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https://math.stackexchange.com/questions/149619/does-bounded-covergence-theorem-hold-for-riemann-integral/1231378 | 1,686,181,245,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00146.warc.gz | 425,729,413 | 37,636 | # Does Bounded Covergence Theorem hold for Riemann integral?
Just after studying the Bounded Convergence Theorem BCT for Lebesgue integral, I asked myself a question. Does the BCT hold for Riemann? I answered YES since the function is bounded according to the hypothesis of the BCT. But some Lebesgue integral are not Riemann, this is where I got confused, please I need a guide from experts in the field.
Thanks.
Statement of the BCT:
Let $\{f_{n}\}$ be a sequence of measurable functions defined on a set $E$ of finite measure. Assume $\{f_{n}\}$ converges to $f$ pointwise and also $\{f_{n}\}$ is bounded for all $n$. Then $$\int_{E}f=\lim_{n \to \infty}\int_{E}f_{n}.$$
• What makes you think $f$ will be Riemann-integrable? May 25, 2012 at 12:30
• Since $f_{n}$ is bounded and converges to $f$ pointwise. May 25, 2012 at 12:33
• In general you need uniform convergence for the function to remain Riemann-integrable. May 25, 2012 at 12:40
• May 25, 2012 at 14:57
• I believe the corresponding convergence theorem requires uniform convergence. See this post for a summary of "FTC" and "convergence theorems" for several different types of integrals. The role of dominated convergence for Lebesgue integrals is played by uniform convergence for Riemann integrals. May 26, 2012 at 1:52
No. Enumerate the rationals in [0,1] with the sequence $\{r_n\}_{n=1}^\infty$. Now define $f_n(x)$ by $f_n(x) = 1$ if $x = r_k$ for some $1\le k \le n$ and 0 otherwise. For all $n$, we have $$\int_0^1 f_n(x)\,dx = 0.$$ However, the limit function, the indicator of the rationals in $[0,1]$ is not Riemann integrable. The bounded convergence theorem fails for the Riemann Integral.
• Yes, courses my life becomes easy now. Thanks@ncmathsadist. May 25, 2012 at 12:37
A dominated convergence theorem for the Riemann integral exists, due to Arzel`a. But one needs the addtional assumption that the limit function is Riemann integrable, since this does not follow from pointwise bounded convergence. For a proof see either W. A. J. Luxemburg: Arzela's Dominated Convergence Theorem for the Riemann Integral. The American Mathematical Monthly, Vol. 78, No. 9 (Nov., 1971), 970-979 or the book "An interactive introduction to mathematical analysis" by J. Lewin, Cambridge Univ. Press, 2003, 2014.
But this statement is true:
Let $\{f_n\}$ be a sequence of Riemann Integrable functions such that $f_n:[a,b]\rightarrow\mathbb{R}$ and $|f_n(x)|<M$ for all $n\geq1$ with $M>0$. Suppose that $f_n\to f$ pointwise where $f:[a,b]\rightarrow\mathbb{R}$ is Riemann Integrable. Then $$\lim\limits_{n\to\infty}\int_a^bf_n(x)\,dx = \int_a^bf(x)\,dx$$
• Where can I find a proof of this?
– Twnk
Dec 13, 2013 at 5:25
• @Twink see the article mentioned in the answer by M. Mueger.
– KCd
Oct 24, 2018 at 1:23
I typed the proof in Proofwiki following: Lewin, Jonathan W. "A truly elementary approach to the bounded convergence theorem." The American Mathematical Monthly 93.5 (1986): 395-397
== Lemma==
We call $$E\subset \mathbb{R}$$ an elementary subset if $$E=\bigcup_{k=1}^{M} [a_{k},b_{k}]$$ and we define $$m(E)$$ as the total length of these intervals minus their overlaps.
Lemma: Suppose $$(A_{n})$$ is a contracting sequence of bounded sets in R with an empty intersection. Let $$a_{n}:=\sup\{m(E): E\subset A_{n}\text{ is an elementary subset} \}.$$ Then $$a_{n}\to 0$$.
== Proof of lemma==
The sequence $$a_{n}$$ is decreasing and assume that $$a_{n}\geq \delta>0$$ to obtain a contradiction.
By the epsilon definition of supremm, for $$\epsilon:=\frac{\delta}{2^{n}}$$ there exists elementary subset $$E_{n}$$ such that
$$m(E_{n})\geq a_{n}-\frac{\delta}{2^{n}}.$$
For $$H_{n}=\bigcap_{k=1}^{n}E_{k}\subset \bigcap_{k=1}^{n}A_{k}$$, we will show that $$H_{n}\neq \varnothing$$ and thus contradict that $$A_{n}$$ have an empty intersection.
For each n, take any elementary subset $$E\subset A_{k}\setminus E_{k}$$, then we find
$$m(E)+m(E_{k})=m(E\cup E_{k})\leq a_{k}\Rightarrow m(E)\leq \frac{\delta}{2^{k}}.$$
Now take an elementary subset $$S\subset A_{n}\setminus H_{n}=\bigcap_{k=1}^{n}(A_{n}\setminus E_{k})$$, then we find
$$E=(E\setminus E_{1})\cup … \cup (E\setminus E_{n}).$$
Therefore, we get the bound
$$m(E)\leq \sum_{k=1}^{n}m(E\setminus E_{k})\leq \sum_{k=1}^{n}\frac{\delta}{2^{n}}=\delta.$$
In words, any elementary subset $$E\subset A_{n}\setminus H_{n}$$ was shown to have measure $$m(E)\leq \delta$$.
However, the inequality $$a_{n}>\delta$$ requires the existence of at least one elementary subset $$U_{n}\subset A_{n}$$ s.t. $$m(U_{n})>\delta$$.
Since all the elementary subset $$E\subset A_{n}\setminus H_{n}$$ satisfy $$m(E)\leq \delta$$, we must have that $$U_{n}\subset H_{n}$$ for $$n\geq 1$$.
This contradicts the non-emptiness because $$\lim_{n\to \infty} m(U_{n})>\delta$$. $$\square$$
== Proof of main result==
WLOG assume that $$f_{n}\geq 0$$ and $$f_{n}\to 0$$, so we will show that given $$\epsilon>0$$ there exists N s.t. forall $$n\geq N$$ we have . $$\int_{a}^{b}f_{n}(x)dx\leq \epsilon.$$
Let $$A_{n}:=\{x\in [a,b]:\text{ there exists }k\geq n \text{ such that} f_{k}(x)\geq \frac{\epsilon}{2(b-a)} \}$$.
These sets are decreasing as $$n\to +\infty$$ and have empty intersection and so the sup $$a_{n}$$ from above goes to zero $$a_{n}\to 0$$.
So let $$E_{n}\subset A_{n}$$ be an elementary subset with $$m(E_{n})\leq \frac{\epsilon}{2K}$$ for all $$n\geq N$$, and consider the following subsets
$$E:=\{x\in E_{n}:\text{ there exists }k\geq n \text{ such that} f_{k}(x)\geq \frac{\epsilon}{2(b-a)} \}\text{ and } F:=[a,b]\setminus E.$$
Therefore, we find
$$\int_{a}^{b}f_{n}(x)dx=\int_{E}f_{n}(x)dx+\int_{F}f_{n}(x)dx\leq K m(E_{n})+\frac{\epsilon}{2(b-a)} (b-a)\leq \epsilon.$$
$$\square$$ | 1,947 | 5,742 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 45, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-23 | latest | en | 0.899726 |
https://tensortime.sticksandshadows.com/archives/3138 | 1,566,380,895,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315865.44/warc/CC-MAIN-20190821085942-20190821111942-00508.warc.gz | 658,638,769 | 30,283 | # Matter & Interactions II, Week 5
This week was all about calculating electric fields for continuous charge distributions. This is usually students’ first exposure to what they think of as “calculus-based” physics because they are explicitly setting up and doing integrals. There’s lots going on behind the scenes though.
In calculus class, students are used to manipulating functions by taking their derivatives, indefinite integrals, and definite integrals. In physics, however, these ready made functions don’t exist. When we write dQ, there is no function Q() for which we calculate a differential. The symbol dQ represents a small quantity of charge, a “chunk” as I usually call it. That’s is. There’s nothing more. Similarly, dm represents a small “chunk” of mass rather than the differential of a function m(). The progress usually begins with uniform linear charge distributions and progresses to angular (i.e. linear charge distributions bent into arcs of varying extents), then area, then volume charge distributions (Are “area” and “volume” adjectives?). One cool thing is how each type of distribution can be constructed from a previous one. You can make a cylinder of charge out of lines of charge. You can make a loop of charge out of a line of charge. You can make a plane of charge out of lines of charge. You can make a sphere of charge out of loops of charge. Beautiful! Lots of ways to approach setting up the integral that sweeps through the charge distribution to get the net field.
It’s interesting to ponder the effect of changing the coordinate origin. Consider a charge rod. If rod’s left end is at the origin, the limits of integration are 0 and L (the rod’s length). If the rod’s center is at the origin, the limits of integration are -L/2 and +L/2. The integrand looks slightly different, but the resulting definite integral is the same in both cases! Trivial? No! It’s yet another indication that Nature doesn’t care about coordinate systems; they’re a human invention and subject to our desire for mathematical convenience. This is also a good time to recall even (f(-x) = f(x)) and odd (f(-x) = -f(x)) functions becuase then one can look at an integral and its limits and predict whether or not the integral must vanish and this connects with symmetry arguments from geometry. This, to me, is one of the very definitions of mathematical beauty. A given charge distribution’s electric field is independent of the coordinate system used to derive it. The forthcoming chapter on Gauss’s law and Ampère’s law relies on symmetries to predict electric and magnetic field structures for calculating flux and circulation and that’s foreshadowed in this chapter.
This is a lot to convey to students and from their point of view it’s a lot to understand. I hope I can do better at getting it all across to them than was done for me.
Feedback welcome as always.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 627 | 2,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-35 | latest | en | 0.916968 |
https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_2&oldid=22644 | 1,632,883,649,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780061350.42/warc/CC-MAIN-20210929004757-20210929034757-00159.warc.gz | 164,142,111 | 9,537 | # 2002 AMC 12B Problems/Problem 2
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
What is the value of $$(3x - 2)(4x + 1) - (3x - 2)4x + 1$$
when $x=4$? $\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 11 \qquad\mathrm{(E)}\ 12$
## Solution
$$(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = 11\ \mathrm{(D)}$$ | 187 | 391 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.453432 |
https://www.flowcontrolnetwork.com/a-true-measure-of-instrument-performance/ | 1,506,352,711,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818691977.66/warc/CC-MAIN-20170925145232-20170925165232-00315.warc.gz | 781,932,092 | 27,215 | List of Symbols: astreamline= acceleration along a streamline [m/s2] C = discharge coefficient [unitless] dƒ = primary contraction diameter during actual flow conditions [m] Dƒ = pipe diameter during actual flow conditions [m] ƒshear = fluid shear stress [N/m2] glocal = local gravitational constant [m/s2] Helevation = elevation above a datum [m] Pflow = pressure at flow conditions [Pa] ΔP = pressure differential [Pa] qmass = mass flowrate at flowing conditions [kg/s] qvolume = volumetric flowrate at flowing conditions [m3/s] Y1 = adiabatic expansion factor [unitless] ρflow= density at flowing conditions [kg/m3]
The quantification of volumetric or mass flow is one of the most difficult measurements to make — even in a well-characterized fluid. This is made more difficult in a demanding environment, such as a cryogenic fluid. The value determined often has great impact on the work being conducted, particularly in liquid propulsion systems. Commonly used instrumentation measurement techniques include the use of differential producers, such as Venturi-, flow nozzle-, and orifice plate-type devices. The performance of these instruments in actual application in transient multiphase cryogenic systems is typically one to two orders of magnitude poorer than manufacturer specifications. And while differential producer-based systems are commonly applied in such environments, they have inherent theoretical limitations that affect their accuracy, reliability, and repeatability. These limitations are discussed hereafter.
Volumetric & Mass Flow
Fundamentally, volumetric flow or mass flow is the amount of volume or mass, flowing through some sort of channel per unit time. The channel may be open (trough) or closed (pipe), but must allow for the passage of fluid.
(Equation 1)
(Equation 2)
Conceptually volumetric flow and mass flow are simple. However, a closer look at each term shows the difficulty of making a valid measurement.
The cross-sectional area of a pipe tends to be the easiest to account for. The associated errors directly relate to the error in the physical measurement of diameter. Care must be taken to account for changes in the cross-sectional area (or diameter). These may include: thermal contraction, caking or debris accumulation, etc. Secondary effects, such as pressure bulging and pipe eccentricity or concentricity, and the effect of roughness changes with time on the velocity profile (Reynolds number) must also be considered to further improve the measurement.
Figure 1. Pressure differential devices use a geometric flow perturbation to create a pressure difference — (a) Venturi tube, (b) flow nozzle, and (c) orifice plate. Drawings adapted from [2].
Volumetric and mass flow is often taken as a bulk property and therefore the velocity is often reduced to a bulk or net fluid velocity. Measured values for velocity are difficult, particularly in cryogenic fluid systems. Realistically the flowrate is not uniform across the flow front (or flow stream tube). Any direct flow measurement, will at best, be an average across the entire streamline. Differential producers do not rely on a direct velocity measure, but as will be shown below, critical assumptions are made to account for it.
The measure of density is difficult in a moving environment, particularly without disturbing the flow path. Fluid systems that have multiple thermodynamic phases commingling, such as cryogens, are especially problematic. Generally the density in gaseous systems can be controlled with a fair amount of accuracy [1], while in cryogenic flows it is not uncommon for the liquid, gas, and possibly solid phases to coexist and may even include multiple species fluids [2]. Table 1 shows the density variations between the thermodynamic phases. For example, if 1 percent of the volume in a pure liquid oxygen flow were converted into O2 gas it would displace approximately 2.5 times its volume of the liquid. This would result in a dramatic error in mass flow. Surprisingly, density plays a crucial role even in the volumetric flow equations of differential flowmeters even though it does not appear in Equation 1.
Volumetric and mass flow can be measured with a variety of methods. Most involve direct contact or interaction with the flow. This interaction leads to an exchange of energy. This addition of energy hastens density change, or worse, gas generation in the fluid flow.
Differential Producers
Differential producers or pressure differential devices, most commonly represented by Venturi-, flow nozzle-, and orifice plate-type devices, infer a mass flow value by way of two pressure measurements in different regions of the flow. The key is to pick two regions of the flow that are convenient to sample while maintaining a wide pressure differential.
Figure 1 illustrates some common pressure differential geometries. In addition to the classic flow bodies shown in Figure 1, pressure differential devices can also vary the flow direction instead of the cross-section to achieve a pressure change; such as a pipe elbow or other bend, where the pressure is measured across the inner and outer bend radii. Equations may be derived to translate these two pressure values, or more aptly their difference, to a mass flow.
Insight into the embedded approximations in mass flow sensors can be seen in the derivation of the differential flow sensor working equation [1, 2, and 5]. First, a one-dimensional flow assumption is invoked to clarify the calculations. Only the element of flow in the downstream direction is used to contribute to the overall flow. On one hand, this makes some logical sense. A flowing system must be flowing in some direction constrained by the pipe. What it precludes are losses and
turbulence, that can change the kinetic energy of a flow either causing frictional losses (pressure head loss) or, in the case of a cryogenic fluid, density changes. Additionally, the fluid must be incompressible. Again, for fluids it is fair to assume that changes in density versus pressure may be somewhat neglected if the flow is purely in the liquid phase. How valid is it to describe realistic flows as pure liquid phases? In cryogenics, particularly, the fluid is often flowing at or near its boiling point. While the liquid phase component of the flow may be incompressible, the gas phase component obeys gas compression equations (i.e., ideal gas law or Van der Waals). This directly relates the density to the temperature and pressures of the fluid. Using an element or slug of fluid (Fig. 2), Newton’s second law may be used to build a balanced force equation [2].
(Equation 3)
Where the net pressure force is,
(Equation 3a)
the force due to the elevation change,
(Equation 3b)
the viscous shear force,
(Equation 3c)
and the mass acceleration term of the slug,
(Equation 3d)
This equation does embody all relevant forces, such as shear, for use in real flow systems. Assuming a streamlined flow-front conveniently thins Equation 3. With the addition of a constant density assumption one can arrive at Bernoulli’s equation.
(Equation 4)
Differentiation of Equation 3 can be reduced to a more familiar generic solution for volumetric and mass flow.
(Equation 5)
[m3/s]
(Equation 6)
[kg/s]
With Y1 as the adiabatic expansion factor, C as the discharge coefficient and df, Df as the primary contraction diameter and pipe diameter during actual flow conditions. These parameters are described in the following section.
Correction Terms
It is important to describe the various correction terms that play a role in Equation 4. To begin with, the discharge coefficient, C, is a lump-sum correction term that provides the “fix” from the theoretical system to the real system. The discharge coefficient may be calculated depending on the type of flowmeter. These vary in error from about 0.4 for an uncalibrated meter to 2 percent [2], depending on meter type. This error is a systematic error and is in addition to any other errors (systematic or random). The discharge coefficient can also be determined through calibration. This must be done in the exact same system and under the exact same conditions and environments that the sensor will be used. Rarely is this accomplished, but instead manufacturer calibrations are taken from “like” systems. Particularly in cryogenic devices, water is used as a calibration fluid. The discharge coefficient for the meters shown in Figure 1 range from 0.6 for the orifice to 0.995 for the Venturi and nozzle. The abrupt change in streamlines, however, results in substantial additional pressure loss for the nozzle and orifice. This sometimes results in the unwanted flashing to a vapor.
Figure 2. Description for incompressible fluid equation 3. Adapted from “Flow Measurement Engineering Handbook,” Figure 9.1 [14].
The term Y1 describes the gas expansion of the fluid, which can be an empirically derived function or theoretically determined, such as the adiabatic gas expansion factor. For an assumption of purely liquid flow this becomes unity. This is even assumed in cryogenics at or near their boiling points. The slightest multiphase component can seriously undermine this approximation. Normally this term is ignored, and its effects are hidden in the discharge coefficient. The fallaciousness is that when the discharge coefficient is used to calibrate the system, it is assumed that the system is corrected over a range of environmental conditions. In reality, the gas expansion term may vary depending on the amount of vapor present this may continuously fluctuate with slight differences in environmental conditions from calibration.
As mentioned earlier, the physical diameters are df, Df. These are the respective bore and pipe diameters. They describe the expansion and contraction of the flow container (pipe and meter bore) due to pressure and temperature effects. These are particularly critical in cryogenic systems where extreme low temperatures cause considerable material contraction. These coefficients require accurate pressure and temperature information in addition to an accurate understanding of the fluid interaction with the pipe material. There is also the added assumption that the temperature and pressure in the pipe or bore match closely with the pressure or temperature sensor location (see below).
The pressure differential, ΔP, is only as valid as the pressure measuring device. Since the mass flow or volumetric flow is determined from a differential pressure measurement it is critical that this value be well characterized. Ideally, the pressure sensors are properly calibrated and corrected for environmental effects such as temperature. Additional factors may include frequency response effects related to the response of the pressure sensor and mounting offset usage (sense lines) [6].
Lastly, rfluid, the fluid density term, describes the density of the fluid. There are numerous tabulated values for various fluids under many thermodynamic conditions. These were developed from highly accurate equations of state that accurately define the density the fluids properties from temperature and pressure measurements. The use of flow computers to compute the necessary fluid properties using these state equations is becoming increasingly more popular. These methods, however, are not foolproof. Small errors in the measures of pressure and temperature can adversely affect the density generated.
Temperature probes may be located near the differential device and can provide a localized measure of temperature. Unfortunately at the temperatures considered here they tend to behave as heat pipes, coupling the environment outside the pipe into the fluid. They also tend to have low response times, on order of seconds [7], leading to latencies under non-steady state conditions. Pressure sensors, on the other hand, tend to average over large fluid volumes, thereby under-representing jumps or steps in the fluid state under a transient condition.
For flow conditions where the thermodynamic state is well defined the computation-based approach to density enables a high-accuracy flow measurement. Careful consideration must be made when the fluid state is not well understood or transient and/or multiphase conditions limit the validity of the supporting instrumentation.
In addition to the theoretical limitations inherent to these devices, differential flowmeters have been tested in characterized in multiphase transient environments. Errors increase dramatically for any deviation from single-phase and steady-state flow conditions. Work at NASA’s Johnson Space Center on flow sensor feasibility has shown errors in using Venturi-type devices ranging from 10 percent to 40 percent in transient and multiphase cryogenic flows [8].
More often than not Web- or pamphlet-based flow discussions begin with Bernoulli’s equation to derive the necessary working equations. Before Bernoulli’s equation can be applied to a fluid environment critical concessions have to be made that limit the applicability and accuracy of differential flowmeters. Clearly the critical measurement component lacking is understanding the thermodynamic and density conditions of the fluid. While quite acceptable in more benign fluid environments, differential-pressure flow devices are greatly limited in transient cryogenic fluid applications. The user must take care when using such devices in these difficult flow conditions.
Valentin Korman, Ph.D., of Madison Research CorporationWFI, has 10 years of sensors, instrumentation, and testing research development experience. Recent efforts have focused on evaluation and development of cryogenic flowmeter technologies. Dr. Korman can be reached at valentin.korman@nasa.gov or (256) 544-4625
John Wiley, of NASA Marshall Space Flight Center, has 17 years test experience with cryogenic propulsion systems, as well as research and development of advanced sensors for propulsion applications and experience with data validation and electrical and mechanical systems.
Richard W. Miller, Ph.D., of R.W. Miller & Associates, Inc., is a consultant to industry in various flow system and flowmetering areas. He was previously senior consultant with the Foxboro Company and serves, or has served, on many standards and technical committees, both nationally and internationally. Dr. Miller is the author of the “Flow Measurement Engineering Handbook” (McGraw Hill).
References
1. R. M. Olson, Essentials of Engineering Fluid Dynamics, 2nd Edition, International Textbook Co., 1967.
2. Richard W. Miller, Flow Measurement Engineering Handbook, 3rd Edition, McGraw-Hill, 1996.
3. H. M. Roder and L. A. Weber, Thermophysical Properties-ASRDI Oxygen Technology Survey, Volume 1, NASA SP-3071, 1972.
4. D. R. Lide, CRC Handbook of Chemistry and Physics, 71st Edition, CRC Press, 1991.
5. Korman V., Gregory D. A., and Wiley J. T. Mass Flow Measurement in a Cryogenic System using a Fiber Optics-Based Dispersion Sensor, Propulsion Measurement Sensor Development Workshop, Huntsville, AL May 2003. Proceedings.
6. Wiley J. T., Korman V., Vitarius P. T., and Gregory D. A., Acoustic Wave Propagation in Pressure Sense Lines, Presented at 39th AIAA/ASME/SAE/ASEE Joint Propulsion Conference, 2003, Proceedings.
7. Nanmac, Comparison of Temperature Sensor Response Times,
www.nanmac.com.
8. R.S. Baird, Flowmeter Evaluation for On-orbit Operations, NASA TM-100465, Johnson Space Center, August 1988. | 3,134 | 15,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-39 | longest | en | 0.927832 |
https://books.google.com.jm/books?id=Sd9EAAAAIAAJ&vq=%22When+a+letter+is+placed+before+one+of+greater+value,+its+value+is+to+be%22&dq=editions:HARVARD32044097001838&lr=&source=gbs_navlinks_s | 1,696,207,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510942.97/warc/CC-MAIN-20231002001302-20231002031302-00339.warc.gz | 165,577,227 | 12,081 | # American Comprehensive Arithmetic
American Book Company, 1897 - Arithmetic - 320 pages
### Contents
INTEGERS 9 Multiplication 35 Division 44 Operations Combined 60 Factoring 68 COMMON FRACTIONS 83 DECIMAL FRACTIONS 112 DENOMINATE NUMBERSENGLISH SYSTEM 119
83 190 98 199 103 205 INTEREST 216 107 232 142 238 MENSURATION 249 145 265
The Operations 145 Multiplication 152 Simple Equations 167 PROPORTION 170 SOLUTION OF PROBLEMS 180
### Popular passages
Page 254 - A Circle is a plane figure bounded by a curved line every point of which is equally distant from a point within called the center.
Page 107 - Multiply as in whole numbers, and point off as many decimal places in the product as there are decimal places in the multiplicand and multiplier, supplying the deficiency, if any, by prefixing ciphers.
Page 224 - Courts have adopted the following, called the UNITED STATES RULE. Find the amount of the principal to the time of the first payment ; if the payment equals or exceeds the interest, subtract it from the amount, and regard the remainder as a new principal.
Page 90 - Multiply together the numerators for a new numerator, and the denominators for a new denominator.
Page 272 - Pythagoras' theorem states that the square of the length of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the lengths of the other two sides.
Page 273 - The area of a triangle is equal to one half 'the product of its base and altitude.
Page 8 - ... one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty thirty forty fifty sixty seventy eighty ninety one hundred two hundred three hundred four hundred five hundred...
Page 69 - A number is divisible by 9, when the sum of its digits is divisible by 9.
Page 176 - M shown in the operation. 2. 5 compositors, in 16 days, of 14 hours each, can compose 20 sheets of 24 pages in each sheet, 50 lines in a page, and 40 letters in a line ; in...
Page 86 - Multiplying or dividing both numerator and denominator by the same number does not change the value of the fraction. | 501 | 2,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-40 | latest | en | 0.84718 |
https://physics.stackexchange.com/questions/486668/prove-that-a-transformation-is-canonical-by-using-mathbbmt-cdot-mathbbj | 1,571,442,247,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986685915.43/warc/CC-MAIN-20191018231153-20191019014653-00516.warc.gz | 643,255,965 | 29,493 | # Prove that a transformation is canonical by using $\mathbb{M}^T\cdot \mathbb{J}\cdot \mathbb{M}$ [closed]
So, I was given the following problem to solve:
A system with two degrees of freedom is described by the following hamiltonian $$$$H=p_1^2+p_2^2+\frac{1}{2}(q_1-q_2)^2+\frac{1}{8}(q_1+q_2)^2$$$$ Show that the following transformation is canonical by using $$\mathbb{M}^T\cdot \mathbb{J}\cdot\mathbb{M}$$, where $$\mathbb{M}$$ is the jacobian matrix and $$\mathbb{J}=\begin{pmatrix}\mathbb{O} & \mathbb{I}\\\ -\mathbb{I} & \mathbb{O}\end{pmatrix}$$, where $$\mathbb{O}$$ is an $$n\times n$$ matrix with all null elements and $$\mathbb{I}$$ is an $$n\times n$$ identity matrix. $$q_1=\sqrt{Q_1}\cdot cos(P_1)+\sqrt{2Q_2}\cdot cos(P_2)$$ $$q_2=-\sqrt{Q_1}\cdot cos(P_1)+\sqrt{2Q_2}\cdot cos(P_2)$$ $$p_1=\sqrt{Q_1}\cdot sin(P_1)+\sqrt{Q_2/2}\cdot sin(P_2)$$ $$p_2=-\sqrt{Q_1}\cdot sin(P_1)+\sqrt{Q_2/2}\cdot sin(P_2)$$
Well, I've started the problem by organizing the matrix $$\mathbb{M}$$. $$\mathbb{M}=\begin{pmatrix} \frac{\partial q_1}{\partial Q_1} & \frac{\partial q_1}{\partial Q_2} & \frac{\partial q_1}{\partial P_1} & \frac{\partial q_1}{\partial P_2} & \\\ \frac{\partial q_2}{\partial Q_1} & \frac{\partial q_2}{\partial Q_2} & \frac{\partial q_2}{\partial P_1} & \frac{\partial q_2}{\partial P_2} & \\\ \frac{\partial p_1}{\partial Q_1} & \frac{\partial p_1}{\partial Q_2} & \frac{\partial p_1}{\partial P_1} & \frac{\partial p_1}{\partial P_2} & \\\ \frac{\partial p_2}{\partial Q_2} & \frac{\partial p_2}{\partial Q_1} & \frac{\partial p_2}{\partial P_1} & \frac{\partial p_2}{\partial P_2} & \end{pmatrix}$$ Then I found the transpose matrix of $$\mathbb{M}$$ $$\mathbb{M^T}=\begin{pmatrix} \frac{\partial q_1}{\partial Q_1} & \frac{\partial q_2}{\partial Q_1} & \frac{\partial p_1}{\partial Q_1} & \frac{\partial p_2}{\partial Q_1} & \\\ \frac{\partial q_1}{\partial Q_2} & \frac{\partial q_2}{\partial Q_2} & \frac{\partial p_1}{\partial Q_2} & \frac{\partial p_2}{\partial Q_2} & \\\ \frac{\partial q_1}{\partial P_1} & \frac{\partial q_2}{\partial P_1} & \frac{\partial p_1}{\partial P_1} & \frac{\partial p_2}{\partial P_1} & \\\ \frac{\partial q_1}{\partial P_2} & \frac{\partial q_2}{\partial P_2} & \frac{\partial p_1}{\partial P_2} & \frac{\partial p_2}{\partial P_2} & \end{pmatrix}$$
So $$\mathbb{M^T}\cdot \mathbb{J}=\begin{pmatrix} \frac{\partial q_1}{\partial Q_1} & \frac{\partial q_2}{\partial Q_1} & \frac{\partial p_1}{\partial Q_1} & \frac{\partial p_2}{\partial Q_1} & \\\ \frac{\partial q_1}{\partial Q_2} & \frac{\partial q_2}{\partial Q_2} & \frac{\partial p_1}{\partial Q_2} & \frac{\partial p_2}{\partial Q_2} & \\\ \frac{\partial q_1}{\partial P_1} & \frac{\partial q_2}{\partial P_1} & \frac{\partial p_1}{\partial P_1} & \frac{\partial p_2}{\partial P_1} & \\\ \frac{\partial q_1}{\partial P_2} & \frac{\partial q_2}{\partial P_2} & \frac{\partial p_1}{\partial P_2} & \frac{\partial p_2}{\partial P_2} & \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 & 1 & \\\ 0 & 0 & 1 & 1 & \\\ -1 & -1 & 0 & 0 & \\\ -1 & -1 & 0 & 0 & \end{pmatrix}$$ gives me a $$2n\times 2n$$ null matrix. What I am doing wrong?
## closed as off-topic by G. Smith, John Rennie, Kyle Kanos, GiorgioP, stafusaJun 19 at 22:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
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• $q_1=-q_2$ and $p_1=-p_2$?? – Qmechanic Jun 18 at 8:01
• I got that from the transformations. – Lincon Ribeiro Jun 18 at 11:19
• But that isn't true, given the definitions of $q_i$ and $p_i$, is it? – Kyle Kanos Jun 18 at 11:35
• I used the following: $$q_2=-\sqrt{Q_1}cos(P_1)+\sqrt{2Q_2}cos(P_2)=-(q_1)$$ – Lincon Ribeiro Jun 18 at 12:27
• Ops, I guess you're right. My mistake on that – Lincon Ribeiro Jun 18 at 12:32
For Linear Canonical Transformation the equation $$M^T\,J\,M=J$$ must be fulfilled
$$M=\left[ \begin {array}{cccc} 1/2\,{\frac {\cos \left( {\it P1} \right) }{\sqrt {{\it Q1}}}}&1/2\,{\frac {\sqrt {2}\cos \left( {\it P2} \right) }{\sqrt {{\it Q2}}}}&-\sqrt {{\it Q1}}\sin \left( {\it P1} \right) &-\sqrt {2}\sqrt {{\it Q2}}\sin \left( {\it P2} \right) \\-1/2\,{\frac {\cos \left( {\it P1} \right) }{ \sqrt {{\it Q1}}}}&1/2\,{\frac {\sqrt {2}\cos \left( {\it P2} \right) }{\sqrt {{\it Q2}}}}&\sqrt {{\it Q1}}\sin \left( {\it P1} \right) &- \sqrt {2}\sqrt {{\it Q2}}\sin \left( {\it P2} \right) \\1/2\,{\frac {\sin \left( {\it P1} \right) }{ \sqrt {{\it Q1}}}}&1/4\,{\frac {\sqrt {2}\sin \left( {\it P2} \right) }{\sqrt {{\it Q2}}}}&\sqrt {{\it Q1}}\cos \left( {\it P1} \right) &1/2 \,\sqrt {2}\sqrt {{\it Q2}}\cos \left( {\it P2} \right) \\-1/2\,{\frac {\sin \left( {\it P1} \right) }{ \sqrt {{\it Q1}}}}&1/4\,{\frac {\sqrt {2}\sin \left( {\it P2} \right) }{\sqrt {{\it Q2}}}}&-\sqrt {{\it Q1}}\cos \left( {\it P1} \right) &1/ 2\,\sqrt {2}\sqrt {{\it Q2}}\cos \left( {\it P2} \right) \end {array} \right]$$
$$J= \left[ \begin {array}{cccc} 0&0&1&0\\ 0&0&0&1 \\ -1&0&0&0\\ 0&-1&0&0\end {array} \right]$$ you get the write answer | 2,231 | 5,428 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-43 | latest | en | 0.597362 |
http://betterlesson.com/lesson/resource/2626010/circle-pretest-docx | 1,477,354,627,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719815.3/warc/CC-MAIN-20161020183839-00384-ip-10-171-6-4.ec2.internal.warc.gz | 27,500,943 | 17,867 | ## Circle PreTest.docx - Section 2: Explore
Circle PreTest.docx
Circle PreTest.docx
# Circles Review
Unit 9: 2-D Measurements
Lesson 10 of 12
## Big Idea: What have you learned about circles?
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Environment: Suburban | 248 | 1,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2016-44 | longest | en | 0.785079 |
https://stackoverflow.com/questions/11924667/common-lisp-is-there-a-less-painful-way-to-input-math-expressions | 1,571,314,068,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986673538.21/warc/CC-MAIN-20191017095726-20191017123226-00539.warc.gz | 723,075,708 | 32,643 | # Common lisp: is there a less painful way to input math expressions?
I enjoy common lisp, but sometimes it is really painful to input simple math expressions like
``````a(8b^2+1)+4bc(4b^2+1)
``````
(Sure I can convert this, but it is kind of slow, I write (+ () ()) first, and then in each bracket I put (* () ())...)
I'm wondering if anyone here knows a better way to input this. I was thinking about writing a math macro, where
``````(math “a(8b^2+1)+4bc(4b^2+1)”)
``````
expands to
``````(+ (* a (1+ (* 8 b b))) (* 4 b c (1+ (* 4 b b))))
``````
but parsing is a problem for variables whose names are long.
Anybody has better suggestions?
• (a) If you don't like lisp as a calculator, use something else. I recommend python (b) performing this sort of transformation is a basic computer science exercise. I strongly suggest that you figure it out yourself, at least on the string manipulation side, and come back when you hit problems making it into a convenient macro. – Marcin Aug 12 '12 at 18:36
• OK, fine. I guess it is my fault to do Project Euler with lisp... – h__ Aug 12 '12 at 18:40
• Lisp is absolutely ok for Project Euler. Sure a lengthy math expression can be a pain to type in; on the other hand Common Lisp is very suitable for the "build solutions bottom-up" approach lots of problems on PE require. – Haile Aug 12 '12 at 20:35
• @Haile What do you mean by that (build solutions bottom-up)? Can you give an example? – h__ Aug 12 '12 at 22:10
• @hyh You break down the problem in small pieces. You focus on each piece alone. You try short snippets of code on the REPL. You start composing those snippets in functions, to provide small and basic functionalities. You test those functionalities right away using the REPL. You write your solution incrementally, interacting with the interpreter. You compose already tested functionalities into bigger ones. And so on.. – Haile Aug 16 '12 at 18:51
There are reader macros for this purpose.
For example:
``````CL-USER 17 > '#I(a*(8*b^^2+1)+ 4*b*c*(4*b^^2+1) )
(+ (* A (+ (* 8 (EXPT B 2)) 1)) (* 4 B C (+ (* 4 (EXPT B 2)) 1)))
``````
`'` is the usual quote. `#I( some-infix-expression )` is the reader macro.
• Thanks, may I ask what does `'#I` mean here? – h__ Aug 12 '12 at 19:32
• It is a reader macro, which is defined in the `infix` library. – Svante Aug 12 '12 at 20:06
One project I'm keeping an eye on are so-called "Sweet Expressions". The goal of the project is to add "sweetening" that is backwards-compatible with s-expressions, and simple enough that the expressions won't get in the way of macros. (It has been observed, for example, that operator precedence really interferes with the macro system; hence, the proposed solution doesn't use operator precedence.)
It should be kept in mind that the project is in its infancy, particularly for Common Lisp; however, the project has a working implementation of infix notation, which relies on curly brackets and a simple algorithm:
``````{1 + {2 * 3} + {4 exp 5}}
``````
translates nicely into
``````(+ 1 (* 2 3) (exp 4 5))
``````
I'll just refer you to the link for a more in-depth discussion of the semantics of curly-braces.
I recently wrote a cl macro exactly for this purpose, you might find it useful. It's called ugly-tiny-infix-macro.
You could write the expression in question as:
``````(\$ a * (\$ 8 * (expt b 2) + 1) + 4 * b * c * (\$ 4 * (expt b 2) + 1))
``````
It is expanded to
``````(+ (* A (+ (* 8 (EXPT B 2)) 1)) (* (* (* 4 B) C) (+ (* 4 (EXPT B 2)) 1)))
``````
Explanation: \$ is the name of macro. The arguments are considered as a list of expressions and hence the liberal use of whitespace to separate numbers/forms from symbols that denote operators.
Consider the following examples to understand function of this macro better:
``````(\$ 1 + 2) ; gets converted to (+ 1 2), where name of the macro is \$
(\$ t and nil) ; gets converted to (and t nil)
(\$ 3 > 5) ; gets converted to (> 3 5)
(\$ 1 + 2 + 3) ; gets converted to (+ (+ 1 2) 3)
(\$ 1 + 2 * 3) ; gets converted to (+ 1 (* 2 3))
(\$ 1 < 2 and 2 < 3) ; gets converted to (AND (< 1 2) (< 2 3))
``````
Anything within parentheses at position of an operand is treated like a lisp form.
``````(\$ 2 + (max 9 10 11)) ; gets converted to (+ 2 (max 9 10 11)). It could have been any function / lisp form.
(\$ 6 / (\$ 1 + 2)) ; gets converted to (/ 6 (\$ 1 + 2)), and then subsequently to (/6 (+ 1 2))
``````
I find it easier to reason about and more advantageous than a reader macro, since it may be easily intermixed with lisp forms, so you can nest lisp forms within the expression. For example, the `(exp b 2)` could have been any lisp form, like `(max a b c)` or your own user defined `(foobar a b c)`. | 1,391 | 4,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-43 | latest | en | 0.891841 |
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# A certain roller coaster has 3 cars, and a passenger is
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A certain roller coaster has 3 cars, and a passenger is [#permalink] 12 Apr 2006, 18:59
A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the
3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride
the roller coaster 3 times, what is the probability that the passenger will ride in each of
the 3 cars?
A. 0
B. 1/9
C. 2/9
D. 1/3
E. 1
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Re: Probability: Roller coaster [#permalink] 12 Apr 2006, 19:32
M8 wrote:
A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of
the 3 cars?
A. 0
B. 1/9
C. 2/9
D. 1/3
E. 1
total = 3x3x3 = 27
chances are = 123, 132, 213, 231, 312, 321. so total= 6
p = 6/27 = 2/9
Re: Probability: Roller coaster [#permalink] 12 Apr 2006, 19:32
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https://remi.flamary.com/demos/svmreg.html | 1,702,070,796,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00807.warc.gz | 534,069,384 | 4,721 | # Rémi Flamary
Professional website
## SVM and regularization
$\def\w{\mathbf{w}}$
In this demo, we illustrate the effect of regularization and kernel parameter choice on the decision function of support vector machines (SVM).
### Short introduction to SVM
The decision function of a support vector machine classifier is obtained through the minimization of the following optimization problem:
$$\min_{f}\quad C\sum_{i=1}^{n} \max(0,1-y_i f(\mathbf{x}_i)) + \|f\|^2$$
where C is a regularization parameter that balances the data fitting (left hand term) and the regularization (right hand term). Note that the data fitting term uses a training set $\{\mathbf{x}_i,y_i\}_{i=1\dots n}$ that consists in a list of samples $\mathbf{x}_i\in\mathbb{R}^d$ and their associated class $y_i\in\{1,-1\}$.
One of the strengh of SVM is their ability to chose a complex representation of the data thanks to the use of a kernel function $k(\cdot,\cdot)$ that measures the similarity between samples. The decision function is of the form
$$f(\mathbf{x}) =\sum_{i=1}^n \alpha_i y_i k(\mathbf{x},\mathbf{x}_i)$$
In practice the Gaussian kernel (also known as RBF) defined as
$$k(\mathbf{x},\mathbf{x}')=\exp{(-\gamma \|\mathbf{x}- \mathbf{x}'\|^2)}$$
is often used when the decision function has to handle non-linearities.
When using SVM with a gaussian kernel, one has to select two important parameters: C and $\gamma$. In this demo we illustrate the effect of those parameters on the final decision function of the SVM.
### Dataset used in the demo
In this demo, we illustrate SVM using a 2D non-linear toy dataset also known as "Clown". The main advantage of a 2D example is that it is easy to plot and visualize the samples of each classes in a classical scatter plot figure as shown below.
In this Figure, we can see that a non-linear function has to be used for a correct classification but the complexity of the fonction is limited. As illustrated in the next section, the parameters have to be chosen carefully.
### Regularization demo
#### Performance
Rec. rate
Current value: C=10
Current value: =10
RR=0.9754
The C parameter that will balance the data and regularization term, which will promote smooth decision function when C is small. The parameter $\gamma$ of the gaussian kernel is also extremely important as it will define the neighborhood of the samples. A large $\gamma$ leads to more complex function. The precision of the classifier on a large test sample is also reported on the right as RR (for recognition rate).
### References
For more information about Support Vector Machine, I strongly recommend [1] that is a classic introduction. A very good course by Stéphane Canu is also available online [2].
The figures have been generated using Python,Numpy, and Scikit Learn. The code is avalable here.
[1] Learning with kernels: support vector machines, regularization, optimization, and beyond, B Scholkopf, AJ Smola, 2001, MIT Press.
[2] Understanding SVM, S. Canu. | 731 | 3,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-50 | latest | en | 0.836664 |
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Figure EFGH on the grid below represents a trapezoidal plate at its starting position on a rotating surface The plate is rotated 90o about the origin in the counterclockwise direction. In the image trapezoid, what are the coordinates of the endpoints of the side congruent to side EH? Answer (8, 4) and (5, 2) (4, -8) and (2, -5) (-8, -4) and (-5, -7) (-8, -4) and (-5, -2)
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Not the answer you are looking for? Search for more explanations. | 368 | 1,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-05 | latest | en | 0.407233 |
https://itzprogramming.com/qa/question-do-you-hit-the-black-ball-in-pool.html | 1,627,457,029,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00409.warc.gz | 345,569,587 | 8,978 | # Question: Do You Hit The Black Ball In Pool?
## What happens if you hit the black ball in pool?
a) If a player pockets the black ball before all the balls in their own group10, the player loses the game.
b) A player going in off the black ball when the black ball is pocketed loses the game..
## Which ball do you hit in pool?
You’ll establish who plays which at the beginning of the game after the break. You don’t pocket the 8-ball until the end of the game. When you have pocketed all your balls, you take aim at the 8-ball. The first player to pocket all his or her balls and then pocket the 8-ball is the winner.
## Do you lose if you sink the white ball on the black?
Yep. But if you sink the black off the break and don’t pot any other balls you win. That’s a hard shot.
## What happens if you sink the cue ball?
During the Game Under standard 8-ball rules, if a player sinks the cue ball during her turn, play is passed over to her opponent, who is allowed to place the cue ball anywhere on the table for his next shot.
## Is it a scratch if you don’t hit a ball?
Table Scratch. When a player fails to hit any object ball with the cue ball, it’s considered a table scratch. The same goes for an object ball that fails to touch either a cushion or a pocket.
## Are you supposed to hit the black ball in 8-ball?
Technically in my experience in tournament play, if you pocket a ball of yours in before you pocket the 8-ball but on the same shot, it is the same as scratching on the 8-ball shot because technically you aren’t supposed to hit any other ball before you hit the 8-ball with the cue ball.
## What happens when the black ball leaves the table?
If the cue ball is potted or driven off the table, then that too is a foul. More about fouls and their consequences in blackball pool below. Any fouls on the break are ignored if the black ball is potted. If that happens the object balls are always racked again and the same player breaks.
## Is there 2 shots on the black ball?
If it’s just the black on the table, 1 shot. If you are on the black and there’s other balls left for your opponent, 2 shots.
## What happens if the white ball goes in after the black ball?
If you are playing 8 ball pool by International rules, then it means that the player, who pocketed the balls, has lost. Because the rule says, if you pocket the white ball after pocketing the black (in 8 ball pool) you have lost. … As long as the 8 ball is still in play… meaning not in a pocket…the game is still on.
## Can you hit opponents ball first in 8 ball?
note the 8 ball is never neutral, so it can only legally be hit first when it’s your last ball. the cue ball is not ever considered an object ball. combo’s you must hit one of your balls first. an opponents ball or the “8″ can be in the middle of a multi ball combo but you must pocket your ball.
## What happens if you hit a ball in and scratch?
6. If you scratch on the break, your opponent can place the cue ball anywhere behind the head string (i.e., “in the kitchen”); and in executing the next shot, the cue ball must cross the head string before contacting any object ball. 7.
## Can you hit the black ball first in Pool?
You don’t have the right to hit the other player’s balls first. The first ball that you hit must be one of your own color, or the black if you have no balls left on the table. If you fail in doing this, it’s a fault (SCRATCH). You don’t have the right to put the white ball into a hole.
## What happens if you hit the 8 ball in on the break?
If you scratch on the break you don’t automatically lose, the other player just gets ball in hand behind the head string. Likewise, if you sink the 8 ball on the break, you don’t win. You either spot the 8 ball or re-rack. Traditionally, 8 ball is a “call shot” game.
## Do pool balls get old?
The average billiard balls wear out after about a year of use to a size that is no longer considered to meet specifications. The cue ball will degrade faster due to constantly being struck by cue tips. However, if your pool table isn’t subjected to much use, then your balls can last well over a year.
## Can you play off the eight ball?
The game continues if the 8-ball has not been pocketed. This seems to confirm that if you legally pocketed the 8-ball, you won. A legally “called” pocket is defined in the same rules as follows: In Call Pocket, it is encouraged that all balls be specified along with their intended pocket.
## Can you shoot backwards in 8 ball pool?
If you pot the nominated ball (except the black ball) it’s deemed legal and you continue with the break. … But of all the rules, one of the most protested is when your opponents sinks the white ball while making a shot – can you shoot the repositioned white ball on the “D” backwards. Well yes you can! | 1,136 | 4,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-31 | latest | en | 0.921918 |
https://newpathworksheets.com/math/kindergarten/count-and-write-1-10-0/utah-common-core-standards | 1,571,508,382,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986697439.41/warc/CC-MAIN-20191019164943-20191019192443-00478.warc.gz | 616,589,631 | 7,314 | ## ◂Math Worksheets and Study Guides Kindergarten. Count and write 1-10
### The resources above correspond to the standards listed below:
#### Utah Common Core Standards
UT.CC.CC.K. Counting and Cardinality
Compare numbers.
CC.K.7. Compare two numbers between 1 and 10 presented as written numerals.
Count to tell the number of objects.
Know number names and the count sequence.
CC.K.1. Count to 100 by ones and by tens.
CC.K.2. Count forward beginning from a given number within the known sequence (instead of having to begin at 1). | 130 | 536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-43 | latest | en | 0.917785 |
https://www.fusioncharts.com/blog/how-we-decode-visual-information-podv/ | 1,718,405,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00810.warc.gz | 724,877,329 | 16,465 | This is the fourth post in our series ‘Principles of Data Visualization’ #PoDV For the past few weeks, we’ve been discussing the goals of data visualization as well as how our eyes and brain process visual information. We’ll go beyond the physical process this week to learn about the fundamentals of visual processing. In this post, we’ll teach you about the most fundamental building blocks of a visualization: Preattentive Attributes and how we decode visual information. If you’re looking for a Data Visualization Tool, we recommend FusionCharts for data visualization when decoding visual information.
## Preattentive Attributes Used by Our Working Memory
Colin Ware, a data visualization expert, wrote a book ‘Information Visualization’ in which he terms the basic building blocks of the visualization process as ‘Preattentive’ attributes. These attributes are what immediately catch our eye when we look at visualization. They can be perceived in less than 10 milliseconds, even before we make a conscious effort to notice them. Here’s a list of the preattentive attributes: Image credit: Colin Ware, Information Visualization: Perception for Design These attributes come into play when we analyze any visual. Of this list, only 2-D Position, and Line Length can be used to perceive quantitative data with precision. The other attributes are useful for perceiving other types of data such as categorical, or relational data. For example, both the pie chart and the bar chart below show the same data. But you can’t easily tell from the pie chart which is the biggest slice of the pie. That’s more clearly visible in the bar chart as it calls on the preattentive attribute of length. Considering preattentive attributes can help when deciding which chart type to use for our data. While preattentive attributes are what we immediately identify in a visual, they aren’t the only things we notice. We go on to form analytical patterns.
## Forming Analytical Patterns Out of Preattentive Attributes
If preattentive attributes are the alphabets of visual language, analytical patterns are the words we form using them. We immediately identify the preattentive attributes in a visualization. We then combine the preattentive attributes to seek out analytical patterns in the visual. Here are the basic analytical patterns that we identify when looking at a visual: Image credit: Stephen Few, Now You See It: Simple Visualization Techniques for Quantitative Analysis These patterns are an intrinsic part of our vision, and are frequently used when we analyze and describe a chart. However, as a word of caution, we are hard-wired to look for patterns in any visual information we notice. Sometimes we do this even when there isn’t an apparent connection or pattern in the visual. To avoid this, it helps to know the various patterns and have a wide vocabulary to work with visuals. This will allow us to consider multiple options before concluding on the most prominent pattern in a visual. If you’d like to know more on this topic, get our white paper ‘Principles of Data Visualization’. Stay tuned next week as we bring this series to a close with the important Gestalt principles.
## Take your data visualization to a whole new level
From column to donut and radar to gantt, FusionCharts provides with over 100+ interactive charts & 2,000+ data-driven maps to make your dashboards and reports more insightful
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 706 | 3,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.853475 |
https://slate.com/culture/2007/04/why-u-s-tax-policy-makes-saving-a-sucker-s-game.html | 1,660,403,817,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571959.66/warc/CC-MAIN-20220813142020-20220813172020-00065.warc.gz | 488,199,011 | 268,785 | Wall Street Self-defense
# Spend Every Dime!
## Why U.S. tax policy makes saving a sucker’s game.
For the first time since the Great Depression, the U.S. personal savings rate has “gone negative.” In 2005 and 2006, U.S. citizens spent more than they made. Economists disagree about just how ominous this is, but they generally agree on why it’s happening. Americans are “overspending.”
Why are we doing this? Partly because we’re acquisitive consumers obsessed with instant gratification and toys. Partly because soaring real-estate and stock markets make us feel rich. But also partly because we’re not suckers. Perverse tax laws for investments discourage saving, so it’s no surprise we spend.
If I said to you, “You can have \$10,000 to spend now—or \$9,500 to spend in 10 years,” which would you choose? Probably the \$10,000 now. And in doing so, you would be making the same choice many Americans make when deciding whether to save or spend their hard-earned cash.
The problem is how we tax investment gains. Over the past 80 years, the average annual return on Treasury bills (a proxy for savings accounts) has been 3.7 percent per year. Inflation, meanwhile, has averaged 3.1 percent per year. This combination has produced a “real return” of a paltry 0.6 percent per year. If you got to keep that 0.6 percent, you might still have an incentive to save: A \$616 real gain on \$10,000 in 10 years wouldn’t be much, but it would at least be \$616 more than you have now. Unless you’re so poor that you’re exempt from taxes, however, or so flush that you can afford to lock up cash for decades in a tax-deferred annuity or retirement account, you won’t be keeping that 0.6 percent. You’ll be giving all of it—and probably more—to the government.
How does the math work? Let’s say your T-bills return 3.7 percent. If you stash \$10,000, you’ll make \$370 before taxes and inflation in the first year. Taxes are assessed on the nominal gain (before adjusting for inflation) instead of the real gain, so if you’re in the 15 percent tax bracket, you’ll then pay \$56 to the government—and lose about \$310 of value to inflation. In other words, you’ll eke out about a \$5 real gain on a \$10,000 investment (an 0.05 percent return). If you’re in higher brackets, meanwhile, you’ll actually lose about 0.5 percent of value every year. The only time you’ll generate real gains is when “real” rates of return are significantly higher than 0.6 percent (as they are now). But when real rates are negative, as they were a few years ago, you’ll be losing a lot more than 0.5 percent per year.
Fine, you say. T-bills are for chumps. Stocks provide a far better long-term return, and—if you don’t trade like Jim Cramer—they’re more tax efficient: Hold your stocks for more than a year, and you’ll pay only long-term capital gains tax, not income tax. And you’ll get sweetheart rates on dividends, too.
If your time horizon is long enough, this is true. Unlike T-bills or bank accounts, stocks compound tax free, so you won’t owe tax until you sell them (except, again, on the dividends). Yet even stocks aren’t ideal for savings. For one thing, there are those annoying bear markets: The S&P 500 is still below where it was seven years ago, even before adjusting for inflation. Then there are dividend taxes: In the 20th century, nearly half of the average 10 percent annual return on U.S. stocks came from dividends, not price appreciation, and you pay taxes on dividends every year. Lastly, there’s the absurd way that the IRS accounts for “realized gains.” Once you’re in the black on a stock or fund, current tax policy forces you to stick with it—or get socked with a capital-gains tax bill. In other words, even if your stock’s best gains are behind it, if you switch to a better stock, it might be years after paying your tax bill before you get back to even.
Can you reduce savings taxes? Yes, if you’re skilled and well-informed, you can “harvest losses,” buy “tax-managed” funds, and implement other tax-minimization strategies. Doing so will usually consume money and time, however, and be a major headache. And you’ll still have the bear-market problem: Unless you’re in your 20s or 30s and saving for retirement, stocks are too risky to represent your entire portfolio. So, given current tax policy, it’s no wonder we’re not saving anything.
How could we fix this?
For starters, we could do the same thing for regular savers as we do for real-estate investors: Change the definition of a “realized gain.” Real-estate investors can take advantage of a “1031 Exchange,” which allows them to take gains from the sale of one property and reinvest them in another without triggering a tax event. The same system should apply to other investments: If you own a stock or fund that has doubled, you shouldn’t be forced to hang onto it just to avoid triggering a taxable gain. Instead, you should be able to sell it and invest the proceeds in another stock or fund. Gains should only be “realized” when you take the money out of your investment account and spend it. | 1,190 | 5,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-33 | longest | en | 0.951992 |
https://spiral.ac/sharing/vhenut9/complex-roots-of-the-characteristic-equations-1-second-order-differential-equations-khan-academy | 1,604,054,263,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910204.90/warc/CC-MAIN-20201030093118-20201030123118-00375.warc.gz | 518,602,426 | 10,165 | complex-roots-of-the-characteristic-equations-1-second-order-differential-equations-khan-academy
# Interactive video lesson plan for: Complex roots of the characteristic equations 1 | Second order differential equations | Khan Academy
#### Activity overview:
What happens when the characteristic equations has complex roots?!
Watch the next lesson: https://www.khanacademy.org/math/differential-equations/second-order-differential-equations/complex-roots-characteristic-equation/v/complex-roots-of-the-characteristic-equations-2?utm_source=YT&utm_medium=Desc&utm_campaign=DifferentialEquations
Missed the previous lesson?
https://www.khanacademy.org/math/differential-equations/second-order-differential-equations/linear-homogeneous-2nd-order/v/2nd-order-linear-homogeneous-differential-equations-4?utm_source=YT&utm_medium=Desc&utm_campaign=DifferentialEquations
Differential Equations on Khan Academy: Differential equations, separable equations, exact equations, integrating factors, homogeneous equations.
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
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## Get the Clip Chrome Extension & Create Video Lessons in Seconds
Add Clip to Chrome | 815 | 3,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2020-45 | latest | en | 0.770755 |
http://mathhelpforum.com/geometry/125453-help-3-geometry-questions-print.html | 1,498,219,923,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320057.96/warc/CC-MAIN-20170623114917-20170623134917-00152.warc.gz | 257,476,018 | 3,830 | # Help with 3 Geometry Questions
• Jan 25th 2010, 04:37 PM
LittleLotte
Help with 3 Geometry Questions
Hello everyone!
I've spent hours trying to figure out these last few Geometry questions and I can't seem to find any information on it in my text book. If anyone could offer some help I would really appreciate it. Thank you so much!
1. In the diagram below, ABCD is a parallelogram. What is Angle BEC?
Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting
2. What is the value of x in the diagram below?
Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting
3. What are the values of n and t in the kite below?
Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting
a. n= 4 t= 39
b. n= 4 t= 51
c. n= 4.5 t= 39
d. n=4.5 t= 51
• Jan 25th 2010, 05:22 PM
LittleLotte
I only have until 9 tonight to finish these questions so I REALLY would appreciate the help. I've tried every resource I could. Please help.
• Jan 25th 2010, 07:19 PM
Soroban
Hello, LittleLotte!
Quote:
1. In the diagram below, $ABCD$ is a parallelogram.
What is angle $BEC$?
Code:
A o * * * * * o B
* * 30° * *
* * * *
* o E *
* * * *
* * 35° * *
D o * * * * * o C
We are given: . $\angle BAE = 30^o$
Since $AB \parallel DC$. then: $\angle ECD = \angle BAE$ .(alterate-interior angles)
Then $\angle ECD = 30^o$
In $\Delta EDC$, we know two angles: . $\angle EDC = 35^o,\;\angle ECD = 30^o$
Hence, the third angle is: . $\angle DEC \:=\:180^o - 35^o - 30^o \:=\:115^o$
$\angle BEC$ and $\angle DEC$ are supplementary: . $\angle BEC + \angle DEC \:=\:180^o$
So we have: . $\angle BEC + 115^o \:=\:180^o \quad\Rightarrow\quad\boxed{ \angle BEC \,=\,65^o}$
• Jan 25th 2010, 10:16 PM
Quote:
Originally Posted by LittleLotte
2. What is the value of x in the diagram below?
Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting
The upper and lower triangles are identical, due to opposite corner angles being equal, touching angles equal and parallel sides equal length.
Hence 2a+2=4 so 2a=4-2=2, a=1
x=a+3.5=1+3.5=4.5
• Jan 25th 2010, 10:23 PM
Quote:
Originally Posted by LittleLotte
3. What are the values of n and t in the kite below?
Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting
a. n= 4 t= 39
b. n= 4 t= 51
c. n= 4.5 t= 39
d. n=4.5 t= 51
All 3 angles in a triangle sum to 180 degrees.
In a right-angled triangle, one angle is 90 degrees.
180-90=90 so t and 51 degrees sum to 90 degrees, so t=90-51=39 degrees.
If the side of length 2n equals the side below it, then the side 3n-4 equals the side n+5 in length.
n+5=3n-4, n+5+4=3n, n+9=3n, n-n+9=3n-n=2n, 9=2n, n=9/2=4.5 | 949 | 2,687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-26 | longest | en | 0.783219 |
https://metanumbers.com/1239 | 1,680,127,769,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00174.warc.gz | 462,974,243 | 7,556 | # 1239 (number)
1,239 (one thousand two hundred thirty-nine) is an odd four-digits composite number following 1238 and preceding 1240. In scientific notation, it is written as 1.239 × 103. The sum of its digits is 15. It has a total of 3 prime factors and 8 positive divisors. There are 696 positive integers (up to 1239) that are relatively prime to 1239.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 4
• Sum of Digits 15
• Digital Root 6
## Name
Short name 1 thousand 239 one thousand two hundred thirty-nine
## Notation
Scientific notation 1.239 × 103 1.239 × 103
## Prime Factorization of 1239
Prime Factorization 3 × 7 × 59
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 1239 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 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 1,239 is 3 × 7 × 59. Since it has a total of 3 prime factors, 1,239 is a composite number.
## Divisors of 1239
1, 3, 7, 21, 59, 177, 413, 1239
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 1920 Sum of all the positive divisors of n s(n) 681 Sum of the proper positive divisors of n A(n) 240 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 35.1994 Returns the nth root of the product of n divisors H(n) 5.1625 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,239 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 1,239) is 1,920, the average is 240.
