into lbs using 41.21 kg vs pound converter. We entered 41.21 as a number of kilograms. This is the result: 41.21 kilogram = 90.8524981702 pounds.
As you see, our 41.21 kg vs lbs converter is intuitive.
Now we can move on to our main issue - how to convert 41.21 kilograms to pounds on your own.
#### 41.21 kg to lbs conversion
We will begin 41.21 kilogram equals to how many pounds conversion with the first formula to get the most exact result. A quick reminder of a formula:
Amount of kilograms * 2.20462262 = 90.8524981702 the outcome in pounds
So what have you do to know how many pounds equal to 41.21 kilogram? Just multiply number of kilograms, this time 41.21, by 2.20462262. It gives 90.8524981702. So 41.21 kilogram is exactly 90.8524981702.
You can also round off this result, for instance, to two decimal places. It gives 2.20. So 41.21 kilogram = 90.6620 pounds.
It is time for an example from everyday life. Let’s calculate 41.21 kg gold in pounds. So 41.21 kg equal to how many lbs? As in the previous example - multiply 41.21 by 2.20462262. It gives 90.8524981702. So equivalent of 41.21 kilograms to pounds, when it comes to gold, is 90.8524981702.
In this example it is also possible to round off the result. It is the result after rounding off, in this case to one decimal place - 41.21 kilogram 90.662 pounds.
Now we can go to examples converted with a short version of a formula.
#### How many 41.21 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 90.662 the outcome in pounds
So 41.21 kg equal to how much lbs? And again, you need to multiply number of kilogram, in this case 41.21, by 2.2. Let’s see: 41.21 * 2.2 = 90.662. So 41.21 kilogram is 2.2 pounds.
Let’s make another calculation using shorer version of a formula. Now calculate something from everyday life, for example, 41.21 kg to lbs weight of strawberries.
So let’s calculate - 41.21 kilogram of strawberries * 2.2 = 90.662 pounds of strawberries. So 41.21 kg to pound mass is exactly 90.662.
If you know how much is 41.21 kilogram weight in pounds and are able to convert it using two different formulas, we can move on. Now we want to show you these results in tables.
#### Convert 41.21 kilogram to pounds
We realize that results shown in tables are so much clearer for most of you. We understand it, so we gathered all these results in charts for your convenience. Thanks to this you can quickly compare 41.21 kg equivalent to lbs results.
Start with a 41.21 kg equals lbs chart for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
41.21 90.8524981702 90.6620
And now let’s see 41.21 kg equal pound chart for the second formula:
Kilograms Pounds
41.21 90.662
As you can see, after rounding off, if it comes to how much 41.21 kilogram equals pounds, the results are not different. The bigger number the more significant difference. Remember it when you need to make bigger number than 41.21 kilograms pounds conversion.
#### How many kilograms 41.21 pound
Now you know how to convert 41.21 kilograms how much pounds but we are going to show you something more. Are you curious what it is? What about 41.21 kilogram to pounds and ounces conversion?
We are going to show you how you can calculate it step by step. Let’s start. How much is 41.21 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, this time 41.21, by 2.20462262. So 41.21 * 2.20462262 = 90.8524981702. One kilogram is 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To calculate how much 41.21 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So final result is equal 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then your outcome is equal 2 pounds and 33 ounces.
As you can see, conversion 41.21 kilogram in pounds and ounces quite easy.
The last conversion which we are going to show you is calculation of 41.21 foot pounds to kilograms meters. Both of them are units of work.
To convert it it is needed another formula. Before we give you it, let’s see:
• 41.21 kilograms meters = 7.23301385 foot pounds,
• 41.21 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to calculate 41.21 foot pounds to kilograms meters you have to multiply 41.21 by 0.13825495. It is equal 0.13825495. So 41.21 foot pounds is equal 0.13825495 kilogram meters.
You can also round off this result, for example, to two decimal places. Then 41.21 foot pounds will be equal 0.14 kilogram meters.
We hope that this calculation was as easy as 41.21 kilogram into pounds conversions.
This article was a huge compendium about kilogram, pound and 41.21 kg to lbs in calculation. Thanks to this calculation you learned 41.21 kilogram is equivalent to how many pounds.
We showed you not only how to make a calculation 41.21 kilogram to metric pounds but also two another conversions - to know how many 41.21 kg in pounds and ounces and how many 41.21 foot pounds to kilograms meters.
We showed you also another solution to do 41.21 kilogram how many pounds calculations, this is using 41.21 kg en pound calculator. This is the best option for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you are able to do 41.21 kilogram equal to how many pounds conversion - on your own or with use of our 41.21 kgs to pounds calculator.
It is time to make your move! Convert 41.21 kilogram mass to pounds in the way you like.
Do you need to do other than 41.21 kilogram as pounds calculation? For example, for 15 kilograms? Check our other articles! We guarantee that calculations for other numbers of kilograms are so simply as for 41.21 kilogram equal many pounds.
### How much is 41.21 kg in pounds
To quickly sum up this topic, that is how much is 41.21 kg in pounds , we gathered answers to the most frequently asked questions. Here you can find all you need to remember about how much is 41.21 kg equal to lbs and how to convert 41.21 kg to lbs . Have a look.
How does the kilogram to pound conversion look? The conversion kg to lb is just multiplying 2 numbers. How does 41.21 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
See the result of the conversion of 41.21 kilogram to pounds. The exact answer is 90.8524981702 lbs.
There is also another way to calculate how much 41.21 kilogram is equal to pounds with another, shortened version of the equation. Have a look.
The number of kilograms * 2.2 = the result in pounds
So now, 41.21 kg equal to how much lbs ? The result is 90.8524981702 lbs.
How to convert 41.21 kg to lbs in an easier way? You can also use the 41.21 kg to lbs converter , which will make whole mathematical operation for you and you will get an exact answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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# Examples for input number questions¶
\begin{question}
\text{\textbf{1. variables: randint, randadjustIf, function; corrector: solution{function with random variables} }\\
\textit{Determine the decimal expansion of $f$ to three decimal places.}}
\explanation{Think about what rounded off to three decimal places means.}
\type{input.number}
\precision{3}
\field{real}
\begin{variables}
\function{f}{a/b}
\randint{a}{11}{16}
\randint{b}{11}{16}
\end{variables}
\solution{f}
\end{question}
\begin{question}
\text{\textbf{2. variables: numbers drawn from pool of variables, function; corrector: solution{variable}}\\
\textit{Calculate $func$ and give the answer as a fraction \frac{a}{b} in its lowest terms.}}
\explanation{Do not forget to do cancellations in the answer.}
\type{input.number}
\field{integer}
\begin{variables}
\function{func}{a/b - c/d}
\end{variables}
\begin{pool}
\begin{variables}
\number{a}{7}
\number{b}{18}
\number{c}{16}
\number{d}{99}
\number{e}{5}
\number{f}{22}
\end{variables}
\begin{variables}
\number{a}{7}
\number{b}{20}
\number{c}{5}
\number{d}{28}
\number{e}{6}
\number{f}{35}
\end{variables}
\end{pool}
\text{a = }
\solution{e}
\text{b = }
\solution{f}
\end{question}
\begin{question}
\textit{What are the roots of f(x)=$f$ ?}}
\explanation{}
\type{input.number}
\field{real}
\begin{variables}
\function[expand]{f}{(x-a)(x-b)}
\randint[Z]{a}{-5}{5}
\randint[Z]{b}{-5}{5}
\end{variables}
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Document
# Connecting Perebor Conjectures: Towards a Search to Decision Reduction for Minimizing Formulas
## File
LIPIcs.CCC.2020.31.pdf
• Filesize: 0.65 MB
• 35 pages
## Acknowledgements
I want to thank Eric Allender, Mathew Katzman, Aditya Potukuchi, Michael Saks, Rahul Santhanam, and Ryan Williams for many helpful discussions regarding this work.
## Cite As
Rahul Ilango. Connecting Perebor Conjectures: Towards a Search to Decision Reduction for Minimizing Formulas. In 35th Computational Complexity Conference (CCC 2020). Leibniz International Proceedings in Informatics (LIPIcs), Volume 169, pp. 31:1-31:35, Schloss Dagstuhl – Leibniz-Zentrum für Informatik (2020)
https://doi.org/10.4230/LIPIcs.CCC.2020.31
## Abstract
A longstanding open question is whether there is an equivalence between the computational task of determining the minimum size of any circuit computing a given function and the task of producing a minimum-sized circuit for a given function. While it is widely conjectured that both tasks require "perebor," or brute-force search, researchers have not yet ruled out the possibility that the search problem requires exponential time but the decision problem has a linear time algorithm. In this paper, we make progress in connecting the search and decision complexity of minimizing formulas. Let MFSP denote the problem that takes as input the truth table of a Boolean function f and an integer size parameter s and decides whether there is a formula for f of size at most s. Let Search- denote the corresponding search problem where one has to output some optimal formula for computing f. Our main result is that given an oracle to MFSP, one can solve Search-MFSP in time polynomial in the length N of the truth table of f and the number t of "near-optimal" formulas for f, in particular O(N⁶t²)-time. While the quantity t is not well understood, we use this result (and some extensions) to prove that given an oracle to MFSP: - there is a deterministic 2^O(N/(log log N))-time oracle algorithm for solving Search-MFSP on all but a o(1)-fraction of instances, and - there is a randomized O(2^.67N)-time oracle algorithm for solving Search-MFSP on all instances. Intriguingly, the main idea behind our algorithms is in some sense a "reverse application" of the gate elimination technique.
## Subject Classification
##### ACM Subject Classification
• Theory of computation → Circuit complexity
• Theory of computation → Problems, reductions and completeness
##### Keywords
• minimum circuit size problem
• minimum formula size problem
• gate elimination
• search to decision reduction
• self-reducibility
## Metrics
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## References
1. Eric Allender. The new complexity landscape around circuit minimization. In Alberto Leporati, Carlos Martín-Vide, Dana Shapira, and Claudio Zandron, editors, Language and Automata Theory and Applications - 14th International Conference, LATA 2020, Milan, Italy, March 4-6, 2020, Proceedings, volume 12038 of Lecture Notes in Computer Science, pages 3-16. Springer, 2020. URL: https://doi.org/10.1007/978-3-030-40608-0_1.
2. Eric Allender, Harry Buhrman, Michal Koucký, Dieter van Melkebeek, and Detlef Ronneburger. Power from random strings. SIAM J. Comput., 35(6):1467-1493, 2006.
3. Eric Allender and Bireswar Das. Zero knowledge and circuit minimization. Inf. Comput., 256:2-8, 2017.
4. Eric Allender, Michal Koucký, Detlef Ronneburger, and Sambuddha Roy. The pervasive reach of resource-bounded kolmogorov complexity in computational complexity theory. Journal of Computer and System Sciences, 77(1):14-40, 2011.
5. David Buchfuhrer and Christopher Umans. The complexity of boolean formula minimization. J. Comput. Syst. Sci., 77(1):142-153, 2011.
6. Marco L. Carmosino, Russell Impagliazzo, Valentine Kabanets, and Antonina Kolokolova. Learning algorithms from natural proofs. In Proceedings of the 31st Conference on Computational Complexity, 2016.
7. Alexander Golovnev, Rahul Ilango, Russell Impagliazzo, Valentine Kabanets, Antonina Kolokolova, and Avishay Tal. Ac^0[p] lower bounds against MCSP via the coin problem. In ICALP, volume 132 of LIPICS, pages 66:1-66:15, 2019.
8. Shuichi Hirahara. Non-black-box worst-case to average-case reductions within NP. In 59th IEEE Annual Symposium on Foundations of Computer Science, FOCS, pages 247-258, 2018.
9. Shuichi Hirahara, Igor Carboni Oliveira, and Rahul Santhanam. NP-hardness of minimum circuit size problem for OR-AND-MOD circuits. In 33rd Computational Complexity Conference, CCC, volume 102, pages 5:1-5:31, 2018.
10. Valentine Kabanets and Jin-Yi Cai. Circuit minimization problem. In Proceedings of the Thirty-Second Annual ACM Symposium on Theory of Computing, STOC ’00, page 73–79, 2000.
11. S. A. Lozhkin. Tighter bounds on the complexity of control systems from some classes. Mat. Voprosy Kibernetiki 6, pages 189-214 (in Russian), 1996.
12. Oleg B. Lupanov. Complexity of formula realization of functions of logical algebra. Problemy Kibernetiki, 3:61-80, 1960.
13. William J. Masek. Some NP-complete set covering problems. Unpublished Manuscript, 1979.
14. Cody D. Murray and R. Ryan Williams. On the (non) np-hardness of computing circuit complexity. Theory of Computing, 13(1):1-22, 2017.
15. Erkki Mäkinen. Generating random binary trees — a survey. Information Sciences, 115(1):123-136, 1999.
16. Nicholas Pippenger. Information theory and the complexity of boolean functions. Mathematical Systems Theory, 10:129-167, January 1977.
17. Alexander A. Razborov and Steven Rudich. Natural proofs. J. Comput. Syst. Sci., 55(1):24-35, 1997.
18. Michael Rudow. Discrete logarithm and minimum circuit size. Inf. Process. Lett., 128:1-4, 2017.
19. Rahul Santhanam. Pseudorandomness and the minimum circuit size problem. In Thomas Vidick, editor, 11th Innovations in Theoretical Computer Science Conference, ITCS, volume 151 of LIPIcs, pages 68:1-68:26, 2020.
20. B. A. Trakhtenbrot. A survey of russian approaches to perebor (brute-force searches) algorithms. IEEE Ann. Hist. Comput., 6(4):384–400, October 1984.
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# Refer to the Metropolis Health System (MHS) comparative financial statements at the back of the Examples and Exercises section.
Question 2:Example 14B: Trend Analysis
Trend analysis allows comparison of figures over time. Reread the chapter text about trend analysis and examine the difference columns shown in Table 14-3.
Practice Exercise 14-2: Trend Analysis
The worksheet below shows the assets of Hospital A over two years.
Perform trend analysis for the assets of Hospital A.
Hospital A
Year 1 Year 2
Current Assets \$1,600,000 \$ 2,000,000
Property, Plant, and Equipment 6,000,000 7,500,000
Other Assets 400,000 500,000
Total Assets \$8,000,000 \$10,000,000
Solution to Practice Exercise 14-2
Hospital A
Year 1 Year 2 Difference
Current Assets \$1,600,000 \$ 2,000,000 \$ 400,000 25%
Property, Plant, and Equipment 6,000,000 7,500,000 1,500,000 25%
Other Assets 400,000 500,000 100,000 25%
Total Assets \$8,000,000 \$10,000,000 \$2,000,000 —
Note: The worksheet below shows Hospital A with both common sizing and trend analysis:
Hospital A
Year 1 Year 2 Difference
Current Assets \$1,600,000 20% \$ 2,000,000 20% \$ 400,000 25%
Property, Plant,
and Equipment 6,000,000 75% 7,500,000 75% 1,500,000 25%
Other Assets 400,000 5% 500,000 5% 100,000 25%
Total Assets \$8,000,000 100% \$10,000,000 100% \$2,000,000 —
Now it's Your Turn: Assignment Exercise 14-2: Trend Analysis
Refer to the Metropolis Health System (MHS) comparative financial statements at the back of the Examples and Exercises section.
Perform trend analysis on the MHS Statement of Revenue and Expenses.
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### Home > PC3 > Chapter 7 > Lesson 7.2.5 > Problem7-132
7-132.
Rewrite the following fraction as a sum of two fractions with linear denominators by using partial fraction decomposition.
$\frac{5x-19}{(x-3)(x-5)}$
$\frac{A}{x-3}+\frac{B}{x-5}=\frac{5x-19}{\left(x-3\right)\left(x-5\right)}$
$\frac{A}{x-3}\cdot\frac{x-5}{x-5}+\frac{B}{x-5}\cdot\frac{x-3}{x-3}=\frac{5x-19}{(x-3)(x-5)}$
$Ax-5A+Bx-3B=5x-19$
Solve the system:
$\qquad A+B=5\\ -5A-3B=-19$
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# How to Calculate How Much Carbohydrates to Consume
#### Step-by-step Instructions
##### Step 1
First off, you will need to know what your daily intake for calories should be. A Recommended Caloric Intake Calculator can be found online through a simple search. By entering your Age, Height (your real height guys!), Gender, and Activity Level (be honest!); you will get an estimate of where your daily caloric intake should be for general health.
##### Step 2
Now that you have your Caloric intake you can find the percent of those calories that should come from carbohydrates. The most recent research states that your diet should contain 45-65% of your total calories coming from carbohydrates. You also need to know that 1 gram of Carbohydrates converts to 4 calories.
##### Step 3
The next step is a simple calculation. For example, if a person’s caloric intake for the day 2,000 total calories and they want to get 60% of those calories from Carbs they would take .6 x 2000 = 1200 is 60% of 2000 and the amount of calories from carbs to take in.
##### Step 4
Now we need to find out how many grams of carbs are in 1200 calories in order to find out how many grams of carbohydrates this person should have daily. We find that by knowing 1g Carbs = 4 calories, so 1200 / 4 = 300 grams of carbohydrates/ day for someone looking to get 60% of their 2000 Daily caloric intake from carbs.
##### Step 5
Why the range? Carbohydrates can be a friend or foe depending on how they are being used. A marathoner’s diet, for example, will crowd on the higher side with 65% of their daily calories being from carbohydrates. This is because the endurance athlete uses these carbs for energy and with daily training burns through them steadily leaving little left at the end of the day to be stored. The average person getting far less activity through the day will not need that amount of carbs for energy. Any additional carbohydrates they have left will be stored in the body as fat (not a desirable outcome). Instead this sedentary to inactive person should look for a lower percent of their diet to come from carbs (while keeping within the healthy recommended range).
#### Special Attention
##### Difficulties people often experience or parts that need special attention to do it right.
All carbohydrates are not created equal. A doughnut is not the same energy source as an apple. For overall health you should look for your carbs to come from sources high in fiber such as fruits and vegetables. Avoiding processed/ refined carbohydrates will help you to get more energy and health benefits from your daily carb intake.
#### Stuff You'll Need
Brand Product Price
Author Title Price
New Rules of Lifting \$12.82
#### This Student Author
Ashley Tryban
##### From
University of South Florida
#### This Student Author's Background
Carbohydrates have gotten a bad rap in today's society; however to an athlete they can mean the difference between gold or the gutter. By learning how to calculate you daily intake needs and selecting healthy sources for your carbs you can help your energy levels whether you are running a marathon or just staying awake during class!
##### When did you first do this & how did you get started?
My sophomore year of high school I collapsed at a track meet after running the 4x800 Relay, the mile, and the 2 mile at our regional meet. All of these events took place because I wasn't taking in the appropriate calories and carbs. Ever since, I have had an interest in nutrition and its effect on athletics.
• ### Earn Money
If you are a college student, you can turn your know-how into cash.
Using our easy "Self-Publishing Tool," students submit "tips" for training and playing selected sports, weightlifting, careers, relationships or any of life’s lessons.
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# Dimensions Mathematics Textbook 8A (Common Core State Standards Edition)
Singaporemath.com Inc / 2012 / Paperback
\$23.00 (CBD Price)
Availability: In Stock
CBD Stock No: WW350627
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## Product Description
The Dimensions Mathematics Common Core Textbooks emphasize the empowerment of students to learn math independently and effectively, a variety of approaches are used to help students master integrated pre-algebra, algebra, geometry, trigonometry, and advanced math topics. This series follows the Singapore Mathematics Framework as well as the topics in the Common Core State Standards. Teacher involvement is required to teach this program. Dimensions Mathematics 8A is designed to be used during the first semester of 8th grade. Softcover.
Features include:
• Worked examples followed by a "Try It!" question to ensure understanding.
• Lesson exercises that include the following exercise categories: Basic Practice (simple questions involving a direct application of the concepts); Further Practice (more challenging questions); Maths@Work (questions that apply mathematical concepts to real-life situations); and Brainworks (questions involving higher order thinking or an open-ended approach to problems).
• Each chapter is followed by a review exercise, an "Extend Your Learning Curve" activity, and questions requiring sentence or paragraph answers that encourage reflection.
The answer key at the back of the book provides answers to the Try It! and the problems in the exercises for the Basic Practice, Further Practice, and Maths@Work questions. It does not include answers to the class activities, Brainworks questions, or the Extend Your Learning Curve activities; those answers are found in the sold-separately Teacher's Guide.
## Product Information
Format: PaperbackVendor: Singaporemath.com IncPublication Date: 2012 ISBN-13: 9789814250627Availability: In Stock
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## Author/Artist Review
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# 97083 (number)
97,083 (ninety-seven thousand eighty-three) is an odd five-digits composite number following 97082 and preceding 97084. In scientific notation, it is written as 9.7083 × 104. The sum of its digits is 27. It has a total of 5 prime factors and 24 positive divisors. There are 52,272 positive integers (up to 97083) that are relatively prime to 97083.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 27
• Digital Root 9
## Name
Short name 97 thousand 83 ninety-seven thousand eighty-three
## Notation
Scientific notation 9.7083 × 104 97.083 × 103
## Prime Factorization of 97083
Prime Factorization 32 × 7 × 23 × 67
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 32361 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 97,083 is 32 × 7 × 23 × 67. Since it has a total of 5 prime factors, 97,083 is a composite number.
## Divisors of 97083
1, 3, 7, 9, 21, 23, 63, 67, 69, 161, 201, 207, 469, 483, 603, 1407, 1449, 1541, 4221, 4623, 10787, 13869, 32361, 97083
24 divisors
Even divisors 0 24 12 12
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 169728 Sum of all the positive divisors of n s(n) 72645 Sum of the proper positive divisors of n A(n) 7072 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 311.581 Returns the nth root of the product of n divisors H(n) 13.7278 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 97,083 can be divided by 24 positive divisors (out of which 0 are even, and 24 are odd). The sum of these divisors (counting 97,083) is 169,728, the average is 7,072.
## Other Arithmetic Functions (n = 97083)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 52272 Total number of positive integers not greater than n that are coprime to n λ(n) 66 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 9329 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 52,272 positive integers (less than 97,083) that are coprime with 97,083. And there are approximately 9,329 prime numbers less than or equal to 97,083.
## Divisibility of 97083
m n mod m 2 3 4 5 6 7 8 9 1 0 3 3 3 0 3 0
The number 97,083 is divisible by 3, 7 and 9.
## Classification of 97083
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
## Base conversion (97083)
Base System Value
2 Binary 10111101100111011
3 Ternary 11221011200
4 Quaternary 113230323
5 Quinary 11101313
6 Senary 2025243
8 Octal 275473
10 Decimal 97083
12 Duodecimal 48223
20 Vigesimal c2e3
36 Base36 22wr
## Basic calculations (n = 97083)
### Multiplication
n×y
n×2 194166 291249 388332 485415
### Division
n÷y
n÷2 48541.5 32361 24270.8 19416.6
### Exponentiation
ny
n2 9425108889 915017846270787 88832677569506814321 8624142836480430054725643
### Nth Root
y√n
2√n 311.581 45.9601 17.6517 9.94097
## 97083 as geometric shapes
### Circle
Diameter 194166 609990 2.96099e+10
### Sphere
Volume 3.83282e+15 1.18439e+11 609990
### Square
Length = n
Perimeter 388332 9.42511e+09 137296
### Cube
Length = n
Surface area 5.65507e+10 9.15018e+14 168153
### Equilateral Triangle
Length = n
Perimeter 291249 4.08119e+09 84076.3
### Triangular Pyramid
Length = n
Surface area 1.63248e+10 1.07836e+14 79267.9
## Cryptographic Hash Functions
md5 de04fd7b8e7499f228171212f3e60307 fafdbdc284c7c8f7170ce45da12a0a8565cd8591 84e9d43ca93e5e2ba6da90df760062185363e919815430a267f8b9b8816b30dd c581defba540e8ec6c4973ccb9933e77ebcd0c9303c6d727cff8d497d994a3b9bddc6ab8eee6493917ad51f8f8042892507874393568c6b6cc369a2b1aca5f42 c3f5a5fe7820e87d4fa0b6d6fb5fca62a3fb2f92
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Generators of p-groups
Let $G$ be a finite $p$-group. Since we can embed $Z_2(G)/Z(G)$ in $Hom(G,Z(G))$, we have $d_2 \leq d(G)d(Z(G))$; where $d_2(G)=d(Z_2(G)/Z(G))$ and $d(G)$ denotes the minimal number of generators of $G$. The question is, does the equality $d_2 = d(G)d(Z(G))$ imply that $Z(G)$ is cyclic?
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I think that the answer is affirmative. If you have not an answer, can you suggest a way to tackle this question? – Yassine Guerboussa Mar 2 '13 at 18:11
But why do you think the answer is yes? – Derek Holt Mar 2 '13 at 22:00
For a cyclic group, isn't $d_2(G)$ zero? – Will Sawin Mar 2 '13 at 23:23
Ok, Prof. Derek Holt; The order of G is greater than the products of the p^{d_i], where d_i is the number of generators of the ith term of the upper central series. Still n is greater than the sum of d_1 and d_2 and the others replaced by 1 ( the last replaced by 2, since the last factor can not be cyclic). Now under our condition, it follows easily that $d_1(d(G)+1) \leq n-c+1$( this inequality is due th A. Abdollahi). Now if G is a counter example, then the coclass of G is at least 5. Also, the class of G is greater than 2(otherwise $d_2 \leq d(G)$). therefore the order of G is at least p^8 – Yassine Guerboussa Mar 3 '13 at 7:32
It is not hard to see that our counter example can not be powerful nor p-central. and with some work it can not be a direct product of two p-groups. – Yassine Guerboussa Mar 3 '13 at 7:35
OK, here is an example of a group $Q$ of class 3 and order $p^{17}$, which will work for $p \ge 5$. We have $d(Q)=5$, $d(Z(Q)) = 2$, $d(Z_2(Q))/Z(Q)) = 10$, with $Z(Q) = \langle Q.16, Q.17 \rangle$ and $Z_2(Q) = \langle Q.6,\ldots,Q.17 \rangle$. All generators have order $p$ - in fact $Q$ has exponent $p$. All pairs of generators commute except for those in the list below. (This is Magma output.)
