content stringlengths 86 994k | meta stringlengths 288 619 |
|---|---|
14: Heat and Heat Transfer Methods (Exercises)
Last updated
Page ID
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\
evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\
newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}
\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}
{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}
[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}
{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\
#3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp
#2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\
wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\
newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}
{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\
bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\
widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
Conceptual Questions
1. How is heat transfer related to temperature?
2. Describe a situation in which heat transfer occurs. What are the resulting forms of energy?
3. When heat transfers into a system, is the energy stored as heat? Explain briefly.
4. What three factors affect the heat transfer that is necessary to change an object’s temperature?
5. The brakes in a car increase in temperature by \(\displaystyle ΔT\) when bringing the car to rest from a speed \(\displaystyle v\). How much greater would \(\displaystyle ΔT\) be if the car
initially had twice the speed? You may assume the car to stop sufficiently fast so that no heat transfers out of the brakes.
6. Heat transfer can cause temperature and phase changes. What else can cause these changes?
7. How does the latent heat of fusion of water help slow the decrease of air temperatures, perhaps preventing temperatures from falling significantly below \(\displaystyle 0ºC\), in the vicinity of
large bodies of water?
8. What is the temperature of ice right after it is formed by freezing water?
9. If you place \(\displaystyle 0ºC\) ice into \(\displaystyle 0ºC\) water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?
10. What effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down?
11. In very humid climates where there are numerous bodies of water, such as in Florida, it is unusual for temperatures to rise above about 35ºC(95ºF). In deserts, however, temperatures can rise far
above this. Explain how the evaporation of water helps limit high temperatures in humid climates.
12. In winters, it is often warmer in San Francisco than in nearby Sacramento, 150 km inland. In summers, it is nearly always hotter in Sacramento. Explain how the bodies of water surrounding San
Francisco moderate its extreme temperatures.
13. Putting a lid on a boiling pot greatly reduces the heat transfer necessary to keep it boiling. Explain why.
14. Freeze-dried foods have been dehydrated in a vacuum. During the process, the food freezes and must be heated to facilitate dehydration. Explain both how the vacuum speeds up dehydration and why
the food freezes as a result.
15. When still air cools by radiating at night, it is unusual for temperatures to fall below the dew point. Explain why.
16. In a physics classroom demonstration, an instructor inflates a balloon by mouth and then cools it in liquid nitrogen. When cold, the shrunken balloon has a small amount of light blue liquid in
it, as well as some snow-like crystals. As it warms up, the liquid boils, and part of the crystals sublimate, with some crystals lingering for awhile and then producing a liquid. Identify the blue
liquid and the two solids in the cold balloon. Justify your identifications using data from Table.
17. What are the main methods of heat transfer from the hot core of Earth to its surface? From Earth’s surface to outer space?
18. Some electric stoves have a flat ceramic surface with heating elements hidden beneath. A pot placed over a heating element will be heated, while it is safe to touch the surface only a few
centimeters away. Why is ceramic, with a conductivity less than that of a metal but greater than that of a good insulator, an ideal choice for the stove top?
19. Loose-fitting white clothing covering most of the body is ideal for desert dwellers, both in the hot Sun and during cold evenings. Explain how such clothing is advantageous during both day and
A jellabiya is worn by many men in Egypt. (credit: Zerida)
20. One way to make a fireplace more energy efficient is to have an external air supply for the combustion of its fuel. Another is to have room air circulate around the outside of the fire box and
back into the room. Detail the methods of heat transfer involved in each.
21. On cold, clear nights horses will sleep under the cover of large trees. How does this help them keep warm?
22. When watching a daytime circus in a large, dark-colored tent, you sense significant heat transfer from the tent. Explain why this occurs.
23. Satellites designed to observe the radiation from cold (3 K) dark space have sensors that are shaded from the Sun, Earth, and Moon and that are cooled to very low temperatures. Why must the
sensors be at low temperature?
24. Why are cloudy nights generally warmer than clear ones?
25. Why are thermometers that are used in weather stations shielded from the sunshine? What does a thermometer measure if it is shielded from the sunshine and also if it is not?
26. On average, would Earth be warmer or cooler without the atmosphere? Explain your answer.
Problems & Exercises
27. On a hot day, the temperature of an 80,000-L swimming pool increases by \(\displaystyle 1.50ºC\). What is the net heat transfer during this heating? Ignore any complications, such as loss of
water by evaporation.
\(\displaystyle 5.02×10^8J\)
28. Show that \(\displaystyle 1cal/g⋅ºC=1kcal/kg⋅ºC\).
29. To sterilize a 50.0-g glass baby bottle, we must raise its temperature from \(\displaystyle 22.0ºC\) to \(\displaystyle 95.0ºC\). How much heat transfer is required?
\(\displaystyle 3.07×10^3J\)
30. The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the
following, originally at \(\displaystyle 20.0ºC\):
(a) water;
(b) concrete;
(c) steel; and
(d) mercury.
31. Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an
average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers.