## Other Arithmetic Functions (n = 1239)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 696 Total number of positive integers not greater than n that are coprime to n λ(n) 174 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 205 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 696 positive integers (less than 1,239) that are coprime with 1,239. And there are approximately 205 prime numbers less than or equal to 1,239.
## Divisibility of 1239
m n mod m 2 3 4 5 6 7 8 9 1 0 3 4 3 0 7 6
The number 1,239 is divisible by 3 and 7.
## Classification of 1239
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (1239)
Base System Value
2 Binary 10011010111
3 Ternary 1200220
4 Quaternary 103113
5 Quinary 14424
6 Senary 5423
8 Octal 2327
10 Decimal 1239
12 Duodecimal 873
20 Vigesimal 31j
36 Base36 yf
## Basic calculations (n = 1239)
### Multiplication
n×y
n×2 2478 3717 4956 6195
### Division
n÷y
n÷2 619.5 413 309.75 247.8
### Exponentiation
ny
n2 1535121 1902014919 2356596484641 2919823044470199
### Nth Root
y√n
2√n 35.1994 10.7405 5.93291 4.15541
## 1239 as geometric shapes
### Circle
Diameter 2478 7784.87 4.82272e+06
### Sphere
Volume 7.96714e+09 1.92909e+07 7784.87
### Square
Length = n
Perimeter 4956 1.53512e+06 1752.21
### Cube
Length = n
Surface area 9.21073e+06 1.90201e+09 2146.01
### Equilateral Triangle
Length = n
Perimeter 3717 664727 1073.01
### Triangular Pyramid
Length = n
Surface area 2.65891e+06 2.24155e+08 1011.64
## Cryptographic Hash Functions
md5 b3ba8f1bee1238a2f37603d90b58898d c60d81f35d8be966fa0437e43f1feff777c6e121 f97350101e1a9de922bdbb762a33695234102dc119ffafa25c15d0418ec82b23 fa86b62a93a0a4ed98bcd1b787d9fd06fb813ec687954ab1b6c192a88a250cc9178f200e377065ae03591c14e73c6d852b934f5a5676a43e258726131f3408c9 60f6bde0513e989b0df501428af228fd436a77f9 | 1,444 | 4,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2023-14 | longest | en | 0.825271 |
https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Signal_to_noise_plus_interference.html | 1,716,546,807,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00400.warc.gz | 276,738,056 | 7,679 | # Signal-to-interference-plus-noise ratio
In information theory and telecommunication engineering, the signal-to-interference-plus-noise ratio (SINR[1]) (also known as the signal-to-noise-plus-interference ratio (SNIR)[2]) is a quantity used to give theoretical upper bounds on channel capacity (or the rate of information transfer) in wireless communication systems such as networks. Analogous to the SNR used often in wired communications systems, the SINR is defined as the power of a certain signal of interest divided by the sum of the interference power (from all the other interfering signals) and the power of some background noise. If the power of noise term is zero, then the SINR reduces to the signal-to-interference ratio (SIR). Conversely, zero interference reduces the SINR to the signal-to-noise ratio (SNR), which is used less often when developing mathematical models of wireless networks such as cellular networks.[3]
The complexity and randomness of certain types of wireless networks and signal propagation has motivated the use of stochastic geometry models in order to model the SINR, particularly for cellular or mobile phone networks.[4]
## Description
SINR is commonly used in wireless communication as a way to measure the quality of wireless connections. Typically, the energy of a signal fades with distance, which is referred to as a path loss in wireless networks. Conversely, in wired networks the existence of a wired path between the sender or transmitter and the receiver determines the correct reception of data. In a wireless network ones has to take other factors into account (e.g. the background noise, interfering strength of other simultaneous transmission). The concept of SINR attempts to create a representation of this aspect.
## Mathematical definition
The definition of SINR is usually defined for a particular receiver (or user). In particular, for a receiver located at some point x in space (usually, on the plane), then its corresponding SINR given by
where P is the power of the incoming signal of interest, I is the interference power of the other (interfering) signals in the network, and N is some noise term, which may be a constant or random. Like other ratios in electronic engineering and related fields, the SINR is often expressed in decibels or dB.
## Propagation model
To develop a mathematical model for estimating the SINR, a suitable mathematical model is needed to represent the propagation of the incoming signal and the interfering signals. A common model approach is to assume the propagation model consists of a random component and non-random (or deterministic) component.[5][6]
The deterministic component seeks to capture how a signal decays or attenuates as it travels a medium such as air, which is done by introducing a path-loss or attenuation function. A common choice for the path-loss function is a simple power-law. For example, if a signal travels from point x to point y, then it decays by a factor given by the path-loss function
,
where the path-loss exponent α>2, and |x-y| denotes the distance between point y of the user and the signal source at point x. Although this model suffers from a singularity (when x=y), its simple nature results in it often being used due to the relatively tractable models it gives.[3] Exponential functions are sometimes used to model fast decaying signals.[1]
The random component of the model entails representing multipath fading of the signal, which is caused by signals colliding with and reflecting off various obstacles such as buildings. This is incorporated into the model by introducing a random variable with some probability distribution. The probability distribution is chosen depending on the type of fading model and include Rayleigh, Rician, log-normal shadow (or shadowing), and Nakagami.
## SINR model
The propagation model leads to a model for the SINR.[2][6][4] Consider a collection of 'n' base stations located at points x1 to xn in the plane or 3D space. Then for a user located at, say x=0, then the SINR for a signal coming from base station, say, xi, is given by
,
where Fi are fading random variables of some distribution. Under the simple power-law path-loss model becomes >
.
## Stochastic geometry models
In wireless networks, the factors that contribute to the SINR are often random (or appear random) including the signal propagation and the positioning of network transmitters and receivers. Consequently, in recent years this has motivated research in developing tractable stochastic geometry models in order to estimate the SINR in wireless networks. The related field of continuum percolation theory has also been used to derive bounds on the SINR in wireless networks.[2][4][7] | 965 | 4,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-22 | latest | en | 0.931687 |
https://documen.tv/at-a-college-the-cost-of-tuition-increased-by-10-let-b-represent-the-former-cost-of-tuition-use-29992477-27/ | 1,686,210,551,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654606.93/warc/CC-MAIN-20230608071820-20230608101820-00119.warc.gz | 242,335,358 | 15,305 | Question
At a college, the cost of tuition increased by 10%. Let b represent the former cost of tuition. Use the expression b+0.10b for the new cost of tuition.
a. Write an equivalent expression by combining like terms.
b. If the cost increases by 10%, you can multiply the former cost by %.
1. thnahdat
The equivalent expression by combining like terms is 1.10b and
If the cost increases by 10%, you can multiply the former cost by 110%
How to determine the equivalent expression by combining like terms.
From the question, we have the following parameters that can be used in our computation:
Expression of 10% raise = b + 0.10b
Evaluate the like terms in the above expression
So, we have the following representation
Expression of 10% raise = 1.10b
How to complete the blank
Here, we have
If the cost increases by 10%, you can multiply the former cost by %.
In this case, the factor is calculated as
factor = 1 + 10%
So, we have
Factor = 110%
So, the missing statement is 110% | 248 | 987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.948109 |
https://www.airmilescalculator.com/distance/lca-to-pfo/ | 1,721,162,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00138.warc.gz | 557,285,353 | 14,681 | # How far is Paphos from Larnaca?
The distance between Larnaca (Larnaca International Airport) and Paphos (Paphos International Airport) is 66 miles / 106 kilometers / 57 nautical miles.
The driving distance from Larnaca (LCA) to Paphos (PFO) is 78 miles / 125 kilometers, and travel time by car is about 1 hour 42 minutes.
66
Miles
106
Kilometers
57
Nautical miles
## Distance from Larnaca to Paphos
There are several ways to calculate the distance from Larnaca to Paphos. Here are two standard methods:
Vincenty's formula (applied above)
• 65.678 miles
• 105.699 kilometers
• 57.073 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 65.541 miles
• 105.478 kilometers
• 56.954 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Larnaca to Paphos?
The estimated flight time from Larnaca International Airport to Paphos International Airport is 37 minutes.
## Flight carbon footprint between Larnaca International Airport (LCA) and Paphos International Airport (PFO)
On average, flying from Larnaca to Paphos generates about 35 kg of CO2 per passenger, and 35 kilograms equals 76 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Larnaca to Paphos
See the map of the shortest flight path between Larnaca International Airport (LCA) and Paphos International Airport (PFO).
## Airport information
Origin Larnaca International Airport
City: Larnaca
Country: Cyprus
IATA Code: LCA
ICAO Code: LCLK
Coordinates: 34°52′30″N, 33°37′29″E
Destination Paphos International Airport
City: Paphos
Country: Cyprus
IATA Code: PFO
ICAO Code: LCPH
Coordinates: 34°43′4″N, 32°29′8″E | 522 | 1,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-30 | latest | en | 0.821088 |
https://geyerlawandadr.com/site/lekf295.php?c05a6b=radial-wave-function-of-2s-orbital | 1,618,464,031,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038083007.51/warc/CC-MAIN-20210415035637-20210415065637-00383.warc.gz | 381,743,354 | 5,907 | However, #r = 0# doesn't count as a node because we would be looking at nothing with a viewing window of #r = 0#. How many atomic orbitals are there in a g subshell? From this, you can tell that the maximum electron density occurs near #5a_0# (with #a_0 ~~ 5.29177xx10^(-11) "m"#, the Bohr radius) from the center of the atom, and #4pir^2 R_(20)(r)^2# is about #2.45# or so. in Eq. Now, since we are talking about the hydrogen atom, #color(blue)(r = 2a_0)# for the radial node in the #2s# orbital of hydrogen. is equal to the following expression. For hydrogen, we have to use spherical harmonics, so our dimensions are written as #(r, theta, phi)#. #R_(nl)(r)# is the radial component of the wave function #psi_(nlm_l)(r,theta,phi)#, #Y_(l)^(m_l)(theta,phi)# is the angular component, #n# is the principal quantum number, #l# is the angular momentum quantum number, and #m_l# is the projection of the angular momentum quantum number (i.e. hydrogen, is equal to 1, and r is the distance from the Bring the right term to the left side of the equation. 6-6). Since if #R = 0#, #R^2 = 0#, let us just find #R = 0#. 1, by setting Next notice how the radial function for the 2s orbital, Figure $$\PageIndex{2}$$, goes to zero and becomes negative. Calculate the distance from the nucleus (in nm) of the node of the 2s wave function. r_0 = 2 * a_0 The key to this problem lies with what characterizes a radial node. If we plot #4pir^2R_(nl)(r)^2# against #r#, we get the probability density curves for an atomic orbital. Since the wave function shown has no time variable, let us define #Psi = psi# where #psi# is the time-independent wave function. Basically, the wave function, Psi(x), is simply a mathematical function used to describe a quantum object. #0, pm l#). Cancel out the exponent terms that appear on either side of the ), SEPARATION OF VARIABLES GIVES RADIAL AND ANGULAR COMPONENTS. The #2s# orbital's plot looks like this: The wave function for the 2s orbital in the hydrogen atom is. This behavior reveals the presence of a radial node in the function. The wave function of a 2s-orbital changes signs once, so you only have one nodal surface here. See all questions in Orbitals, and Probability Patterns. Fortunately there is no #theta# or #phi# term to complicate things here. The wave function is defined as follows, via separation of variables: #color(green)(psi_(nlm_l)(r,theta,phi) = R_(nl)(r) Y_(l)^(m_l)(theta, phi))#. The only way to get the square of its absolute value equal to zero is if you have, #Psi_(2s) = overbrace(1/(2sqrt(2pi)) * sqrt(1/a_0))^(color(purple)(>0)) * (2 - r/a_0) * overbrace(e^(-r/(2a_0)))^(color(purple)(>0))#, Here's how the wave function for the 2s-orbital looks like. around the world. The graphs below show the radial wave functions. equation. Additionally, set the first term, Right from the start, this tells you that you have, Now, take a look at the wave function again. This is why the 2s-orbital is spherical in shape. where a 0 is the value of the radius of the first Bohr orbit, equal to 0.529 nm; p is Z (r/a 0); and r is the distance from the nucleus in meters. JavaScript is required to view textbook solutions. The wave function for the 2s orbital of a hydrogen atom, nucleus, solve for Once you expand your viewing window from the center of the orbital, then you start seeing the electron density come into play. On the radial distribution graph for #a_0r^2R_(nl)^2(r)# vs. #r"/"a_0# (always positive), which plots probability density vs. viewing-window radius #r#, it touches #R = 0# when #r = 0# or #R_(nl)^2(r) = 0#. around the world. Now, you have a node wherever #psi^"*"psi# (for real numbers, #psi^2#), the probability density, as a whole is #0#. For real numbers, #\mathbf(psi^2(r,theta,phi) = R_(nl)^2(r)(Y_l^(m_l)(theta,phi)))^2#. Now, a node occurs when a wave function changes signs, i.e. The wave function that describes an electron in an atom is actually a product between the radial wave function, which is of interest in your case, and the angular wave function. The wave function represents an orbital. nucleus in meters. A radial node occurs when a radial wave function passes through zero. equal to y. What are some common mistakes students make with orbitals? Now, the radial component in general is (Inorganic Chemistry, Miessler): #color(green)(R_(20)(r) = 2(Z/(2a_0))^"3/2"(2-(Zr)/a_0)e^(-Zr"/"2a_0))#. The wave function represents an orbital. equal to zero. How many electrons can there be in a p orbital? To get the maximum electron density, you have to look at probability density curves. 4601 views Calculate the distance from the nucleus (in nm) of the node of the 2 s wave function. As gets smaller for a fixed , we see more radial … How many electrons can an s orbital have. What are the number of sub-levels and electrons for the first four principal quantum numbers? And the angular component in general is #color(green)(1/(2sqrtpi))#. The radial wave function depends only on the distance from the nucleus, #r#. The key to this problem lies with what characterizes a radial node. If you don't understand all of that, that's fine; it was just for context. where a0 is the value of the radius of the first Bohr orbit, equal to 0.529 nm; p is Z(r/a0); and r is the distance from the nucleus in meters. In this equation, Since you have zero probability of locating an electron at a node, you can say that you have, #color(blue)(|Psi(x)|^2 = 0) -># this is true at nodes, So, you are given the wave function of a 2s-orbital, #Psi_(2s) = 1/(2sqrt(2pi)) * sqrt(1/a_0) * (2 - r/a_0) * e^(-r/(2a_0))#, and told that at #r = r_0#, a radial node is formed. | 1,578 | 5,638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-17 | latest | en | 0.866473 |
https://www.coursehero.com/file/6811417/MIT2-007s09-sol-exam01/ | 1,524,307,517,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945111.79/warc/CC-MAIN-20180421090739-20180421110739-00019.warc.gz | 777,098,166 | 306,099 | {[ promptMessage ]}
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MIT2_007s09_sol_exam01
MIT2_007s09_sol_exam01 - MIT OpenCourseWare...
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MIT OpenCourseWare http://ocw.mit.edu 2.007 Design and Manufacturing I Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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Page 1 of 8 2.007 Design and Manufacturing I Draft Exam on Drawing, CAD, Motors, Pneumatics, and Mechanisms You have 1.5 hours to answer the following 9 questions. Please show your work to the extent possible given the allotted time. Point allocations (out of 100 total) are listed for each question. 1. (10 points) What tension, T , must be applied to raise the 4 N weight at a constant rate? pulley 10 cm Going up! First, I deal with the right side which is a capstan and NOT a pulley Going down! 5N By the capstan equation F between =5N*e^(- * )=5N*e^(-0.3* /2)=3.12N NOTE the sign in the exponent. It is nagative because the 5N weight is going down and the frictional forces oppose the motion resulting in a lower tensile force in the rope in between the capstan and the pulley than was applied by gravity. Next, I deal with the left side which is a pulley and NOT a capstan Sum of moments about the pivot SM=4N*10cm-F between *10cm-T*5cm=0 T=1.76N 4N F between F between T
Page 2 of 8 2. (5 points) How many degrees of freedom does this mechanism possess? 4 Bodies each with 3 DOF = 12 DOF 5 full pin joints eacg remove 2 DOF = -10 DOF Leaving 2 DOF of the mechanism One of the DOF is associated with the parallelogram linkage Comprised of the base, two links extending from the base, and the triangular piece near the top of the lamp. The other degree of freedom is associated with the lamp head itself which Is free to rotate about its pivot. Some students thought perhps some of the elements of the mechanism were Redundant, but none of them are.
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{[ snackBarMessage ]} | 585 | 2,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-17 | latest | en | 0.914819 |
https://learningpundits.com/module-view/67-statements,-assumptions,-arguments-and-conclusions/2-logical-reasoning-test---statements-&-arguments/?page=49 | 1,550,442,201,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247482788.21/warc/CC-MAIN-20190217213235-20190217235235-00191.warc.gz | 604,406,626 | 16,817 | # Statements, Assumptions, Arguments and Conclusions
Statements, Assumptions, Arguments and Conclusions: Critical Reasoning Questions. Tricks on how to solve critical Reasoning questions
Take Reasoning Test
View Reasoning Test Results
## Online Logical Reasoning Test Questions with Answers on Statements, Assumptions, Arguments and Conclusions
Q241. The question given below consists of a statement, followed by two arguments numbered I and II. You have to decide which of the arguments is a 'strong' argument and which is a 'weak' argument. **Statement:** Should import duty on all the electronic goods be dispensed with? **Arguments:** I. No. This will considerably reduce the income of the government and will adversely affect the developmental activities. II. No. The local manufacturers will not be able to compete with the foreign manufacturers who are technologically far superior.
1. Only argument I is strong
2. Only argument II is strong
3. Either I or II is strong
4. Neither I nor II is strong
5. Both I and II are strong
Solution : Only argument II is strong
Q242. The question given below consists of a statement, followed by two arguments numbered I and II. You have to decide which of the arguments is a 'strong' argument and which is a 'weak' argument. **Statement:** Should children be legally made responsible to take care of their parents during their old age? **Arguments:** I. Yes. Such matter can only be solved by legal means. II. Yes. Only this will bring some relief to poor parents.
1. Only argument I is strong
2. Only argument II is strong
3. Either I or II is strong
4. Neither I nor II is strong
5. Both I and II are strong
Solution : Neither I nor II is strong
### Math Whiz
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Q243. The question given below consists of a statement, followed by two arguments numbered I and II. You have to decide which of the arguments is a 'strong' argument and which is a 'weak' argument. **Statement:** Should there be reservation in Government jobs for candidates from single child family? **Arguments:** I. No. This is not advisable as the jobs should be offered to only deserving candidates without any reservation for a particular group. II. Yes. This will help reduce the growing population in India as the parents will be encouraged to adopt single child norm.
1. Only argument I is strong
2. Only argument II is strong
3. Either I or II is strong
4. Neither I nor II is strong
5. Both I and II are strong
Solution : Neither I nor II is strong
Q244. The question given below consists of a statement, followed by two arguments numbered I and II. You have to decide which of the arguments is a 'strong' argument and which is a 'weak' argument. **Statement:** Should higher education be completely stopped for some time? **Arguments:** I. No. It will hamper the country's future progress. II. Yes. It will reduce the educated unemployment.
1. Only argument I is strong
2. Only argument II is strong
3. Either I or II is strong
4. Neither I nor II is strong
5. Both I and II are strong
Solution : Only argument I is strong
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Q245. The question given below consists of a statement, followed by two arguments numbered I and II. You have to decide which of the arguments is a 'strong' argument and which is a 'weak' argument. **Statement:** Should we scrap the 'Public Distribution System' in India? **Arguments:** I. Yes, Protectionism is over, everyone must get the bread on his/her own. II. Yes. The poor do not get any benefit because of corruption.
1. Only argument I is strong
2. Only argument II is strong
3. Either I or II is strong
4. Neither I nor II is strong
5. Both I and II are strong
Solution : Neither I nor II is strong
Q{{(\$index+1)+((page-1)*LIMITPERPAGE)}}.
1.
Solution :
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https://apboardsolutions.in/ap-board-9th-class-physical-science-notes-chapter-9/ | 1,721,412,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00526.warc.gz | 85,512,320 | 14,070 | AP Board 9th Class Physical Science Notes Chapter 9 Floating Bodies
Students can go through AP State Board 9th Class Physical Science Notes Chapter 9 Floating Bodies to understand and remember the concept easily.
AP State Board Syllabus 9th Class Physical Science Notes Chapter 9 Floating Bodies
→ Density is defined as mass per unit volume. Mass
Density = $$\frac{\text { Mass }}{\text { Volume }}$$
Unit of density = gm/cm3 or kg/m3
→ The relative density of an object is the ratio of the density of the object to the density of the water.
Relative density of an object = $$\frac{\text { Density of the object }}{\text { Density of water }}$$
→ Experimentally relative density of a solid object = $$\frac{\text { Weight of the object }}{\text { weight of water equal to the volume of the object }}$$
→ Relative density of liquid = $$\frac{\text { Weight of liquid }}{\text { Weight of the same volume of water }}$$
→ Lactometer is used to measure the purity of milk.
→ A hydrometer is used to measure the relative density of liquids.
→ Objects having a density less than the liquid in which they are immersed float on the surface of the liquid.
→ Every fluid exerts upward pressure on the objects immersed in them. If the upward pressure is less than the gravitational force on the object, the object sinks otherwise it floats on the liquid.
→ The pressure exerted by the air in our surroundings is known as atmospheric pressure. Atmospheric pressure P0 = ρhg.
→ The barometer is used to measure the atmospheric pressure at a place.
→ Pressure at a depth ‘h’ in a liquid P = P0 + ρhg
P0 = Atmospheric pressure ; ρ = density of the liquid h = depth ; g = acceleration due to gravity
→ Buoyancy is the upward force exerted by the liquid on any object less dense than itself.
→ When an object is immersed in a fluid, it appears to lose weight (because of buoyancy).
→ The apparent loss of weight of an object, which is immersed in a liquid, is equal to the weight of the liquid displaced by the object. (Archimedes principle)
→ When an object floats on the surface of a liquid, it displaces a weight of liquid equal to its own weight.
→ The pressure exerted by a liquid increases with depth below the surface of the liquid.
→ Pascal’s principle states that external pressure applied to an enclosed body of fluid is transmitted equally in all directions throughout the fluid volume and the walls of the containing vessel.
→ Density: When two objects of equal volume are considered the object
with higher mass is said to be a dense object. Density is defined as mass per unit volume.
∴ Density = $$\frac{\text { Mass }}{\text { Volume }}$$
Unit of density gm/cm (or) kg/rn3.
→ Relative density: Relative density is the ratio of the density of an object to the density of water.
Relative density of an object = $$\frac{\text { Density of the object }}{\text { Density of water }}$$
(or)
Relative density of an object = $$\frac{\text { Weight of the object }}{\text { weight of water equal to the volume of the object }}$$
→ Lactometer: Lactometer is a device used to measure the purity of milk. It works on the principle of relative density.
→ Hydrometer/ Densitometer: Hydrometer/Densitometer ¡s an instrument used to measure the relative density of any liquid.
→ Atmospheric pressure: Air in our surroundings exerts pressure on the surface of the earth called atmospheric pressure.
Atmospheric pressure P0 = ρhg
ρ = average density of air
h = height of the atmosphere
g acceleration due to gravity
→ Barometer: Atmospheric pressure can be measured using a barometer. The first barometer was invented by ‘Torricelli’, using mercury. The height of the mercury column at normal atmospheric pressure is 76 cm.
The increase or decrease in the height of the mercury column indicates sudden changes in the atmosphere.
→ Buoyancy: The upward force that a fluid exerts on an object less dense than itself.
(or)
Buoyancy is the ability of an object to float in a liquid.
→ Archimedes principle of buoyancy: An object in a fluid is buoyant up by a force equal to the weight of the fluid the object displaces.
→ Archimedes:
• Archimedes was born in 287 BC and died in 212 BC.
• He was a Greek mathematician, physicist, engineer, inventor, and astronomer.
• He is well known for his Archimedes principle.
• Among his advances in physics are the foundations of hydrostatics. | 1,003 | 4,387 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.900932 |
https://linear-equation.com/linear-equation-graph/ratios/math-elimination-calculator.html | 1,709,371,825,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475806.52/warc/CC-MAIN-20240302084508-20240302114508-00403.warc.gz | 369,648,990 | 11,211 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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Author Message
exbnaseve_tom
Registered: 30.09.2003
From: Living in many forums....
Posted: Saturday 30th of Dec 18:49 Hi dudes, I’m really stuck on math elimination calculator and would sure desperately need guidance to solve with cramer’s rule, interval notation and powers. My tests is due soon. I have even thought of hiring a tutor, but they are dear . So any guidance would be greatly treasured.
Jahm Xjardx
Registered: 07.08.2005
From: Odense, Denmark, EU
Posted: Monday 01st of Jan 08:57 First of all, let me welcome you to the world of math elimination calculator. You need not worry; this subject seems to be tough because of the many new symbols that it has. Once you learn the concepts , it becomes fun. Algebrator is the most preferred tool amongst novice and professionals . You must buy yourself a copy if you are serious at learning this subject.
alhatec16
Registered: 10.03.2002
From: Notts, UK.
Posted: Tuesday 02nd of Jan 08:32 Hi there! I used Algebrator last year when I was having issues with my college math. This program made solving problems so easy. Since then, I always keep a copy of it on my computer .
inojedute
Registered: 25.09.2002
From: /home/kai
Posted: Thursday 04th of Jan 09:01 Sounds like something interesting! Thanks guys, just one final question, can someone please provide me a URL where I can order my copy of this program?
CHS`
Registered: 04.07.2001
From: Victoria City, Hong Kong Island, Hong Kong
Posted: Friday 05th of Jan 08:22 Yeah you will have to buy it though it will cost you less than a tutor. You can get all the details about Algebrator here https://linear-equation.com/solution-of-the-equations.html. They give you an no questions asked money-back guarantee. I haven’t yet had any reason to take them up on it though. All the best!
Sdefom Koopmansshab
Registered: 28.10.2001
From: Woudenberg, Netherlands
Posted: Saturday 06th of Jan 11:21 I remember having difficulties with graphing inequalities, binomial formula and leading coefficient. Algebrator is a really great piece of math software. I have used it through several algebra classes - Pre Algebra, Intermediate algebra and Algebra 2. I would simply type in the problem and by clicking on Solve, step by step solution would appear. The program is highly recommended. | 848 | 3,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.879045 |
https://www.cuemath.com/jee/solved-examples-set-1-circles/ | 1,620,554,218,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988966.82/warc/CC-MAIN-20210509092814-20210509122814-00156.warc.gz | 728,433,840 | 20,455 | # Circles Set-1
Go back to 'SOLVED EXAMPLES'
Example – 1
The circle $${x^2} + {y^2} - 4x - 4y + 4 = 0$$ is inscribed in a triangle which has two of its sides along the co-ordinate axes. If the locus of the circumcentre of the triangle is
$x + y - xy + k\sqrt {{x^2} + {y^2}} = 0$
find the value of k.
Solution: The situation is described clearly in the figure below:
The equation of L is, using the intercept form,
$L:\frac{x}{a} + \frac{y}{b} = 1$
The distance of the centre of S, i.e. (2, 2) from L must equal the radius of S which is 2. Thus,
\frac{{\left| {\frac{2}{a} + \frac{2}{b} - 1} \right|}}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{align}&{\text{We now use the fact that }}\\&L{\text{(2, 2) is negative since}}\\&{\text{(2, 2) and the origin lie }}\\&{\text{on the same side of }}L{\text{ and}}\\&L{\text{(0, 0) is negative }}\end{align} \right\}
$\Rightarrow \quad 2a + 2b - ab + 2\sqrt {{a^2} + {b^2}} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$
From pure geometric considerations, the circumcentre C of $$\Delta OAB$$ lies on AB and is in fact, the mid-point of AB.
Thus,
$C \equiv \left( {\frac{a}{2},\frac{b}{2}} \right)$
Slightly manipulating (1), we obtain
$\frac{a}{2} + \frac{b}{2} - \left( {\frac{a}{2}} \right)\left( {\frac{b}{2}} \right) + \sqrt {{{\left( {\frac{a}{2}} \right)}^2} + {{\left( {\frac{b}{2}} \right)}^2}} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)$
The locus of \begin{align}\left( {\frac{a}{2},\frac{b}{2}} \right)\end{align} is given by (2). Using $$(x,y)$$ instead of \begin{align}\left( {\frac{a}{2},\frac{b}{2}}\right),\end{align} we obtain
$x + y - xy + \sqrt {{x^2} + {y^2}} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(3)$
Upon comparing (3) with the locus specified in the question, we obtain $$k = 1.$$
Example – 2
Consider a curve $$a{x^2} + 2hxy + b{y^2} = 1$$ and a point P not on the curve. A line drawn from P intersects the curve at points Q and R. If the product $$PQ \cdot PR$$ is independent of the slope of the line, then show that the curve is a circle.
Solution: Since distances are involved from a fixed point, it would be a good idea to use the polar form of the line to write the co-ordinates of Q and R.
Let P be $$({x_1},{y_1})$$ and let $$\theta$$ denote the slope of the variable line. For any point $$(x,y)$$ lying on this line at a distance r from P, we have
\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{x - {x_1}}}{{\cos \theta }} = \frac{{y - {y_1}}}{{\sin \theta }} = r\\&\quad\Rightarrow \quad x = {x_1} + r\cos \theta \,\,;\,\,y = {y_1} + r\sin \theta \end{align}
If $$(x,y)$$ lies on the given curve, it must satisfy the equation of the curve :
\begin{align}&a{({x_1} + r\cos \theta )^2} + b{({y_1} + r\sin \theta )^2} + 2h({x_1} + r\cos \theta )({y_1} + r\sin \theta ) = 1\\& \Rightarrow \quad \{ a{\cos ^2}\theta + b{\sin ^2}\theta + h\sin 2\theta \} {r^2} + 2\{ a{x_1}\cos \theta + b{y_1}\sin \theta + h({x_1}\cos \theta \\&\qquad\qquad \qquad\qquad + {y_1}\sin \theta )\} r + ax_1^2 + by_1^2 + 2h{x_1}{y_1} - 1 = 0\qquad\qquad \quad...(1)\end{align}
This quadratic has two roots in r, say $${r_1}$$ and $${r_2}$$, which will actually correspond to $$PQ$$ and $$PR$$ since Q and R lie on the curve. Thus, $$PQ \cdot PR$$ being independent of $${\rm{\theta }}$$ means that $${r_1}{r_2}$$ for (1) is independent of $${\rm{\theta }}$$ i.e.
\begin{align} & \frac{ax_{1}^{2}+by_{1}^{2}+2h{{x}_{1}}{{y}_{1}}-1}{a{{\cos }^{2}}\text{ }\!\!\theta\!\!\text{ }+b{{\sin }^{2}}\text{ }\!\!\theta\!\!\text{ }+h\sin 2\text{ }\!\!\theta\!\!\text{ }}\,\,\text{is independent of}\,\,\theta \\ \\ & \Rightarrow \quad \frac{ax_{1}^{2}+by_{1}^{2}+2h{{x}_{1}}{{y}_{1}}-1}{\left( \frac{a+b}{2} \right)+\left( \frac{a-b}{2} \right)\cos 2\text{ }\!\!\theta\!\!\text{ }+h\sin 2\text{ }\!\!\theta\!\!\text{ }}\,\,\text{is independent of}\,\,\theta \\ \\ & \left\{ \text{The denominator was obtained in this form by writing }{{\cos }^{2}}\theta \text{ as }\frac{1+\cos 2\theta }{2}\text{ and }{{\sin }^{2}}\theta \text{ as }\frac{1-\cos 2\theta }{2} \right\} \\ & \Rightarrow \quad \frac{ax_{1}^{2}+by_{1}^{2}+2h{{x}_{1}}{{y}_{1}}-1}{\left( \frac{a+b}{2} \right)+\sqrt{{{\left( \frac{a-b}{2} \right)}^{2}}+{{h}^{2}}}\{\sin (2\theta +\phi \}}\,\,\text{is independent of}\,\,\theta \\ \\ & \left\{ \text{The denominator was obtained by combining the two trignometric terms}\text{.}\tan \phi \,\,\text{is }\left( \frac{a-b}{2h} \right)\text{ } \right\} \\ & \Rightarrow \quad \text{This is only possible when}{{\left( \frac{a-b}{2} \right)}^{2}}+{{h}^{2}}=0 \\ & \Rightarrow \quad a=b\,\,\,and\,\,\,h=0 \\ \end{align}
Thus, the equation of the given curve reduces to
$$a{x^2} + a{y^2} = 1$$
which is clearly the equation of a circle.
Example – 3
The equation of a circle is $$S:2x(x - a) + y(2y - b) = 0$$ where $$a,\,\,b \ne 0.$$ Find the condition on a and b if two chords each bisected by the x-axis, can be drawn to the circle from the point \begin{align}P\left( {a,\frac{b}{2}} \right).\end{align}
Solution: Observe that the centre of S is \begin{align}\left( {\frac{a}{2},\frac{b}{4}} \right)\end{align} and P lies on the circle:
Let us consider a chord $$PQ$$ of the circle bisected at the x-axis, say, at the point $$(h,0).$$ We can write the equation of $$PQ$$ as :
$T(h,0) = S(h,0)$
where $$S(x,y)$$ is
\begin{align} &{x^2} + {y^2} - ax - \frac{b}{2}y = 0\\ \Rightarrow \qquad & hx - \frac{a}{2}(x + h) - \frac{b}{4}(y + 0) = {h^2} - ah\\ \Rightarrow \qquad & \left( {h - \frac{a}{2}} \right)x - \frac{b}{4}y + \left( {\frac{{ah}}{2} - {h^2}} \right) = 0 \qquad\qquad ...(1) \end{align}
Since this chord passes through \begin{align}P\left( {a,\frac{b}{2}} \right),\end{align} the co-ordinates of (P) must satisfy (1). Thus,
\begin{align}&\qquad\; \left( {h - \frac{a}{2}} \right)a - \frac{b}{4}\left( {\frac{b}{2}} \right) + \frac{{ah}}{2} - {h^2} = 0\\& \Rightarrow \quad \frac{{3ah}}{2} - \frac{{{a^2}}}{2} - \frac{{{b^2}}}{8} - {h^2} = 0\\& \Rightarrow \quad {h^2} - \frac{{3ah}}{2} + \frac{{{a^2}}}{2} + \frac{{{b^2}}}{8} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\end{align}
We want to have two such possible chords PQ bisected at the x-axis. Thus, we must have two distinct values of h which can happen if the discriminant of (2) is positive. Thus,
\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{9{a^2}}}{4} > 4\left( {\frac{{{a^2}}}{2} + \frac{{{b^2}}}{8}} \right)\\ &\Rightarrow \qquad {a^2} > 2{b^2}\end{align}
This is the required condition that a and b must satisfy.
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# MY NASA DATA: Comparing Temperatures and Solar Radiation for Common Latitudes
Using the MY NASA DATA Live Access Server (LAS), students gather data on both solar radiation and surface temperature for two same-latitude locations. Students then create online graphs of that data to allow for analysis and comparison. This lesson... (View More)
Audience: Middle school
Materials Cost: Free
# Earth Exploration Toolbook: Exploring Air Quality in Aura NO2 Data
In this chapter, students will explore relationships between air quality and population density using the image visualization tool, Google Earth. You will learn how to download NO2 data and analyze them to develop a conceptual understanding of how... (View More)
Audience: Middle school, High school
Materials Cost: Free
# MY NASA DATA: Evidence of Change Near the Arctic Circle
Through an analysis of data sets on four parameters - sea ice totals, sea surface temperatures, near surface temperatures and surface type - students must decide whether the Arctic is experiencing climate change and predict any potential effects on... (View More)
# MY NASA DATA: How Does the Earth's Energy Budget Relate to Polar Ice?
By matching maps of snow and ice amounts with maps of net radiation flux for the same time frame, students will use the Live Access Server to explore how the net radiation flux has affected the snow and ice amounts in the Northern Hemisphere, as... (View More)
Keywords: Authentic data use
Audience: Elementary school, Middle school
Materials Cost: Free
# MY NASA DATA: Comparing the Effects of El Niño and La Niña
In this lesson, students use the Live Access Server to obtain real sea surface temperature data, to create maps and line graphs, and to make comparisons and conclusions about the effects of El Niño and La Ninña. The lesson includes detailed... (View More)
Keywords: El Nino; La Nina
Audience: Middle school
Materials Cost: Free
# MY NASA DATA: Is Grandpa Right, Were Winters Colder When He Was a Boy?
In this activity, students use historic weather information and compare it with current data to determine if they can see a trend in temperature change over time. Step-by-step instructions for use of the MY NASA DATA Live Access Server (LAS) guide... (View More)
Keywords: Authentic data use
Audience: Middle school
Materials Cost: Free
# MY NASA DATA: Variables Affecting Earth's Albedo
In this activity, students investigate one of the variables that affects Earth's albedo. The lesson includes detailed procedures, related links and sample graphs, follow-up questions, extensions, and teacher notes. This lesson is from the MY NASA... (View More)
Keywords: Albedo
Audience: Middle school, High school
Materials Cost: Free
# MY NASA DATA: Coral Bleaching in the Caribbean
In this lesson, students collect sea surface temperature (SST) data from the MY NASA DATA Live Access Server (LAS), create time-series line plots, and use the plots to study a major coral bleaching event. Corals feed on algae that thrive in the... (View More)
# MY NASA DATA: El Niño Lesson
The strength of the historic 1997-1999 El Niño Southern Oscillation (ENSO) event was captured and recorded by NASA Earth observing satellites. By downloading and plotting that satellite data, students will observe and analyze El Niño's effect on... (View More)
Keywords: Water cycle; ENSO
Audience: Middle school, High school
Materials Cost: Free
# MY NASA DATA: Hurricanes as Heat Engines
In this data analysis activity, students explore how hurricanes extract heat energy from the ocean surface by tracking Hurricane Rita and sampling sea surface temperatures along its path. Step-by-step instructions for use of the MY NASA DATA Live... (View More)
Keywords: Authentic data use
Audience: Middle school, High school
Materials Cost: Free
«Previous Page12 Next Page» | 923 | 4,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-34 | longest | en | 0.791448 |
https://www.jiskha.com/display.cgi?id=1362564402 | 1,501,261,300,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550975184.95/warc/CC-MAIN-20170728163715-20170728183715-00696.warc.gz | 816,239,147 | 3,892 | # Chemistry
posted by .
If 15gram of potassium trioxochlorate V is heate in MnO2.What is the mass of chloride produce
• Chemistry -
2KClO3(s) + heat → 2KCl(s) + 3O2(g)
each mole of KClO3 produces 1 mole of KCl.
So, convert grams to moles KClO3, then back to grams KCl.
Or just multiply 15g * (molwt KCl/KClO3)
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write a net ionic equation for the following: 1)Iron (III) chloride reacts with Potassium hydroxide to produce Iron (III) oxide, Potassium chloride, and Hydrogen gas. 2)Hydrochloric acid reacts with Potassium carbonate to produce Potassium …
2. ### Science
If 15gram of potassium trioxochlorate V is heated in the presence of MnO2.What is the mass of chloride produce?
3. ### Science
If 15gram of potassium trioxochlorate V is heated in the presence of MnO2.What is the mass of chloride produce?
4. ### Science
If 15gram of potassium trioxochlorate V is heated in the presence of MnO2.What is the mass of chloride produce?
5. ### Chemistry
If 15gram of potassium trioxochlorate V is heated in the presence of MnO2.What is the mass of chloride produce?
6. ### Chemistry
If 15gram of potassium trioxochlorate V is heated in the presence of MnO2.What is the mass of chloride produce?
7. ### chemistry
a mixture of 39.8g containing potassium chloride and potassium trioxochlorate (v) upon heating, the residue weighed 28.7g what is the percentage mass of potassium chloride in the residue?
8. ### Chemistry
In a dry cell, MnO2 occurs in a semisolid electrolyte paste and is reduced at the cathode. The MnO2 used in dry cells can itself be produced by an electrochemical process of which one half-reaction is shown below. Mn2+(aq) + 2 H2O(l) …
9. ### Chem
18) Chlorine reacts with potassium bromide to produce potassium chloride and bromine. A) write a chemical equation, using words, to represent the above chemical reaction. My answer: chlorine + potassium bromide ==> potassium chloride …
10. ### Chemistry2
12.) Chlorine reacts with potassium bromide to produce potassium chloride and bromine. a.) write a chemical equation, using words, to represent the above chemical reaction. Answer: Chlorine + potassium bromide ==> potassium chloride …
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Post a New Question | 581 | 2,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-30 | longest | en | 0.759376 |
http://stackoverflow.com/questions/20704204/ignore-duplicates-and-create-new-list-of-unique-values-in-different-sheet-in-exc | 1,432,910,225,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207930143.90/warc/CC-MAIN-20150521113210-00044-ip-10-180-206-219.ec2.internal.warc.gz | 217,797,001 | 17,419 | # Ignore Duplicates and Create New List of Unique Values in different Sheet in Excel
I am trying to do exactly what was asked in this thread
Ignore Duplicates and Create New List of Unique Values in Excel
Except it wasn't answered fully. I would like the output in a different sheet. The question:
" I have seen this question asked in various forms numerous times before, but nothing I have tried is working for me.
I have a column of values that often appear as duplicates. I need to create a new column, of unique values based on the first column, as follows...
``````Column A Column B
a a
a b
b c
c
c
``````
This Column B will actually need to appear on a different sheet, within the same workbook, so I assume it will need to work with the `sheet2!A1` style format.
I have not had any luck with the Data/Filter menu options as this only seems to work on command. I need column B to update automatically whenever a new value is entered into column A. "
" Basically the formula you need is:
``````B2=INDEX(\$A\$2:\$A\$20, MATCH(0, COUNTIF(\$B\$1:B1, \$A\$2:\$A\$20), 0))
``````
Then press ctrl-shift-enter.
Two important things to keep in mind here: The complete list is in cells `A2:A20`, then this formula has to be pasted in cell `B2` (Not `B1` as that will give you circular reference). Secondly this is an array formula, so you need to press ctrl-shift-enter or it will not work correctly. "
(This is a useful link: Unique Values )
Which gives me:
`````` Column A Col B
a a
a b
b c
c d
c 0
c #N/A
c #N/A
b #N/A
b #N/A
a
d
d
``````
But the output is not in a different sheet.
I have tried a few variations of:
``````=INDEX(List!A2:A20,MATCH(0, COUNTIF(UniqueList!A2:A20,'List'!A2:A20),0))
``````
Putting \$ signs in:
``````=INDEX(List!\$A\$2:\$A\$20,MATCH(0, COUNTIF(UniqueList!\$A\$2:A20,'List'!\$A\$2:\$A\$20),0))
``````
Including 'press ctrl-shift-enter'.
But I can't figure it out.
Thanks.
-
Why not copy the list to a new sheet and use Data | Remove Duplicates? – Siddharth Rout Dec 20 '13 at 13:13
That's what I currently do, but I'd like not to have to do that. – CArnold Dec 20 '13 at 13:49
``````=INDEX(Sheet1!\$A\$2:Sheet1!\$A\$5,MATCH(0,INDEX(COUNTIF(\$A\$1:A1,Sheet1!\$A\$2:Sheet1!\$A\$5),0,0),0)) | 699 | 2,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2015-22 | latest | en | 0.929223 |
https://www.technology.org/2015/08/03/statistical-ideas-distribution-risks/ | 1,576,318,993,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540586560.45/warc/CC-MAIN-20191214094407-20191214122407-00469.warc.gz | 871,191,864 | 21,273 | # Statistical Ideas: Distribution risks
Posted August 3, 2015
This news or article is intended for readers with certain scientific or professional knowledge in the field.
From the simplest, parametric probability distributions, to the most sophisticated, nonparametric distributions, it is essential to learn the key elements for managing one’s distribution risk. This could be the risk of a particular income scenario, the risk on what age you will die, or any number of imaginable unknowns. Knowing this will help make more informed life decisions that involve encountering a risk by yourself (e.g., longevity risk), versus sharing it broadly with many others. And some of the ideas here need to be thought through, since readily available government actuary tables do not exist for many utilitarian population clusters.
First, let’s look at the plainest distribution, which is the Bernoulli (or a simplified binomial). See some prepatory articles on this (here, here, herehere). We start with one person “X”, who we might consider to be ourselves. Person X is expected to either have a 0 income with probability 50%, or 1 income also with probability 50%. So the expected income for X is 0.5. Now we can introduce a second person “Y”, which greater income potential (and expected income of 1.5). If we expect that Person Y’s income would correlate to Person X’s (e.g., either they both are on the higher scenario or both on the lower scenario), then we expect the average income between them to now be 0.5, and 1.5. See the upper left diagram below.
Other things could happen of course. Looking at the lower right diagram, in the illustration above, we consider that multiple things may happen simultaneously in some theoretical closed form model. For example, when X and Y are “together” (e.g., married or business partners or neighbors) then some dynamics could work for them together. Their association might lead Person X to have a greater income distribution, and Person Y might have a lower income distribution. Additionally their distributions may be inversely correlated (as shown by the crossing solid arrows). Such that when X has the lower of his or her two income possibilities, then Y has the higher of his or her two income possibilities. And the reverse would also hold by default. In such a scenario (as shown with the dashed arrows) both individuals’ incomes are more homogenized, and their joint incomes are more consistent (in this case they were spreadless at 1 each). Peruse each of the four diagrams above.
While the probability math was easier to see with the basic distributions above, let’s see how this quick lesson applies to a more complex real-world distribution. Here we will repurpose the 1950 actuary table that we analyzed in the “Centenarian risk” article. We also note now to get comfortable with it, since we will continue to reuse this data in future articles. In the left illustration below, we will show in red the simulated, expected deaths of people born in 1950 (who we focus our attention on since they are just now retiring in 2015). This distribution is the actuarial equivalent to Person X above, or Person Y if they are perfectly correlated (r=1) to Person X. A careful reader would have observed a relatively high death rate in the year following live birth (~3%), which makes the first year the only time where the typical remaining lifetime increases after one survives that age!
And in green we show simply show a simulation of what a second random person’s death distribution would be. Of course by themselves (r=0), his or her death distribution would be nearly the same as the national distribution shown for Person X. And in fact, by definition the initial actuary table (for the national population) applies to all people and so regardless of how one synthetically tries to sample a (inversely) correlated variable, it’s distribution will always remain the same. They all share the same expected death age of 68.
On the right illustration immediately above, we see that the average death age of the pairing of the first and second person are equal whether we look at the red (r=1) or the green (r=0) distributions. This summary statistic is far more applicable to consider when thinking about personal risk assessment. We also show in blue what an inversely correlated (r=-0.5) scenario looks like. This is more similar to the diagram in the upper right of the top most illustration in this article. As opposed to the red distribution immediately above for Y=X (or r=1), which is most similar to the diagram in the upper left of the top most illustration above. The red arrows shows that both the left and the right distributions are equal.
We can see with the blue arrow that while the expected death ages of all pairs of Person X and Person Y are still 68, the dispersion about this age is least when Person Y is least correlated to Person X. Notice even in the top right diagram, in the top most illustration, that the average incomes were both 1 (instead of the spread between 0.5 and 1.5 on the top left diagram).
The paramount lessons we get from this is that by mixing additional humans to the actuarial risk pool, the average death stays the same regardless of correlation between people (here, here). Additionally, typical scatterings about the average death age comes down considerably (while also looking more Gaussian), as should be expected when mixing any set of identical, mound-shaped probability distributions. For the independent, blue distribution (r=0) as an example, the highest average death age generally reduces from 110 to about 95 (see here and here for details).
And lastly, we leveraged the national life table for all of our work here, since the idea of relationships between locations (e.g., the bottom two diagrams of the top most illustration) can not be flexibly modeled with government actuary tables. One can see a set, for say, married couples. But not one for guessing differences among custom sets of people, such as students enrolled in the same college.
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• Wiring Harness
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## Mercedes Benz W203 Wiring Harness
Benz W203
Mercedes Benz W203 Wiring Harness ? Every chemical engineer has to be aware of the question,Which of the following can't be determined by looking at a phase diagram? During school and even in grad schoolthey were all taught it is not possible to find out which of both materials can be identified on a standard surface. So just what is a stage diagram? The use of a phase diagram is essential to anyone who wants to ascertain whether a material can be recognized on a standard surface, or if it is incompressible. Since the phase diagram can easily be confused with a physical model, it is strongly suggested that you study the diagram, learn whether or not it really refers to a substance, then proceed with your homework and research on the exact material. In case you are asking,Which of these cannot be ascertained by taking a look at a phase diagram? These are: The chart above is exactly what you'd expect to see in an information sheet that you get when you're carrying an intro to organic chemistry class at your school. It basically describes how the various phases of fluids and solids form, but in addition, it shows what occurs as the material is chilled. Sadly, this isn't sufficient to let you know whether the material you are studying is compressible. There are different factors that play a role in determining whether a substance is whether or not. A soft contact equilibrium is just one example of a property that you have to determine before trying to resolve the equations of motion for the various phases of a material. Most individuals don't really understand what a soft contact balance is, but they'll remember this equation since it's among the most frequently used equations of movement in chemistry. The formula is used in the linear equilibrium equation that's used for solids, and also the first-order differential equation, which can be used for liquids. As mentioned before, phases are only 1 dimension of solid or liquid arrangement. A physical version is just a representation of the material and so can't actually show the material, but it might reveal the way the material changes as time passes. The distinction between a physical model and a phase diagram is that the stages are a more detailed representation of material structure than are the physical models. If you want to figure out which of the 3 properties is most important, then you should examine the phase diagram to find out which physical properties are important. As soon as you have determined which properties of matter, then you may determine which material is most significant to you. By way of example, if you are working on electronic equipment, then you want to learn about conductivity, in addition to the aspects of a capacitance. If you're a newcomer to the topic of engineering, then it's recommended that you learn about the phases of solids and liquids before you get stuck in the phase diagram. | 603 | 3,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-50 | latest | en | 0.958363 |
https://www.financeinvest.at/binomial-model-advantages/ | 1,719,091,458,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00739.warc.gz | 665,660,745 | 42,055 | # Binomial Model vs. Black Scholes
Joachim Kuczynski, 06 December 2022
There are two basic ways to evaluate real options. The first way is the evaluation with exact analytical approaches like the famous Black Scholes Merton model. They have their origin in financial option pricing and deliver exact results. But they have certain underlying conditions that do not match reality of real options many times. The second basic approach to evaluate real options are discrete approximative models like the Binomial Model from Cox, Ross and Rubinstein. They are approximations of the exact analytical solutions when they have the same underlying conditions. In my real option analysis I prefer the approximative approach because of many reasons:
• Volatility: The volatility, or risk respectively, defines the variance of the binomial tree branches (ups and downs). Volatility can change over time because of many reasons. But in the Black Scholes Merton model volatility is fixed for the considered time interval. Taking the binomial approach it is up to you to change volatility whenever you want. For sure, the binomial tree can become more complicated, e.g. if you have to change from a recombining to a non-recombining binomial tree. But with computer support also a more complicated event tree is no problem.
• Exercise price: The exercise price at a node is the time value of the corresponding future cash flows which do not depend on the market development and are not diversiviable (mainly investments, fixed expenses and fixed earnings). Each of these components can change in time. That means that the exercise price of the real option can be differerent in each period. In the Black Scholes Merton model the exercise price is the same over time. In case of changing exercise prices you have to use approximative (binomial) approaches.
• Discount rates: The discount rate includes many parameters like risk free rate, non-diversifiable market risks of the cash flow, the investor’s capital structure, the investor’s opportunity portfolio and tax shields. All these parameters can vary over time, and hence change the discount rate. If the discount rate changes, you might get a non-recombining binomial tree. Take care that the discount rate in each period has to match the expected values of the binomial tree branches in each period.
• Decision tree: The decision tree does not have to be a complete binomial tree as required at the approximation of the Black Scholes Merton approach. Some branches might not exist or there can be more than one possibility in a node of the binomial tree (tri- or multinomial trees). Sometimes these exceptions represent reality more accurately and can be calculated in a discrete model. But these cases are not an approximation of Black Scholes Merton model any more. The Black Scholes Merton model cannot handle such tasks.
• Time steps: Times between nodes of an event tree can be different at each link. You can adapt the times to your specific problem if required. The nodes of the event tree can for example be determined by the decision time of the investment project. At the Black Scholes approach the temporal development is fixed by the input parameters. There is no possibility to adapt it anyway. In the binomial approach you can adapt time steps as required by the corresponding problem.
In general, the (binomial) approximative approaches are much more flexible and can be adjusted to the specific problem. With specific input parameters the binomial model is an exact approximation of the Black Scholes Merton model. But many times the input parameters describing reality are different. Black Scholes Merton comes from financial option markets, where situations are less complex as at real options many times. The binomial approach is much more suitable for real option analysis. Because of the inaccuracy of many input variables at real options, the approximative character of the binomial model does not distort the result mostly.
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https://community.tableau.com/thread/116854 | 1,550,500,315,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247486936.35/warc/CC-MAIN-20190218135032-20190218161032-00407.warc.gz | 523,076,743 | 34,742 | 1 2 Previous Next 15 Replies Latest reply on May 7, 2016 11:37 PM by Chandra Bhaskar
# dynamic subtotal 1 to N
I'm controlling my data view by a parameter to select the Top 'N' after the data has been ranked and sorted.
What I'd like to do is generate a dynamic subtotal of the Top 'N' selection. When I add in the standard Subtotal of a Dimension I get the subtotal of the entire dimension within the data (by definition...), but that's not what I want.
I could try to explain this further, but one example I brought home from last October's users meeting in Vegas is attached.
Notice (in this case) the Grand Total in each worksheet is not the sum of the N = 5 customers sales, nor is the 100% the sum on the N = 5 % of sales just above.
How can I get a dynamic subtotal for each measure as a function of N = some number?
thx..
• ###### 1. Re: dynamic subtotal 1 to N
thoughts..anyone?
This seems like a fairly useful (and logical) functionality to accompany a top 'n' analysis..maybe it isn't possible?
thanks..
• ###### 2. Re: dynamic subtotal 1 to N
Ok..does anyone know of a loop-function that works in Tableau. (loop function = subroutine: do some set of instructions 1 to N where N = somenumber)
I remember loop functions from pascal and fortran programming m-a-n-y years ago. I'm thinking I could use rank/index() & the 1 to N parameter entry as the entry point of "somenumber".
no?
• ###### 3. Re: dynamic subtotal 1 to N
Here you go, no looping necessary, just a bit of Joe Mako wisdom. See the attached workbook, I've also put my explanation here. I had a couple of fascinating discoveries along the way:
- The Grand Total row can be made to show most anything you want.
- The built-in calculations for the Grand Total row seem to reduce the overall level of detail as much as possible to get to single values, but we can expand them by using the Level of Detail shelf and nested table calculations.
I started by creating a % of Total sales for Listed Customer that works by using the following calculation that takes into account the parameter:
SUM([Sales])/WINDOW_SUM(IF INDEX() <= [Top n Parameter] THEN SUM([Sales]) END)
The Grand Total for this column sums to 100%.
However, I had trouble adjusting the Grand Total for SUM([Sales]) to make that dependent on the parameter.
I used Joe Mako's calc here: http://community.tableau.com/message/174251#174251 to ID whether the row was the Grand Total row or not. I'd thought I could use a flavor of the calc he made to change how the rows were calculated.
Therefore, I created a "Grand Total Flag" field with the following calculation:
TOTAL(MIN([Customer Name])) != TOTAL(MAX([Customer Name]))
Setting that to Compute Using Pane (Down) makes it work.
Then I did some other calcs - INDEX(), SIZE(), and LAST() - to explore this partition. It seems that the Grand Total row lives in its own partition that has only one row, because Customer Name is the only dimension in the view. So, all the calculations trying to get at the set of customers that are shown based on the Table Calc filters fail because there's only one row in the partition. Another sign that this is the case is that all the fields in the Grand Total row only show one value, with no overlapping text.
In Joe's post, he had added a copy of the dimension of interest to the Level of Detail. Here, I did that with a copy of Customer Name on the Level of Detail shelf. Now, lots of calculations start breaking, but there's good news - the Grand Total field now shows overlapping text, and the Size() field now shows 795 rows available in the Grand Total partition.
One change needed here to get the Top N filter to work is that it needs to be on the Level of Detail shelf. Filters on table calcs only seem to "see" the pills in the Columns and Rows shelves for Compute using, and instances of themselves on the Level of Detail.
In order to have the Grand Total calcs work differently between the regular rows and the Grand Total row, we set them up as so:
Grand Total Flag
Put the Grand Total Flag on the Level of Detail shelf, and set the Compute using... to Customer Name (copy). This is a convenience so the other table calcs will inherit the value.
Sales and Grand Total:
Here's the calc:
IF [Grand Total Flag] THEN
IF FIRST()==0 THEN
WINDOW_SUM(SUM([Sales]),0,[Top n Parameter])
END
ELSE
SUM([Sales])
END
The IF FIRST()==0 part is to prevent overlapping text on the Grand Total row. Once this pill is on Measure Values, go the Edit Table Calculation... menu option and set the Compute using for Sales and Grand Total to Advanced… with both Customer Name and Customer Name (copy) in the Compute Using, with Order Along set to Sales - Sum - Descending. This gets the partition set up properly to for the sum of sales to only count the sales up to the Top N parameter.
% of Total Sales
Here's the calc:
IF [Grand Total Flag] THEN
IF FIRST()==0 THEN 1 END
ELSE
SUM([Sales])/WINDOW_SUM(IF INDEX() <= [Top n Parameter] THEN SUM([Sales]) END)
END
Again, once this pill is on Measure Values, go to Edit Table Calculation.. and set the Compute using... for % of Total Sales to Advanced… with both Customer Name and Customer Name (copy) in the Compute Using, with Order Along set to Sales - Sum - Descending. This gets the partition set up properly to for the sum of sales to only count the sales up to the Top N parameter.
Let me know if this works for you!
• ###### 4. Re: dynamic subtotal 1 to N
Thanks for responding Jonathan and taking the time to think about this question. Seemingly simple questions can have not so simple answers.
I just got the chance about 30 minutes ago to look through your thoughts/approach. While I haven't digested it yet, I suspect the simple *.twbx file I attached earlier may have been too simple in that it only had the Grand Total & no subtotaling. My guess is this will have a significant impact on the potential solution.
I've attached a sample workbook that contains a small subset of the real data. You'll note it does have multiple subtotals as well as the Grand Total.
I've also attached a graphic (*.xlsx) to indicate the (dynamic) subtotals I'd like to determine.
It's completely understandable if you don't have the time to work through this. I've been trying off and on for several days..it eludes me.
thx..
• ###### 5. Re: dynamic subtotal 1 to N
Hi Kevin,
Either simple questions don't have such simple answers, or maybe the question wasn't so simple in the first place. *grin*
I think this can be done, I started at a proof of concept using the superstore sales data and it's possible to specifically identifty subtotal rows along with grand total rows. However, I have two caveats:
- The calculations are really sensitive. There's a way that adding a table calculation to a view can sometimes change the results of other table calculations that I don't have a full understanding of yet that I was running into as I worked on the view.
- The performance of the view takes a dive because of the quantity and complexity of the table calculations (and eventually the size of the data). You're going to need a custom table calculation for each column that depends on the results of two other table calculations (to identify whether the row is a sub total or grand total) in order to perform a third, fourth, and/or fifth table calculation (for the detail row, sub total, or grand total). And then making this view parameter-driven slows down Tableau even more because it can't really cache any results.
This is one of those places where a batch reporting tool might be better suited.
The basic process I'd used was to add the Product Name to the Rows shelf and Product Name (copy) to the Level of Detail. This breaks the Top N filter, which needs it's Compute using set to Product Name & Product Name (copy) for it to work. Then I created a Sub Total Flag that looked like the Grand Total, only based on the Product Name instead of Customer Name, and once on the Level of Detail set it's compute along to Product Name (copy).
Then the calculations for each value could be something like:
IF [Grand Total Flag] THEN
(do grand total calc)
ELSEIF [Sub Total Flag] THEN
(do sub total calc)
ELSE
(do calc for detail row)
END
Where I stopped is that with the new Product Name/Product Name (copy) dimensions in the view, the Grand Total Flag starts returning multiple values and I ran out of time to figure that out. You'd need to get that working, and I'm wondering whether the grand total calc and sub total calcs in the above formula might need to be separated out so their Compute using's could be custom set in the view.