Q.2^Q.1 = Q.2 * Q.6,
Q.3^Q.1 = Q.3 * Q.7,
Q.3^Q.2 = Q.3 * Q.8,
Q.4^Q.1 = Q.4 * Q.9,
Q.4^Q.2 = Q.4 * Q.10,
Q.4^Q.3 = Q.4 * Q.11,
Q.5^Q.1 = Q.5 * Q.12,
Q.5^Q.2 = Q.5 * Q.13,
Q.5^Q.3 = Q.5 * Q.14,
Q.5^Q.4 = Q.5 * Q.15,
Q.6^Q.1 = Q.6 * Q.16,
Q.7^Q.1 = Q.7 * Q.17,
Q.8^Q.2 = Q.8 * Q.16,
Q.9^Q.2 = Q.9 * Q.17,
Q.10^Q.1 = Q.10 * Q.17,
Q.10^Q.3 = Q.10 * Q.16,
Q.11^Q.2 = Q.11 * Q.16,
Q.11^Q.3 = Q.11 * Q.17,
Q.12^Q.4 = Q.12 * Q.16,
Q.13^Q.4 = Q.13 * Q.17,
Q.14^Q.5 = Q.14 * Q.16,
Q.15^Q.1 = Q.15 * Q.16,
Q.15^Q.2 = Q.15 * Q.17,
Q.15^Q.5 = Q.15 * Q.17.
The conditions that you listed on such an example just mean that examples are moderately large, and so are more difficult to construct. They do not provide any genuine evidence that there are no such examples. There was a conjecture about $p$-groups called the class-breadth conjecture that was open for a long time, but as soon as it became possible to use computers to study larger groups, it became relatively easy to find counterexamples.
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@Derek: Is it possible to have a 2-group of class at least 4 satisfying the required condition? Do you expect that such a 2-group exists? Only I would like to know your opinion? Thanks in advance. – Alireza Abdollahi Mar 3 '13 at 15:43
Wow, I have thinked about this problem for more than a year. I am really grateful Prof. Derek. unfortunately, I know almost any thing about Magma ( also Gap), I try to check your example without computer. – Yassine Guerboussa Mar 3 '13 at 17:38
Dear Prof. Abdollahi, Now I think that there is a great possibility to find a counter example to the noninner conjecture. And why not among those in small libraries of groups. I don't think that are all checked. – Yassine Guerboussa Mar 3 '13 at 17:43
@Yassine: Derek's groups are of exponent $p$ and so they do not provide any counterexample to the noninner conjecture: Every non-abelian finite $p$-group has a noninner automorphism of order $p$. Every possible counterexample to this conjecture satisfies the property mentioned by Yassine. – Alireza Abdollahi Mar 3 '13 at 19:43
@Yassine: Another point is that as you mentioned above a possible counterexample must have order at least $p^8$, in this case GAP cannot help as the small group library is limited to groups of order at most $p^6$. But for the case of 2-groups of orders 512 or 1024 I do not know as some experts have access to the classification of these groups. – Alireza Abdollahi Mar 3 '13 at 19:47
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• ### 院校库
ydyxj
GRE数学！求解~
In a distribution of 8,500 different measurements of the variable x, 26.5 is the 56th percentile and 37.1 is the 78th percentile. Which of the following is closest to the number of measurements of x that are in the distribution such tha... 1 ? A. 3,500 D. 4,750 E,200 C,500 different measurements of the variable x, 26.5 is the 56th percentile and 37In a distribution of 8. Which of the following is closest to the number of measurements of x that are in the distribution such that 26.1 is the 78th percentile. 1,850 B. 2.5 ≤ x ≤37 展开
Phe_lix
（78-56）/100*8500比2200小。所以是A
• 最新问答
• 最新GRE资讯
GRE课程
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# NSTSE (National Science Talent Search Exam- Unified Council) Class 3: Questions 384 - 390 of 2216
Access detailed explanations (illustrated with images and videos) to 2216 questions. Access all new questions- tracking exam pattern and syllabus. View the complete topic-wise distribution of questions. Unlimited Access, Unlimited Time, on Unlimited Devices!
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## Question 384
### Question
MCQ▾
Each number in set S is related in the same way to the number below it in set T.
Set S 1 3 5 Set T 5 15 25
If the number in set S is 7, then find the number related to it in set T?
### Choices
Choice (4)Response
a.
49
b.
56
c.
40
d.
35
## Question 385
### Question
MCQ▾
Look at the following magic square and answer the following questions:
Which number is least number in the given square?
### Choices
Choice (4)Response
a.
4
b.
1
c.
5
d.
2
## Question 386
### Question
MCQ▾
Jay is taller than Sunil. Raju is smaller than Manu but taller than jay. Which figure represents Manu?
### Choices
Choice (4)Response
a.
A
b.
B
c.
D
d.
C
## Question 387
### Question
MCQ▾
What numeral means the same as ?
### Choices
Choice (4)Response
a.
425
b.
490
c.
400
d.
495
## Question 388
MCQ▾
is equal to:
### Choices
Choice (4)Response
a.
55
b.
65
c.
75
d.
52
## Question 389
### Question
MCQ▾
Look at the symbols used for each place and find out the hidden number?
### Choices
Choice (4)Response
a.
231
b.
312
c.
321
d.
213
## Question 390
### Question
MCQ▾
Which figure is on the 9th place from Start?
### Choices
Choice (4)Response
a.
books
b.
helicopter
c.
bike
d.
sun
Developed by:
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## DEV Community
Berk Hakbilen
Posted on • Updated on
# Evaluation for Regression Models in Machine Learning
Model evaluation is very important since we need to understand how well our model is performing. In comparison to classification, performance of a regression model is slightly harder to determine because, unlike classification, it is almost impossible to predict the exact value of a target variable. Therefore, we need a way to calculate how close our prediction value is to the real value.
There are different model evaluation metrics that are used popularly for regression models which we are going to dive into in the following sections.
## Mean Absolute Error
Mean absolute error is a very intuitive and simple technique, therefore also popular. It is basically the average of the distances between the predicted and the true values. Basically the distances between the predicted and the real values are also the error terms. The overall error for the whole data is the average of all prediction error terms. We take the absolute of the distances/errors to prevent negative and positive terms/errors from cancelling each other.
• MAE is not sensitive to outliers. Use MAE when you do not want outliers to play a big role in error calculated.
• MAE is not differentiable globally. This is not convenient when we use it as a loss function, due to the gradient optimization method.
## Mean Squared Error (MSE)
MSE is one of widely used metrics for regression problems. MSE is the the measure of average of squared distance between the actual values and the predicted values. Squared terms help to also take into consideration of negative terms and avoid cancellation of the total error between positive and negative differences.
• Graph of MSE is differantiable which means it can be easily used as a loss function.
• MSE can be decomposed into variance and bias squared. This helps us understand the effect of variance or bias in data to the overall error.
• The value calculated MSE has a different unit than the target variable since it is squared. (Ex. meter → meter²)
• If there exists outliers in the data, then they are going to result in a larger error. Therefore, MSE is not robust to outliers (this can also be an advantage if you are looking to penalize outliers).
Root Mean Squared Error (RMSE)
As the name already suggests, in RMSE we take the root of the mean of squared distances, meaning the root of MSE. RMSE is also a popularly used evaluation metric, especially in deep learning techniques.
• The error calculated has the same unit as the target variables making the interpretation relatively easier.
• Just like MSE, RMSE is also susceptible to outliers.
## R-Squared
R square is a different metric compared to the ones we have discussed until now. It does not directly measure the error of the model.
R-squared evaluates the scatter of the data points around the fitted regression line. It is the percentage of the target variable variation which the model considers compared to the actual target variable variance. It is also known as the “coefficient of determination” or goodness of fit.
As we can see above, R-squared is calculated by dividing the sum of squared error of predictions by the total sum of square, where predicted value is replaced by the mean of real values.
R-squared is always between 0 and 1. 0 indicates that the model does not explain any of the variation in the target variable around its mean value. The regression model basically predicts the mean of the parget variable. A value of 1 indicates, that the model explains all the variance in the target variable around its mean.
A larger R-squared value usually indicates that the regression model fits the data better. However, a high R-square model does not necessarily mean a good model.
``````import matplotlib.pyplot as plt
from sklearn.datasets import make_regression
from sklearn.linear_model import LinearRegression
import seaborn as sns; sns.set_theme(color_codes=True)
X, y = make_regression(n_samples = 80, n_features=1,
n_informative=1, bias = 50, noise = 15, random_state=42)
plt.figure()
ax = sns.regplot(x=X,y=y)
model = LinearRegression()
model.fit(X, y)
print('R-squared score: {:.3f}'.format(model.score(X, y)))
``````
``````X, y = make_regression(n_samples = 80, n_features=1,
n_informative=1, bias = 50, noise = 200, random_state=42)
plt.figure()
ax = sns.regplot(x=X,y=y)
model = LinearRegression()
model.fit(X, y)
print('R-squared score: {:.3f}'.format(model.score(X, y)))
``````
• R-square is a handy, and an intuitive metric of how well the model fits the data. Therefore, it is a good metric for a baseline model evaluation. However, due to the disadvantages we are going to discuss now, it should be used carefully.
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Mathematics Goes to the Movies
by Burkard Polster and Marty Ross
Death of a Neapolitan Mathematician (1992)
This movie was inspired by the mathematician Renato Caccioppoli (1904-1959). Two other mathematicians feature prominently in the movie. Pietro, one of Renato’s students who is about to become a professor himself and Father Simplicio, who is also a priest.
8:00
Renato drawing something mathematical on the wall.
13:00
Renato lecturing. He is proving the Sandwich Theorem from calculus
(three continuous functions f,g,and h. If f\leq g \leq h and \lim_{x\to x_0} f(x) = \lim_{x\to x_0} h(x)=l, then \lim_{x]to \x_0} g(x)=l. )
STUDENT: Professor, may I ask an off-subject question?
RENATO: Of course
STUDENT: Why was Einstein opposed to quantum theory?
RENATO: According to Einstein God doesn’t play dice, and then you see, the world of truth in physics as the one in mathematics is as closed as a sphere. If a new vision is profound, it’s an escape from this sort of prison. You can have some resistance to escape or you cannot see the reason for it.
14:30
Some of Renato’s students and colleagues talking at lunch.
FATHER SIMPLICIO: Did you read the Mathematical Review?
PIETRO: Yes, I saw it.
FATHER SIMPLICIO: The sixth slashing is a row.
PIETRO: Renato asks for it. He goes to international meetings and mocks the scientific world. Americans, Swiss….
26:00
Renato has checked some of Pietro’s mathematics
PIETRO: You’ve checked everything. You did it very quickly, but it took me forever
RENATO: One shouldn’t love math too much. It should love you…
PIETRO: Actually, you did it all over again.
RENATO: Yes, but selfishly. I should have underlined the errors and make you come to the solution. It would have taken time
….
Turn to formula 25. The exponent is four, not three.
30:00
Some of Renato’s scribblings
32:00
Renato doing some calculations in his apartment.
35:00
blackboard in his apartment
41:00
Pietro talking to father Simplicio.
PIETRO: So this is my work for the chair competition. Renato had to check it, but started from scratch. I’m grateful. It’s planned in a better manner, but you know Renato, he rushes sometimes. You have patience, Father Simplicio and you’ll be discreet about all this. Will you help me check it through?
FATHER SIMPLICIO: I would give you a hand but it’s too late. Send your own work in then.
PIETRO: There’s an extension. I phoned Rome this morning.
FATHER SIMPLICIO: Let’s see. A mathematician’s career starts too early. One, two analysis exams and one almost has to discover a theorem.
PIETRO: You know one doesn’t learn mathematics. It’s an eye you have inside. It shows you a field of vision.
Immediately. You start young.
FATHER SIMPLICIO: I have no field of vision. I know you have to take it a step at a time.
52:00
Father Simplicio talking to Renato.
FATHER SIMPLICIO: They’ve pointed out important passages of your work. They’re delighted. If they can handle it Look: “Thanks to his sets we’ll have a geometrical measure theory.”
1:01:00
Another lecture theatre scene: Blackboards not clear enough for deciphering … some integral formula
1:09:00
Student writing on the blackboard. Taylor expansion … remainder tends to 0…
RENATO: It’s nice to see that woman have brains, too. A five minute suspension for this young lady.
1:11:00
Renato examining another student. Actually, Father Simplicio is asking all the questions.
Father Simplicio asking a question: Give me an example where the hypotheses are not fulfilled, but where the function is implicit.
RENATO: Lauretano, 26.
STUDENT (protesting): You didn’t follow my exam!
RENATO: Did you want a 29? Lauretano, 29. End of the movie.
Obituary:… Renato was the first to study problems in functional analysis which in the thirties received a great boost with Schauder’s research in Poland and Leray’s in France.
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# Ch6. Triangles
In this online course, you will learn Definitions, examples, counter examples of similar triangles. 1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. 2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side. 3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar. 4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar. 5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar. 6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other. 7. (Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. 8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. 9. (Prove) In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angles opposite to the first side is a right angle. For further understanding of concepts and for examination preparation, this course has explanation of all NCERT Exercise questions and important NCERT example. This course also has self practice assignment for students to practice that have questions from Board’s Question Bank, RD Sharma, NCERT Exemplar etc. instead of only one book. The PDF of assignment can be downloaded within the course.
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Home / Length Conversion / Convert Break to Inch (US Survey)
# Convert Break to Inch (US Survey)
Please provide values below to convert break to inch (US survey) [in], or vice versa.
From: break To: inch (US survey)
### Break to Inch (US Survey) Conversion Table
BreakInch (US Survey) [in]
0.01 break3.937E+30 in
0.1 break3.937E+31 in
1 break3.937E+32 in
2 break7.874E+32 in
3 break1.1811E+33 in
5 break1.9685E+33 in
10 break3.937E+33 in
20 break7.874E+33 in
50 break1.9685E+34 in
100 break3.937E+34 in
1000 break3.937E+35 in
### How to Convert Break to Inch (US Survey)
1 break = 3.937E+32 in
1 in = 2.5400050800102E-33 break
Example: convert 15 break to in:
15 break = 15 × 3.937E+32 in = 5.9055E+33 in
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Physics Work,Power & Energy Level: High School
A 60kg skydiver moving at terminal speed falls 50m in 1 sec. What power is the skydiver expending in air?
45
Physics Simple Machines Level: High School
A lever is used to lift a heavy load. When a 50N force pushes one end of the lever down 1.2m, the load rises 0.2 m.Calculate the weight of the load.
44
Physics Work,Power & Energy Level: High School
A person dives from atop a flagpole into a swimming pool below. His potential energy at the top is 10,000J. What is his kinetic energy when his potential energy reduces to 1000J?
43
Physics Rotational Mechanics Level: High School
Consider a too-small space habitat that consists of a rotating cylinder of radius of 4m. If a man standing inside is 2m tall and his feet are at 1g, what is the g force at the elevation of his head?
42
Physics Force & Laws Of Motion Level: High School
A 15.0 kg uniform board 2.4 m long is pivoted so that it can rotate at a horizontal axis through the end (point A). If a 100-kg mass and is suspended 1.80 m from the pivoted end, what is the tension in the string ?
41
Physics Force & Laws Of Motion Level: High School
Two masses are suspended from a frictionless pulley as shown in "click here " The pulley itself if has a mass of 10.kg and its radius r = 10 cm.
(a) what is the magnitude of the acceleration of the suspended masses is m1 = 8 kg
(b) what are the tension in the strings? (Neglect the mass of the string.)
40
Physics Force & Laws Of Motion Level: High School
Two blocks are connected by a cable via a pulley which are massless and frictionless. Given the following: m=300.0 kg, m2= 100.0 kg, theta = 40.0 degree, micro = 0.350, determine:
a. The direction of acceleration (mathematically no guesses)
b. The value of acceleration
c. The tension in the cable
29
Physics Force & Laws Of Motion Level: High School
A box of mass 8.3 kg placed on a smooth horizontal table is acted on by two horizontal forces . The first of magnitude 0.3 Newton's acts due East , and the second of magnitude 0.3 Newton's acts due North . Find the velocity of the box after 14.9 seconds .
26
Physics Force & Laws Of Motion Level: High School
A stone of m = 0.8 kg is projected across a sheet of ice with an initial speed of 2.40m s^-1 . The coefficient of sliding friction between the stone and the ice is 0.0250 . How far will the stone travel before it come to rest ? Take g = 9.81m s^-2 .
24
Physics Force & Laws Of Motion Level: High School
Blocks A and B are moving towards each other along the X axis. A has a mass of 4.00 kg and a velocity of 30.0 m/s ( in the positive X direction), While B has a mass of 2.00 kg and a velocity of -25.0 m/s (in the negative direction). A) Assume they sugffer an elastic head-on collision and move off along the X axis. What will be the velocity of A and B , respectively , after the collision ? B) Using the information from a) and using impulse-momentum theorem, calculate the force acting on block A due to block B if the collision took a time of 0.256s C) If these same two blocks were to have a perfectly inelastic head-on collision, what would the velocity of the combined block system be and what would be the loss in KE?
23
Physics Force & Laws Of Motion Level: High School
A girl and her bicycle have a total mass of 40.0 kg. At the top of a hill her speed is 5.00 m/s the hill is 10.0 m high vertically and 100.0 m along the incline use work energy methods A) Take friction to be acting . If her speed was 10.0 m/s as she arrives at the bottom of the hill , how much mechanical energy is lost to friction? Note girl doesn't pedal . B) From your answer to part A calculate the force of the frictional resistance of the air , etc , that is removing useful mechanical energy from the system as the girl rolls down the hill . C) Now if the girl and her bike coast along level ground , how far will she coast if the coefficient of the Kinetic Friction is 0.300 ?
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2012-12-07 07:53:53
White_Owl
Full Member
Offline
### Distance to the wall of the box
The problem:
We have a squared box of size w*w.
Inside the box, we have a point at (x,y).
From the point we draw a line at some angle d.
If x, y, d, and w are known, find the distance to the closest wall of the box.
x and y are always positive and less then w (point is always inside the box). The d can be anything.
On the attached image I put two points. I need to know the length of green lines.
From a simple definitions of sin and cos, I have four distances to the four walls of the box
dl = (0 - x) / cos d
db = (0 - y) / sin d
dr = (w - x) / cos d
dt = (w - y) / sin d
From the set {dl, db, dr, dt} remove negative values (those are distances along the same line but in opposite directions).
And the final answer is the minimal from two remaining values.
Looks like the final answer is correct, but solutiona is fairly complicated. Maybe I am not seeing the obvious, but is it possible to reduce this algorithm to a one formula?
## #2 2012-12-07 17:23:53
John E. Franklin
Star Member
Offline
### Re: Distance to the wall of the box
Hi my friend. If you are allowed to use the absolute value operator, I do believe an elegant solution will arrive. For starters, let's couple together the dbottom with the dtop in an expression to find the one that is positive. And also let's couple together the dleft with the dright in an expression to find the one that is positive meaning non-negative. In these two cases, we simply need a "maximum finder" equation that uses the absolute value signs. After these two terms are resolved, they can be embedded in the larger single equation in which the "minimum of two values" is found, and that is the answer. Now let me give you an idea as to how one would find the max value of two numbers with the absolute value sign. See next post for more details...
Imagine for a moment that even an earthworm may possess a love of self and a love of others.
## #3 2012-12-07 17:29:47
John E. Franklin
Star Member
Offline
### Re: Distance to the wall of the box
max value of a and b = (a + b + |a - b|) / 2
minValue(C,D)= (C + D - |C - D|) / 2
Put it all together in one large equation and get...
minValue(maxValue(dTop,dBottom),maxValue(dLeft,dRight))
Now just expand this with you trigonometric equations you have
mentioned in your original post and I believe it is possible.
I hope you are allowed to use the absolute value operator though!!!!
Have a super week...
Imagine for a moment that even an earthworm may possess a love of self and a love of others.
## #4 2012-12-07 18:11:58
John E. Franklin
Star Member
Offline
### Re: Distance to the wall of the box
Here's an awkward way to avoid absolute value signs:
Abs(X) = (x**2)**(1/2) = (x^2)^(1/2)
Basically square root the square of the interested value.
Imagine for a moment that even an earthworm may possess a love of self and a love of others.
## #5 2012-12-08 04:42:15
White_Owl
Full Member
Offline
### Re: Distance to the wall of the box
#### John E. Franklin wrote:
minValue(maxValue(dTop,dBottom),maxValue(dLeft,dRight))
Yes, I am allowed to use min() and max() functions, and right now I actually have an equation similar to the one you showed here. It works perfectly but I still believe there is a way to avoid individual calculations of these distances.
I think, the ray tracing algorithms are dealing with the similar problem. Do we have any experts in 2.5D graphics here? I remember reading some articles on that subject, but it was too many years ago.
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This site uses cookies. By continuing to browse the ConceptDraw site you are agreeing to our Use of Site Cookies.
# ConceptDraw Solution Park
ConceptDraw Solution Park collects graphic extensions, examples and learning materials
## Contoh Flowchart
A Flowchart is a graphical representation of the algorithm, process or the step-by-step solution of the problem. There are ten various types of Flowcharts. The ConceptDraw allows you to draw the flowchart of any type.
The Contoh Flowchart included to Flowcharts solution are professional looking practical samples and you can quick and easy modify them, print, or publish on web.
## Flowcharts
The Flowcharts solution for ConceptDraw DIAGRAM is a comprehensive set of examples and samples in several varied color themes for professionals that need to represent graphically a process. Solution value is added by the basic flow chart template and shapes' libraries of flowchart notation. ConceptDraw DIAGRAM flow chart creator lets one depict the processes of any complexity and length, as well as design the Flowchart either vertically or horizontally.
## Food Court
Use the Food Court solution to create food art. Pictures of food can be designed using libraries of food images, fruit art and pictures of vegetables.
## Bar Graphs
The Bar Graphs solution enhances ConceptDraw DIAGRAM functionality with templates, numerous professional-looking samples, and a library of vector stencils for drawing different types of Bar Graphs, such as Simple Bar Graph, Double Bar Graph, Divided Bar Graph, Horizontal Bar Graph, Vertical Bar Graph, and Column Bar Chart.
## Line Graphs
How to draw a Line Graph with ease? The Line Graphs solution extends the capabilities of ConceptDraw DIAGRAM with professionally designed templates, samples, and a library of vector stencils for drawing perfect Line Graphs.
## Picture Graphs
Typically, a Picture Graph has very broad usage. They many times used successfully in marketing, management, and manufacturing. The Picture Graphs Solution extends the capabilities of ConceptDraw DIAGRAM with templates, samples, and a library of professionally developed vector stencils for designing Picture Graphs.
## Health Food
The Health Food solution contains the set of professionally designed samples and large collection of vector graphic libraries of healthy foods symbols of fruits, vegetables, herbs, nuts, beans, seafood, meat, dairy foods, drinks, which give powerful possi
## Workflow Diagrams
Workflow Diagrams solution extends ConceptDraw DIAGRAM software with samples, templates and vector stencils library for drawing the work process flowcharts.
## Flow Chart Symbols
Flowcharts are used in designing and documenting simple processes or programs. Like other types of diagrams, they help visualize what is going on and thereby help understand a process, and perhaps also find flaws, bottlenecks, and other less-obvious features within it. There are many different types of flowcharts, and each type has its own repertoire of boxes and notational conventions.
Flowchart diagrams consists of symbols of process, decision, data and document, data base, termination or initiation processes, processing loops and conditions. To create an drawn flowchart use professional flowchart maker of ConceptDraw DIAGRAM.
## Basic Line Graphs
This solution extends the capabilities of ConceptDraw DIAGRAM (or later) with templates, samples and a library of vector stencils for drawing Line Graphs.
## Pie Charts
Pie Charts are extensively used in statistics and business for explaining data and work results, in mass media for comparison (i.e. to visualize the percentage for the parts of one total), and in many other fields. The Pie Charts solution for ConceptDraw DIAGRAM offers powerful drawing tools, varied templates, samples, and a library of vector stencils for simple construction and design of Pie Charts, Donut Chart, and Pie Graph Worksheets.
## Basic Bar Graphs
This solution enhances ConceptDraw DIAGRAM (or later) with templates, samples and a library of vector stencils for drawing Bar Graphs.
## Safety and Security
Safety and security solution extends ConceptDraw DIAGRAM software with illustration samples, templates and vector stencils libraries with clip art of fire safety, access and security equipment.
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# Live Trailing Stop
Questions about MultiCharts and user contributed studies.
duration
Posts: 175
Joined: 20 Dec 2005
### Live Trailing Stop
Hello,
I am using a trailing stop of 5 points, with a trailing percentage of 10.
It looks like it works on backtesting, but not live.
The screenshot is a trade from the backtest: it enters long 1 contract at 1.5121, then exit at 1.5131.
But the same trade live is totally different: it still enters long at 1.5121, but then generates two trailing stop orders at 1.5124, exit with one, but go short with the other!
1. why did the trailing stop not exit at 1.5131 like in the backtest version?
2. why did it generate two orders?
They are programmed as follow:
Code: Select all
inputs: profitgt(50),pctrl(10), stopls(300),stoptype(1);
if stoptype = 1 then
begin
setstopposition;
setpercenttrailing(profitgt,pctrl);
setstoploss(stopls);
end;
if stoptype = 2 then
begin setstopcontract;
setpercenttrailing(profitgt,pctrl);
setstoploss(stopls);
end;
Attachments
trailing stop.png (8.45 KiB) Viewed 609 times
ct3232
Posts: 16
Joined: 25 Nov 2010
Has thanked: 1 time
### Re: Live Trailing Stop
I am having the same problem. I'm using "setpercenttrailing" and "setstoploss" and when applying the strategy to the MC charts, all entries and exits are calculated and shown correctly.
"inputs: Num(1),AvgLen(5),Threshold(250),PerCentTrail(10),StopAmt(700);
vars: MP(0);
Condition1 = ......................;
Condition2 = .......................;
If Condition1 and MP < 1 Then
Buy ("LE") Num contracts next bar at market;
If Condition2 and MP = 1 Then
Sell ("LX") Num contracts next bar at market;
setstopposition;
if MP <> 0 then
setpercenttrailing(Threshold,PerCentTrail);
setstoploss(StopAmt);"
However, in auto-trading with IB, I find the entries are occuring correctly, however the trailing stop does not even seem to be sent to IB. Funnily enough, the "setstoploss" order is sent.
The setstoploss order is obviously entered at the time the buy or sell signal is activated, whereas the setpercenttrailing order would only be entered when the trade has a certain accumulated profit.
Is there a setting I need to fix on the strategy properties, to allow for this subsequent trail order to be sent to IB. What am I missing? Any help would be much appreciated.
bomberone1
Posts: 189
Joined: 02 Nov 2010
Has thanked: 14 times
Been thanked: 12 times
### Re: Live Trailing Stop
I use two data set to avoid this.
data1 where order are sent at 1 minute
data2 where i do calculation at 60 minite.
Hope that this could help.
ct3232
Posts: 16
Joined: 25 Nov 2010
Has thanked: 1 time
### Re: Live Trailing Stop
Thanks Bomber - I worked around it by inserting specific sell and buytocover orders in my code - i cannot use setpercenttrailing.
TJ
Posts: 6873
Joined: 29 Aug 2006
Location: Global Citizen
Has thanked: 984 times
Been thanked: 1976 times
### Re: Live Trailing Stop
note:
EasyLanguage Essentials Programmers Guide
Built-in Stops.......... pg. 90
Built-in Stops
EasyLanguage includes built-in exit commands that may be included directly in your strategies. These special commands will be active even on the bar of entry; that is, they are evaluated on each tick (regardless of any setting for Intrabar Order Generation on the Calculation tab of the Format Strategy dialog). These stops are not written in EasyLanguage, but are part of the strategy engine.
The built-in stop commands are:
SetBreakEven - sets an exit stop at the entry price, after a minimum profit is achieved.
SetDollarTrailing - sets an exit stop a fixed number of dollars away from the peak profit.
SetPercentTrailing - sets an exit stop a fixed percent of the peak profit away from the peak profit, after a minimum profit is achieved.
SetProfitTarget - sets an exit order at a fixed dollar profit target.
SetStopLoss - sets a stop loss order at a fixed dollar risk from entry.
ct3232
Posts: 16
Joined: 25 Nov 2010
Has thanked: 1 time
### Re: Live Trailing Stop
TJ
Thanks - but as far as I can see, I've coded the trailing stop correctly. As you can see above, I have a relatively simple strategy, that does display 100% correctly on the MC chart.
However, IB seems to be the problem. They do get the initial entry orders processed quickly and the setstoploss order is entered. But the setpercenttrail stop is never sent to IB, or if it is, it is never entered as an order on the IB interface.
Its really becoming annoying; I am autotrading on the 24hr currency futures and I need to be sure that the trailing stop is working.
Another odd thing is that although my simple strategy calls for one contract to be entered, IB always enters two. Has anyone else had this problem.
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×
Ch-ch-ch-changes! Mathalicious is now Citizen Math . Please visit www.CitizenMath.com for the real-world math lessons you know and love. Alternatively, you can continue to access Mathalicious.com until the end of the school year, at which point Mathalicious will ride off into the sunset. For more details about the transition to Citizen Math, please click here.
Why do different jobs pay so differently? top-paid teacher earns in a year. On one hand, this seems unfair. On the other hand...it still seems unfair. But there’s a reason some jobs pay so much better than others.
In this lesson, students use unit rates to compare how much different professions make per year/day/hour and discuss ways to possibly equate compensation with social contribution.
### Students will
• Use ratio and rate reasoning to solve real world problems
• Calculate daily and hourly unit rates
• However, in other professions that generate value for society at large – e.g. a teacher increases students’ future earnings, the President boosts economic performance – that value is often not reflected in compensation.
### Before you begin
Students will find rates throughout this lesson. If they aren't familiar with the concept of a unit rate, they can reason their way through answers; simply adjust the timings to allow for lots of discussion of the mathematical reasoning.
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##### Assignment Instructions
Discuss and analyze the following transactions for X Ltd, using the concept of accounting equation (Assets, Liabilities and Equities)..
Internal Assignment Applicable for June 2021 Examination
Assignment Marks: 30
Instructions:
• All Questions carry equal marks.
• All Questions are compulsory
• All answers to be explained in not more than 1000 words for question 1 and 2 and for question 3 in not more than 500 words for each subsection. Use relevant examples, illustrations as far as possible.
• All answers to be written individually. Discussion and group work is not advisable.
• Students are free to refer to any books/reference material/website/internet for attempting their assignments, but are not allowed to copy the matter as it is from the source of reference.
• Students should write the assignment in their own words. Copying of assignments from other students is not allowed.
• Students should follow the following parameter for answering the assignment questions.
Assessment Parameter Weightage Assessment Parameter Weightage
Introduction 20% Understanding and usage of the formula 20%
Concepts and Application related to the question 60%
Procedure / Steps 50%
Conclusion 20%
1. Discuss and analyze the following transactions for X Ltd, using the concept of accounting equation (Assets, Liabilities and Equities).
1. Purchased Furniture for Rs675000
2. Capital Introduced by the business Owner by depositing 12 Lakhs in the bank account
3. Goods purchased on credit from Aman Enterprises for Rs105000
Internal Assignment Applicable for June 2021 Examination
4. Goods sold on credit for Rs 400000. The cost of the goods sold was Rs 300000
5. Purchased goods from Sneha Enterprises for Rs 600000 and made the payment from the business’s bank account
(5*2 = 10 Marks)
2. Love Doddle is a gifting enterprise of Ms. Dorati. The enterprise generates inflows by arranging gift hampers for the customer’s loved ones. The inflows arises from the sale of gift hampers Rs 505000 and from bank interest, dividend receipt Rs4200. Ms. Dorati is confused on how to record these inflows. She would like to understand from you about the concepts Revenue from operation and other income, so that she can record the information so as to prepare the profit and loss statement of the enterprise. Define, share examples, and elaborate on your understanding towards the terms Revenue from Operation and Other Income
(10 Marks)
3. The following information is given with respect to the ratio’s of two companies
Aman Ltd Roger Ltd
Current ratio 2:01 1.60:1
Quick Ratio 1.35:1 1:01
Return on investment 15% 13%
Debt Equity Ratio 2.5:1 1:01
a. Define the concepts of Current and Quick ratio’s and also, reflect on your understanding towards the financial performance of the companies by looking to the above information (2marks for defining and 3 marks for interpretation and reasoning) (5 Marks)
Internal Assignment Applicable for June 2021 Examination
b. Define the terms- Return on Investment and Debt equity ratio and also, reflect on your understanding towards the financial performance of the companies (2marks for defining and 3 marks for interpretation and reasoning) (5 Marks)
**********
Discuss and analyze the following transactions for X Ltd, using the concept of accounting equation (Assets, Liabilities and Equities).
Price (USD)
\$
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# Year 11 Permutations (arrangements in a circle) (1 Viewer)
#### hs17
##### New Member
3) Bob, Betty, Ben, Brad and Belinda are to be seated at a round table. In how many ways can this be done:
a if there are no restrictions,
b if Betty sits on Bob’s right-hand side,
c if Brad is to sit between Bob and Ben,
d if Belinda and Betty sit apart,
e if Ben and Belinda sit apart, but Betty sits next to Bob?
#### Pedro123
##### Member
Firstly, label them A,B,C,D,E. The question is intentionally pissing you off. I assume it means the seats at the table are not unique, that is A,B,C,D and E is the same as B,C,D,E and A.
a) This is a simple 4!, since position doesn't matter.
b) By this, I assume it means directly right. As such, there should be 3! ways to do it.
c) If the 3 people are defined, there are only 2 other people to be selected. but, brad and bob can be on either side, so there is actually 2*2 = 4.
d) With this, you can use b). If there are 3! for right, there are 3! for left. This is how many ways someone can sit next to someone. Take this away from 4!
d) Consider here the gaps. Between A and B, there must be one 1 person gap and 1 2 person gap. The 2 person gap must be C and D, which have 2 ways to be positioned. The other gap must only have E. As such, there are 2*2
Take with a pinch of salt though - I may have just been entirely wrong.
Last edited:
#### ultra908
##### Member
a) This is a simple 5!, since position doesn't matter.
Its 4!, because the five rotations of the table count as one arrangement.
#### Pedro123
##### Member
Its 4!, because the five rotations of the table count as one arrangement.
There it is. Changed now
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Published: 30 September 2016
# Lateral and torsional vibrations of cable-guided hoisting system with eccentric load
Jinjie Wang1
Guohua Cao2
Zhencai Zhu3
Weihong Peng4
Jishun Li5
1, 2, 3School of Mechatronic Engineering, China University of Mining and Technology, Xuzhou, China
2, 3Jiangsu Key Laboratory of Mine Mechanical and Electrical, Xuzhou, China
4School of Mechanics and Civil Engineering, China University of Mining and Technology, Xuzhou, China
5State Key Laboratory of Heavy Mining Equipment, Luoyang, China
Corresponding Author:
Guohua Cao
Views 101
#### Abstract
Theoretical investigation of the lateral and torsional vibrations of the hoisting cage in the cable-guided hoisting system caused by the eccentric load and the flexibility of the guiding cable is presented in this paper. The assumed modes method (AMM) is adopted to discretize the hoisting cable and two guiding cables, then Lagrange equations of the first kind are used to derive the equations of motion, while the geometric relationships between the hoisting cage and the cables are accounted for by the Lagrangian multiplier. Considering all the geometric matching conditions are approximately linear, the differential algebraic equations (DAEs) are transformed to the ordinary differential equations (ODEs). The dynamic responses of the hoisting cage are calculated, and especially the lateral displacements of the guiding cable and the constraints forces at the interfaces are obtained. Preload plays a vital role in affecting the cage vibration, thus, the effects of the total preload and the tension difference are analyzed. The numerical results indicate increasing the total preload can decrease the vibration displacements, while the tension difference has little impact on the vibration but can obviously change the constraint forces. In addition, the vibration displacements are directly proportional to the eccentric load, but less sensitive to the hoisting mass.
## 1. Introduction
Cables, due to their high strength and light weight, have been extensively applied to not only the hoisting system, such as mobile cranes [1], elevators [2] and mine hoisting [3], but also the other engineering including suspension bridges [4] and cable-driven parallel robots [5]. The cable-guided bucket hoisting system [6] is usually exerted to sink the deep vertical shaft, and then temporarily converted into one cable-guided cage hoisting system for transportation of equipment. In the typical cage hoisting system shown in Fig. 1(a), cables play an important role in guiding the cage when compared with the rail-guided hoisting system. The system is mainly composed of one hoisting cable, two guiding cables, a hoisting cage and a set of tensioning device, where the cage is held by the hoisting cable and supported by the two guiding cables. It is worth noting that the cage is connected to the guiding cables by four guide sleeves, and both ends of the guiding cables are completely restricted by the tensioning device. Unfortunately, cable is subjected to three-dimensional vibrations because of the external excitation and the cage undergoes the lateral vibration and even rotation for the high flexibility of the guiding cable. Therefore, it is necessary to predict the vibration displacements to avoid the danger of collision, and probe further into the relationship between the vibrations and the preload to improve the performance of the cable-guided cage hoisting system.
Many researchers have concentrated on the lateral vibrations of the hoisting cable or container for decades. Deb [7] studied the lateral vibration of a fixed-length string impacted by an elastic load at any point using. Fung et al. [8] analyzed the lateral vibration of an elevator cable characterized by time-varying length. Zhu and Xu [9] described the lateral behaviors for the stationary and moving hoisting cables with the initial displacement and presented the optimal suspension stiffness and damping coefficient. Kaczmarczyk and Ostachowicz [10, 11] investigated the lateral in-plane and out-of-plane responses of a catenary cable in the mine hoisting system and a compensation cable in a high-rise high-speed elevator. Zhu and Chen [12] conducted innovative experiments on a scaled elevator to validate the theoretical predictions for the uncontrolled and controlled lateral responses of a moving cable. Kimura et al. [13] performed a theoretical solution to the forced vibration of an elevator cable where both ends are excited and clarified the effect of the rate of change of cable length and the damping factor on the maximum lateral displacement. Bao et al. [14] theoretically and experimentally studied the nonlinear lateral vibration of a moving elevator cable and analyzed the passage through resonances by calculating the natural frequencies of the system. Ren and Zhu [15] presented the longitudinal and lateral vibrations of a moving two-cable one-rigid-body-car system, in which the rotation of the car is considered. For the spatial discretization of a cable, assumed modes method (AMM) [9, 14] and finite element method [16, 17] are commonly adopted. For the equation of motion, Lagrange’s equations [18, 19] or Hamilton’s principle can be selected.
However, until now little research has focused on the dynamic responses of the cable-guided hoisting system, especially the lateral and torsional vibrations of the hoisting cage caused by the eccentric load. In general, the guiding cables are simplified to a spring-mass system with a single degree of freedom and the high-order modes are omitted by the linear deformation assumption [6], thus, it is difficult to describe the whole vibration of the guiding cable. But in this paper, several challenging problems, such as the lateral displacement of the guiding cable and the constraint forces of the guide sleeves, are worked out. Considering the complicated boundary conditions between the hoisting cage and the guiding cables, the AMM and Lagrange’s equations with Lagrangian multipliers are combined to establish the vibration model.
Fig. 1Cable-guided cage hoisting system with eccentric load
a) Cage hoisting system
b) Schematic diagram
## 2. Lateral and torsional vibrations
For simplicity and without loss of generality, the cable-guided hoisting system (Fig. 1(a)) can be described as the vertically translating model shown in Fig. 1(b). This model is composed of one hoisting cable of length $l\left(t\right)$ at time instant $t$, two guiding cables of length $L$ and one cage attached to the lower end of the hoisting cable (${h}_{0}={d}_{0}$). The cage of height 2$h$ is supported by the two guiding cables through four guide sleeves (${h}_{i}={d}_{i}$, $i=\mathrm{}$1-4). The hoisting cable has a vertically translating velocity $v\left(t\right)=\stackrel{˙}{l}\left(t\right)$ and acceleration $a\left(t\right)=\stackrel{¨}{l}\left(t\right)$, where the over dot denotes time differentiation. In Fig. 1(b), the upper ends of two guiding cables are fixed to the derrick and the lower ends are constrained laterally and tensioned by the preloads ${T}_{b1}$ and ${T}_{b2}$; the displacements ${x}_{c}$, ${y}_{c}$ and $\theta$ represent the longitudinal, lateral and torsional vibrations of the hoisting cage, respectively; ${F}_{a}$ and $b$ are the eccentric load and the offset distance.
### 2.1. Spatial discretization
With reference to Fig. 1(b), the kinetic energy ${K}_{e}$ associated with the longitudinal, lateral and torsional vibrations is:
1
${K}_{e}=\frac{1}{2}{\int }_{0}^{l\left(t\right)}{\rho }_{1}{\left(\frac{Du}{Dt}+v\right)}^{2}dx+\frac{1}{2}{\int }_{0}^{L}\sum _{i=1}^{2}{\rho }_{2}{\left(\frac{\partial {y}_{i}}{\partial t}\right)}^{2}dx+\frac{1}{2}J{\stackrel{˙}{\theta }}^{2}+\frac{m}{2}{\left({\stackrel{˙}{x}}_{c}+v\right)}^{2}+\frac{m}{2}{\stackrel{˙}{y}}_{c}^{2},$
where ${\rho }_{1}$ and ${\rho }_{2}$ are the mass per unit length of the hoisting cable and the guiding cable; $u$ and ${y}_{i}$ denote the longitudinal vibration of the hoisting cable and the lateral vibration of the guiding cables; $m$ and $J$ are the mass and moment of inertia of the hoisting cage, respectively.
The static tension in the cable is much larger than the vibratory one, as demonstrated in [15], thus the tension can be approximated by the static tension. The total potential energy ${V}_{e}$ of all the cables and the hoisting cage is:
2
${V}_{e}={\int }_{0}^{l\left(t\right)}\left[\frac{1}{2}EA{\left(\frac{\partial u}{\partial x}\right)}^{2}+{T}_{0}\left(x,t\right)\frac{\partial u}{\partial x}\right]dx+\sum _{i=1}^{2}{\int }_{0}^{L}\frac{1}{2}{T}_{i}\left(x\right){\left(\frac{\partial {y}_{i}}{\partial x}\right)}^{2}dx$
$\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}-{\int }_{0}^{l\left(t\right)}{\rho }_{1}gudx-mg\left({x}_{c}+l\right),$
where $EA$ is the axial stiffness of the hoisting cable, ${T}_{0}\left(x,t\right)$ and ${T}_{i}\left(x\right)$ are the static tensions in the hoisting and guiding cables at position $x$ due to the gravitational acceleration g and given as:
3
${T}_{0}\left(x,t\right)=\left[m+{\rho }_{1}\left(l\left(t\right)-x\right)\right]g,\mathrm{}\mathrm{}\mathrm{}\mathrm{}{T}_{i}\left(x\right)={T}_{bi}+{\rho }_{2}g\left(L-x\right).$
The dissipated energy ${R}_{e}$ of the cables due to the damping is:
4
${R}_{e}=\frac{1}{2}{\int }_{0}^{l\left(t\right)}{\mu }_{1}{\stackrel{˙}{u}}^{2}dx+\sum _{i=1}^{2}\frac{1}{2}{\int }_{0}^{L}{\mu }_{2}{\stackrel{˙}{y}}_{i}^{2}dx,$
where ${u}_{1}$ and ${u}_{2}$ are the distributed damping coefficients of the hoisting and guiding cables, respectively.
For the hoisting cable and the guiding cables, the boundary conditions at both ends are:
5
$u\left(0,t\right)=0,\mathrm{}\mathrm{}\mathrm{}{y}_{i}\left(0,t\right)=0,\mathrm{}\mathrm{}\mathrm{}{y}_{i}\left(L,t\right)=0,\mathrm{}\mathrm{}\mathrm{}\mathrm{}\left(i=1,\mathrm{}2\right).$
The geometric matching condition at the interface between the hoisting cable and the hoisting cage is:
6
${g}_{0}=u\left(l,t\right)-{x}_{c}=0,$
while the geometric matching conditions at four contact points between the hoisting cage and the guiding cables are:
7
$\left\{\begin{array}{l}{g}_{1}={y}_{1}\left(l,t\right)+h\mathrm{s}\mathrm{i}\mathrm{n}\theta -d\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{t}\mathrm{a}\mathrm{n}\theta -{y}_{c}=0,\\ {g}_{2}={y}_{2}\left(l,t\right)+h\mathrm{s}\mathrm{i}\mathrm{n}\theta -d\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{t}\mathrm{a}\mathrm{n}\theta -{y}_{c}=0,\\ {g}_{3}={y}_{2}\left(l+2h\mathrm{c}\mathrm{o}\mathrm{s}\theta ,t\right)-h\mathrm{s}\mathrm{i}\mathrm{n}\theta +d\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{t}\mathrm{a}\mathrm{n}\theta -{y}_{c}=0,\\ {g}_{4}={y}_{1}\left(l+2h\mathrm{c}\mathrm{o}\mathrm{s}\theta ,t\right)-h\mathrm{s}\mathrm{i}\mathrm{n}\theta +d\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{t}\mathrm{a}\mathrm{n}\theta -{y}_{c}=0.\end{array}\right\$
Since the rotation of the cage $\theta$ is generally small, the approximations $\mathrm{s}\mathrm{i}\mathrm{n}\theta \approx \theta$, $\mathrm{c}\mathrm{o}\mathrm{s}\theta \approx 1$ and $\mathrm{s}\mathrm{i}\mathrm{n}\theta \mathrm{t}\mathrm{a}\mathrm{n}\theta \approx 0$ can be adopted. Therefore, Eq. (7) becomes:
8
$\left\{\begin{array}{l}{g}_{1}={y}_{1}\left(l,t\right)+h\theta -{y}_{c}=0,\\ {g}_{2}={y}_{2}\left(l,t\right)+h\theta -{y}_{c}=0,\\ {g}_{3}={y}_{2}\left(l+2h,t\right)-h\theta -{y}_{c}=0,\\ {g}_{4}={y}_{1}\left(l+2h,t\right)-h\theta -{y}_{c}=0.\end{array}\right\$
For simplicity, herein two new dimensionless parameters $\xi =x/l\left(t\right)$ and $\eta =x/L$ are introduced and the time-varying domain $\left[0,l\left(t\right)\right]$ and the time-invariant $\left[0,L\right]$ for $x$ are both transformed to a time-invariant one [0, 1] for $\xi$ and $\eta$. Hence, the dependent variable $u\left(x,t\right)$ and ${y}_{i}\left(x,t\right)$ become $\stackrel{-}{u}\left(\xi ,t\right)$ and ${\stackrel{-}{y}}_{i}\left(\eta ,t\right)$, respectively. Further, the partial derivatives of $u\left(x,t\right)$with respect to $x$ and $t$ are related to those of ${\stackrel{^}{y}}_{0}\left(\xi ,t\right)$ with respect to $\xi$ and $t$:
9
${u}_{x}=\frac{1}{l\left(t\right)}{\stackrel{-}{u}}_{\xi },\mathrm{}\mathrm{}\mathrm{}\mathrm{}{u}_{t}={\stackrel{-}{u}}_{t}-\frac{v\xi }{l\left(t\right)}{\stackrel{-}{u}}_{\xi }.$
The similar derivations are presented as:
10
${y}_{i,x}=\frac{1}{L}{\stackrel{-}{y}}_{i,\eta },\mathrm{}\mathrm{}\mathrm{}\mathrm{}{y}_{i,t}={\stackrel{-}{y}}_{i,t},$
${T}_{0}\left(x,t\right)={\stackrel{-}{T}}_{0}\left(\xi ,t\right)=\left[m+{\rho }_{1}l\left(t\right)\left(1-\xi \right)\right]g,$
${T}_{i}\left(x\right)={\stackrel{-}{T}}_{i}\left(\eta \right)={T}_{bi}+{\rho }_{2}gL\left(1-\eta \right).$
Accordingly, the boundary conditions in Eq. (5) become:
11
$\stackrel{-}{u}\left(0,t\right)=0,\mathrm{}\mathrm{}\mathrm{}{\stackrel{-}{y}}_{i}\left(0,t\right)=0,\mathrm{}\mathrm{}\mathrm{}{\stackrel{-}{y}}_{i}\left(1,t\right)=0.$
The lateral displacements can be approximated by expansions of a complete set of trial functions and expressed as:
12
$\stackrel{-}{u}\left(\xi ,t\right)=\sum _{i=1}^{n}{U}_{0,i}\left(\xi \right){q}_{0,i}\left(t\right),\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\stackrel{-}{y}}_{1}\left(\eta ,t\right)=\sum _{i=1}^{n}{U}_{1,i}\left(\eta \right){q}_{1,i}\left(t\right),\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\stackrel{-}{y}}_{2}\left(\eta ,t\right)=\sum _{i=1}^{n}{U}_{2,i}\left(\eta \right){q}_{2,i}\left(t\right),$
where $n$ represents the number of included modes; ${q}_{j,i}\left(t\right)$ are the generalized coordinates, in which $j=\mathrm{}$0, 1, 2; ${U}_{j,i}$ are the trial functions and should satisfy the homogeneous boundary conditions in Eq. (11), which can be expressed as:
13
${U}_{0,i}\left(\xi \right)=\sqrt{2}\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{2i-1}{2}\pi \xi \right),\mathrm{}\mathrm{}\mathrm{}{U}_{1,i}\left(\eta \right)={U}_{2,i}\left(\eta \right)=\sqrt{2}\mathrm{s}\mathrm{i}\mathrm{n}\left(i\pi \eta \right).$
Substituting Eqs. (1), (2) and (4) into Lagrange equations of the first kind [18]:
14
$\frac{d}{dt}\frac{\partial {K}_{e}}{\partial {\stackrel{˙}{q}}_{i}}-\frac{\partial {K}_{e}}{\partial {q}_{i}}+\frac{\partial {V}_{e}}{\partial {q}_{i}}+\frac{\partial {R}_{e}}{\partial {\stackrel{˙}{q}}_{i}}=\sum _{j=1}^{5}{\lambda }_{j}\frac{\partial {g}_{j}}{\partial {q}_{i}},$
yields the equations of motion:
15
$\left\{\begin{array}{l}\mathbf{M}\stackrel{¨}{\mathbf{q}}+\mathbf{C}\stackrel{˙}{\mathbf{q}}+\mathbf{K}\mathbf{q}=\mathbf{F}+{\mathbf{G}}^{T}\lambda ,\\ \mathbf{g}\left(\mathbf{q},t\right)=0.\end{array}\right\$
where $\mathbf{q}=\left({q}_{0,1},{q}_{1,1},{q}_{1,2},{q}_{2,1},{q}_{2,2},{q}_{0,2},{q}_{0,3},{q}_{1,3},{q}_{2,3},\dots ,{q}_{0,n},{q}_{1,n},{q}_{2,n},{q}_{3n+1},{q}_{3n+2},{q}_{3n+3}{\right)}^{T}$ is the vector of generalized coordinates; ${q}_{3n+1}$, ${q}_{3n+2}$, ${q}_{3n+3}$ denote the displacements ${x}_{c}$, ${y}_{c}$ and $\theta$, respectively. It should be noted that these (3$n+$3) generalized coordinates are not independent, however, the geometric matching conditions Eqs. (6) and (8) yield the holonomic constraints $\mathbf{g}$ of the generalized coordinates, where $\mathbf{g}=\left({g}_{1},{g}_{2},{g}_{3},{g}_{4},{g}_{5}{\right)}^{T}$ is a vector including all the constraint conditions; $\mathbf{G}=\partial \mathbf{g}/\partial \mathbf{q}$ is the Jacobian matrix of the constraint Eqs. (6) and (8), which is a 5×(3$n+$3) matrix; and $\lambda =\left({\lambda }_{1},{\lambda }_{2},{\lambda }_{3},{\lambda }_{4},{\lambda }_{5}{\right)}^{T}$ are called the Lagrangian multipliers [20], which denote the constraint forces between the hoisting cage and the three cables.