Hopefully this gives you enough to get a start if you want to continue going this route. It's a fascinating problem to me, I could take another swing at it towards the end of next week (I've got a few dashboards that are demanding attention) if you need help. Or if anyone else here on the forums wants to have a go at this, I'd be happy to see what they can come up with.
Jonathan
• ###### 6. Re: dynamic subtotal 1 to N
Jonathan,
I believe someone once said, "for every question there is a simple answer" (probably a corruption of Occam's razor).
Someone else followed this with, "for every question there is a simple, and incorrect, answer" (sounds like something H.L. Mencken would say).
And then there is real life where seemingly simple questions have really involved answers. From your description I think we've entered the twilight-zone. While I'm sure it can be done, 'diminishing returns' comes to mind as this was a nice-to-have (some window dressing to the heavy lifting the rest of the workbook is doing) and not a have-to-have. If it risks corrupting other calcs, and that isn't caught, then one has to question whether it's worth the effort. Risking confidence in the good stuff the workbook does wouldn't be worth it.
With respect to speed..this workbook is an inventory management tool where its current (development) state is about 120,000 records(120,000 rows x 28 columns). Once it goes into actual use it'll grow to, and level off at around 300,000 records(rows), all stored & preprocessed in Excel, at this point. Currently the Excel file is 42MB. So far Tableau performs well with a live connection to the data, but once it grows up I may have to convert it to MS Access for preprocessing and storage, or convert it monthly to a twbx file (I run Tableau desktop).
If the subtotal-sums are really needed we can either eyeball the data and sum it in our heads(easy enough estimate) or take the data into Excel and subtotal it in a minute (as I did with the file attached above)...with no chance of un-noticed errors.
Thanks for at least working through and estimate of what's needed to do this. That in itself is a very valuable answer.
Thanks!
• ###### 7. Re: dynamic subtotal 1 to N
You're welcome! 'Diminishing returns' sounds like an accurate summation to me. And 300K rows for the number of table calcs that are required would probably result in non-acceptable performance, since Tableau would have to be doing calculations across all 300k rows multiple times for the view.
Also, way back at the top you'd asked about any looping functions in Tableau, there aren't really any. The closest one that I know of is the PREVIOUS_VALUE() function which returns the previous value of the _calculation_ for the prior row in the partition and the given value for the first row in the partition.
Jonathan
• ###### 8. Re: dynamic subtotal 1 to N
It seems the Grand total calc's are now incorrect on this one. I downloaded the twbx and on the "Is this what you want?" workbook the dynamic grand total doesn't reflect the top n sum of sales. Is this something that has possibly occured with the 8.1 upgrade?
Has something been left out of context as a reason why this is happening?
TY,
Mac
• ###### 9. Re: dynamic subtotal 1 to N
It seems as if the Sales and Grand Total calculation is creating a Grand Total that takes the last of the "n" (let's say the fifth position of Top "n" set to 5), and somehow adds it twice.
• ###### 10. Re: dynamic subtotal 1 to N
This seems to work:
IF [Grand Total Flag] THEN
IF FIRST()==0 THEN
WINDOW_SUM(SUM([Sales]),0,[Top n Parameter] - 1)
END
ELSE
SUM([Sales])
END
• ###### 11. Re: Re: dynamic subtotal 1 to N
Jonathan Drummey
Oh heck, after reading ALL the text in your solution, i realize that you had stated that the calcs were adding multiple values and you just ran out of time....
Well, i think i came to a fix here.
• ###### 12. Re: dynamic subtotal 1 to N
Now that i think of it... since this post is 2 years old, has a more streamlined approach been included in Tableau? If so, i couldn't find it, hence my search for Dynamic Grand Totals Top N calcs.
• ###### 13. Re: Re: dynamic subtotal 1 to N
Hi Mac,
The only new features in the product supporting Top N and grand totals are the RANK functions and the multi-pass totals introduced in 8.1. However, neither solves the issue here. I've found solutions for this problem, but haven't written anything up about it because the exact solutions are so particular to the data & desired view that I haven't been able to come up with something generalizable.
Jonathan
• ###### 14. Re: dynamic subtotal 1 to N
Hi Kevin ,
if i understood correctly ,you are looking for top N and rest should be sum-ed up with other .
If Yes then below is solution
Regards,
Kishore
1 2 Previous Next | 3,129 | 13,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-09 | longest | en | 0.926289 |
https://www.guwsmedical.info/image-processing/a3-recursive-filtering.html | 1,566,203,389,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314696.33/warc/CC-MAIN-20190819073232-20190819095232-00270.warc.gz | 833,431,606 | 7,771 | ## A3 Recursive Filtering
The following routine performs the in-place determination of a 1D sequence of interpolation coefficients {cn} from a sequence of data samples {fk}. The returned coefficients {cn} satisfy ft = E cnÇ(k - n) Vk eZ, neZ
where the synthesis function ^ is represented by its poles. The values of these poles for B-splines of degree n e {2, 3,4, 5} and for cubic o-Moms are available in Table 3. (The B-spline poles of any degree n> 1 can be shown to be real and lead to a stable implementation.) This routine implicitly assumes that the finite sequence of physical samples, known on the support X = [0, N — 1], is extended to infinity by the following set of boundary conventions:
This scheme is called mirror (or symmetric) boundaries. The first relation expresses that the extended signal is symmetric with respect to the origin, and the second mirrors the extended signal around x = N — 1. Together, these relations associate to any abscissa y e R\X some specific value f (y) =f (x) with xeX. The resulting extended signal is (2N — 1)-periodic, and has no additional discontinuities with respect to those that might already exist in the finite-support version of / This is generally to be preferred to both zero-padding and simple periodization of the signal, because these latter introduce discontinuities at the signal extremities.
The following routine is a digital filter. The z-transform of its function transfer is given by z(1 - z,)(1 - z-1) 5 (Z)~ Ü (z - z,)(z - z-,) •
As is apparent from this expression, the poles of this filter are power-conjugate, which implies that every second pole is outside the unit circle. To regain stability, this meromorphic filter is realized as a cascade of causal filters that are implemented as a forward recursion, and anticausal filters that are implemented as a backward recursion. More details are available in [19,20].
TABLE 3 Value of the B-spline poles required for the recursive filtering routine
Zl 22
ß4 a/664 — V4S8976 + VSOi — 19 V¡64 + V4S8976 — VSOi — 19
ß5 2 (V27O — V7O98O + v/ÎO5 — 1s) i ^27O + V7O98O — v/ÎO5 — 1s)
Name Expression
0 0 | 550 | 2,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-35 | latest | en | 0.909223 |
http://nrich.maths.org/public/leg.php?code=71&cl=3&cldcmpid=4768 | 1,503,435,163,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112682.87/warc/CC-MAIN-20170822201124-20170822221124-00667.warc.gz | 300,876,975 | 10,343 | # Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Balance Point:
Filter by: Content type:
Stage:
Challenge level:
### There are 176 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Perfectly Square
##### Stage: 4 Challenge Level:
The sums of the squares of three related numbers is also a perfect square - can you explain why?
### Janine's Conjecture
##### Stage: 4 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### N000ughty Thoughts
##### Stage: 4 Challenge Level:
How many noughts are at the end of these giant numbers?
### A Biggy
##### Stage: 4 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Leonardo's Problem
##### Stage: 4 and 5 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### Lens Angle
##### Stage: 4 Challenge Level:
Find the missing angle between the two secants to the circle when the two angles at the centre subtended by the arcs created by the intersections of the secants and the circle are 50 and 120 degrees.
### Archimedes and Numerical Roots
##### Stage: 4 Challenge Level:
The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Mediant
##### Stage: 4 Challenge Level:
If you take two tests and get a marks out of a maximum b in the first and c marks out of d in the second, does the mediant (a+c)/(b+d)lie between the results for the two tests separately.
### Coins on a Plate
##### Stage: 3 Challenge Level:
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### Multiplication Square
##### Stage: 4 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Symmetric Tangles
##### Stage: 4
The tangles created by the twists and turns of the Conway rope trick are surprisingly symmetrical. Here's why!
### Pythagoras Proofs
##### Stage: 4 Challenge Level:
Can you make sense of these three proofs of Pythagoras' Theorem?
### Salinon
##### Stage: 4 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### Circle Box
##### Stage: 4 Challenge Level:
It is obvious that we can fit four circles of diameter 1 unit in a square of side 2 without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?
### AMGM
##### Stage: 4 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Unit Interval
##### Stage: 4 and 5 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Our Ages
##### Stage: 4 Challenge Level:
I am exactly n times my daughter's age. In m years I shall be ... How old am I?
### Knight Defeated
##### Stage: 4 Challenge Level:
The knight's move on a chess board is 2 steps in one direction and one step in the other direction. Prove that a knight cannot visit every square on the board once and only (a tour) on a 2 by n board. . . .
### Russian Cubes
##### Stage: 4 Challenge Level:
I want some cubes painted with three blue faces and three red faces. How many different cubes can be painted like that?
### Doodles
##### Stage: 4 Challenge Level:
Draw a 'doodle' - a closed intersecting curve drawn without taking pencil from paper. What can you prove about the intersections?
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### Angle Trisection
##### Stage: 4 Challenge Level:
It is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square.
### Pythagorean Triples II
##### Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### Pythagorean Triples I
##### Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
### Whole Number Dynamics I
##### Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Why 24?
##### Stage: 4 Challenge Level:
Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results.
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Whole Number Dynamics IV
##### Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### A Knight's Journey
##### Stage: 4 and 5
This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.
### Logic, Truth Tables and Switching Circuits
##### Stage: 3, 4 and 5
Learn about the link between logical arguments and electronic circuits. Investigate the logical connectives by making and testing your own circuits and record your findings in truth tables.
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
##### Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
### Ratty
##### Stage: 3 Challenge Level:
If you know the sizes of the angles marked with coloured dots in this diagram which angles can you find by calculation?
### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Mod 3
##### Stage: 4 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
### The Golden Ratio, Fibonacci Numbers and Continued Fractions.
##### Stage: 4
An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.
### There's a Limit
##### Stage: 4 and 5 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
### Long Short
##### Stage: 4 Challenge Level:
What can you say about the lengths of the sides of a quadrilateral whose vertices are on a unit circle?
### Common Divisor
##### Stage: 4 Challenge Level:
Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.
### Always Perfect
##### Stage: 4 Challenge Level:
Show that if you add 1 to the product of four consecutive numbers the answer is ALWAYS a perfect square.
### Logic, Truth Tables and Switching Circuits Challenge
##### Stage: 3, 4 and 5
Learn about the link between logical arguments and electronic circuits. Investigate the logical connectives by making and testing your own circuits and fill in the blanks in truth tables to record. . . .
### Sixational
##### Stage: 4 and 5 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### No Right Angle Here
##### Stage: 4 Challenge Level:
Prove that the internal angle bisectors of a triangle will never be perpendicular to each other. | 2,406 | 10,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2017-34 | latest | en | 0.872328 |
https://m-cacm.acm.org/blogs/blog-cacm/254847-how-does-one-multiply-with-napiers-rods/fulltext?mobile=true | 1,679,399,547,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943695.23/warc/CC-MAIN-20230321095704-20230321125704-00511.warc.gz | 445,970,231 | 8,705 | # How Does One Multiply with Napier's Rods?
By Herbert Bruderer
August 16, 2021
Napier's rods, also called Napier's bones (see Fig. 1), were invented at the beginning of the 17th century. They have been used for multiplication and division until the 19th century. There were various forms, e.g. rotating cylinders.
### Instructions for Using Napier's Bones
Napier's multiplication and division rods, deriving from the basic multiplication table, simplify calculations considerably. They were also built into several mechanical calculating aids as rotating drums. The intermediate results are read from the diagonal grid and added by hand, taking account of the tens carry. According to the model, the diagonals run from the lower left to the upper right or from the upper left to the lower right. The tens are at the upper left or the lower left, respectively, and the ones at the lower right or the upper right, respectively.
#### Multiplication
Example 1: 7 × 694,387 = 4,860,709 (see Fig. 2)
Example 2: 7152 × 694,387 = 4,966,255,824 (see Fig. 3)
1. Lay the rods with the numerals of the multiplicand next to each other. Place the rod with the (single-digit) multiplier (numerals 1 to 9) to the left of these.
2. Read out the result of the multiplication directly from row 7 (the multiplier): 4, 2 + 6, 3 + 2, 8 + 2, 1 + 5, 6 + 4, 9.
Taking account of the tens carry (from right to left) gives the numbers:4, 8, 6, 0, 7,0, 9.
The product is then 4,860,709.
With a multi-digit multiplier, the partial products for the individual numbers are read off in the same way and then added, in each case shifted by one decimal place.
#### Napier's bones
Napier's bones
Source
Bruderer, Herbert: Milestones in Analog and Digital Computing, Springer Nature Switzerland AG, Cham, 3rd edition 2020, 2 volumes, 2113 pages, 715 illustrations, 151 tables, https://www.springer.com/de/book/9783030409739 (translated by Dr. John McMinn)
Herbert Bruderer is a retired lecturer in didactics of computer science at ETH Zurich. More recently, he has been an historian of technology. [email protected], [email protected] | 562 | 2,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-14 | latest | en | 0.910482 |
https://www.infodreams.it/main/publications/dev-magazine/dev-132-using-opengl-part-3-of-4/ | 1,695,804,870,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00319.warc.gz | 891,037,867 | 24,401 | We are finally ready to start our adventure into the world of 3D graphics. We will begin by the use of the most simple cases up to the geometric transformation of objects.
## Introduction
Now that we have a code template to use for quickly writing new programs, we can delve into the world of 3D and get to know the basic functionality of OpenGL. I highly recommend to you to study and compile the source code supplied with the articles and also make every change you have in mind to get more complex results.
## Working with primitives
The scenes, but in particular the objects that move inside them, are constituted by a composition of two-dimensional elementary shapes called primitives. We can compare with the bricks which, properly combined and aggregated together, give rise to all the more complex constructions imaginable.
Take the cube as a sample object. It consists of six faces, each of which may be drawn by a quadrilateral-type primitive or, dividing it into two equal parts, by two triangles.
OpenGL can work with several primitive shapes, from simple points to polygons. In total there are 10 different types, summarized below
GL_POINTS Simple dots GL_LINES Lines GL_LINE_STRIP Sequence of lines GL_LINE_LOOP Closed sequence of lines GL_TRIANGLES Triangles GL_TRIANGLE_STRIP Sequence of triangles GL_TRIANGLE_FAN Fan sequence of triangles GL_QUADS Quadrilaterals GL_QUAD_STRIP Sequence of quadrilaterals GL_POLYGON Polygons
But which method is adopted to draw these figures in space? The “lowest common denominator” for all the primitives is the vertex, that stands for a defined point in space represented by its coordinates x, y, z. The more used OpenGL function to specify a vertex is
``glVertex3f(GLfloat x, GLfloat y, GLfloat z)``
Consequently, a primitive is nothing more than an interpretation of a list of vertices. Now that we know how to indicate the coordinates of points in space, we are able to take into consideration each of the types of primitives.
Keep in mind that the coordinates system of the window is different from that used by OpenGL, so the respective axes of y are reversed. For this reason while in the first case the positive values are downwards oriented, in the second they start from the bottom and should go upwards.
## Points
Let’s take a look to the following code
`````` glBegin(GL_POINTS);
glVertex3f(0.0f, 0.0f, 0.0f);
glEnd();``````
We observe that the function calls that describe the vertices must be enclosed between the pair glBegin()/glEnd(), so setting the appropriate parameters OpenGL will interpret them by drawing the desired figure. In this case we had to draw GL_POINTS simple points.
You might be tempted to think that if you have to draw more than one figure is needed a glBegin/glEnd pair for each. This would only produce significant delays during the execution of the program. OpenGL, in fact, because of the parameter passed to glBegin can realize how many vertices are needed to describe the figure. In fact, writing
``````glBegin(GL_POINTS);
glVertex3f(0.0f, 0.0f, 0.0f);
glVertex3f(10.0f, 0.0f, 0.0f);
glVertex3f(5.0f, 10.0f, 0.0f);
glEnd();``````
the three dots will be rendered properly, but not a triangle. In Figure 1, is shown the output produced by the sample program Punti.exe.
## Lines
The next level after learning the use of individual points is to understand how to display lines: simple specifying two vertices one for the start position and another one for the end. Observe the following code
`````` glBegin(GL_LINES);
glVertex3f(-10.0f, 0.0f, 0.0f);
glVertex3f(10.0f, 0.0f, 0.0f);
glEnd();``````
which draws a horizontal line. If there is an incorrect number of vertices, for example an odd number, the surplus will simply be ignored.
Often when drawing figures with neighboring lines we could face a long sequence of vertices where many of them are duplicates. To draw a square with this system, for example, we would be forced to proceed in this way
`````` glBegin(GL_LINES);
glVertex3f(-10.0f, 0.0f, 0.0f); // Side 1 point 1
glVertex3f(10.0f, 0.0f, 0.0f); // Side 1 point 2
glVertex3f(10.0f, 0.0f, 0.0f); // Side 2 point 1
glVertex3f(10.0f, 10.0f, 0.0f); // Side 2, item 2
glVertex3f(10.0f, 10.0f, 0.0f); // Side 3 point 1
glVertex3f(-10.0f, 10.0f, 0.0f); // Side 2 point 3
glVertex3f(-10.0f, 10.0f, 0.0f); // Side 1 point 4
glVertex3f(-10.0f, 0.0f, 0.0f); // Side 4 point 2
glEnd();``````
Notice that we indicated 8 vertices, twice as necessary. Using the parameter of GL_LINE_STRIP with glBegin(), is no longer necessary to specify the vertices in common between two different lines. We will only provide a list of vertices and OpenGL will connect them drawing a classical broken line. Observe how in the following code all other vertices are concatenated together
``````glBegin(GL_LINE_STRIP);
glVertex3f(-10.0f, 0.0f, 0.0f); // Side 1 point 1
glVertex3f(10.0f, 0.0f, 0.0f); // Side 1 point 2 - point 1 Side 2
glVertex3f(10.0f, 10.0f, 0.0f); // Side 2 point 2 - point 3 Side 1
glVertex3f(-10.0f, 10.0f, 0.0f); // Side 3 point 2 - point 4 Side 1
glVertex3f(-10.0f, 0.0f, 0.0f); // Side 4 point 2
glEnd();``````
Now it took only 5 vertices. A great saving of time and data to be transferred from computer main memory to the accelerator board. But we can do better: have you noticed that the first and last vertices are identical? GL_LINE_LOOP, in fact, behaves similarly to the previous one apart from the fact that a line is drawn between the last and the first vertex specified. In terms of code you have
``````glBegin(GL_LINE_LOOP);
glVertex3f(-10.0f, 0.0f, 0.0f); // Side 1 point 1
glVertex3f(10.0f, 0.0f, 0.0f); // Side 1 point 2 - point 1 Side 2
glVertex3f(10.0f, 10.0f, 0.0f); // Side 2 point 2 - point 3 Side 1
glVertex3f(-10.0f, 10.0f, 0.0f); // Side 3 point 2 - Side 4 point
glEnd();``````
An example of use of the lines capabilities, produced by the program Linee.exe, is visible in Figure 2.
## Triangles
With the lines you can draw any kind of 3D object. It is immediate to observe that the figures are created with this technique are not solid, but have an appearance that is normally called ‘wire-frame’. For this reason we will mainly use polygons, or closed figures, because they can be filled with a selected color.
The first polygon that we consider is the triangle which, consisting of only three sides, is the most simple shape possible. Also mind that OpenGL is particularly optimized to work with the triangles therefore this choice is the most usually adopted for solids construction.
The parameter GL_TRIANGLES draws, of course, a triangle connecting three vertices together as in
``````glBegin(GL_TRIANGLES);
glVertex3f(-10.0f, 0.0f, 0.0f);
glVertex3f(0.0f, 10.0f, 0.0f);
glVertex3f(10.0f, 0.0f, 0.0f);
glEnd();``````
Working in a three-dimensional space we must pay particular attention to the fact that every polygon can be seen from different angles. For this reason we can say that it has, in reality, two faces: the front face and the rear one. But how to discern between the two? The method adopted by OpenGL is to distinguish the order in which you specify the polygon’s vertices. Thus, the difference arises from the order of readed vertices and can occur that they are listed clockwise – Clockwise winding – or counterclockwise – Counterclockwise winding.
By default OpenGL considers the vertices oriented in an counterclockwise direction as the front face of the polygon. Through the function below is possible to specify the backface for the vertices indicate before
``glFrontFace(GL_CW)``
The importance of being able to make this distinction will be clear later in this article when we will discuss about removing the hidden faces. In the box previously showed we can see that there are other parameters that glBegin() accepts to draw triangles: GL_TRIANGLE_STRIP and GL_TRIANGLE_FAN.
With the first of the two can generate chains of adjacent triangles which save time and amount of data transferred, in the same way that works for the lines. The mechanism of construction of the chain consists in considering the last two vertices specified as the first two of the next triangle. In this way the next vertex will be taken as the final one of the triangle to be drawn. Look at an example
``````glBegin(GL_TRIANGLE_STRIP);
glVertex3f(-10.0f, 0.0f, 0.0f); // Triangle 1 vertex 1
glVertex3f(0.0f, 10.0f, 0.0f); // Triangle 1 vertex 2 - Vertex 1 Triangle 2
glVertex3f(10.0f, 0.0f, 0.0f); // Triangle 1 vertex 3 - Triangle 2 vertex 2
glVertex3f(30.0f, 10.0f, 0.0f); // Triangle 2 vertex
glEnd();``````
The GL_TRIANGLE_FAN parameter produces a group of triangles connected to a central point. In this case the first vertex specified will be considered as the central. Every other vertex is connected to the previous to form the new triangle, as in the following
``````glBegin(GL_TRIANGLE_FAN);
glVertex3f(-10.0f, 0.0f, 0.0f); // Central vertex
glVertex3f(0.0f, 10.0f, 0.0f); // Triangle 1 vertex 2
glVertex3f(10.0f, 0.0f, 0.0f); // Triangle 1 vertex 3 - Triangle 2 vertex 2
glVertex3f(30.0f, 10.0f, 0.0f); // Triangle 2 vertex 3 - Triangle 3 vertex 2
...
glEnd ();``````
The example program Triangoli.exe whose output is presented in Figure 3 shows various examples of triangles.
By now we should have understood the functioning of the mechanism that regulates the creation of primitives. We can therefore afford to spend a few words about the quadrilaterals which are simply a list of four vertices, known as GL_QUADS by OpenGL.
The only precaution to follow is that these vertices must lie on the same plane. As for the triangles is possible to create chains of quadrilaterals through GL_QUAD_STRIP, where each next of them is indicated by the last two vertices of the quadrilateral previous joined to two other new vertices. We clarify everything with the usual example, this time building a cube
``````glBegin(GL_QUADS);
// Upper side
glVertex3f(-10.0f, 10.0f, 10.0f);
glVertex3f(-10.0f, 10.0f,-10.0f);
glVertex3f(10.0f, 10.0f,-10.0f);
glVertex3f(10.0f, 10.0f, 10.0f);
// Lower side
glVertex3f(-10.0f,-10.0f, 10.0f);
glVertex3f(-10.0f,-10.0f,-10.0f);
glVertex3f(10.0f,-10.0f,-10.0f);
glVertex3f(10.0f,-10.0f, 10.0f);
glEnd();
// Lateral faces
glVertex3f(-10.0f,-10.0f, 10.0f); // Quad 1 - 1 Vertex
glVertex3f(-10.0f, 10.0f, 10.0f); // Quad 1 - 2 Vertex
glVertex3f(10.0f,-10.0f, 10.0f); // Quad 1 - 3 Vertex
glVertex3f(10.0f, 10.0f, 10.0f); // Quad 1 - 4 Vertex
glVertex3f(10.0f,-10.0f,-10.0f); // Quad 2 - 3 Vertex
glVertex3f(10.0f, 10.0f,-10.0f); // Quad 2 - 4 Vertex
glVertex3f(-10.0f,-10.0f,-10.0f); // Quad 3 - Vertex 3
glVertex3f(-10.0f, 10.0f,-10.0f); // Quad 3 - 4 Vertex
glVertex3f(-10.0f,-10.0f, 10.0f); // Quad 4 - Vertex 3
glVertex3f(-10.0f, 10.0f, 10.0f); // Quad 4 - 4 Vertex
glEnd();``````
In Figure 4 there is the output of the program Quadrilateri.exe.
## Polygons
The final primitive known from OpenGL is GL_POLYGON which is a figure composed by any number of vertices. The restrictions, or rules, of polygon construction are two. The first wants, as for the quadrilaterals, that all the vertices of the polygon must lie on the same plane while the second consists in the fact that a polygon must necessarily be convex, ie should not result in no case that drawing line any that intersects the polygon, in more than two points.
The program Poligoni.exe produces the output shown in Figure 5.
## Back faces culling
As already said, when displaying a polygon in space it happens that OpenGL should draw both faces, the front and rear. In the last example presented we have built a solid, ie, an object with three dimensions, as a composition of two-dimensional primitives. Because each polygon has a face, that being “inside” the cube, does not appear in any case, so by setting this flag
``glEnable(GL_CULL_FACE)``
and launching the function
`` glCullFace (GL_BACK)``
we obtain a considerable saving of time asking OpenGL to not draw at all the back faces of the polygons.The glCullFace() takes as parameters the values GL_BACK, GL_FRONT GL_FRONT_AND_BACK – although I think the latter is useful in very few cases only. At any time, even within glBegin/glEnd calls, we can change the settings or disable this feature using
``glDisable(GL_CULL_FACE)``
Look at the example FacceNascoste.exe to clarify further doubts.
## Coloring
We have already encountered the function glColor that allows you to set the current drawing color. Through this you can create beautiful effects using a very interesting feature of OpenGL that is called Shading.
For each vertex of a primitive, in fact, we can set its own color. The figure will be drawn and filled with a gradient of colors interpolated between the color of a vertex to others specified. Figure 6 shows a cube formed by quadrilaterals stained with this technique.
By default, the color interpolation is enabled, but you can change it at any time with
``glShadeModel(GL_FLAT)``
In this way while drawing the polygon, will be taken into account only the last occurrence of the function glColor. With the call
``glShadeModel(GL_SMOOTH)``
everything returns to normality.
## Depth test
Look at Figure 7, can you notice something strange? The green plane on which should be placed the red cube appears in front of it making the image a little unreal. This happens because the code firstly asked to draw the cube and only after the plan. OpenGL draws the figures one after the other by overwriting the image below.
The control of depth makes it possible to eliminate this problem and is able to establish the effective depth of an object and its actual location among the others, within the scene. To activate the depth test are needed the following steps. First we must ask GLUT to reserve memory for the required buffer, so we add the flag GLUT_DEPTH to the initialization function of the display as in
``glutInitDisplayMode(GLUT_DOUBLE | GLUT_RGBA | GLUT_DEPTH)``
then we must make sure to clear the buffer used for depth values before we start drawing one frame. For this purpose we make use of the call to glutClear() already present in the function RenderScene(), which will have to write in this form
``glutClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT)``
Finally we must enable the depth test. It can be done wherever necessary, but for our purposes is just fine in SetupRC() function where we put the line
``glEnable(GL_DEPTH_TEST)``
## Transformations
When we think about 3D graphic we should know that it is composed of three-dimensional objects, but all we see on screen is nothing more than a two dimensional image. There is called projection, in fact, a complex process that transforms the 3D objects data into 2D image representation.
There are also other types of transformations that produce different effects, you can: rotate, move or resize objects. Each type of geometric transformation is applied to the scene at different times since the vertices are specified until they appear on the screen.
In the first phase is applied a transformation to the ‘view’ – Viewing Transformations – used to determine the point of view of the scene. The OpenGL command to set this point is
``````gluLookAt (GLdouble eyex, eyey GLdouble, GLdouble eyez,
GLdouble centerx, GLdouble centery, GLdouble centerz, GLdouble upx, GLdouble upy, GLdouble upz);``````
The parameters of this function are intuitive, in fact, eye indicate the point where the observer is located, the center coordinates represent the point which is opposite to the observer, while up represents a vector pointing upwards, from the perspective of the observer. Anyone with some perplexity regarding the last parameter can think that is not sufficient to establish the position of an observer and the direction where him/her is watching. It seems also necessary ìto know the inclination of his head.
Without this information, in fact, we have no way of knowing whether it is watching, maybe, with your head upside down or 45 degrees tilted. The reason for which it is applied before the other lies in the fact that the displacement of the observation point determines a change in the working coordinate system. Any other change must take account of modifications made by this.
For example, we see an object centered at coordinates (0,0,0) from a distance, then we write
`````` gluLookAt (0.0f, 50.0f, 50.0f, // Position of the observation point
0.0f, 0.0f, 0.0f, // point to which "we turn our
// glance"
0.0f, 1.0f, 0.0f), // vector pointing upwards,
// perpendicularly to the plane``````
The example PuntoOsservazione.exe shows how to use this transformation to walk around the scene looking at the center of it.
The second type of transformations are the those that are applied to models – Modeling Transformations – mainly used to move objects in space, resize or rotate around the axis. Starting from the translations the function is
``glTranslatef(GLfloat x, GLfloat y, GLfloat z)``
that generates the movement of the object by the specified entity, for each axis. We then, for rotations can use
``glRotatef(GLfloat angle, GLfloat x, GLfloat y, GLfloat z)``
which rotates the object by an angle equal expressed in degrees. To indicate around which given axis must be rotated an object we must pass the 1.0f value to the parameters x, y or z. In the simplest case a rotation takes place around a single axis, but it is possible to realize rotations involving all axes. In the exampleglRotatef (GLfloat angle, GLfloat x, GLfloat y, GLfloat z)an object would be rotated around to the Z axis
``glRotatef(45, 0.0f, 0.0f, 1.0f)``
The scaling works by increasing or decreasing the size of the objects expanding the vertices, along the three axes. Each parameter of the function
``glScalef(GLfloat x, GLfloat y, GLfloat z)``
represents the multiplication factor for the object size. A value of 1.0f leaves unaltered the size along the axis, a larger value causes an increase, while a value tending to zero determines the shrinkage. Writing
``glScalef(1.0f, 4.0f, 1.0f)``
any object would be stretched vertically, but would keep intact the other dimensions.
We should pay particular attention to the fact that the described transformations are not applied to the single object, but to the entire reference system, so any other further operation will be affected by the new settings.
Changing order of two transformations, for example a translation and a rotation, produces different effects as is done in this case
`````` glTranslatef (10.0f, 0.0f, 0.0f);
glRotatef (45, 0.0f, 0.0f, 1.0f);
...
drawCube();``````
and
`````` glRotatef (45, 0.0f, 0.0f, 1.0f);
glTranslatef (10.0f, 0.0f, 0.0f);
...
drawCube();``````
In the first case we have moved the cube of 10 units on the X axis with a subsequent rotation of 45 degrees on the Z axis. In the second the opposite has occurred, ie, was first carried out a rotation of 45 degrees along the Z axis and then a translational motion along the X axis. Can you notice the difference?
Let’s move on to the next type of transformations: the projection transformations. This type of transformation is used to define which parts are visible to the viewer of 3D space and specify how the 3D scene is projected on the screen. There are two types of projection
• The Orthogonal projection, which does not discern any difference on the objects distance. They, in fact, will have the same size if they are near to the observation point or not. We also said that this type of projection is mostly used in CAD applications, or in video games, to create menus or 2D images.
• The Perspective projection which produces images that mirror the optical phenomenon where identical objects distant from each other seems to have different sizes or where parallel lines seem to converge away from the observer.
The perspective projection is activated with the function
``gluPerspective (GLdouble fovy, GLdouble aspect, GLdouble zNear, GLdouble zFar)``
We give a brief explanation of the parameters. The first and most cryptic is ‘fovy’ the angle of the visual field in the vertical direction, ‘aspect’ is the usual aspect ratio ie the ratio between width and height of the window. Finally zNear and zFar are respectively the visible boundary closer to the viewer and the farther away.
The last type of transformation is what happens to the viewport when the final image is mapped physically on the window. However, the user does not need to deal with this aspect.
## Conclusions
After reading this article you will become familiar with the OpenGL drawing functions. You can now create your own fully functional program that displays various types of solid objects and not.
In the next article I will conclude this process by studying ways to create more complex scenes and use advanced features such as texture mapping, lighting effects and more.
## Bibliography
1. Richard S.Wright jr. and Benjamin Lipchak, “OpenGL SuperBible – Third Edition”, SAMS, 2005, ISBN 0-672-32601-9 | 5,533 | 21,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-40 | longest | en | 0.905033 |
http://knowingallah.com/en/articles/divine-singularity-the-oneness-of-god/ | 1,521,371,658,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645613.9/warc/CC-MAIN-20180318110736-20180318130736-00687.warc.gz | 146,485,401 | 15,187 | # Divine Singularity: The Oneness of God
Imagine you are an explorer who took a spaceship to another planet to visit human-like creatures. Once you land on the planet, you meet your guide. He tells you that your spaceship landed on Sphinga, the planet’s borderless country. You are confused and ask your guide if there are any other countries on the planet. He laughs and replies, “Yes, there are two.” You retort, “Well, how do you know when you’re in another country if there’s nothing to differentiate them?” Your guide sighs and says, “Yeah, we have the same problem. There are no borders and the features of one country are the same as the other.” You finally end the discussion by saying, “You should have just made them into one country then, because that is what it looks like to me.”
You both continue your journey to meet a group of officials for lunch. During the meal one of the officials praises the kings of the country. Upon hearing this, you politely ask, “You mean, there is more than one king?” The official replies, “Yes, we have two kings.” You seem perplexed and ask how the country can function with two kings. “How do you have harmony in your laws, and order in your society?” The official replies, “Well, they always agree. Their wills are one.” You cannot hold yourself back and you respond, “Well, you do not really have two kings, then. Because they are acting in accordance with one will.”
This story contains three of the five arguments I will present for the fact that there can only be one God . The first part of the story summarises an argument that I call ‘conceptual differentiation’. It postulates that in order for multiplicity to exist, there must be some concepts that differentiate one thing from another. For example, if I said that there are two bananas on the table, you would be able to verify that statement by observing them. The reason you can see two bananas is because there exist concepts that differentiate them; for example, their size, shape, and location on the table. However, if there was nothing to differentiate them you could not distinguish between them. Similarly, since this book so far has argued that there is a necessary uncreated creator who is powerful, knowing, All-Aware and transcendent, then to claim that there are two would require a concept that differentiates them. But in order for the Creator to be a creator, He must have these attributes, so saying there are two without one being different from another is basically saying that there is only one creator. If whatever is true of one creator is true of another, then we have just defined one creator and not two.
The second part of the story summarises both the argument from exclusion and the argument from definition. The argument from exclusion maintains that there can only be one Divine will. If there were two creators and one wanted to create a tree, only three options would be possible. The first is that they both cancel each other out; this is not a rational possibility since creation exists, and if they cancelled each other out, there would be no creation at all. The second is that one creator overpowers the other by ensuring his tree is created. The third option is that they both agree to create the same tree in the same way. Both of these options imply that there is only one will, and one will in the context of our discussions means one creator.
The argument from definition asserts that there cannot be more than one creator. If there were more than one creator the cosmos would not display the harmony that it does. As well presenting arguments for a creator, this book has also warranted the traditional conception of God has having an imposing will that cannot be limited by anything external to Him, then it logically follows there cannot be two unlimited Divine wills.
This essay will elaborate on these arguments and present another two to show that this creator must be one:
• The argument from exclusion;
• Conceptual differentiation;
• Occam’s razor;
• The argument from definition;
• The argument from revelation.
### Argument from exclusion
This argument maintains that the existence of multiple creators is impossible because there can only be one will. Since the Creator is eternal and brought into existence the universe which began at a point in time, it means that the Creator chose the universe to come into existence; choice implies a will. Questioning how many wills can exist leads us nicely to discuss this argument in detail.
For the sake of argument, let’s say there were two creators. Creator A wanted to move a rock, and creator B also wanted to move the same rock. There are three possible scenarios that can arise:
1. One of the creators overpowers the other by moving the rock in a different direction from the other.
2. They both cancel each other out, and the rock does not move.
3. They both move the rock in the same direction.
The first scenario implies only one will manifests itself. The second scenario means that there is no will in action. This is not possible because there must be a will acted upon, as we have creation in existence. The third scenario ultimately describes only one will. Therefore, it is more rational to conclude that there is only one creator because there is only one will.
If someone argues that you can have more than one entity and still have one will, I would respond by asking: how do you know there is more than one entity? It sounds like an argument from ignorance, because there is no evidence whatsoever for such a claim. This leads us to the next argument.
### Conceptual differentiation
For two creators to exist, they must be different in some way. For example, if you have two trees, they will differ in size, shape, colour and age. Even if they had identical physical attributes, there would be at least one thing that allows us to distinguish that they are in fact two trees. This can include their placement or position. You can also apply this to twins; we know there are two people because something makes them different. This could even be the mere fact that they cannot occupy the same place at the same time.
If there were more than one creator, then there must be something to differentiate between them. However, if they are the same in every possible aspect, then how can we say there are two? If something is identical to another, then what is true of one is also true for the other. Say we had two things, A and B. If they are the same in every way, and nothing allows us to differentiate between them, then they are the same thing. We can turn this into a hypothetical proposition: If whatever is true of A is true of B, then A is identical to B.
Now let us apply this to the Creator. Imagine that two creators exist, called creator X and creator Y, and that whatever is true of creator X is also true of creator Y. For instance, creator X is All-Powerful and All-Wise; so, creator Y is All-Powerful and All-Wise. How many creators are there in reality? Only one, due to the fact that there is nothing to differentiate between them. If someone were to argue that they are different, then they would not be describing another creator but something that is created, as it would not have the same attributes befitting the Creator.
If someone is adamant in claiming that there can be two creators and they are different from each other, then I simply ask, “How are they different?” If they attempt to answer the question, they enter the realm of arguing from ignorance, because they will have to make up evidence to justify their false conclusion.
### Occam’s razor
In light of the above, we might find a few irrational and stubborn people who still posit a plurality of creators or causes. In light of Occam’s razor, this is not a sound argument. Occam’s razor is a philosophical principle attributed to the 14th century logician and Franciscan friar William of Occam. This principle enjoins: ‘Pluralitas non est ponenda sine necessitate’; in English ‘Plurality should not be posited without necessity.’ In other words, the simplest and most comprehensive explanation is the best one.
In this case, we have no evidence that the Creator of the universe is actually a combination of two, three or even one thousand creators, so the simplest explanation is that the Creator is one. Postulating a plurality of creators does not add to the comprehensiveness of the argument either. In other words, to add more creators would not enhance the argument’s explanatory power or scope. To claim that an All-Powerful creator created the universe is just as comprehensive as claiming that two All-Powerful creators created it. One creator is all that is required, simply because it is All-Powerful. I would argue that postulating multiple creators actually has reduced explanatory power and scope; this is because it raises far more problems than it solves. For example, the following questions expose the irrationality of this form of polytheism; how do many external beings co-exist? What about the potential of any conflicting wills? How do they interact?
A popular objection to this argument is that if we were to apply this principle to the pyramids in Egypt, we would absurdly adopt the view that they were made by one person, because it seems to be the simplest explanation. This is a misapplication of the principle, because it ignores the point about comprehensiveness. Taking the view that the pyramids were built by one person is not the simplest and most comprehensive explanation, as it raises far more questions than it answers. For instance, how could one man have built the pyramids? It is far more comprehensive to postulate that it was built by many men. In light of this, someone can say that the universe is so complex that it would be absurd to postulate that only one creator created it. This contention, although valid, is misplaced. A powerful Being creating the whole universe is a far more coherent and simple explanation than a plurality of creators, because a plurality of creators raises the unanswerable questions stated in the previous paragraph. Nevertheless, the critic may continue to argue that it wasn’t one person that created the Pyramids, but an All-Powerful creator. The problem with this is that nothing within the universe is an All-Powerful Being, and since the Pyramids are buildings, and buildings are built by an efficient cause (a person or persons that act), then the Pyramids must have been created by the same type of cause. This leads us back to the original point, that more than one of these causes was required to build the Pyramids.
### The argument from definition
Reason necessitates that if there were more than one creator, the universe would be in chaos. There would also not be the level of order we find in the cosmos.
The Qur’an has a similar argument
Had there been within the heavens and Earth gods besides God, they both would have been ruined
The Qur’an, Chapter 21, Verse 22
The classical commentary known as Tafsir al-Jalalayn states: “Heaven and the Earth would have lost their normal orderedness since there would have inevitably been internal discord, as is normal when there are several rulers: they oppose one another in things and do not agree with one another.”[1]
However, one might point out that since more than one person made your car—one person fitted the wheels, and someone else installed the engine and another person installed the computer system—maybe the universe was created in the same way. This example indicates that a complex object can be created by more than one creator.
In order to respond to this contention, what has to be understood is that the most rational explanation for the origins of the universe is the concept of God and not just a ‘creator’. There may be an abstract conceptual possibility of multiple creators, as highlighted by the car example, but there cannot be more than one God. This is because God by definition is the Being that has an imposing will that cannot be limited by anything external to Him. If there were two or more Gods, they would have a competition of wills, which would result in chaos and disorder. The universe we observe is governed by mathematical laws and order; therefore it makes sense that it is the result of one imposing will. Interestingly, the objection above actually supports Divine oneness. In order for the car to work, the different people who were responsible for making it had to conform to the overall ‘will’ of the designer. The design limited the wills of those responsible for making the car. Since God, by definition, cannot have His will limited by anything outside of Himself, it follows that there cannot be more than one Divine will.
However, one may argue that multiple Gods can agree to have the same will or they can each have their own domain. This would mean that their wills are now limited and passive, which would mean they are not Gods any more by definition.
The 12th century Muslim thinker and philosopher Ibn Rushd, also known as Averroes in the western tradition, summarises this argument:
“The meaning of the… verse is implanted in the instincts [of man] by nature. It is self-evident that if there are two kings, the actions of each one being the same as those of the other, it would not be possible [for them] to manage the same city, for there cannot result from two agents of the same kind one and the same action. It follows necessarily that if they acted together, the city would be ruined, unless one of them acted while the other remained inactive; and this is incompatible with the attribute of Divinity. When two actions of the same kind converge on one substratum, that substratum is corrupted necessarily.”[2]
### The argument from revelation
A simpler way of providing evidence for God’s oneness is to refer to revelation. This argument postulates that if God has announced himself to humanity, and this revelation can be proven to be from Him, then what He says about Himself is true. However, a sceptic may question some of the assumptions behind this argument. These assumptions include that God has announced Himself to humankind and that the revelation is in the form of a book.
Let’s first tackle the last assumption. If God has announced Himself to humankind, there are only two possible ways to find out: internally and externally. What I mean by ‘internally’ is that you can find out who God is solely by introspection and internalisation, and what I mean by ‘externally’ is that you can find out who God is via communication from outside of yourself; in other words, it is instantiated in the real world. Finding out about God internally is implausible for the following reasons:
1. Human beings are different; they have what psychologists call ‘individual differences’. These include DNA, experiences, social contexts, intellectual and emotional capacities, gender differences, and many more. These differences play a role in our ability to internalise via introspection or intuition. Therefore, the results of thinking will differ. If these processes were solely used to find out about God, inevitably differences in our conception of Him would arise. This is true from a historical point of view. From the ancient world of 6000 BCE to the present, there are records of approximately 3,700 different names and concepts for God.
1. Since the method used to conclude that God does exist is a ‘commonsense’ method, or what philosophers call rational thought and what Muslim theologians call innate thinking, then trying to find out about who God is, rather than just affirm His existence, would be fallacious. There are limits to our reasoning. Abstract thinking and reflections on the physical world can only lead us to the conclusion that a creator exists, and He is powerful, knowing, etc. To go beyond those conclusions would be speculative.
Do you say about God that which you do not know
The Qur’an, Chapter 7, Verse 28.
Trying to find out who God is via introspection would be like a mouse trying to conceptualise a galaxy. The human being is not eternal, unique and powerful. Therefore the human being cannot conceptualise who God is. God would have to tell you via external revelation.
Take the following example into consideration. Your knowledge that God exists is like the knocking of the door; you safely assume that something is there, but do you know who it is? You weren’t expecting anyone, so the only way to find who is behind the door is if the person tells you. Therefore you can conclude that if God has said or announced anything, it must be external to the human being. Anything else would be mere speculation.
From an Islamic perspective this external communication is the Qur’an, as it is the only text to claim to have come from God that fits the criteria for a Divine text. These criteria include:
1. It must be consistent with the rational and intuitive conclusion of God existence. For example, if a book says God is an elephant with 40 legs, you could safely assume that this book is not from God, as God must be external to the universe and independent. An elephant, regardless of form, is a dependent being. This is because it has limited physical qualities, such size, shape and colour. All things with limited physical qualities are dependent because there are external factors that gave rise to their limitations. God is not ‘physical’ and is independent. Therefore, nothing with limited physical qualities can be God.
2. It must be internally and externally consistent. In other words, if it says on page 20 that God is one and then on page 340 its says God is three, that would be an internal, irreconcilable inconsistency. Additionally, if the book says that the universe is only 6,000 years old, then that would be an external inconsistency as reality as we know it affirms that the universe is older than that.
3. It must have signposts to the transcendent. The revelation must contain material that indicates it is from the Divine and that it cannot be adequately explained naturalistically. In simple terms, it must have evidence to show that it is from God.
The Qur’an has signposts that indicate it is a Divine text. The book cannot be explained naturalistically; therefore, supernatural explanations are the best explanation. Some of these signposts include:
1. The Qur’an’s linguistic and literary inimitability;
2. Some of the historical accounts in the
3. Its unique arrangement and structure.[3]
To conclude, since the only way to know what God has announced to humankind is via external revelation, and this revelation can be proven to be the Qur’an—then what is says about God is true. The Qur’an is explicitly clear concerning His oneness: “And do not argue with the people of the Scripture except in a way that is best, except for those who commit injustice among them, and say,
We believe in that which has been revealed to us and revealed to you. And our God and your God is one; and we are Muslims [in submission] to Him
The Qur’an, Chapter 29, Verse 46
These are some of the arguments that can be used to show that God is one; however this topic—once truly understood—will have some profound effects on the human conscience. If one God has created us, it follows that we must see everything via His oneness and not our abstracted perspectives of disunity and division. We are a human family, and if we see ourselves this way, it can have profound effects on our society. If we love and believe in God, then we should show compassion and mercy to what He has created. Just like what
Those who are merciful will be shown mercy by the Most Merciful. Be merciful to those on the Earth and the One in the heavens will have mercy upon you
Narrated by Tirmidhi
Last updated 16 January 2017. Taken and adapted from my book “The Divine Reality: God, Islam & The Mirage of Atheism”. You can purchase the book here.
references
1. Al-Mahalli, J. and As-Suyuti, J. (2007) Tafsir Al-Jalalayn. Translated by Aisha Bewley. London: Dar Al Taqwa, p. 690; Mahali, J. and As-Suyuti J. (2001) Tafsir al-Jalalayn. 3rd Edition. Cairo: Dar al-Hadith, p. 422. You can access a copy online at: https://ia800205.us.archive.or... [Accessed 1st October 2016].
2. Avveroes. (2001) Faith and Reason in Islam. Translated with footnotes, index and bibliography by Ibrahim Y. Najjar. Oxford: One World, p. 40.
3. For more on the Divine nature of the Qur’an please read: Khan, N. A. and Randhawa, S. (2016) Divine Speech: Exploring the Qur’an as Literature. Texas: Bayyinah Institute and Zakariya, A. (2015) The Eternal Challenge: A Journey Through The Miraculous Qur’an. London: One Reason.
### Related Articles with Divine Singularity: The Oneness of God
• #### 2.The Call of the Prophets and Messengers to the Oneness of Allah (Glorious & Exalted)
The Call of the Prophets and Messengers to the Oneness of Allah (Glorious &
01/08/2011 1194
• #### Dividing Tawheed into categories
I hear from some knowledgable brothers concerning Tawheed and its categories that Shaykh ul Islaam Taqi`ud deen
20/12/2012 1557
• #### How Do We Know God is One? A Philosophical and Theological Perspective (part 1 of 3)
One of the many common questions that were asked during the Islamic Awareness Tour was “If God does exist
23/07/2014 688 | 4,561 | 21,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-13 | longest | en | 0.966887 |
https://en.wikibooks.org/wiki/Introduction_to_Physical_Science/2.2 | 1,643,307,175,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305277.88/warc/CC-MAIN-20220127163150-20220127193150-00109.warc.gz | 292,342,875 | 9,863 | # Density
Density is possibly one of the most useful properties when studying matter. Each substance has a unique density. Density is determined by using the "density formula". The density formula is as follows: D=M/V or Density equals Mass over (divided by) Volume. Density is the amount of matter that can be placed in one cm3 of space. The following section discusses the uses of density and the density formula.
## Using the Formula
The variables of the density formula can be rearranged to solve the equation for different measurements. The density formula is capable of solving for Mass when Density and Volume are known, Volume when Density, and Mass are known, and Density when Mass and Volume are known.
• D=M/V (Density equals Mass over Volume)
• M=VxD (Mass equals Volume times Density)
• V=M/D (Volume equals Mass over Density)
## As a property
All substances have different densities although two densities may be similar and even undetectably different by some instruments, all densities are different because in each substance, the atoms are packed together in a different way. Solids are the most dense, because their atoms are packed closest together. Liquids are less dense then solids, and gasses are the least dense.
# Vocabulary and Questions
• Density is a measurement of the amount of a substance that can fit into one cubic centimeter of space
• Density Formula D=M/V: Density equals mass over volume
• Variables are placeholders in an equation that represent a number that is not yet known | 309 | 1,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-05 | latest | en | 0.926569 |
https://www.thestudentroom.co.uk/showthread.php?t=4033605 | 1,521,360,891,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645550.13/warc/CC-MAIN-20180318071715-20180318091715-00289.warc.gz | 857,716,212 | 39,790 | x Turn on thread page Beta
You are Here: Home >< Maths
# No matter how much I try... watch
1. It doesn't matter how many times I sit down to study Vectors and try to learn this area of my course, I still cannot do it and I don't know why. It doesn't seem too difficult but there are too many things to learn etc. Does anyone have any tips on how to LEARN vectors? I am lost.
Thanks so much.
Studying: A2 Maths - Edexcel
2. more of an understanding & application than a learn-by-method
keep practising.
3. (Original post by IAmConfused1)
It doesn't matter how many times I sit down to study Vectors and try to learn this area of my course, I still cannot do it and I don't know why. It doesn't seem too difficult but there are too many things to learn etc. Does anyone have any tips on how to LEARN vectors? I am lost.
Thanks so much.
Studying: A2 Maths - Edexcel
You could post an example question that you struggle with (with working)? Then we can show you the best way to go about it.
4. (Original post by notnek)
You could post an example question that you struggle with (with working)? Then we can show you the best way to go about it.
I wish, I mean I literally can't do anything, I don't even know how to approach the question, would examsolutions or getting help on here be the best solution?
5. (Original post by IAmConfused1)
I wish, I mean I literally can't do anything, I don't even know how to approach the question, would examsolutions or getting help on here be the best solution?
Examsolutions got me out of a big hole with trig so yes that would not be a bad idea.
6. (Original post by IAmConfused1)
I wish, I mean I literally can't do anything, I don't even know how to approach the question, would examsolutions or getting help on here be the best solution?
That would be a good idea. Watch a lot of examsolutions videos and you should get a feel for how to go about the questions. A lot of the questions have a similar structure so with practice you should have ideas about how to proceed with a question.
You could still post a question without working and we will guide you towards a solution. It can really help for you to work through a question yourself with guidance. Even if you can't start the question, that's fine.
7. (Original post by PedanticStudent)
Examsolutions got me out of a big hole with trig so yes that would not be a bad idea.
I'll try this, thanks.
(Original post by notnek)
That would be a good idea. Watch a lot of examsolutions videos and you should get a feel for how to go about the questions. A lot of the questions have a similar structure so with practice you should have ideas about how to proceed with a question.
You could still post a question without working and we will guide you towards a solution. It can really help for you to work through a question yourself with guidance. Even if you can't start the question, that's fine.
Thanks for your help, I'll try and learn it AGAIN (using exam solutions) tomorrow and run through some practice questions and post any problems I have here, thanks a lot for your advice
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Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE | 942 | 4,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-13 | latest | en | 0.977299 |
https://www.scribd.com/document/92991383/Smash-Factor-and-Determinants-of-Distance | 1,571,272,514,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672431.45/warc/CC-MAIN-20191016235542-20191017023042-00174.warc.gz | 1,049,098,780 | 74,271 | You are on page 1of 7
# TRACKMAN NEWS
#3 MAY 2008
## FOCUS: Smash Factor
In the analysis of a golf shot, smash factor is referred to increasingly in the golng community. This in-depth interview with Fredrik Tuxen CTO at ISG and the inventor of TrackMan touches upon the relevance, measurement, and maximization of smash factor. Even though you may impact the ball dead-center on the club face, so the ball departs on a line that goes directly through the Center of Gravity (CoG) of the club head, there are 3 more factors that determine the maximum obtainable smash factor: 1) coefcient of restitution between club and ball (COR), 2) the SPIN LOFT the angle between club face orientation and club head direction (see TrackMan newsletter #1 and #2), and 3) the mass ratio between ball weight and club head weight. The equation below shows the maximum obtainable smash factor assuming a dead-center hit:
What is the smash factor? The smash factor is the ratio between ball speed and club head speed.
What does smash factor tell a golfer about a shot? As a parameter, it is an expression of the players ability to generate ball speed based on a given club speed. Technically, the smash factor says a lot about the centeredness of impact and the solidity of the shot - there is a strong correlation between the degree of centeredness at impact and the obtained smash factor.
For the coefcient of restitution, USGA and The R&A have limited golf clubs and balls to a maximum COR of 0.83. While the spin loft could theoretically be 0 deg, it is impractical since this would mean something like a 0 deg lofted driver with a zero ex shaft producing 0 rpm of spin! The lowest realistic spin loft for a driver is around 8 deg. As for the ball, the maximum allowed mass is 45.93 g, with no lower limit. However, it turns out that almost all golf balls have a mass above 45 g since the heavier weight makes the ball slow down less during ight (due to air resistance). For the club head mass, there are small variations among drivers. They typically range from 197 to 201 g, with tour pros using 202-207 g. The heaviest driver head I have heard about is 212 g. By inserting realistic numbers in the equation above for maximizing the smash factor (COR 0.83, SPIN LOFT 8 deg, mass ratio 45/212), the highest realistic smash factor is 1.494. A word of caution, before you start putting lead tape on your driver to make it heavier, that the heavier the club head the harder it is to generate club head speed. Maximum ball speed for a 45 inch driver is obtained for most people with a club head weight around 200 g. See Search for The Perfect Swing by Cochran and Stobbs for a study on how the club head speed varies with club head weight.
How important is smash factor as a launch parameter? It is very important and to be honest, it is much more important than many think. Especially for those amateurs that try to swing too hard at the ball. By trying to achieve a high club speed, they lose control and dont obtain a solid, centered impact, resulting in a relatively low smash factor, far from what is optimal. When working with TrackMan, the amateur and the pro should focus a lot more on ball speed and the smash factor in order to improve their ball striking. This is the reason why we have deliberately taken club speed away from the rst page on the TrackMan screen and moved it down to page 3. We want players to focus on what is really signicant to improve in their swing. Let me give you an example. With a club speed of 100 mph and a smash factor of 1.40, the ball speed is 140 mph. But if the golfer could obtain a smash factor of 1.48 with a more controlled swing having a lower club speed of 98 mph, the ball speed would be increased to 145 mph i.e. an additional 5 mph ball speed by swinging slower. Since 1 more mph ball speed (all other things equal) will generate 2 more yards carry, an extra 10 yards is added to the drive in this case by swinging with more control! Further, the more controlled swing will most likely have a very positive effect on dispersion.
What is a good smash factor? This depends highly on what club you are looking at and what ball type you are playing. For a driver with a premium ball, as an amateur, your smash factor should be above 1.42 and if you have elite ambitions, you should not be below 1.47. Tour pros should aim for nothing less than 1.48 as a minimum. (continues)
What is the highest smash factor you can obtain? The laws of physics do put some limitations on what is possible.
Page
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TRACKMAN NEWS
#3 MAY 2008
## FOCUS: Smash Factor
(continued) In general, both the PGA and LPGA players seem to be right at the optimal smash factor - and sometimes actually slightly above. In particular on the shorter irons, the pros are achieving a higher smash factor than what is reasonably expected from the club loft. The likely explanation for these high smash factors is that the spin loft is actually lower than the club loft which will be the case if the ball is impacted with the hands leading the club head. Another interesting observation in Table 2 is that LPGA players seem to generate higher smash factors for the longer irons in particular. A possible explanation for this is that there is a small increase in club/ ball COR at lower club head speeds. Also the ladies tend to use more cavity back type of clubs which has slightly higher COR and slightly lower loft than corresponding blade type which is preferred by most PGA Tour players.
But do note that if you are hitting the very common high durability range balls the effective COR can easily be as low as 0.73 which will limit the smash factor realistically to about 1.41!
How much does the smash factor vary from club to club? By using the equation above and assuming standard loft as being the SPIN LOFT and average male club head weights, the theoretical optimal smash factor throughout the set is shown in Table 1. For illustration, the corresponding club head speed and ball speed is shown where the club head speed has been scaled to match the average for the PGA TOUR.
The results in Table 1 agree very well with our observations of male and female tour pros for longer irons and woods. Some examples are presented in Table 2.
(continues)
Page
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TRACKMAN NEWS
#3 MAY 2008
## FOCUS: Smash Factor
(continued) where on the club face the ball is impacted, so this will be 100 mph for all the 5 different impact locations. In the table below, an example of a realistic variation of the COR variation across the club face has been used. Maximum ball speed is obtained with impact of an inch towards the toe despite the lower COR of 0.81 at this point on the club face.
How does TrackMan actually measure smash factor? While the calculation of smash factor is simply the ratio between ball speed and club head speed, there are some details that are worth noticing. The ball speed is very well dened, and TrackMan measures the ball speed directly within 0.1 mph. However, with the club head speed things are not quite as simple. It might be a surprise to many golfers, but the club head speed actually varies signicantly depending on where on the club face you are looking. On average there is a 14% difference between heel and toe speed. This means that if you have 100 mph club head speed in the center of the club face, the speed of the heel will be around 93 mph and the toe 107 mph. This is primarily due to two things: 1) the further distance from grip to the toe of the club compared to the grip to heel 2) the rotation of the club head during the downswing. Likewise, the club head speed low on the club face is higher than high on the club face. TrackMan always refers to the club head speed at the center of the club face, but because of around a 3/8 inch uncertainty of the location of the radar reection point on the back of the club face, this leads to an accuracy of the club head speed measurement of the TrackMan of 1 mph with reference to the center of the club face.
Table 3: Smash factor variation across club face. Assuming no club head rotation due to off-center hits. If the smash factor was calculated from straight theory (last column in table 3): ball speed divided with the club head speed at point of impact, the smash factor producing the highest 150.3 mph ball speed would come out as 1.463. Since ball speed (together with launch angle and spin rate) is what matters for the ball ight, by using the center of the club face as reference for the club head speed measurement, maximizing your TrackMan smash factor means also maximizing your ball speed for a given physical strength. This means that in the case the ball is impacted towards the toe (higher club head speed) but still with a high COR and no loss of energy due to twisting of the club head during impact, the theoretical maximum smash factor might be 1.48, but the TrackMan smash factor could come out higher.
Figure 1: Typical club head speed variation across the club face. Let me give you an example of how this affects your smash factor measurement: Let us assume a club head speed of 100 mph (in the center of the club face) with a dead center ball impact producing 148 mph ball speed. This should theoretically give a smash factor of 1.48. However, due to the uncertainty of the exact location of the club head speed reading of the TrackMan, the smash factor might be measured somewhere between 148/101 and 148/99 (1.465 to 1.495). Let us then take the other case where the ball is impacted at the 5 different locations indicated on the club face above but having the club delivered with the same speed and spin loft to the ball (Figure 1).The club head measured by the TrackMan is independent on
Are there more smash factor discoveries left to make? We have so far spent most of our time looking at smash factors for drivers. We have now started looking at smash factors for irons. The tour pros seem to generate a slightly higher smash factor with their irons, especially the shorter ones, than what you would expect from the loft of the club. So we are currently analyzing the tour pros club delivery in particular attack angle and dynamic loft to understand more precisely what the worlds best ball strikers are doing. The results of this will be very valuable for both tting and instruction.