For notational convenience, a transformation matrix ${\mathbf{T}}_{r}$ is introduced:
16
$\mathbf{q}={\mathbf{T}}_{r}\cdot \mathbf{p},$
where $\mathbf{p}=\left({q}_{0,1},\dots ,{q}_{0,n},{q}_{1,1},\dots ,{q}_{1,n},{q}_{2,1},\dots ,{q}_{2,n},{q}_{3n+1},{q}_{3n+2},{q}_{3n+3}{\right)}^{T}$ is another sequence of the generalized coordinates. The matrices $\mathbf{M}$, $\mathbf{C}$, $\mathbf{K}$ and $\mathbf{F}$ are expressed as:
17
$\mathbf{M}={\mathbf{T}}_{r}\stackrel{-}{\mathbf{M}}{\mathbf{T}}_{\mathrm{r}}^{T},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathbf{C}={\mathbf{T}}_{r}\stackrel{-}{\mathbf{C}}{\mathbf{T}}_{r}^{T},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathbf{K}={\mathbf{T}}_{r}\stackrel{-}{\mathbf{K}}{\mathbf{T}}_{r}^{T},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathbf{F}={\mathbf{T}}_{r}\stackrel{-}{\mathbf{F}}{\mathbf{T}}_{r}^{T},$
where:
${\stackrel{-}{\mathbf{M}}}_{ij}=\left\{\begin{array}{l}{\int }_{0}^{1}{\rho }_{1}l{U}_{0,i}\left(\xi \right){U}_{0,j}\left(\xi \right)d\xi ,\left(i,j\le n\right),\\ {\int }_{0}^{1}{\rho }_{2}L{U}_{1,i}\left(\eta \right){U}_{1,j}\left(\eta \right)d\eta ,\left(n
${\stackrel{-}{\mathbf{C}}}_{ij}=\left\{\begin{array}{l}{\rho }_{1}v{\int }_{0}^{1}\left(1-\xi \right)\left[{U}_{0,i}^{\mathrm{\text{'}}}\left(\xi \right){U}_{0,j}\left(\xi \right)-{U}_{0,i}\left(\xi \right){U}_{0,j}^{\mathrm{\text{'}}}\left(\xi \right)\right]d\xi \\ +\left({\rho }_{1}v+{u}_{1}l\right){\int }_{0}^{1}{U}_{0,i}\left(\xi \right){U}_{0,j}\left(\xi \right)d\xi ,\left(i,j\le n\right),\\ {u}_{2}L{\int }_{0}^{1}{U}_{1,i}\left(\eta \right){U}_{1,j}\left(\eta \right)d\eta ,\left(n
${\stackrel{-}{\mathbf{K}}}_{ij}=\left\{\begin{array}{l}\frac{EA}{l}{\int }_{0}^{1}{U}_{0,i}^{\text{'}}\left(\xi \right){U}_{0,j}^{\text{'}}\left(\xi \right)d\xi +{\rho }_{1}a{\int }_{0}^{1}\left(1-\xi \right){U}_{0,i}^{\text{'}}\left(\xi \right){U}_{0,j}\left(\xi \right)d\xi -{u}_{1}v{\int }_{0}^{1}\xi {U}_{0,i}^{\text{'}}\left(\xi \right){U}_{0,j}\left(\xi \right)d\xi \\ +\frac{{\rho }_{1}{v}^{2}}{l}{\int }_{0}^{1}\left[\xi {U}_{0,i}^{\text{'}}\left(\xi \right){U}_{0,j}\left(\xi \right)-\left(1-\xi {\right)}^{2}{U}_{0,i}^{\text{'}}\left(\xi \right){U}_{0,j}^{\text{'}}\left(\xi \right)\right]d\xi ,\left(i,j\le n\right),\\ \frac{1}{L}{\int }_{0}^{1}{\stackrel{^}{T}}_{1}\left(\eta \right){U}_{1,i}^{\text{'}}\left(\eta \right){U}_{1,j}^{\text{'}}\left(\eta \right)d\eta ,\left(n
${\stackrel{-}{\mathbf{F}}}_{i}=\left\{\begin{array}{l}{\rho }_{1}{v}^{2}{\int }_{0}^{1}\left[\left(1-\xi \right){U}_{0,i}^{\text{'}}\left(\xi \right)-{U}_{0,i}\left(\xi \right)\right]d\xi +{\rho }_{1}l{\int }_{0}^{1}\left(g-a\right){U}_{0,i}\left(\xi \right)d\xi \\ -{\int }_{0}^{1}\left[mg+{\rho }_{1}gl\left(1-\xi \right)\right]{U}_{0,i}^{\text{'}}\left(\xi \right)d\xi ,\left(i\le n\right),\\ 0,\left(n
### 2.2. DAEs to ODEs for solution
Considering the constraint conditions in Eqs. (6) and (8) are all linear algebraic equations, Eq. (15), a system of DAEs, could be transformed to ODEs. Thus, the second equation in Eq. (15) could be expressed in the form [21]:
18
$\mathbf{g}\left(\mathbf{q},t\right)=\mathbf{G}\left(t\right)\mathbf{q}+{\mathbf{g}}_{r}\left(t\right)=0,$
where ${\mathbf{g}}_{r}\left(t\right)=0$. Without loss of generality, suppose:
19
$\mathbf{G}\left(t\right)=\left(\begin{array}{ll}{\mathbf{G}}_{0}& {\mathbf{G}}_{1}\end{array}\right),\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathbf{q}=\left[{\mathbf{q}}_{0}^{T}{\mathbf{q}}_{1}^{T}{\right]}^{T},$
where ${\mathbf{G}}_{0}$ must be a nonsingular 5×5 matrix, because its inverse matrix will be used subsequently. By Eqs. (6), (8) and (12), one can obtain:
$\mathbf{G}\left(t\right)=\left[\begin{array}{ccccc}{U}_{0,1}\left(1\right),0,0& 0,0,{U}_{0,2}\left(1\right)& \cdots & {U}_{0,n}\left(1\right),0,0& -1,0,0\\ 0,{U}_{1,1}\left(\frac{l}{L}\right),{U}_{1,2}\left(\frac{l}{L}\right)& 0,0,0& \cdots & 0,{U}_{1,n}\left(\frac{l}{L}\right),0& 0,-1,h\\ 0,0,0& {U}_{2,1}\left(\frac{l}{L}\right),{U}_{2,2}\left(\frac{l}{L}\right),0& \cdots & 0,0,{U}_{2,n}\left(\frac{l}{L}\right)& 0,-1,h\\ 0,0,0& {U}_{2,1}\left(\frac{l+2h}{L}\right),{U}_{2,2}\left(\frac{l+2h}{L}\right),0& \cdots & 0,0,{U}_{2,n}\left(\frac{l+2h}{L}\right)& 0,-1,-h\\ 0,{U}_{1,1}\left(\frac{l+2h}{L}\right),{U}_{1,2}\left(\frac{l+2h}{L}\right)& 0,0,0& \cdots & 0,{U}_{1,n}\left(\frac{l+2h}{L}\right),0& 0,-1,-h\end{array}\right],$
${\mathbf{G}}_{0}=\left[\begin{array}{ccccc}{U}_{0,1}\left(1\right)& 0& 0& 0& 0\\ 0& {U}_{1,1}\left(\frac{l}{L}\right)& {U}_{1,2}\left(\frac{l}{L}\right)& 0& 0\\ 0& 0& 0& {U}_{2,1}\left(\frac{l}{L}\right)& {U}_{2,2}\left(\frac{l}{L}\right)\\ 0& 0& 0& {U}_{2,1}\left(\frac{l+2h}{L}\right)& {U}_{2,2}\left(\frac{l+2h}{L}\right)\\ 0& {U}_{1,1}\left(\frac{l+2h}{L}\right)& {U}_{1,2}\left(\frac{l+2h}{L}\right)& 0& 0\end{array}\right],$
${\mathbf{G}}_{1}=\left[\begin{array}{ccccc}{U}_{0,2}\left(1\right)& {U}_{0,3}\left(1\right),0,0& \cdots & {U}_{0,n}\left(1\right),0,0& -1,0,0\\ 0& 0,{U}_{1,3}\left(\frac{l}{L}\right),0& \cdots & 0,{U}_{1,n}\left(\frac{l}{L}\right),0& 0,-1,h\\ 0& 0,0,{U}_{2,3}\left(\frac{l}{L}\right)& \cdots & 0,0,{U}_{2,n}\left(\frac{l}{L}\right)& 0,-1,h\\ 0& 0,0,{U}_{2,3}\left(\frac{l+2h}{L}\right)& \cdots & 0,0,{U}_{2,n}\left(\frac{l+2h}{L}\right)& 0,-1,-h\\ 0& 0,{U}_{1,3}\left(\frac{l+2h}{L}\right),0& \cdots & 0,{U}_{1,n}\left(\frac{l+2h}{L}\right),0& 0,-1,-h\end{array}\right],$
${\mathbf{q}}_{0}=\left[{q}_{0,1},{q}_{1,1},{q}_{1,2},{q}_{2,1},{q}_{2,2}{\right]}^{T,}$
${\mathbf{q}}_{1}=\left[{q}_{0,2},{q}_{0,3},{q}_{1,3},{q}_{2,3},\cdots ,{q}_{2,n},{q}_{3n+1},{q}_{3n+2},{q}_{3n+3}{\right]}^{T}.$
It should be noted that the sequence of generalized coordinates is of great importance, because ${\mathbf{G}}_{0}$ will be a singular matrix if $\mathbf{p}$ rather than $\mathbf{q}$ is selected in Eq. (15).
By Substituting Eq. (19) into Eq. (18), one has:
20
$\mathbf{q}=\mathrm{\Phi }\left(t\right){\mathbf{q}}_{1},\mathrm{}\mathrm{}\mathrm{}\mathrm{\Phi }\left(t\right)=\left[\begin{array}{c}-{\mathbf{G}}_{0}^{-1}{\mathbf{G}}_{1}\\ \mathbf{I}\end{array}\right],$
where ${\mathbf{q}}_{1}\text{,}$ a (3$n-$2) vector, become the new generalized coordinates, which are linearly independent; $\mathbf{I}$ is a (3$n-$2) identity matrix.
Differentiating Eq. (20) twice yields:
21
$\stackrel{˙}{\mathbf{q}}=\mathrm{\Phi }\left(t\right){\stackrel{˙}{\mathbf{q}}}_{1}+\stackrel{˙}{\mathrm{\Phi }}\left(t\right){\mathbf{q}}_{1},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\stackrel{¨}{\mathbf{q}}=\mathrm{\Phi }\left(t\right){\stackrel{¨}{\mathbf{q}}}_{1}+2\stackrel{˙}{\mathrm{\Phi }}\left(t\right){\stackrel{˙}{\mathbf{q}}}_{1}+\stackrel{¨}{\mathrm{\Phi }}\left(t\right){\mathbf{q}}_{1}.$
Eq. (19) multiplied by Eq. (20) can yield $\mathbf{G}\mathrm{\Phi }=0$. Substituting Eq. (21) into the first equation in Eq. (15), premultiplying by ${\mathrm{\Phi }}^{T}$ and applying ${\mathrm{\Phi }}^{T}{\mathbf{G}}^{T}=\left(\mathbf{G}\mathrm{\Phi }{\right)}^{T}=0$ yield:
22
${\mathbf{M}}_{G}{\stackrel{¨}{\mathbf{q}}}_{1}+{\mathbf{C}}_{G}{\stackrel{˙}{\mathbf{q}}}_{1}+{\mathbf{K}}_{G}{\mathbf{q}}_{1}={\mathbf{F}}_{G},$
with:
23
${\mathbf{M}}_{G}={\mathrm{\Phi }}^{T}\mathbf{M}\mathrm{\Phi },\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\mathbf{C}}_{G}=2{\mathrm{\Phi }}^{T}\mathbf{M}\stackrel{˙}{\mathrm{\Phi }}+{\mathrm{\Phi }}^{T}\mathbf{C}\mathrm{\Phi },\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\mathbf{K}}_{\mathrm{G}}={\mathrm{\Phi }}^{T}\mathbf{M}\stackrel{¨}{\mathrm{\Phi }}+{\mathrm{\Phi }}^{T}\mathbf{C}\stackrel{˙}{\mathrm{\Phi }}+{\mathrm{\Phi }}^{T}\mathbf{K}\mathrm{\Phi },\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\mathbf{F}}_{\mathrm{G}}={\mathrm{\Phi }}^{T}\mathbf{F}.$
Eq. (22) can be solved by an ODE solver. $\mathbf{q}$ can be assembled as:
24
$\mathbf{q}=\left[\begin{array}{c}-{\mathbf{G}}_{0}^{-1}{\mathbf{G}}_{1}{\mathbf{q}}_{1}\\ {\mathbf{q}}_{1}\end{array}\right].$
## 3. Examples and discussion
The parameters of the cable-guided hoisting system used in the numerical simulation are as follows: ${\rho }_{1}=$3 kg/m, ${\rho }_{2}=$2.2 kg/m, $m=$800 kg, $h=$8 m, $J=$870 kg·m2, $EA=\mathrm{}$3.2×107 N and $L=$400 m. The damping coefficients are ${u}_{1}=$30 kg/(m∙s) and ${u}_{2}=$0.3 kg/(m∙s). The eccentric load and the offset distance are ${F}_{a}=$1000 N and $b=$1 m. The preloads in two guiding cables are ${T}_{b1}={T}_{b2}=$3×104 N. The downward movement profile is plotted in Fig. 2, where the maximum velocity and acceleration are 5 m/s and 1 m/s2, respectively. The initial and final lengths of the hoisting cable are 50 m and 300 m, respectively. The external excitation is $e\left(t\right)=$0.01$\mathrm{s}\mathrm{i}\mathrm{n}\left(\pi t\right)$ m. The total simulation time is 60 s and the time step size is 0.01 s. The number of the modes is $n=$50.
Fig. 2Downward movement profiles
a) Displacement $l\left(t\right)$
b) Velocity $v\left(t\right)$
c) Acceleration $a\left(t\right)$
### 3.1. Dynamic responses
The longitudinal, lateral and torsional displacements of the hoisting cage can be obtained by solving Eq. (22) and the results are shown in Fig. 3. It is interesting to describe the shape of the guiding cable at different times. The lateral displacements of two guiding cables are identical because of the same preloads, thus, only one of two guiding cables is analyzed and the results at $t=$18, 36 and 60 s are shown in Fig. 4.
Fig. 3The a) longitudinal, b) lateral and c) torsional displacements of the hoisting cage
a)
b)
c)
Fig. 4The lateral displacements of one guiding cable at different times
In Fig. 4, there is one short segment in each broken line, which is the position of the hoisting cage at that moment and results from the rotation of the cage. The length of the short segment is consistent with the height of the hoisting cage. It should be noted that the high-order modes of the guiding cable are considered, thus, the proposed method could present the whole vibration of the guiding cable.
The Lagrangian multipliers $\lambda$ can be obtained by substituting the solution Eq. (24) into Eq. (15), where ${\lambda }_{i}$ denote the constraint forces between ${h}_{i}$ and ${d}_{i}$. The results shown in Fig. 5 indicate ${\lambda }_{1}={\lambda }_{2}$ and ${\lambda }_{3}={\lambda }_{4}$. However, ${\lambda }_{1}$ and ${\lambda }_{3}$ are approximately equal, because the height of the hoisting cage is too small when compared with the length of the guiding cable, but in opposite direction.
Fig. 5The constraint forces between the hoisting cage and two guiding cables
### 3.2. Effect of preload on system response
Preload in the guiding cable plays an important role in affecting the lateral and torsional vibrations, thus, here will analyze the effect from two aspects: total preload and tension difference.
In case 1, the total preload ${T}_{b1}+{T}_{b2}=\mathrm{}$4.8×104, 5.4×104, 6×104 N and ${T}_{b1}={T}_{b2}$. The effect on the lateral and torsional vibrations of the hoisting cage is shown in Fig. 6. It can be observed that the lateral and torsional displacements decrease with the increase of the total preload but the frequencies of both vibrations increase instead. The constraint force ${\lambda }_{1}$ is shown in Fig. 7, which indicates that the frequency increases in the uniform velocity stage (first 50 s) and the amplitude decreases in the deceleration stage (last 10 s) as the total preload increases.
Fig. 6The a) lateral and b) torsional displacements with different total preloads
a)
b)
Fig. 7The constraint forces with different total preloads
In case 2, the total preload is constant ${T}_{b1}+{T}_{b2}=\mathrm{}$6×104 N and the tension difference $\mathrm{\Delta }T={T}_{b2}-{T}_{b1}=\mathrm{}$0, 5000, 10000 N. The effect of tension difference on the lateral and torsional vibrations of the hoisting cage is shown in Fig. 8. As can be seen in Fig. 8, in the uniform velocity stage the tension difference has little impact on the vibrations, but in the deceleration stage the displacements increase with the increase of the tension difference. Fig. 9 is presented to describe the effect on the constraint forces between the hoisting cage and two guiding cables. In Fig. 9, the constraint forces ${\lambda }_{1}$ and ${\lambda }_{4}$ decrease with the increase of the tension difference but ${\lambda }_{2}$ and ${\lambda }_{3}$ increase instead, which is reasonable because the lateral stiffness of the guiding cable increases with increase of the preload. In addition, the effect on the bottom (${\lambda }_{3}$ and ${\lambda }_{4}$) is more remarkable than the top (${\lambda }_{1}$ and ${\lambda }_{2}$).
Fig. 8The a) lateral and b) torsional displacements with different tension differences
a)
b)
Fig. 9The constraint forces with different tension differences
### 3.3. Effect of other parameters
The system responses are affected by many factors besides the preload, here will concentrate on the effect of the eccentric load ${F}_{a}$ and the mass of the hoisting cage $m$ on the lateral and torsional vibrations of the hoisting cage. Let the eccentric load ${F}_{a}=\mathrm{}$1000, 2000, 3000 N and fix other parameters and the results are shown in Fig. 10. The lateral and torsional displacements are both directly proportional to the eccentric load, so is the constraint force, as shown in Fig. 11.
Fig. 10The a) lateral and b) torsional displacements with different eccentric loads
a)
b)
Fig. 11The constraint forces with different eccentric loads
Considering the moment of inertia has a linear relation with the mass, the vibrations with $J=\mathrm{}$870, 1305, 1740 kg∙m2, corresponding to $m=\mathrm{}$800, 1200, 1600 kg are analyzed. As shown in Fig. 12(a), increasing the hoisting mass has little effect on the lateral vibration of the hoisting cage, but does decrease the frequency of vibration. In Fig. 12(b), increasing the hoisting mass has little impact on the amplitude and frequency of the torsional vibration in the uniform velocity stage, but decreases the amplitude in the deceleration stage. The effect on the constraint force is shown in Fig. 13. It can be observed from Fig. 13 that the amplitude remains the same and the frequency decreases with the increase of the hoisting mass in uniform velocity stage, but in the deceleration stage the amplitude increases.
Fig. 12The a) lateral and b) torsional displacements with different hoisting mass
a)
b)
Fig. 13The constraint forces with different hoisting mass
## 4. Conclusions
In this paper, the lateral and torsional vibrations of the hoisting cage induced by the eccentric load and the flexibility of the guiding cable are investigated. The cables are spatially discretized using the AMM and the equations of motion are established by Lagrange equations of the first kind. However, the introduction of the Lagrangian multiplier could deal with the complex geometric constraint conditions in the cable-guided hoisting system, but generates the DAEs. Therefore, the DAEs are transformed to the ODEs for solution, in which the permutation of the generalized coordinates is of great importance. The dynamic responses of the hoisting cage are calculated, and the lateral displacements of the guiding cable and the constraints forces at the interfaces are obtained. Because the high-order modes of the guiding cable are considered, the position and height of the hoisting cage is accurately presented in lateral vibration of the guiding cable. The theoretical model can not only predict the response of the cage to optimize the design of the hoistway, but also determine the suitable preload in the guiding cable.
Preload plays a vital role in the cable-guiding hoisting system, thus, the effects of the total preload and the tension difference on the cage vibration are analyzed. The numerical results indicate increasing the total preload can decrease the vibration displacements but increase the vibration frequency, while the tension difference has little impact on the vibrations but can obviously change the constraint forces. In addition, both the vibration displacements and the constraint forces are directly proportional to the eccentric load, but the displacements are less sensitive to the hoisting mass.
#### References
• Qian S., Zi B., Zhang D., Zhang L. Kinematics and error analysis of cooperative cable parallel manipulators for multiple mobile cranes. International Journal of Mechanics and Materials in Design, Vol. 10, Issue 4, 2014, p. 395-409.
• Arrasate X., Kaczmarczyk S., Almandoz G., Abete J. M., Isasa I. The modelling, simulation and experimental testing of the dynamic responses of an elevator system. Mechanical Systems and Signal Processing, Vol. 42, Issues 1-2, 2014, p. 258-282.
• Wang D. G., Zhang D. K., Ge S. R. Effect of terminal mass on fretting and fatigue parameters of a hoisting rope during a lifting cycle in coal mine. Engineering Failure Analysis, Vol. 36, 2014, p. 407-422.
• Wang Z. Q., Kang H. J., Sun C. S., Zhao Y. B., Yi Z. P. Modeling and parameter analysis of in-plane dynamics of a suspension bridge with transfer matrix method. Acta Mechanica, Vol. 225, Issue 12, 2014, p. 3423-3435.
• Du J. L., Bao H., Cui C. Z. Stiffness and dexterous performances optimization of large workspace cable-driven parallel manipulators. Advanced Robotics, Vol. 28, Issue 3, 2014, p. 187-196.
• Wang J. J., Cao G. H., Zhu Z. C., Wang Y. D., Peng W. H. Lateral response of cable-guided hoisting system with time-varying length: theoretical model and dynamics simulation verification. Proceedings of the Institution of Mechanical Engineers, Part C: Journal of Mechanical Engineering Science, 2015.
• Deb K. K. Dynamics of a string and an elastic hammer. Journal of Sound and Vibration, Vol. 40, Issue 2, 1975, p. 243-248.
• Fung R. F., Lin J. H., Yao C. M. Vibration analysis and suppression control of an elevator string actuated by a PM synchronous servo motor. Journal of Sound and Vibration, Vol. 206, Issue 3, 1997, p. 399-423.
• Zhu W. D., Xu G. Y. Vibration of elevator cables with small bending stiffness. Journal of Sound and Vibration, Vol. 263, Issue 3, 2003, p. 679-699.
• Kaczmarczyk S., Ostachowicz W. Transient vibration phenomena in deep mine hoisting cables. Part 1: Mathematical model. Journal of Sound and Vibration, Vol. 262, Issue 2, 2003, p. 219-244.
• Kaczmarczyk S., Picton P. The prediction of nonlinear responses and active stiffness control of moving slender continua subjected to dynamic loadings in a vertical host structure. International Journal of Acoustics and Vibration, Vol. 18, Issue 1, 2013, p. 39-44.
• Zhu W. D., Chen Y. Theoretical and experimental investigation of elevator cable dynamics and control. Journal of Vibration and Acoustics, Transactions of the ASME, Vol. 128, Issue 1, 2006, p. 66-78.
• Kimura H., Ito H., Fujita Y., Nakagawa T. Forced vibration analysis of an elevator rope with both ends moving. Journal of Vibration and Acoustics, Vol. 129, 2007, p. 471-477.
• Bao J. H., Zhang P., Zhu C. M., Sun W. Transverse vibration of flexible hoisting rope with time-varying length. Journal of Mechanical Science and Technology, Vol. 28, Issue 2, 2014, p. 457-466.
• Ren H., Zhu W. D. An accurate spatial discretization and substructure method with application to moving elevator cable-car systems. Part II: Application. Journal of Vibration and Acoustics, Transactions of the ASME, Vol. 135, Issue 5, 2013, p. 51037.
• Wang P. H., Fung R. F., Lee M. J. Finite element analysis of a three-dimensional underwater cable with time-dependent length. Journal of Sound and Vibration, Vol. 209, Issue 2, 1998, p. 223-249.
• Du J. L., Cui C. Z., Bao H., Qiu Y. Y. Dynamic analysis of cable-driven parallel manipulators using a variable length finite element. Journal of Computational and Nonlinear Dynamics, Vol. 10, Issue 1, 2015, p. 11013.
• Dvorak R., Freistetter F. Chaos and Stability in Planetary Systems. Springer-Verlag, Berlin, Germany, 2005.
• Zhu W. D., Ren H. An accurate spatial discretization and substructure method with application to moving elevator cable-car systems. Part I: Methodology. Journal of Vibration and Acoustics, Transactions of the ASME, Vol. 135, Issue 5, 2013, p. 051036.
• Lanczos C. The Variational Principles of Mechanics. Dover Publications, New York, 1986.
• Ilchmann A., Reis T. Surveys in Differential-Algebraic Equations II. Springer-Verlag, Berlin, Germany, 2015.
19 April 2016
Accepted
28 June 2016
Published
30 September 2016
SUBJECTS
Mechanical vibrations and applications
Keywords
lateral response
hoisting cable
guiding cable
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### Primary 4 Problem Sums/Word Problems - Try FREE
Score :
(Single Attempt)
#### Question
Mark and Linda have 832 stamps in total.
He has 4/9 as many stamps as Linda.
Linda gives Mark some stamps such that both of them had an equal number of stamps in the end.
How many stamps did Linda give to Mark?
The correct answer is : 160
stamps
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This week's book giveaway is in the Testing forum.We're giving away four copies of The Way of the Web Tester: A Beginner's Guide to Automating Tests and have Jonathan Rasmusson on-line!See this thread for details.
Win a copy of The Way of the Web Tester: A Beginner's Guide to Automating Tests this week in the Testing forum!
Rashid Ali
Ranch Hand
Posts: 349
I received 2nd nitpicked reply from Ms Marilyn. She asked me to do more hardwork on the above assignment and turning me to the proper way to complete the job and i think in that way my source code length will reduce.
I made my program using arrays, substring and case statements which works fine but its length is more than 100 lines. But she wants me to make the assignment using wacky maths functions and if-else statments.
So could u please let me know what is that simple maths function to dig out the right string from the 2 arrays in which i stored my counting in strings (1 to 20 in 1 string and 30 to 90 decimal-wise in another string) and now i hv to use math methods to perform the job. Please suggest me.
~Rashid
Johannes de Jong
tumbleweed
Bartender
Posts: 5089
Rashid,
There is no magical math method/class than will solve this for you.
You must use division (/) and remainder (%) operators.
The way to take solve this "problem" is to sit down and write how the different possible numbers that can be entered are represented as "words".
IE :
21 twenty-one, 1 twenty and one seperated by a -
56 fifty-six, 1 fifty and a six seperated by a -
Another way to look at it is :
21 = 2 tens with a remainder of 1
56 = 5 tens with a remainder of 6
But we dont say : two tens and one or five tens and six.
So you are on the right track with you two seperate arrays
Also look ahead to Java-4b. You are not coding it yet but you are building the foundation for the solution for Say b.
See if your solution for the "small" part of big numbers will work.
723 788, seven hundred and twenty-three, seven hundred and eighty-eight
Notice how the solution for Java-4a comes back in big numbers Thats why they make us to it in two phases.
Hope I helped you a bit i find it hard to explain something without giving away to much, actually I find it hard to explain anything .
But Rashid the less one gets spoon fed the more one will learn, good luck.
[This message has been edited by Johannes de Jong (edited April 14, 2001).]
Rashid Ali
Ranch Hand
Posts: 349
Thank u very much for your tips Johannes i'll try my best to do that using your guidelines during this weekend.
wish u a good weekend
kind regards
Rashid
Marilyn de Queiroz
Sheriff
Posts: 9067
12
How about this idea, if the number is less than 20, use an array of 0 to 19 to show the results.
Otherwise:� show the tens stuff and if there is a ones digit, show a hyphen and the one digit.
Rashid Ali
Ranch Hand
Posts: 349
How about this idea, if the number is less than 20, use an array of 0 to 19 to show the results.
Otherwise: show the tens stuff and if there is a ones digit, show a hyphen and the one digit.
Dear Marilyn, my code does work on the first part of your suggestion but it is getting hard to code the other part i.e. to link the 2nd array to 1st array but without "zero" and the logic is not getting cleared as yet.
Dear Johannes, how could i use (%) and (/) operators to serve the purpose. please suggest some examples.
Please note that this program works fine with case statements but nitpicker asks me to do this using if-else statements.
Or is there anyother way to make this program.
[This message has been edited by Rashid Ali (edited April 18, 2001).]
Johannes de Jong
tumbleweed
Bartender
Posts: 5089
Hi Rashid,
If you take 56.
j = 56 / 10 ; // j = 5
k = 56 % 10 ; // k = 6
Use these numbers to point to the correct array entry.
Rashid Ali
Ranch Hand
Posts: 349
Thanks Johannes
But the problem is that i have to use some iteration process which automatically check the number from command line and after calculating pickup the right word from the both arrays list.
is some other example for this purpose. i think i am bothering u much but be patient.
Thanks for all your messages sir
Rashid
pete hesse
Ranch Hand
Posts: 44
Rashid,
You don't need to use a case statement. An if...else will work fine.
The suggestions that have been made up to now, combined, will get you there. (You stated yourself that you understood one part!) Just be patient and keep working. You'll get it.
Pete
[This message has been edited by pete hesse (edited April 19, 2001).]
Marilyn de Queiroz
Sheriff
Posts: 9067
12
Get and parse the number from the command line
Get the tens digit by calculating
Get the ones digit by calculating
If
the number is less than 20, pick the right word from the array containing the ones and teens
Otherwise
Pick the right word from the array containing the tens
If the ones digit is not zero, print a hyphen and pick the right word from the array containing the ones and teens
Rashid Ali
Ranch Hand
Posts: 349
Thanks Marilyn and Johannes for your kind messages and help in finalizing my assignment.
Pete, when i reviewed my program with patience yesterday, then i could be able to understand the logic involved and finalized my assignment successfully in the same way as you mentioned but it spread over 107 lines and it works fine.
Thanks again to all of you
Rashid
Rashid Ali
Ranch Hand
Posts: 349
Again here :
I made my program Java-4a using the code such as:
The above is a piece of code just to give you an idea of logic.
May i know that is this okay or i will have to made it using & and % math tricks.
FYI, This program work fine as per the desires of this assignments.
Please advise to complete it in the right way. But using the above code logic.
Thanks
Rashid
[This message has been edited by Rashid Ali (edited April 27, 2001).]
[This message has been edited by Rashid Ali (edited April 27, 2001).]
[This message has been edited by Rashid Ali (edited April 27, 2001).]
Umesh Khosla
Greenhorn
Posts: 10
Dear Rashid ,
I think I can be of help try this code . this is for 4b think u will be able to find a soln for 4a also. Rest of us pls give suggestions. the identation is getting all lost suggest how to put code in aproper manner.
Marilyn added code tags and removed some code to protect future participants.
[This message has been edited by Marilyn deQueiroz (edited April 26, 2001).]
Richard Boren
Ranch Hand
Posts: 233
Umesh, DO NOT PUT COMPLETE CODE IN YOU POST. Only pseudo code and code snippets. Otherwise you will ruin the learning experience for others. Please remove the code form your posts, else a sheriff will.
It is good to help others, but bad to do it for them.
[This message has been edited by Richard Boren (edited April 26, 2001).]
Marilyn de Queiroz
Sheriff
Posts: 9067
12
Rest of us pls give suggestions. the identation is getting all lost suggest how to put code in aproper manner.
You need to put it in code tags that look like [ code ] [ /code ]
(without the spaces).
Echoing Richard's statement, please do not post code here except for very small snippets or pseudocode. Don't rob other students of an education.
Umesh Khosla
Greenhorn
Posts: 10
Sorry, People would not happen again , I am a bit over enthusiastic about things...
Richard Boren
Ranch Hand
Posts: 233
That's cool
Rashid Ali
Ranch Hand
Posts: 349
My question is still there. could my code be accepted ?
thanks.
Rashid
Johannes de Jong
tumbleweed
Bartender
Posts: 5089
Rashid look at your code carefully you repeat a lot of code.
If twenty , if thirty etc. If you divide the number by 10 you get 2,3,4, etc. you can use that to point to the correct position in the specific array.
Read the piece by Paul http://www.javaranch.com/drive/baby.html and you might understand what I mean.
Good luck
[This message has been edited by Johannes de Jong (edited April 28, 2001).]
Rashid Ali
Ranch Hand
Posts: 349
Get and parse the number from the command line
Get the tens digit by calculating
Get the ones digit by calculating
If
the number is less than 20, pick the right word from the array containing the ones and teens
Otherwise
Pick the right word from the array containing the tens
If the ones digit is not zero, print a hyphen and pick the right word from the array containing the ones and teens
Thanks very much Marilyn and Johannes for your great support in finallizing my 4a assignment which i yesterday completed as per the above instruction from Marilyn. Both Marilyn and Johannes helped me a lot !
Thank you.
Rashid Ali
Ranch Hand
Posts: 349
In response to my 4a attmpts Marilyn asks me to use some proper name for variable which i used in calculation using '/' and '%' operators instead of i, j, k, to pick out the right variable from arrays.
These names are not coming in my mind what to use in this program. Can i be suggested for the same if possible.
Thanks
Rashid
Johannes de Jong
tumbleweed
Bartender
Posts: 5089
Thanks for your kind words Rashid
When you use a variable name think of what is is used for. In this case your i,j,k are used as index's. Ask yourself what is in the the array you are indexing ie. what is it used for..
Rashid Ali
Ranch Hand
Posts: 349
Marilyn suggest:
"Notice how you print tensName[ divArgs ] ) in both
the else if block and the else block.
Maybe you can think of a way to consolidate this code."
My 4a is in the final stage, only the above clarification is required and after that i am sure it would be accepted.
Can I have a suggestion that how can i consolidate my code as desired by Marilyn. Hope from the above suggestion you could understand the scenario of my program flow.
I don't show my code here as it is against the forum requirement, but hope for your understanding.
Rashid
Johannes de Jong
tumbleweed
Bartender
Posts: 5089
Rashid, Marilyn suggests you look at what the else if & else have in common. Try and put that in a method and call the metod.
Amber Woods
Ranch Hand
Posts: 111
Hi Rashid,
If you are printing the same line in both the if/else block. Why not just print it once and then do a check for the rest of the number (ones digit).
Hope that helps and doesn't give away too much.
Amber
Rashid Ali
Ranch Hand
Posts: 349
Thanks Johannes and Amber for your suggestions.
Now my program works fine in that way.
Rashid
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# Falling Physics In Game
1. Feb 2, 2017
### Lotuschild
I will ask the question first and give the reason why I'm asking after.
Question: Approximately how fast can a 5'11" tall man weighing 195 lbs. fall towards the ground if he's 3 miles above the surface and has control over the winds of a category 5 hurricane storm to propell him downwards?
Reason: I'm playing a game of Dungeons and Dragons and find myself in control of the weather and able to fly and such. I'm in the middle of a battle when my character decides to use both himself and the winds to push an enemy into the ground. He grabs hold of his target, which is flying 3 miles above ground and proceeds to push him down into the ground, using gavity and the winds at his disposal to push past average terminal velocity. I was told that he would be traveling less than 250 mph, I just thought he'd be going faster. I'm no good at physics, so I will greatly appreciate any answer and explaination I can get. Thanks!
2. Feb 2, 2017
### Staff: Mentor
Welcome to the PF.
Hurricane winds go mostly sideways, don't they?
http://s.hswstatic.com/gif/hurricane-orig.jpg
3. Feb 2, 2017
### Lotuschild
In the game I can control them to go any direction I'd like.
4. Feb 2, 2017
### Staff: Mentor
Then I guess I would just add the normal terminal velocity of a skydiver to the downward velocity of your wind.
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# Number 646
## Description of the number 646
646 is a natural number (hence integer, rational and real) of 3 digits that follows 645 and precedes 647.
646 is an even number, since it is divisible by 2.
The number 646 is a unique number, with its own characteristics that, for some reason, has caught your attention. It is logical, we use numbers every day, in multiple ways and almost without realizing it, but knowing more about the number 646 can help you benefit from that knowledge, and be of great use. If you keep reading, we will give you all the facts you need to know about the number 646, you will see how many of them you already knew, but we are sure you will also discover some new ones.
## how to write 646 in letters?
Number 646 in English is written as six hundred forty-six
The number 646 is pronounced digit by digit as (6) six (4) four (6) six.
## What are the divisors of 646?
The number 646 has 8 divisors, they are as follows:
The sum of its divisors, excluding the number itself is 434, so it is a defective number and its abundance is -212
## Is 646 a prime number?
No, 646 is not a prime number since it has more divisors than 1 and the number itself
## What are the prime factors of 646?
The factorization into prime factors of 646 is:
21*171*191
## What is the square root of 646?
The square root of 646 is. 25.416530054278
## What is the square of 646?
The square of 646, the result of multiplying 646*646 is. 417316
## How to convert 646 to binary numbers?
The decimal number 646 into binary numbers is.1010000110
## How to convert 646 to octal?
The decimal number 646 in octal numbers is1206
## How to convert 646 to hexadecimal?
The decimal number 646 in hexadecimal numbers is286
## What is the natural or neperian logarithm of 646?
The neperian or natural logarithm of 646 is.6.4707995037826
## What is the base 10 logarithm of 646?
The base 10 logarithm of 646 is2.8102325179951
## What are the trigonometric properties of 646?
### What is the sine of 646?
The sine of 646 radians is.-0.92000241195914
### What is the cosine of 646?
The cosine of 646 radians is. 0.39191269689737
### What is the tangent of 646?
The tangent of 646 radians is.-2.3474677376938
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# [Zener Diode] What is Resistance Rz (Impedance Zz)?
In the zener diode, when the zener current $$I_Z$$ changes, the zener voltage $$V_Z$$ changes slightly. The slope of the Zener voltage $$V_Z$$ and the Zener current $$dV_Z/dI_Z$$ is called the resistance (or impedance) of zener diode.
Now, I will explain "Resistance (Impedance) of Zener Diode" in detail using diagrams.
## Resistance (Impedance) of Zener Diode
The above figure shows the "Zener current $$I_Z$$-Zener voltage $$V_Z$$ characteristic" of the zener diode.
Ideally, the zener voltage $$V_Z$$ of the zener diode should remain constant even when the zener current $$I_Z$$ changes.
In reality, however, the Zener voltage is not perfectly constant, and as the Zener current $$I_Z$$ changes $$dI_Z$$, the Zener voltage $$V_Z$$ changes $$dV_Z$$. Therefore, the "Zener current $$I_Z$$ - Zener voltage $$V_Z$$ characteristic" has a slope.
This slope is the resistance $$R_Z$$ of the zener diode. The resistance $$R_Z$$ of the zener diode is expressed by the following equation.
\begin{eqnarray}
R_Z=\frac{dV_Z}{dI_Z}
\end{eqnarray}
The resistance $$R_Z$$ of the zener diode is called its dynamic resistance (or operating resistance).
Supplement
• The symbols for the resistance of the zener diode are commonly represented by $$R_Z$$, $$Z_Z$$, $$R_{ZT}$$, $$Z_{ZT}$$, and $$r$$.
• Ideally, the zener diode has a constant zener voltage $$V_Z$$ ($$dV_Z = 0$$) even when the zener current $$I_Z$$ changes, so the ideal value of the resistance $$R_Z$$ of the zener diode is zero.
• The smaller the resistance $$R_Z$$ of the zener diode, the smaller the change in zener voltage $$V_Z$$ when the zener current $$I_Z$$ changes, and thus the higher the performance of the zener diode.
• The resistance $$R_Z$$ of the zener diode varies with the zener current flowing through it. Therefore, when a certain amount of current is applied, the resistance $$R_Z$$ becomes smaller, resulting in better stability of the zener voltage $$V_Z$$.
• The resistance $$R_Z$$ of the zener diode tends to increase at the rise of the zener diode (when current begins to flow). The operating resistance at the rise is called "Rise Dynamic Resistance (Rise Operating Resistance) $$R_{ZK}$$". Rise operating resistance will be explained later.
### Equivalent Circuit of Zener Diode
Using the resistance $$R_Z$$ of the zener diode, the equivalent circuit of the zener diode is shown in the figure above.
The equivalent circuit of the zener diode can be represented by a series connection of the zener voltage $$V_Z$$ and the resistor $$R_Z$$. In the equivalent circuit, the zener voltage $$V_Z$$ is assumed to be constant, and the change in zener voltage $$V_Z$$ due to zener current $$I_Z$$ is represented by the resistor $$R_Z$$.
## Rise Dynamic Resistance (Rise Operating Resistance) of Zener Diode
The resistance at the rising edge of the zener diode is called 'Rise Dynamic Resistance (Rise Operating Resistance) $$R_{ZK}$$' and is listed on the datasheet.
The above figure shows a part of the datasheet for Panasonic DZ2J062x0L. It can be seen that the rise dynamic resistance (rise operating resistance) $$R_{ZK}$$ is 100Ω when the zener current $$I_Z = 0.5{\mathrm{mA}}$$ is applied.
It can also be seen that as the zener current increases and $$I_Z = 5{\mathrm{mA}}$$, the resistance $$R_Z$$ of the zener diode becomes 30 Ω.
Supplement
• The symbols for the rise dynamic resistance of the zener diode are commonly represented by $$R_{ZK}$$ or $$Z_{ZK}$$.
#### Summary
In this article, the following information on the "Resistance (Impedance) of Zener Diodes" was explained.
• Resistance (Impedance) of Zener Diode
• Rise Dynamic Resistance (Rise Operating Resistance) of Zener Diode
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When Writing Coordinates Which Comes First
Last Updated on September 6, 2022 by amin
Contents
When plotting or finding coordinates which do you find first?
The order in which you write x- and y-coordinates in an ordered pair is very important. The x-coordinate always comes first followed by the y-coordinate.
Here are the characteristics for each of the four coordinate plane quadrants:
• Quadrant I: positive x and positive y.
• Quadrant II: negative x and positive y.
• Quadrant III: negative x and negative y.
• Quadrant IV: positive x and negative y.
See also who was the first person to walk around the world
How do you find the coordinates on a graph?
To identify the x-coordinate of a point on a graph read the number on the x-axis directly above or below the point. To identify the y-coordinate of a point read the number on the y-axis directly to the left or right of the point. Remember to write the ordered pair using the correct order (x y) . See also Which Of The Following Changes May Occur During Metamorphism? What Changes Occur During Metamorphism?
What is the set of the first numbers of the ordered pairs?
domainA relation is any set of ordered pairs. The set of all first components of the ordered pairs is called the domain. The set of all second components is called the range.
What is quadrant 3 on a graph?
Quadrant III: The third quadrant is in the bottom left corner of the plane. Furthermore both x and y have negative values in this quadrant.
What is XY coordinate system?
An x-y axis also known as a cartesian coordinate system or a coordinate plane is a two-dimensional plane of points defined uniquely by a pair of coordinates. … The horizontal line then is known as the x axis and measures the distance left or right from the vertical line.
What is significant about two points that have the same first coordinate or the same second coordinate?
Points that have the same x-coordinate and opposite y-coordinates will always be the same distance but on opposite sides of the x-axis because having the same x-coordinate means they will be the same distance along the x-axis and having opposite y-coordinates means they will be on opposite sides of y = 0.
What are the 4 coordinates?
The axes of a two-dimensional Cartesian system divide the plane into four infinite regions called quadrants each bounded by two half-axes. These are often numbered from 1st to 4th and denoted by Roman numerals: I (where the signs of the (x y) coordinates are I (+ +) II (− +) III (− −) and IV (+ −).
When Writing Coordinates Which Comes First?
The order in which you write x- and y-coordinates in an ordered pair is very important. The x-coordinate always comes first followed by the y-coordinate. As you can see in the coordinate grid below the ordered pairs (3 4) and (4 3) are two different points!Oct 22 2021
What is the set of all second coordinates of order pairs?
the range of the relationThe set of all second numbers of the ordered pairs in a relation is called the range of the relation. The values in the domain and range are usually listed from least to greatest.
The decision that a circle should be labelled counterclockwise was made by the ancient Babylonians long before any surviving mathematical works so nobody really knows why it was made.
How do you find three ordered pairs?
Quadrants are named using the Roman numerals I II III and IV beginning with the top right quadrant and moving counter clockwise. Locations on the coordinate plane are described as ordered pairs.
What is the position of 1st quadrant?
The 1st quadrant is in which position? Explanation: The position of reference planes will be similar to quadrants in x y plane co-ordinate system. As the 1st quadrant lies above the x-axis and in front of the y-axis here also the 1st quadrant is above H.P in front of V.P.
What comes first domain or range?
In the set of ordered pairs the Domain is the set of the first number in every pair (those are the x-coordinates). The Range is the set of the second number of all the pairs (those are the y-coordinates).
What is the 1st quadrant in a graph?
Quadrant I: The first quadrant is in the upper right-hand corner of the plane. Both x and y have positive values in this quadrant. Quadrant II: The second quadrant is in the upper left-hand corner of the plane. X has negative values in this quadrant and y has positive values.
Which ordered pair is Quadrant 2?
The point with coordinates (-3 4) is located in quadrant II.
How do you write XY coordinates?
Coordinates are written as (x y) meaning the point on the x axis is written first followed by the point on the y axis. Some children may be taught to remember this with the phrase ‘along the corridor up the stairs’ meaning that they should follow the x axis first and then the y.
How many quadrants are in a coordinate plane?
fourThe coordinate axes divide the plane into four regions called quadrants (or sometimes grid quadrants or Cartesian coordinate quadrants).
What order should coordinates go?
Coordinates are ordered pairs of numbers the first number number indicates the point on the x axis and the second the point on the y axis. When reading or plotting coordinates you always go across first and then up (a good way to remember this is: ‘across the landing and up the stairs’).
What are the XYZ axis?
A. X. A three-dimensional structure. The x-axis and y-axis represent the first two dimensions the z-axis the third dimension. In a graphic image the x and y denote width and height the z denotes depth.
What is Cartesian coordinates with example?
The Cartesian coordinate system uses a horizontal axis that is called the x-axis and a vertical axis called the y-axis. Equations for lines in this system will have both the x and y variable. For example the equation 2x + y = 2 is an example of a line in this system.
How do you teach coordinates to plot?
The Trick. In the alphabet the letter /x/ comes before /y/ and the letter /o/ comes before /u/. If you can remember the phrase “over and up ” you will be plotting coordinate points correctly in no time!! Go “over” (because /o/ comes first in the alphabet) and then “up” (because /u/ comes second).
How do you find two ordered pairs?
To figure out if an ordered pair is a solution to an equation you could perform a test. Identify the x-value in the ordered pair and plug it into the equation. When you simplify if the y-value you get is the same as the y-value in the ordered pair then that ordered pair is indeed a solution to the equation.
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# How many calories does a mental breakdown burn?
Contents
Crying is thought to burn roughly the same amount of calories as laughing – 1.3 calories per minute, according to one study . That means that for every 20-minute sob session, you’re burning 26 more calories than you would have burned without the tears. It’s not much.
## How many calories do you burn screaming?
Oh, and when you start yelling because no one in your family offers to help you, well, that’ll burn up to 16 calories per hour.
## Do you burn more calories when you are stressed?
A rush of epinephrine activates the body’s fight-or-flight response, which prepares a person to flee or fight off an impending threat. Epinephrine causes the heart to beat faster and breathing to speed up, which can burn calories.
## How many calories do you naturally burn in a day?
This means that, at rest, they’ll burn approximately 1,829.8 calories in a day (equation: 66 + (6.2 x 180) + (12.7 x 72) – (6.76 x 40) = 1,829.8). For females, use the following equation: 655.1 + (4.35 x weight) + (4.7 x height) – (4.7 x age) = BMR for females.
## How many calories do you burn if you stay in bed all day?
The amount of calories burned increases according to body weight. So, a person who weighs 150 pounds might burn 46 calories an hour or between 322 and 414 calories a night. And a person who weighs 185 pounds might burn around 56 calories or between 392 and 504 calories for a full night of sleep.
## Do you lose calories when you cry?
Crying is thought to burn roughly the same amount of calories as laughing – 1.3 calories per minute, according to one study . That means that for every 20-minute sob session, you’re burning 26 more calories than you would have burned without the tears. It’s not much.
## Does crying count as exercise?
“People who cry exhibit elevated heart rates and increased sweating. In this sense, crying is a ‘workout’ for the body.
## Does anxiety increase metabolism?
Now, new research by Baoji Xu and colleagues has revealed a link between anxiety and hyper-metabolism. The authors found that increases in the activation of anxiogenic circuits can reduce body weight via the promotion of adaptive thermogenesis and basal metabolism.
## Does anxiety burn fat?
JUPITER, FL – Scripps Research scientists have published a study revealing a shared mechanism for both anxiety and weight loss. Their research, published in the journal Cell Metabolism, describes a key molecule that triggers anxiety in the brain, while also increasing metabolism and fat burning.
## Can stress speed up metabolism?
Your body’s “fight or flight” response can speed up your metabolism. When you’re stressed, your body goes into “fight or flight” mode. Also known as the “acute stress response,” this physiological mechanism tells your body it must respond to a perceived threat.
IT IS INTERESTING: Does losing weight help fibromyalgia?
## Is burning 2000 calories a day too much?
She recommends burning 2,000 calories per week by exercising, and then trimming 1,500 calories a week from your diet, which breaks down to about 214 fewer calories per day. A general rule is to aim to burn 400 to 500 calories, five days a week during your workouts.
## Can you burn 3500 calories in a day?
Can you really lose a whole pound in one day?” The answer is yes—and this is why. As a general rule, a deficit of 3,500 calories will lead to a pound of fat loss. Cut 3,500 calories out of your day, and you will lose weight quickly. Try these 50 ways to lose weight without a lick of exercise.
## How can I burn 1000 calories a day?
How Much Exercise Do You Have To Do To Burn 1000 Calories In A Day?
1. General weight lifting – 112.
2. Vigorous weight lifting – 223.
3. Water aerobics – 149.
4. Low impact aerobics (e.g walking) – 205.
5. High impact aerobics (e.g running) – 260.
6. Moderate stationary bicycling – 260.
7. Vigorous stationary bicycling – 391.
## Is it OK to go to bed hungry?
Going to bed hungry can be safe as long as you’re eating a well-balanced diet throughout the day. Avoiding late-night snacks or meals can actually help avoid weight gain and an increased BMI. If you’re so hungry that you can’t go to bed, you can eat foods that are easy to digest and promote sleep.
## What happens if you eat less than 1000 calories?
If you take in fewer calories than needed, you will lose weight. Restricting intake to fewer than 1,000 calories daily can slow down your metabolic rate and lead to fatigue since you’re not taking in enough calories to support even the basic functions that keep you alive.
IT IS INTERESTING: How can I lose weight if I don't like vegetables?
## Where do you start losing fat first?
For some people, the first noticeable change may be at the waistline. For others, the breasts or face are the first to show change. Where you gain or lose weight first is likely to change as you get older. Both middle-aged men and postmenopausal women tend to store weight around their midsections.
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# Find the relation
we have to find the value of 'a' & 'b'
Given:
f(x) is continuous at x = 3
If f(x) is be continuous at x = 3,then,f(3) = f(3) + = f(3)
LHL = f(3) =
3a – 0 × a + 1
3a + 1 .....(1)
LHL = f(3) + =
3b – 0 × b + 3
3b + 3 ...(2)
Since ,f(x) is continuous at x = 3 and From (1) & (2),we get
3a + 1 = 3b + 3
3a + 3b = 3 – 1
3a + 3b = 2
3(a + b) = 2
(a + b) =
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m119+q5+spr2010
# m119+q5+spr2010 - g 4 3 2 − ⋅ = 3 3 2 4 x x y − = 4...
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M119 Quiz 5 Spring 2010 Name: _____________________________ Clearly print first and last name No graphing calculators . Find the derivative of each function. 1. 3 4 ) 1 ( 5 + = x y 2. t e t t
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Unformatted text preview: g 4 3 2 ) ( − ⋅ = 3. 3 2 4 x x y − = 4. If x y ln = , find 2 2 dx y d 5. Find an equation for the line tangent to the graph of 1 7 3 2 + − = x x y at 1 = x...
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## This note was uploaded on 10/23/2011 for the course MATH-M 119 taught by Professor Rainey during the Spring '10 term at IUPUI.