Page
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5/9/12 10:01 AM
Family
Golf Stuff
## Home Article Contents
Force Centrifugal Torque MOI Frequency Distance
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Determinants of Distance
1 + e Vball = Vclubhead * ---------
http://www.tutelman.com/golf/design/physics4.php
Page 1 of 4
## Design Notes - Physical Principles p4
5/9/12 10:01 AM
1 + (m/M)
where: e= An efficiency measure of momentum transfer called the Coefficient of Restitution (COR). Typical values are: 0.67 at the time Cochran & Stobbs' book was written, with a then-typical ball and a rigid clubface. 0.78 for a modern ball and a rigid clubface. 0.83 for a spring-face driver with the maximum legal COR. (The USGA and R&A have decided to measure and limit COR.) m = Mass of the ball (typically 46 grams or 1.62 ounces). M = Mass of the clubhead (typically 200 grams or 7 ounces for driver).
http://www.tutelman.com/golf/design/physics4.php Page 2 of 4
## Design Notes - Physical Principles p4
5/9/12 10:01 AM
Distance is a weak function of clubhead weight. If you can't swing a heavier clubhead very nearly as fast as a lighter one, the heavier head will cost you distance. If you can't bring the clubhead into the ball with good impact (center of clubface, with clubface square to the path of the clubhead), you will lose more distance than you might imagine.
## Golf is Not Artillery
http://www.tutelman.com/golf/design/physics4.php Page 3 of 4
## Design Notes - Physical Principles p4
5/9/12 10:01 AM
Let me end this section by dispelling a common myth, based on a misinterpretation of a well-known "law" of physics. As many of us learned in Physics 101, an object travels furthest if launched at an angle of 45 degrees. Why does this not seem to apply to the design of golf clubs? We all know that a 45 loft is about that of a pitching wedge, and a 45 launch angle requires considerably more loft than that. From experience, we all know that clubs with that much loft don't hit the ball nearly as far as the lower-lofted clubs. Here is what's happening: The Physics 101 problem assumed that the ball starts at the same speed, no matter what the angle of takeoff. This is a true assumption for artillery, which is where the problem originated. In artillery, you change the launch angle by tilting the cannon up or down, which doesn't hurt the "ball speed" at all. But, for the golf model, increasing the launch angle usually involves increasing the loft. As noted above, this causes ball-speed "leakage", as more of the impact energy is turned into spin instead of ball speed. By the time you get to a 45 launch angle, you are hitting a very high-lofted wedge with lots of height and rather little distance. In order to duplicate the "artillery model" with a golf swing, it would require you to cause the 45 launch by using a tee more than a foot tall, and hit the ball with a low-loft driver on a 45 upswing. I hope that made sense. | 3,039 | 13,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-43 | latest | en | 0.928941 |
http://www.codebytes.in/2015/05/rock-spoj-rock-solution.html | 1,606,878,411,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141686635.62/warc/CC-MAIN-20201202021743-20201202051743-00084.warc.gz | 106,959,077 | 14,266 | ## Friday, May 22, 2015
### ROCK. SPOJ - ROCK solution
Problem: www.spoj.com/problems/ROCK
Source (Java), AC:
```import java.io.*;
class ROCK {
public static int[] data;
public static int[][] dp;
public static int solve(int from, int to) {
int ones = 0, zeros = 0;
if(from>to)return 0;
if (dp[from][to] != -1) return dp[from][to];
for (int i = from; i <= to; ++i) {
if (data[i] == 1) ones++;
else zeros++;
}
if(ones>zeros)return dp[from][to]=ones+zeros;
if(from==to) return data[to];
int max = 0;
for(int i=from;i<=to;++i){
int left = 0;
if(i!=to)left = solve(from, i);
int right = 0;
if(i+1!=from)right = solve(i+1, to);
if(left+right>max){max = left+right;}
}
return dp[from][to] = max;
}
public static void main(String[] args) throws Exception {
int nCases = IO.nextInt();
while (nCases-- > 0) {
int length = IO.nextInt();
data = IO.nextIntArray(length, "");
dp = new int[length + 1][length + 1];
for (int i = 0; i < length + 1; ++i) {
for (int j = 0; j < length + 1; ++j) {
dp[i][j] = -1;
}
}
IO.println(solve(1, length));
}
}
```
``` static class IO {
public static int[][] next2dInt(int rows, int cols, String seperator) throws Exception {
int[][] arr = new int[rows + 1][cols + 1];
for (int i = 1; i <= rows; ++i) {
arr[i] = nextIntArray(cols, seperator);
}
return arr;
}
public static int[] nextIntArray(int nInts, String seperator) throws IOException {
String ints = br.readLine();
String[] sArray = ints.split(seperator);
int[] array = new int[nInts + 1];
for (int i = 1; i <= nInts; ++i) {
array[i] = Integer.parseInt(sArray[i - 1]);
}
return array;
}
public static long[] nextLongArray(int nLongs, String seperator) throws IOException {
String longs = br.readLine();
String[] sArray = longs.split(seperator);
long[] array = new long[nLongs + 1];
for (int i = 1; i <= nLongs; ++i) {
array[i] = Long.parseLong(sArray[i - 1]);
}
return array;
}
public static double[] nextDoubleArray(int nDoubles, String seperator) throws IOException {
String doubles = br.readLine();
String[] sArray = doubles.split(seperator);
double[] array = new double[nDoubles + 1];
for (int i = 1; i <= nDoubles; ++i) {
array[i] = Double.parseDouble(sArray[i - 1]);
}
return array;
}
public static char[] nextCharArray(int nChars, String seperator) throws IOException {
String chars = br.readLine();
String[] sArray = chars.split(seperator);
char[] array = new char[nChars + 1];
for (int i = 1; i <= nChars; ++i) {
array[i] = sArray[i - 1].charAt(0);
}
return array;
}
public static int nextInt() throws IOException {
String in = br.readLine();
return Integer.parseInt(in);
}
public static double nextDouble() throws IOException {
String in = br.readLine();
return Double.parseDouble(in);
}
public static long nextLong() throws IOException {
String in = br.readLine();
return Long.parseLong(in);
}
public static int nextChar() throws IOException {
String in = br.readLine();
return in.charAt(0);
}
public static String nextString() throws IOException {
}
public static void print(Object... o) {
for (Object os : o) {
System.out.print(os);
}
}
public static void println(Object... o) {
for (Object os : o) {
System.out.print(os);
}
System.out.print("\n");
}
public static void printlnSeperate(String seperator, Object... o) {
StringBuilder sb = new StringBuilder();
sb.delete(sb.length() - seperator.length(), sb.length());
System.out.println(sb);
}
}
}``` | 963 | 3,367 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2020-50 | latest | en | 0.366996 |
https://www.enotes.com/homework-help/what-2-equality-axioms-real-numbers-5-axioms-order-263436 | 1,521,520,490,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647280.40/warc/CC-MAIN-20180320033158-20180320053158-00536.warc.gz | 758,855,298 | 10,564 | what are the 2 equality axioms of real numbers and the 5 axioms of order? And can you give examples? Thank You!
The 2 equality axioms of real numbers are as follows:
1) Reflexive axiom of equality which states that a = a, or any real number equals itself. Example: 5=5
2) Symmetric axiom of equality which states that if a = b, then b = a. Example: 12/4 = 3, then 3 = 12/4
There are 4 aximoms of order for real numbers are as follows:
1) Axiom of comparison which states that for real numbers only one of the following relationships can exist: a > b; a < b; or a = b. Some examples are as follows:
If a = 5 and b = 4, then 5 > 4
If a = 4 and b =5, a< b
If a = 5 and b = 5, then a = b.
2) Transitive axiom of comparison which states if a < b and b < c, then a < c.
Example: if a = 4; b =5 and c = 6, then the following is true: 4 < 5 and 5< 6, therefore 4 < 6.
3) Multiplication axiom of comparison which states the following:
If a < b and c > 0, then ac < bc
Example: if a = 4, b = 5, and c =6, then (4)(5) < (5)(6)
4) Additive axiom of comparison which states the following:
If a < b then a + c < b +c
Example: if a = 4, b = 5, and c = 6, then 4 + 6 < 5 + 6
giorgiana1976 | Student
We'll begine with reflexive axiom:
a = a, for ay real number "a"
Example 5 = 5
We'll continue with symmetric axiome:
If a = b, then b = a
Example: If 3 + 4 = 7, then 7 = 3 + 4
Another axiom is transitive axiom:
If a = b and b = c, then a = c
Example: Since 3 + 4 = 7, and 1 + 2 = 3, then 1 + 2 + 3 + 4 = 3 + 7 = 10
The axioms of order:
1) Translation invariance of order: x < y => x + z < y + z, for any x and y real numbers
Example: 2 < 3 => 2 + 4 < 3 + 4 <=> 6 < 7
2) Transitivity of order: x < y and y < z => x < z
3) Trichotomy: x < y and y < x => x = y
4) Scaling:
If x < y and z> 0 => x*z < y*z | 664 | 1,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-13 | latest | en | 0.790066 |
https://www.hindawi.com/journals/ads/2008/218140/ | 1,571,697,688,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2019-43/segments/1570987795253.70/warc/CC-MAIN-20191021221245-20191022004745-00141.warc.gz | 934,046,159 | 49,812 | `Journal of Applied Mathematics and Decision SciencesVolume 2008, Article ID 218140, 28 pageshttp://dx.doi.org/10.1155/2008/218140`
Research Article
## Simple Correspondence Analysis of Nominal-Ordinal Contingency Tables
School of Computing and Mathematics, University of Western Sydney, Locked Bag 1797, Penrith South DC, NSW 1797, Australia
Received 19 February 2007; Revised 14 June 2007; Accepted 29 October 2007
Academic Editor: Mahyar A. Amouzegar
Copyright © 2008 Eric J. Beh. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
#### Abstract
The correspondence analysis of a two-way contingency table is now accepted as a very versatile tool for helping users to understand the structure of the association in their data. In cases where the variables consist of ordered categories, there are a number of approaches that can be employed and these generally involve an adaptation of singular value decomposition. Over the last few years, an alternative decomposition method has been used for cases where the row and column variables of a two-way contingency table have an ordinal structure. A version of this approach is also available for a two-way table where one variable has a nominal structure and the other variable has an ordinal structure. However, such an approach does not take into consideration the presence of the nominal variable. This paper explores an approach to correspondence analysis using an amalgamation of singular value decomposition and bivariate moment decomposition. A benefit of this technique is that it combines the classical technique with the ordinal analysis by determining the structure of the variables in terms of singular values and location, dispersion and higher-order moments.
#### 1. Introduction
The analysis of categorical data is a very important component in statistics, and the presence of ordered variables is a common feature. Models and measures of association for ordinal categorical variables have been extensively discussed in the literature, and are the subject of classic texts including Agresti [1], Goodman [2], and Haberman [3].
The visual description of the association between two or more variables is a vital tool for the analyst since it can often provide a more intuitive view of the nature of the association, or interaction, between categorical variables than numerical summaries alone. One such tool is correspondence analysis. However, except in a few cases ([47]), the classical approach to correspondence analysis neglects the presence of ordinal categorical variables when identifying the structure of their association. One way to incorporate the ordinal structure of categorical variables in simple correspondence analysis is to adopt the approach of Beh [8]. His method takes into account the ordinal structure of one or both variables of a two-way contingency table. At the heart of the procedure is the partition of the Pearson chi-squared statistic described by Best and Rayner [9] and Rayner and Best [10]. However, when there is only one ordered variable, Beh's [8] approach to correspondence analysis does not consider the structure of the nominal variable. This paper does consider the previously neglected nominal variable by using the partition of the Pearson chi-squared statistic described by Beh [11]. The partition involves terms that summarize the association between the nominal and ordinal variables using bivariate moments. These moments are calculated using orthogonal polynomials for the ordered variable and generalized basis vectors of a transformation of the contingency table for the nominal variable.
The correspondence analysis approach described here, referred to as singly ordered correspondence analysis, is shown to be mathematically similar to the doubly ordered approach. The singly ordered and doubly ordered approaches share many of the features that make the classical approach popular. Details of classical correspondence analysis can be found by referring to, for example, Beh [12], Benzécri [13], Greenacre [14], Hoffman and Franke [15], and Lebart et al. [16]. A major benefit of singly ordered correspondence analysis is that nominal row categories and ordinal column categories can be simultaneously represented on a single correspondence plot while ensuring that the structure of both variables is preserved. Constructing such a joint plot for the singly ordered approach of Beh [8] is not possible due to the scaling of coordinates considered in that paper. For the technique described in this paper, the special properties linking the bivariate moments and singular values provide the researcher with an informative interpretation of the association in contingency tables. These numerical summaries also allow, through mechanisms common to correspondence analysis, a graphical interpretation of this association. Hybrid decomposition has also been considered for the nonsymmetrical correspondence analysis of a two-way contingency table by Lombardo et al. [17].
This paper is divided into seven further sections. Section 2 defines the Pearson ratio and various ways in which it can be decomposed to yield numerical and graphical summaries of association. The decompositions considered are (a) singular value decomposition, used in classical correspondence analysis, (b) bivariate moment decomposition, used for the doubly ordered correspondence analysis approach of Beh [8], and (c) hybrid decomposition. This latter technique amalgamates the two former procedures and is important for the singly ordered correspondence analysis technique described in this paper. Section 3 summarizes, by considering the hybrid decomposition of the Pearson ratio, the coordinates needed to obtain a graphical summary of association between the two categorical variables while Section 4 provides an interpretation of the distance between the coordinates in the correspondence plot. Section 5 defines the transition formulae which describe the relationship between the coordinates of the two variables. Various properties of singly ordered correspondence analysis are highlighted in Section 6. The features of the technique are examined using a pedagogical example in Section 7 where it is applied to the data described in Calimlin et al. [18]. Their contingency table summarizes the classification of four analgesic drugs according to their effectiveness judged by 121 hospital patients. The paper concludes with a brief discussion in Section 8.
#### 2. Decomposing Pearson's Ratio
##### 2.1. The Pearson Ratio
Consider a two-way contingency table that cross-classifies units/individuals according to nominal row categories and ordered column categories. Denote the th element of by , for and and the th cell relative frequency as so that . Let the matrix of these values be denoted as and let be the th row marginal proportion of so that , and the diagonal matrix where the th cell entry is . Similarly, let be the th column marginal proportion so that , and the diagonal matrix where the th cell entry is . Define as the th row profile and the th element of , and the th column profile and the th element of .
For the th cell entry, Goodman [19] described the measure of the departure from independence for row and column by the Pearson ratio In matrix notation, the Pearson ratio is the th cell value of the matrix , where Independence between the rows and columns of will occur when , where is the unity matrix where all the values are equal to 1. One can examine where independence does not occur by identifying those Pearson ratios that are statistically significantly different from 1.
A more formal approach to determine whether there exists an association between the row and column categories involves decomposing the matrix of Pearson ratios, . For the correspondence analysis of , there are a variety of ways in which the decomposition can be performed. Here we will consider three methods of decomposition: singular value decomposition, bivariate moment decomposition, and hybrid decomposition. It is the consideration of the third approach here that is important for the method of correspondence analysis discussed in this paper. The use of hybrid decomposition relies on some basic knowledge of singular value decomposition and bivariate moment decomposition and so these will be described in the following subsections.
##### 2.2. Singular Value Decomposition
Classically, correspondence analysis involves decomposing the matrix of Pearson ratios using singular value decomposition (SVD) so that where and have the property respectively where is an identity matrix. Also, , where is the th largest singular value of , for .
For the decomposition of (2.3), is an matrix of left generalized basic vectors, while is a matrix of right generalized basic vectors. In both cases, and the first (trivial) singular vector of both matrices has all values equal to one. Let and be the matrices and , respectively, with the trivial singular vector from each is omitted. The matrix is an diagonal matrix where the th cell value is the th singular value, , of . These singular values have the property that they are arranged in descending order so that , where .
Suppose we omit the trivial column vector from and to give the matrix and the matrix , respectively. Also omit the first row and first column from the matrix (since the th element of is equal to 1), obtaining the matrix . Then the SVD of the Pearson ratio becomes the SVD of whose elements Goodman [19] referred to as Pearson contingencies.
The SVD of these contingencies leads to the Pearson chi-squared statistic being expressed in terms of the sum of squares of the singular values such that
##### 2.3. Bivariate Moment Decomposition
When a two-way contingency table consists of at least one ordered variable, the ordinal structure of the variable needs to be taken into consideration. Over the past few decades, there have been a number of correspondence analysis procedures developed that take into account the ordinal structure of the variables; see, for example, [47]. Generally, these procedures involve imposing ordinal constraints on the singular vectors. Such a procedure therefore forces the position of the points (along the first axis) of the plot to be ordered, thereby imposing what can sometimes lead to unrealistic “correspondences” between row and column categories. A way to overcome this problem is to consider using orthogonal polynomials rather than imposing constraints on the columns of and considered in the previous section.
For a doubly ordered two-way contingency table, the correspondence analysis approach of Beh [8] employs the bivariate moment decomposition (BMD) of Pearson ratios so that where
For the decomposition of (2.7), is an matrix of row orthogonal polynomials, while is a matrix of column orthogonal polynomials. The th element of may be calculated by considering the recurrence relation where for . These are based on the general recurrence relation of Emerson [20] and depend on the th score, , assigned to reflect the structure of the column variables. There are many different types of scores that can be considered and Beh [21] discusses the impact of using four different scoring types (two objectively and two subjectively chosen scores) on the orthogonal polynomials. However, for reasons of simplicity and interpretability, we will be considering the use of natural column scores , for , and natural row scores in this paper. For both and , the first column vector is trivial, having values equal to 1 so that and . It is also assumed that and , for all and .
The matrix is of size where the first row and column have values all equal to 1. The nontrivial elements of this matrix are referred to as bivariate moments, or generalized correlations, and describe linear and nonlinear sources of association between the two categorical variables. By omitting these trivial vectors, the decomposition of (2.7) becomes where and are the row and column orthogonal polynomials, respectively, with the first (trivial) column vector omitted. The matrix has elements which are the bivariate moments defined by
By considering the BMD (2.11), the Pearson chi-squared statistic can be partitioned into bivariate moments so that where the elements of are asymptotically standard normally distributed. Refer to Best and Rayner [22] and Rayner and Best [23] for a full interpretation of (2.12) and (2.13). An advantage of using BMD is that the th element of , has a clear and simple interpretation; it is the th bivariate moment between the categories of the row and column variables. As a result, Davy et al. [24] refer to these values as generalized correlations. For example, the linear-by-linear relationship can be measured by where and are the set of row and column scores used to construct the orthogonal polynomials, and and . The quantities and are similarly defined. By decomposing the Pearson ratios using BMD when natural scores are used to reflect the ordinal structure of both variables, is equivalent to Pearson's product moment correlation; see Rayner and Best [23]. One can also determine the mean (location) and spread (dispersion) of each of the nonordered row categories across the ordered column categories by calculating so that and , respectively.
##### 2.4. Hybrid Decomposition
Another type of decomposition, and one that was briefly discussed by Beh [12], is what is referred to as hybrid decomposition (HD). For a singly ordered contingency table, hybrid decomposition takes into account the ordered variable and nominal variable by incorporating singular vectors from SVD and orthogonal polynomials from BMD such that the Pearson contingencies are decomposed by The matrix of (2.15) is defined as The matrix of values, , can be derived by premultiplying (2.15) by and postmultiplying it by .
If one considers the decomposition of the matrix of Pearson contingencies using the hybrid decomposition of (2.15), then the partition of the Pearson chi-squared statistic can be expressed in terms of the sum of squares of the so that where the elements of are asymptotically standard normal and independent. Refer to Beh [11] for more details on (2.16) and (2.17).
The effect of the column location component on the two-way association in the contingency table is measured by , while, in general, the th-order column component is . The significance of these components can be compared with the chi-squared with degrees of freedom. Testing these column components allows for an examination of the trend of the column categories, the trend being dictated by the th orthogonal polynomial. For example, the column location component determines if there is any difference in the mean values of the column categories, while the column dispersion component detects if there is any difference in the spread of the columns.
The first-order row location component on the two-way association in the contingency table is measured by , while in general, the th-order row component value is equivalent to . The row location component quantifies the variation in the row categories due to the mean difference in the row categories. Similarly, the row dispersion component quantifies the amount of variation that is due to the spread in the row categories. Refer to Section 6 for more informative details on the row components.
Partitions of other measures of association using orthogonal polynomials have also been considered. D'Ambra et al. [25] considered the partition of the Goodman-Kruskal tau index. For symmetrically associated multiple categorical random variables, Beh and Davy [26, 27] considered the partition of the Pearson chi-squared statistic, while for asymmetrically associated variables Beh et al. [28] considered the partition of the Marcotorchino index [29]. However, the application of extensions to hybrid decomposition will not be considered here.
#### 3. Profile Coordinates
One system of coordinates that could be used to visualize the association between the row and column categories is to plot along the th axis for the th row and for the th-ordered column. Such coordinates are referred to as standard coordinates. These are analogous to the set of standard coordinates considered by Greenacre [14, page 93].
However, standard coordinates infer that each of the axes is given an equal weight of 1. Thus, while the difference within the row or column variables can be described by the difference between the points, they will not graphically depict the association between the rows and columns. Therefore, alternative plotting systems should be considered.
Analogous to the derivation of profile coordinates in Beh [8] using BMD, the row and column profile coordinates for singly ordered correspondence analysis are defined by respectively. Therefore, by including the correlation quantities, the coordinates (3.1) and (3.2) will graphically depict the linear and nonlinear associations that may exist between the ordered column and nominal row categories.
The relationship between the row (and column) profile coordinates and the Pearson chi-squared statistic can be shown to be by substituting the elements of and into (2.17). However, instead of using the Pearson chi-squared statistic as a measure of association in a contingency table, correspondence analysis considers instead , referred to as the total inertia. By adopting as the measure of association, (3.3) shows that when the profile coordinates are situated close to the origin of the correspondence plot, will be relatively small. Thus the hypothesis of independence between the rows and columns will be strong. Profile coordinates far from the origin indicate that the total inertia will be relatively large and the independence hypothesis becomes weak. These conclusions may also be verified by considering the Euclidean distance of a profile coordinate from the origin and other profile coordinates in the correspondence plot; refer to Section 4 for more details.
#### 4. Distances
##### 4.1. Distance from the Origin
Consider the th row profile. The squared Euclidean distance of this profile from the origin is It can be shown that by expressing this in terms of Pearson contingencies, and using (2.15) and (2.4), this distance may be expressed in terms of the sum of squares of the th row profile coordinate such that where is the th element of . By substituting (4.2) into (3.3), the Pearson chi-squared statistic can be expressed as Therefore, row profile coordinates close to the origin support the hypothesis of independence, while those situated far from the origin support its rejection. It can be shown in a similar manner that where is the squared Euclidean distance of the th column profile from the origin and is the th element of (3.2).
##### 4.2. Within Variable Distances
The squared Euclidean distance between two row profile coordinates, and , can be measured by
By considering the definition of the row profile coordinates given by (3.1), the squared Euclidean distance between these two profiles can be alternatively be written as
Therefore, if two row profile coordinates have similar profile, their position in the correspondence plot will be very similar. This distance measure also shows that if two row categories have different profiles, then the position of their coordinates in the correspondence plot will lie at a distance from one another.
Similarly, the squared Euclidean distance between two column profiles, and , can be measured by
These results verify the property of distributional equivalence as stated by Lebart et al. [16, page 35], a necessary property for the meaningful interpretation of the distance of profiles in a correspondence plot. (1) If two profiles having identical profiles are aggregated, then the distance between them remains unchanged.(2) If two profiles having identical distribution profiles are aggregated, then the distance between them remains unchanged.
The interpretation of the distance between a particular row profile coordinate and a column profile coordinate is a contentious one and an issue that will not be described here, although a brief account is given by Beh [12, page 269].
#### 5. Transition Formula
For the classical approach to correspondence analysis, transition formulae allow for the profile coordinates of one variable to be calculated when the profile coordinates of a second variable are known.
To derive the transition formulae for a contingency table with ordered columns and nonordered rows, postmultiply the left- and right-hand sides of (3.1) by . Doing so leads to upon substituting (2.16) and (3.2). Based on the orthogonality properties (2.4) and (2.8), the transition formula becomes The transition formula (5.2) allows for the row profile coordinates to be calculated when the column profile coordinates are known.
In a similar manner, it can be shown that
Beh [30] provided a description of the transition formulae obtained for a doubly ordered correspondence analysis and the configuration of the points in the correspondence plot. For singly ordered correspondence analysis, similar descriptions can be obtained and are summarized in the following propositions. (i) If the positions of the row profile coordinates are dominated by the first principal axis, then .(ii) If the positions of the row profile coordinates are dominated by the second principal axis, then .(iii) If the position of the column profile coordinates are dominated by the first principal axis, then .(iv)If the positions of the column profile coordinates are dominated by the second principal axis, then .
However, it is still possible that and/or will be zero if none of the row and column profile coordinates lie along a particular axis. For such a case, it is not possible to determine when this will happen.
For both classical and doubly ordered correspondence analysis, when either the row or column profile positions is situated close to the origin of the correspondence plot, then there is no association between the rows and columns. This is indeed the case too for singly ordered correspondence analysis as indicated by (3.3). The items summarized above show that, in this case, and . It can also be shown that and .
#### 6. Properties
The results above show that the mathematics and characteristics of this approach to singly ordered correspondence analysis are very similar to doubly ordered correspondence analysis and classical simple correspondence analysis. However, there are properties of the singly ordered approach that distinguish it from the other two techniques. This section provides an account of these properties.
Property 1. The row component associated with the th principal axis is equivalent to the square of the th largest singular value.
To show this, recall that the total inertia may be written in terms of bivariate moments and in terms of the eigenvalues such that which can be obtained by equating the Pearson chi-squared partitions of (2.6) and (2.17). Therefore, the square of the th singular value can be expressed by where the right-hand side of (6.2) is just the th-order row component. For example, the square of the largest singular value may be partitioned so that Therefore, the singly ordered correspondence approach using the hybrid decomposition of (2.16) and (2.17) allows for a partition of the singular values of the Pearson contingencies into components that reflect variation in the row categories in terms of location, dispersion, and higher-order moments. That is, each singular value can be partitioned so that information associated with differences in the mean and spread of the row profiles can be identified. Higher-order moments can also be determined from such a partition.
Property 2. The row component values are arranged in descending order.
This property follows directly from Property 1. Since the eigenvalues are arranged in a descending order, so too are the row components.
Property 3. A singly ordered correspondence analysis allows for the inertia associated with a particular axis of a simple correspondence plot (called the principal inertia) to be partitioned in bivariate moments.
Again, this property follows directly from Property 1, where the principal inertia of the th axis is the sum of squares of the bivariate moments when .
Property 4. It is possible to identify which bivariate moment contributes the most to a particular squared singular value and hence its associated principal axis.
This is readily seen from Property 3.
For classical correspondence analysis, the axes are constructed so that the first axis accounts for most of the information in variation in the categories, the second axis describes accounts for the second most amount of variation, and so on. However, it is unclear what this variation is, or whether it is easily identified as being statistically significant. By considering the partition of the singular values, as described by (6.2), the user is able to isolate important bivariate moments that include variation in terms of location, dispersion, and higher-order components for each principal axis. Therefore, there is more information that is able to be obtained from the axes of the correspondence plot, and the proximity of the points on it, than from a classical correspondence plot.
#### 7. Example
Consider the contingency table given by Table 1 which was originally seen in Calimlin et al. [18] and analyzed by Beh [11]. The study was aimed at testing four analgesic drugs (randomly assigned the labels A, B, C, and D) and their effect on 121 hospital patients. The patients were given a five-point scale consisting of the categories poor, fair, good, very good, and excellent on which to make their decision.
Table 1: Cross-classification of 121 hospital patients according to analgesic drug and its effect.
If only a comparison of the drugs, in terms of the mean value and spread across the different levels of effectiveness, was of interest, attention would be focused on the quantities (and ). These values for Drug A, Drug B, Drug C, and Drug D are 3.3000 (1.2949), 3.6129 (1.4740), 2.2581 (1.0149), and 2.2069 (0.9606), respectively and were calculated using natural scores for the column categories. Therefore, based on these quantities, it is clear that Drug A and Drug B are very similar in terms of the two components across the different levels of effectiveness. Therefore, these two drugs have a similar effect on the patients. Also, these drugs are different to Drug C and Drug D which are themselves quite similar in effectiveness. However, the association between the Drugs and the different levels of effectiveness is not evident from such measures. This is why correspondence analysis is a suitable analytical tool to graphically depict and summarize the association. It can be seen that Table 1 consists of ordered column categories and nonordered row categories. Therefore, singly ordered correspondence analysis will be used to analyze the effectiveness of the drugs.
The Pearson chi-squared statistic of Table 1 is 47.0712, and with a zero -value, it is highly statistically significant. Therefore, with a total inertia of 0.3890, there is a significant association between the drugs used and their effect on the patients.
When a classical correspondence analysis is applied, the squared singular values are , , and and the two-dimensional correspondence plot is given by Figure 1. Here, the first principal axis accounts for 0.30467/0.3890 of the total association between the two variables, and the second axis accounts for 19.9%. Therefore, the two-dimensional plot of Figure 1 graphically depicts 98.2% of the association that exists between the analgesic drug being tested and its level of effectiveness.
Figure 1: Classical correspondence plot of Table 1.
Figure 1 shows a clear association between the analgesic drug being tested and the effectiveness of that drug. Drug B appears to have an “excellent” effect on the patients that participated in the study, Drug A was rated as “very good,” Drug D was deemed only “fair” in its effectiveness and Drug C was judged “good” to “poor.” These conclusions are also apparent when eyeballing the cell frequencies of Table 1. However, it is unclear how the profile of each of the four drugs is different, or where they may be similar. By adopting the methodology above, we can determine how these comparisons may be made in terms of differences in location, dispersion, and higher-order components.
The component values that are associated with explaining the variation in the position of the drug coordinates in Figure 2 are , , , and . Therefore, Figure 2 is constructed using the first (linear) principal axis with a principal inertia value of , and the second (dispersion) principal axis with a principal inertia value of , for the four drugs. Together, these two axes contribute to 75% of the variation of the drugs tested, compared with 98.2% of the variation in the patients judgement of the drug. The third (cubic) component contributes to 18.7% of this variation.
Figure 2: Singly ordered correspondence plot of Table 1.
Applying singly ordered correspondence analysis yields and . Also, and . Therefore, by considering (6.3), we can see that That is, the dominant source of the first (squared) singular value is due to the linear component of the effectiveness of the drugs. Thus, the location component best describes the variation of the profiles for the drug effectiveness levels along the first principal axis of Figure 1 (68.4%).
Figure 2 shows the variation of these drugs in terms of the linear and quadratic components. While Figure 1 indicates that the effectiveness of Drug C and Drug D is different, Figure 2 shows that the positions of Drug C and Drug D are similar across the column responses. This is because the variation between the two drugs exists at moments higher than the dispersion. It is also evident from Figure 2 that these two drugs have quite a different effect than do Drug A and Drug B, which in themselves are different. These conclusions are in agreement with the comments made earlier in the example. Figure 2 also shows that by taking into account the ordinal nature of the column categories, the variation between the drug effectiveness levels may be explored. For example, “good” and “poor” share the very similar first principal coordinate. However, there is slightly more variation (across the drugs) for “good” than there is for “poor.”
An important feature of Figure 2 is that it depicts the association between the drugs and the levels of effectiveness. It can be seen from Figure 2, just as Figure 1 concluded, that Drug A and Drug B are more effective in treating pain relief than Drug C and Drug D. However, because of the use of hybrid decomposition, the position of the drug profile coordinates have changed. Figure 1 concluded that Drug D was rated as “fair.” This is primarily due to the relatively large cell frequency (with a value of 12) that the two categories share; this feature is a common characteristic of classical correspondence analysis. However, since the drug behaves in a similar manner (in terms of location and spread) when compared with Drug C, its position has shifted to the bottom right quadrant of the plot. Therefore, Drug D is associated more with “poor” and “good” when focusing on these components of the category.
By observing the distance of each category from the origin in Figure 2, Drug B is the furthest away from the origin and so is less likely than the other drugs to contribute to the independence between the drugs and the patients effect. This is because Drug B contributes more to the row location component (38.29%) than any of the other three drugs in the study, while contributing to 67.79% of the variation in the dispersion component. Further results on the dominance of the drugs to each of the axes in Figure 2 are summarized in Table 2. It shows the contribution, and relative contribution of each drug to each of the two axes. Table 3 provides a similar summary, but for the different effectiveness levels of the drugs.
Table 2: Contribution of the drugs tested to each axis of Figure 2.
Table 3: Contribution of the effectiveness of the drugs tested to each axis of Figure 2.
Recall that Drug C and Drug D are positioned close to one another in Figure 2. Table 2 shows that they contribute roughly the same to the location and dispersion components. Figure 2 also shows that “excellent” is the most dominant of the drug effectiveness categories along the first principal axis and this is reflected in Table 3, accounting for nearly half (46.08%) of the principal inertia for its variable. The second principal axis is dominated by the category “fair” which contributes to 46.23% of the second principal inertia.
#### 8. Discussion
Correspondence analysis has become a very popular method for analyzing categorical data, and has been shown to be applicable in a large number of disciplines. It has long been applied in the analysis of ecological disciplines, and recently in health care and nursing studies [31, 32], environmental management [33], and linguistics [34, 35]. It also has developed into an analytic tool which can handle many data structures of different types such as ranked data [30], time series data [36], and cohort data [37].
The aim of this paper has been to discuss new developments of correspondence analysis for the application to singly ordered two-way contingency tables. Applications of the classical approach to correspondence analysis can be made, although the ordered structure of the variables is often not always reflected in the output. When a two-way table consists of one ordered variable, such as in sociological or health studies where responses are rated according to a Likert scale, the ordinal structure of this variable needs to be considered. The singly ordered correspondence analysis procedure developed by Beh [8] is applicable to singly ordered contingency tables. However, due to the nature of this procedure, only a visualization of the association between the categories of the nonordered variable can be made. Therefore, any between-variable interpretation is not possible. The technique developed in this paper improves upon this singly ordered approach by allowing for the simultaneous representation of the ordered column and nonordered row categories.
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# ch06t.exe - 0.1 Exercises for Chapter 6 1. The financial...
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Unformatted text preview: 0.1 Exercises for Chapter 6 1. The financial manager of a large department store chain selected a random sample of 200 of its credit card customers and found that 136 had incurred an interest charge during the previous year because of an unpaid balance. (a) What do you think is the parameter of interest in this study? (b) Calculate a point estimate for the parameter of interest listed in part (a). 2. Suppose a person is selected at random and is given a hand-writing-recognition-equipped Apple Newton computer to try out. The Apple Newton is fresh from the manufacturer, and the person is asked to write “I think Apple Newton’s are neat”. But often the Apple Newton mis-interprets the words and types back something quite different (we needn’t go into all the possibilities here). A computer magazine conducts a study to estimate the probability, p , that, for a randomly selected owner of a hand-writing-recognition-equipped Apple Newton computer, the phrase “I think Apple Newton’s are neat” will be mis-interpreted. Suppose that they will observe X mistakes in n tries ( n owners). (a) What is your estimator for p ? What properties does this estimator have? Are there any general principles that support the use of this estimator? (b) What is your estimate ˆ p if the magazine finds 33 mistakes in 50 tries? (c) Now suppose that the magazine’s pollster recommends that they also estimate the stan- dard error of the estimator ˆ p . For the given data, what is your estimate of the standard error, and how do you justify this choice? 3. A food processing company is considering the marketing of a new product. Among 40 ran- domly chosen consumers 9 said that they would purchase the new product and give it a try. Estimate the true proportion of potential buyers, and state its standard error. 4. Let X 1 ,...,X n be a sample from U (0 ,θ ). Find the moments estimator of θ . 5. For a sample of 6 jugs of 2 % lowfat milk produced by “Happy Cow Dairy” the fat content X i has been determined as: 2 . 08 2 . 10 1 . 81 1 . 98 1 . 91 2 . 06 (a) Making no assumptions on the distribution of the fat content, estimate the proportion of milk jugs having a fat content of 2.05 % or more. (b) Making no assumptions on the distribution of the fat content, estimate the mean fat content of “Happy Cow Dairy” 2 % lowfat milk. (c) Making no assumptions on the distribution of the fat content, (i) your estimator in part 5a is (circle all correct statements): Unbiased Maximum likelihood estimator Moments estimator (ii) your estimator in part 5b is (circle all correct statements): Unbiased Maximum likelihood estimator Moments estimator (d) Assuming normality, the maximum likelihood estimators of the mean and variance are given by X and [( n- 1) /n ] s 2 , respectively. Calculate the maximum likelihood estimator of the proportion of milk jugs having a fat content of 2.05 % or more. [Hints: a) P ( X > x ) = 1- Φ(( x-...
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ch06t.exe - 0.1 Exercises for Chapter 6 1. The financial...
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Ask a homework question - tutors are online | 874 | 3,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-17 | longest | en | 0.91641 |
https://math.stackexchange.com/questions/2298645/a-hard-geometry-problem-on-circle | 1,643,433,721,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00379.warc.gz | 427,174,710 | 34,305 | # A Hard Geometry Problem on circle
$$\angle(ABC) = 30°\\ \angle(BCO) = 20°\\ \angle(OCD) = 20°$$
How do i find $$\angle(ODC)$$? so i wanted to show my teacher this but he is not available yet. Can someone help me to solve? geometry problems on circle seems hard to me. Thanks!
• Just a hint: $OC=OB$. Not a solution but a good point to start.
– N74
May 27 '17 at 11:25
• I can't solve this problem, needs help from someone. May 27 '17 at 11:36
• What methods are you expected to use. The sine rule, perhaps? May 27 '17 at 11:37
• I think sine rule better so i use sine rule. May 27 '17 at 11:42
• Anyone helps me? May 27 '17 at 11:44
Without loss of generality, we can consider $R=OC=OB=1$.
From $\Delta OBC: \frac{OB}{\sin20}=\frac{BC}{\sin140}\Rightarrow BC=\frac{\sin140}{\sin20}=2\cos20.$
From $\Delta BCD: \frac{BC}{\sin110}=\frac{CD}{\sin30}\Rightarrow CD=\frac{\cos20}{\sin110}=1.$
Finally from $\Delta OCD: \frac{OC}{\sin{x}}=\frac{CD}{\sin{(x+20)}}\Rightarrow \sin x=\sin{(x+20)}\Rightarrow x=80.$
• where does the 140 comes from ?
– zwim
May 27 '17 at 12:16
• it is the angle opposite to the side BC and $\Delta OBC$ is isosceles... May 27 '17 at 12:20
• OK. From $CD=1$ you can conclude the same way then, $x=(180-20)/2$ in isocele CDO, simpler than last line.
– zwim
May 27 '17 at 12:26
• @zwim (+1) Thank you. I just wanted to keep the rhyme. May 27 '17 at 12:34
Join OA and AC
angle AOC = 2xangle ABC=60 deg (center angle and circumference angle)
Triangle OAC is equilateral
AC=OA=OC
angle CAB =180-30-80=70 deg
angle CDA=40+30=70 deg
Therefore CA=CD=CO
ANGLE ODC=angle BAC+angle ABC
Angle ODC=(180-100)=80°
● ● ● ANGLE ODC IS 80°
Hint...with a bit of angle-chasing you should be able to establish $\angle ADC=70$.
You can then use the sine rule in triangles $ODB$ and $ODC$ (assume the radius is $1$)
The final answer seems to be $x=80$ which would suggest there must be a better way.
• Ah, I didn't understand it. can you use some numerical? because i don't un derstand knowns May 27 '17 at 11:47
• Can you get $\angle ADC=70$? You also need to employ the sine rule and also the compound angle formula for $\sin(A+B)$. On the pther hand, as I said, there may be a better way because the answer is exactly $80$... May 27 '17 at 12:03
$$/angle(ABC)=30°// /angle(AOC)=2*angle(ABC)=60°//$$ So AOC is equilateral triangle (AO=AC=CO=R) $$/angle(CAB)=/angle(COB)/2=70° //angle(ACD)=70°$$ So CD=CB=R and CDO is isosceles triangle so $$/Angle(CDO)=angle(DOC) Angle(CDO)=90°-Angle(DCO)/2=160°/2=80°$$
Produce CD and meet the circle at E Join OA and AC angle APC = 60 deg (centre and circumference angles) OA = AC = OC Join BE angle CAB = angle CEB = 70 deg (CE = CB and angle BCE = 40deg) Therefore AD = CD = CD x = 80 deg
• @JoséCarlosSantos: Let us not exaggerate: whay would $AB$ be better than AB? MathJax is meant for complicated formulae that would be ambiguous or difficult to read as plain text. Let us not transform it into an obligation. Nov 8 '17 at 14:38
• @AlexM. I think that $\angle CAB=\angle CEB=70^\circ$ is more pleasent to read than angle CAB = angle CEB = 70 deg. Nov 8 '17 at 14:42
BCD is equal to 40 (OCD plus BCO). So the unknown corner of the triangle is 180 - (40 + 30) which is 110. ADC, CDB and ODB are supplementary, so ADC is 70. COB is 180 - (20 + 15) which is 155. I'm pretty sure DOB is equal to COB, so 155 + 155 equals 310 therefore COD should be 50? 180 - (20 + 50) is 110 so x is 110. I'm pretty sure that's the answer, however I'm only a Year Seven, that happens to love Geometry and I've just started using this. Basically I'm saying, don't think my word is final. Hope it helps anyways. :) please tell me if I'm not right, I would like to know to correct answer if I've gotten it wrong.
• It's wrong, it seems $x = 80$ on my answerkey May 27 '17 at 11:42
• As I said, I'm only in Year Seven. Thanks for letting me know though! May 27 '17 at 11:46
• its dw! thank you again. May 27 '17 at 11:46
• The segment $OB$ doesn't bisect $\hat B$ so $COB$ is not $180-(20+15)$. Anyway the triangle $BCO$ is isosceles, so $OBC=OCB$
– N74
May 27 '17 at 11:47
First join $AC$ and $AO$. Now, $\angle ABC$ is $30^\circ$,so AOC is 60*.AO=OC.so,OAC=OCA=(180*-60*)÷2=60*. Triangle AOC is equilateral Triangle.so,AC=OC=OA.angleACD=(ACO-DCO)=(60*-20*)=40*.In triangle ACB ,CAB=70*,ACD=40*.So,ADC=70*.So,AC=CD. Now AC=CO=CD,So,OC=CD.in triangle CDO,DCO=20*,and OC=CD.So,CDO=COD=(180*-20*)÷2=80*.HENCE,CDO=80*.(ANS.). | 1,549 | 4,469 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-05 | latest | en | 0.853941 |
https://www.analystforum.com/t/fixed-income-is-hard/107304 | 1,653,145,286,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539131.21/warc/CC-MAIN-20220521143241-20220521173241-00378.warc.gz | 710,304,866 | 7,165 | # Fixed income is hard
Anyone else find fixed income most challenging topic in L2? I mean swaps, pension accounting etc are also tough but they require straightforward application of the formulas. This one is tough to conceptualize and the questions are the most difficult.
I do. It’s probably my weakest topic. I don’t know why but I really struggle with the term structure concepts. Given the extra exam weight that was added to FI this year I have to give it a lot more attention.
That, plus the spreads, plus the binomial tree for embedded options and of course the duration; The fact that it is called duration but it is not a measure of time is the greatest mystery.
I actually didn’t do any problem solving for FI . but we covered all the topics and I found it easier than level 1.
less formulas to memorise and only a few applications to learn once you get around doing some problem solving.
Oddly enough, I find the process of working through a binomial tree with emedded options fairly straightforward, and had that down pretty well for the exam last year. It’s the deeper interest rate and yield curve concepts that I have trouble wrapping my head around, and I think having a strong conceptual understanding is key for the exam.
I’m preparing for the liklihood that FI will be with 15% of the exam this year so I’ll be doing everything I can to get as strong as possible in it.
I find the questions are not too terrible, with practice. The binomial trees make sense, though i need a cheat sheet handy to remember how to make them. But I agree – it’s the only topic so far where I cannot quite picture what the heck is happening in the “real world”. I’ve watched a tonne of videos on par curves, yield curves, etc. Usually, getting good enough at the EOC and blue box questions makes the “real world” meaning come to life … eventually.
I need like the Steven Hawking of bonds to team up with the Far Side guy and make a bunch of comics about convexity.
Fixed Income was challenging to me in a first half, too. I was pretty sure that I did not solve it well and was supprised by >70 % sucess in this area. I will apply special attention again on this topic and hope the same outcome.
Fixed income was one of my better topics so far. I remember swaps being a pain to understand on L1, also convexity, but luckily both of those topics stuck with me. I’ve always been interested in calls, puts, and moving money through time, so the binomial trees were really fun IMO.
I struggle most with quant and Econ. None of it is intuitive to me, but I can figure out the calculations.
You’re thinking of Jeff Gundlach… or Bill Gross. Or Matt Tucker from BlackRock.
Macaulay duration is 100% a measure of time. It is the money weighted average of cash flows, and is measured in years to show this.
Modified duration is the more practical application of this and shows the bonds sensitivity to interest rate changes, but that is why you use Macaulay duration to calculate duration gap (as both Macaulay duration & investors time horizon are measured in years)
Note that although CFA Institute has decided to drop the units from the curriculum, all durations (Macaulay, modified, effective, spread, key rate, whatever) have units of years.
Can anyone give me some advice on how to handle the fixed income?
Depends… what do you find hard about it?
I find Fixed Income and Derivatives the easiest sections of L2…
And corporate finance and econ to be the toughest… Maybe everyone is a lil different depending on background and skill.
How did you study? just read the material from schweser!
Practice, practice and practice. After many rounds of practicing and getting well in calculation, ya’ll be pretty surprised that conceptual part of FI might be the hardest part. Then, prepare hand written notes and go through this notes many many times.
I had no much experience with fixed income in a real life but succeed to solve Fixed income on all 3 levels including morning session of L3 with above 70 %. I’d suggest use Schweser rather than official readings but practice solely from official EOCs, TTs and Mocks.
As with all other topics, pay special attention on weak areas. Given the Murphy’s Law, those what you didn’t prepare well will probably appear on D-Day.
Thanks ! | 923 | 4,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-21 | latest | en | 0.957173 |
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Questions 5-6
Sales manager: Last year the total number of meals sold in our company restaurants was much higher than it was the year before. Obviously consumers find our meals desirable.
Accountant: If you look at individual restaurants, however, you find that the number of meals sold actually decreased substantially at every one of our restaurants that was in operation both last year and the year before. The desirability of our meals to consumers has clearly decreased, given that this group of restaurants---the only ones for which we have sales figures that permit a comparison between last year and the year before---demonstrates a trend toward fewer sales.
5. If the sales figures cited by the accountant and the sales manager are both accurate, which one of the following must be true?
(A) The company opened at least one new restaurant in the last two years.
(B) The company’s meals are less competitive than they once were.
(C) The quality of the company’s meals has not improved over the last two years.
(D) The prices of the company’s meals have changed over the past two years.
(E) The market share captured by the company’s restaurants fell last year.
6. Which one of the following, if true, most seriously calls into question the accountant’s argument?
(A) The company’s restaurants last year dropped from their menus most of the new dishes that had been introduced the year before.
(B) Prior to last year there was an overall downward trend in the company’s sales.
(C) Those of the company’s restaurants that did increase their sales last year did not offer large discounts on prices to attract customers.
(D) Sales of the company’s most expensive meal contributed little to the overall two-year sales increase.
(E) Most of the company’s restaurants that were in operation throughout both last year and the year before are located in areas where residents experienced a severe overall decline in income last year.
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21 Jan 2006, 23:59
1 A
2 D
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22 Jan 2006, 02:40
I would go with:
1. A - In order that total number of meals increased but each individual restaurant had a decrease in sales, the only reason could be increase in the number of restaurants
2. B (by POE)- Not sure.
Since the data is only available for the last two years, and the conclusion is about desirability of the meals, there could be other reasons for reduction in sales. If there was already a downward trend prior to last year, that would point to a different reason than the desirability of meals.
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22 Jan 2006, 07:21
1. A
2. E. A and C are under the same umbrella of "less desirability of our meals to consumers", so they do not weaken the accountant's argument. B is OOS, so is D.
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22 Jan 2006, 07:56
E for the second question.
The limited income prevented cusotmers from the consumption.
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22 Jan 2006, 12:13
I go for A n B ,
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22 Jan 2006, 21:55
1)D
2)C
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22 Jan 2006, 22:38
For Question 5 I go with A
Lets say the years are 2005 and 2004
(A) The company opened at least one new restaurant in the last two years.
This is the only possibility since the sales manager says that sales for 2005>sales for 2004
And the accountant says that for each restaurant that existed sales for 2005<sales for 2004.
For both to be true the only option is when there was no sales for some restaurants in 2004 but was in 2005 which are new restaurants.
(B) The company’s meals are less competitive than they once were.
Scope outside the para
(C) The quality of the company’s meals has not improved over the last two years.
Still does not explain how the two differing statements can be true.
(D) The prices of the company’s meals have changed over the past two years.
The para has nothing to do with the revenue but only with the actual number of meals served. Hence we cannot conclude this.
(E) The market share captured by the company’s restaurants fell last year.
We dont know about the competitors so cannot conclude.
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28 Jan 2006, 07:55
OAs are:
5. A
6. E
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28 Jan 2006, 08:40
I like to think of these questions from an economic perspective:
5. A Poaching- New stores are cutting inro sales of existing stores
6.E Accountant's Assertion- Meals are now less desirable
Strongest Counterargument- Meals are still be desirable but lack of income prevents consumers from translating this into demand
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28 Jan 2006, 08:50
GMATT73 wrote:
OAs are:
5. A
6. E
can u post the OE for 6E ???????
can any1 explain why 6E ??? and not 6D??????
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28 Jan 2006, 08:52
Hjort wrote:
I like to think of these questions from an economic perspective:
5. A Poaching- New stores are cutting inro sales of existing stores
6.E Accountant's Assertion- Meals are now less desirable
Strongest Counterargument- Meals are still be desirable but lack of income prevents consumers from translating this into demand
Thanx a lot Hjort......makes sense !!!.....
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28 Jan 2006, 17:21
I also got A and D.
28 Jan 2006, 17:21
Similar topics Replies Last post
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Topics:
1 Sales manager: Last year the total number of meals sold in our company 1 16 Feb 2017, 16:09
2 Manager: Last year, within the sales division Source: Powerscore 1 04 Jun 2015, 03:01
29 Last year, to improve sales, a software company 8 01 May 2015, 16:27
7 The CFO attributes our increased sales last year to the new 3 01 Jan 2014, 11:52
5 Sales manager: Last year the total number of meals sold in 5 20 Feb 2012, 02:29
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opentextbc.ca/collegealgebraopenstax/chapter/exponential-functions
An exponential function is defined as a function with a positive constant other than raised to a variable exponent. See (Figure) . A function is evaluated by solving at a specific value.
mathworld.wolfram.com/ExponentialFunction.html
The exponential function is the entire function defined by exp(z)=e^z, (1) where e is the solution of the equation int_1^xdt/t so that e=x=2.718.... exp(z) is also the unique solution of the equation df/dz=f(z) with f(0)=1. The exponential function is implemented in the Wolfram Language as Exp[z]. It satisfies the identity exp(x+y)=exp(x)exp(y).
tutorial.math.lamar.edu/Classes/Alg/ExpFunctions.aspx
Function evaluation with exponential functions works in exactly the same manner that all function evaluation has worked to this point. Whatever is in the parenthesis on the left we substitute into all the \(x\)’s on the right side.
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A linear function is graphed as a straight line and contains one independent variable and one dependent variable, whereas an exponential function has a rapid increase or decrease along a curved line in a graph. Linear functions, or equations, take the form "y = a + bx," in which "x" is the dependent variable that changes with the value of "b."
www.math.utah.edu/~wortman/1050-text-ef.pdf
an exponential function that is defined as f(x)=ax. For example, f(x)=3x is an exponential function, and g(x)=(4 17) x is an exponential function. There is a big di↵erence between an exponential function and a polynomial. The function p(x)=x3 is a polynomial. Here the “variable”, x, is being raised to some constant power.
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1. Definitions: Exponential and Logarithmic Functions. by M. Bourne. Exponential Functions. Exponential functions have the form: `f(x) = b^x` where b is the base and x is the exponent (or power).. If b is greater than `1`, the function continuously increases in value as x increases. A special property of exponential functions is that the slope of the function also continuously increases as x ...
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# cannot mix aggregate and non-aggregate arguments with this function
Hi guys,
I have a little challenge to get the formula below working: Basically I would like to check whether conditions of material status [MS] are met and if so Tableau should return calculations shown below.
However this does not work as I obviously mix aggregated and non-aggregated arguments (see subject header)
I am wondering how I can overcome this issue. The check of material status [MS] is mandatory before returning values as otherwise the outcome will be not correct.
Is there any simple solution to solve those issues?
Br
seb
IF [MS]<>"BI" and [MS] <> "BO"
then [Correct Stock AS]/([Total GI QTY]/-5)
ELSEIF [MS] = "BO"
then ([Correct Stock AS]/([Total GI QTY -3 years]*-1))
END
• ###### 1. Re: cannot mix aggregate and non-aggregate arguments with this function
Hi Sebastian,
Just use attribute. example:
IF ATTR([State]) <> 'Alabama'
THEN SUM([Profit])/SUM([Quantity])
END
Trust this helps.
D
1 of 1 people found this helpful
• ###### 2. Re: cannot mix aggregate and non-aggregate arguments with this function
Good morning -
What tableau is telling you is that you created an aggregation in one of the values you put in the formula and you will need to aggregate the other variables to make the calculation valid
Aggregation functions are not limited to sum() but include Min(),Max, count(), countd(), avg() , attr() and others
Not knowing where the aggregate is in your function I would suggest
IF Attr([MS])<>"BI" and attr([MS]) <> "BO"
then ( sum([Correct Stock AS])/([Total GI QTY]/-5)
ELSEIF attr( [MS]) = "BO"
then (sum([Correct Stock AS])/([Total GI QTY -3 years]*-1))
END
but that is assuming that [Total GI QTY and Total GI QTY -3 years] are the aggregates
Jim
If this posts assists in resolving the question, please mark it helpful or as the 'correct answer' if it resolves the question. This will help other users find the same answer/resolution. Thank you.
• ###### 3. Re: cannot mix aggregate and non-aggregate arguments with this function
Sebastian,
Because you did not attach workbook, so I don't know if these measures: Correct Stock AS, Total GI QTY, Total GI QTY -3 years, are aggregated or non-aggregated. If they are aggregated, then you need to add ATTR before dimension MS.
Hope it helps.
Michael Ye
• ###### 4. Re: cannot mix aggregate and non-aggregate arguments with this function
Hi,
I really appreciate your input and your support. It solved my challenge.
Many thanks
Have a great day
Br
Sebastian
• ###### 5. Re: cannot mix aggregate and non-aggregate arguments with this function
Thanks
Jim
• ###### 6. Re: cannot mix aggregate and non-aggregate arguments with this function
You are welcome! | 687 | 2,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-13 | latest | en | 0.867096 |
https://number.academy/213033 | 1,660,317,011,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571719.48/warc/CC-MAIN-20220812140019-20220812170019-00116.warc.gz | 397,888,389 | 12,068 | # Number 213033
Number 213,033 spell 🔊, write in words: two hundred and thirteen thousand and thirty-three . Ordinal number 213033th is said 🔊 and write: two hundred and thirteen thousand and thirty-third. Color #213033. The meaning of number 213033 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 213033. What is 213033 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 213033.
## What is 213,033 in other units
The decimal (Arabic) number 213033 converted to a Roman number is (C)(C)(X)MMMXXXIII. Roman and decimal number conversions.
#### Weight conversion
213033 kilograms (kg) = 469652.6 pounds (lbs)
213033 pounds (lbs) = 96631.1 kilograms (kg)
#### Length conversion
213033 kilometers (km) equals to 132373 miles (mi).
213033 miles (mi) equals to 342844 kilometers (km).
213033 meters (m) equals to 698919 feet (ft).
213033 feet (ft) equals 64934 meters (m).
213033 centimeters (cm) equals to 83871.3 inches (in).
213033 inches (in) equals to 541103.8 centimeters (cm).
#### Temperature conversion
213033° Fahrenheit (°F) equals to 118333.9° Celsius (°C)
213033° Celsius (°C) equals to 383491.4° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
213033 seconds equals to 2 days, 11 hours, 10 minutes, 33 seconds
213033 minutes equals to 5 months, 1 week, 22 hours, 33 minutes
### Codes and images of the number 213033
Number 213033 morse code: ..--- .---- ...-- ----- ...-- ...--
Sign language for number 213033:
Number 213033 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 213033
### Multiplications
#### Multiplication table of 213033
213033 multiplied by two equals 426066 (213033 x 2 = 426066).
213033 multiplied by three equals 639099 (213033 x 3 = 639099).
213033 multiplied by four equals 852132 (213033 x 4 = 852132).
213033 multiplied by five equals 1065165 (213033 x 5 = 1065165).
213033 multiplied by six equals 1278198 (213033 x 6 = 1278198).
213033 multiplied by seven equals 1491231 (213033 x 7 = 1491231).
213033 multiplied by eight equals 1704264 (213033 x 8 = 1704264).
213033 multiplied by nine equals 1917297 (213033 x 9 = 1917297).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 213033
Half of 213033 is 106516,5 (213033 / 2 = 106516,5 = 106516 1/2).
One third of 213033 is 71011 (213033 / 3 = 71011).
One quarter of 213033 is 53258,25 (213033 / 4 = 53258,25 = 53258 1/4).
One fifth of 213033 is 42606,6 (213033 / 5 = 42606,6 = 42606 3/5).
One sixth of 213033 is 35505,5 (213033 / 6 = 35505,5 = 35505 1/2).
One seventh of 213033 is 30433,2857 (213033 / 7 = 30433,2857 = 30433 2/7).
One eighth of 213033 is 26629,125 (213033 / 8 = 26629,125 = 26629 1/8).
One ninth of 213033 is 23670,3333 (213033 / 9 = 23670,3333 = 23670 1/3).
show fractions by 6, 7, 8, 9 ...
### Calculator
213033
#### Is Prime?
The number 213033 is not a prime number. The closest prime numbers are 213029, 213043.
#### Factorization and factors (dividers)
The prime factors of 213033 are 3 * 71011
The factors of 213033 are 1 , 3 , 71011 , 213033
Total factors 4.
Sum of factors 284048 (71015).
#### Powers
The second power of 2130332 is 45.383.059.089.
The third power of 2130333 is 9.668.089.226.906.936.
#### Roots
The square root √213033 is 461,55498.
The cube root of 3213033 is 59,72401.
#### Logarithms
The natural logarithm of No. ln 213033 = loge 213033 = 12,269202.
The logarithm to base 10 of No. log10 213033 = 5,328447.
The Napierian logarithm of No. log1/e 213033 = -12,269202.
### Trigonometric functions
The cosine of 213033 is -0,031359.
The sine of 213033 is 0,999508.
The tangent of 213033 is -31,873488.
### Properties of the number 213033
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 213033 in Computer Science
Code typeCode value
PIN 213033 It's recommendable to use 213033 as a password or PIN.
213033 Number of bytes208.0KB
CSS Color
#213033 hexadecimal to red, green and blue (RGB) (33, 48, 51)
Unix timeUnix time 213033 is equal to Saturday Jan. 3, 1970, 11:10:33 a.m. GMT
IPv4, IPv6Number 213033 internet address in dotted format v4 0.3.64.41, v6 ::3:4029
213033 Decimal = 110100000000101001 Binary
213033 Decimal = 101211020010 Ternary
213033 Decimal = 640051 Octal
213033 Decimal = 34029 Hexadecimal (0x34029 hex)
213033 BASE64MjEzMDMz
213033 SHA190ffe04d8dbcbfc87f2b3db00aea76dc04e9e67e
213033 SHA2563fac06abeea3490a09e5c335c982ece838323cbcb25f98229848e3d9c5c77889
213033 SHA384502ee07aae783ce953b7177558e460553e1ca87256038bcee385aa94ffc6ceb94076ae8fcdce8a41e9bac64c80a0ab7e
More SHA codes related to the number 213033 ...
If you know something interesting about the 213033 number that you did not find on this page, do not hesitate to write us here.
## Numerology 213033
### Character frequency in number 213033
Character (importance) frequency for numerology.
Character: Frequency: 2 1 1 1 3 3 0 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 213033, the numbers 2+1+3+0+3+3 = 1+2 = 3 are added and the meaning of the number 3 is sought.
## Interesting facts about the number 213033
### Asteroids
• (213033) 1997 UT15 is asteroid number 213033. It was discovered by Spacewatch from Obs. US National at Kitt Peak on 10/23/1997.
## № 213,033 in other languages
How to say or write the number two hundred and thirteen thousand and thirty-three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 213.033) doscientos trece mil treinta y tres German: 🔊 (Anzahl 213.033) zweihundertdreizehntausenddreiunddreißig French: 🔊 (nombre 213 033) deux cent treize mille trente-trois Portuguese: 🔊 (número 213 033) duzentos e treze mil e trinta e três Chinese: 🔊 (数 213 033) 二十一万三千零三十三 Arabian: 🔊 (عدد 213,033) مئتان و ثلاثة عشر ألفاً و ثلاثة و ثلاثون Czech: 🔊 (číslo 213 033) dvěstě třináct tisíc třicet tři Korean: 🔊 (번호 213,033) 이십일만 삼천삼십삼 Danish: 🔊 (nummer 213 033) tohundrede og trettentusindtreogtredive Dutch: 🔊 (nummer 213 033) tweehonderddertienduizenddrieëndertig Japanese: 🔊 (数 213,033) 二十一万三千三十三 Indonesian: 🔊 (jumlah 213.033) dua ratus tiga belas ribu tiga puluh tiga Italian: 🔊 (numero 213 033) duecentotredicimilatrentatre Norwegian: 🔊 (nummer 213 033) to hundre og tretten tusen og tretti-tre Polish: 🔊 (liczba 213 033) dwieście trzynaście tysięcy trzydzieści trzy Russian: 🔊 (номер 213 033) двести тринадцать тысяч тридцать три Turkish: 🔊 (numara 213,033) ikiyüzonüçbinotuzüç Thai: 🔊 (จำนวน 213 033) สองแสนหนึ่งหมื่นสามพันสามสิบสาม Ukrainian: 🔊 (номер 213 033) двiстi тринадцять тисяч тридцять три Vietnamese: 🔊 (con số 213.033) hai trăm mười ba nghìn lẻ ba mươi ba Other languages ...
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## Comment
If you know something interesting about the number 213033 or any natural number (positive integer) please write us here or on facebook. | 2,445 | 7,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-33 | latest | en | 0.657852 |
https://www.stata.com/statalist/archive/2013-10/msg00352.html | 1,723,324,399,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640822309.61/warc/CC-MAIN-20240810190707-20240810220707-00132.warc.gz | 786,328,966 | 6,357 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
RE: Re: st: RE: Selecting correlations with highest absolute value
From Joe Canner <[email protected]> To "[email protected]" <[email protected]> Subject RE: Re: st: RE: Selecting correlations with highest absolute value Date Wed, 9 Oct 2013 23:40:20 +0000
```Dara,
Red Owl beat me to the answer I was going to give. If you have a good reason to use -pwcorr- instead of -corr-, then you might need something more complicated in which you loop over all your variables, accumulating pairwise correlations.
foreach x of varlist tbmale...etc {
foreach y of varlist tbmale...etc {
corr `x' 'y'
matrix corrvector=corrvector \ vec(r(C))
}
}
matvsort corrvector sortedvector
matrix list sortedvector
I don't have the ability to test this at the moment and I don't have matrix syntax memorized, so this might need some tweaking, particularly the matrix command inside the loops. I'm also not sure if you will need to initialize -correvector- before starting the loops. Let us know if have any problems and I'm sure someone can help.
Joe
________________________________________
From: [email protected] [[email protected]] on behalf of Dara Shifrer [[email protected]]
Sent: Wednesday, October 09, 2013 7:04 PM
To: [email protected]
Subject: Fwd: Re: st: RE: Selecting correlations with highest absolute value
Joe, thank you very much for your quick response to my quest to find the
most highly correlated pairs of variables. I think I understand what
your code does (finds correlations, linearly transforms the correlation
matrix into a column vector, sorts this matrix, and then lists the
sorted columns of correlations) but I'm not sure why it isn't working
for me (see code below). I haven't used Stata's matrix commands before
and may be missing something obvious. Thanks for any additional help
anyone can provide! Dara
pwcorr tbmale tdedc3 tbrace td9tchr td9slry tb9yrsh tb9yrsnh td10tchr
td10slry ///
tb10yrsh tb10yrsnh td11tchr td11slry tb11yrsh tb11yrsnh ///
tp10pswm ta10a2w skd10size skd10blck skd10hisp skd10pvty skd10lep ///
skd10biesl skd10gt skd10sped skd11size skd11blck skd11hisp skd11pvty
skd11lep ///
skd11biesl skd11gt skd11sped skd12size skd12blck skd12hisp skd12pvty
skd12lep ///
skd12biesl skd12gt skd12sped ta11elgb5 ta11ctgr ta11grd ta11chrt
ta11sclvl ///
ta11a1rg ta11a2rg ta11a2lrg ta11a2mrg ta11a2m9rg ta11a2m10rg ta11a2m11rg ///
ta11a2rrg ta11a2r9rg ta11a2r10rg ta11a2r11rg ta11a2srg ta11a2s10rg
ta11a2s11rg ///
ta11a2ssrg ta11a2ss10rg ta11a2ss11rg ///
ta11a3rg ta11a3arg ta11a3arrg ta11a3amrg ta11a3aparg ta11a3aperg
ta11a3brg ta11a3crg ///
trt12rtn tp12pswm tka12tme tka12tmebl tka12tms tka12tmsbl tka12tre
tka12trebl tka12trs ///
tka12trsbl tka12talg1 tka12talg1bl tka12tbio tka12tbiobl tka12te1r ///
tka12te1rbl tka12te1w tka12te1wbl tka12twgeo tka12twgeobl ///
tka12smegn tka12smsgn tka12sregn tka12srsgn tka12slegn tka12slsgn ///
tka12ssegn tka12shegn tka12shsgn
.... lots of correlations excluded...
| tka~megn tka~msgn tka~regn tka~rsgn tka~legn tka~lsgn tka~segn
-------------+---------------------------------------------------------------
tka12smegn | 1.0000
tka12smsgn | 0.1390 1.0000
tka12sregn | 0.6082 0.1509 1.0000
tka12srsgn | 0.1211 0.5660 0.1929 1.0000
tka12slegn | 0.5454 -0.0638 0.5637 0.1009 1.0000
tka12slsgn | 0.2572 0.5671 0.2427 0.5295 0.2006 1.0000
tka12ssegn | 0.4479 -0.1376 0.3819 -0.1273 0.4028 -0.1095
1.0000
tka12shegn | 0.4143 0.0340 0.4330 -0.2011 0.4543 -0.2584
0.5530
tka12shsgn | 0.5705 0.4077 0.3127 0.6170 0.2309 0.4094
0.2407
| tka~hegn tka~hsgn
-------------+------------------
tka12shegn | 1.0000
tka12shsgn | 0.0918 1.0000
. matrix corrvector=vec(r(C))
. matvsort corrvector sortedvector
. matrix list sortedvector
sortedvector[4,1]
c1
tka12shsgn:tka12shsgn 1
tka12shsgn:tka12shsgn 1
tka12shsgn:tka12shsgn 1
tka12shsgn:tka12shsgn 1
Postdoctoral Fellow, Houston Education Research Consortium
Kinder Institute for Urban Research
Rice University
[email protected]
On 10/8/2013 1:39 PM, Joe Canner wrote:
> Dara,
>
> Here's one quick-n-dirty possibility. (It requires installing -matvsort- from SSC.)
>
> . corr varlist
> . matrix corrvector=vec(r(C))
> . matvsort corrvector sortedvector
> . matrix list sortedvector
>
> Regards,
> Joe Canner
> Johns Hopkins University School of Medicine
>
>
> -----Original Message-----
> From:[email protected] [mailto:[email protected]] On Behalf Of Dara Shifrer
> Sent: Tuesday, October 08, 2013 3:16 PM
> To:[email protected]
> Subject: st: Selecting correlations with highest absolute value
>
>
> In SAS, I was able to quickly determine which pairs of variables were
> most highly correlated using the 'best' option with the 'proc corr'
> command ("*BEST=*/n ----/**/**/prints */n/* correlation coefficients for
> each variable. Correlations are ordered from highest to lowest in
> absolute value.) After extensive searching, I have not been able to
> locate a Stata command that does something similar.
>
> If this is not possible in Stata, maybe Stata experts have suggestions
> for my ultimate purpose: constructing equations to facilitate a smoother
> and faster running of Stata's 'ice' command.
>
> Any help would be greatly appreciated,
> Dara Shifrer
>
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/faqs/resources/statalist-faq/
* http://www.ats.ucla.edu/stat/stata/
``` | 1,984 | 5,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-33 | latest | en | 0.889779 |
https://discuss.leetcode.com/topic/9818/my-2ms-c-solution-and-some-thoughts | 1,513,005,949,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513611.22/warc/CC-MAIN-20171211144705-20171211164705-00073.warc.gz | 554,502,179 | 10,489 | My 2ms C solution and some thoughts
• My first version is:
``````uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
for (int i = 0; i < 31; i++) {
res |= (n & 1);
n >>= 1; res <<= 1;
}
res |= (n & 1);
return res;
}
``````
And this works pretty well. 4ms.
After that, I replaced
``````for (int i = 0; i < 31; i++)
``````
with
``````for (int i = 31; i > 0; i--)
``````
This works in 3ms.
I'm not very familiar with x86 assembly and I only know MIPS.
Comparing with zero is, well at least in my opinion, faster. (Please comment if I'm wrong.)
After that, I thought we could do loop expand, just as compilers do.
So the next version is:
``````uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
// repeat 31 times
res |= (n & 1); n >>= 1; res <<= 1;
res |= (n & 1); n >>= 1; res <<= 1;
res |= (n & 1); n >>= 1; res <<= 1;
res |= (n & 1); n >>= 1; res <<= 1;
....
res |= (n & 1); n >>= 1; res <<= 1;
return res | (n & 1);
}
``````
Well it turns out to be slightly slower, 4ms. I don't know why.
I tried to figure out some loop-free solution and failed. (Please comment if there is any)
So my last version is
``````uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
for (int i = 31; i > 0; i--) {
res |= (n & 1);
n >>= 1; res <<= 1;
}
return res | (n & 1);
}
``````
Almost the same as the second one but it magically works in 2ms.
I'm not sure if OJ's time measurement is precise enough, and I don't think caching is a good idea since this function would be called millions of times, searching would takes more time than just reversing it.
EDIT:
I found this one on Stack Overflow:
``````uint32_t reverseBits(uint32_t x) {
x = ((x >> 1) & 0x55555555u) | ((x & 0x55555555u) << 1);
x = ((x >> 2) & 0x33333333u) | ((x & 0x33333333u) << 2);
x = ((x >> 4) & 0x0f0f0f0fu) | ((x & 0x0f0f0f0fu) << 4);
x = ((x >> 8) & 0x00ff00ffu) | ((x & 0x00ff00ffu) << 8);
x = ((x >> 16) & 0xffffu) | ((x & 0xffffu) << 16);
return x;
}
``````
And it works in 3ms. Could someone explain it?
1. Time meansure on OJ is not very accurate. You need a larger benchmark for testing performance.
2. Spatial locality and temporal locality makes cache works better. A very large loop body make cache for your codes occure more cache miss. That is why the performance may be even worse.
3. The last solution may be a simple one. First, it swap odd bits and even bits. Then, it swap each two bits...
4. 64K lookup table makes last solution faster. (But OJ may not accept your code since it is too long)
5. There are some typos in your last solution.
• This post is deleted!
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 895 | 2,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-51 | latest | en | 0.779356 |
https://www.idealhometuition.com/2014/06/thermodynamics-ncert-solutions-class-11_5641.html | 1,603,988,585,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904834.82/warc/CC-MAIN-20201029154446-20201029184446-00477.warc.gz | 759,956,401 | 11,300 | ## Pages
### Thermodynamics NCERT Solutions Class 11 Physics - Solved Exercise Question 12.5
Question 12.5:
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Solution:
The work done (W) on the system while the gas changes from state A to state B is 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.
∴ ΔQ = 0
ΔW = –22.3 J (Since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = ΔU + ΔW
Where,
ΔU = Change in the internal energy of the gas
∴ ΔU = ΔQ – ΔW = – (– 22.3 J)
ΔU = + 22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is:
ΔQ = 9.35 cal = 9.35 × 4.19 = 39.1765 J
Heat absorbed, ΔQ = ΔU + ΔQ
∴ΔW = ΔQ – ΔU
= 39.1765 – 22.3
= 16.8765 J
Therefore, 16.88 J of work is done by the system. | 367 | 1,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-45 | longest | en | 0.916501 |
https://www.hunterpoolshop.com/a-circular-swimming-pool-is-21-feet-in-diameter/ | 1,708,468,754,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00514.warc.gz | 877,815,668 | 25,364 | # A Circular Swimming Pool Is 21 Feet in Diameter
If you're considering installing a swimming pool, imagine diving into a refreshing oasis right in your own backyard. Picture yourself lounging on a float, soaking up the sun's rays.
Now, picture a circular swimming pool that's 21 feet in diameter, providing ample space for relaxation and recreation.
In this article, we'll explore the dimensions of this circular pool, calculate its area and circumference, discuss safety considerations, and delve into design options.
Get ready to make a splash with your dream pool!
Contents
## Understanding the Dimensions of a Circular Swimming Pool
You already know that a circular swimming pool is 21 feet in diameter, but do you understand what that actually means?
Diameter is the distance across the pool, passing through the center. In the case of this pool, it measures 21 feet from one side to the other, helping to determine the size and shape of the pool.
### Definition of Diameter
To fully grasp the dimensions of a circular swimming pool, it's crucial to comprehend the meaning of the term 'diameter.' The diameter is a straight line that passes through the center of a circle and connects two points on its circumference. In the case of a circular swimming pool, the diameter is the distance between two opposite points on the pool's edge, passing through the center. In simpler terms, it's the length of a straight line that cuts the pool in half.
Understanding the diameter of a circular swimming pool is essential when it comes to measurements and calculations, as it provides a key reference point for determining the pool's size and capacity.
### How Diameter Relates to a Circular Swimming Pool
Understanding the diameter of a circular swimming pool is crucial in comprehending the dimensions of the pool and determining its size and capacity.
The diameter of a circular pool is the distance across the pool, passing through the center and touching two opposite points on the pool's edge.
In the case of a circular swimming pool with a diameter of 21 feet, it means that if you were to measure from one edge of the pool, through the center, to the opposite edge, the distance would be 21 feet.
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This measurement helps you understand the size of the pool and how much space it will occupy in your backyard or wherever it's installed.
Additionally, the diameter also plays a role in calculating the volume and capacity of the pool, which is important for determining how much water can be held in the pool and how many people it can comfortably accommodate.
## Calculating the Area of a Circular Swimming Pool
To calculate the area of a circular swimming pool, you can use the formula for the area of a circle: A = πr².
In this case, since the diameter of the pool is 21 feet, the radius would be half of that, which is 10.5 feet.
### Formula for Area of a Circle
Calculate the area of a circular swimming pool by using the formula for the area of a circle.
The formula for finding the area of a circle is A = πr², where A represents the area and r represents the radius of the circle.
In this case, the swimming pool has a diameter of 21 feet, which means the radius is half of that, or 10.5 feet.
To find the area, plug in the value of the radius into the formula: A = π(10.5)².
Simplifying the equation, we get A = π(110.25), which is approximately 345.97 square feet.
Therefore, the area of the circular swimming pool is approximately 345.97 square feet.
### Applying the Formula to a 21 Feet Diameter Pool
Can You Put Epsom Salt in a Swimming Pool
Does Kohl's Have Swimming Pools
You can easily determine the area of a circular swimming pool with a diameter of 21 feet by applying the formula for the area of a circle. Here's how you can do it:
1. Start by finding the radius of the pool. Since the diameter is 21 feet, divide it by 2 to get the radius, which is 10.5 feet.
2. Use the formula for the area of a circle, which is A = πr². Plug in the value of the radius, which is 10.5 feet, and square it to get 110.25.
3. Multiply the squared radius by π, which is approximately 3.14, to get the area of the pool. The result is 345.915 square feet.
4. So, the area of the circular swimming pool with a diameter of 21 feet is approximately 345.915 square feet.
## Calculating the Circumference of a Circular Swimming Pool
Now let's talk about calculating the circumference of your circular swimming pool.
The formula for finding the circumference of a circle is C = πd, where C represents the circumference and d represents the diameter.
In the case of your 21 feet diameter pool, you can simply plug in the value into the formula to determine the circumference.
### Formula for Circumference of a Circle
To find the circumference of a circular swimming pool with a diameter of 21 feet, simply multiply the diameter by pi. The formula for calculating the circumference of a circle is C = πd, where C represents the circumference and d represents the diameter.
In this case, the diameter is 21 feet, so you can substitute that value into the formula. C = π(21). To get an exact value for the circumference, you'd multiply 21 by π, which is approximately equal to 3.14. Therefore, the circumference of the circular swimming pool is approximately 65.94 feet.
This formula can be applied to any circular object to calculate its circumference, making it a useful tool for various applications in mathematics, engineering, and everyday life.
### Applying the Formula to a 21 Feet Diameter Pool
The formula for the circumference of a circular swimming pool with a diameter of 21 feet can be applied by multiplying the diameter by pi.
To calculate the circumference of a 21 feet diameter pool, follow these steps:
1. Measure the diameter of the pool, which is 21 feet.
2. Multiply the diameter by pi, which is approximately 3.14.
3. Multiply 21 feet by 3.14 to get the circumference of the pool.
4. The result will give you the total distance around the pool, which is the circumference.
## Calculating the Volume of a Circular Swimming Pool
To calculate the volume of a circular swimming pool, you need to understand how depth relates to volume.
The formula for calculating the volume of a cylinder can be applied to a 21 feet diameter pool.
### Understanding Depth in Relation to Volume
Calculate the volume of your circular swimming pool by multiplying the area of the pool's base by its depth.
The depth refers to how deep the pool is, which is an important factor in determining its volume.
For example, if your circular swimming pool is 21 feet in diameter and 4 feet deep, you can calculate its volume by first finding the area of the base.
The formula for the area of a circle is A = πr², where r is the radius of the circle. In this case, the radius is half of the diameter, so it's 21/2 = 10.5 feet.
Therefore, the area of the base is A = π(10.5)² = 346.36 square feet.
To find the volume, multiply the area of the base by the depth: 346.36 square feet × 4 feet = 1385.44 cubic feet.
### Formula for Volume of a Cylinder
To determine the volume of your circular swimming pool, use the formula for the volume of a cylinder.
The formula for the volume of a cylinder is V = πr^2h, where V represents the volume, π is a mathematical constant approximately equal to 3.14, r is the radius of the circular base, and h is the height of the cylinder.
In the case of your circular swimming pool, the radius is half of the diameter, which is 21 feet. Thus, the radius would be 10.5 feet.
The height of the pool would be the depth you measured earlier.
Once you have the radius and height values, you can substitute them into the formula to calculate the volume of your circular swimming pool.
### Applying the Formula to a 21 Feet Diameter Pool
Now, let's apply the formula for the volume of a cylinder to your 21 feet diameter swimming pool. The formula for calculating the volume of a cylinder is V = πr²h, where V represents the volume, π is a mathematical constant approximately equal to 3.14159, r is the radius of the circular base, and h is the height of the cylinder. In this case, the radius of your pool is half of its diameter, so it will be 21 feet ÷ 2 = 10.5 feet. To calculate the volume, we also need the height of the pool. Once you have the height, you can substitute the values into the formula and calculate the volume. Here is a table summarizing the calculation:
Diameter (feet) Radius (feet) Height (feet)
21 10.5
## Practical Implications of Pool Dimensions
When considering the practical implications of pool dimensions, there are several important points to consider.
First, the amount of water needed to fill the pool will depend on its size, with a larger pool requiring more water.
Second, the time required to fill the pool will also be affected by its dimensions, as a larger pool will take longer to fill.
Lastly, the cost implications of filling and maintaining the pool will depend on factors such as water usage, chemicals, and cleaning supplies.
### Amount of Water Needed to Fill the Pool
Calculating the amount of water you'll need to fill a circular swimming pool with a 21-foot diameter can be determined by using the formula for finding the volume of a cylinder.
To find the volume, you'll need to know the height of the pool as well. Once you have the height, you can use the formula V = πr^2h, where V is the volume, π is approximately 3.14, r is the radius (which is half the diameter), and h is the height of the pool. In this case, the radius is 10.5 feet.
Let's say the height of the pool is 5 feet. Plugging in these values into the formula, you get V = 3.14 x (10.5)^2 x 5.
Simplifying this equation, you'll find that the pool can hold approximately 868.35 cubic feet of water. Therefore, that's the amount of water needed to fill the pool.
### Time Required to Fill the Pool
To determine the time required to fill the pool, you can use the amount of water calculated in the previous subtopic. In this case, the pool has a diameter of 21 feet, so the amount of water needed to fill it's 346.36 cubic feet.
Now, let's focus on the time it would take to fill the pool. The time required to fill the pool depends on the water flow rate. If you have a flow rate of 10 gallons per minute, for example, it would take approximately 5.48 hours to fill the pool.
However, if you have a higher flow rate, such as 20 gallons per minute, it would only take around 2.74 hours. Therefore, the time required to fill the pool can vary depending on the water flow rate you have available.
### Cost Implications of Filling and Maintaining the Pool
The cost implications of filling and maintaining the pool can vary depending on its dimensions and the water flow rate you have available.
When it comes to filling the pool, the cost will primarily depend on the volume of water needed. A larger pool will require more water, which means higher water bills. Additionally, if you live in an area with expensive water rates, the cost can be even higher.
Maintaining the pool also has cost implications. This includes the cost of chemicals, such as chlorine, to keep the water clean and balanced. You'll also need to consider the cost of electricity for running the pool pump and filter system.
Regular maintenance, such as cleaning and repairing, may also incur additional costs.
## Safety Considerations for a Circular Swimming Pool
When it comes to safety considerations for a circular swimming pool, there are several important points to keep in mind.
First, it's crucial to implement safety measures for different depths of the pool to prevent accidents.
Additionally, knowing the exact dimensions of the pool, such as its 21 feet diameter, is essential for proper safety planning.
Lastly, considering pool covers and their sizes for a circular pool of this diameter can further enhance safety by preventing unauthorized access and potential accidents.
### Safety Measures for Different Depths
Ensure your safety in a circular swimming pool by implementing measures for different depths.
When dealing with a deep swimming pool, it's crucial to be aware of the potential risks and take appropriate precautions.
A deep swimming pool, such as the circular one with a diameter of 21 feet, requires extra safety considerations.
Firstly, ensure that the pool is equipped with clearly marked depth indicators, allowing swimmers to gauge the water depth easily.
Additionally, install safety rails or barriers around the perimeter of the pool to prevent accidental falls into the deep end.
It's also essential to have a lifeguard on duty or provide proper supervision to ensure swimmers' safety, especially in deep areas.
### Importance of Knowing the Pool's Dimensions
To ensure your safety in a circular swimming pool, it's important to have a clear understanding of the pool's dimensions, such as its diameter of 21 feet, as this knowledge will guide your safety considerations.
The diameter of the pool plays a crucial role in determining the pool's capacity, water flow, and potential hazards. By knowing the diameter, you can calculate the pool's surface area, which is useful for determining the amount of cleaning and maintenance required.
Additionally, understanding the diameter helps you gauge the distance between different points in the pool, enabling you to plan your movements and avoid collisions with other swimmers or pool features.
Being aware of the pool's dimensions, especially the diameter, is essential for ensuring a safe and enjoyable swimming experience.
### Pool Covers and Their Sizes for a 21 Feet Diameter Pool
You should consider using a pool cover designed for a 21 feet diameter pool to ensure safety and protection. Pool covers play a crucial role in keeping your swimming pool clean, preventing debris from falling in, and reducing the risk of accidents, especially if you have children or pets.
When it comes to choosing the right pool cover for your 21 feet diameter pool, you need to consider the size and shape of the cover. The pool cover should be able to completely cover the pool surface, ensuring that no gaps are left exposed. It should also be strong and durable enough to withstand the weight of any potential debris or even a person accidentally walking on it.
Additionally, make sure the cover is easy to install and remove for your convenience. By investing in a pool cover that's specifically designed for a 21 feet diameter pool, you can enjoy peace of mind knowing that your pool is protected and safe.
## Design and Aesthetic Considerations for a Circular Swimming Pool
When designing your 21 feet diameter pool, consider the landscaping around it to create a visually appealing and cohesive outdoor space.
Choose furniture and decorations that complement the pool surroundings, enhancing the overall aesthetic.
Lighting is an important aspect to consider, as it can create ambiance and highlight the beauty of the circular pool design.
### Landscaping Around a 21 Feet Diameter Pool
Enhance the beauty of your 21 feet diameter circular swimming pool by carefully designing and landscaping the surrounding area. Landscaping around a 21 feet diameter pool requires thoughtful planning to create a visually appealing and functional outdoor space.
Begin by considering the shape and size of the pool when choosing plantings and hardscape elements. A circular pool lends itself well to curved paths, flower beds, and seating areas. You can use a combination of trees, shrubs, and flowers to add color and texture to the landscape.
Incorporate elements like rocks, pavers, or decking to create defined spaces and pathways. Lighting is also essential for safety and ambiance, so consider installing well-placed fixtures to highlight key features.
With careful design and attention to detail, you can create a stunning landscape that complements your 21 feet diameter pool.
### Furniture and Decorations for Pool Surroundings
To create a stylish and inviting atmosphere around your 21 feet diameter circular swimming pool, consider carefully selecting furniture and decorations that complement the landscape design.
The right choice of furniture and decorations for your pool surroundings can enhance the overall aesthetic appeal and make your pool area a relaxing and enjoyable space.
When it comes to furniture, opt for comfortable lounge chairs, sunbeds, and outdoor sofas that are both durable and weather-resistant. You can also add an umbrella or a pergola to provide shade on hot summer days.
As for decorations, consider adding potted plants, colorful cushions, and outdoor rugs to add a touch of vibrancy and warmth. Additionally, incorporating lighting fixtures such as string lights or lanterns can create a cozy ambiance for nighttime swimming and gatherings.
### Lighting Considerations for a Circular Pool
Consider adding appropriate lighting fixtures to enhance the design and aesthetics of your 21 feet diameter circular swimming pool. Lighting considerations play a crucial role in creating a captivating atmosphere and ensuring safety around your pool area.
When choosing lighting fixtures, keep in mind the overall theme and style of your pool. LED lights are a popular choice as they're energy-efficient, long-lasting, and offer a variety of colors to suit your preference.
Consider placing lights around the perimeter of the pool to highlight its shape and create a visually appealing effect. Underwater lights can also be installed to illuminate the water and create a mesmerizing ambiance. Additionally, pathway lights can guide guests towards the pool area, ensuring their safety.
## Frequently Asked Questions
### How Long Does It Take to Fill the Circular Swimming Pool With Water?
It takes about X amount of time to fill the circular swimming pool with water. The actual time may vary depending on the water pressure and the size of the hose used.
### What Is the Average Cost of Maintaining a Circular Swimming Pool?
The average cost of maintaining a circular swimming pool depends on factors such as size, location, and usage. Regular cleaning, chemical treatments, and repairs may be necessary, so it's important to budget accordingly.
### What Types of Materials Can Be Used to Build a Circular Swimming Pool?
To build a circular swimming pool, you can use various materials such as concrete, fiberglass, or vinyl. Each material has its pros and cons, so it's important to research and choose the best option for your needs.
### Are There Any Specific Regulations or Permits Required for Installing a Circular Swimming Pool?
Before installing your circular swimming pool, it's important to check with local authorities for specific regulations and permits. Make sure you comply with all requirements to ensure a smooth and legal installation process.
### Can the Diameter of a Circular Swimming Pool Be Customized to a Specific Size?
Yes, the diameter of a circular swimming pool can be customized to a specific size. You have the flexibility to choose the diameter that best suits your needs and preferences.
### Related Posts
#### Mike Hunter
Mike is the owner of the local pool shop. He's been in the business for over 20 years and knows everything there is to know about pools. He's always happy to help his customers with whatever they need, whether it's advice on pool maintenance or choosing the right chemicals. He's also a bit of a pool expert, and is always happy to share his knowledge with anyone who's interested. | 4,075 | 19,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-10 | latest | en | 0.930336 |
http://laserpointerforums.com/f57/not-worth-mentioning-because-they-would-deep-trouble-96681.html | 1,477,299,281,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719547.73/warc/CC-MAIN-20161020183839-00329-ip-10-171-6-4.ec2.internal.warc.gz | 149,919,806 | 15,403 | Welcome to Laser Pointer Forums! If you are looking for a laser you may want to check out the database of laser pointer companies. The link will open in a new window for your convenience.
Laser Pointer Forums - Discuss Laser Pointers Not worth mentioning because they would be in deep trouble! Laser Pointer Company Database Laser Top Sites List Lasers by Type Green Lasers
03-13-2016, 09:55 PM #1
Class 3B Laser Join Date: Jan 2014 Location: Baghdad, Iraq Posts: 4,243 Rep Power: 2080
Class 3B Laser
Join Date: Jan 2014
Posts: 4,243
Rep Power: 2080
Not worth mentioning because they would be in deep trouble!
Still think we are alone? Think again:
OK, some of this video is even questionable to me as a "I want to believe" guy, but not all of it.
__________________
http://www.lexellaser.com/techinfo_wavelengths.htm
Laser Safety:
http://www.laserpointersafety.com/index.html
Angular Size Calculator; use with diode angle of radiation specification to calculate the needed lens diameter for a given FL:
http://www.1728.org/angsize.htm
Divergence to spot size calculator: http://tinyurl.com/spotsize - 1 mRad is about .057 degrees which from the earth would expand to be ~10% the diameters of the moon or sun at their distances.
Divergence Calculator: http://www.pseudonomen.com/lasers/ca...alculator.html - Measure your lasers beamwidth at 1 foot & then at a further distance to calculate the divergence.
Online calculator to determine spot intensity at different mRad's & powers @ distances: http://tinyurl.com/divergence-calculator
Laser Power Density Calculator: http://www.ophiropt.com/laser-measur...ity-calculator
How to build a laser beam expander to reduce divergence: http://tinyurl.com/BeamExpander
RHD's Relative Perceived Brightness Calculator. Compare brightness @nm: http://lsrtools.1apps.com/relativebrightness
YAG Power Calculator, i.e. convert ns @ mJ to peak power in watts http://tinyurl.com/YAG-Pulse
The forum costs more to run than donations received, if you wish to help click this link: http://laserpointerforums.com/donations.htm
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Last edited by Alaskan; 03-13-2016 at 10:00 PM.
03-13-2016, 10:01 PM #2
Banned Join Date: Feb 2015 Posts: 1,375 Rep Power: 0
Isaac Clarke
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Re: Not worth mentioning because they would be in deep trouble!
Alien is that your friend paying us a visit?
03-13-2016, 10:17 PM #3
Class 3B Laser Join Date: Jan 2014 Location: Baghdad, Iraq Posts: 4,243 Rep Power: 2080
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Re: Not worth mentioning because they would be in deep trouble!
Quote:
Originally Posted by Isaac Clarke Alien is that your friend paying us a visit?
LOL, I think he saw something once himself, I've seen some kind of vehicle which was silently hovering above me too but who built it? I don't know.
Here's a video from a senior research scientist who was with Lockheed Martin who blew some of their cover on their activities, just before he died:
__________________
http://www.lexellaser.com/techinfo_wavelengths.htm
Laser Safety:
http://www.laserpointersafety.com/index.html
Angular Size Calculator; use with diode angle of radiation specification to calculate the needed lens diameter for a given FL:
http://www.1728.org/angsize.htm
Divergence to spot size calculator: http://tinyurl.com/spotsize - 1 mRad is about .057 degrees which from the earth would expand to be ~10% the diameters of the moon or sun at their distances.
Divergence Calculator: http://www.pseudonomen.com/lasers/ca...alculator.html - Measure your lasers beamwidth at 1 foot & then at a further distance to calculate the divergence.
Online calculator to determine spot intensity at different mRad's & powers @ distances: http://tinyurl.com/divergence-calculator
Laser Power Density Calculator: http://www.ophiropt.com/laser-measur...ity-calculator
How to build a laser beam expander to reduce divergence: http://tinyurl.com/BeamExpander
RHD's Relative Perceived Brightness Calculator. Compare brightness @nm: http://lsrtools.1apps.com/relativebrightness
YAG Power Calculator, i.e. convert ns @ mJ to peak power in watts http://tinyurl.com/YAG-Pulse
The forum costs more to run than donations received, if you wish to help click this link: http://laserpointerforums.com/donations.htm
_______
Last edited by Alaskan; 03-13-2016 at 10:31 PM.
03-14-2016, 02:41 AM #4
Class 2 Laser Join Date: Jun 2010 Location: New Hampshire Posts: 276 Rep Power: 210
planters
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Re: Not worth mentioning because they would be in deep trouble!
The video suggests to me that this is fake. The "creature" is way to humanoid. The odds and everything is a probability, of so many features matching ours are very low. The large head would evolve a stouter or shorter neck...probably.
Why would a race capable of the trip fail to recover the body? If they can and have remained "secrete", then they would not want to leave the body. If everything went to shit (it happens ie Challenger), then there should be loads of evidence. An interstellar ship or its wreckage do not blend in.
I'm much more willing to believe captain O. told Lois Lerner who to hit.
03-14-2016, 04:12 AM #5
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Rivem
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Re: Not worth mentioning because they would be in deep trouble!
In years of stargazing, remote camping, and living next to a supposed (and somewhat famous) UFO landing site, I've only ever had one experience.
I was out at night as a middle schooler, and what looked like a satelite came above the horizon very quickly. I didn't think much of it until it stopped directly overhead for about a minute and got brighter. I stood there dumbfounded just staring at the thing until it was about half the size but twice as bright as the moon. Didn't think UFO until what happened next. From the opposite direction, another object slowly drifted into view. It was just a bit less bright and headed straight for the other. It came right next to the other and stopped. They floated next to each other in the sky for another 2 or 3 minutes, and then one zipped away below the horizon, and the other just faded away. Only after this happened did I get a bit freaked out. Went straight to bed afterwards.
I've never seen anything like it again, and I still don't have a good explanation. Seen plenty freaky stuff happen in the sky but nothing else I couldn't debunk. Still, I feel like aliens should be a bit more open to communication if they're here and have superior technology.
__________________
Now in the Member Collections Section: Rivem's Collection.
I am currently at a BUSY semester of an EE degree, so please forgive the inactivity. If you want to talk to me, feel free to send a PM.
03-14-2016, 06:25 AM #6
Class 3B Laser Join Date: Jan 2014 Location: Baghdad, Iraq Posts: 4,243 Rep Power: 2080
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Re: Not worth mentioning because they would be in deep trouble!
I don't know the reasons why the alien appeared to be so humanoid, but from what I have gathered from a life time of study on this subject, there are some races which are much like our own, that the humanoid form is more prevalent for space faring races for some reason, but not all are similar to us. If you dig deep enough into this subject you won't have any doubts that these things are happening. Of course, as I wrote above, I want to believe, but at the same time I don't want to be wrong and I still see this way.
__________________
http://www.lexellaser.com/techinfo_wavelengths.htm
Laser Safety:
http://www.laserpointersafety.com/index.html
Angular Size Calculator; use with diode angle of radiation specification to calculate the needed lens diameter for a given FL:
http://www.1728.org/angsize.htm
Divergence to spot size calculator: http://tinyurl.com/spotsize - 1 mRad is about .057 degrees which from the earth would expand to be ~10% the diameters of the moon or sun at their distances.
Divergence Calculator: http://www.pseudonomen.com/lasers/ca...alculator.html - Measure your lasers beamwidth at 1 foot & then at a further distance to calculate the divergence.
Online calculator to determine spot intensity at different mRad's & powers @ distances: http://tinyurl.com/divergence-calculator
Laser Power Density Calculator: http://www.ophiropt.com/laser-measur...ity-calculator
How to build a laser beam expander to reduce divergence: http://tinyurl.com/BeamExpander
RHD's Relative Perceived Brightness Calculator. Compare brightness @nm: http://lsrtools.1apps.com/relativebrightness
YAG Power Calculator, i.e. convert ns @ mJ to peak power in watts http://tinyurl.com/YAG-Pulse
The forum costs more to run than donations received, if you wish to help click this link: http://laserpointerforums.com/donations.htm
_______
03-14-2016, 07:25 AM #7
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Vision
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Re: Not worth mentioning because they would be in deep trouble!
Quote:
Originally Posted by Alaskan I don't know the reasons why the alien appeared to be so humanoid, but from what I have gathered from a life time of study on this subject, there are some races which are much like our own, that the humanoid form is more prevalent for space faring races for some reason, but not all are similar to us. If you dig deep enough into this subject you won't have any doubts that these things are happening. Of course, as I wrote above, I want to believe, but at the same time I don't want to be wrong and I still see this way.
VFX artist here. Looks worse than what a noobie just starting a vfx course would do. I know 3d sculptors than can do better than this.
You want real ufos, check nasa's old footage. One big reason they cancelled live broadcasts is that people were seeing too many things up there.
Move to 1:40, watch the objects to your left side. Watch a "laserrrrr!!@!@!@" almost hit them coming from earth and see how they evade and fly away. Nasa's explanation? They were dust particles and that the thrusters caused this. Except. The horizon line does not move. If it was the thrusters, the craft would have moved. Nice try. This ain't 1930, where people believe everything you say.
And if you want the best.
Try and deny this NASA.
Last edited by Vision; 03-14-2016 at 07:27 AM.
03-15-2016, 12:02 AM #8
Class 2 Laser Join Date: Jun 2010 Location: New Hampshire Posts: 276 Rep Power: 210
planters
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Re: Not worth mentioning because they would be in deep trouble!
Quote:
there are some races which are much like our own, that the humanoid form is more prevalent for space faring races for some reason, but not all are similar to us.
Sorry, but this is too presumptive. The humanoid form is more prevalent ? So, now you are suggesting there is enough evidence that a curve can be generated?
To me, this as scientific as claims that we can group angels into different categories.
Quote:
Why would a race capable of the trip fail to recover the body? If they can and have remained "secrete", then they would not want to leave the body. If everything went to shit (it happens ie Challenger), then there should be loads of evidence. An interstellar ship or its wreckage do not blend in.
I'm going to ask this again.
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-- DarkShadows V5 -- Responsive LPF -2562016 -- Default Style Contact Us - Laser Pointer Forums - Archive - Top | 3,034 | 12,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-44 | latest | en | 0.825024 |
https://web2.0calc.com/questions/7x-x-2-what-is-the-value-of-x | 1,545,230,871,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376832330.93/warc/CC-MAIN-20181219130756-20181219152756-00471.warc.gz | 788,443,600 | 5,941 | +0
7x = x^2 What is the value of x?
0
586
3
7x = x^2 What is the value of x?
Guest May 20, 2015
#3
+5
This is the person who asked the question and I just wanted to say thank you for the help and explanations!
Guest May 21, 2015
#1
0
$${\frac{{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{x}}}{{\mathtt{x}}}} = {\frac{{{\mathtt{x}}}^{{\mathtt{2}}}}{{\mathtt{x}}}}$$
$${\mathtt{7}} = {\mathtt{x}}$$
Guest May 20, 2015
#2
+93038
+5
Let's be careful with Anonymous' "method".......this is what is known as "thtowing away a root"
7x = x^2 subtract 7x from both sides
x^2 - 7x = 0 factor
x (x -7) = 0 and setting both factors to 0, we have that x = 0 and x = 7....two solutions....!!!!
CPhill May 20, 2015
#3
+5 | 290 | 734 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-51 | latest | en | 0.900571 |
https://meta.stackoverflow.com/questions/321194/marking-a-question-as-a-duplicate-of-a-question-whose-body-contains-the-answer | 1,713,200,174,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00589.warc.gz | 365,858,577 | 32,099 | # Marking a question as a duplicate of a question whose body contains the answer
Let's suppose we have a question Q1 like:
I know that when we do X, Y happens, and so you get Z. So, How can I use X to get W?
This question contains answers on how to get W by using X.
Now, we have a question Q2 like:
Why do I get Z on doing X?
Because Y happens.
So, is it OK to close Q2 as a duplicate of Q1 when the question contains the answer?
A similar thing happened to me a few months ago, when my question was closed like this.
Since I see close votes saying "unclear what you are asking", I shall now explain with a "real-world example" as DavidPostill says.
Suppose there is a question Q1 :
I know that when we use `str.split("[0-9]")` on a string, it gets split up into parts, if there are any numbers in the string, as [0-9] is a regular expression which matches any single digit. Here, it matches 0 and 9, thereby splitting the String into three parts, i.e, `"a["`, `"-"`, and `"]b"`. So, how can I use `split()` to split a String with the exact sequence "[0-9]". Like:
`"a[0-9]b".split(??)` => `{"a","b"}`
You can do: `"a[0-9]b".split("\\Q[0-9]\\E")`
Now there is a question Q2 like:
When I do `"a[0-9]b".split("[0-9]")`, why do I get `{"a[","-","]b"}` instead of `{"a","b"}` ?
And the answer to this one would be :
This is because `[0-9]` is a regular expression which matches any single digit. Here, it matches 0 and 9, and splits your String into three parts, i.e, `"a["`, `"-"`, and `"]b"`.
So, now you can see that answers of Q1 do not answer Q2, but the question Q1 does answer Q2.
So, would it be right to close Q2 as a duplicate of Q1?
• (For your example, it was closed as the wrong duplicate. I left a comment there) Apr 17, 2016 at 13:45
• Your example is hard to understand (all those W, X, Y and Zs). Can you provide a real world example of Q1 and Q2? Your example of Why does a HTTP URL in Java compile? appears to be a Q1 that has been correctly closed as a Q2 which is the other way around from what you say in this question. I'm very confused and now my brain hurts :/ Apr 17, 2016 at 15:15
• @Tunaki Yes. I agree with that one. And so, since you are a gold badge user, can you not make it correct? Apr 18, 2016 at 11:28
• @DavidPostill See the edit. Apr 18, 2016 at 11:49
• With your real example, then "So, would it be right to close Q2 as a duplicate of Q1?" is No (Q1 does not answer the Why of Q2). Apr 18, 2016 at 11:55
• IMHO if there isn't an answer on the question you are using to close as a dupe then it is not appropriate. For me it goes along the same rational that you need an up voted answer on the dupe target in order to use it. The question needs useful answer that answers the question. Apr 18, 2016 at 11:58
• "So, is it OK to close Q2 as a duplicate of Q1 when the question contains the answer?" Your question was hard to understand until this line, consider putting it (or a variation of it) at the top and/or as the question title. Apr 19, 2016 at 0:27
• I'd suggest editing the title to `Marking a question as a duplicate of a question when the other question is itself the answer` as 'contains' makes it sound like the QA contains an answer, which is what a normal duplicate generally is... Apr 19, 2016 at 9:00
• You are looking for this meta.stackoverflow.com/q/292329/792066 Apr 20, 2016 at 14:40
I'm fairly sure this has been answered before. But no, for a question A to be marked as duplicate of question B, the questions A and B must be more or less the same, and the answer(s) to question B must answer the question from question A.
I believe we should be more careful about closing questions as duplicates.
I've seen several times (I'll try to find a specific example) where a questioner gets referred to a similar question whose answer, with just a little interpretation, answers the OP's question. But I feel this is fallacious, and is often a disservice to the OP.
For almost any question anyone can ask, on almost any topic, there's probably some information out there on the net somewhere which, with a little interpretation, answers the question. But people who ask questions aren't looking for puzzles, they're looking for answers! And although it may be obvious to an expert how a referenced answer can also solve a related question, it may be a complete mystery to the less-experienced user asking the related question.
So, please, if it's really exactly the same question, go ahead and mark it as a dupe, but otherwise, maybe it's okay to spin up some new answers tailored specifically to the variation on the question that's just been asked.
One of the huge consideration imo is searchability and also the semantic of the site. When you are asking a question or checking if questions haqve been asked you use the question. You cannot seriously expect someone to look in the body of every vaguley related question and spend time parsing complex logic to se that it has already been asked.
Id say that these shuoldn't be marked as dupes because you can argue that a ton of this stuff is known a priori, which is basically where this could go unchecked. Most questions could already be closed as dupes of RTFM
Just because one question contains an answer to another due to inverse - or worse itself being a compound question doen not mean that the inverse or the single question shouldnt be answered on its own. If anything the original question shuold be cut into parts.
• In your first paragraph, are you answering the question "should it be closed as a dupe", or the question "should it be deleted as a dupe"? What you say about about searchability and checking other questions would mean we should close as a dupe, because a closed dupe has a direct link to the specific question that answers it. Thus someone who happens upon it is directed to their answer, without having to search further. Apr 20, 2016 at 14:44
• yes fair comment on close vs delete. You are right then imo. Apr 20, 2016 at 17:38
You should use "Close as duplicate" if the banner text "Your question has already been answered " is correct.
Which, when another question contains the answer (including sufficient explanation) for the newer question, is met.
Note that someone else having not been confused by this isn't sufficient, the other question has to contain enough explanation of the behavior being asked about to be considered an answer to the newer question. But then, yes you can flag as a duplicate.
The questions don't necessarily need to match. Quite often, you get questions of the nature:
"What is wrong with programming feature x? When I use it, the program crashes."
But then it turns out that the actual bug is not related to feature x at all, but rather to feature y. The correct answer is: "x misbehaves because you don't understand how to use y. Use y like this: ...". Almost every crappy beginner mistake FAQ looks like this.
It is then perfectly fine to close it as a duplicate to the canonical FAQ, even though the questions are completely different.
Real world example with C++:
"What is wrong with cout? I try to print an integer with it but it prints unexpected garbage."
``````i=i++;
std::cout << i;
``````
And then another question goes like:
"This code gives `i` value 5 but I arrogantly demand that it gives value 6! Is my compiler broken?"
`````` int i=5;
i=++i;
``````
It would be correct and encouraged moderator practice to close both of these questions as canonical duplicates to Undefined behavior and sequence points. As you can see, the two made-up questions above are quite different from the one in the canonical duplicate.
• This can be the right course of action but it's rare. One example is the popular .NET NullReferenceException question. All dupes are different but the issue is the same. That's good enough to get this crap out of sight and close it. Really, all these questions are what we formerly closed as "too narrow" before that was disallowed. Apr 19, 2016 at 14:11
• @boot4life Rare? This is extremely common, there's probably hundreds of such crap beginner FAQ posted every day. Instead of finding some obscure, similar-sounding question, use the canonical duplicate, which will have good and correct answers. And closing questions with canonical duplicates is how you build up the "frequent" tab of each tag to something useful, where questions often used to close other questions as duplicates gain rank. Crude as that might be, that's the SO FAQ system per current design. Apr 19, 2016 at 14:15
• "It is then perfectly fine to close it as a duplicate to the canonical FAQ" - that is not the consensus. You may have missed the past few discussions on this, but you can't state it like that. Apr 19, 2016 at 15:19 | 2,158 | 8,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.954992 |
https://stats.stackexchange.com/questions/380254/glm-and-implementation-of-poisson-regression-model-in-r-by-hand | 1,582,554,640,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145960.92/warc/CC-MAIN-20200224132646-20200224162646-00321.warc.gz | 549,448,258 | 33,213 | # GLM and implementation of Poisson regression model in R by hand
first of, this is not my school exercise but a given example that I'd like to convert from Stan to my own code. I am very much a pragmatic learner so doing this helps me a lot to visualize the problem.
The problem is the following:
#The data give numbers of fatal accidents on scheduled airline flights per year over a ten-year
#period. Assume that the number of fatal accidents in year t follows a Poisson distribution with
#mean theta where log(theta)=a+bt.
#Year 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985
#accidents 24 25 31 31 22 21 26 20 16 22
#theta[i] # true' mean number of fatal accidents
y = c(24, 25, 31, 31, 22, 21, 26, 20, 16, 22) # number of fatal accidents
t = c(1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985) # year
#Using noninformative priors on a and b, obtain the posterior distributions of a and b. Plot
#the approximate probability density for the expected number of fatal accidents in year 1988.
#Obtain the predictive interval for the the number of fatal accidents in 1988.
Which is run with the following Stan model:
//Poisson_regression_model
data{
int N; //the number of observations
real x[N]; //years, real because of mean
int y[N]; // accidents
}
parameters{
real a_star;
real b;
}
transformed parameters{
vector[N] log_theta;
vector[N] theta;
real a;
a=a_star-b*mean(x);
for( i in 1 : N ) {
log_theta[i] = a_star+b*(x[i]-mean(x));
}
theta=exp(log_theta);
}
model{
//priors
a_star ~ normal(0, 10^3);
b ~ normal(0, 10^3);
//likelihood
for( i in 1 : N ) {
//see Stan reference manual page 520
y[i] ~ poisson_log(log_theta[i]);
}
}
Full code and which I'd like to convert into my own, very naive implementation. So this is how far I got but I must have some mistake calculating my likelihood/log_theta.
a_param_space <- seq(0.1, 20, length=100)
b_param_space <- seq(0.1, 20, length=100)
a_prior <- function(x) dnorm(x, 0, 1e3)
b_prior <- function(x) dnorm(x, 0, 1e3)
theta_likelihood <- function(x, theta) dpois(x, theta, log = T)
y = c(24, 25, 31, 31, 22, 21, 26, 20, 16, 22)
t = c(1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985)
posterior <- function() {
plen <- length(a_param_space)
res <- matrix(rep(0, plen*2), ncol=2)
for(i in 1:plen) {
for(j in 1:plen) {
a <- a_param_space[i]
b <- b_param_space[j]
log_theta <- a + b*(t - mean(t))
print(log_theta)
ab_likelihood <- theta_likelihood(y, log_theta)
print(ab_likelihood)
ab_prior_likelihood <- sum(a_prior(a) + b_prior(b) + ab_likelihood)
print(ab_prior_likelihood)
res[i, j] = ab_prior_likelihood
}
}
a_sum <- sum(res[,1])
b_sum <- sum(res[,2])
normalized <- matrix(rep(0, plen*2), ncol=2)
normalized[,1] <- res[,1]/a_sum
normalized[,2] <- res[,2]/b_sum
return(normalized)
}
Also I'm not sure have I named my variables correctly, it's very hard to me to convert the mathematical formula into programming terms. Say as another example I have a likelihood function $$y_i|\theta_i \sim Bin(n_i, \theta_i)$$ where $$logit(\theta_i)=\alpha + \beta x_i$$ with the link function $$logit(\theta_i)=log(\theta_i/(1-\theta_i)$$.
So this would mean I have to pass as a parameter to binomial density the values y-vector, n-vector and theta, where theta is computed from two parameters (with some values from parameter space), alpha and beta. Those two are then computed from the linear regression model: alpha + beta * x-vector which are inputted to the logit function (alpha + beta * x-vector)/(1-alpha + beta * x-vector) to get as a result a vector of log-odds? Am I almost correct?
Thank you in advance for answering, I am very much a caricature programmer who knows how to code but can't understand math. No need to rub it in.
• Hi TeemuK - have you looked into the use of the rstanarm package in R? The reason I ask is that you might find it simpler to specify your model in R using the built functions related to setting priors for your model parameters and for specifying your likelihood function in that syntax format first. You can have the function write the stan code to file for inspection. Also you are using pretty diffuse priors - the Stan manual recommends weakly informative priors whenever possible. – Matt Barstead Dec 8 '18 at 16:39
• You might also find it easier to specify the linear predictor first in your transformed parameters statement, then exponentiate, and then model that term as Poisson-distributed. See this link for an example: rpubs.com/kaz_yos/stan-pois1 – Matt Barstead Dec 8 '18 at 17:11
Well I did manage to create the Poisson regression using MAP estimation (not MLE :)). Although I'm sure there is still some parts of it wrong, I'm happy that I managed to pull it off. It has pretty funky looking posterior distribution. However, I didn't figure out how to do the binomial regression. Dunno will I bother looking into it.
a_param_space <- seq(2, 4, length=100)
b_param_space <- seq(-1, 1, length=100)
a_prior <- function(x) dnorm(x, 0, 1e3)
b_prior <- function(x) dnorm(x, 0, 1e3)
poisson_log_ll <- function(y, x, a, b) {
theta = exp(a + b*x) # link function
return(sum(dpois(y, theta), log=TRUE))
}
y = c(24, 25, 31, 31, 22, 21, 26, 20, 16, 22)
t = c(1976, 1977, 1978, 1979, 1980, 1981, 1982, 1983, 1984, 1985)
tt = t - 1976 # Normalize to 0
posterior <- function() {
plen <- length(a_param_space)
res <- matrix(rep(0, plen*plen), ncol=plen)
for(i in 1:plen) {
for(j in 1:plen) {
a <- a_param_space[i]
b <- b_param_space[j]
ab_likelihood <- poisson_log_ll(y, tt, a, b)
ab_prior_likelihood <- sum(a_prior(a) + b_prior(b) + ab_likelihood)
res[i, j] = ab_prior_likelihood
}
}
normalized <- matrix(rep(0, plen*plen), ncol=plen)
normalized <- res/sum(res)
return(normalized)
}
map <- function(post) {
res = which(post==max(post), arr.ind = TRUE)
return(c(a_param_space[res[1]], b_param_space[res[2]]))
}
pois_post <- posterior()
print(map(pois_post))
print(glm(y ~ tt, family=poisson)\$coefficients)
contour(y=a_param_space, x=b_param_space, z=pois_post)
image(y=a_param_space, x=b_param_space, z=pois_post)
persp(y=a_param_space, x=b_param_space, z=pois_post, theta = 25, phi = 35)
persp(y=a_param_space, x=b_param_space, z=pois_post, theta = 0, phi = 90)
` | 1,890 | 6,243 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-10 | latest | en | 0.72713 |
https://robotics.stackexchange.com/questions/20210/how-to-find-the-optimal-path-for-a-line-following-maze-robot | 1,725,881,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651098.19/warc/CC-MAIN-20240909103148-20240909133148-00145.warc.gz | 468,951,817 | 41,813 | # How to find the optimal path for a line following maze robot?
First let me explain the problem.
This is a maze made only out of black lines on a white surface. The robot has only a few IR sensors which can sense the position of the line. No other sensory input is available.
This is a maze with many self-loops and so a simple LSRB or equivalent algorithm wont work. The robot is supposed to learn the maze and then solve it as optimally as possible .
The above figure represent the possible intersections as all the paths are at 90deg to each other.
As far as my understanding goes, the robot will first scan how many nodes there are, and what their connections to each other is, thus effectively constructing the graph. Next, implement any shortest path algorithm and make your robot follow it.
However, the main problem that i cant get my head around is this:
How will this blind robot know it isnt viewing the same node multiple times if it keeps coming back to the same point after getting caught in a loop?
Also, please suggest good approaches to solving this problems along with any experience anybody has. How does one shot searching methods like DFS , Iterative deepeing DFS , hill climbing work in such scenarios ?
The arrow indicates the direction of the robot and the black lines represent the track, which the robot has to follow .
How will this blind robot know it isn't viewing the same node multiple times if it keeps coming back to the same point after getting caught in a loop?
If you can assume the robot moves at constant velocity, you can measure the elapsed-time between intersections, and use this travel-time as a stand-in for segment length. A grid-representation should allow you to determine whether the robot has arrived at a node that is already present on the map.
Once you have the complete map of node connections, you can easily find a path to minimize travel distance.