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Class 10
MATHS
Geometry
# In the diagram , O is the centre of the circle and angle OPA=30^@, Find angle ACB and angle ADB respectively. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PS_MATH_X_C13_E06_001_Q01.png" width="80%">
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120^@,60^@ 60^@,120^@ 75^@,105^@ 35^@, 145^@
B
Solution :
(i) Tangents drawn form an external point to the same circle are equal, i.e, PA=PB. <br> (ii) angle APO=angle OPB and angle AOB +angle APB=180^@ <br> (iii) angle ACB=(1)/(2)angle AOB. <br> (iv) ACBD is a cyclic quadrilateral.
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# Tricky questions
### Orange Kangaroo In Denmark
Effect: A person is given a few instructions with a seemingly unlimited choice of answers. They are instructed not to reveal their chosen answers. You should then be able to predict what that person had been thinking of, to everybody's amazement.
Method : Read out the following instructions to your friend and get them to think of the answer as soon as possible. ( They may want to scribble some notes on paper to make the task easier. )
• Think of a number from 1 to 10.
• Multiply that number by 9.
• If the number is a 2-digit number, add the digits together.
• Now subtract 5.
• Determine which letter in the alphabet corresponds to the number you ended up with (example: 1=a, 2=b, 3=c, e.t.c.)
• Think of a country that starts with that letter.
• Remember the last letter of the name of that country.
• Think of the name of an animal that starts with that letter.
• Remember the last letter in the name of that animal.
• Think of the name of a fruit that starts with that letter.
You can then ask them if they are thinking of a Kangaroo in Denmark eating an Orange?
This is a really cool prediction trick that should work most of the time. The secret is in the instructions. By instruction 3 your friend will always have come up with the number 9. Subtracting 5, leaves you with he the number 4. The number 4 is then converted to the letter D.
From here it is down to a question of probability, as Denmark is by far the most common choice that people tend to think of. Further on when they are trying to think of an animal beginning with K, again Kangaroo is by far the most common choice although you may very occasionally be caught out with a Koala or Kiwi as an answer.
This cool magic prediction trick is meant as a light hearted bit of amusement and can really catch people out, leaving them wondering if you can actually read their mind.
### 3 glasses and one is upside down
Turn then all up in 3 moves by flipping 2 glasses at a time
### Lily in the pond
In a pond, there is a patch of lily pads. Each day, the lily pads double in size. If it takes the lily pads 48 days to cover the whole pond, how many days does it take to cover half the pond?
Answer: 47 days. Whatever day the lily pads covered half the pond, the whole pond would be covered the next day.
### Arm coordination
Twist arms and lock them in front of you. Turn the arms lock to be in front of your face.
Now move one finger after another.
### General coordination
Touch your pointer fingers in front of you.
Now close one eye and do the same.
### How many holes does the t-shit have?
Add the hole at the front side which is the same hole on the back side. Usually person is forgetting 4 extra holes for arms, head and body.
### Old lady, friend and girl on bus stop
You’re driving down the road in your car on a wild and stormy night. The weather is like a hurricane, with heavy rains, high winds, and lightning flashing constantly. While driving, you come across a partially-covered bus stop, and you can see three people waiting for a bus:
• An old woman who looks as if she is about to die.
• An old friend who once saved your life.
• The perfect partner you have been dreaming about (your “soulmate”).
Knowing that you only have room for one passenger in your car (it’s a really small car), which one would you choose to offer a ride to? And why?
Answer: “I would give the car keys to my old friend, and let him take the old woman to the hospital. Then I would stay behind and wait for the bus with the partner of my dreams.”
### Trolley problem
You see a runaway trolley moving toward five tied-up (or otherwise incapacitated) people lying on the tracks. You are standing next to a lever that controls a switch. If you pull the lever, the trolley will be redirected onto a side track, and the five people on the main track will be saved. However, there is a single person lying on the side track. You have two options:
• Do nothing and allow the trolley to kill the five people on the main track.
• Pull the lever, diverting the trolley onto the side track where it will kill one person.
Which is the more ethical option?
### Do circle with one hand and square with another.
It is impossible. The left side of the brain gives commands to right hand and vice versa. So tho brains are conflicting.
### Say 50 words in 20 seconds without letter A
Say all numbers. First number with A is thousand.
### Men needs to cross the river on boat
Fox, chicken and grain.
Boat can fit only man and one item.
You cant leave fox and chicken, or chicken and grain together.
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# Demystifying the LOWER Function in Google Sheets: A Comprehensive Guide
Table of Content
Are you tired of staring at your spreadsheet, wondering why you can't make heads or tails of that LOWER function? Fear not, for today we embark on a journey of enlightenment. Together, we will unravel the secrets of the LOWER function in Google Sheets. Prepare to become a LOWER master!
## Mastering the LOWER Function
Before we can conquer the LOWER function, we must first understand its syntax. In its simplest form, the LOWER function converts text to lowercase. Just imagine it as a magical spell that transforms your uppercase nightmares into lowercase dreams.
Now that we have the basics down, let's dive into some practical examples. Brace yourself for a wild ride through the wonderful world of LOWER.
### Understanding the Syntax of the LOWER Formula
So how does one concoct this spell of lowercase wizardry? It's as easy as pie. Simply use the following syntax:
``=LOWER(text)``
Replace "text" with the cell reference or the actual text you want to convert. Voila! The LOWER function will work its charm and create lowercase magic.
But what if you want to convert multiple cells at once? Fear not, for the LOWER function can handle that too. Just select the range of cells you want to convert and apply the LOWER function to the first cell. The magic will spread like wildfire, transforming all the selected cells into lowercase.
### Practical Examples of Using the LOWER Function
Let's say you have a list of names that are written in all caps. It's time to unleash the power of LOWER and bring order to the chaos! By applying the LOWER function to the list, you can transform "JOHN," "JANE," and "JACK" into the more elegant "john," "jane," and "jack." Talk about a lowercase makeover!
But wait, there's more! You can also use the LOWER function to extract hidden gems from a mix of lowercase and uppercase. Imagine a scenario where you have a jumbled mess of text, with certain keywords hidden within. By lowering everything down with the LOWER function, you can spot those elusive keywords and conquer the data jungle like the spreadsheet hero you are.
Now, let's explore another practical example. Imagine you have a column of email addresses, some of which are in uppercase. By applying the LOWER function to the column, you can ensure that all email addresses are in lowercase. This can be particularly useful when you need to compare email addresses or perform any case-insensitive operations.
### Tips and Tricks for Maximizing the Potential of LOWER
As with any magical spell, there are always tips and tricks to take your powers to the next level. Here are some golden nuggets of wisdom:
1. Combine with other functions: Don't be afraid to team up the LOWER function with other formulas. The possibilities are endless! Concatenate it with the mighty CONCATENATE function or nest it within the powerful IF function. You can create complex formulas that manipulate text in ways you never thought possible.
2. Watch out for non-alphabetic characters: The LOWER function works its magic on alphabets, but other characters might remain unaffected. That tricky exclamation mark won't be converted to lowercase, so be wary! If you want to ensure that all characters are converted, you may need to use additional functions or formulas.
3. Embrace cell references: Instead of typing out the actual text, use cell references for flexibility. This way, you can change the text in one cell and watch the magic ripple through your spreadsheet. It also makes it easier to apply the LOWER function to multiple cells or ranges without having to manually type the text each time.
4. Combine with conditional formatting: Conditional formatting is a powerful tool that allows you to format cells based on specific criteria. By combining the LOWER function with conditional formatting, you can highlight cells that contain uppercase text or apply custom formatting to lowercase text. This can be especially useful when dealing with large datasets or when you want to visually identify uppercase or lowercase text.
## Avoiding Common Mistakes with the LOWER Formula
Even the most skilled wizards stumble occasionally. Here are some common mistakes to watch out for when dealing with the LOWER function:
Forgetting to close brackets: The LOWER function, like any good spell, requires proper spellcasting etiquette. Remember to close those brackets, or your formula might just whimper in confusion.
Imagine this scenario: you're working on a magical spreadsheet, trying to transform a column of names into lowercase. You confidently type in the LOWER function, but in your haste, you forget to close the brackets. As you cast the spell, the formula fails to work its magic. The names remain in their original form, mocking your oversight. Remember, even the most powerful wizards can make simple mistakes, so always double-check your brackets to ensure the LOWER function works its spellbinding charm.
Using the wrong data type: The LOWER function is designed for text, not numbers. Trying to convert a number with the LOWER function will result in a syntax error. Keep your text and numbers in their respective realms to maintain harmony in your spreadsheet kingdom.
Picture this: you're working on a mystical spreadsheet, and you come across a column of numbers that you want to transform into lowercase. Without thinking twice, you apply the LOWER function to the entire column. Suddenly, your spreadsheet rebels, displaying a syntax error instead of the expected lowercase numbers. What went wrong? The LOWER function is a wizard of text manipulation, not number transformation. To avoid such mishaps, always remember to use the LOWER function exclusively for text-related tasks. Keep your numbers in their own realm, allowing the LOWER function to work its magic on the appropriate data type.
As you navigate the enchanting world of spreadsheet sorcery, be mindful of these common mistakes with the LOWER function. By closing your brackets and using the function on the correct data type, you'll ensure a seamless and spellbinding experience in your spreadsheet kingdom.
## Troubleshooting: Why Isn't My LOWER Formula Working?
Oh no, the LOWER function is not cooperating! When faced with such a predicament, fear not. We're here to troubleshoot and get that formula back on track. Here are a few common issues:
• Hidden extra characters: Sometimes, invisible gremlins sneak into your text and wreak havoc on your LOWER formula. Check for hidden spaces or special characters that might be causing confusion.
• Incorrect cell references: Double-check that you've referenced the correct cells. A simple typo can throw your formula off balance.
Now, let's dive deeper into these issues to better understand how they can affect your LOWER formula.
Hidden extra characters: While invisible to the naked eye, hidden spaces or special characters can cause unexpected results in your LOWER formula. These sneaky gremlins can find their way into your text when copying and pasting from different sources or when importing data from external files. To identify and eliminate these hidden culprits, you can use the TRIM function to remove any leading or trailing spaces. Additionally, you can use the CLEAN function to remove non-printable characters, such as line breaks or tabs, that might be causing confusion.
Incorrect cell references: One tiny mistake in referencing the wrong cells can lead to frustration when your LOWER formula doesn't produce the desired outcome. It's crucial to double-check that you've accurately specified the cell range or individual cell reference in your formula. Take a moment to review the cell addresses and ensure they match the intended data. Remember, even a simple typo, like using a lowercase "a" instead of an uppercase "A," can throw off your formula's balance.
By paying attention to these common issues and taking the necessary steps to address them, you'll be well on your way to resolving the troubles with your LOWER formula. Remember, troubleshooting is an essential part of the formula-building process, and with a little patience and perseverance, you'll conquer any formula-related challenge that comes your way!
## Exploring Related Formulae to LOWER
The LOWER function is just the tip of the magical iceberg. There are other formulae waiting to be discovered. Dive in and explore the wonders of PROPER, UPPER, and TRIM functions. Combine them together, and you shall unlock a formulae symphony!
Let's take a closer look at the PROPER function. This amazing formula capitalizes the first letter of each word in a given text. Imagine having a long list of names or titles that are not properly capitalized. With the PROPER function, you can easily transform them into a more professional and visually appealing format. It's like giving your text a makeover!
Now, let's move on to the UPPER function. This powerful formula converts all the letters in a text to uppercase. Whether you want to make a bold statement or ensure consistency in your data, the UPPER function has got you covered. It's like shouting your text from the rooftops, demanding attention and authority!
Lastly, we have the TRIM function. This handy formula removes any extra spaces from a text. Have you ever copied and pasted data from different sources, only to find unwanted spaces messing up your formatting? The TRIM function comes to the rescue, eliminating those pesky spaces and leaving your text clean and tidy. It's like decluttering your data, making it easier to read and work with!
Now that you hold the key to LOWER greatness, go forth and conquer your spreadsheet kingdom. Transform uppercase nightmares into lowercase dreams with a single formula. The power is in your hands!
But wait, there's more! As you delve deeper into the world of Excel formulas, you'll discover a vast array of functions waiting to be explored. From mathematical calculations to date and time manipulation, Excel offers a treasure trove of possibilities. So don't stop at LOWER, keep expanding your formula repertoire and unlock the true potential of your spreadsheets.
Remember, practice makes perfect. The more you experiment with different formula combinations, the more confident and proficient you'll become. So don't be afraid to dive in, make mistakes, and learn from them. With each formula you master, you'll become a more efficient and effective spreadsheet wizard.
So go ahead, embrace the world of Excel formulas and let your creativity soar. Unleash the power of PROPER, UPPER, TRIM, and many more functions to transform your data and elevate your spreadsheets to new heights. The possibilities are endless, and the rewards are immense.
###### Simon Taylor
Hi there!
I'm Simon, your not-so-typical finance guy with a knack for numbers and a love for a good spreadsheet. Being in the finance world for over two decades, I've seen it all - from the highs of bull markets to the 'oh no!' moments of financial crashes. But here's the twist: I believe finance should be fun (yes, you read that right, fun!).
As a dad, I've mastered the art of explaining complex things, like why the sky is blue or why budgeting is cool, in ways that even a five-year-old would get (or at least pretend to). I bring this same approach to THINK, where I break down financial jargon into something you can actually enjoy reading - and maybe even laugh at!
So, whether you're trying to navigate the world of investments or just figure out how to make an Excel budget that doesn’t make you snooze, I’m here to guide you with practical advice, sprinkled with dad jokes and a healthy dose of real-world experience. Let's make finance fun together!
### Related Articles:
Your navigator through the financial jungle. Discover helpful tips, insightful analyses, and practical tools for taxes, accounting, and more. Empowering you to make informed financial decisions every step of the way.
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Πολεοδομικά Έργα
25 Νοε 2013 (πριν από 4 χρόνια και 5 μήνες)
151 εμφανίσεις
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM
FOR THE TEST ENGINEER
Christopher C. Flanigan
Quartus Engineering Incorporated
San Diego, California USA
18th International Modal Analysis Conference (IMAC
-
XVIII)
San Antonio, Texas
February 7
-
10, 2000
Quartus Engineering
Quartus Engineering Incorporated, 2000.
QUARTUS ENGINEERING WEB SITE
http://www.quartus.com
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM PEOPLE ARE REALLY SMART
Or so they would have you believe!
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM for the Test Engineer
TOPICS
There’s reality, and then there’s FEM
FEM in a nutshell
FEM strengths and challenges
Pretest analysis
Model reduction
Sensor placement
Posttest analysis
Correlation
Model updating
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Quartus Engineering
Quartus Engineering Incorporated, 2000.
There’s Reality, and Then There’s FEM
REALITY IS VERY COMPLICATED!
Many complex subsystems
Unique connections
Nonlinearities
Flight
-
to
-
flight variability
Chaos
Extremely high order behavior
Quartus Engineering
Quartus Engineering Incorporated, 2000.
There’s Reality, and Then There’s FEM
FEM ATTEMPTS TO
SIMULATE REALITY
Fortunately, reality is
surprisingly linear
Material properties (
v献s
)
Tension vs. compression
Small deflections (sin
)
Allows reasonable
opportunity simulate reality
using FEM
-1
-0.75
-0.5
-0.25
0
0.25
0.5
0.75
1
-1
-0.5
0
0.5
1
Quartus Engineering
Quartus Engineering Incorporated, 2000.
There’s Reality, and Then There’s FEM
REMEMBER THAT FEM
ONLY APPROXIMATES REALITY
Reality has lots of hard challenges
Nonlinearity, chaos, etc.
FEM limited by many factors
Engineering knowledge and capabilities
Basic understanding of mechanics
Computer and software power
But it’s the best approach we have
Experience shows that FEM works well when used properly
FEM
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM Strengths and Challenges
TEST IS NOT REALITY EITHER!
Test article instead of flight article
Mass simulators, missing items, boundary conditions
Excitation limitations
Load level, spectrum (don’t break it!)
Nonlinearities
Testing limitations
Sensor accuracy and calibration
Data processing
But it’s the best “reality check” available
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM
in a Nutshell
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM for the Test Engineer
FEM IN A NUTSHELL
Divide and conquer!
Shape functions
Elemental stiffness and mass matrices
Assembly of system matrices
Solving
Related topics
Element library
Superelements
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM in a Nutshell
CLOSED FORM SOLUTIONS, ANYONE?
Consider a building
Steel girders
Concrete foundation
Can you write an equation to
fully describe the building?
I can’t!
Even if possible, probably not
the best approach
Very time consuming
One
-
time solution
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM in a Nutshell
DIVIDE AND CONQUER!
Behavior of complete
structure is complex
Example: membrane
Divide the membrane
into small pieces
Buzzword: “element”
Feasible to calculate
properties of each piece
Collection of pieces
represents structure
1
3
5
7
9
11
13
15
17
19
S1
S3
S5
S7
S9
S11
S13
S15
S17
S19
-1.00
-0.80
-0.60
-0.40
-0.20
0.00
0.20
0.40
0.60
0.80
1.00
0.80-1.00
0.60-0.80
0.40-0.60
0.20-0.40
0.00-0.20
-0.20-0.00
-0.40--0.20
-0.60--0.40
-0.80--0.60
-1.00--0.80
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM in a Nutshell
SHAPE FUNCTIONS ARE THE
FOUNDATION OF FINTE ELEMENTS
Shape function
Assumed shape of element when deflected
Some element types are simple
Springs, rods, bar
Other elements are more difficult
Plates, solids
But that’s what Ph.D.’s are for!
Extensive research
Still evolving (MSC.NASTRAN V70.7)
Spring
F = K X
F
X
K
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM in a Nutshell
ELEMENT STIFFNESS MATRIX
FORMED USING SHAPE FUNCTIONS
Element stiffness matrix
Relates deflections of elemental DOF
Forces at element DOF when unit
deflection imposed at DOF
i
and
other DOF
j
are fixed
Example: linear spring (2 DOF)
Spring
F = K X
F
X
K
K
K
K
K
K
spring
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM in a Nutshell
ELEMENT MASS MATRIX
HAS TWO OPTIONS
Lumped mass
Apply 1/N of the element mass to each node
Consistent mass
Called “coupled mass” in NASTRAN
Use shape functions to generate mass matrix
In practice, usually little difference
between the two methods
Consistent mass more accurate
Lumped mass faster
M
5
.
0
0
0
M
5
.
0
M
spring
1/4
1/4
1/4
1/4
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM in a Nutshell
SYSTEM MATRICES FORMED
FROM ELEMENT MATRICES
K = 2
K = 5
K = 1
M = 1
M = 2
M = 3
2
2
2
2
K
1
5
5
5
5
K
2
1
1
1
1
K
3
1
1
0
0
1
6
5
0
0
5
7
2
0
0
2
2
K
5
.
1
0
0
0
0
5
.
2
0
0
0
0
5
.
1
0
0
0
0
5
.
0
M
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM in a Nutshell
CALCULATE SYSTEM STATIC
AND DYNAMIC RESPONSES
Static analysis
Normal modes analysis
Transient analysis
P
q
K
q
C
q
M
T
T
T
T
0
M
K
i
i
X
K
P
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM in a Nutshell
COMMERCIAL FEM ISSUES
Element libraries
Springs, rods, beams, shells, solids, rigids, special
Linear and parabolic (shape functions, vertex nodes)
Commercial codes
NASTRAN popular for linear dynamics (aero, auto)
ABAQUS and ANSYS popular for nonlinear
Superelements (substructures)
Simply a collection of finite elements
Special capabilities to reduce to boundary nodes
Assemble system by addition I/F nodes
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM in a Nutshell
HONORARY DEGREE IN FEM
-
OLOGY!
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM for the Test Engineer
FEM STRENGTHS AND CHALLENGES
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM Strengths and Challenges
FEM IS VERY POWERFUL FOR
WIDE ARRAY OF STRUCTURES
Regular structures
Fine mesh
Sturdy connections
Seam welds
Well
-
defined mass
Smooth distributed
Small lumped masses
Linear response
Small displacements
General Dynamics
Control
-
Structure Interaction Testbed
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM Strengths and Challenges
FEM HAS MANY CHALLENGES
Mesh refinement
How many elements required?
Material properties
A
-
basis, B
-
basis, etc.
Composites
Dimensions
Tolerances, as
-
manufactured
Joints
Fasteners, bonds, spot welds
continued...
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM Strengths and Challenges
FEM HAS MANY CHALLENGES
Mass modeling
Accuracy of mass prop DB
Difficulty in test/weighing
Secondary structures
Avionics boxes, batteries
Wiring harnesses
Shock mounts
Nonlinearities
(large deformation, slop, yield, etc.)
Pilot error!
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM Strengths and Challenges
IN H/W AND S/W POWER
Computers
Moore’s law for CPU
Disk space, memory
Software
Sparse, iterative
Lanczos eigensolver
Domain decomposition
Pre
-
and post
-
processing
Increasing resolution
Closer to reality
Moravec, H., “When Will Computer Hardware Match the Human Brain?”
Robotics Institute Carnegie Mellon University
http://www.transhumanist.com/volume1/moravec.htm
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM Strengths and Challenges
FEM CONTINUES TO IMPROVE
ABILITY TO SIMULATE REALITY
Model resolution
Local details
Some things still
very difficult
Joints
Expertise
Mesh size, etc.
FEM is not exact
Big models do not guarantee accurate models
That’s why testing is still required!
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM for the Test Engineer
PRETEST ANALYSIS
Develop
FEM
Pretest
Analysis
Test
Posttest
Correlation
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
MODAL SURVEY OFTEN PERFORMED
TO VERIFY FINITE ELEMENT MODEL
Must be confident that structure will survive
operating environment
Unrealistic to test flight structure to flight loads
Alternate procedure
Test structure under controlled conditions
Correlate model to match test results
Use test
-
correlated model to predict operating responses
Modal survey performed to verify analysis model
“Reality check”
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM
TEST AND ANALYSIS DATA HAVE
DIFFERENT NUMBER OF DOF
Model sizes
FEM = 10,000
-
1,000,000 DOF
Test = 50
-
500 accelerometers
Compare test results to
analysis predictions
Need a common basis for
comparison
M
Ortho
T
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM
TEST
-
ANALYSIS MODEL (TAM)
PROVIDES BASIS FOR COMPARISON
Test
-
analysis model (TAM)
Mathematical reduction of finite element model
Master DOF in TAM corresponds to accelerometer
Transformation (condensation)
Many methods to perform reduction transformation
Transformation method and sensor selection critical
for accurate TAM and test
-
analysis comparisons
ga
gg
T
ga
aa
ga
gg
T
ga
aa
T
M
T
M
T
K
T
K
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
GUYAN REDUCTION IS THE
INDUSTRY STANDARD METHOD
Robert Guyan, Rockwell, 1965
Pronounced “Goo
-
yawn”, not “Gie
-
yan”
Implemented in many commercial software codes
NASTRAN, I
-
DEAS, ANSYS, etc.
Assume forces at omitted DOF are negligible
a
o
a
o
aa
ao
oa
oo
P
P
U
U
K
K
K
K
0
P
o
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
GUYAN REDUCTION IS A
SIMPLE METHOD TO IMPLEMENT
Solve for motion at omitted DOF
Rewrite static equations of motion
Transformation matrix for Guyan reduction
a
oa
1
oo
o
U
K
K
U
a
aa
oa
1
oo
a
o
U
I
K
K
U
U
aa
oa
1
oo
Guyan
I
K
K
T
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
TRANSFORMATION VECTORS
ESTIMATE MOTION AT “OTHER” DOF
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1
2
3
4
Node ID
Displacement
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
TRANSFORMATION VECTORS CAN
REDUCE OR EXPAND DATA
TAM
Display
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
DISPLAY MODEL RECOVERED USING
TRANSFORMATION VECTORS
-1.00
-0.75
-0.50
-0.25
0.00
0.25
0.50
0.75
1
2
3
4
Node ID
Enhanced Display
-1.00
-0.75
-0.50
-0.25
0.00
0.25
0.50
0.75
1
2
3
4
Standard Display
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
FIRST ORDER MASS CORRECTION
Guyan neglects mass effects at omitted DOF
IRS adds first order approximation of mass effects
aa
IRS
Guyan
Guyan
I
G
G
T
oa
1
oo
Guyan
K
K
G
aa
1
aa
Guyan
oo
oa
1
oo
IRS
K
M
G
M
M
K
G
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
DYNAMIC REDUCTION ALSO
Replace eigenvalue with constant value
L
Equivalent to Guyan reduction if
L
㴠=
i
a
o
aa
ao
oa
oo
i
i
a
o
aa
ao
oa
oo
M
M
M
M
K
K
K
K
L
L
aa
oa
oa
1
oo
oo
d
Re
Dyn
I
M
K
M
K
T
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
MODAL TAM BASED ON
FEM MODE SHAPES
Partition FEM mode shapes
Pseudo
-
inverse to form transformation matrix
o
o
U
a
a
U
aa
T
a
1
a
T
a
o
Modal
I
T
a
al
mod
o
U
T
U
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
EACH REDUCTION METHOD HAS
STRENGTHS AND WEAKNESSES
Easy to use, efficient
Limited accuracy
Guyan
Works well if good A-set
Widely accepted
Unacceptable for high M/K
Better than Guyan
Requires DMAP alter
IRS
Errors if poor A-set
Better than Guyan
Requires DMAP alter
Dynamic
Choice of Lamda?
Limited experience
Exact within freq. range
Requires DMAP alter
Modal
Hybrid TAM option
Sensitivity
Limited experience
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
TAM Transformation Methods
STANDARD PRACTICE FAVORS
GUYAN REDUCTION
Guyan reduction used most often
Easy to use and commercially available
Computationally efficient
Widely used and accepted
Good accuracy for many/most structures
Use other methods when Guyan is inadequate
Modal TAM very accurate but sensitive to FEM error
IRS has 1st order mass correction but can be unstable
Dynamic reduction seldom used (how to choose
L
)
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
Sensor Placement
SENSOR PLACEMENT IMPORTANT
FOR GOOD TAM AND TEST
Optimize TAM
Minimize reduction error
Optimize test
Get as much independent data as possible
Focus on uncertainties
High confidence areas need only modest instrumentation
More instrumentation near critical uncertain areas (joints)
Common sense and engineering judgement
General visualization of mode shapes
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
Sensor Placement
MANY ALGORITHMS FOR
SENSOR PLACEMENT
Kinetic energy
Retain DOF with large kinetic energy
Mass/stiffness ratio
Retain DOF with high mass/stiffness ratio
Iterated K.E. and M/K
Remove one DOF per iteration
Effective independence
Retain DOF that maximize observability of mode shapes
Genetic algorithm
Survival of the fittest!
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
Sensor Placement
SENSOR PLACEMENT ALGORITHM
Guyan or IRS reduction
Must retain DOF with large mass
Iterated K.E. or M/K
Mass
-
weighted effective independence
Modal or Hybrid reduction
Effective independence
Genetic algorithm offers best of all worlds
Examine tons of TAMs!
Seed generation from other methods
Cost function based on TAM method
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Pretest Analysis
-
Sensor Placement
PRETEST ANALYSIS ASSISTS
PLANNING AND TEST
Best estimate of modes
Frequencies, shapes
Accelerometer locations
Optimized by sensor placement
studies
TAM mass and stiffness
Real
-
time ortho and x
-
ortho
Frequency response functions
Dry runs/shakedown prior to test
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM for the Test Engineer
TEST CONSIDERATIONS
Develop
FEM
Pretest
Analysis
Test
Posttest
Correlation
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Test Considerations
PRETEST DATA ALLOWS
REAL
-
TIME CHECKS OF RESULTS
What if test accuracy goals aren’t met?
Keep testing (different excitement levels, locations, types)
Stop testing (FEM may be incorrect!)
Decide based on test quality checks
Experienced test engineer extremely valuable!
test
TAM
T
test
M
ORTHO
test
TAM
T
TAM
M
XORTHO
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM for the Test Engineer
POSTTEST CORRELATION
Develop
FEM
Pretest
Analysis
Test
Posttest
Correlation
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Posttest Correlation
CORRELATION MUST BE FAST!
FEM almost always has some differences vs. test
Very limited opportunity to do correlation
After structural testing and data processing complete
Before operational use of model
First flight of airplane
Need methods that are fast!
Maximum insight
Accurate
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Posttest Correlation
NO UNIQUE SOLUTION FOR
POSTTEST CORRELATION
More “unknowns” than “knowns”
Knowns
Test data (FRF, frequencies, shapes at
test DOF, damping)
Measured global/subsystem weights
Unknowns
FEM stiffness and mass (FEM DOF)
No unique solution
Seek “best” reasonable solution
When you
have
eliminated
the
impossible,
whatever
remains,
however
improbable,
must be
the truth
.”
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Posttest Correlation
MANY CORRELATION METHODS
Trial
-
and
-
error
Stop doing this! It's (almost)
the new millenium!
Too slow for fast
-
paced projects
Not sufficiently insightful for
complex systems
FEM matrix updating
FEM property updating
Error localization
FEM
Test
OK?
Done
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Posttest Correlation
MATRIX UPDATE METHODS
ADJUST FEM K AND M ELEMENTS
Objective
Identify changes to FEM K and M so that analysis
matches test
Baruch and Bar
-
Itzhack (1978, 1982)
Berman (1971, 1984)
Kabe (1985)
Kammer (1987)
Smith and Beattie (1991)
… and many others
1
1
0
0
1
6
5
0
0
5
7
2
0
0
2
2
K
5
.
1
0
0
0
0
5
.
2
0
0
0
0
5
.
1
0
0
0
0
5
.
0
M
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Posttest Correlation
MATRIX UPDATE METHODS
HAVE LIMITATIONS
Lack of physical insight
What do changes in K, M coefficients mean?
Lack of physical plausibility
Baruch/Berman method doesn't enforce connectivity
Limitations for large problems
Great for small “demo” models, but ...
“Smearing" caused by Guyan reduction/expansion
What if test article different than flight vehicle?
Requires very precise mode shapes (unrealistic)
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Posttest Correlation
PROPERTY UPDATE METHODS
Objective
Identify changes to element and material
properties so that FEM matches test
Hasselman (1974)
Chen (1980)
Flanigan (1987, 1991)
Blelloch (1992)
Smith (1995)
… and many others
design sensitivity and optimization
FEM
Test
OK?
Done
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Posttest Correlation
COMMERCIAL SOFTWARE
FOR CORRELATION
SDRC/MTS
I
-
DEAS Correlation (MAC, ortho, x
-
ortho, mapping)
LMS
MSC
SOL 200 design optimization (modes, FRF)
Dynamic Design Solutions (DDS)
FEMtools (follow
-
on to Systune)
Others (SSID, ITAP, etc.)
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Posttest Correlation
MODE SHAPE EXPANSION
FOR CORRELATION IMPROVEMENT
TAM
Display
Quartus Engineering
Quartus Engineering Incorporated, 2000.
Posttest Correlation
SHAPE EXPANSION IS AN
ALTERNATIVE TO MATRIX REDUCTION
Expand test mode shapes to FEM DOF
Expansion and reduction give same results if same
matrices used
Dynamic expansion based on eigenvalue equation
Computationally intensive
But computers are getting faster all the time!
a
ga
g
U
T
U
i
a
oa
i
oa
oo
i
oo
i
o
M
K
M
K
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM for the Test Engineer
SUMMARY
FEM is a simple yet powerful method
Complex structures from simple building blocks
FEM must make many assumptions
Joints, tolerances, linearity, mass, etc.
Big models do not guarantee accuracy
Testing provides a valuable “reality check”
Within limits of test article, excitation levels, etc.
FEM can work closely with test for mutual benefit
Pretest analysis to optimize sensor locations
TAM for providing test
-
analysis comparison basis
Correlation and model updating for validated model
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM PEOPLE REALLY ARE SMART!
And maybe test people are smart too!
Quartus Engineering
Quartus Engineering Incorporated, 2000.
FEM for the Test Engineer
Finite element method
Concepts and Applications of Finite Element Analysis
, 3rd ed.; Cook,
Robert D./Plesha, Michael E./Malkus, David S.; John Wiley & Sons; 1989
Finite Element Procedures
, Klaus
-
Jurgen Bathe; Prentice Hall; 1995
Correlation and model updating
Finite Element Model Updating in Structural Dynamics
; M. I. Friswell,
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CC-MAIN-2018-17
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https://www.jobilize.com/course/section/basic-integration-formulas-by-openstax?qcr=www.quizover.com
| 1,568,762,259,000,000,000
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crawl-data/CC-MAIN-2019-39/segments/1568514573124.40/warc/CC-MAIN-20190917223332-20190918005332-00306.warc.gz
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5.4 Integration formulas and the net change theorem
Page 1 / 8
• Apply the basic integration formulas.
• Explain the significance of the net change theorem.
• Use the net change theorem to solve applied problems.
• Apply the integrals of odd and even functions.
In this section, we use some basic integration formulas studied previously to solve some key applied problems. It is important to note that these formulas are presented in terms of indefinite integrals. Although definite and indefinite integrals are closely related, there are some key differences to keep in mind. A definite integral is either a number (when the limits of integration are constants) or a single function (when one or both of the limits of integration are variables). An indefinite integral represents a family of functions, all of which differ by a constant. As you become more familiar with integration, you will get a feel for when to use definite integrals and when to use indefinite integrals. You will naturally select the correct approach for a given problem without thinking too much about it. However, until these concepts are cemented in your mind, think carefully about whether you need a definite integral or an indefinite integral and make sure you are using the proper notation based on your choice.
Basic integration formulas
Recall the integration formulas given in [link] and the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.
Integrating a function using the power rule
Use the power rule to integrate the function ${\int }_{1}^{4}\sqrt{t}\left(1+t\right)dt.$
The first step is to rewrite the function and simplify it so we can apply the power rule:
$\begin{array}{cc}{\int }_{1}^{4}\sqrt{t}\left(1+t\right)dt\hfill & ={\int }_{1}^{4}{t}^{1\text{/}2}\left(1+t\right)dt\hfill \\ \\ & ={\int }_{1}^{4}\left({t}^{1\text{/}2}+{t}^{3\text{/}2}\right)dt.\hfill \end{array}$
Now apply the power rule:
$\begin{array}{cc}{\int }_{1}^{4}\left({t}^{1\text{/}2}+{t}^{3\text{/}2}\right)dt\hfill & ={\left(\frac{2}{3}{t}^{3\text{/}2}+\frac{2}{5}{t}^{5\text{/}2}\right)|}_{1}^{4}\hfill \\ & =\left[\frac{2}{3}{\left(4\right)}^{3\text{/}2}+\frac{2}{5}{\left(4\right)}^{5\text{/}2}\right]-\left[\frac{2}{3}{\left(1\right)}^{3\text{/}2}+\frac{2}{5}{\left(1\right)}^{5\text{/}2}\right]\hfill \\ & =\frac{256}{15}.\hfill \end{array}$
Find the definite integral of $f\left(x\right)={x}^{2}-3x$ over the interval $\left[1,3\right].$
$-\frac{10}{3}$
The net change theorem
The net change theorem considers the integral of a rate of change . It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.
Net change theorem
The new value of a changing quantity equals the initial value plus the integral of the rate of change:
$\begin{array}{}\\ \\ F\left(b\right)=F\left(a\right)+{\int }_{a}^{b}F\text{'}\left(x\right)dx\hfill \\ \hfill \text{or}\hfill \\ {\int }_{a}^{b}F\text{'}\left(x\right)dx=F\left(b\right)-F\left(a\right).\hfill \end{array}$
Subtracting $F\left(a\right)$ from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.
The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement .
We looked at a simple example of this in The Definite Integral . Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in [link] .
I don't understand the set builder nototation like in this case they've said numbers greater than 1 but less than 5 is there a specific way of reading {x|1<x<5} this because I can't really understand
x is equivalent also us 1...
Jogimar
a < x < b means x is between a and b which implies x is greater than a but less than b.
Bruce
alright thanks a lot I get it now
Ivwananji
I'm trying to test this to see if I am able to send and receive messages.
👍
Jogimar
It's possible
joseph
yes
Waidus
Anyone with pdf tutorial or video should help
joseph
maybe.
Jogimar
hellow gays. is it possible to ask a mathematical question
Michael
What is the derivative of 3x to the negative 3
-3x^-4
Mugen
that's wrong, its -9x^-3
Mugen
Or rather kind of the combination of the two: -9x^(-4) I think. :)
Csaba
-9x-4 (X raised power negative 4=
Simon
-9x-³
Jon
I have 50 rupies I spend as below Spend remain 20 30 15 15 09 06 06 00 ----- ------- 50 51 why one more
Pls Help..... if f(x) =3x+2 what is the value of x whose image is 5
f(x) = 5 = 3x +2 x = 1
x=1
Bra
can anyone teach me how to use synthetic in problem solving
Mark
x=1
Mac
can someone solve Y=2x² + 3 using first principle of differentiation
ans 2
Emmanuel
Yes ☺️ Y=2x+3 will be {2(x+h)+ 3 -(2x+3)}/h where the 2x's and 3's cancel on opening the brackets. Then from (2h/h)=2 since we have no h for the limit that tends to zero, I guess that is it....
Philip
Correct
Mohamed
thank you Philip Kotia
Rachael
Welcome
Philip
g(x)=8-4x sqrt of 3 + 2x sqrt of 8 what is the answer?
Sheila
I have no idea what these symbols mean can it be explained in English words
which symbols
John
How to solve lim x squared two=4
why constant is zero
Rate of change of a constant is zero because no change occurred
Highsaint
What is the derivative of sin(x + y)=x + y ?
_1
Abhay
How? can you please show the solution?
Frendick
neg 1?
Frendick
find x^2 + cot (xy) =0 Dy/dx
Continuos and discontinous fuctions
integrate dx/(1+x) root 1-x square
Put 1-x and u^2
Ashwini
hi
telugu
d/dxsinh(2xsquare-6x+4)
how
odofin
how to do trinomial factoring?
Give a trinomial to factor and I'll show you how
Bruce
i dont know how to do that either. but i really wants to know know how
Bern
Do either of you have a trinomial to present that needs factoring?
Bruce
5x^2+11x+2 and 2x^2+7x-4
Samantha
Thanks, so I'll show you how to do the 1st one and then you can try to do the second one. The method I show you is a general method you can use to factor ANY factorable polynomial.
Bruce
ok
Samantha
I have to link you to a document on how to do it since this chat does not have LaTeX
Bruce
Give me a few
Bruce
Sorry to keep you waiting. Here it is: ***mathcha.io/editor/ZvZkJUdrHpVHXwhe7
Bruce
You are welcome to ask questions if you have them. Good luck
Bruce
Can I ask questions with photos ?
Lemisa
If you want
Bruce
I don't see an option for photo
Lemisa
Do you know Math?
Lemisa
You have to post it as a link
Bruce
Can you tell me how to. Post?
Lemisa
Yes, you can upload your photo on imgur and post the link here
Bruce
thanks it was helpful
Samantha
I'm glad that was able to help you. Seriously, if you have any questions, please ask. That is not easy to understand at first so I anticipate you will have questions. It would be best to convince that you understood to attempt factoring the second trinomial you posted.
Bruce
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| 4.9375
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CC-MAIN-2019-39
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longest
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https://www.unitconverter.dev/volume/imperialquart/usliquidgallon/50/
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|
# 🧮 Volume
## Imperial Quart to US Liquid Gallon
The volume conversion of 50 imperial quart is ~15.010507355149 us liquid gallon.
Imperial Quart
to
US Liquid Gallon
Imperial Quart US Liquid Gallon
0.01 ~0.0030021014710297
0.05 ~0.015010507355149
0.1 ~0.030021014710297
0.25 ~0.075052536775743
1 ~0.30021014710297
5 ~1.5010507355149
10 ~3.0021014710297
20 ~6.0042029420594
50 ~15.010507355149
100 ~30.021014710297
### Volume
Volume is the quantity of three-dimensional space enclosed by a closed surface, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains. Volume is often quantified numerically using the SI derived unit, the cubic metre. The volume of a container is generally understood to be the capacity of the container; i. e., the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces. Three dimensional mathematical shapes are also assigned volumes. Volumes of some simple shapes, such as regular, straight-edged, and circular shapes can be easily calculated using arithmetic formulas. Volumes of complicated shapes can be calculated with integral calculus if a formula exists for the shape's boundary. One-dimensional figures (such as lines) and two-dimensional shapes (such as squares) are assigned zero volume in the three-dimensional space.
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CC-MAIN-2024-33
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4 Replies Latest reply on Sep 16, 2015 8:35 AM by John Sobczak
# Fundamental Question About Co-horts Using Sets
This should be low hanging fruit for many of you sharp folks
In the attached mock workbook I'm trying to get a set of Insured for the records New or Renewal = 'New' and Effective Year = '2015' to further use in cohort analysis. I'm not quite able to get this as it is picking up one extra record for 2015 in which New or Renewal = "Renewal' because the Insured member 'L' has both 'New' AND 'Renewal' values in 2015. The Premium for this record is 140 which is bringing the set premium for 2015 up to 480 (it should be 340).
I'm using the below logic in defining my set but have also tried other varying combinations of sets as included in the mock workbook.
sum(if [Effective Year]=2015 and [New or Renewal]='New' then 1 else 0 end)>0sum(if [Effective Year]=2015 and [New or Renewal]='New' then 1 else 0 end)>0
I've seen some other posts on how to create these calculations, but admittedly since I don't fully intuitively understand the aggregated logic required for sets I am trying by example. I'm probably one combination away from getting it right
I appreciate any help that anyone can offer!
• ###### 1. Re: Fundamental Question About Co-horts Using Sets
John,
I hope I understood this correctly maybe you could just create a LOD calc like this?
Basically at a fixed level of insured and year, if the count of new or renewal is 1 then count or don't. See sheets 3 and 4 in the attached.
• ###### 2. Re: Fundamental Question About Co-horts Using Sets
Thanks Pooja,
I need more than just a number or count as I need the specific Insured Members and associated Premium as defined by a Set to use in cohort analysis to compare this group (set) across any year.
• ###### 3. Re: Fundamental Question About Co-horts Using Sets
Hi John,
You can still use the result of the LOD calc to further analyze. You will slightly need to change the calcs depending on vizLOD if you are using other LOD calcs like include/exclude. Although, I tried this with sets too and it seems to be working fine, except the way I did it with sets wouldn't be dynamic (I think). Because I brought the fields on to the view and then created a set, so if something was to change they will need to be recreated which is why I think LOD would work better.
• ###### 4. Re: Fundamental Question About Co-horts Using Sets
Thanks Pooja.
Yes the method of Sets needs to be formula based and thus dynamic, rather than a fixed member set.
Can you show me using the LOD method in the attached modified workbook on 2nd tab how I can show the bar in 2014 represent the premium in 2014 for the Insured members identified in calculation 1 for 2015? It should be only 1 insured member of L and the premium for that member is 140 in 2014. This would be the cohort portion of 2014 premium. I realize I have new or renewal on the color shelf and the premium is NOT new for this cohort in 2014, but what logic would I use to represent it on the bars?
Thanks.
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# Finance
posted by .
Can anyone explain to me how to use this formula?
Cost of failing to
take a cash discount
Disc
ount percent
100 percent Discount percent
3
60
Final due date Discount period
I am trying to calculate Assume the proceeds from the loan with the compensating balance requirement will be used to take cash discounts. Disregard part b about installment payments. and us the loan costs from part a. If the terms of the cash discount are 2/10, net 60, should the firm borrow the funds to take the discount? I cannot understand how I am supposed to do this. The figures I am using come from the following:
The Watson Corporation is negotiating a loan from PNC Bank. The company needs to borrow \$300,000.
The bank offers a rate of 7% with a 30% compensating balance requirement, or as an alternative, 10% with additional fees of \$4,700 to cover the services the bank is providing. In either case, the rate on the loan is floating (changes as the prime interest rate changes), and the loan would be for one year
## Similar Questions
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Compute the present value of a \$100 cash flow for the following combinations of discount rates and times: a. r = 8 percent. t = 10 years. b. r = 8 percent. t = 20 years. c. r = 4 percent. t = 10 years. d. r = 4 percent. t = 20 years
2. ### finance
compute the present value of a \$100 cash flow for the following combinations of discount rates and times. a. r=8 percent. t= 10 years b. r=8 percent. t= 20 years c. r=4 percent. t= 10 years d. r=4 percent. t= 20 years
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(Mix and Match) Instructions: choose an item from (a), (b), (c), (d) or (e) that best matches the given problem. Enter your response in the space provided. (a) single equivalent discount (b) amount of discount (c) 3 / 10 , n / 30 (d) …
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RECENT C & G 2382 question Goto page Previous 1, 2, 3 ... 5, 6, 7 ... 9, 10, 11 Next
Author Message
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Posted: Mon Apr 04, 2011 10:19 pm no i got some questions wrong not just one or two
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Posted: Mon Apr 04, 2011 10:25 pm
ebee wrote: What is the name of the line that separates a circle in to two equal halves?
Is it Michael?
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Posted: Mon Apr 04, 2011 10:26 pm
RF Lighting wrote: I suppose technically it's D, but in reality it's going to be B.
No - it's B.
B gets you to Table 54.8, and 54.8 tells you that you might need a different size than the ones in the table if the local DNO says so.
So that covers off all of the concerns that D might be the right answer because in theory the DNO make the final call - choosing B means recognising that the DNO might require something special.
ebee wrote: It really was a question I got last month in the exam so I answered D ... I knew they wanted B.
Then WTF did you answer D?
ebee wrote: YES it is a question about BS 7671 and BS7671 tells us the answer is B unless the DNO overrule it then you must follow the DNO, therefore following the DNO is explicit in BS 7671 so the answer can only be D
No - BS 7671, in Table 54.8, tells us that for these supply neutral sizes you use a minimum of those main bonding conductor sizes, but that the local distributor's network conditions may require a larger conductor.
The DNO would not be "overruling" B - B is the answer which means "use the values from the table unless the DNO say otherwise".
RF Lighting wrote: B is wrong because you can't just size the PEB conductor from table 54.8, unless you know for certain that the DNO has no overriding requirements, which you couldn't know with out consulting them.
Yes you can because Table 54.8 tells you that the DNO may have overriding requirements and therefore to size your main bonding cable "in accordance with" 54.8 (the wording in the question, please note) you'll be selecting it in accordance with the fact that the DNO might have their own requirements and that therefore unless you know they don't you should contact them.
And so on and so forth to everyone who thinks that D is the answer because the DNO might require a non-standard value.
dingbat is absolutely right - just read what is in front of you.
But this is not the first time, even today, and nor will it be the last, that I've encountered a truly mind-boggling inability to simply read what's written.
And those 50% of tutors who got it wrong?
They can't read either. There's nothing wrong with the question - the source of their confusion and the cause of their failure is their poor reading ability.
The following 2 users say thank you to ban-all-sheds for this useful post: dingbat (5 Apr 2011), B67BU (4 Apr 2011) Alert Moderators
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Posted: Tue Apr 05, 2011 3:33 am "ebee wrote: It really was a question I got last month in the exam so I answered D ... I knew they wanted B. Then WTF did you answer D? icon_rolleyes.gif" 'cos I'm an awkward git . And I still say D/ because the question does not mention the note as an exception. Therefore you can only determine the size after you have consulted them not before. Poorly worded question I says. If I answered the question in a way that I believe is wrong (even if that belief is 100% deluded & insane) would you have more respect for my answer than if I deliberately answered it as I believe is wrong just to receive an extra mark?
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Posted: Tue Apr 05, 2011 3:50 am
EFLImpudence wrote:
ebee wrote: What is the name of the line that separates a circle in to two equal halves?
Is it Michael?
No it's not, but I'm not giving the answer until I see someone give the answer that Tarant's lot gave or the answer the answer I gave and I also have removed the temporary ban on BAS answering just to give others chance to catch up (Central Line - I did not see that one coming! )
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Posted: Tue Apr 05, 2011 8:52 am
ban-all-sheds wrote: dingbat is absolutely right - just read what is in front of you. But this is not the first time, even today, and nor will it be the last, that I've encountered a truly mind-boggling inability to simply read what's written. And those 50% of tutors who got it wrong? They can't read either. There's nothing wrong with the question - the source of their confusion and the cause of their failure is their poor reading ability.
Halle-bleedin'-lujah!
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Posted: Tue Apr 05, 2011 9:15 am ..Last edited by holmslaw on Mon Apr 09, 2012 8:20 pm, edited 1 time in total
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Posted: Tue Apr 05, 2011 1:49 pm "Exactly the same as giving someone a restaurant menu with a note at the top saying, 'The prices on this menu may increase when we make up your bill, please confirm price before ordering" ie as with table 54.8 the menu prices are meaningless." Nice one Holmslaw. I see no one has answered the circle question so I'll lift the Ban on ban (Bet he knows the answer)
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Posted: Tue Apr 05, 2011 2:04 pm Go on, I'll take the bait - put us out of our misery....! The diameter of a circle separates the circle into two parts of equal area. What were the other options? Disclaimer: I know that wasn't the exact question. It is not the only line which will do this.
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Posted: Tue Apr 05, 2011 4:13 pm
echoes wrote: The diameter of a circle separates the circle into two parts of equal area.
Wrong!
A circle does not posess an area. It encloses an area. Perhaps you are thinking of a disc?
Another example of a badly drafted question. Maybe it should have read What separates the circumference of a circle into two equal lengths?
.
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Posted: Tue Apr 05, 2011 4:24 pm Diameter, good answer, very good answer. It was in fact the answer they were looking forand one of the choices of 4 given. It's wrong though. LOL. Any line intersecting an arc (including a circle) in two places is a "Chord". The chord of that particular question would have a length equal to the diameter of that circle (between its intersecting points) But the question they asked was the "Name" of the line not the "length" of it.
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Posted: Tue Apr 05, 2011 4:30 pm D is just wrong, wrong, diddly-wrong-wrong. Its B, taken from 544.1.1 second paragraph. simples, bosh, done, end of, JD, next question..... Its definately not the diameter, thats a description of the largest measurement between two points on the circumference of a circle. Lets try 'the bisect', whats the prize?
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Posted: Tue Apr 05, 2011 4:43 pm Line of symmetry?
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Posted: Tue Apr 05, 2011 4:46 pm ..Last edited by holmslaw on Mon Apr 09, 2012 8:21 pm, edited 1 time in total
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Posted: Tue Apr 05, 2011 4:58 pm
ebee wrote: Any line intersecting an arc (including a circle) in two places is a "Chord". The chord of that particular question would have a length equal to the diameter of that circle (between its intersecting points) But the question they asked was the "Name" of the line not the "length" of it.
Goodness, what on earth are we doing discussing semantics again in an 'Electrics' forum?
Anyway, I'm afraid you are wrong. As well as being the length of the chord which passes through the centre, "diameter" is also the name of that line. The OED defines "diameter" as (amongst other things) "Straight line passing from side to side of any body or figure through the centre". "Diameter" is therefore a perfectly correct answer to the question.
In passing, note that is doesn't only apply to circles, nor even only to 2-D figures - which is something that many people don't realise.
Kind Regards, John
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Welcome to Doom9's Forum, THE in-place to be for everyone interested in DVD conversion. Before you start posting please read the forum rules. By posting to this forum you agree to abide by the rules.