You have already suggested that a simple LSRB or equivalent won't work but could you please refer to this article, Coding a Line Follower Robot for Maze using LSRB Algorithm and finding its Shortest Path, which suggest a method for reducing the LSRB path with final optimisation. I am quoting the suggested shortest path finding step here,
LBR = B
LBS = R
RBL = B
SBL = R
SBS = B
LBL = S | 485 | 2,311 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-38 | latest | en | 0.947399 |
http://symplio.com/aafmw/dynamic-programming-on-trees-10da54 | 1,624,603,490,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00284.warc.gz | 43,298,113 | 12,734 | independent set. Writing code in comment? This constraint can be satisfied by iteratively finding the subsolutions from sure it has been computed beforehand and its solution stored in $D$. be improved by making use of the tree structure as the memoization matrix Greedy vs. Dynamic Programming on Trees Rachit Jain; 6 videos; 10,346 views; Last updated on Feb 11, 2019; Join this playlist to learn three types of DP techniques on Trees data structure. Dynamic Programming : Both techniques are optimization techniques, and both build solutions from a collection of choices of individual elements. Though I went on to implement this approach, and it did work, all memoization arrays $D$ and $\dbar$ is stored in the tree alongside the node it This post starts with a brief overview on dynamic programming, and ends with an $D_2$ up to $D_{k-1}$. Given a tree with N nodes and N-1 edges, find out the maximum height of tree when any node in the tree is considered as the root of the tree. sense there commonly exists – although not necessarily – a time-space 64-bit long long int would represent. memorizing previous answers and systematically accessing them later we can get problem in LeetCode. maximizes the sum of its weights. tradeoff when implementing a dynamic programming algorithm. The above problem can be solved by using Dynamic Programming on Trees. algorithm execution by solving a problem with table lookups instead of set is actually known to be medium in difficulty by the website. gist. On the other hand $\dbar_2$ is right children of the $k$-th node, we can know the maximum-weight independent We can also use DP on trees to solve some specific problems. where L(m) is the number of nodes in the left-sub-tree of m and R(m) is the number of nodes in the right-sub-tree of m. (a) Write a recurrence relation to count the number of semi-balanced binary trees with N nodes. define $D_k$ as, Similarly, $\dbar_k$ does not contain the $k$-th node, thus, it may or may not an algorithm design technique in which a problem is solved by combining stored rid of the two recursive function calls altogether. Dynamic programming did not play a crucial role in the above-mentioned problems until a ⦠Now notice how the solution of a subproblem $D_k$ requires This solution for node 2 is $D_2 = 5 + 3 + 0 = 8$. Our algorithm supports constraints on the depth of the tree and number of nodes and we argue it can be extended with other requirements. on dynamic programming and search. We all know of various problems using DP like subset sum, knapsack, coin change etc. programming memoization based on arrays. let’s have a deeper look into the House Robber III problem and independent sets Dynamic Programming Problems Time Complexity; Longest Common Subsequence (LCS) O ( M * N ).M and N are the lengths of the first and second sequence respectively. recursion tree has only logarithmic depth and a polynomial number of nodes. first strategy when designing an algorithm. That would grant us an $O(n)$ additional space for the memory array. The maximum height is 3. The problem of finding the maximum-weight independent along the way I felt like there was more going on with my program than was $w_l$ is the weight of the $l$-th node. Characteristics of the underlying data structure being applied at An easy inductive ... name âdynamic programmingâ to hide the mathematical character of his work Given a graph $G=(V,E)$, an independent set = 0$and$D_1 = 1. independent set of a graph is a subset of its vertices in which no two solutions of smaller subproblems. How can we make this less complex? Recursively deï¬ne the value of an optimal solution based on optimal solutions of subproblems 3. The primary topics in this part of the specialization are: greedy algorithms (scheduling, minimum spanning trees, clustering, Huffman codes) and dynamic programming (knapsack, sequence alignment, optimal search trees). Trees (basic DFS, subtree definition, children etc.) The other direction is to move to the parent(call it parent2 to avoid confusion) of the parent(call it parent1) of node i. One will be the maximum height while traveling downwards via its branches to the leaves. contain its children. brightness_4 Dynamic programming is Dynamic Programming(DP) is a technique to solve problems by breaking them down into overlapping sub-problems which follow the optimal substructure. Dynamic programming is an optimization technique. Thus the full recursion tree generally has polynomial depth and an exponential number of nodes. But if the graph was a Tree, that means if it had (n-1) nodes where n is the number of edges and there are no cycle in the graph, we can solve it using dynamic programming. Recently I came by the House Robber III memoization matrices don’t necessarily have to be implemented as actual attention at the subtree rooted at node 2 for a moment. Optimal Substructure:If an optimal solution contains optimal sub solutions then a problem exhibits optimal substructure. Attention reader! The discussion above illustrates how the idea of We all know of various problems using DP like subset sum, knapsack, coin change etc. By the end realization that enables dynamic programming to be applied in this problem. pretty bad. solution. recomputation. Assumingn$is the number of nodes in the tree, suppose we For There are various problems using DP like subset sum, knapsack, coin change etc. alongside tree nodes, actual computation related to the problem solution can member of the Fibonacci In case of multiple branches of a parent, take the longest of them to count(excluding the branch in which the node lies). memoization when solving the House Robber III problem. Looking back at the solution scheme described in the previous section we These bounds can be further In this case, our longest path will be maximum2. The traditional naive recursive solution in C++ is. 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In contrast, in a typical dynamic programming formulation, a problem is reduced to subproblems that are only slightly smallerÅ for instance, L(j) relies on L(j 1). Find$n$, the size of the tree, so that the$D$and$\dbar$memoization pointer implementation tend not to work well with the traditional dinamic computed, and the algorithm takes$O(n)$time to solve the maximum-weight Dynamic Programming (DP) is a technique to solve problems by breaking them down into overlapping sub-problems which follows the optimal substructure. This way memoization matrix access is done implicitly, as opposed to There are various problems using DP like subset sum, knapsack, coin change etc. So the maximum height of both has been taken to count in such cases when parent and branches exist. In the image above, values of in[i] have been calculated for every node i. Elements of dynamic programming Optimal substructure A problem exhibits optimal substructure if an optimal solution to the problem contains within it optimal solutions to subproblems.. Overlapping subproblems The problem space must be "small," in that a recursive algorithm visits the same sub-problems again and again, rather than continually generating new subproblems. 1. In case you’re interested this first implementation can be The maximum height upwards via parent2 is out[parent1] itself. This prevents bloat in the base Dynamic Trees mod which only includes vanilla Minecraft trees. Traverse the tree using DFS and calculate in[i] as max(in[i], 1+in[child]) for every node. (b) Provide a Dynamic Programming algorithm for computing the recurrence in (a). Such a pattern characterizes an$O(2^n)$know which entry of the memoization arrays correspond to a given node. Dynamic programming is both a mathematical optimization method and a computer programming method. Video created by Stanford University for the course "Greedy Algorithms, Minimum Spanning Trees, and Dynamic Programming". defined above. There are various problems using DP like subset sum, knapsack, coin change etc. recursion tree for RF as a binary tree of additions, with only 0s and 1s at the leaves. Different tree data structures allow quicker and easier access to the data as it is a non-linear data structure. This was my found in this Perspective . of this process the$n$-th member of the Fibonacci sequence will be stored in generate link and share the link here. In this tree the outlined independent set has total weight the one from last section, except that now the information from the More simply put, an The success of our approach is attributed to a series of quickly notice that in order to implement it the traditional dynamic The above problem can be solved by using Dynamic Programming on Trees. in order of discovery. Notice this algorithm now requires Please use ide.geeksforgeeks.org, quickly realized that the algorithm scheme showed in the previous section could A gain in time can The solution$D_k$has to contain the$k$-th node, thus, by Create a mapping of tree nodes to integers in the interval$[0, n)$, so we Now we’re on the same page with respect to the dynamic programming technique, In the above diagram, when 2 is considered as root, then the longest path found is in RED color. Improved memoization by storing subsolutions in a payload. Let’s start off this new approach by defining our memoization matrix. which can be done in$O(1)$time. In this Or, do we absolutely need arrays at all? for our purposes here. in trees. require$O(n)$time, which won’t increase the overall complexity of the solution in half the number of lines. vertices and asked to find an independent through all possible solutions without having to repeat computations. of the weights of its vertices. Given a leaf node$l$we have that$D_l = w_l$and$\dbar_l = 0$, where Each of the additional steps In order to perform any operation in a linear data structure, the time complexity increases with the increase in the data size. begin right away. Recently I came by the House Robber III problem in LeetCode. been solved. From the base cases of the problem we know$D_0 set that F_{n-1} + F_{n-2}$, with$F_0 = 0$and$F_1 = 1$. This implementation runs instantaneously for values of$n$way past what a C++ Third Application: Optimal Binary Search Trees. From the definitions of$D$and$\dbar$we see that solving the subproblem for the maximum-weight independet set of the subtree rooted at the$k$-th node that Since the eventual output is F n, exactly F n of the leaves must have value 1; these leaves represent the calls to RR(1). This is a dynamic programming problem rated medium in difficulty by the website.$D_k$, corresponds to the$k$-th member of the Fibonacci sequence. computer by adding up the two last answers with a calculator. We know$D_2$will be : Longest Increasing Subsequence (LIS) O ( N ^ 2 ).N is the number of elements in the sequence. memoization array. Dynamic programming is breaking down a problem into smaller sub-problems, solving each sub-problem and storing the solutions to each of these sub-problems in an array (or similar data structure) so each sub-problem is only calculated once. Add 1 for the edge between parent and subtree. independent set problem on trees. But, it is not acceptable in today's computational world. actually necessary. 2. I was patient enough to run this algorithm in my machine up to input$n=45$, at Dynamic Programming Memoization with Trees 08 Apr 2016. : Matrix Chain Multiplication can be done along the traversal in the previous requirement by numbering nodes This constraint can be satisfied by finding subsolutions from the the sum of the maximum of the solutions of its children. My problem, and the reason I decided to write this post, was that trees on a among the simplest dynamic programming examples one can find, it serves well improved to constant space while maintaining$O(n)$time by realizing that only Let’s focus our A naive approach will be to traverse the tree using DFS traversal for every node and calculate the maximum height when the node is treated as the root of the tree. Dynamic Programming on Trees - In Out DP! maximum among$D_r$and$\dbar_r$, where$r$is the node that represent the For more explanation about dynamic programming and other algorithm design Tree DP Example Problem: given a tree, color nodes black as many as possible without coloring two adjacent nodes Subproblems: â First, we arbitrarily decide the root node r â B v: the optimal solution for a subtree having v as the root, where we color v black â W v: the optimal solution for a subtree having v as the root, where we donât color v â Answer is max{B includes (excludes) the$k$-th node. By storing memoization as a payload Let B(S,i,j) denote the size of the largest independent subset I of Di such that Iâ©Xiâ©Xj=S, where Xi and Xj are adjacent pair of nodes and Xi is farther from the root than Xj. Although the actual Like divide-and-conquer method, Dynamic Programming solves problems by combining the solutions of subproblems. sets on the children of$k$that do not include them. Manual we have an array$D_{0..n}$of size$n+1$, where its$k$-th entry, denoted One will be the maximum height while traveling downwards via its branches to the leaves. Both$D_k$and$\dbar_k$can be computed The maximum height of tree when node i is considered as root will be max(in[i], out[i]). This solution requires us to store two arrays of size$n$each, corresponding Besides, this led to a more elegant, and more readable Add-ons are mods that do the work of including modded trees in a more modular and maintainable fashion using the Dynamic Trees API. Let A(S,i) denote the size of the largest independent subset I of Di such that Iâ©Xi=S. I can answer this faster than my computer. In the following section we explore implementation details of the algorithm Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Essentially the concept of the solution algorithm here is the same scheme as The above diagram explains the calculation of out[i] when 2 is considered as the root of the tree. to$O(n)$words of extra memory space. Lecture 10: Dynamic Programming ⢠Longest palindromic sequence ⢠Optimal binary search tree ⢠Alternating coin game. DP can also be applied on trees to solve some specific problems. Overlapping subproblems:When a recursive algorithm would visit the same subproblems repeatedly, then a problem has overlapping subproblems. corresponds to the addition$w_k + \dbar_l + \dbar_r$. complexity algorithm. DP notions.$O(n)$solution. After the arrays$D$and$\dbar$programming way we will need to: Only after these two steps are done we would be able to compute the memoization This way whenever we need a previous solution we can be I. create a mapping of nodes to integers. The base case of this dynamic programming solution are the leaves of the Below is the implementation of the above idea : edit dynamic programming problem, is probably the problem of finding the$n$-th The above diagram represents a tree with 11 nodes and 10 edges, and the path which gives us the maximum height when node 1 is considered as root. In this tutorial we will be discussing dynamic programming on trees, a very popular algorithmic technique that solves many problems involving trees. corresponds to. No need to store all the lengths of branches. Dynamic Programming Optimal Binary Search Trees Section 3.5 . Who Should Enroll Learners with at least a little bit of programming experience who want to learn the essentials of algorithms. DP can also be applied on trees to solve some specific problems. In both contexts it refers to simplifying a complicated problem by breaking it down into simpler sub-problems in a recursive manner. From the parent of node i, there are two ways to move in, one will be in all the branches of the parent. We'll be learning this technique by example. sequence defined by$F_n = If a problem has overlapping subproblems, then we can improve on a recurs⦠have been entirely computed, the answer of the problem will correspond to the Some redefinitions of BST ⢠The text, âFoundations of Algorithmsâ defines the level, height and depth of a tree a little differently than Carrano/Prichard ⢠The depth of a node is the number of edges in the path from the root to the node â This is also the level of the node tree. I was pretty bad at DP when i started training for the ICPC (I think i've improved a little :D), also read CLRS, Topcoder and USACO tutorials. Provided leaves up to the root, which can be fulfilled in either depth-first or From now on I will keep in mind that the concept of dynamic programming matrices. code, Time Complexity : O(N) Auxiliary Space : O(N). A(S,i)=|S|+âj(B(Sâ©Xj,j,i)âw(Sâ©Xj))B(S,i,j)=maxA(Sâ²,i)whereSâ²âXiandS=Sâ²â©Xj If the first maximum path thus obtained is same as in[i], then maximum1 is the length of the branch in which node i lies. This is a dynamic programming problem rated Let’s have a look at an example to illustrate the idea. If a problem has optimal substructure, then we can recursively define an optimal solution. As stated earlier, although the $n$-th member of the Fibonacci sequence is systematically storing answers in a memoization matrix can help you speed up \dbar_5 + D_3$, which corresponds to$3 + 3 = 6$.$\max(D_l,\dbar_l) + \max(D_r, \dbar_r)$. Both options are allowed so we choose whichever is the definition of independent sets, it can’t contain either of his children. in constant time. The definition of this have to implement, a function that returns the weight of its maximum-weight Dynamic Programming works when a problem has the following features:- 1. Computing one entry of the arrays is Moving up, in this case, the parent of 2 i.e., 1 has no parent.$w_2 = 5$plus the solutions of its children that do not contain its children. The dynamic programming version computes both VC(root, false) and VC(root, true) simultaneously, avoiding the double call for each child. Unlike Factorial example, this time each recursive step recurses to two other smaller sub-problems. Optimal Substructure : When node i is considered as root, in[i] be the maximum height of tree when we travel downwards via its sub-trees and leaves.Also, out[i] be the maximum height of the tree while traveling upwards via its parent. The basic idea in this problem is youâre given a binary tree with weights on its vertices and asked to find an independent set that maximizes the sum of its weights. Oct 24, 2019 Consider the following problem - Given a tree, for each node, output the distance to the node farthest from it. In this problem we are asked to find an independent set that maximizes the sum arrays systematically up to the tree root and solve the problem. root of the tree. Since, algorithm used is based on DFS, all the branches connected to parent will be considered, including the branch which has the node. its size, so this requires a full tree traversal. At the general case we wish to solve the maximum-weight independent set of Trees(basic DFS, subtree definition, children etc.) While the other will be the maximum height when traveling upwards via its parent to any of the leaves. larger, which means$\dbar_k$corresponds to the computation of anecdote on how I tried two different implementations of dynamic programming be achieved by referring to precomputed solutions instead of repeating$(u,v) \in E$, either$u \notin S$or$v \notin S$. memozation matrices when entries are the nodes of a tree led to considerable Characterize the structure of an optimal solution 2. dynamic programming type approach to deal with a variety of constraint types on laminar cut families of small width, with applications to chain-constrained spanning trees, path TSP and beyond. The parent of node 10, i.e., 7 has a parent and a branch(precisely a child in this case). which point execution was so slow I could answer for$n=46$faster than my that the previous subproblems$D_{k-1}$and$D_{k-2}have already been solved. typically defined by the TreeNode C++ struct. Other data structures such as arrays, linked list, stack, and queue are linear data structures that store data sequentially. However, in House Robber III we happen to be dealing strictly with trees. accomplished with no more than a few integer summations and array accesses, Advanced dynamic programming: the knapsack problem, sequence alignment, and optimal binary search trees. The rob function is what we by Prof. Steven S. Skiena. Recurrence relation of in[i] and out[i] : in[i] = max(in[i], 1 + in[child]) out[i] = 1 + max(out[parent of i], 1 + longest path of all branches of parent of i). By using our site, youD$($\dbar$), denoted$D_k$($\dbar_k$), corresponds to the total weight of Moreover, Dynamic Programming algorithm solves each sub-problem just once and then saves its answer in a table, thereby avoiding the work of re-computing the answer every time. D_0 = 0$ and $D_1 = 1$ the problem we know $D_0 = 0$ $! Define such functions recursively on the depth of the solutions of smaller subproblems plus! Such functions recursively on the other will be the maximum height while moving.... Trees to solve problems by breaking them down into simpler sub-problems in a recursive algorithm would the! Of$ N $substructure: If an optimal solution based on optimal solutions of subproblems are linear structure... Numbering nodes in order to perform any operation in a linear data structure, the time complexity for DFS of! This implementation neither there are various problems using DP like subset sum, knapsack, coin change.. 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Different tree data structures allow quicker and easier access to the leaves individual elements also define functions... Follows the optimal substructure such functions recursively on the nodes of a binary tree typically. In which no two vertices are adjacent need to follow two strategies: dynamic programming for... Satisfied by iteratively finding the subsolutions from $D_2$ will be discussing dynamic programming algorithm for the... Not necessarily – a time-space tradeoff when implementing a dynamic programming: techniques... Its maximum-weight independent set has total weight 13, as computed from the memoization! Sub solutions then a problem has the following algorithm calculates the MIS problem in LeetCode is number! Can recursively define an optimal solution to construct a DP solution, we need to all! Above a node is its $D_k$ and $D_1 = 1$ of! Tree data structures allow quicker and easier access to the problem we know D_0! Know of various problems using DP like subset sum, knapsack, coin change etc ). Vertices are adjacent solution contains optimal sub solutions then a problem exhibits substructure... Given a tree decomposition with treewidth k. the algorithm design Manual by Prof. Steven S... Recursion tree generally has polynomial depth and a branch ( precisely a child in this implementation neither there various... Structures allow quicker and easier access to the parent of node 10 i.e.! When traveling upwards via parent2 is out [ i ], move upwards to the parent of i. Queue are linear data structures allow quicker and easier access to the parent 2! Payload alongside tree nodes, actual computation related to the data size alignment, and more readable in. Into simpler sub-problems in a recursive manner optimal substructure, then a problem has optimal substructure then a problem the! At an example to illustrate the idea it is not acceptable in today 's computational world various problems using like. At an example to illustrate the idea illustrate the idea programming solves problems by it... Of Di such that Iâ©Xi=S contain its children while $\dbar_k$ is the exact realization that enables dynamic algorithm... Our algorithm supports constraints on the other will be maximum2 number above a node is its ... 1S at the $k$ -th member of the subtree rooted at the general we. Number below = 0 $and$ \dbar_k $can be found in this case our. Memoization as a dynamic programming: both techniques are optimization techniques, and dynamic programming is algorithm! Other algorithm design technique in which a problem exhibits optimal substructure: an! Such cases when parent and a polynomial algorithm does exists algorithm design technique which. Commonly exists – although not necessarily – a time-space tradeoff when implementing a dynamic programming on trees solve! Solutions then a problem has the node are considered while calculating the maximum height while traveling via. Re interested this first implementation can be satisfied by iteratively finding the independent! For people wanting to get started at competitive programming and get good at it all... Begin right away are optimization techniques, and dynamic programming is both a mathematical optimization method and a computer method..., then a problem has overlapping subproblems: when a problem has overlapping subproblems problems! Learn the essentials of algorithms end of this dynamic programming ( DP ) is a dynamic programming is also in... From a collection of choices of individual dynamic programming on trees$ \dbar_k $is the root of a binary tree typically! Has the following features: - 1 recursive manner know its size, this., which corresponds to$ D_ { k-1 } $difficulty by the House Robber III problem LeetCode... A branch ( precisely a child in this problem decomposition with treewidth k. the algorithm design technique in no!$ w_k + \dbar_l + \dbar_r $set is actually known to be$ w_2 = 5 $the. Follows the optimal substructure: If an optimal solution contains optimal sub solutions then a has. Binary tree of additions, with only 0s and 1s at the subtree rooted at the subtree at! The lengths of branches the book the algorithm uses dynamic programming problem rated medium in difficulty by website. The$ N $way past what a C++ 64-bit long long int represent! Many problems involving trees to$ D_ { k-1 } $all know of various problems DP! The definition of this problem people wanting to get started at competitive programming and algorithm. Has a parent and subtree visit the same subproblems repeatedly, then problem! Base dynamic trees mod which only includes vanilla Minecraft trees, generate link share... Finding the subsolutions from$ D_2 \$ will be the maximum of tree!
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# Problem 42875. Assignment Problem
Solution 2502602
Submitted on 9 Jun 2020 by Jens Kjærgaard Boldsen
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [2,12,8,5; 4,4,5,12; 3,10,8,3; 12,0,1,2]; y_correct1 = [1,2,4,3]; y_correct2 = [1,4,2,3]; test1 = isequal(Assignment(x),y_correct1); test2 = isequal(Assignment(x),y_correct2); assert(test1||test2)
2 Pass
x = [7,5,3,2,6; 6,3,6,6,5; 3,3,9,4,7; 4,5,3,3,4; 2,3,4,3,5]; y_correct = [5,3,4,1,2]; assert(isequal(Assignment(x),y_correct))
3 Pass
x = eye(4); x(x==0) = 3; y_correct = 1:4; assert(isequal(Assignment(x),y_correct))
4 Pass
x = 5 * ones(5); x(1,4) = 1; x(2,2) = 1; x(3,5) = 1; x(4,1) = 1; x(5,3) = 1; y_correct = [4,2,5,1,3]; assert(isequal(Assignment(x),y_correct))
5 Pass
x = 1:4; x = [x; circshift(x,1); circshift(x,2); circshift(x,3)]; y_correct = 1:4; assert(isequal(Assignment(x),y_correct))
6 Pass
x = [7,5,4; 14,16,2; 7,1,9]; y_correct = [1,3,2]; assert(isequal(Assignment(x),y_correct))
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June 24, 2022, 04:50 Boundary condition for temperature inlet with sinusoidal motion #1 New Member Giulia Join Date: Feb 2022 Posts: 19 Rep Power: 2 Hello, I am trying to simulate a tube (solid region) in which a fluid is flowing inside (fluid region). I am using openFoam version 6 and the chtmultiregionfoam solver. The fluid enters the tube at the inlet with a certain initial temperature, flows through the tube, reaches the outlet of the tube, and comes back in a sinusoidal fashion. So it means that once the fluid enters the tube, it remains inside and it moves back and forth. It interacts with the tube walls (which are heated) so the fluid will heat up and increase its temperature. The fluid has a certain inlet temperature and inlet velocity but with time it should absorb the heat generated by the solid region. However, it seems that the at every repetition of the sinusoidal velocity cycle (at 0.1 Hz), the water inlet is always coming back to the initial input temperature. What I want is rather the evolution of the water temperature depending on the solid heat source, given an initial inlet temperature (occurring at t=0 only). How can I impose that? I hope I am being clear. This is the temperature for the water: Code: ```dimensions [ 0 0 0 1 0 0 0 ]; internalField uniform 293.15; boundaryField { #includeEtc "caseDicts/setConstraintTypes" minX { type fixedValue; value \$internalField; } maxX { type zeroGradient; } "(min|max)(Z)" { type empty; ~value; } "water_to_.*" { type compressible::turbulentTemperatureCoupledBaffleMixed; value \$internalField; Tnbr T; kappaMethod fluidThermo; } }``` And this is the code for the water velocity: Code: ```dimensions [ 0 1 -1 0 0 0 0 ]; internalField uniform (0 0 0); boundaryField { #includeEtc "caseDicts/setConstraintTypes" minX { type uniformFixedValue; uniformValue sine; uniformValueCoeffs { frequency 0.1; amplitude 0.015; scale (0.7 0 0); level (0 0 0); t0 0; } } maxX { type pressureInletOutletVelocity; value \$internalField; } "(min|max)(Z)" { type empty; ~value; } "water_to_.*" { type noSlip; } }``` Thank you for your help! | 576 | 2,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-49 | longest | en | 0.834614 |
https://www.coursehero.com/file/44515346/ECON-404-Homework-1pdf/ | 1,600,638,599,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198652.6/warc/CC-MAIN-20200920192131-20200920222131-00118.warc.gz | 829,989,330 | 229,267 | ECON 404 - Homework 1.pdf - Mia Ha ECON 404 Summer 2019 Quarter Honewort 1(a(6 points Suppose the cost function is given by T C(q = F cq where F c > 0
# ECON 404 - Homework 1.pdf - Mia Ha ECON 404 Summer 2019...
This preview shows page 1 - 7 out of 15 pages.
(a) (6 points) Suppose the cost function is given by T C (q) = F + cq where F; c > 0: At what output level is the average cost minimized? (6 points) Consider a monopoly market. Demand is given by Q(p) = 100 Mia Ha ECON 404 Summer Quarter 2019 Honewort 1 . - TC Cg ) = F t Cq ( F , c > O ) ACC g) = Ff t c . If we want to minimize the average eositgthen :O q has to be a really large number . q = to quantity Qcp ) = 100 p p = IN - g TR = 100g qd MRCq ) = too -2g MCC g) = MR (g) 10 = 100 - Iq g = 45 p = too -45=55 maximizing a = 45 maximizing I = 55 DWL = Ig x 45×45 = 1012 . 5 p ^ CS = ¥45 x 45 = 1012 'S too , yes 55 Hi . . . sa : : : :c :& : : : : : : : - ions : : MR l D = Its d 90 45 I
(c) (6 points) Consider a monopoly market. Demand is given by Q(p) = 100 - p and the total cost function of the monopolist is TC(q) = 10q+800: If the government imposes the average cost pricing restriction, how much would the price and dead-weight loss be? Qlp )= 100 - p p = 100 - g TCC g) = log -1800 ACG ) = lot 8,01 Mug ) = 10 ACC g) = 100 - q to +891=100-9 8090-+9 - 90=0 gd - 90g -1800=0 { 9g to III. ' 00-10=90 IN 80=20 p n Quantity is 9=80 price is p = 20 AC DWL - - Lx 10×10=50 §¥w So the dead weight loss is 50 - i i 80190 Q 10
Tel get g) = I 491 -192 ) & MC (g) = 91 -19£ Formarketti Qe Cp ) = 100 - Lp g , = 100 - Ip p = 50 - TRE p . q= 50g , - O . 599 Mrfq ) = 50 9 , Let MR , Cgt = McCoy 't 50 91 = 91 -192 50 - 2g , = q , 91=25-0.5927 From the calculation in market I 92=20-1*921 q , = 25 - 0.5 ( IO - ¥91 ) 9 , = 25 - IO t ¥9 , 91 = 15 + Eg 91 ¥91 = 15 q*_ Demand Elasticity for market : Hp (G) = . If = JO - = 25 / = - o ' ' ' ET , PY- = - 0.5 Inelastic ble - LL - 0.5 CO
For . Market 2 : Q2 ( p ) = 50 - 0.5 p p = 5%92 = 100 - Iga TRL 92 ) = p.gg = 10092 - 2922 MR (g) = 100 - 492 MRC q ) = MC Cgs ) I 00 - 492 = 9 , t 92 100 - 9 , = 592 92=20-1*921 From the calculation on market 2 : 91=25-0.59292 = 20 - ¥ ( 25 -0.5g ) 92 = 20 - 5 + Is 92 92 = 5 t Es 92 ¥92 = 5 gI= Pat = 100 - 29£ = 100 - 2x = 2530 ¥-::i¥¥÷÷ = - I It's elastic because - I L - I
Fix this one , tcfq ) - - 91 tccqg ) - 492 p= 10 - q Firmtthaxttecaf ) =p Castaway - Tccae ) = ( to - 91 - 92791 - 91 In-car ) 297 = 10-291 - ga - t O = g - 291-92 92=9-291 92 = 9 - 2×4=1 192=17 e- Firms : Mae takes )= play -19192 - Tccqa ) Iska ) CIO q - g) Cgs ) - 4qa 2£91 ) = to - ga . - Iga 4 292 0 = 6 - q , - Iqs 91=6 - 292 91=6 - 219 - Ige ) q , = 6 - 18 -149 , 391=12 191=47 -
PL q ) = 100 - q , TC Cq ) = Iq Mc Cq ) = 2 p = I TR ( q ) = 100g - Iqd MRL g) = 100 49 profit It = TR ( q ) - TC ( q ) = 100g - Ig d - Iq Icq ) = 100 9 q = too - p = 100 - 2 = 98 It = p .
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https://tcnmicrosites.com/just-what-lotto-methods-can-be-often-the-greatest-as-soon-as-picking-lottery-figures-your-own-personal-method-is-usually-everything/ | 1,627,348,616,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152168.38/warc/CC-MAIN-20210727010203-20210727040203-00438.warc.gz | 559,081,253 | 8,670 | # Just what Lotto Methods Can be often the Greatest? As soon as Picking Lottery Figures Your own personal Method is usually Everything
At any time question oneself, ‘Which lottery methods are the ideal for my lottery?’
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The key to understanding this entire process is unveiled when we reply this issue, ‘Better than what?’ In other terms, we have to have a reference stage. As soon as we have it, then all of our methods are calculated from that reference. And, when taking part in the lottery, the reference stage is constantly located the identical way.
How well would we do if we randomly picked the quantities?
Here’s an illustration. ผล สลาก am going to use the Mega Hundreds of thousands lottery, a 5/fifty six sport, to show. If we randomly decide on 5 numbers to perform, that represents eight.nine% of the fifty six variety pool. For that reason, in excess of numerous drawings of the lotteries historical past, we would anticipate to get 8.9% of the winning numbers proper on the typical. This signifies we would typical .forty five right figures per lottery drawing by guessing. If you randomly picked 10 figures to perform, you would regular .ninety proper numbers and so on.
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The Best Lottery Predictions Method
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# Edexcel A2 C4 Mathematics June 2015 - Official Thread Watch
1. (Original post by Pineappleface317)
Hey I'm actually OCR but I couldn't find an OCR C4 2015 thread but I'm guessing you guys do vectors??
Could someone help me with this question:
Find a unit vector that is perpendicular to both 4i + 10j - 3k and 5i - 9j + 7k
All I have so far is:
4a + 10b - 3c = 0
and
5a - 9b + 7c = 0
(ai + bj + ck being my unknown unit vector)
Thanks
Hint: it says "unit" vector
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2. (Original post by Pineappleface317)
Hey I'm actually OCR but I couldn't find an OCR C4 2015 thread but I'm guessing you guys do vectors??
Could someone help me with this question:
Find a unit vector that is perpendicular to both 4i + 10j - 3k and 5i - 9j + 7k
All I have so far is:
4a + 10b - 3c = 0
and
5a - 9b + 7c = 0
(ai + bj + ck being my unknown unit vector)
Thanks
Maybe use c=1 and solve
Posted from TSR Mobile
3. (Original post by aersh8)
Hint: it says "unit" vector
Posted from TSR Mobile
A very poor hint - if it could be considered a hint at all.
4. (Original post by Medicjug)
Maybe use c=1 and solve
Posted from TSR Mobile
Probably better to use c=0
But c=1 would also give a perpendicular vector - just easier when using c=0
5. (Original post by Medicjug)
Maybe use c=1 and solve
Posted from TSR Mobile
just cross them!!!!!!
6. (Original post by Medicjug)
Maybe use c=1 and solve
Posted from TSR Mobile
So that's just because each component in i k and k is one unit long? So I could make anything one?
Sorry for some reason I can't get my head around vectors ..
7. (Original post by Tow)
Probably better to use c=0
But c=1 would also give a perpendicular vector - just easier when using c=0
How come you can use c=0 ?
8. (Original post by Pineappleface317)
So that's just because each component in i k and k is one unit long? So I could make anything one?
Sorry for some reason I can't get my head around vectors ..
No, think of it this way if som vector is parralel to both it will be a mutiple of some direction vector. We can factor out a multiple such that c=1 leaving a,b not determined as of yet and then adjust the actual vector in the end.
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9. (Original post by Tow)
Probably better to use c=0
But c=1 would also give a perpendicular vector - just easier when using c=0
You cant use Z=0 this results in the vector 0,0,0.
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10. (Original post by physicsmaths)
You cant use Z=0 this results in the vector 0,0,0.
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Oh yeah.
I assumed it would be easier - haven't done this question since we went through the book back in Feb!
But yeah you use c=1 sorry about that xD
It can be anything BUT 0 and obviously 1 would be the easiest choice.
11. (Original post by Tow)
A very poor hint - if it could be considered a hint at all.
To be fair that can be used to form 3 simultaneous equations but that'll just be harder to solve
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12. (Original post by Tow)
Oh yeah.
I assumed it would be easier - haven't done this question since we went through the book back in Feb!
But yeah you use c=1 sorry about that xD
It can be anything BUT 0 and obviously 1 would be the easiest choice.
Correct.
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13. useful material for the upcoming exams
http://www.thestudentroom.co.uk/show....php?t=3353445
http://www.thestudentroom.co.uk/show....php?t=3361867
http://www.thestudentroom.co.uk/show....php?t=3361905
14. (Original post by Tow)
Yeah definitely.
But no question should require you to use the 'Loop' method.
Integration by parts like this is fine:
x^2sin(x)
But integration by parts of this shouldn't come up:
Sin(3x)cos(x)
Unless you can rearrange to make it so that you can integrate it with using the loop method.
I don't know the integration by loops method, what should I use to work this out?
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15. (Original post by lam12)
I don't know the integration by loops method, what should I use to work this out?
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Work what out?
16. (Original post by Tow)
Work what out?
The integration of the sin2x whatever it was :-)
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17. (Original post by lam12)
The integration of the sin2x whatever it was :-)
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I think a previous post said that you can rearrange to integrate so try that.
I can't do it right now so if you still don't understand I'll do it when I'm free.
18. do we need to know the integration of tanx cotx secx and cosecs?
19. (Original post by crepole)
do we need to know the integration of tanx cotx secx and cosecs?
They are in the formula book. You could be asked to differentiate one of the ln ones to get to tanx or something along those lines.
Posted from TSR Mobile
20. Does anyone know why you need a minus sign infront of the constant when setting up a diff. equation and its decreasing? thanks
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# Circle Equations
## A circle is easy to make:
Draw a curve that is "radius" away
from a central point.
And so:
All points are the same distance
from the center.
In fact the definition of a circle is
Circle: The set of all points on a plane that are a fixed distance from a center.
Let us put that center at (a,b).
So the circle is all the points (x,y) that
are "r" away from the center (a,b).
Now we can work out exactly where all those points are!
We simply make a right-angled triangle (as
shown), and then use Pythagoras (a
2
+ b
2
=
c
2
):
(x-a)
2
+ (y-b)
2
= r
2
And that is the "Standard Form" for
the equation of a circle!
You can see all the important information at a glance: the center (a,b) and the
General Form
But you may see a circle equation and not know it!
Because it may not be in the neat "Standard Form" above.
As an example, let us put some values to a, b and r and then expand it
2
+ (y-b)
2
= r
2
Set (for example) a=1, b=2, c=3: (x-1)
2
+ (y-2)
2
= 3
2
Expand: x
2
- 2x + 1 + y
2
- 4y + 4 = 9
Gather like terms: x
2
+ y
2
- 2x - 4y + 1 + 4 - 9 = 0
And we end up with this:
x
2
+ y
2
- 2x - 4y - 4 = 0
It is a circle equation, but "in disguise"!
So when you see something like that think "hmm ... that might be a circle!"
In fact we can write it in "General Form" by putting constants instead of the
numbers:
x
2
+ y
2
+ Ax + By + C = 0 | 481 | 1,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-26 | latest | en | 0.855103 |
https://aakashsrv1.meritnation.com/ask-answer/question/a-train-140m-long-running-at-speed-of-60km-h-in-how-much-tim/direct-and-inverse-proportions/1913339 | 1,669,936,159,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710870.69/warc/CC-MAIN-20221201221914-20221202011914-00764.warc.gz | 103,048,682 | 8,833 | # A train 140m long running at speed of 60km/h. In how much time will it pass a platform 260m long?
Length of the train = 140 m
Length of the platform = 260 m.
Total distance covered by the train to pass the platform = 140 m + 260 m = 400 m
Speed of the train = 60 km/hr
Time taken by the train to pass the platform
Thus, the time taken by train to pass the platform is 24 sec.
• -1
What are you looking for? | 118 | 415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-49 | latest | en | 0.957758 |
https://docs.unity3d.com:443/ja/current/ScriptReference/Mathf.Clamp.html | 1,669,987,205,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710902.80/warc/CC-MAIN-20221202114800-20221202144800-00747.warc.gz | 249,740,802 | 5,876 | Version: 2021.3
# Mathf.Clamp
マニュアルに切り替える
public static float Clamp (float value, float min, float max);
## パラメーター
value The floating point value to restrict inside the range defined by the minimum and maximum values. min The minimum floating point value to compare against. max The maximum floating point value to compare against.
## 戻り値
float The float result between the minimum and maximum values.
## 説明
Clamps the given value between the given minimum float and maximum float values. Returns the given value if it is within the minimum and maximum range.
Returns the minimum value if the given float value is less than the minimum. Returns the maximum value if the given value is greater than the maximum value. Use Clamp to restrict a value to a range that is defined by the minimum and maximum values.
Note: if the minimum value is is greater than the maximum value, the method returns the minimum value.
```using UnityEngine;// Mathf.Clamp example.
//
// Animate a cube along the x-axis using a sine wave.
// Let the minimum and maximum positions on the x-axis
// be changed. The cube will be visible inside the
// minimum and maximum values.public class ExampleScript : MonoBehaviour
{
private float xMin = -0.5f, xMax = 0.5f;
private float timeValue = 0.0f; void Update()
{
// Compute the sin position.
float xValue = Mathf.Sin(timeValue * 5.0f); // Now compute the Clamp value.
float xPos = Mathf.Clamp(xValue, xMin, xMax); // Update the position of the cube.
transform.position = new Vector3(xPos, 0.0f, 0.0f); // Increase animation time.
timeValue = timeValue + Time.deltaTime; // Reset the animation time if it is greater than the planned time.
if (xValue > Mathf.PI * 2.0f)
{
timeValue = 0.0f;
}
} void OnGUI()
{
// Let the minimum and maximum values be changed
xMin = GUI.HorizontalSlider(new Rect(25, 25, 100, 30), xMin, -1.0f, +1.0f);
xMax = GUI.HorizontalSlider(new Rect(25, 60, 100, 30), xMax, -1.0f, +1.0f); // xMin is kept less-than or equal to xMax.
if (xMin > xMax)
{
xMin = xMax;
} // Display the xMin and xMax value with better size labels.
GUIStyle fontSize = new GUIStyle(GUI.skin.GetStyle("label"));
fontSize.fontSize = 24; GUI.Label(new Rect(135, 10, 150, 30), "xMin: " + xMin.ToString("f2"), fontSize);
GUI.Label(new Rect(135, 45, 150, 30), "xMax: " + xMax.ToString("f2"), fontSize);
}
}
```
public static int Clamp (int value, int min, int max);
## パラメーター
value The integer point value to restrict inside the min-to-max range. min The minimum integer point value to compare against. max The maximum integer point value to compare against.
## 戻り値
int The int result between min and max values.
## 説明
Clamps the given value between a range defined by the given minimum integer and maximum integer values. Returns the given value if it is within min and max.
Returns the min value if the given value is less than the min value. Returns the max value if the given value is greater than the max value. The min and max parameters are inclusive. For example, Clamp(10, 0, 5) will return a maximum argument of 5 and not 4.
```using UnityEngine;// Mathf.Clamp integer example.
//
// Add or subtract values from health.
// Keep health between 1 and 100. Start at 17.public class ExampleScript : MonoBehaviour
{
public int health = 17;
private int[] healthUp = new int[] {25, 10, 5, 1};
private int[] healthDown = new int[] {-10, -5, -2, -1}; // Width and height for the buttons.
private int xButton = 75;
private int yButton = 50; // Place of the top left button.
private int xPos1 = 50, yPos1 = 100;
private int xPos2 = 125, yPos2 = 100; void OnGUI()
{
GUI.skin.label.fontSize = 20;
GUI.skin.button.fontSize = 20; // Generate and show positive buttons.
for (int i = 0; i < healthUp.Length; i++)
{
if (GUI.Button(new Rect(xPos1, yPos1 + i * yButton, xButton, yButton), healthUp[i].ToString()))
{
health += healthUp[i];
}
} // Generate and show negative buttons.
for (int i = 0; i < healthDown.Length; i++)
{
if (GUI.Button(new Rect(xPos2, yPos2 + i * yButton, xButton, yButton), healthDown[i].ToString()))
{
health += healthDown[i];
}
} // Show health between 1 and 100.
health = Mathf.Clamp(health, 1, 100);
GUI.Label(new Rect(xPos1, xPos1, 2 * xButton, yButton), "Health: " + health.ToString("D3"));
}
}
``` | 1,172 | 4,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-49 | latest | en | 0.535941 |
http://www.jiskha.com/display.cgi?id=1351815256 | 1,498,301,504,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320257.16/warc/CC-MAIN-20170624101204-20170624121204-00317.warc.gz | 562,349,746 | 3,581 | # math
posted by .
Dave swam 6 laps and did 9 dives at the swimming pool.crystal did 4 drives and swam 8 laps.how many more laps than drives did crystal and Dave do?
• math -
Laps: 6 + 8 = 14
Dives: 9 + 4 = 13
They did one lap more than dives.
### Answer This Question
First Name: School Subject: Answer:
### Related Questions
More Related Questions
Post a New Question | 107 | 381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-26 | latest | en | 0.960632 |
https://www.indiabix.com/verbal-reasoning/direction-sense-test/discussion-26 | 1,712,955,119,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816070.70/warc/CC-MAIN-20240412194614-20240412224614-00402.warc.gz | 752,630,190 | 7,778 | # Verbal Reasoning - Direction Sense Test - Discussion
Discussion Forum : Direction Sense Test - Direction Sense 1 (Q.No. 15)
15.
Some boys are sitting in three rows all facing North such that A is in the middle row. P is just to the right of A but in the same row. Q is just behind of P while R is in the North of A. In which direction of R is Q?
South
South-West
North-East
South-East
Explanation:
Q is in South-East of R.
Discussion:
19 comments Page 1 of 2.
Alan said: 6 months ago
@All.
P is in not in the same row as A. It could be R P, then A and Q.
P Gayathri said: 1 year ago
I think North-East is the right answer.
(2)
Poonam said: 1 year ago
North East is the right answer.
Mohammad saqib said: 2 years ago
Moreen said: 2 years ago
I agree South-East is right. As per the question which direction of R is Q. This means on R's POV where can Q be seen.
Roshna said: 3 years ago
As per the question, they asked in which direction R is to Q. So R is in North and West of Q so North-West.
(1)
ASHOK KUMAR said: 5 years ago
Here, A and P should not be in the same row. So in first row R and P, A and Q should be in a 2nd row.
Khalid gowhar said: 5 years ago
As per my knowledge, the question is asking direction of R in respective of Q. So, the answer should be North-West.
Shiv said: 6 years ago
Please tell me what is the meaning of "just behind" in the question?
Srinivas said: 6 years ago
How did you place A and P in the same row? I am not getting it. Please explain me. | 437 | 1,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-18 | latest | en | 0.962989 |
http://mxplx.com/memelist/taxonomy=games | 1,532,358,062,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596542.97/warc/CC-MAIN-20180723145409-20180723165409-00186.warc.gz | 241,997,604 | 6,265 | 10 MAR 2017 by ideonexus
Four Game Mechanics
Agon: This ancient Greek word—meaning “struggle” or “contest”— defines those games in which some aspect of a player’s or team’s skill is measured against another player or team. Any game that is based on skill and eliminates luck is a game of agon. The best examples of this type of game are athletic games such as wrestling and baseball. The games of chess and checkers are also classic examples of agon. Contemporary abstract strategy games, such as those in the Project GIPF ser...
Folksonomies: games gaming mechanics
Folksonomies: games gaming mechanics
10 MAR 2017 by ideonexus
What We Learn from Games
What also went unremarked was how much I was learning by playing these games: basic ideas such as taking turns and developing patience while others completed their turns, the strengthening of simple memory, improved physical coordination, an ability to recognize and act on patterns, the capacity to see what might happen in a few turns if I took one move as opposed to another, resilience when losing, and the kind of strategic thinking that emerges once you realize that Scrabble is both a game ...
02 SEP 2016 by ideonexus
Math Exercise: Multiple Approaches to Problem-Solving
For example, if the problem was to fi nd the answer to 8 × 6, students may suggest three options: memorizing the multiplication table for 6, knowing that 8 × 5 = 40 and adding another 8 to equal 48, or adding a column of six 8s. Allowing students to personally choose among approaches all confi rmed as correct and to support their choice will increase their comfort levels. Th is process also builds math logic, intuition, and reasoning skills that extend into other academic subjects and real-...
Folksonomies: education games math exercises
Folksonomies: education games math exercises
1 notes
02 SEP 2016 by ideonexus
"This Is Not a..." Game
A game called “Th is Is Not a...” encourages multiple solutions and is played in a relaxed environment that encourages creativity. Students pass around an object—such as a toy telephone—and say, “This is not a....” Younger students name an object that is not a toy telephone (for example, “This is not a pencil.”). Older students continue and say, “This is not a toy telephone, it is a...,” and they gesture or mime to suggest the object that they are pretending the toy teleph...
Folksonomies: education games
Folksonomies: education games
1 notes
02 SEP 2016 by ideonexus
Math Games
Buzz. An example of a low-stress, win-win game is Prime Number Buzz. Students stand in a circle or at their desks and go around the room in order, saying either the next sequential number if it is a composite or “buzz” if it is a prime. If they are incorrect, they sit down, but they keep listening and when they catch another student’s error, they stand up and rejoin the game. (The same game format works for Multiples Buzz, using multiples of, for example, 3, 4, and so on.) Telephone. T...
Folksonomies: education games math
Folksonomies: education games math
1 notes
02 SEP 2016 by ideonexus
Math Exercise: Comparisons
Select two boxes or cans of food that weigh 8 ounces and 16 ounces, respectively. Have students hold each as you tell them (or they read) the weights of the containers. Give students a box or can with the weight covered and have them compare the weight of the new package to the weight of the 8- and 16-ounce samples. Th ey can then estimate whether the new item’s weight is closer to 8 or 16 ounces. As students become more successful, they may want to predict a more specifi c weight. Ask them...
Folksonomies: education games math
Folksonomies: education games math
1 notes
03 JUN 2016 by ideonexus
"No Man's Sky" as Humanist Adventure
The true value of No Man’s Sky lies in something both incredibly simple and breathtaking. The point of the game is to discover and share knowledge with the other inhabitants of the universe. It’s almost as if the developers took the Enlightenment-era Encycloédie and turned it into a science fiction video game; a true testament to the best qualities and powers of the Information Age. While the sheer size may overwhelm some or risk boredom for others, players shouldn’t ignore the larger...
09 AUG 2014 by ideonexus
The Race to 100
The children take turns rolling the dice, which are labeledu00a0zero 1s tou00a0five 1su00a0onu00a0the first die andu00a0zero 10s tou00a0five 10s on the second die. Afteru00a0a studentu00a0rolls theu00a0two dice, he takes theu00a0rodsu00a0and cubes representing the number of 10s and 1s he rolled and puts them on his mat. It is then the next player's turn to roll. When a player has ten or more 1s cubes on his mat, he must replace ten 1s with a 10s rod before he hands over the dice for the nex...
Folksonomies: education games math
Folksonomies: education games math
29 JAN 2014 by ideonexus
1/9998 Produces Binary Output
The pattern will break down once you get past 8192, which is 2^13. That means that the pattern continues for an impressive 52 significant figures (well, it actually breaks down on the 52nd digit, which will be a 3 instead of a 2). The reason it works is that 9998 = 10^4 - 2. You can expand as 1 / (10^n - 2) = 1/10^n * 1/(1 - 2/10^n) = 1/10^n * (1 2/10^n 2^2 /10^2n 2^3 /10^3n ...) which gives the observed pattern. It breaks down when 2^k has more than n digi...
Folksonomies: games math puzzles
Folksonomies: games math puzzles
30 DEC 2013 by ideonexus
The Tapping Game
The tapping game is when I tap one time, you tap two times, and when I tap two times, you tap one time. Children play this game sixteen times, mixing up the times that children are asked to tap once when the experimenter taps twice, and tap twice when the experimenter taps once. In other words, the rules of the game keep changing, and the children need to apply their focus and attention to follow what’s going on. Blair says: What happens with four-year-olds is, in general, they’ll hang in...
Folksonomies: games parenting
Folksonomies: games parenting
1 notes
A game for children to learn focus. | 1,557 | 6,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-30 | latest | en | 0.954777 |
http://www.jiskha.com/display.cgi?id=1319257833 | 1,495,777,591,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608642.30/warc/CC-MAIN-20170526051657-20170526071657-00070.warc.gz | 687,420,282 | 4,237 | # science
posted by on .
A 4kg block is sitting on the floor.
(a) How much potential energy does it have? (b) How much kinetic energy does it have?
(c) The block is raised to 2m high. How much potential energy does it have? (d) The block is raised to 4m high. How much potential energy does it have?
(e) The block is raised to 45m high. How much potential energy does it have? (f) The block is dropped to the ground. How fast is it traveling when it hits the ground? (remember chapter 2?)
(g) How much kinetic energy does it have when it hits the ground?
• science - ,
(a)
recall that potential energy is stored energy, and is given by the formula:
PE = mgh (units in Joules)
where
m = mass (in kg)
g = acceleration due to gravity = 9.8 m/s^2
h = height (in m)
if the reference position is at the floor (that is, h = 0), the PE is equal to
PE = 4*9.8*0
PE = 0
(b)
recall that kinetic energy is energy in motion and is given by the formula:
KE = (1/2)*m*v^2
where
v = velocity (in m/s)
since it's not moving (v = 0),
KE = 0
(c)
at 2 m high,
PE = 4*9.8*2
PE = 78.4 J
(d)
at 4 m high,
PE = 4*9.8*4
PE = 158.8 J
(e)
at 45 m high,
PE = 4*9.8*45
PE = 1764 J
(f)
recall that the motion of the block is uniformly accelerated motion, and we can therefore use the formula:
(v,f)^2 - (v,o)^2 = 2gh
where
v,f = final velocity (in m/s)
v,o = initial velocity (in m/s)
since it is dropped from rest, v,o = 0:
v,f^2 = 2gh
v,f = sqrt(2gh)
v,f = sqrt(2*9.8*45)
v,f = 29.7 m/s
(g)
KE = (1/2)mv^2
KE = (1/2)*4*29.7^2
KE = 1764.18 J
this actually shows the law conservation of energy, which is ΔPE = -ΔKE
hope this helps~ :) | 553 | 1,616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-22 | latest | en | 0.947269 |
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# Search results
23 records were found.
## Sufficient Sample Sizes for Multilevel Modeling
An important problem in multilevel modeling is what constitutes a sufficient sample size for accurate estimation. In multilevel analysis, the major restriction is often the higher-level sample size. In this paper, a simulation study is used to determine the influence of different sample sizes at the group level on the accuracy of the estimates (regression coefficients and variances) and their standard errors. In addition, the influence of other factors, such as the lowest-level sample size and different variance distributions between the levels (different intraclass correlations), is examined. The results show that only a small sample size at level two (meaning a sample of 50 or less) leads to biased estimates of the second-level standard errors. In all of the other simulated conditions the estimates of the regression coefficients, the...
## Robustness issues in multilevel regression analysis
A multilevel problem concerns a population with a hierarchical structure. A sample from such a population can be described as a multistage sample. First, a sample of higher level units is drawn (e.g. schools or organizations), and next a sample of the sub-units from the available units (e.g. pupils in schools or employees in organizations). In such samples, the individual observations are in general not completely independent. Multilevel analysis software accounts for this dependence and in recent years these programs have been widely accepted. Two problems that occur in the practice of multilevel modeling will be discussed. The first problem is the choice of the sample sizes at the different levels. What are sufficient sample sizes for accurate estimation? The second problem is the normality assumption of the level-2 error distributio...
## The accuracy of multilevel structural equation modeling with pseudobalanced groups and small samples
Hierarchical structured data cause problems in analysis, because the usual assumptions of independently and identically distributed variables are violated. Muthn (1989) described an estimation method for multilevel factor and path analysis with hierarchical data. This article assesses the robustness of the method with unequal groups, small sample sizes at both the individual and the group level, in the presence of a low or a high intraclass correlation (ICC). The within-groups part of the model poses no problems. The most important problem in the between-groups part of the model is the occurrence of inadmissible estimates, especially when group level sample size is small (50) while the intracluster correlation is low. This is partly compensated by using large group sizes. When an admissible solution is reached, the factor loadings are ...
## Optimal experimental designs for multilevel logistic models with two binary predictors.
Many experiments aim at populations with persons nested within clusters. In the design stage of such experiments one has to decide whether to randomize complete clusters or persons within clusters to treatment conditions. Furthermore, the optimal sample sizes have to be calculated. In this article these two design issues will be dealt with for logistic models with a binary treatment condition and a binary covariate. The multilevel model is used to relate treatment condition and the covariate to the binary outcome. The optimal design is analytically derived for first order Marginal Quasi Likelihood (MQL) by linearizing the model using a Taylor series expansion. A simulation study shows results for second order Penalized Quasi Likelihood, (PQL), which is known to produce less biased estimates. The results show that person level randomiza... | 723 | 3,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-34 | longest | en | 0.912748 |
https://advancesindifferenceequations.springeropen.com/articles/10.1186/s13662-022-03723-7 | 1,670,552,086,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00248.warc.gz | 103,142,311 | 81,786 | Theory and Modern Applications
Dynamics analysis and optimal control of SIVR epidemic model with incomplete immunity
Abstract
In this paper, we establish an SIVR model with diffusion, spatially heterogeneous, latent infection, and incomplete immunity in the Neumann boundary condition. Firstly, the threshold dynamic behavior of the model is proved by using the operator semigroup method, the well-posedness of the solution and the basic reproduction number $$\Re _{0}$$ are given. When $$\Re _{0}<1$$, the disease-free equilibrium is globally asymptotically stable, the disease will be extinct; when $$\Re _{0}>1$$, the epidemic equilibrium is globally asymptotically stable, the disease will persist with probability one. Then, we introduce the patient’s treatment into the system as the control parameter, and the optimal control of the system is discussed by applying the Hamiltonian function and the adjoint equation. Finally, the theoretical results are verified by numerical simulation.
Introduction
The SARS in 2003, the Zika virus (ZIKV) invasion in 2013, the H7N9 avian influenza pandemic, and the emergence of the Dengue virus in the world, these recurrent infectious diseases and various emerging infectious diseases have been challenging modern life and medical standards [1]. For example, COVID-19, which broke out in 2019, is still affecting the world. As of August 24, 2021, the cumulative number of COVID-19 cases and deaths has reached 212,357,898 and 4,439,843. Therefore, how to prevent and control the occurrence and spread of infectious diseases is one of the hot issues today.
From the perspective of mathematics, the study of infectious diseases usually starts according to the transmission mechanism of diseases, which is analyzed by establishing mathematical models. The earliest epidemic model was established by Kermack and Mckendrick. They established the plague susceptibility infection removal model (SIR) [2] in 1927 and the plague susceptible infected susceptible model (SIS) [3] in 1932, respectively. Since the establishment of SIR and SIS models, most of the subsequent research is based on the standard SIR model. Among the existing prevention and treatment methods for infectious diseases, vaccine injection is one of the fast and effective methods. For example, in the prevention and control of COVID-19, vaccine injection can reduce the infection rate of the Delta variant virus to a certain extent. Therefore, an increasing number of researchers take vaccine injection into account in the process of modeling infectious diseases to make the model close to the actual situation.
Among them, Chen et al. in [4] established the susceptibility vaccination–infection isolation–recovery (SVIQR) model and susceptibility–vaccination–infection isolation (SVIQS) model, respectively. The basic reproduction number of the two models was given. Furthermore, the global attractivity and the global asymptotic stability of the solutions were proved by the Lyapunov function method. And the existence of backward bifurcation was also proved. In [5], Kribs-Zaleta and Velasco-Hernández studied a simple SIV model with inoculation and demonstrated the backward bifurcation of solutions to some parameter values. At the same time, complete bifurcation analysis of the model was given under the condition that the vaccine reduces the basic reproduction number. Liu et al. in [6] studied the following SIVR system:
$$\textstyle\begin{cases} S_{t}=\lambda -\beta _{1}SI-(\mu +\alpha )S, \\ V_{t}=\alpha S-\beta _{2}VI-(\mu +\gamma )E, \\ I_{t}=\beta _{1}SI+\beta _{2}VI-(\mu +\delta )I, \\ R_{t}=\gamma V+\delta I-\mu R. \end{cases}$$
(1.1)
Here, λ is the constant update rate of the susceptible host and α is the rate at which susceptible individuals are vaccinated. $$\beta _{1}$$ and $$\beta _{2}$$ are the transmission rates of infected people in contact with susceptible groups and vaccinated groups, respectively. Since vaccinated individuals may have partial immunity during vaccination, it is assumed that $$\beta _{1}>\beta _{2}$$. μ is the host mortality per compartment. γ and δ are the recurrence rates of vaccinated people and infected people, respectively. All of these parameters are assumed to be positive. In [6], the authors gave the threshold dynamics of system (1.1) by using the basic reproduction number, showing that reducing the number of infected individuals by vaccination can control the disease. In addition, many researchers have studied infectious disease models with immunization from the perspective of age structure [7] and pulse vaccination [8].
The above models are all established in a homogeneous space environment. However, in practice, the transmission of some diseases is often related to spatial location. For example, the transmission rate of COVID-19 in Asia is different from that in North America. In [9], Wu et al. discussed a class of spatially heterogeneous host-pathogen models. The authors used the basic reproduction number to discuss the threshold dynamic behavior of the models and gave the asymptotic behavior of the models. In [10], a reaction-diffusion model of SVIR infection in a spatially heterogeneous environment was proposed. The authors gave the proof of the extinction and persistence of the disease by giving a basic reproduction number. In [11], the authors established an SIVS epidemic model with a degree-dependent transmission rate and incomplete vaccination on a scale-free network. The global asymptotic stability of the equilibrium and the global attractivity of the unique endemic equilibrium were proved. In addition, the effects of various immunization programs such as unified immunization, target immunization, and acquaintance immunization were studied and compared.
Motivated by the recent development of epidemic modeling, the optimal control problem is often discussed in some cases, optimal control theory is one of the important branches of mathematical optimization, which is often used to study how to find a control for a dynamic system in a period of time to optimize the objective function. Thus we consider two different models based on (1.1). The first is a direct extension of (1.1). A reaction-diffusion SIVR model is established based on spatial heterogeneity with incomplete immunity. The well-posedness of the system is discussed by using the operator semigroup method. At the same time, the global dynamic behavior of the system solution is discussed by analyzing the basic reproduction number. In the second model, as [12, 13], it is assumed that the spread of a disease can be influenced by decision-makers. That is, decision-makers can control the response rate to a certain extent by increasing the treatment ability or the efficiency of drug treatment. Therefore, by further expanding the model, we obtain the control system under the assumption of limited control resources. Considering the progress of medical technology, the targeted treatment for patients will be gradually developed, so we will consider the targeted treatment for patients as a control parameter in the system and discuss its optimal control problem. In addition, we analyze the optimal control of the system by using the Hamilton equation and the adjoint equation. In the proof, we may encounter the following problems:
• How to determine the basic regeneration number of the system. In the presence of the diffusion term, it is necessary to select an appropriate method to represent the basic reproduction number $$\Re _{0}$$, which is an important prerequisite for discussing the dynamic behavior of the system by using $$\Re _{0}$$ as the threshold value.
• Can the existence of optimal control be obtained? Because of the existence of diffusion terms, it is difficult to define the adjoint equation and the Hamiltonian function of the control system. At the same time, there are some requirements for the selection of parameters in the numerical simulation.
In view of the above problems, this article is organized as follows. In Sect. 2, an SIVR model with incomplete immunity and spatial heterogeneity is established. Furthermore, the well-posedness of the model is derived, meanwhile, the global existence and global attractiveness of the solution are proved. Section 3 is devoted to studying the threshold dynamic behavior of the system. The extinction or persistence of diseases is analyzed by using the basic reproductive number as the threshold. In Sect. 4, the optimal control of the control system is analyzed by taking the treatment for the patient as the control parameter in the system. Meanwhile, the optimal control problem is discussed by using the Hamiltonian function and adjoint equation. Finally, in Sect. 5, the corresponding results are verified by numerical simulation.
Model formulation and well-posedness
In this paper, the spatial heterogeneity of the spread of infectious diseases and spatial diffusion is considered. In addition, for vaccines, we consider vaccination rates in susceptible individuals and the effectiveness of the vaccine. Based on model (1.1), we can establish the following epidemic model of SVIR with incomplete immunity. The meanings of parameters in the system (2.1) are shown in Table 1.
$$\textstyle\begin{cases} \frac{\partial S}{\partial t}=D_{1}\Delta S+\Lambda (x)-r(x)S-(1-r(x)) \beta (x)SI-d_{1}(x)S, \\ \frac{\partial I}{\partial t}=D_{2}\Delta I+(1-r(x))\beta (x)SI+(1- \eta (x))\frac{\alpha (x)VI}{K(x)+I}-(\gamma (x)+d_{2}(x))I, \\ \frac{\partial V}{\partial t}=D_{3}\Delta V+r(x)S-(1-\eta (x)) \frac{\alpha (x)VI}{K(x)+I}-(\eta (x)+d_{3}(x))V, \\ \frac{\partial R}{\partial t}=D_{4}\Delta R+\gamma (x)I+\eta (x)V-d_{4}(x)R. \end{cases}$$
(2.1)
Remark 1
We considered that there is a vaccine coverage rate $$r(x)$$ for susceptible path S, and unvaccinated susceptible persons will be injected into the infected path with a transmission rate $$\beta (x)$$. The susceptible person who has been vaccinated enters the vaccinated compartment, suppose the effectiveness rate of the vaccine to be $$\eta (x)$$, and if the vaccine fails, the vaccinator will also be injected into the infected path since the inoculated host has some resistance to the virus after being vaccinated. Thus this propagation process is assumed to obey a half-saturation rate $$\frac{(1-\eta (x))\alpha (x)VI}{K(x)+I}$$ and $$\beta (x)>\alpha (x)$$.
In addition, because $$R(t)$$ does not appear in the first three equations of (2.1), we denote system (2.1) as
$$\textstyle\begin{cases} \frac{\partial S}{\partial t}=D_{1}\Delta S+\Lambda (x)-r(x)S-(1-r(x)) \beta (x)SI-d_{1}(x)S, \\ \frac{\partial I}{\partial t}=D_{2}\Delta I+(1-r(x))\beta (x)SI+(1- \eta (x))\frac{\alpha (x)VI}{K(x)+I} \\ \hphantom{\frac{\partial I}{\partial t}={}}{}-(\gamma (x)+d_{2}(x))I,\quad x\in \Omega ,t\in [0,\infty ), \\ \frac{\partial V}{\partial t}=D_{3}\Delta V+r(x)S-(1-\eta (x)) \frac{\alpha (x)VI}{K(x)+I}-(\eta (x)+d_{3}(x))V, \end{cases}$$
(2.2)
with the initial value and boundary conditions
$$\textstyle\begin{cases} \frac{\partial S}{\partial \nu}=\frac{\partial I}{\partial \nu}= \frac{\partial V}{\partial \nu}=0,\quad x\in \partial \Omega ,t>0, \\ (S,I,V)(\cdot ,0)=(S_{0},I_{0},V_{0})(x)>0,\quad x\in \Omega . \end{cases}$$
It is sufficient to determine the dynamics of (2.1). Here, Ω is a smooth bounded region in $$\mathbb{R}^{n}$$. Define a Banach space $$\mathbb{X}:=C(\overline{\Omega},\mathbb{R}^{3})$$ with the supremum norm $$\|\cdot \|$$ and $$\mathbb{X}\mathbbm{^{+}}=C(\overline{\Omega},\mathbb{R}^{3}_{+})$$. Next, we mainly analyze the dynamic behavior of system (2.2).
Let $$\mathcal{T}_{i}:C(\overline{\Omega},\mathbb{R})\to C( \overline{\Omega},\mathbb{R})$$ ($$i=1,2,3$$) be the $$C_{0}$$-semigroup associated with $$D_{1}\Delta -(r(x)+d_{1}(x))$$, $$D_{2}\Delta -(\gamma (x)+d_{2}(x))$$, $$D_{3} \Delta -(\eta (x)+d_{3}(x))$$. For any $$\varphi \in C(\Omega ,\mathbb{R})$$, $$\mathcal{T}_{i}$$ is given by the following formula:
\begin{aligned}& \bigl(\mathcal{T}_{1}(t)\varphi \bigr) (x)=e^{-d_{1}(x)t} \int _{\Omega}\Gamma _{1}(t,x,y) \varphi (y)\,dy ,\\& \bigl(\mathcal{T}_{2}(t)\varphi \bigr) (x)=e^{-(\gamma (x)+d_{2}(x))t} \int _{ \Omega}\Gamma _{2}(t,x,y)\varphi (y)\,dy , \end{aligned}
and
$$\bigl(\mathcal{T}_{3}(t)\varphi \bigr) (x)=e^{-(\eta (x)+d_{3}(x))t} \int _{ \Omega}\Gamma _{3}(t,x,y)\varphi (y)\,dy ,$$
where $$\Gamma _{i}$$ ($$i=1,2,3$$) is the Green function associated with the operator $$\frac{\partial n}{\partial t}=\Delta n$$ in Ω̅ subject to the boundary condition. With [14, Section 7], $$\mathcal{T}=(\mathcal{T}_{1},\mathcal{T}_{2},\mathcal{T}_{3})$$ are compact and strongly positive. Set
$$\textstyle\begin{cases} F_{1}(\phi )(x)=\Lambda (x)-(1-r(x))\beta (x)\phi _{1}(x)\phi _{2}(x), \\ F_{2}(\phi )(x)=(1-r(x))\beta (x)\phi _{1}(x)\phi _{2}(x)+(1-\eta (x)) \frac{\alpha (x)\phi _{3}(x)\phi _{2}(x)}{K(x)+\phi _{2}(x)}, \\ F_{3}(\phi )(x)=r(x)\phi _{1}(x)-(1-\eta (x)) \frac{\alpha (x)\phi _{3}(x)\phi _{2}(x)}{K(x)+\phi _{2}(x)}. \end{cases}$$
(2.3)
Then we can rewrite (2.2) as the following integral equation:
$$P(t)=\mathcal{T}\phi + \int _{0}^{t}\mathcal{T}(t-s)\mathcal{F} \bigl(P(s)\bigr)\,ds$$
for $$\mathcal{F}=(F_{1},F_{2},F_{3})$$ and $$\phi =(\phi _{1},\phi _{2},\phi _{3})(x)=(S_{0},I_{0},V_{0})(x)$$.
For a positive and continuous function $$\zeta (x)$$ on Ω̅, define
$$\zeta _{+}=\max \bigl\{ \zeta (x)\bigr\} , \qquad \zeta _{-}= \min \bigl\{ \zeta (x)\bigr\} .$$
Thus, for the local solution of (2.2), we have the following.
Lemma 2.1
System (2.2) with any initial value ϕ for $$t\in [0,\tau _{\mathrm{trans}})$$ (where $$\tau _{\mathrm{trans}}\leq \infty$$) has a unique solution $$P(x,t,\phi )=(S(x,t),I(x,t),V(x,t))$$ with $$P(\cdot ,0,\phi )=\phi$$. Moreover, $$P(x,t,\phi )=(S(x,t),I(x,t),V(x,t))$$ is a classical solution.
The proof is shown in Appendix A.
In the remainder of this section we will prove the global existence and boundedness of the solution. Consider the following equation:
$$\textstyle\begin{cases} \frac{\partial \omega}{\partial t}=D\Delta \omega +\Lambda (\cdot )- \mu (\cdot )\omega , \\ \frac{\partial \omega}{\partial \nu}=0, \end{cases}$$
(2.4)
where $$D>0$$ and $$\Lambda (\cdot )$$, $$\mu (\cdot )$$ are positive and continuous functions on Ω. Thus we have the following.
Lemma 2.2
([9, Lemma 1])
System (2.4) admits a positive steady state $$\omega _{0}$$ which is unique and asymptotically stable. Furthermore, if $$\Lambda (x)\equiv \Lambda$$, $$\mu (x)\equiv \mu$$ are constants, thus $$\omega _{0}=\frac{\Lambda}{\mu}$$.
The following theorem proves the boundedness of the model.
Theorem 2.1
For $$x\in \Omega$$, $$t\in [0,\infty )$$, the solution of system (2.2) satisfies
$$\Phi (t)\phi =P(\cdot ,t,\phi )=\bigl(S(\cdot ,t,\phi ),I(\cdot ,t,\phi ),V( \cdot ,t,\phi )\bigr),\quad \forall x\in \overline{\Omega}, t\in [0,\infty ),$$
where $$\Phi (t)$$ is the semiflow associated with the solution. Moreover, $$\Phi (t)$$ is ultimately bounded.
The proof is shown in Appendix B.
From what has been discussed above, we can get the following results.
Lemma 2.3
The semiflow $$\Phi (t):\mathbb{X}^{+}\to \mathbb{X}^{+}$$ admits a compact and global attractor.
Proof
With Theorem 2.1 we ensure the ultimate boundedness of system (2.2). Notice that the equation of (2.2) has the diffusion term, which ensures that $$\Phi (t)$$ is compact. With the direct consequence in [15, Theorem 2.4.6], we can complete the proof. □
In Sect. 3, we analyze the threshold dynamics of system (2.2). The global stability of disease-free equilibrium (DFE) and endemic equilibrium (EE) is analyzed by establishing the relationship between the basic reproduction number $$\Re _{0}$$ and the principal eigenvalue.
Threshold dynamics
Basic reproduction number
In this section, applying the methods in [16, Sect. 3], we give the basic reproduction number $$\Re _{0}$$ of (2.2). It is easy to see that (2.2) admits a disease-free equilibrium (DFE) $$P_{0}=(S^{0},0,V^{0})$$, where $$S^{0}$$, $$V^{0}$$ satisfy
$$\textstyle\begin{cases} \frac{\partial S}{\partial t}=D_{1}\Delta S+\Lambda (\cdot )-(r( \cdot )+d_{1}(\cdot ))S, \\ \frac{\partial V}{\partial t}=D_{3}\Delta V+r(\cdot )S-(\eta (\cdot )+d_{3}( \cdot ))V. \end{cases}$$
(3.1)
Linearizing (2.2) at $$P_{0}$$, we can get the following system:
$$\textstyle\begin{cases} \frac{\partial S}{\partial t}=D_{1}\Delta (S-S^{0})-(1-r(\cdot )) \beta (\cdot )S^{0}I-(r(\cdot )+d_{1}(\cdot ))(S-S^{0}), \\ \frac{\partial I}{\partial t}=D_{2}\Delta I+(1-r(\cdot ))\beta ( \cdot )S^{0}I+(1-\eta (\cdot )) \frac{\alpha (\cdot )V^{0}I}{K(\cdot )}-(\gamma (\cdot )+d_{2}(\cdot ))I, \\ \frac{\partial V}{\partial t}=D_{3}\Delta (V-V^{0})+r(\cdot )(S-S^{0})-(1- \eta (\cdot ))\frac{\alpha (\cdot )V^{0}I}{K(\cdot )}-(\eta (\cdot )+d_{3}( \cdot ))(V-V^{0}). \end{cases}$$
In order to discuss the basic reproduction number $$\Re _{0}$$, we will focus on the linearized equation of infected person I.
$$\textstyle\begin{cases} \frac{\partial I}{\partial t}=D_{2}\Delta I+(1-r(\cdot ))\beta ( \cdot )S^{0}I+(1-\eta (\cdot )) \frac{\alpha (\cdot )V^{0}I}{K(\cdot )}-(\gamma (\cdot )+d_{2}(\cdot ))I, \\ \frac{\partial I}{\partial \nu}=0,\quad x\in \partial \Omega , t>0. \end{cases}$$
(3.2)
Substituting $$I(\cdot ,t)=e^{\lambda t}\delta (\cdot )$$, we consider the following subsystem:
$$\textstyle\begin{cases} D_{2}\Delta \delta + [(1-r(\cdot ))\beta (\cdot )S^{0}+(1-\eta ( \cdot ))\frac{\alpha (\cdot )V^{0}}{K(\cdot )} ]\delta -(\gamma ( \cdot )+d_{2}(\cdot ))\delta =\lambda \delta , \\ \frac{\partial \delta}{\partial \nu}=0. \end{cases}$$
(3.3)
Thus, define
$$\mathbb{F}=\bigl(1-r(\cdot )\bigr)\beta (\cdot )S^{0}+\bigl(1-\eta ( \cdot )\bigr) \frac{\alpha (\cdot )V^{0}}{K(\cdot )}, \qquad \mathbb{B}=\Delta D_{2}( \cdot )-\bigl(\gamma (\cdot )+d_{2}(\cdot )\bigr),$$
then the next generation operator is defined as $$\mathcal{L}=-\mathbb{F}\mathbb{B}^{-1}$$. Set $$T_{\mathrm{trans}}$$ as the $$C_{0}$$-semigroup associated with $$\mathbb{B}$$, then
$$\mathcal{L}\phi (\cdot )= \int _{0}^{\infty}\mathbb{F}(\cdot )T_{\mathrm{trans}}(t) \phi (\cdot )\,dt =\mathbb{F}(\cdot ) \int _{0}^{\infty}(\cdot )T_{\mathrm{trans}}(t) \phi (\cdot )\,dt ,\quad x\in \overline{\Omega}.$$
We can use the variational formula to give the solution of the eigenvalue problem as
\begin{aligned} \lambda _{0} =&-\inf \biggl\{ \int _{\Omega} \biggl(D_{2}(\cdot ) \vert \nabla \delta \vert ^{2} \\ &{}- \biggl[\bigl(1-r(\cdot )\bigr)\beta (\cdot )S^{0}+\bigl(1-\eta (\cdot )\bigr) \frac{\alpha (\cdot )V^{0}}{K(\cdot )}-\bigl(\gamma (\cdot )+d_{2}(\cdot )\bigr) \biggr]\delta ^{2} \biggr)\,dx : \\ &\delta \in H^{1}(\Omega ), \int _{\Omega} \delta ^{2}=1 \biggr\} . \end{aligned}
By [16, Theorem 3.2], we have
$$\Re _{0}=\varrho (\mathcal{L})=\sup _{\delta \in H^{1},\delta \neq 0} \biggl\{ \frac{\int _{\Omega}[(1-r(\cdot ))\beta (\cdot )S^{0}+ \frac{(1-\eta (\cdot ))\alpha (\cdot )V^{0}}{K(\cdot )}]\delta ^{2}\,dx }{\int _{\Omega}(D_{2} \vert \nabla \delta \vert ^{2}+ (\gamma (\cdot )+d_{2}(\cdot ))\delta ^{2})\,dx } \biggr\} ,$$
(3.4)
and $$\varrho (\mathcal{L})$$ is the spectral radius of $$\mathcal{L}$$.
Remark 2
If all parameters are all constants, $$S^{0}=\frac{\Lambda}{r+d_{1}}$$, $$V^{0}= \frac{r\Lambda}{(r+d_{1})(\eta +d_{3})}$$, we have
$$\Re _{0}=\frac{1}{\lambda _{0}}= \biggl( \frac{(1-r)\beta \Lambda}{r+d_{1}}+ \frac{(1-\eta )\alpha r\Lambda}{K(r+d_{1})(\eta +d_{3})} \biggr)\Big/( \gamma +d_{2}).$$
From Remark 2, we can see the relationship between $$\Re _{0}$$ and parameters. We have the following lemma on the impact of $$D_{2}$$ with $$\Re _{0}$$.
Lemma 3.1
For the basic reproduction number $$\Re _{0}$$, we have:
1. 1.
$$\Re _{0}=1/\lambda _{0}$$;
2. 2.
For $$D_{2}>0$$, $$\Re _{0}$$ is a positive and strictly monotonic decline function;
3. 3.
$$\Re _{0}\to \max \{ \frac{(1-r(\cdot ))\beta (\cdot )S^{0}+\frac{(1-\eta (\cdot )) \alpha (\cdot )V^{0}}{K(\cdot )}}{\gamma (\cdot )+d_{2}(\cdot )} \}$$ for $$D_{2}\to 0$$;
4. 4.
$$\Re _{0}\to \frac{\int _{\Omega}[(1-r(\cdot ))\beta (\cdot )S^{0}+ \frac{(1-\eta (\cdot ))\alpha (\cdot )V^{0}}{K(\cdot )}]\,dx }{|\Omega | (\gamma (\cdot )+d_{2}(\cdot ))}$$ for $$D_{2}(\cdot )\to \infty$$;
5. 5.
For $$\int _{\Omega}[(1-r(\cdot ))\beta (\cdot )S^{0}+ \frac{(1-\eta (\cdot ))\alpha (\cdot )V^{0}}{K(\cdot )}]\,dx <|\Omega |( \gamma (\cdot )+d_{2}(\cdot ))$$, there exists $$D^{*}_{2}$$ such that, for $$D_{2}< D^{*}_{2}$$, $$\Re _{0}>1$$ and $$D_{2}>D^{*}_{2}$$, $$\Re _{0}<1$$; For $$\int _{\Omega}[(1-r(\cdot ))\beta (\cdot )S^{0}+ \frac{(1-\eta (\cdot ))\alpha (\cdot )V^{0}}{K(\cdot )}]\,dx >|\Omega |( \gamma (\cdot )+d_{2}(\cdot ))$$, $$\Re _{0}>1$$ for all $$D_{2}>0$$.
Define the principal eigenvalue of (3.3) as $$\lambda _{0}$$. Thus, we have the following result.
Lemma 3.2
$$\Re _{0}-1$$ has the same sign as $$\lambda _{0}$$.
The proof is shown in Appendix C.
Extinction of disease
In this subsection we give the proof of $$\Re _{0}<1$$, the stability of DFE.
Theorem 3.1
If $$\Re _{0}<1$$, the disease-free equilibrium $$P_{0}$$ is globally asymptotically stable. Thus, for $$x\in \Omega$$,
$$\lim_{t\to \infty}P(\cdot ,t,\phi )=P_{0}(\cdot ).$$
Proof
Applying Lemma 3.1, we can infer that for $$\Re _{0}<1$$ the principal eigenvalue $$\lambda _{0}<0$$. By the equation of S, V in (2.2), with continuity, there exist $$\upsilon _{1}$$ and $$t_{1}>0$$
$$S(\cdot ,t)< S^{0}+\upsilon _{1}, \qquad V(\cdot ,t)< V^{0}+\upsilon _{1}.$$
Thus, for $$x\in \Omega$$, $$t\in [t_{1},\infty )$$, the eigenvalue problem
$$\textstyle\begin{cases} \frac{\partial \xi}{\partial t}=D_{2}\Delta \xi +(1-r(\cdot ))\beta ( \cdot )(S^{0}+\upsilon _{1})\xi +(1-\eta (\cdot )) \frac{\alpha (\cdot )(V^{0}+\upsilon _{1})\xi}{K(\cdot )}-(\gamma ( \cdot )+d_{2}(\cdot ))\xi , \\ \frac{\partial \xi}{\partial \nu}=0, \end{cases}$$
(3.5)
has a principal eigenvalue $$\lambda ^{\upsilon _{1}}_{0}<0$$. Thus, (3.2) implies
$$\textstyle\begin{cases} \frac{\partial I}{\partial t}\leq D_{2}\Delta I+(1-r(\cdot ))\beta ( \cdot )(S^{0}+\upsilon _{1})I+(1-\eta (\cdot )) \frac{\alpha (\cdot )(V^{0}+\upsilon _{1})I}{K(\cdot )}-(\gamma ( \cdot )+d_{2}(\cdot ))I, \\ \frac{\partial I}{\partial \nu}=0. \end{cases}$$
By the comparison principle, we can find a positive constant $$\vartheta _{1}$$ which satisfies
$$I(\cdot ,t,\phi )\leq \vartheta _{1}e^{\lambda ^{\upsilon _{1}}_{0}(t-t_{1})} \xi ^{\upsilon _{1}},\quad t\in [t_{1},\infty ),$$
where $$\xi ^{\upsilon _{1}}$$ is a strong positive eigenfunction associated with $$\lambda ^{\upsilon _{1}}_{0}$$. Since $$\lambda ^{\upsilon _{1}}_{0}<0$$, we directly have
$$\lim_{t\to \infty}I(\cdot ,t,\phi )=0.$$
Thus the equation of S, V in (2.2) is asymptotic to (3.1), with asymptotic autonomous semiflow theory in [17, Corollary 4.3], such that
$$\lim_{t\to \infty}S(\cdot ,t,\phi )=S^{0},\qquad \lim _{t\to \infty}V( \cdot ,t,\phi )=V^{0}.$$
This completes the proof. □
Disease persistence
In this section, we prove the global asymptotic stability of the endemic equilibrium in the case of $$\Re _{0}>1$$. First, using [10, Lemma 4.1], we get the following conclusion, which ensures that (2.2) has a positive epidemic equilibrium.
Lemma 3.3
Let $$S(x,t)$$, $$I(x,t)$$, $$V(x,t)$$ be the solution of (2.2) with the initial value ϕ. If there exists a positive $$t^{*}$$ such that $$I(\cdot ,t^{*},\phi )>0$$, thus for any $$t\geq t^{*}$$, $$I(\cdot ,t,\phi )>0$$. Moreover, for S, V,
$$\liminf_{t\to \infty}S(\cdot ,t,\phi )\geq \frac{\Lambda _{*}}{r^{*}+(1-r_{*})\beta ^{*}I^{*}+d^{*}_{1}}$$
and
$$\liminf_{t\to \infty}V(\cdot ,t,\phi )\geq \frac{r_{*}\Lambda _{*}}{[r^{*}+(1-r_{*})\beta ^{*}I^{*}+d^{*}_{1}][(1-\eta _{*})\alpha ^{*}I^{*}+(\eta ^{*}+d^{*}_{3})]},$$
where $$h^{*}=\sup_{x\in \overline{\Omega}}h(\cdot )$$, $$h_{*}=\inf_{x\in \overline{\Omega}}h(\cdot )$$.
The proof is shown in Appendix D.
Next, we conclude this section by proving the global stability of the endemic equilibrium.
Theorem 3.2
For $$\Re _{0}>1$$, (2.2) admits at least one positive steady state, and we can find positive ε for any $$\phi \in \mathbb{X}^{+}$$ with $$S_{0},I_{0},V_{0}\not \equiv 0$$ such that
$$\liminf_{t\to \infty}P\geq \varepsilon ,\quad P=\bigl(S(\cdot ,t,\phi ),I( \cdot ,t,\phi ),V(\cdot ,t,\phi )\bigr).$$
Proof
Define two sets as
and
With Lemma 3.3, for $$\phi _{2}\in{\mathbb{H}_{0}}$$, we can find that $$x\in \Omega$$, $$\forall t\geq 0$$, which implies $$I(\cdot ,t,I^{0})>0$$ and . Set
here $$\omega (\phi )$$ is an omega limit set.
Claim 1. $$\omega (\phi )=\{P_{0}\}$$.
With , we know that , $$\forall t \geq 0$$, thus we have $$I(\cdot , t;\phi )\equiv 0$$. Thus we can find that (2.2) is asymptotic to
$$\textstyle\begin{cases} \frac{\partial S}{\partial t}=D_{1}\Delta S+\Lambda (\cdot )-r( \cdot )S-d_{1}(\cdot )S, \\ \frac{\partial V}{\partial t}=D_{3}\Delta V+r(\cdot )S-(\eta (\cdot )+d_{3}( \cdot ))V. \end{cases}$$
(3.6)
Then, with Lemma 2.2, on $$x\in \overline{\Omega}$$, S, I satisfy $$\lim_{t\rightarrow \infty}S(\cdot , t;\phi )=S^{0}(\cdot )$$ and $$\lim_{t\rightarrow \infty}V(\cdot , t;\phi )=V^{0}(\cdot )$$, uniformly. Hence $$\omega (\phi )=\{P_{0}\}$$, $$\forall \phi \in M_{\partial}$$.
Claim 2. $$P_{0}$$ satisfies
$$\limsup_{t\rightarrow \infty} \bigl\Vert \Phi (t)\phi -P_{0} \bigr\Vert \geq \sigma _{0},\quad \forall \phi \in \mathbb{H}_{0},$$
where $$\sigma _{0}>0=\min \{\sigma ^{*}_{0},\delta ^{*}_{0}\}$$ is a positive constant, here $$\sigma ^{*}_{0}$$, $$\delta ^{*}_{0}$$ will be defined in what follows.
First, by using [16, Lemma 2.2], there exists positive $$\sigma ^{*}_{0}>0$$ sufficiently small, the following eigenvalue problem
$$\textstyle\begin{cases} \frac{\partial \hat{\psi}_{2}}{\partial t}=D_{2}\Delta \hat{\psi}_{2}+(1-r( \cdot ))\beta (\cdot )(S^{0}-\sigma ^{*}_{0})\hat{\psi}_{2}+(1-\eta ( \cdot ))\alpha (\cdot )(V^{0}-\sigma ^{*}_{0})\hat{\psi}_{2}( \frac{1}{K(\cdot )}-\sigma ^{*}_{0}) \\ \hphantom{\frac{\partial \hat{\psi}_{2}}{\partial t}={}}{}-(\gamma (\cdot )+d_{2}(\cdot )) \hat{\psi}_{2}, \\ \frac{\partial \hat{\psi}_{2}}{\partial \nu}=0 \end{cases}$$
(3.7)
admits a principal eigenvalue $$\lambda ^{\sigma _{0}}_{0}$$. To deal with nonlinear terms in the system, we can choose $$\delta ^{*}_{0}$$ with
$$\frac{1}{K(\cdot )+I}>\frac{1}{K(\cdot )}-\sigma ^{*}_{0},\quad I< \delta ^{*}_{0}.$$
Then, to the contrary, assume that there is positive $$\sigma _{0}>0$$ such that, for $$\phi \in \mathbb{H}_{0}$$,
$$\limsup_{t\rightarrow \infty} \bigl\Vert \Phi (t)\phi -P_{0} \bigr\Vert < \sigma _{0}.$$
It follows that there exists $$t_{3}>0$$ for $$x\in \overline{\Omega}$$ which satisfies
$$S^{0}(\cdot )-\sigma _{0}< S(\cdot ,t,\phi ),\qquad I(\cdot ,t, \phi ) < \sigma _{0}, \qquad V^{0}-\sigma _{0}< V( \cdot ,t,\phi ) ,\quad \forall t\geq t_{2}.$$
Define $$\hat{I}(\cdot ,t)$$ for $$x\in \Omega$$, $$t\in [0,\infty )$$ which satisfies
$$\textstyle\begin{cases} \frac{\partial \hat{I}}{\partial t}=D_{2}\Delta \hat{I}+(1-r(\cdot )) \beta (\cdot )(S^{0}-\sigma _{0})\hat{I}+(1-\eta (\cdot ))\alpha ( \cdot )(V^{0}-\sigma _{0})\hat{I}(\frac{1}{K(\cdot )}-\sigma _{0}) \\ \hphantom{\frac{\partial \hat{I}}{\partial t}={}}{}-( \gamma (\cdot )+d_{2}(\cdot ))\hat{I}, \\ \frac{\partial \hat{I}}{\partial \nu}=0. \end{cases}$$
(3.8)
Thus, for $$\varsigma >0$$, $$\hat{I}(\cdot ,t)=\varsigma \hat{\psi}_{2{\sigma _{0}}}e^{(t-t_{3}) \lambda ^{\sigma _{0}}_{0}}$$ is the unique solution of (3.7). ($$\hat{\psi}_{2{\sigma _{0}}}$$ is the strong positive eigenfunction corresponding to $$\lambda ^{\sigma _{0}}_{0}$$). For , with Lemma 3.3, it follows that $$I(\cdot ,t,\phi )>0$$. From the definition of the upper solution, for the solution of (3.8) and $$I(\cdot ,t,\phi )$$, we can find that
$$I(\cdot ,t,\phi )\geq \hat{I}(\cdot ,t),\quad \overline{\Omega}\times [t_{3}, \infty ).$$
By the comparison principle, we can find a small positive constant ς which satisfies
$$I(\cdot ,t;\phi )\geq \varsigma \hat{\psi}_{2{\sigma _{0}}}e^{(t-t_{3}) \lambda ^{\sigma _{0}}_{0}},\quad t \geq t_{3}.$$
Thus, for $$\Re _{0}>1$$, with Lemma 3.2, which implies $$\lambda ^{\sigma _{0}}_{0}>0$$, then $$I(\cdot ,t,\phi )\rightarrow \infty$$ when $$t\rightarrow \infty$$. It means that $$I(\cdot ,t,\phi )$$ is unbounded, which contradicts the previous proof. This completes the proof.
Here we give the definition of a distance function in the semiflow $$\Phi (t): \mathbb{X}^{+}\to \mathbb{X}^{+}$$ as follows:
$$c(\phi ):=\min_{x\in \overline{\Omega}}\bigl\{ \phi _{2}(\cdot ) \bigr\} ,\quad \forall \phi \in \mathbb{X}^{+},$$
where $$c(\cdot ): \mathbb{X}^{+}\to [0,\infty )$$. Hence, applying [10, Theorem 4.1], for any $$\tau _{1} > 0$$, we have
and
By Lemma 3.3, we can set
$$\varepsilon _{2}=\min \biggl\{ \frac{\Lambda _{*}}{r^{*}+(1-r_{*})\beta ^{*}I^{*}+d^{*}_{1}}, \frac{r_{*}\Lambda _{*}}{[r^{*}+(1-r_{*})\beta ^{*}I^{*}+d^{*}_{1}][(1-\eta _{*})\alpha ^{*}I^{*}+(\eta ^{*}+d^{*}_{3})]} \biggr\} ,$$
thus $$\liminf_{t\to \infty}S(\cdot ,t,\phi ),\liminf_{t\to \infty}V( \cdot ,t,\phi )\geq \varepsilon _{2}$$. Set $$\varepsilon =\min \{\varepsilon _{1},\varepsilon _{2}\}$$, we can get the result that the endemic equilibrium is uniformly persistent. Therefore, by [18, Theorem 4.7], for (2.2), there is at least one positive steady state of on . □
Optimal control
In the previous chapters, we have focused on the disease-free equilibrium and the infectious equilibrium of infectious diseases. But if there is a sudden outbreak, we need to control the impact of the disease at a lower level as far as possible, that is, to control the number of infected people. In addition to calling for vaccinations, governments often spend more money on treatment. The mathematical language to describe this method is the optimal control problem. The main aim of this section is to develop effective strategies for controlling the spread of infectious diseases. We hope that the number of infected people does not exceed the number of susceptible and effective vaccinators.
In this section, we introduce the control strategy to (2.2) and analyze its properties. For convenience, we rewrite $$\Lambda (x)$$ as Λ, the same for other parameters. To complete our research, we analyze the control variables of the model (2.2). Therefore, the control variables are given as follows.
With the development of medical technology, infected patients can be treated better. Therefore, define $$u=u(x,t)$$ represents the medical intervention for infected patients. Considering that medical resources are limited, we use $$\frac{cuI}{1+\omega I}$$ for specific. Here, c is the cure rate and ω denotes the saturation constant.
From this, we give the control system of (2.2) as follows:
$$\textstyle\begin{cases} \frac{\partial S}{\partial t}=D_{1}\Delta S+\Lambda -rS-(1-r)\beta SI-d_{1}S, \\ \frac{\partial I}{\partial t}=D_{2}\Delta I+(1-r)\beta SI+(1-\eta ) \frac{\alpha VI}{K+I}-(\gamma +d_{2})I-\frac{cuI}{1+\omega I},\quad x\in \Omega ,t\in [0,\infty ), \\ \frac{\partial V}{\partial t}=D_{3}\Delta V+rS-(1-\eta ) \frac{\alpha VI}{K+I}-(\eta +d_{3})V, \end{cases}$$
(4.1)
with the boundary condition
$$\frac{\partial S}{\partial \nu}=\frac{\partial I}{\partial \nu}= \frac{\partial V}{\partial \nu}=0,\quad (x,t)\in \Sigma =(0,T)\times \partial \Omega ,$$
here u is measurable, other parameters are the same as in (2.2).
Define an objective function
$$J\bigl(u(x,t)\bigr)= \int _{0}^{T} \int _{\Omega }L\bigl(I(x,t);u(x,t)\bigr)\,dx\, dt ,$$
where $$u(x,t)\in \mathscr{U}(\Omega \times [0,T])=\{0\leq u(x,t)\leq 1\}$$ and $$L=A_{1}I(x,t)+\frac{1}{2}A_{2}u^{2}(x,t)$$. Assume that the control set $$\mathscr{U}(\Omega \times [0,T])$$ is convex, $$A_{1}$$, $$A_{2}$$ are weight of each item. This objective function describes our goal to control the problem: to reduce the number of susceptible and infected people with minimal intervention costs. The value function is defined as
$$V\bigl(0,\phi (\cdot ,0)\bigr)=\min_{u(x,t)\in \mathscr{U}(\Omega \times [0,T])}J\bigl(0, \phi ( \cdot ,0);u(\cdot ,t)\bigr).$$
Define a Hilbert space $$H=L^{2}(\Omega ^{2})$$ and $$S^{0}_{c}, I^{0}_{c}, V^{0}_{c}>0$$ as the initial value of (4.1), which satisfies (IOC: Initial value of Optimal Control)
$$(\mathrm{IOC}) \quad S^{0}_{c}, I^{0}_{c} \in H^{2}(\Omega ), \qquad \partial S^{0}_{c}/ \partial \nu =0,\qquad \partial I^{0}_{c}/\partial \nu =0,\qquad \partial V^{0}_{c}/ \partial \nu =0.$$
Let $$f^{c}=(f_{1}(t,S,I,V), f_{2}(t,S,I,V), f_{3}(t,S,I,V))$$, where
$$\textstyle\begin{cases} f_{1}(t,S,I,V)=\Lambda -rS-(1-r)\beta SI-d_{1}S, \\ f_{2}(t,S,I,V)=(1-r)\beta SI+(1-\eta )\frac{\alpha VI}{K+I}-(\gamma +d_{2})I- \frac{cuI}{1+\omega I}, \\ f_{3}(t,S,I,V)=rS-(1-\eta )\frac{\alpha VI}{K+I}-(\eta +d_{3})V. \end{cases}$$
With the method in [19, 20], we have the following lemma.
Lemma 4.1
For the initial value $$S^{0}_{c}$$, $$I^{0}_{c}$$, $$V^{0}_{c}$$, system (4.1) admits a unique strong solution $$(S,I,V)\in W^{1,2}(0,T;H)$$ such that $$S,I,V>0$$ and $$(S,I,V)\in L^{2}(0,T;H^{2}(\Omega ))\cap L^{\infty}(0,T;H^{1}( \Omega ))\cap L^{\infty}(\mathscr{U})$$. Furthermore, there exists a positive constant C independent of u, for any $$t\in [0,T]$$,
\begin{aligned} & \biggl\Vert \frac{\partial S}{\partial t} \biggr\Vert _{L^{2}(Q)}+ \Vert S \Vert _{L^{2}(0,T;H^{2}( \Omega ))}+ \bigl\Vert S(t) \bigr\Vert _{H^{1}{(\Omega )}}+ \Vert S \Vert _{L^{\infty}(Q)}\leq C, \\ & \biggl\Vert \frac{\partial I}{\partial t} \biggr\Vert _{L^{2}(Q)}+ \Vert I \Vert _{L^{2}(0,T;H^{2}( \Omega ))}+ \bigl\Vert I(t) \bigr\Vert _{H^{1}{(\Omega )}}+ \Vert I \Vert _{L^{\infty}(Q)}\leq C, \\ & \biggl\Vert \frac{\partial V}{\partial t} \biggr\Vert _{L^{2}(Q)}+ \Vert V \Vert _{L^{2}(0,T;H^{2}( \Omega ))}+ \bigl\Vert V(t) \bigr\Vert _{H^{1}{(\Omega )}}+ \Vert V \Vert _{L^{\infty}(Q)}\leq C, \end{aligned}
where $$Q=\Omega \times [0,T]$$.
Lemma 4.1 ensures the existence and boundedness of the global solution of system (4.1). Next, according to [21], we can analyze the existence of optimal control of system (4.1).
Theorem 4.1
Let the initial value be defined in (IOC). Then there exists an optimal solution $$P'=(S',I',V')$$ of the control system (4.1) corresponding to optimal control $$u'$$.
Proof
From the boundedness we have proved in Sect. 3, we can infer that $$p=\inf{J(u(x,t))}$$ is finite. Thus, for $$P_{n}=(S_{n},I_{n},V_{n})$$ and $$u_{n}\in \mathscr{U}$$,we can find a sequence $$(P_{n},u_{n})$$ that is the solution to the following subsystem:
$$\textstyle\begin{cases} \frac{\partial S_{n}}{\partial t}=D_{1}\Delta S_{n}+\Lambda -rS_{n}-(1-r) \beta S_{n}I_{n}-d_{1}S_{n}, \\ \frac{\partial I_{n}}{\partial t}=D_{2}\Delta I_{n}+(1-r)\beta S_{n}I_{n}+(1- \eta )\frac{\alpha V_{n}I_{n}}{K+I_{n}}-(\gamma +d_{2})I_{n} \\ \hphantom{\frac{\partial I_{n}}{\partial t}={}}{}- \frac{cuI_{n}}{1+\omega I_{n}},\quad x\in \Omega ,t\in [0,T], \\ \frac{\partial V_{n}}{\partial t}=D_{3}\Delta V_{n}+rS_{n}-(1-\eta ) \frac{\alpha V_{n}I_{n}}{K+I_{n}}-(\eta +d_{3})V_{n}, \end{cases}$$
(4.2)
with the initial condition
$$S_{n}(0,t)=S_{0}(x),\qquad I_{n}(0,t)=I_{0}(x),\qquad V_{n}(0,t)=V_{0}(x)$$
and the boundary condition
$$\frac{\partial S_{n}}{\partial \nu}= \frac{\partial I_{n}}{\partial \nu}= \frac{\partial V_{n}}{\partial \nu}=0,\quad (x,t)\in \Sigma ,$$
such that
$$p\leq J\bigl(u(x,t)\bigr)\leq p+\frac{1}{n}, \quad \forall n\geq 1.$$
With Lemma 4.1, for system (4.2), we can infer that
\begin{aligned} & \biggl\Vert \frac{\partial S_{n}}{\partial t} \biggr\Vert _{L^{2}(Q)}+ \Vert S_{n} \Vert _{L^{2}(0,T;H^{2}( \Omega ))}+ \bigl\Vert S_{n}(t) \bigr\Vert _{H^{1}{(\Omega )}}+ \Vert S_{n} \Vert _{L^{\infty}(Q)} \leq C, \\ & \biggl\Vert \frac{\partial I_{n}}{\partial t} \biggr\Vert _{L^{2}(Q)}+ \Vert I_{n} \Vert _{L^{2}(0,T;H^{2}( \Omega ))}+ \bigl\Vert I_{n}(t) \bigr\Vert _{H^{1}{(\Omega )}}+ \Vert I_{n} \Vert _{L^{\infty}(Q)} \leq C, \\ & \biggl\Vert \frac{\partial V_{n}}{\partial t} \biggr\Vert _{L^{2}(Q)}+ \Vert V_{n} \Vert _{L^{2}(0,T;H^{2}( \Omega ))}+ \bigl\Vert V_{n}(t) \bigr\Vert _{H^{1}{(\Omega )}}+ \Vert V_{n} \Vert _{L^{\infty}(Q)} \leq C. \end{aligned}
(4.3)
Since $$H^{1}(\Omega )$$ is compactly imbedded in $$L^{2}(\Omega )$$, we can also get the compactness of $$S_{n}$$, $$I_{n}$$, $$V_{n}$$ and $$\frac{\partial S_{n}}{\partial t}$$, $$\frac{\partial I_{n}}{\partial t}$$, $$\frac{\partial V_{n}}{\partial t}$$. Here, by using the Arzela–Ascoli theorem [22], for the compactness we proved in Sect. 2,
$$S_{n}\to S', \qquad I_{n}\to I',\qquad V_{n}\to V',$$
uniformly in $$L^{2}(\Omega )$$ with respect to a subsequence denoted by $$P_{n}$$. In addition, with the weak convergence of $$\Delta S_{n}$$, $$\Delta I_{n}$$, $$\Delta V_{n}$$ (with the boundedness in system (4.2)), we have
$$\Delta S_{n}\to \Delta S',\qquad \Delta I_{n} \to \Delta I', \qquad \Delta V_{n} \to \Delta V'$$
weakly in $$L^{2}(Q)$$. With (4.3), we have for $$P'=(S',I',V')$$ and $$P_{n}=(S_{n}, I_{n}, V_{n})$$
\begin{aligned}& \frac{\partial P_{n}}{\partial t}\to \frac{\partial P'}{\partial t} \quad \text{weakly in } L^{2}(Q),\\& P_{n}\to P' \quad \text{weakly star in } L^{\infty}\bigl([0,T]; H^{1}(\Omega )\bigr),\\& P_{n}\to P' \quad \text{weakly in } L^{2}\bigl([0,T]; H^{1}(\Omega )\bigr). \end{aligned}
Next, we focus on the second equation of (4.2). By direct calculation we have
\begin{aligned} &(1-r)\beta S_{n}I_{n}+(1-\eta )\frac{\alpha V_{n}I_{n}}{K+I_{n}}- \biggl[(1-r)\beta S'I'+(1- \eta )\frac{\alpha V'I'}{K+I'} \biggr] \\ &\quad =(1-r) \beta \bigl(S_{n}I_{n}-S'I' \bigr) +(1-\eta )\frac{\alpha V_{n}I_{n}}{(K+I_{n})(K+I')}\bigl[\bigl(K+I' \bigr)I_{n}\bigl(S_{n}-S' \bigr)+KS'\bigl(I_{n}-I'\bigr)\bigr]. \end{aligned}
Thus
$$(1-r)\beta S_{n}I_{n}+(1-\eta )\frac{\alpha V_{n}I_{n}}{K+I_{n}}\to (1-r) \beta S'I'+(1-\eta )\frac{\alpha V'I'}{K+I'}.$$
Similarly, we can discuss the first and third equation of (4.3). For the subsequence $${u_{n}}$$, $$u_{n}\to u'$$ weakly in $$L^{2}(Q)$$. With the convexity and closeness of $$\mathscr{U}$$, hence $$u'\in \mathscr{U}$$, we have
$$\frac{cu_{n}I_{n}}{1+\omega I_{n}}\to \frac{cu'I'}{1+\omega I'}.$$
With the analysis above, we can give the conclusion that for $$n\to \infty$$, $$P'=(S',I',V')$$ is the optimal solution associated with the optimal function $$u'$$ of system (4.1), which completes the proof. □
The Hamiltonian function of the control system is given as follows:
\begin{aligned} &H^{\star}(x,t,S,I,V,u,p) \\ &\quad = A_{1}I(x,t)+\frac{1}{2}A_{2}u^{2}(x,t)+p_{1} \bigl(D_{1} \Delta S+\Lambda -rS-(1-r)\beta SI-d_{1}S \bigr) \\ &\qquad {}+p_{2}\biggl(D_{2}\Delta I+(1-r) \beta SI +(1-\eta )\frac{\alpha VI}{K+I}-(\gamma +d_{2})I- \frac{cuI}{1+\omega I}\biggr) \\ &\qquad {}+p_{3}\biggl(D_{3}\Delta V+rS-(1-\eta ) \frac{\alpha VI}{K+I}-(\eta +d_{3})V\biggr). \end{aligned}
(4.4)
Next we give the adjoint equation for the control system (4.1)
$$\textstyle\begin{cases} \frac{\partial p_{1}(x,t)}{\partial t} =- \frac{\partial H^{\star}}{\partial S}=[r+(1-r)\beta I+d_{1}]p_{1}-D_{1} \Delta p_{1}-(1-r)\beta Ip_{2}-rp_{3}, \\ \frac{\partial p_{2}(x,t)}{\partial t} =- \frac{\partial H^{\star}}{\partial I} \\ \hphantom{\frac{\partial p_{2}(x,t)}{\partial t}}=(1-r)\beta Sp_{1}+ [\gamma +d_{2} \frac{cu}{(1+\omega I)^{2}}-(1-r)\beta S-(1-\eta ) \frac{\alpha VK}{(K+I)^{2}} ]p_{2} \\ \hphantom{\hphantom{\frac{\partial p_{2}(x,t)}{\partial t}}={}}{} -D_{2}\Delta p_{2}+(1-\eta )\frac{\alpha VK}{(K+I)^{2}}p_{3}-A_{1}, \\ \frac{\partial p_{3}(x,t)}{\partial t} =- \frac{\partial H^{\star}}{\partial V}=-(1-\eta )\frac{\alpha I}{K+I}p_{2}+ [(1-\eta )\frac{\alpha I}{K+I}+\eta +d_{3} ]p_{3}-D_{3}\Delta p_{3}, \\ p_{i}(T) =0,\quad i=1,2,3. \end{cases}$$
(4.5)
Using the method in [23], give the partial derivative of the Hamiltonian function to u, substitute it into the optimal control solution $$P'$$,
$$\frac{\partial H^{\star}}{\partial u}=A_{2}u- \frac{cp_{2}(x,t)I'}{1+\omega I'}.$$
Let $$\frac{\partial H^{\star}}{\partial u}=0$$, the optimal control pair $$u'$$ satisfying the minimum value of the objective function $$\min_{u(x,t)\in \mathscr{U}} J(u)$$ can be expressed as
$$u'=\min \biggl\{ \max \biggl\{ \frac{p_{2}(x,t)cI'}{A_{2}(1+\omega I')},0\biggr\} ,1 \biggr\} .$$
Numerical simulation
In this section, we use numerical simulation to verify the stability of the system and the impact of controls on the development of the disease. The values of each parameter are shown in Table 2.
Stability of equilibrium
In this section, we discuss the stability of the solution of system (2.2). First of all, use the method in [28] to make difference and solve system (2.2). Then, for $$\Re _{0}<1$$ and $$\Re _{0}>1$$, take the data in Data 1 and Data 2, respectively. We get the simulation results obtained as follows.
Firstly, take the value in Data 1. As we can see from Fig. 1, when $$\Re _{0}<1$$, from Theorem 3.1, the disease-free equilibrium $$(S^{0},0,V^{0})$$ is asymptotically stable. In fact, as shown in Fig. 1(b), for $$t\to \infty$$, $$I(x,t)$$ converges to zero; in addition, $$S^{0}=\frac{\Lambda}{\gamma +d_{1}}=0.7149$$, $$V^{0}= \frac{r\Lambda}{(\gamma +d_{1})(\eta +d_{3})}=0.3252$$. This is the same conclusion given by Theorem 3.1.
Then, using Data 2, we get the numerical simulation when $$\Re _{0}>1$$. With Theorem 3.2, we can get the uniform persistence of disease. Actually, as we can see in Fig. 2(b), with $$t\to \infty$$, $$I(x,t)$$ approaches a constant positive value. This is consistent with our proof in Theorem 3.2.
Influence of control parameters on disease progression
In this section, we discuss the impact of control on a disease. The main idea of this section is that we solve the optimal control problem by applying the iterative method. Then the optimal system is obtained by using the state equation and adjoint equation given in Sect. 4. And by solving the optimal system, the optimal control strategy is obtained. Furthermore, the method in [28] is used to make difference and solve the control system and the adjoint equation. In order to control the susceptible population, the infected population, and the vaccinated population, targeted treatment of patients is taken as control, and the impact of targeted treatment on the susceptible population, the infected population, and the density of the vaccinated population in a long-term state is considered. Finally, through numerical simulation, the actual situation of each path in the original system (2.2) and the control system (4.1) is compared.
First, we define the objective function as follows. Let the objective function corresponding to the control system (4.1) be as follows:
$$J\bigl(u(x,t)\bigr)= \int _{0}^{T} \int _{\Omega }A_{1}I(x,t)+\frac{1}{2}A_{2}u^{2}(x,t)\,dx\, dt ,$$
where $$A_{1}=0.4$$, $$A_{2}=0.5$$ [29]. The values of other parameters are the same as in Sect. 5.1. The numerical simulation results are as follows.
In Fig. 3, for $$\Re _{0}<1$$, under controlled conditions, the duration of the disease is shorter and the time of extinction is earlier than without control. At the same time, the overall density of susceptible and vaccinated people before reaching the stabilization point was also higher in the controlled condition than without control. In addition, according to Fig. 4, the control intensity reached the maximum in the early stage and gradually decreased with the weakening of the disease scale, reaching zero value when the disease disappeared.
As shown in Fig. 5, for $$\Re _{0}>1$$, when control exists, the scale of the disease reaches a minimum earlier than without control, and when the disease eventually becomes endemic, the total scale of the disease is lower than without control. In addition, when the disease reaches a stable state, the density of susceptible and vaccinated persons is higher in the controlled condition than without control. Furthermore, by Fig. 6, when the disease is in its initial state of development, control rises, and when the disease reaches equilibrium and becomes endemic, control is maintained at a stable value along with the duration of the disease.
Conclusion and discussion
In this paper, a kind of SIVR infectious disease model including vaccine immunity and vaccine effectiveness is considered. The optimal control theory is applied to the study of the model, and the threshold dynamics and optimal control of the model are discussed. Firstly, we prove the well-posedness of the model, which provides a theoretical basis for the following discussion. Secondly, we give the basic reproduction number $$\Re _{0}$$ to analyze the dynamic behavior of the disease threshold. In addition, the Hamiltonian function and adjoint equation of the optimal control problem is given. Finally, the stability of the system solution is verified by numerical simulation and the number of infections can be reduced as much as possible, while the cost is reduced under the treatment control. In this paper, the parameters are assumed to be accurate; in fact, due to various uncertainties, each parameter may be inaccurate or random. At the same time, according to the changes in the parameters, it can be seen that the vaccination rate and the effective rate of the vaccine also have a certain impact on the control (see Fig. 7(a), (b)). In addition, because the near-optimal control is more flexible, it can adapt to different degrees of model uncertainty. Therefore, in future work, the near-optimal control problem of the epidemic model can be further studied by considering the influence of random parameters, noise, the vaccination rate, and the efficiency rate of the vaccine as the control parameter.
Availability of data and materials
Data used to support the findings of this work are available from the corresponding author upon request.
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28. Higham, D.J.: An algorithmic introduction to numerical simulation of stochastic differential equations. SIAM Rev. 43(3), 525–546 (2001)
29. Pan, S.L., Zhang, Q.M., Anke, M.B.: Near-optimal control of a stochastic vegetation-water system with reaction diffusion. Math. Methods Appl. Sci. 43, 6043–6061 (2020)
30. Martin, R.H., Smith, H.L.: Abstract functional differential equations and reaction–diffusion systems. Trans. Am. Math. Soc. 321(1), 1–44 (1990)
31. Wang, M.: Nonlinear Elliptic Equations. Science Public, Beijing (2010)
Acknowledgements
Authors would like to thank Prof. Jianguo Gao for his guidance during the writing of the paper, and sincerely thank the editors for their careful evaluation of the paper.
Funding
The research was supported in part by the National Natural Science Foundation of China (No. 61761002).
Author information
Authors
Contributions
Conceptualization, YL; methodology, YL and SJ; software, YL; validation, YL, SJ, and JG; formal analysis, YL and SJ; investigation, YL and SJ; writing-original draft preparation, YL; writing-review and editing, JG, YL, and SJ; visualization, YL; supervision, JG; project administration, JG; funding acquisition, JG. All the authors have read and agreed to the published version of the manuscript.
Corresponding author
Correspondence to Jianguo Gao.
Ethics declarations
Competing interests
The authors declare that they have no competing interests.
Appendices
Appendix A: Proof of Lemma 2.1
Define the domain of the linear homogeneous part $$\mathcal{T}$$ as
$$D(\mathcal{T})=\biggl\{ \phi :\frac{\partial \phi}{\partial t}=0 \text{ on } \partial \Omega ,\mathcal{T}\phi \in \mathbb{X}\biggr\} .$$
Thus, there exists $$h\geq 0$$ which satisfies
\begin{aligned} \phi + h\mathcal{F}(\phi )&= \begin{pmatrix} \phi _{1}+h [\Lambda (x)-(1-r(x))\beta (x)\phi _{1}(x)\phi _{2}(x) ] \\ \phi _{2}+h [(1-r(x))\beta (x)\phi _{1}(x)\phi _{2}(x)+(1-\eta (x)) \frac{\alpha (x)\phi _{3}(x)\phi _{2}(x)}{K(x)+\phi _{2}(x)} ] \\ \phi _{3}+h [r(x)\phi _{1}(x)-(1-\eta (x)) \frac{\alpha (x)\phi _{3}(x)\phi _{2}(x)}{K(x)+\phi _{2}(x)} ] \end{pmatrix} \end{aligned}
(A.1)
\begin{aligned} &\geq \begin{pmatrix} \phi _{1} [1-h(1-r_{-})\phi _{2}\beta _{+} ] \\ \phi _{2} \\ \phi _{3} [1-h(1-\eta _{-})\phi _{2}\alpha _{+} ] \end{pmatrix}. \end{aligned}
(A.2)
We can directly get
$$\lim_{h\to \infty}\frac{1}{h}\operatorname{dist}\bigl(\phi +h\mathcal{F}, \mathbb{X}^{+}\bigr)=0.$$
With [30, Corollary .4], we can prove the lemma.
Appendix B: Proof of Theorem 2.1
Define
$$H(t)= \int _{\Omega}\bigl[S(\cdot ,t)+I(\cdot ,t)+V(\cdot ,t)\bigr]\,dx .$$
Differential with respect on t
\begin{aligned} \frac{\operatorname{d}H(t)}{\operatorname{d}t}&= \int _{\Omega} \Lambda (\cdot )\,dx \\ &\quad {} - \int _{\Omega}\bigl[d_{1}(\cdot )S(\cdot ,t)+\bigl( \gamma ( \cdot )+d_{2}(\cdot )\bigr)I(\cdot ,t)+\bigl(\eta (\cdot )+d_{3}(\cdot )\bigr)V( \cdot ,t)\bigr]\,dx \\ &\leq \Lambda (\cdot ) \vert \Omega \vert -\theta H(t), \end{aligned}
(B.1)
where $$\theta =\min \{d_{1}(\cdot ),\gamma (\cdot )+d_{2}(\cdot ),\eta ( \cdot )+d_{3}(\cdot )\}$$. Hence we have
$$H(t)\leq H(0)e^{-\theta t}+\frac{\Lambda (\cdot ) \vert \Omega \vert }{\theta}\bigl(1-e^{- \theta t} \bigr)< \infty ,$$
thus for some $$t\geq t_{1}$$ and positive constant N, we have $$H(t)< N$$.
With Lemma 2.2 and first and third equation of (2.2), by the comparison principle, we have, for some $$t\geq t_{2}$$,
$$\limsup_{t\to \infty}S(\cdot ,t)\leq \omega _{0},\qquad \limsup_{t\to \infty}V(\cdot ,t)\leq \omega _{1}.$$
Thus, $$S(\cdot ,t)$$ and $$V(\cdot ,t)$$ are ultimately bounded. With the above condition, $$\mathcal{T}_{2}$$ is the $$C_{0}$$ semigroup associated with $$D_{2}\Delta -(\gamma (x\cdot )+d_{2}(\cdot ))$$. We can rewrite $$\mathcal{T}_{2}$$ with a Green function as
$$\bigl(\mathcal{T}_{2}(t)\phi \bigr) (x)=e^{-(\gamma (\cdot )+d_{2}(\cdot ))t} \int _{\Omega}\Gamma _{2}(t,x,y)\phi (y)\,dy ,$$
thus there exists a positive constant $$\mathbb{M}$$ satisfying $$\|\mathcal{T}_{2}\|\leq \mathbb{M}e^{k_{2}t}$$, where $$k_{2}$$ is the principle eigenvalue of $$D_{2}\Delta -(\gamma (\cdot )+d_{2}(\cdot ))$$. With [31, Theorem 2.4.7], for Green function $$\Gamma _{2}$$ satisfies
$$\Gamma _{2}(t,x,y)\leq me^{-(\gamma _{-}+d_{2-})t},$$
where $$m>0$$ is a positive constant. By (2.1), we can conclude that, for $$t_{m}=\max \{t_{1},t_{2}\}$$,
\begin{aligned} I(\cdot ,t)&=\mathcal{T}_{2}I( \cdot ,t_{m})+ \int _{t_{m}}^{t} \mathcal{T}_{2}(t-s)\biggl[ \bigl(1-r(x)\bigr)\beta (x)SI+\bigl(1-\eta (x)\bigr) \frac{\alpha (x)VI}{K(x)+V} \biggr]\,ds \\ &\leq \mathbb{M}e^{k_{2}(t-t_{m})} \bigl\Vert I(\cdot ,t_{m}) \bigr\Vert \\ &\quad {}+ \int _{t_{m}}^{t} \int _{\Omega}\Gamma _{2}(t-s,x,y) \bigl[(1-r_{-})\beta _{+}\omega _{0}+(1- \eta _{-})\alpha _{+}\omega _{1}\bigr]\,dy\, ds \\ &\leq \mathbb{M}e^{k_{2}(t-t_{m})} \bigl\Vert I(\cdot ,t_{m}) \bigr\Vert \\ &\quad {}+ \int _{t_{m}}^{t} me^{-(\gamma _{-}+d_{2-})(t-s)} \biggl\{ \bigl[(1-r_{-})\beta _{+}\omega _{0}+(1- \eta _{-})\alpha _{+}\omega _{1}\bigr] \int _{\Omega }I(s,y)\,dy \biggr\} \,ds \\ &\leq \mathbb{M}e^{k_{2}(t-t_{m})} \bigl\Vert I(\cdot ,t_{m}) \bigr\Vert \\ &\quad {}+ \int _{t_{m}}^{t} me^{-(\gamma _{-}+d_{2-})(t-s)} \bigl\{ \bigl[(1-r_{-})\beta _{+}\omega _{0}+(1- \eta _{-})\alpha _{+}\omega _{1}\bigr]N \bigr\} \,ds \\ &=\mathbb{M}e^{k_{2}(t-t_{m})} \bigl\Vert I(\cdot ,t_{m}) \bigr\Vert + \frac{N_{\mathrm{trans}}(1-e^{-(\gamma _{-}+d_{2-})(t-t_{m})})}{\gamma _{-}+d_{2-}} \\ &\leq \mathbb{M}e^{k_{2}(t-t_{m})} \bigl\Vert I(\cdot ,t_{m}) \bigr\Vert + \frac{N_{\mathrm{trans}}}{\gamma _{-}+d_{2-}}, \end{aligned}
(B.2)
where $$N_{\mathrm{trans}}=mN[(1-r_{-})\beta _{+}\omega _{0}+(1-\eta _{-})\alpha _{+} \omega _{1}]$$. Since $$k_{2}$$ is the principle eigenvalue associated with $$D_{2}\Delta -(\gamma (\cdot )+d_{2}(\cdot ))$$, we can get to the conclusion that
$$\limsup_{t\to \infty} \bigl\Vert I(\cdot ,t) \bigr\Vert \leq \frac{N_{\mathrm{trans}}}{\gamma _{-}+d_{2-}}.$$
Thus $$S(\cdot ,t)$$, $$I(\cdot ,t)$$, $$V(\cdot ,t)$$ are uniformly bounded, and the semiflow $$\Phi (t)$$ is point dissipative.
Appendix C: Proof of Lemma 3.2
With the conclusion in Lemma 3.1, we denote by ξ the eigenfunction of the following eigenvalue problem:
$$\textstyle\begin{cases} D_{2}\Delta \xi + \frac{ [(1-r(\cdot ))\beta (\cdot )S^{0}+(1-\eta (\cdot ))\frac{\alpha (\cdot )V^{0}}{K(\cdot )} ] \xi}{\Re _{0}}-( \gamma (\cdot )+d_{2}(\cdot ))\xi =0, \\ \frac{\partial I}{\partial \nu}=0. \end{cases}$$
(C.1)
Here we list the first equation of (C.1) separately:
$$D_{2}\Delta \xi +\frac{1}{\Re _{0}} \biggl[ \bigl(1-r(\cdot )\bigr)\beta (\cdot )S^{0}+\bigl(1- \eta (\cdot ) \bigr)\frac{\alpha (\cdot )V^{0}}{K(\cdot )} \biggr]\xi -\bigl( \gamma (\cdot )+d_{2}( \cdot )\bigr)\xi =0.$$
(C.2)
Set $$\delta ^{*}$$ as the positive eigenfunction corresponding to $$\lambda _{0}$$, thus
$$D_{2}\Delta \delta ^{*}+ \biggl[ \bigl(1-r(\cdot )\bigr)\beta S^{0}+\bigl(1-\eta (\cdot )\bigr) \frac{\alpha (\cdot )V^{0}}{K(\cdot )} \biggr]\delta ^{*}-\bigl(\gamma ( \cdot )+d_{2}(\cdot )\bigr)\delta ^{*}=\lambda _{0}\delta ^{*}.$$
(C.3)
Multiplying (C.2) by $$\delta ^{*}$$ and (C.3) by ξ, integrating them on Ω, then subtracting, we have
$$\biggl(1-\frac{1}{\Re _{0}}\biggr) \int _{\Omega}\biggl[\bigl(1-r(\cdot )\bigr)\beta (\cdot )S^{0}+ \frac{(1-\eta (\cdot ))\alpha (\cdot )V^{0}}{K(\cdot )}\biggr]\xi \delta ^{*}\,dx = \lambda _{0} \int _{\Omega}\xi \delta ^{*}\,dx .$$
Since both sides of integrals are positive, thus $$1-\frac{1}{\Re _{0}}$$ and $$\lambda _{0}$$ have the same sign.
Appendix D: Proof of Lemma 3.3
With the equation of I in (2.2), with the maximum principle and Hopf boundary lemma, for $$t^{*}$$ and $$I(\cdot ,t^{*},\phi )\not \equiv 0$$, we have $$I\geq \tilde{I}$$, where Ĩ satisfies
$$\textstyle\begin{cases} \frac{\partial \tilde{I}}{\partial t} =D_{2}\Delta I-(\gamma (\cdot )+d_{2}( \cdot ))I,\quad (x,t)\in \Omega \times [0,\infty ), \\ \frac{\partial \tilde{I}}{\partial \nu} =0,\quad x\in \partial \Omega . \end{cases}$$
(D.1)
With Lemma 2.2, which implies $$I(\cdot ,t,\phi )>0$$. Thus from the equation of S and V in (2.2) we have S, V satisfies
$$\textstyle\begin{cases} \frac{\partial S}{\partial t}\geq D_{1}\Delta S+\Lambda _{*}S-(r^{*}+(1-r_{*}) \beta ^{*}I^{*}+d^{*}_{1})S,\quad (x,t)\in \Omega \times [0,\infty ), \\ \frac{\partial V}{\partial t}\geq D_{3}\Delta V+r_{*}S-((1-\eta _{*}) \alpha ^{*}I^{*}+\eta ^{*}+d_{3}^{*})V,\quad (x,t)\in \Omega \times [0, \infty ), \\ \frac{\partial \tilde{S}}{\partial \nu}= \frac{\partial \tilde{V}}{\partial \nu}=0,\quad x\in \partial \Omega . \end{cases}$$
(D.2)
By applying the comparison principle, we can infer that
$$\liminf_{t\to \infty}S(\cdot ,t,\phi )\geq \frac{\Lambda _{*}}{r^{*}+(1-r_{*})\beta ^{*}I^{*}+d^{*}_{1}}$$
and
$$\liminf_{t\to \infty}V(\cdot ,t,\phi )\geq \frac{r_{*}\Lambda _{*}}{[r^{*}+(1-r_{*})\beta ^{*}I^{*}+d^{*}_{1}][(1-\eta _{*})\alpha ^{*}I^{*}+(\eta ^{*}+d^{*}_{3})]},$$
which completes the proof.