Doom9's Forum Multiple statements on a single line
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26th March 2013, 09:13 #1 | Link fvisagie Registered User Join Date: Aug 2008 Location: Isle of Man Posts: 588 Multiple statements on a single line This is probably old hat to the old-timers here, but I stumbled across something quite exciting. Avisynth does support multiple statements on a single line, in contradiction to Multiple Avisynth statements on a single line can only be achieved in the context of OOP notation or embedding filters as parameters of another function. This seems to be supported for normal statements, by Eval() etc. and opens up a whole new world (for me at least ) of things like simple conditional block statements. This is with 2.6.0, I haven't tested with earlier versions. Here's an example of a trivial function returning multiple values Code: ```#------------------------------------------- # ColourCoefficients 1.0.0 # Generates specified colour matrix coefficient values # # Useful when # # Doing YUV<->RGB colourspace calculations or conversions # Doing colour coefficient calculations or conversions # # ColourCoefficients(string "matrix") # # Colour coefficient matrix which to generate coefficients for - "Rec601" / "Rec709" / "FCC", default "Rec601" # # Usage # # coeffs = ColourCoefficients("Rec709") # ColourCoefficients returns a multi-statement string with matrix coefficient value assignments # Eval(coeffs) # This defines Kr containing Red channel coefficient, Kg with Green channel coefficient and Kb for Blue channel # y = Kr*r + Kg*g + Kb*b # ... # # References # # http://avisynth.org/mediawiki/Color_conversions # # Version history # # 1.0.0 Francois Visagie # function ColourCoefficients(string "matrix") { matrix = Default(matrix, "Rec601") matrixnum = (matrix == "Rec601") ? 0 : \ (matrix == "Rec709") ? 1 : \ (matrix == "FCC" ) ? 2 : 3 Assert(matrixnum < 3, """'matrix' value """" + matrix + """" is invalid""") # Channel coefficients Rec.601 Rec.709 FCC Kr = Select(matrixnum, 0.299, 0.2125, 0.3 ) Kg = Select(matrixnum, 0.587, 0.7154, 0.59) Kb = Select(matrixnum, 0.114, 0.0721, 0.11) return("Kr=" + string(Kr) + " Kg=" + string(Kg) + " Kb=" + string(Kb)) }``` To what extent has anyone else played with this, and what gotchas have been discovered so far?
26th March 2013, 12:09 #2 | Link Gavino Avisynth language lover Join Date: Dec 2007 Location: Spain Posts: 3,412 Yes, that's right. Newline (unless accompanied by a '\') always terminates the current statement, but it is not strictly necessary. The Avisynth parser recognises it has reached the end of a statement when the next symbol is not a valid continuation of what it already has. So Code: `x = 1 y = 2 z = 3` is parsed as three separate (assignment) statements. Although it is conventional to use newlines to separate statements (and good practice for readability), the syntax is such that it is only strictly necessary if the following statement starts with a unary minus (or plus) operator - not very common, but can occur in cases like (silly example!): Code: ```function f(int x, int y) { z = x + y - z # meaning return -z }``` Here '- z' must appear on a separate line, since the previous statement could validly be continued with a '-'. __________________ GScript and GRunT - complex Avisynth scripting made easier
26th March 2013, 13:55 #3 | Link fvisagie Registered User Join Date: Aug 2008 Location: Isle of Man Posts: 588 Thanks for the explanation, Gavino.
26th March 2013, 18:32 #4 | Link TheFluff Excessively jovial fellow Join Date: Jun 2004 Location: rude Posts: 1,098 You could just have put newlines in the returned string, you know. There are also several better ways to "return" multiple values from a function. Using a callback function immediately comes to mind, but I'd say even global variables is cleaner than this solution.
27th March 2013, 08:36 #5 | Link
fvisagie
Registered User
Join Date: Aug 2008
Location: Isle of Man
Posts: 588
Quote:
Originally Posted by TheFluff You could just have put newlines in the returned string, you know.
I also thought so, but in this case I've found no way of writing the multi-line return string in a natural way that doesn't cause Eval() to fail. I've tried with literal "\n", variables by that value, triple quotes etc.
The only other syntax I found that works is
Code:
```return("Kr=" + string(Kr) + Chr(10) + "Kg=" + string(Kg) + Chr(10) + "Kb=" + string(Kb))
# or Chr(13)```
although that defeats the purpose even further, so to speak.
27th March 2013, 09:25 #6 | Link
fvisagie
Registered User
Join Date: Aug 2008
Location: Isle of Man
Posts: 588
Quote:
Originally Posted by TheFluff Using a callback function immediately comes to mind
How is this done in Avisynth (without using globals)?
27th March 2013, 09:26 #7 | Link
StainlessS
HeartlessS Usurer
Join Date: Dec 2009
Location: Over the rainbow
Posts: 9,263
I see nothing wrong with either space or Chr(10) string concatenation on a 'once in a while' usage, eg
clipwise return string, but not too good an idea on framewise or pixelwise usage, mainly due to fact that
there is no dead string garbage collection and freeing of memory.
The Gavino noted ' - z' thing would not come into it.
EDIT:
Quote:
How is this done in Avisynth (without using globals)?
No Idea (In script).
__________________
I sometimes post sober.
StainlessS@MediaFire ::: AND/OR ::: StainlessS@SendSpace
"Some infinities are bigger than other infinities", but how many of them are infinitely bigger ???
Last edited by StainlessS; 27th March 2013 at 09:30.
27th March 2013, 10:40 #8 | Link
Gavino
Avisynth language lover
Join Date: Dec 2007
Location: Spain
Posts: 3,412
Quote:
Originally Posted by fvisagie I've found no way of writing the multi-line return string in a natural way
You can use newlines directly inside a string literal, eg
Code:
```multiline = "first line
second line"```
__________________
GScript and GRunT - complex Avisynth scripting made easier
27th March 2013, 10:45 #9 | Link
TheFluff
Excessively jovial fellow
Join Date: Jun 2004
Location: rude
Posts: 1,098
Quote:
Originally Posted by fvisagie How is this done in Avisynth (without using globals)?
Code:
```function foo(int x, string callback_name) {
apply(callback_name, x)
}
function bar(int z) {
# do something with z
}
y = 5
foo(y, 'bar')```
9th April 2013, 10:19 #10 | Link
fvisagie
Registered User
Join Date: Aug 2008
Location: Isle of Man
Posts: 588
Quote:
Originally Posted by fvisagie things like simple conditional block statements
For the record, it seems this is a bit of a moot point (as you guys probably knew all along ). As far as I can determine the only expressions accepted by the ternary operator are function calls, be it internal, plugin or script functions. It does not seem to accept assignment statements, unless wrapped in Eval(), which kind of defeats the purpose from this point of view .
9th April 2013, 11:56 #11 | Link StainlessS HeartlessS Usurer Join Date: Dec 2009 Location: Over the rainbow Posts: 9,263 Code: ``` x = (z == true) ? Q * 17 : Q * 21 # assignment (z == false) ? Do_somthing(q) : Do_something_else(q) # select one of two funcs (z == true) ? Do_anything(q) : Nop # conditional do something``` __________________ I sometimes post sober. StainlessS@MediaFire ::: AND/OR ::: StainlessS@SendSpace "Some infinities are bigger than other infinities", but how many of them are infinitely bigger ??? Last edited by StainlessS; 9th April 2013 at 12:15.
9th April 2013, 14:17 #12 | Link
Gavino
Avisynth language lover
Join Date: Dec 2007
Location: Spain
Posts: 3,412
Quote:
Originally Posted by fvisagie As far as I can determine the only expressions accepted by the ternary operator are function calls, be it internal, plugin or script functions. It does not seem to accept assignment statements...
From http://avisynth.org/mediawiki/Grammar:
Quote:
All basic AviSynth scripting statements have one of these forms: 1. variable_name = expression 2. expression 3. return expression ... An expression can have one of these forms: 1. numeric_constant, string_constant or boolean_constant 2. variable_name or clip_property 3. Function(args) 4. expression.Function(args) 5. expression1 operator expression2 6. bool_expression ? expression1 : expression2
The expressions accepted by the ternary operator (ie in expression form 6) are any of the forms 1-6, so are not limited to function calls (forms 3 and 4).
However, an assignment is not an expression, so can appear only at the start of a statement (statement form 1).
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9th April 2013, 14:46 #13 | Link fvisagie Registered User Join Date: Aug 2008 Location: Isle of Man Posts: 588 Thanks, guys. Now to read and apply documentation as attentively as you do...
23rd January 2014, 01:37 #14 | Link
ChiDragon
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Posts: 608
Quote:
Originally Posted by fvisagie Avisynth does support multiple statements on a single line, in contradiction to Multiple Avisynth statements on a single line can only be achieved in the context of OOP notation or embedding filters as parameters of another function.
Could someone who understands this issue in-depth update the Wiki to be more accurate? I was recently led astray.
I would change the line myself but I don't want to make it wrong in a different way...
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# Lindblad Master Equation Solver
## Unitary evolution
The dynamics of a closed (pure) quantum system is governed by the Schrödinger equation
(1)$i\hbar\frac{\partial}{\partial t}\Psi = \hat H \Psi,$
where $$\Psi$$ is the wave function, $$\hat H$$ the Hamiltonian, and $$\hbar$$ is Planck’s constant. In general, the Schrödinger equation is a partial differential equation (PDE) where both $$\Psi$$ and $$\hat H$$ are functions of space and time. For computational purposes it is useful to expand the PDE in a set of basis functions that span the Hilbert space of the Hamiltonian, and to write the equation in matrix and vector form
$i\hbar\frac{d}{dt}\left|\psi\right> = H \left|\psi\right>$
where $$\left|\psi\right>$$ is the state vector and $$H$$ is the matrix representation of the Hamiltonian. This matrix equation can, in principle, be solved by diagonalizing the Hamiltonian matrix $$H$$. In practice, however, it is difficult to perform this diagonalization unless the size of the Hilbert space (dimension of the matrix $$H$$) is small. Analytically, it is a formidable task to calculate the dynamics for systems with more than two states. If, in addition, we consider dissipation due to the inevitable interaction with a surrounding environment, the computational complexity grows even larger, and we have to resort to numerical calculations in all realistic situations. This illustrates the importance of numerical calculations in describing the dynamics of open quantum systems, and the need for efficient and accessible tools for this task.
The Schrödinger equation, which governs the time-evolution of closed quantum systems, is defined by its Hamiltonian and state vector. In the previous section, Using Tensor Products and Partial Traces, we showed how Hamiltonians and state vectors are constructed in QuTiP. Given a Hamiltonian, we can calculate the unitary (non-dissipative) time-evolution of an arbitrary state vector $$\left|\psi_0\right>$$ (psi0) using the QuTiP function qutip.sesolve. It evolves the state vector and evaluates the expectation values for a set of operators e_ops at the points in time in the list times, using an ordinary differential equation solver.
For example, the time evolution of a quantum spin-1/2 system with tunneling rate 0.1 that initially is in the up state is calculated, and the expectation values of the $$\sigma_z$$ operator evaluated, with the following code
>>> H = 2*np.pi * 0.1 * sigmax()
>>> psi0 = basis(2, 0)
>>> times = np.linspace(0.0, 10.0, 20)
>>> result = sesolve(H, psi0, times, e_ops=[sigmaz()])
See the next section for examples on how dissipation is included by defining a list of collapse operators and using qutip.mesolve instead.
The function returns an instance of qutip.Result, as described in the previous section Dynamics Simulation Results. The attribute expect in result is a list of expectation values for the operators that are included in the list in the fifth argument. Adding operators to this list results in a larger output list returned by the function (one array of numbers, corresponding to the times in times, for each operator)
>>> result = sesolve(H, psi0, times, e_ops=[sigmaz(), sigmay()])
>>> result.expect
[array([ 1. , 0.78914057, 0.24548559, -0.40169513, -0.8794735 ,
-0.98636142, -0.67728219, -0.08258023, 0.54694721, 0.94581685,
0.94581769, 0.54694945, -0.08257765, -0.67728015, -0.98636097,
-0.87947476, -0.40169736, 0.24548326, 0.78913896, 1. ]),
array([ 0.00000000e+00, -6.14212640e-01, -9.69400240e-01, -9.15773457e-01,
-4.75947849e-01, 1.64593874e-01, 7.35723339e-01, 9.96584419e-01,
8.37167094e-01, 3.24700624e-01, -3.24698160e-01, -8.37165632e-01,
-9.96584633e-01, -7.35725221e-01, -1.64596567e-01, 4.75945525e-01,
9.15772479e-01, 9.69400830e-01, 6.14214701e-01, 2.77159958e-06])]
The resulting list of expectation values can easily be visualized using matplotlib’s plotting functions:
>>> H = 2*np.pi * 0.1 * sigmax()
>>> psi0 = basis(2, 0)
>>> times = np.linspace(0.0, 10.0, 100)
>>> result = sesolve(H, psi0, times, [sigmaz(), sigmay()])
>>> fig, ax = plt.subplots()
>>> ax.plot(result.times, result.expect[0])
>>> ax.plot(result.times, result.expect[1])
>>> ax.set_xlabel('Time')
>>> ax.set_ylabel('Expectation values')
>>> ax.legend(("Sigma-Z", "Sigma-Y"))
>>> plt.show()
If an empty list of operators is passed to the e_ops parameter, the qutip.sesolve and qutip.mesolve functions return a qutip.Result instance that contains a list of state vectors for the times specified in times
>>> times = [0.0, 1.0]
>>> result = sesolve(H, psi0, times, [])
>>> result.states
[Quantum object: dims = [[2], [1]], shape = (2, 1), type = ket
Qobj data =
[[1.]
[0.]], Quantum object: dims = [[2], [1]], shape = (2, 1), type = ket
Qobj data =
[[0.80901699+0.j ]
[0. -0.58778526j]]]
## Non-unitary evolution
While the evolution of the state vector in a closed quantum system is deterministic, open quantum systems are stochastic in nature. The effect of an environment on the system of interest is to induce stochastic transitions between energy levels, and to introduce uncertainty in the phase difference between states of the system. The state of an open quantum system is therefore described in terms of ensemble averaged states using the density matrix formalism. A density matrix $$\rho$$ describes a probability distribution of quantum states $$\left|\psi_n\right>$$, in a matrix representation $$\rho = \sum_n p_n \left|\psi_n\right>\left<\psi_n\right|$$, where $$p_n$$ is the classical probability that the system is in the quantum state $$\left|\psi_n\right>$$. The time evolution of a density matrix $$\rho$$ is the topic of the remaining portions of this section.
## The Lindblad Master equation
The standard approach for deriving the equations of motion for a system interacting with its environment is to expand the scope of the system to include the environment. The combined quantum system is then closed, and its evolution is governed by the von Neumann equation
(2)$\dot \rho_{\rm tot}(t) = -\frac{i}{\hbar}[H_{\rm tot}, \rho_{\rm tot}(t)],$
the equivalent of the Schrödinger equation (1) in the density matrix formalism. Here, the total Hamiltonian
$H_{\rm tot} = H_{\rm sys} + H_{\rm env} + H_{\rm int},$
includes the original system Hamiltonian $$H_{\rm sys}$$, the Hamiltonian for the environment $$H_{\rm env}$$, and a term representing the interaction between the system and its environment $$H_{\rm int}$$. Since we are only interested in the dynamics of the system, we can at this point perform a partial trace over the environmental degrees of freedom in Eq. (2), and thereby obtain a master equation for the motion of the original system density matrix. The most general trace-preserving and completely positive form of this evolution is the Lindblad master equation for the reduced density matrix $$\rho = {\rm Tr}_{\rm env}[\rho_{\rm tot}]$$
(3)$\dot\rho(t)=-\frac{i}{\hbar}[H(t),\rho(t)]+\sum_n \frac{1}{2} \left[2 C_n \rho(t) C_n^\dagger - \rho(t) C_n^\dagger C_n - C_n^\dagger C_n \rho(t)\right]$
where the $$C_n = \sqrt{\gamma_n} A_n$$ are collapse operators, and $$A_n$$ are the operators through which the environment couples to the system in $$H_{\rm int}$$, and $$\gamma_n$$ are the corresponding rates. The derivation of Eq. (3) may be found in several sources, and will not be reproduced here. Instead, we emphasize the approximations that are required to arrive at the master equation in the form of Eq. (3) from physical arguments, and hence perform a calculation in QuTiP:
• Separability: At $$t=0$$ there are no correlations between the system and its environment such that the total density matrix can be written as a tensor product $$\rho^I_{\rm tot}(0) = \rho^I(0) \otimes \rho^I_{\rm env}(0)$$.
• Born approximation: Requires: (1) that the state of the environment does not significantly change as a result of the interaction with the system; (2) The system and the environment remain separable throughout the evolution. These assumptions are justified if the interaction is weak, and if the environment is much larger than the system. In summary, $$\rho_{\rm tot}(t) \approx \rho(t)\otimes\rho_{\rm env}$$.
• Markov approximation The time-scale of decay for the environment $$\tau_{\rm env}$$ is much shorter than the smallest time-scale of the system dynamics $$\tau_{\rm sys} \gg \tau_{\rm env}$$. This approximation is often deemed a “short-memory environment” as it requires that environmental correlation functions decay on a time-scale fast compared to those of the system.
• Secular approximation Stipulates that elements in the master equation corresponding to transition frequencies satisfy $$|\omega_{ab}-\omega_{cd}| \ll 1/\tau_{\rm sys}$$, i.e., all fast rotating terms in the interaction picture can be neglected. It also ignores terms that lead to a small renormalization of the system energy levels. This approximation is not strictly necessary for all master-equation formalisms (e.g., the Block-Redfield master equation), but it is required for arriving at the Lindblad form (3) which is used in qutip.mesolve.
For systems with environments satisfying the conditions outlined above, the Lindblad master equation (3) governs the time-evolution of the system density matrix, giving an ensemble average of the system dynamics. In order to ensure that these approximations are not violated, it is important that the decay rates $$\gamma_n$$ be smaller than the minimum energy splitting in the system Hamiltonian. Situations that demand special attention therefore include, for example, systems strongly coupled to their environment, and systems with degenerate or nearly degenerate energy levels.
For non-unitary evolution of a quantum systems, i.e., evolution that includes incoherent processes such as relaxation and dephasing, it is common to use master equations. In QuTiP, the function qutip.mesolve is used for both: the evolution according to the Schrödinger equation and to the master equation, even though these two equations of motion are very different. The qutip.mesolve function automatically determines if it is sufficient to use the Schrödinger equation (if no collapse operators were given) or if it has to use the master equation (if collapse operators were given). Note that to calculate the time evolution according to the Schrödinger equation is easier and much faster (for large systems) than using the master equation, so if possible the solver will fall back on using the Schrödinger equation.
What is new in the master equation compared to the Schrödinger equation are processes that describe dissipation in the quantum system due to its interaction with an environment. These environmental interactions are defined by the operators through which the system couples to the environment, and rates that describe the strength of the processes.
In QuTiP, the product of the square root of the rate and the operator that describe the dissipation process is called a collapse operator. A list of collapse operators (c_ops) is passed as the fourth argument to the qutip.mesolve function in order to define the dissipation processes in the master equation. When the c_ops isn’t empty, the qutip.mesolve function will use the master equation instead of the unitary Schrödinger equation.
Using the example with the spin dynamics from the previous section, we can easily add a relaxation process (describing the dissipation of energy from the spin to its environment), by adding np.sqrt(0.05) * sigmax() in the fourth parameter to the qutip.mesolve function and moving the expectation operators [sigmaz(), sigmay()] to the fifth argument.
>>> times = np.linspace(0.0, 10.0, 100)
>>> result = mesolve(H, psi0, times, [np.sqrt(0.05) * sigmax()], e_ops=[sigmaz(), sigmay()])
>>> fig, ax = plt.subplots()
>>> ax.plot(times, result.expect[0])
>>> ax.plot(times, result.expect[1])
>>> ax.set_xlabel('Time')
>>> ax.set_ylabel('Expectation values')
>>> ax.legend(("Sigma-Z", "Sigma-Y"))
>>> plt.show()
Here, 0.05 is the rate and the operator $$\sigma_x$$ (qutip.sigmax) describes the dissipation process.
Now a slightly more complex example: Consider a two-level atom coupled to a leaky single-mode cavity through a dipole-type interaction, which supports a coherent exchange of quanta between the two systems. If the atom initially is in its groundstate and the cavity in a 5-photon Fock state, the dynamics is calculated with the lines following code
>>> times = np.linspace(0.0, 10.0, 200)
>>> psi0 = tensor(fock(2,0), fock(10, 5))
>>> a = tensor(qeye(2), destroy(10))
>>> sm = tensor(destroy(2), qeye(10))
>>> H = 2 * np.pi * a.dag() * a + 2 * np.pi * sm.dag() * sm + 2 * np.pi * 0.25 * (sm * a.dag() + sm.dag() * a)
>>> result = mesolve(H, psi0, times, [np.sqrt(0.1)*a], e_ops=[a.dag()*a, sm.dag()*sm])
>>> plt.figure()
>>> plt.plot(times, result.expect[0])
>>> plt.plot(times, result.expect[1])
>>> plt.xlabel('Time')
>>> plt.ylabel('Expectation values')
>>> plt.legend(("cavity photon number", "atom excitation probability"))
>>> plt.show()
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Congruent Angles
In congruent angles we will learn how to recognize that when two angles are congruent.
In case of two angles having the same vertex:
Suppose ∠LOM = ∠PON having same vertex O. Now we will verify their congruence.
A straight line QOR is drawn through point O and a perpendicular xy is drawn on QOR at O.
With respect to the axis of reflection xy, ∠P’ON’ is the image of ∠PON. Now with centre of rotation O, OP’ is rotated through such an angle in anticlockwise direction, so that OP’ coincides with OM. Since ∠P’ON’ is rigid figure and equal to ∠LOM, ON’ falls on OL.
Therefore, ∠LOM ≅ ∠N’OP’ ≅ ∠NOP = ∠PON
With O as center of rotation, OP is rotated through such an angle (in anticlockwise direction) so that OP lies on OL. In the same manner ON being rotated equally, falls on OM.
Therefore, ∠LOM ≅ ∠PON
If two equal angles are at different positions but lie on the same plane.
∠LMN and ∠PQR are two equal angles at different positions but lie on the same plane.
Taking xy, the perpendicular bisector of MQ as the axis of reflection, the image of ∠PQR is ∠P’MR’.
Therefore, ∠P’MR’ ≅ ∠PQR
Now observe that ∠LMN and ∠P’MR’ are two equal angles sharing common vertex M.
`
Congruent Line-segments
Congruent Angles
Congruent Triangles
Conditions for the Congruence of Triangles
Side Side Side Congruence
Angle Side Angle Congruence
Angle Angle Side Congruence
Pythagorean Theorem
Converse of Pythagorean Theorem
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# determing the cores necessary
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October 11, 2013, 11:54 determing the cores necessary #1 New Member lisa Join Date: Apr 2009 Posts: 17 Rep Power: 10 Sponsored Links Hi, I would like to know on how i could determine the no of cores which are necessary for my simulation. What kind of parametric study that i have to do. Thanks Lisa
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October 14, 2013, 03:26 #2 Senior Member Philipp Join Date: Jun 2011 Location: Germany Posts: 1,297 Rep Power: 20 What do you mean by "cores"? CPU cores? __________________ The skeleton ran out of shampoo in the shower.
October 14, 2013, 03:27 #3 New Member lisa Join Date: Apr 2009 Posts: 17 Rep Power: 10 Yes, i mean cpu cores.
October 14, 2013, 03:28 #4 Senior Member Philipp Join Date: Jun 2011 Location: Germany Posts: 1,297 Rep Power: 20 This doesn't make any sense. "Necessary" is a single core. If you have more than one, you will be faster. Why would you really need more than one core? __________________ The skeleton ran out of shampoo in the shower.
October 14, 2013, 03:39 #5 New Member lisa Join Date: Apr 2009 Posts: 17 Rep Power: 10 I do understand that. I have a mesh sizes 1million, 2 Million elements. Just do a bit of maths on how many cores should i choose for such a simulation. Not always more cores mean more faster. Thats why wanted to know if there is any parameteric study on calculate the no of cores for simulations , like 4 core or 8 core.
October 14, 2013, 03:42 #6 Senior Member Philipp Join Date: Jun 2011 Location: Germany Posts: 1,297 Rep Power: 20 1million cells will be much faster on 8 cores than on 4 cores. But you need 8 "real" cores, not 4 cores with hyperthreading. __________________ The skeleton ran out of shampoo in the shower.
October 14, 2013, 03:51 #7 New Member lisa Join Date: Apr 2009 Posts: 17 Rep Power: 10 What do you mean on real cores. What is the difference of real cores and cores with hyperthreading. Can you please explain. Thanks
October 14, 2013, 03:57 #8 Senior Member Philipp Join Date: Jun 2011 Location: Germany Posts: 1,297 Rep Power: 20 On your normal PC-CPU (at least if you have Intel, such as i7-xxxx), you have just 4 cores, but each of them can handle two threads. Windows / linux shows 8 CPUs (4x2). At least for my cases, they don't get faster when you use more than 4 of them. So it looks like that Fluent and OpenFoam can't gain any speed-up by hyperthreading, just by "real" CPUs. As far as I know, Intel only offers more than 4 cores, when you get a Xeon. __________________ The skeleton ran out of shampoo in the shower.
October 14, 2013, 04:14 #9 New Member lisa Join Date: Apr 2009 Posts: 17 Rep Power: 10 Yes i do have an i7 But i am planning to run on CLuster machines in Uni which has got supercomputers. So thought to check on this if there is amths behind checking the cores.
October 14, 2013, 04:19 #10 Senior Member Philipp Join Date: Jun 2011 Location: Germany Posts: 1,297 Rep Power: 20 Here are my results, Fluent, all run the same number of iterations. If I recall correctly, the grid has about 1mio cells. 1 core: 881 sec. 2 cores: 485 sec. 3 cores: 377 sec. 4 cores: 320 sec. 6 cores: 357 sec. Its an i7-3770 with windows. I know, that on a linux server (with many cores) the simulation gets much faster even for 8 cores. __________________ The skeleton ran out of shampoo in the shower.
October 16, 2013, 13:20 #11 Senior Member Reza Join Date: Mar 2009 Location: Appleton, WI Posts: 116 Rep Power: 10 Just to add to what Philipp said, you should ideally gain speed increase as long as the load on CPUs are balanced, and the interaction between the CPUs is not more time intensive than the computations on each CPU. So, for example, if you have a pipe geometry with 1 million elements, and if you partition the geometry only in the axial direction: By load balancing I mean: every CPU should get almost equal number of elements assigned to it. So if you want to use 8 CPUs, divide the pipe into 8 equal segments when you partition the domain. The interaction between the cores is going to be only on a cross-section of the pipe, so by minimizing CPU-CPU interaction I mean: make sure the number of elements in a partition (1/8th of the pipe volume in this case, so around 125,000 elements) is much larger than the number of faces in a single cross section of the pipe (depends on the pipe diameter, and the mesh resolution, but let's say 5000 faces). When you do partition any case for parallel processing, most partitioners will give you some statistics on these, so you can decide if you want to re-partition or just go ahead.
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