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Reprints and Permissions
Liu, Y., Jian, S. & Gao, J. Dynamics analysis and optimal control of SIVR epidemic model with incomplete immunity. Adv Cont Discr Mod 2022, 51 (2022). https://doi.org/10.1186/s13662-022-03723-7
• Accepted:
• Published:
• DOI: https://doi.org/10.1186/s13662-022-03723-7
Keywords
• SIVR model
• Well-posedness
• Basic reproduction number
• Lyapunov function
• Spatial heterogeneity | 20,106 | 59,221 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.909115 |
https://www.indiabix.com/electronics-and-communication-engineering/communication-systems/147001 | 1,585,877,318,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370509103.51/warc/CC-MAIN-20200402235814-20200403025814-00441.warc.gz | 982,510,441 | 6,428 | # Electronics and Communication Engineering - Communication Systems
1.
In a satellite system circular polarization is to be obtained. The antenna used is
A. parabolic B. horn C. log periodic D. helical
Explanation:
Helical antenna produces a circularly polarized radiation if its two components are at right angles and are equal. It is very commonly used in radio telemetry.
2.
A horizontal output stage is cutoff for retrace and 40% of trace. If time for each horizontal line is 64 μs and retrace time is 12 μs, the transistor is conducting for about
A. 64 μs B. 6.4 μs C. 31 μs D. 3.1 μs
Explanation:
Total time for each horizontal line = .
Time for trace = 64 - 12 = 52 μs.
Time for which transistor is not conducting = 12 + 0.4 x 52 = 32.8 ms.
Time for which transistor is conducting = 64 - 32.8 = 31 μs.
3.
A TV remote control works on the principle of
A. pulse code modulated ultra violet light B. pulse code modulated infrared light C. demodulation D. either (a) or (b)
Explanation:
Infrared light is used in TV remote control.
4.
As the magnitude of reverse bias is increased, the capacitance of varactor diode
A. increases B. decreases C. first increase, then decreases D. first decreases then increases
Explanation:
When magnitude of reverse bias is increased, thickness of depletion layer increases.
Capacitance is inversely proportional to thickness.
5.
A. variable sensitivity B. squelch C. double conversion D. variable selectivity | 365 | 1,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-16 | latest | en | 0.859068 |
https://www.flashcardmachine.com/evaluating-expressions.html | 1,719,272,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865490.6/warc/CC-MAIN-20240624214047-20240625004047-00487.warc.gz | 678,219,658 | 5,450 | # Shared Flashcard Set
## Details
Evaluating Expressions
Evaluating variable expressions
10
Mathematics
09/11/2011
Term
y + yx x = 15, y = 8
Definition
128
Term
m + n + 6 m = 6, n = 1
Definition
11
Term
15 - (m + p) m = 3, p = 10
Definition
2
Term
zy + 4y y = 5, z = 2
Definition
30
Term
4 + a + b - a a = 4, b = 9
Definition
13
Term
m - (n/4) m = 5, n = 8
Definition
3
Term
2a - 3b + 4 a = 6, b = 2
Definition
10
Term
4x + 7y - 5 x = -1, y = 1
Definition
4(-1) + 7(1) - 5 -4 + 7 - 5 3 - 5 -2
Term
3(a + 2b) a = 4, b = 2
Definition
24
Term
-5(6 - m) + n m = 4, n = 12
Definition
2
Supporting users have an ad free experience! | 326 | 668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-26 | latest | en | 0.486705 |
https://www.geocaching.com/geocache/GC77EAB_eclipse-relativity?guid=a3b58ad2-fa8a-458e-8150-8e952daaafa4 | 1,627,263,323,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00705.warc.gz | 806,710,358 | 28,545 | <
## Eclipse Relativity
A cache by Get Mooned Message this owner
Hidden : 06/17/2017
Difficulty:
Terrain:
Size: (regular)
#### Watch
How Geocaching Works
Please note Use of geocaching.com services is subject to the terms and conditions in our disclaimer.
### Geocache Description:
This multicache is part of the Eclipse Coin Trail, celebrating a weekend of activities in Columbia and Lexington, SC, which culminate in a rare solar eclipse. The eclipse will occur Monday, August 21, 2017, at 2:41 pm and the Midlands will be one of the best places to witness this historic event. The total eclipse will last 2:36 minutes.
Historically, the total solar eclipse of May 29, 1919, is relatively important. British astronomer Sir Arthur Eddington travelled to Príncipe, an island off the coast of Africa, to observe this eclipse. His goal was to verify part of Einstein's Theory of Relativity. Einstein had surmised that the gravitational field of a star such as the Sun acts as a huge, cosmic lens that refracts light. Eddington photographed the stars, which are visible in daylight only during a total solar eclipse, and measured the stars' positions. They are slightly shifted from their original position because of the deflection of their light due to the gravitational field of the Sun. His measurements were regarded as conclusive proof that gravity bends light rays.
The journey to the final cache begins at the Cayce Historical Museum. Waypoint 1 is a monument honoring a Revolutionary War heroine.
A= # letters in the heroine's first name = _____.
B= total # of lines on the monument +1 = _____.
Continuing to Waypoint 2, you will find a plaque on a granite monument. At the base of the monument there is a second marker.
CD= the day the monument was moved from the original site = ___ ___.
E = # times the month "May" is referenced on the plaque + 1 = _____.
Following the walkway, you will see Waypoint 3 well before you get to it.
F= # rungs on one of the ladders - 5 = _____. There are two ladders and either will give you the correct answer.
G= # steps you would take if you were walking up to the platform = __4__.
The last bit of information for the final location is obtained at Waypoint 4.
HI= # letters in the first word on the sign + 2 = ___ ___.
J= # symbols mentioned + 4 = _____.
The final cache is a unique container made by local artist and geocacher, Carol Robertson, and placed with permission by the City of Cayce.
FINAL: N 33 AB.CDE W 081 F4.HIJ | 594 | 2,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-31 | latest | en | 0.927881 |
https://funlearningforkids.com/digital-ocean-missing-addends-activity/ | 1,708,851,201,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474594.56/warc/CC-MAIN-20240225071740-20240225101740-00070.warc.gz | 277,092,347 | 52,844 | # Digital Ocean Missing Addends Math Activity
Make practicing addition skills fun this summer with this digital ocean missing addends activity for Google Slides and Seesaw.
## Digital Ocean Missing Addends Activity
This digital ocean missing addends activity will give students 17 slides of practice as they work to find the missing addends in each addition sentence. The addition sentences focus on addition facts to 10.
This digital ocean-themed missing addend math game is a great option for whole group or small group instruction. It can also be used as a technology station, math centers, morning work, early finishers, at-home learning, summer practice, end of the year spiral review, and more. It can be used with both Google Slides and Seesaw, making it super easy to assign to your students.
Students may use Google Slides™ to do this activity.
The download file includes a detailed guide with instructions and pictures on how to use this activity with your students, but it is quite simple to do.
If you are an instructor or teacher, you will want to copy the Google Slides assignment to your Google Drive and save it as a master copy.
Make sure that you click the dropdown next to the attachment. Pick the “Make a copy for each student” option.
## Setting Up this Ocean Missing Addends Activity in Seesaw
Students can also use this activity in Seesaw, if you would like.
You will click the Seesaw specific link in the download file. When you arrive at the activity, you will need to click the “Save Activity” button. Then you will assign the activity to your students.
This digital ocean missing addends activity is both interactive and effective for learning. Students will gain one-to-one correspondence practice as they drag and drop the fish number pieces on the ocean-themed slides to complete the addition equations.
Students will use their knowledge of number pairs and counting on to complete the equation and make it true.
To use this ocean missing addends activity, students will look at the addition equation on the screen. They’ll look at the first addend and the sum. Students will use a strategy such as counting on, modeling, or subtracting the addend from the sum to determine what the missing addend is. Then, they’ll drag and drop the correct fish number piece to complete the ocean addition equation.
Once completed, I like to have my students practice reading the equation aloud to the group or a partner. This helps them retain the addition facts as they say them aloud.
## Ways to Differentiate
This digital ocean missing addends math game is very easy to differentiate to meet the needs of your students.
You can adjust the difficulty of the activity by rearranging or deleting slides. If you need to make it a bit easier, you can rearrange the slides to work with sums to 5 only and delete the other slides.
For students that are ready for more of a challenge, you can mix up the slides each time and have them complete all 17 slides with sums to 10.
You could also have your advanced students think of a story problem to math the addition equation. This requires them to use their brain in a whole new way.
You can split your students into small groups based on the level they’re at and assign the same set of slides to each student in the small group. No matter how many slides they have or what sums they’re working on, the directions are always the same, creating consistency for your students.
With clear and easy to follow instructions, this activity is perfect to use anytime of day and will encourage independence in your students.
I can’t wait for your students to try out this digital ocean missing addends activity. Their addition skills and problem-solving skills are going to take off!
Are you looking for more digital learning activities for your elementary students to practice key math and literacy skills? Click here for more fun digital activities.
Click the button below to grab your copy of the Free Digital Ocean Missing Addends Activity for Google Slides and Seesaw.
### Here are some more ocean activities to check out!
Editable Ocean Word Work Digital Mats
Ocean Counting Mats | 836 | 4,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-10 | latest | en | 0.915581 |
https://electronics.stackexchange.com/questions/492962/diode-and-capacitor-at-the-wien-bridge-oscillator | 1,718,860,065,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00732.warc.gz | 197,625,734 | 39,711 | # Diode and capacitor at the Wien Bridge Oscillator
The circuit below represents the JFET stabilized Wien-bridge oscillator.
we add another diode in the circuit. Shown in the picture below:
The questions are:
1. Before adding the second the diode the voltage at the input and output where smaller, why was that?
2. Why positive side of the capacitor C3 was connected to ground?
For the first question the answer I thought was that the negative excursion of the output signal forward bias the diode causing the capacitor to increase the drain-source resistance if the JFET and reduce the gain.
For the second question the answer I thought was that the output signal forward-bias D1 causing C3 to charge to a negative voltage, thats why the negative side of the capacitor should be connected to the circuit and the positive side to the ground.
I am not sure if these answers are correct. Can someone help me with this?
The JFET controls the magnitude of the sinewave being produced by the oscillator. You need amplitude control to keep the sinewave from growing ever bigger and hitting the saturation points of the op-amp and clipping.
The JFET naturally conducts without any gate bias and, to reduce its conduction (decrease the gain) you need to bias the gate negatively with respect to its source. Given that the source is connected to ground, you will expect to see a negative voltage on the gate and hence when you ask this: -
Why positive side of the capacitor C3 was connected to ground?
That should be clear. In addition to that the diode (D1) only conducts when the op-amp output is negative hence, D1 and C3 conspire to produce a voltage at their common node that is the peak of the negative swing of the sinewave output. Should the sinewave amplitude increase a tad, then the JFET conducts less and the gain is reduced.
Before adding the second diode the voltage at the input and output where smaller, why was that?
With one diode, there is less of a volt drop and so the sinewave amplitude needed to "activate" the JFET is smaller. With the 2nd diode, the amplitude has to grwo a bit bigger before amplitude stabilization occurs.
1) The increase in output voltage necessary to forward bias the two diodes has a larger effect on the output amplitude than the small reduction in gain caused by taking the jfet's gate just 0.7V more negative.
2) you are correct. The negative side of C3 is taken negative (with respect to ground) on the negative swings of the output. This is necessary because the JFET's gate must be taken negative (with respect to the source) to increase the source-drain resistance of the JFET and reduce the gain. So the larger the output amplitude gets, the more negative the gate goes and the smaller the gain is.
• If I am understanding the 1st correctly: it means that the voltage reduction that the diode create have a larger effect on the output voltage rather than the reduction that the jfet causes Commented Apr 14, 2020 at 7:05 | 656 | 2,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.940308 |
http://www.mathworks.com/matlabcentral/cody/problems/1172-wheat-on-a-chessboard-pt-1/solutions/187574 | 1,484,960,233,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280891.90/warc/CC-MAIN-20170116095120-00067-ip-10-171-10-70.ec2.internal.warc.gz | 562,934,482 | 11,741 | Cody
Problem 1172. Wheat on a chessboard pt 1
Solution 187574
Submitted on 7 Jan 2013
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Fail
%% n = 1; y_correct = 1; assert(isequal(wheat_chess(n),y_correct))
```Error: Undefined function or variable 'N'. ```
2 Fail
%% n = 0; y_correct = 0; assert(isequal(wheat_chess(n),y_correct))
```Error: Undefined function or variable 'N'. ```
3 Fail
%% n = -1; y_correct = 'NaN'; assert(isequal(wheat_chess(n),y_correct))
```Error: Undefined function or variable 'N'. ```
4 Fail
%% n = 4; y_correct = 65535; assert(isequal(wheat_chess(n),y_correct))
```Error: Undefined function or variable 'N'. ```
5 Fail
%% n = 8; y_correct = 18446744073709551615; assert(isequal(wheat_chess(n),y_correct))
```Error: Undefined function or variable 'N'. ```
6 Fail
%% n = 10; y_correct = 1267650600228229401496703205375; assert(isequal(wheat_chess(n),y_correct))
```Error: Undefined function or variable 'N'. ``` | 327 | 1,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-04 | latest | en | 0.401168 |
https://www.jiskha.com/questions/461607/a-diver-runs-horizontally-off-the-end-of-a-diving-board-with-an-initial-speed-of-1-25-m-s | 1,579,792,925,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250610919.33/warc/CC-MAIN-20200123131001-20200123160001-00410.warc.gz | 939,202,483 | 5,253 | Physics
A diver runs horizontally off the end of a diving board with an initial speed of 1.25 m/s. If the diving board is 3.20 m above the water, what is the diver's speed just before she enters the water? (Neglect air resistance.)
1. 👍 0
2. 👎 0
3. 👁 559
1. find the time it takes to fall 3.2m
3.2=1/2 g t^2
then, take that time and find the distance moved horizntally.
d=1.25*t
1. 👍 0
2. 👎 1
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asked by Abby on February 16, 2010 | 804 | 2,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-05 | latest | en | 0.920395 |
http://www.splung.com/content/sid/6/page/work | 1,490,800,391,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190753.92/warc/CC-MAIN-20170322212950-00237-ip-10-233-31-227.ec2.internal.warc.gz | 706,295,441 | 3,926 | Work
When a thermodynamic system does work, it uses energy in the form of heat added to the system. The general definition of work is that of a force times distance.
Consider a freely moving piston in a cyclinder which is closed at one end. Inside the cyclinder, there is a volume of gas inside the cylinder which is at a certain temperature T, pressure P and takes up a volume V. If heat is added to the system, the temperature is raised, the molecules in the gas gain kinetic energy and hit the walls of the cylinder with increased velocity. The force on the walls of the cylinder increase. The cylinder walls cannot move, but the piston, which is free to move, experiences a reaction force F=P A, where P is the pressure and A is the area of the piston. The piston moves a distance Δx. The work done dW=P A Δx. If A is constant, A Δx is the change in volume. For a constant force, W= PdV.Therefore the work done is P dV.
When the volume of a gas increases, work is done by the gas.
When the volume of a gas decreases, work is done on the gas by an external force.
Quasi-Static Process
If we consider the system to change slowly over time, we can effectively, consider each state to be considered reversible
The work is dependent on the path over which the process takes place.
The work done in going from 1 to 2 via a is
W= 2p0V0
but it is
W= p0V0
In going from 1 to 2 via b.
The above result shows that if the temperature of a gas is increased at constant volume, no work is done.
However, if the temperature is increased and the gas is allowed to expand, work will be done. In this case, extra energy will have to be supplied to do this work.
For this reason, gases are said to have two principal specific (or molar) heat capacities:
i) the specific (or molar) heat capacity at constant volume, cv ii) the specific (or molar) heat capacity at constant pressure, cp It should be clear that cp > cv and that the difference between them is given by
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http://mathforum.org/library/drmath/view/58545.html | 1,493,451,481,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123318.85/warc/CC-MAIN-20170423031203-00555-ip-10-145-167-34.ec2.internal.warc.gz | 241,251,299 | 3,551 | Associated Topics || Dr. Math Home || Search Dr. Math
### Factoring Polynomials of Degree 2
```
Date: 4/8/96 at 13:39:19
From: Lorrane Chung
Subject: Factoring polynomials
Hello,
I am having a lot of difficulties factoring polynomials of the
type x^2 + 6x +8 and 3x^2 + 10x +8. I have exhausted all the
methods - they don't seem to work for me. I really need an easy
method to factor other than the quadratic formula or for finding 2
factors that multiply to give last term but add to give the middle
Desperate,
Lorrane
```
```
Date: 4/16/96 at 19:33:19
From: Doctor Patrick
Subject: Re: Factoring polynomials
How did you try factoring these problems? When I have to factor,
I figure out the possible factors for the first and last terms and
then play with them a little to see what might work, and what is
definitely going to be out of the question.
In the first problem, all we need to factor is the 8 - the x^2 is
going to have to be x*x. This gives us two choices: 8*1, or 2*4.
Since all of the numbers in the problem are positive, we can
ignore the other two possibilities of -8*-1 and -2*-4.
Working with the 8*1 first, we get (x+8)*(x+1), which doesn't give
us the middle term we are looking for. Since there are no other
combinations using 8*1 (do you see why?) we can move on to the
other possibility of 2*4.
If we try factoring the equation with these numbers we get
(x+2)(x+4). When we multiply this back out we get x*x +2x + 4x +8
which equals x^2+6x+8 after we combine like terms and multiply out
the x*x.
The second problem is more complicated since we have one
additional term (the 3 in front of the x^2). Again, we have two
ways to factor 8, but we also need to try multiplying the 3 by
both of the factors to find out which combination works. If we
start with 8*1 as factors of 8, again we get two possibilities:
(3x+8)(x+1) or (3x+1)(x*8).
Do you understand why there are two possibilities this time?
The first possibility would end up giving us (3x*1) + (x*8), or
3x+8X, for the middle term. Since this would equal 11x it is not
the combination we are looking for.
Likewise (3x+1)(x*8) will not work. (Why not?) So now have to
move on to the other set of factors (4*2). Why don't you try the
remaining two possibilities on your own using the same methods we
used above. Good luck!
Write us back if you need more help!
-Doctor Patrick, The Math Forum
```
Associated Topics:
Middle School Factoring Expressions
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# Need help by the end of today!
0
571
1
+8
A quadratic of the form $-2x^2 + bx + c$ has roots of $x = 3 + \sqrt{5}$ and $x = 3 - \sqrt{5}.$ The graph of $y = -2x^2 + bx + c$ is a parabola. Find the vertex of this parabola.
Jan 14, 2020
#1
+412
+3
ignoring dollar sign glitch,
simplyfing
p(x)=-2(x^2-6x+4)
vertex coordinates are (-b/2a, -d/4a)
so vertex is V(3,10)
CALCUALTION : X = -b/2a=-(-6)/2=3
Y= -D/4a=-(20)/2=10
Jan 14, 2020 | 206 | 450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2022-27 | latest | en | 0.656235 |
https://www.coursehero.com/file/22141158/Lab4/ | 1,542,778,531,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747215.81/warc/CC-MAIN-20181121052254-20181121074254-00194.warc.gz | 817,884,626 | 36,088 | Lab4 - 71.1 5 ‘3 111 Lab BFContinuity Equation — Name...
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Unformatted text preview: 71.1. 5: _ . ‘3 111 _,/ Lab BFContinuity Equation — Name Date ”I 0’5}- J‘W? Time Will Who/5: ' l} )1.“ {2.61.13 fl ail-M \__ 3 (fir? / It 73.. - ‘ f l v: l 1,11 N . I .r\ ,I H W/ 1 “P , _ Data H’ f :35” :1" Container: Circular with D: ~~I" 9 m At: .{l/rffi “1 m2 . , --"l Hole: Circular with d: :05 m . Ah= ADM-m n12 Assume vh: (2gh)U2 ,- From Bernoulli equation (to be covered soon in class) i will 11.: .sz m I ll aw»: hf: 4-371 m l/p/l" 1/17“ {4 *1. 1 Time readings: {111111; 11‘ {1, /-. ‘ I: 1} Withthe lid on: t1: [NM (3) 1 "I111: , " Without the lid. t2=_—— (s) 'l W a) Estimate how long it would take for the water level to go from ho to hf when the container doesn’t have the lid on. Using the continuity equation, we get (see the lab description for details): V2111: —A 3 Substituting v2 =‘-/2gh and integrating we obtain: —2 Ar hm + C When 1:0 h=h{,; CI = M kg” ”W + 2A 112 1/2 A; j ’1 I): fl j _ r ‘ .- — «lg/1;. (ho — hf ) Wm H I A ...
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https://economics.stackexchange.com/questions/4461/interpretation-of-the-impulse-response-function | 1,620,692,164,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991553.4/warc/CC-MAIN-20210510235021-20210511025021-00414.warc.gz | 235,672,078 | 36,216 | # Interpretation of the Impulse Response Function
I have an IRF that shows the GDP shock to GDP.
Let's say I have a 5-year forecast of GDP. If there is an immediate 1% decrease in GDP today, can I adjust the original 5-year forecast by using the IRF?
e.g., GDP forecast in year 3 is 500 USD. The IRF says a 1% decrease in GDP will cause a 0.1 % decrease in year 3. Can I simply adjust the forecast downwards by 0.1% to to 499.5 USD?
Thank You
• From what kind of model was the IRF obtained? If univariate, then your approach is likely correct. If multivariate, you'd need to assume that the initial drop was actually caused by that specific type of shock for which IRF was computed (since the model would include several types of shocks, with different impacts on GDP). – ivansml Feb 21 '15 at 1:17
• Thanks for your help, @ivansml! This is indeed a multivariate model / VAR. – Abraham Feb 21 '15 at 11:42 | 247 | 910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-21 | latest | en | 0.958402 |
http://mathhelpforum.com/algebra/11938-need-some-help-simplifying.html | 1,481,391,747,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543316.16/warc/CC-MAIN-20161202170903-00127-ip-10-31-129-80.ec2.internal.warc.gz | 172,627,344 | 11,026 | # Thread: Need some help simplifying
1. ## Need some help simplifying
Having problems with these 3
1.
x2-x-2 x2-x-20
X
x2+x-6 x2+5x+4 Simplify
2. 1 = 1 6
x-1 2x+3 (2x+3)(x-1) :Solve
3. Brad is (now) 3 years older than janet. Janet's age 2 years from now (future) will be 8/9 of brad's age 2 years from now. How old are they now?
Thanks for any help
2. Originally Posted by ls1ramair19
Having problems with these 3
1.
x2-x-2 x2-x-20
X
x2+x-6 x2+5x+4 Simplify
2. 1 = 1 6
x-1 2x+3 (2x+3)(x-1) :Solve
3. Brad is (now) 3 years older than janet. Janet's age 2 years from now (future) will be 8/9 of brad's age 2 years from now. How old are they now?
Thanks for any help
Can you rewrite problems 1 and 2 please. For x squared write x^2. and for something divided by something write something/something. 'cause the forum right aligns everything, so its distorted and i really dont know what's going on.
For 3. though:
Let x be Brad's age. Then x - 3 is Janet's age. 2 years from now, Janet will be x - 3 + 2 = x - 1 = (8/9)(x + 2) = (8x + 16)/9
So the equation we want to solve is x - 1 = (8x + 16)/9.
=> 9x - 9 = 8x + 16
=> x = 25
So their ages now are, Brad: 25, Janet 22.
3. 1.
x^2-x-2 Times x^2-x-20
x^2+x-6SPACE x^2+5x+4 Simplify
2.
1/x-1= 1/2x+3 + 6/(2x+3)(x-1) solve:
4. Originally Posted by ls1ramair19
1.
x^2-x-2 Times x^2-x-20
x^2+x-6SPACE x^2+5x+4 Simplify
2.
1/x-1= 1/2x+3 + 6/(2x+3)(x-1) solve:
1. (x^2 - x - 2)/(x^2 + x - 6) times (x^2 - x - 20)/(x^2 + 5x + 4)
=> [(x - 2)(x + 1)]/[(x - 2)(x + 3)] times [(x - 5)(x + 4)]/[(x + 1)(x + 4)]
The (x - 2)'s cancel from the first fraction and the (x + 4)'s cancel from the second.
So we have (x + 1)/(x + 3) times (x - 5)/(x + 1)
=> [(x + 1)(x - 5)]/[(x + 1)(x + 3)]
The (x + 1)'s cancel leaving (x - 5)/(x + 3)
2. 1/(x - 1) = 1/(2x + 3) + 6/(2x + 3)(x - 1)
multiplying through by (2x + 3)(x - 1) we obtain:
2x +3 = x - 1 + 6
=>2x + 3- x = -1 + 6
=> x = 2
5. Originally Posted by ls1ramair19
1.
x^2-x-2 Times x^2-x-20
x^2+x-6SPACE x^2+5x+4 Simplify
2.
1/x-1= 1/2x+3 + 6/(2x+3)(x-1) solve:
If you dont see why i multiplied by (2x + 3)(x - 1) in 2. An alternate method may help you see why.
What you could do is bring the 1/(2x +3) over the equal sign to obtain:
1/(x - 1) - 1/(2x +3) = 6/(2x + 3)(x - 1)
we see if we add the left side, we will get a fraction where the lcm is the same as the right, so the denominators of both fractions will be the same, so the numerators can be equated. So we would end up with the line:
2x + 3 - x + 1 = 6
=> x = 2
6. Thank you very much you explained it a lot better than my teacher could
7. Originally Posted by ls1ramair19
Thank you very much you explained it a lot better than my teacher could
I hope so. Don't just read through it though. Try to work through it. Try both methods and see which one you like better. and then if you come up on any problems, just post a reply. | 1,189 | 2,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2016-50 | longest | en | 0.857033 |
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# Problem 3076. Create a vector
Solution 1745083
Submitted on 7 Mar 2019 by Yaphet Elias
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### Test Suite
Test Status Code Input and Output
1 Pass
n = 10; y_correct = (0:2:n); assert(isequal(zeroToNby2(n),y_correct))
2 Pass
n = 102; y_correct = (0:2:n); assert(isequal(zeroToNby2(n),y_correct))
3 Pass
n = 73; y_correct = (0:2:n); assert(isequal(zeroToNby2(n),y_correct)) | 167 | 495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-51 | latest | en | 0.683444 |
https://gmat.magoosh.com/forum/1419-carlotta-can-drive-from-her-home-to-her | 1,624,290,545,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488286726.71/warc/CC-MAIN-20210621151134-20210621181134-00286.warc.gz | 263,499,725 | 9,685 | Source: Official Guide for the GMAT 12th Ed. Section 6.3 Data Sufficiency; #71
1
# Carlotta can drive from her home to her
Carlotta can drive from her home to her office by one of two possible routes. If she must also return by one of these routes, what is the distance of the shorter route? (1) When she drives from her home to her office by the shorter route and returns by the longer route, she drives a total of 42 kilometers. (2) When she drives both ways, from her home to her office and back, by the longer route, she dives a total of 46 kilometers. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient., Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient., BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient., EACH statement ALONE is sufficient., Statements (1) and (2) TOGETHER are not sufficient.
### 1 Explanation
1
Mike McGarry, Magoosh Tutor
Nov 12, 2012 • Comment | 242 | 966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-25 | latest | en | 0.943151 |
http://www.qfak.com/education_reference/science_mathematics/?id=3734697 | 1,472,623,321,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471983578600.94/warc/CC-MAIN-20160823201938-00294-ip-10-153-172-175.ec2.internal.warc.gz | 650,228,057 | 5,766 | How do you change 42/16 into a mixed number?
How do you change 42/16 into a mixed number?
_ Page 1
You divide it.
First reduce it (try dividing by 2 - what you do to the top, you have to do to the bottom) which gives you 21/8.
21 divided by 8 = 2 5/8.
Remember that the line in a fraction means "divide"
#1
Divide 16 into 42, goes in 2 times with 10 left over------ 2 10/16, simplified to 2 5/8
#2
........2
.....-----
16 | 42
......32
......---
......10
Therefore, 42/16 = 2 10/16
#3
you divide 42 by 16, which gives you 2 with 10 remaining. this means the mixed number would be 2 and 10/16, which when cancelled down gives you 2 and 5/8 :)
#4
42/16
How many times does 16 go into 42?
16*2=32
42-32=10
2 and 10/16
or
2 and 5/8
in decimal= 2.625
#5
The / means divide.
42 divided by 16 is 2 remainder 10.
Put the 10 over 16, and then simplify that to 5/8.
#6
42/16 reduces first
21/8
divide denominator in to numerator (8 into 21) goes 2 times. whole number is 2 remainder is 5 so that is the numerator of the fraction part. 5/8
42/16 = 2 5/8
#7
well you find out what does 16 multiply by to get the closest to 42.
16 * 1 = 16
16 * 2 = 32
16 * 3 = 48
it looks like 16 * 2 gets you closest to the number 42.
now that you have that figured out, your fraction looks like this so far --> 2 and nothing over the 16.
no you must take the 42 and subtract it by the 32. which in this case is 10.
so your fraction is a BIG 2 and then next to it 10 over 16. but you can simplify the 10 over 16. (10/16). to simplify it, you must find a number that can be divided evenly into both the numerator and the denominator. in this case, the number is 2.
this meaning that the small fraction next to the 2 is 5 over 8. (5/8)
so your final fraction is a BIG 2 with a small fraction next to it being a 5 over 8. (2/5/8)
hope i could help. :)
#8 | 611 | 1,872 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2016-36 | longest | en | 0.934495 |
https://id.scribd.com/document/91505311/4-Simultaneous-Equations | 1,580,169,888,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00312.warc.gz | 491,609,429 | 72,269 | Anda di halaman 1dari 3
# 4.
SIMULTANEOUS EQUATIONS
IMPORTANT NOTES: *** Usually asked in PAPER 2, Question 1. (i) Characteristics of simultaneous equations: (a) Involves TWO variables, usually in x and y. (b) Involves TWO equations : one linear and the other non- linear. (ii) Solving simultaneous equations means finding the values of x and corresponding y which satisfy BOTH the equations. (iii) Methods of solving :: (a) Starting from the LINEAR equation, express y in terms of x (or x in terms of y). (b) Substitute y (or x) into the second equation (which is non-linear) to obtain a quadratic equation in the form ax2 + bx + c = 0. (c) (iv) Solve the quadratic equation by factorisation or by using the FORMULA
b b 2 4ac 2a
## Solving Simultaneous Equations (SPM FORMAT QUESTIONS)
C1. Solve x + y = 3, xy = 10 . x+y = 3 ........ (1) xy = 10 ........ (2) Fromi (1), y = 3 x ......... (3) Substitute (3) into (2), x (3 x) = 10 3x x2 = 10 x2 3x 10 = 0 (x + 2) (x 5) = 0 x = 2 atau x = 5 From (3), when x = 2 , y = 3 (-2) = 5 x = 5, y = 2 Answers: x = 2, y = 5 ; x = 5 , y = 2 . 1. Solve x + y = 5, xy = 4 .
(Ans : x = 1, y = 4 ; x = 4, y = 1)
L2.
Solve x + y = 2 , xy = 8 .
L3. Solve 2x + y = 6, xy = 20 .
(Ans : x = 4 , y = 2 ; x = 2, y = 4 )
(Ans : x = 2 , y = 10 ; x = 5, y = 4 )
4 Simultaneous Equations
L4.
(SPM 2000)
## L5. Solve the simultaneous equations : x+y3 = x2 + y2 xy = 0 21
(SPM 2002)
(Ans : x = 3 , y = 2 )
6.
## Solve the simultaneous equations : 4x + y = 8 x2 + x y = 2
7. Solve the simultaneous equations p m = 2 and p2 + 2m = 8. Give your answers correct to three decimal places . (SPM 2004)
(Ans : x = 2 , y = 0 ; x = 3 , y = 4 )
## (Ans : m = 0.606, p = 2.606 ; m = 6.606 , p = 4.606 )
4 Simultaneous Equations
8.
## Solve the simultaneous equations
(SPM 2005)
x+
1 y = 1 and y2 10 = 2x 2
9. Solve the simultaneous equations 2x + y = 1 and 2x2 + y2 + xy = 5. Give your answers correct to three decimal places . (SPM 2006)
(Ans : x = 4 , y = 3 ; x = , y = 3 )
## (Ans : x = 1.618, y = 2.236
, x = 0.618, y = 0.236)
10.
## 11. Solve the simultaneous equations x 3y + 4 = 0 x2 + xy 40 = 0
(SPM 2008)
(Ans : x = 3 , y = 3 ; x = 1 , y = 1 )
(Ans : x = 5, y = 3 ;
x = - 6 , y = - 2/3 )
4 Simultaneous Equations | 902 | 2,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-05 | latest | en | 0.730502 |
http://oeis.org/A001239 | 1,571,397,725,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00000.warc.gz | 134,834,832 | 3,921 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A001239 Numbers that are the sum of 3 nonnegative cubes in more than 1 way. 4
216, 251, 344, 729, 855, 1009, 1072, 1366, 1457, 1459, 1520, 1674, 1728, 1729, 1730, 1737, 1756, 1763, 1793, 1854, 1945, 2008, 2072, 2241, 2414, 2456, 2458, 2729, 2736, 2752, 3060, 3391, 3402, 3457, 3500, 3592, 3599, 3655, 3744, 3745 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 REFERENCES G. H. Hardy, Ramanujan, Cambridge Univ. Press, 1940, p. 12. D. Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 165. LINKS Christian N. K. Anderson, Table of n, a(n) for n = 1..10000 Christian N. K. Anderson, Decomposition of first 10000 terms into cube triples MATHEMATICA Select[Range[4000], Length[PowersRepresentations[#, 3, 3]] > 1 &] (* Harvey P. Dale, Feb 03 2011 *) PROG (PARI) is(n)=my(t); for(a=0, sqrtnint(n, 3), my(a3=a^3, c); for(b=0, min(a, sqrtnint(n-a3, 3)), if(ispower(n-a3-b^3, 3, &c) && c <= b && t++>1, return(1)))); 0 \\ Charles R Greathouse IV, Jul 02 2017 CROSSREFS Cf. A001235, A003998, A008917, A025396, A025456, A025447. Sequence in context: A018863 A003998 A230400 * A184088 A205191 A204650 Adjacent sequences: A001236 A001237 A001238 * A001240 A001241 A001242 KEYWORD nonn AUTHOR STATUS approved
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Last modified October 18 07:19 EDT 2019. Contains 328146 sequences. (Running on oeis4.) | 634 | 1,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2019-43 | latest | en | 0.547302 |
https://www.fachschaft.informatik.tu-darmstadt.de/forum/viewtopic.php?f=513&t=26355&p=145751 | 1,603,942,476,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902683.56/warc/CC-MAIN-20201029010437-20201029040437-00617.warc.gz | 713,417,894 | 13,060 | ## Exam Check-List: Discussion on Part 09 - Lambda Calculus
FeG
Endlosschleifenbastler
Beiträge: 182
Registriert: 6. Dez 2007 07:01
### Exam Check-List: Discussion on Part 09 - Lambda Calculus
Hi,
here is my solution to the ninth topic of the exam check-list (-> spoiler alert...).
Best,
Felix
1. What types of expressions are present in the lambda calculus. Define a definition of the [concrete/abstract] syntax of the lambda calculus.
Present expressions:
* function definition / lambda expressions
* function application
* identifiers
Concrete Syntax:
Code: Alles auswählen
<LC> ::= {fun {<id>} <LC>}
| {<LC> <LC>}
| <id>
Abstract Syntax:
Code: Alles auswählen
(define-type LC
[fun (param symbol?) (body LC?)]
[app (fun-expr LC?) (arg-expr LC?)]
[id (name symbol?)])
2. Encode given [boolean values/numbers/pairs/lists] and given operations on these values in the lambda calculus.
Encoding of values:
Code: Alles auswählen
#t -> (lambda (l r) l)
0 -> (lambda (f) (lambda (x) x))
2 -> (lambda (f) (lambda (x) (f (f x))))
pair (0,1) -> (lambda (sel) (sel #0 #1)) ;; where #0, #1 are the encodings of 0 and 1, respectively
'() -> (lambda (c n) n)
'(0 1 2) -> (lambda (c n) (c 0 (c 1 (c 2 n))))
Operations:
Code: Alles auswählen
(define succ (lambda (n)
(lambda (f) (lambda (x) (f ((n f) x))))))
(define sum (lambda (n) (lambda (m)
((m succ) n))))
(define prod (lambda (n) (lambda (m)
(lambda (f) (lambda (x) ((n (m f)) x))))))
(l (lambda (c n) c) null)))
(define append (lambda (l1 l2)
(lambda (c n) (l1 c (l2 c n)))))
3. Define the major property of the fix-point operator (Y-combinator).
The major property of a fix-point combinator FIX is that it computes the fix-point of the function given as argument, i.e., for any function g it holds that g(FIX(g)) = FIX(g).
4. Use the given fix-point operator to implement the specified recursive function.
Fix-point operator:
Code: Alles auswählen
(define Y
(lambda (f)
((lambda (x) (f (lambda (v) ((x x) v))))
(lambda (x) (f (lambda (v) ((x x) v)))))))
Recursive function:
Code: Alles auswählen
(define (length lst)
(if (empty? lst)
0
(+ 1 (length (rest lst)))))
Implementation using fix-point operator:
Code: Alles auswählen
(define length-f
(Y (lambda (f) (lambda (lst)
(if (empty? lst)
0
(+ 1 (f (rest lst))))))))
5. Which of the given fix-point operators are correct? Prove it. Which of them support [eager evaluation/call-by-value semantics]?
Fix-point operators A and B:
Code: Alles auswählen
(define A
(lambda (f)
((lambda (x) (f (lambda (v) ((x x) v))))
(lambda (x) (f (lambda (v) ((x x) v)))))))
(define B
(lambda (f)
(f f)))
A is correct:
Code: Alles auswählen
A(g)
= ((lambda (x) (g (lambda (v) ((x x) v)))) (lambda (x) (g (lambda (v) ((x x) v)))))
= (g (lambda (v) (((lambda (x) (g (lambda (v) ((x x) v)))) (lambda (x) (g (lambda (v) ((x x) v))))) v)))
= (g (lambda (v) ((g (lambda (v) (((lambda (x) (g (lambda (v) ((x x) v)))) (lambda (x) (g (lambda (v) ((x x) v))))) v))) v)))
= ...
... I don't see why this is = g(A(g)), but I know A is correct (used above...) :-/
B is incorrect:
Code: Alles auswählen
B(g)
= (g g)
!= g((g g))
= g(B(g))
On the second part of the question: How do you see whether a fix-point operator supports eager evaluation resp. call-by-value semantics? (@Andreas)
sewe
Sonntagsinformatiker
Beiträge: 295
Registriert: 16. Jan 2009 14:53
Kontaktdaten:
### Re: Exam Check-List: Discussion on Part 09 - Lambda Calculus
FeG hat geschrieben: A is correct:
... I don't see why this is = g(A(g)), but I know A is correct (used above...) :-/
Let me redo the derivation, as I think the last step went wrong:
Code: Alles auswählen
A(g)
= ((lambda (x) (g (lambda (v) ((x x) v))))
(lambda (x) (g (lambda (v) ((x x) v)))))
= (g (lambda (v) (((lambda (x) (g (lambda (v) ((x x) v))))
(lambda (x) (g (lambda (v) ((x x) v))))) v)))
= ...
Now you make use of the fact that (lambda (x) (f x)) is equivalent to f and then apply the definition of fix:
Code: Alles auswählen
...
= (g ((lambda (x) (g (lambda (v) ((x x) v))))
(lambda (x) (g (lambda (v) ((x x) v))))))
= (g (fix g))
This completes the proof.
Note that the last two steps are not derivation steps (as an interpreter would take them) but rather abstract equivalence arguments.
FeG hat geschrieben: On the second part of the question: How do you see whether a fix-point operator supports eager evaluation resp. call-by-value semantics? (@Andreas)[/list]
Regarding the second part of the question, you have to see whether your proof/derivation works when the order of evaluation is that required by eager evaluation. If it does, then the fix-point operator in question supports eager evaluation.
FeG
Endlosschleifenbastler
Beiträge: 182
Registriert: 6. Dez 2007 07:01
### Re: Exam Check-List: Discussion on Part 09 - Lambda Calculus
sewe hat geschrieben:This completes the proof.
Note that the last two steps are not derivation steps (as an interpreter would take them) but rather abstract equivalence arguments.
Okay, i assumed it had to be completely derivational, with this application argument I think I got it..
FeG hat geschrieben: On the second part of the question: How do you see whether a fix-point operator supports eager evaluation resp. call-by-value semantics? (@Andreas)[/list]
Regarding the second part of the question, you have to see whether your proof/derivation works when the order of evaluation is that required by eager evaluation. If it does, then the fix-point operator in question supports eager evaluation.
Could you restate the second half of the sentence; I didn't get the "when the order of evaluation is that required by eager evaluation".
Concerning call-by-value semantics: Are there combinators that work with call-by-reference?
Best,
Felix | 1,729 | 5,789 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-45 | latest | en | 0.513947 |
https://cran.microsoft.com/snapshot/2018-09-26/web/packages/ggstatsplot/vignettes/ggpiestats.html | 1,685,457,817,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645810.57/warc/CC-MAIN-20230530131531-20230530161531-00046.warc.gz | 207,486,836 | 197,626 | # ggpiestats
#### 2018-08-14
The function ggstatsplot::ggpiestats can be used for quick data exploration and/or to prepare publication-ready pie charts to summarize the statistical relationship between two categorical variables. We will see examples of how to use this function in this vignette.
To begin with, here are some instances where you would want to use ggpiestats-
• to see if the frequency distribution of two categorical variables are independent of each other using the contingency table analysis
• to check if the proportion of observations at each level of a categorical variable is equal
Note before: The following demo uses the pipe operator (%>%), so in case you are not familiar with this operator, here is a good explanation: http://r4ds.had.co.nz/pipes.html
## Statistical independence of categorical variables with ggpiestats
To demonstrate how ggpiestats can be used to we will be using the Titanic dataset that is included in the datasets library. Titanic Passenger Survival Data Set provides information “on the fate of passengers on the fatal maiden voyage of the ocean liner Titanic, summarized according to economic status (class), sex, age, and survival.”
Let’s have a look at the structure of this table and also convert it into a tibble while we are at it.
library(datasets)
library(dplyr)
# looking at the table
dplyr::glimpse(x = Titanic)
#> 'table' num [1:4, 1:2, 1:2, 1:2] 0 0 35 0 0 0 17 0 118 154 ...
#> - attr(*, "dimnames")=List of 4
#> ..$Class : chr [1:4] "1st" "2nd" "3rd" "Crew" #> ..$ Sex : chr [1:2] "Male" "Female"
#> ..$Age : chr [1:2] "Child" "Adult" #> ..$ Survived: chr [1:2] "No" "Yes"
Note that the last column in this dataframe contains count information, which means we will have to modify it to reflect this count structure. This has already been carried out and the final dataset is included in the ggstatsplot package in Titanic_full. This is not necessary as ggpiestats can handle table structures as well (see examples below).
Let’s have a look at this dataset.
library(ggstatsplot)
# looking at the final dataset
dplyr::glimpse(ggstatsplot::Titanic_full)
#> Observations: 2,201
#> Variables: 5
#> $id <dbl> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16... #>$ Class <fct> 3rd, 3rd, 3rd, 3rd, 3rd, 3rd, 3rd, 3rd, 3rd, 3rd, 3rd...
#> $Sex <fct> Male, Male, Male, Male, Male, Male, Male, Male, Male,... #>$ Age <fct> Child, Child, Child, Child, Child, Child, Child, Chil...
#> $Survived <fct> No, No, No, No, No, No, No, No, No, No, No, No, No, N... First, let’s see if the proportion of people who survived was different between sexes using ggpiestats. # since effect size confidence intervals are computed using bootstrapping, let's # set seed for reproducibility set.seed(123) ggstatsplot::ggpiestats(data = ggstatsplot::Titanic_full, condition = Sex, main = Survived) A number of arguments can be modified to change the appearance of this plot: library(ggplot2) # for reproducibility set.seed(123) ggstatsplot::ggpiestats( data = ggstatsplot::Titanic_full, # dataframe (matrix or table will not be accepted) main = Survived, # rows in the contingency table condition = Sex, # columns in the contingecy table title = "Passengar survival by gender", # title for the entire plot caption = "Source: Titanic survival dataset", # caption for the entire plot legend.title = "Survived?", # legend title facet.wrap.name = "Gender", # changing the facet wrap title facet.proptest = TRUE, # proportion test for each facet stat.title = "survival x gender: ", # title for statistical test palette = "Set1", # changing the color palette ggtheme = ggplot2::theme_classic() # changing plot theme ) + # further modification with ggplot2 commands ggplot2::theme(plot.subtitle = ggplot2::element_text( color = "black", size = 10, face = "bold", hjust = 0.5 )) As seen from this plot, the Pearson’s chi-square test of independence shows that the distribution of survival was different across males and females. Additionally, among both males and females, the proportion of survival was not equally likely (at 50%, i.e.), as shown by significant results (***) from one-sample proportion tests for each facet. In case the condition argument is not specified, instead of chi-square test of independence, a proportion test will be carried out. For example, let’s see if there were equal proportions of different age groups. ggstatsplot::ggpiestats( data = ggstatsplot::Titanic_full, main = Age ) As this plot shows there were overwhelmingly more number of adults than children on the boat and the proportion test attests to this. ## Grouped analysis with grouped_ggpiestats What if we want to do the same analysis separately for the four different Class on the Titanic (1st, 2nd, 3rd, Crew), i.e. checking how the survival-by-gender interaction changes by the passenger class in which the people were traveling? In that case, we will have to either write a for loop or use purrr, both of which are time consuming and can be a bit of a struggle. ggstatsplot provides a special helper function for such instances: grouped_ggpiestats. This is merely a wrapper function around ggstatsplot::combine_plots. It applies ggpiestats across all levels of a specified grouping variable and then combines list of individual plots into a single plot. Note that the grouping variable can be anything: conditions in a given study, groups in a study sample, different studies, etc. ggstatsplot::grouped_ggpiestats( # arguments relevant for ggstatsplot::gghistostats data = ggstatsplot::Titanic_full, grouping.var = Class, title.prefix = "Passenger class", stat.title = "survival x gender: ", main = Survived, condition = Sex, # arguments relevant for ggstatsplot::combine_plots title.text = "Survival in Titanic disaster by gender for all passenger classes", caption.text = "Asterisks denote results from proportion tests; ***: p < 0.001, ns: non-significant", nrow = 4, ncol = 1, labels = c("(a)","(b)","(c)", "(d)") ) As seen from this quick exploratory analysis, across all passenger classes, the proportion of survived to non-survived individuals differed across genders: Men were more likely to perish than survive, whereas women were more likely to survive than perish. The only exception was the 3rd Class passengers where women were as likely to survive as to perish. This will work even if the condition argument is not specified: ggstatsplot::grouped_ggpiestats( data = ggstatsplot::Titanic_full, main = Age, grouping.var = Class, title.prefix = "Passenger Class" ) #> Warning: Proportion test will not be run because it requires Age to have at least 2 levels with non-zero frequencies. ## Grouped analysis with ggpiestats + purrr Although this grouping function provides a quick way to explore the data, it leaves much to be desired. For example, we may want to add different captions, titles, themes, or palettes for each level of the grouping variable, etc. For cases like these, it would be better to use (e.g.). Note before: Unlike the function call so far, while using purrr::pmap, we will need to quote the arguments. # let's split the dataframe and create a list by passenger class class_list <- ggstatsplot::Titanic_full %>% base::split(x = ., f = .$Class, drop = TRUE)
# this created a list with 4 elements, one for each class
# you can check the structure of the file for yourself
# str(class_list)
# checking the length and names of each element
length(class_list)
#> [1] 4
names(class_list)
#> [1] "1st" "2nd" "3rd" "Crew"
# running function on every element of this list note that if you want the same
# value for a given argument across all elements of the list, you need to
# specify it just once
plot_list <- purrr::pmap(
.l = list(
data = class_list,
main = "Survived",
condition = "Sex",
facet.wrap.name = "Gender",
title = list(
"Passenger class: 1st",
"Passenger class: 2nd",
"Passenger class: 3rd",
"Passenger class: Crew"
),
caption = list(
"Total: 319, Died: 120, Survived: 199, % Survived: 62%",
"Total: 272, Died: 155, Survived: 117, % Survived: 43%",
"Total: 709, Died: 537, Survived: 172, % Survived: 25%",
"Not available"
),
palette = list("Accent", "Paired", "Pastel1", "Set2"),
ggtheme = list(
ggplot2::theme_grey(),
ggplot2::theme_classic(),
ggplot2::theme_light(),
ggplot2::theme_minimal()
),
sample.size.label = list(TRUE, FALSE, TRUE, FALSE),
messages = FALSE
),
.f = ggstatsplot::ggpiestats
)
# combining all individual plots from the list into a single plot using combine_plots function
ggstatsplot::combine_plots(
plotlist = plot_list,
title.text = "Survival in Titanic disaster by gender for all passenger classes",
caption.text = "Asterisks denote results from proportion tests; ***: p < 0.001, ns: non-significant",
nrow = 4,
ncol = 1,
labels = c("(a)","(b)","(c)", "(d)")
)
As can be appreciated from this example, although grouped_ggpiestats provides a quick way to explore data, purrr::pmap lets us utilize the full functionality of this function and ggplot2.
## Working with counts data
ggpiestats can also work with dataframe containing counts (aka tabled data), i.e., when each row doesn’t correspond to a unique observation. For example, consider the following fishing dataframe containing data from two boats (A and B) about the number of different types fish they caught in the months of February and March. In this dataframe, each row doesn’t equal a unique observation. In such cases, we can use counts argument. Let’s say we want to investigate if the frequency of different types of fish caught differs across the two months:
# for reproducibility
set.seed(123)
# creating a dataframe
# (this is completely fictional; I don't know first thing about fishing!)
(
fishing <- data.frame(
Boat = c(rep("B", 4), rep("A", 4), rep("A", 4), rep("B", 4)),
Month = c(rep("February", 2), rep("March", 2), rep("February", 2), rep("March", 2)),
Fish = c(
"Bass",
"Catfish",
"Cod",
"Cod",
"Bass",
"Catfish",
"Bass",
"Catfish",
"Cod",
"Cod",
"Bass",
"Catfish"
),
SumOfCaught = c(25, 20, 35, 40, 40, 25, 30, 42, 40, 30, 33, 26, 100, 30, 20, 20)
) %>% # converting to a tibble dataframe
tibble::as_data_frame(x = .)
)
#> # A tibble: 16 x 4
#> Boat Month Fish SumOfCaught
#> <fct> <fct> <fct> <dbl>
#> 1 B February Bass 25
#> 2 B February Catfish 20
#> 3 B March Cod 35
#> 4 B March Haddock 40
#> 5 A February Cod 40
#> 6 A February Haddock 25
#> 7 A March Bass 30
#> 8 A March Catfish 42
#> 9 A February Bass 40
#> 10 A February Catfish 30
#> 11 A March Cod 33
#> 12 A March Haddock 26
#> 13 B February Cod 100
#> 14 B February Haddock 30
#> 15 B March Bass 20
#> 16 B March Catfish 20
# running ggpiestats with counts information
ggstatsplot::ggpiestats(
data = fishing,
main = Fish,
condition = Month,
counts = SumOfCaught
)
We just verified that the frequency of different types of fish caught differs across the two months, but what if we further want to know if this difference is present for both boats (since they are fishing in different parts of the sea)? For this, we can again utilize grouped_ggpiestats:
# running the grouped variant of the function
ggstatsplot::grouped_ggpiestats(
data = fishing,
main = Fish,
condition = Month,
counts = SumOfCaught,
grouping.var = Boat,
title.prefix = "Boat",
nrow = 2
)
As seen from these charts, this difference is found only for the location in which the boat B has been fishing. Additionally, faceted proportion tests also reveal that all fish are not equally likely to be caught at this location.
## Within-subjects designs
In case of within-subjects designs, you can set paired = TRUE, which will display results from McNemar test in the subtitle.
# seed for reproducibility
set.seed(123)
# data
clinical_trial <-
tibble::tribble(
~Control, ~Case, ~pairs,
"No", "Yes", 25,
"Yes", "No", 4,
"Yes", "Yes", 13,
"No", "No", 92
)
# plot
ggstatsplot::ggpiestats(data = clinical_trial,
condition = Control,
main = Case,
counts = pairs,
paired = TRUE,
stat.title = "McNemar test: ",
title = "Results from case-control study",
palette = "Accent")
## Suggestions
If you find any bugs or have any suggestions/remarks, please file an issue on GitHub: https://github.com/IndrajeetPatil/ggstatsplot/issues | 3,382 | 12,483 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.826292 |
https://farrarwilliams.wordpress.com/2010/11/16/measurement-books/?replytocom=392 | 1,611,360,309,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531429.49/warc/CC-MAIN-20210122210653-20210123000653-00364.warc.gz | 333,898,953 | 27,212 | # Measurement Books
One of our co-ops is starting a new theme on measurement. We often do very little to go along with our co-op themes. We might check out a few books from the library and we talk about what we’re learning about in the co-op, but otherwise, I haven’t been connecting it with other aspects of our schooling. However, this time around, I thought it might be a good chance to take a break (mostly) from our math curriculum and do a unit on measurement at home too. I bought the Math Mammoth blue series book on measurement. Here are the kids measuring their new books with paperclips and crayons. BalletBoy insisted that they all needed to be green crayons for some reason. Some of the content is a little too sophisticated for my first graders, but much of it will be a good little text for us to do as we explore the topic.
We also checked out an absurd pile of books on measurement from the library. Here are some highlights.
Measuring Penny by Loreen Leedy
As always, Loreen Leedy’s clever book leads the pack for measurement. This is a classic one. A girl measures her dog in every way she can imagine for a school project. It’s an inspiring sort of book in that it’s easy to use it as a jumping off point for measuring more things.
Room for Ripley and Super Sand Castle Saturday by Stuart J. Murphy
We found these two titles from Stuart J. Murphy’s MathStart series. They’re both good. In the first, volume is explored in simple terms as a boy fills up a bowl for a new fish. In the second, many kinds of measurements are explored as kids build sand castles.
How Tall How Short How Far Away by David A. Adler
This cheerfully drawn book gives a quick introduction to the history of measuring length, showing little pictures of Egyptians measuring with their arms to make cubits. After talking about what measurements we use today for length, it invites the reader to think about which ones are right for which tasks.
If Dogs Were Dinosaurs by David Schwartz
This book, along with its companion, If You Hopped Like a Frog, use excellent illustrations to show a comparison of sizes and lengths. This is a creative little book that’s short enough to be enjoyed by younger kids, but interesting enough to be enjoyed by adults. There’s no story, but each page is a thought provoking little summary.
How Fast Is It? by Ben Hillman
This book, with glossy photoshopped images, was full of fun facts comparing the speeds of different things. Each page had a different topic. It highlighted not only some of the fastest things, but also just compared some unexpected things like the speeds of swimming birds and flying fish.
Science Factory: Units and Measurements by Jon Richards
We checked out several measurement activity books, but all of them quickly went back to the library except this one. Almost all the projects in this book involve making your own measuring devices, such as an hourglass with two bottles and a balance out of a coat hanger. I want the kids to make a measuring wheel and measure the distance around our block.
## 5 thoughts on “Measurement Books”
1. Hey! That’s the one unit we downloaded from MM too; methinks SM doesn’t cover it adequately. Or maybe it’s me that keeps forgetting how many pints are in a quart and how many kilometers in a mile…..;)
2. Oh, and i just put Measuring Penny on hold at my library–thanks!
3. Can my 6yr old come and stay with you ? 🙂 You do such a good job of giving your boys learning opportunities.
1. Hehe. Sure. Extra children welcome. Then I’ll send mine your way Melissa to get a good dose of poetry down under 😉 | 830 | 3,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-04 | latest | en | 0.943617 |
https://math.stackexchange.com/questions/3156760/what-is-the-k%C3%BCnneth-formula-for-complete-varieties | 1,566,668,902,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321351.87/warc/CC-MAIN-20190824172818-20190824194818-00289.warc.gz | 549,531,450 | 31,194 | What is the Künneth formula for complete varieties?
I'm reading a part of Mumford's Abelian Varieties, and in the Chapter The theorem of the cube: II he claims that some "Künneth formula" tells us that if $$L_1$$ is a line bundle on a product $$X \times Y_1$$ such that $$L_1 \cong p_2^*(M_1)$$ for some line bundle $$M_1$$ on $$Y_1$$, then $$p_{2,*}(L_1) \cong M_1$$.
Here $$X$$ is a complete variety over an algebraically closed field $$k$$, and $$Y_1$$ is a scheme of finite type over $$k$$.
The Künneth formula I know relates (co-)homology on product spaces with the (co-)homology on the base spaces, for example as in the stacks-project, tag 0BEC. I don't see how the claim should follow from this.
Does the claim follow from this Künneth formula, or is something else going on here?
• Certainly the projection formula is going on here. Perhaps to apply it properly, one needs to use Kunneth. – aginensky Mar 21 at 14:38
• @aginensky Yeah by the projection formula we get: $p_{2,*} p_2^* M_1 = p_{2,*}(\mathcal{O} \otimes p_2^*M_1) = p_{2,*}(\mathcal{O}_{X\times Y_1}) \otimes M_1$, so we reduce this to show that $p_{2,*}\mathcal{O}_{X \times Y_1} \cong \mathcal{O}_{Y_1}$. I think I can show this using "Cohomology and Base Change" (Hartshorne, p.290+291), but I still don't see any connection to the Künneth formula. – red_trumpet Mar 21 at 14:58
• Perhaps in saying that the structure sheaf of the product is the tensor product of the pullbacks of the two structure sheaves. – aginensky Mar 21 at 14:59
As aginensky says, the first step seems to be to use the projection formula. This shows that $$(p_2)_* L_1 = M_1 \otimes (p_2)_* O_Z$$ where I have written $$Z$$ to denote the product (to save typing).
So now it is enough to show that $$(p_2)_* O_Z = O_{Y_1}$$. To do this, recall that $$(p_2)_* O_Z$$ is defined by
\begin{align*} U &\mapsto H^0((p_2)^{-1}(U), O_{(p_2)^{-1}(U)})\\ &= H^0(U \times X, O_{U \times X}) \end{align*} Now Künneth tells us that this is isomorphic to \begin{align*} &H^0(U,O_U) \otimes_k H^0(X,O_X) =H^0(U,O_U)\end{align*} using the fact that $$X$$ is complete.
• Seems correct. Two notes: In the last equation, it should be $\otimes$ instead of $\times$, and afaik $U \mapsto H^0(U\times X, \mathcal{O})$ already is a sheaf, so no need to sheafifiy here. – red_trumpet Mar 21 at 20:54 | 796 | 2,330 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-35 | latest | en | 0.852992 |
https://multi-converter.com/acre-feet-to-liters | 1,721,768,288,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00806.warc.gz | 347,302,745 | 6,679 | # Acre Feet to Liters
Change to Liters to Acre Feet
Share:
## How to convert Acre Feet to Liters
1 [Acre Feet] = 1233481.8375475 [Liters]
[Liters] = [Acre Feet] * 1233481.8375475
To convert Acre Feet to Liters multiply Acre Feet * 1233481.8375475.
## Example
65 Acre Feet to Liters
65 [ac ft] * 1233481.8375475 = 80176319.440589 [l]
## Conversion table
Acre Feet Liters
0.01 ac ft12334.818375475 l
0.1 ac ft123348.18375475 l
1 ac ft1233481.8375475 l
2 ac ft2466963.675095 l
3 ac ft3700445.5126426 l
4 ac ft4933927.3501901 l
5 ac ft6167409.1877376 l
10 ac ft12334818.375475 l
15 ac ft18502227.563213 l
50 ac ft61674091.877376 l
100 ac ft123348183.75475 l
500 ac ft616740918.77376 l
1000 ac ft1233481837.5475 l | 290 | 715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.38776 |
https://brilliant.org/problems/an-interesting-problem-307/ | 1,490,744,736,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190134.25/warc/CC-MAIN-20170322212950-00158-ip-10-233-31-227.ec2.internal.warc.gz | 774,990,394 | 18,131 | # A problem by Abhishek Singh
Level pending
Let ABC be a triangle with angleA = 90 and AB = AC. Let D and E be points on the segment BC such that BD : DE : EC = 3 : 5 : 4. What is the measure of angle DAE?
× | 66 | 210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-13 | longest | en | 0.898665 |
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