\(\displaystyle 0.171ºC\)
32. A 0.250-kg block of a pure material is heated from \(\displaystyle 20.0ºC\) to \(\displaystyle 65.0ºC\) by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the
substance of which it is most likely composed.
33. Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water?
34. (a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in
a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a \(\displaystyle 54.9ºC\) temperature increase?
(b) Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent.
35. Following vigorous exercise, the body temperature of an 80.0-kg person is \(\displaystyle 40.0ºC\). At what rate in watts must the person transfer thermal energy to reduce the the body
temperature to \(\displaystyle 37.0ºC\) in 30.0 min, assuming the body continues to produce energy at the rate of 150 W? (1 watt = 1 joule/second or 1 W = 1 J/s).
617 W
36. Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat
transfer causes a rapid increase in temperature if the cooling system fails (1 watt = 1 joule/second or 1 W = 1 J/s and 1 MW = 1 megawatt).
(a) Calculate the rate of temperature increase in degrees Celsius per second (\(\displaystyle ºC/s\)) if the mass of the reactor core is \(\displaystyle 1.60×10^5kg\) and it has an average specific
heat of \(\displaystyle 0.3349 kJ/kgº⋅C\).
(b) How long would it take to obtain a temperature increase of \(\displaystyle 2000ºC\), which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature
increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the \(\displaystyle
5×10^5-kg\) steel containment vessel would also begin to heat up.)
37. How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package of frozen vegetables originally at \(\displaystyle 0ºC\) if their heat of fusion is the same as that of water?
35.9 kcal
38. A bag containing \(\displaystyle 0ºC\) ice is much more effective in absorbing energy than one containing the same amount of 0ºC water.
a. How much heat transfer is necessary to raise the temperature of 0.800 kg of water from \(\displaystyle 0ºC\) to \(\displaystyle 30.0ºC\)?
b. How much heat transfer is required to first melt 0.800 kg of \(\displaystyle 0ºC\) ice and then raise its temperature?
c. Explain how your answer supports the contention that the ice is more effective.
39. (a) How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from \(\displaystyle 30.0ºC\) to the boiling point and then boil away 0.750
kg of water?
(b) How long does this take if the rate of heat transfer is 500 W 1 watt = 1 joule/second (1 W = 1 J/s)?
(a) 591 kcal
(b) \(\displaystyle 4.94×10^3s\)
40. The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.00 g of condensation forms on a glass containing both water and 200 g of ice, how
many grams of the ice will melt as a result? Assume no other heat transfer occurs.
41. On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at \(\displaystyle 0ºC\) and
completely melts to \(\displaystyle 0ºC\) water in exactly one day 1 watt = 1 joule/second (1 W = 1 J/s)?
13.5 W
42. On a certain dry sunny day, a swimming pool’s temperature would rise by \(\displaystyle 1.50ºC\) if not for evaporation. What fraction of the water must evaporate to carry away precisely enough
energy to keep the temperature constant?
43. (a) How much heat transfer is necessary to raise the temperature of a 0.200-kg piece of ice from \(\displaystyle −20.0ºC\) to \(\displaystyle 130ºC\), including the energy needed for phase
(b) How much time is required for each stage, assuming a constant 20.0 kJ/s rate of heat transfer?
(c) Make a graph of temperature versus time for this process.
(a) 148 kcal
(b) 0.418 s, 3.34 s, 4.19 s, 22.6 s, 0.456 s
44. In 1986, a gargantuan iceberg broke away from the Ross Ice Shelf in Antarctica. It was approximately a rectangle 160 km long, 40.0 km wide, and 250 m thick.
(a) What is the mass of this iceberg, given that the density of ice is \(\displaystyle 917 kg/m^3\)?
(b) How much heat transfer (in joules) is needed to melt it?
(c) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of \(\displaystyle 100 W/m^2\), 12.00 h per day?
45. How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee from \(\displaystyle 95.0ºC\) to \(\displaystyle 45.0ºC\)? You may assume the coffee has the
same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 cal/g). (You may neglect the change in mass of the coffee as it cools, which will give you an answer that
is slightly larger than correct.)
33.0 g
46. (a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases \(\displaystyle 2.80×10^7J\) of energy when burned. To illustrate this difficulty,
calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water has its temperature raised from \(\displaystyle 20.0ºC\) to \
(\displaystyle 100ºC\), it boils, and the resulting steam is raised to \(\displaystyle 300ºC\).
(b) Discuss additional complications caused by the fact that crude oil has a smaller density than water.
(a) 9.67 L
(b) Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less
efficient in absorbing the heat generated by the oil.
47. The energy released from condensation in thunderstorms can be very large. Calculate the energy released into the atmosphere for a small storm of radius 1 km, assuming that 1.0 cm of rain is
precipitated uniformly over this area.
48. To help prevent frost damage, 4.00 kg of \(\displaystyle 0ºC\) water is sprayed onto a fruit tree.
(a) How much heat transfer occurs as the water freezes?
(b) How much would the temperature of the 200-kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be \(\displaystyle 3.35 kJ/kg⋅ºC\), and assume that no phase
change occurs.
a) 319 kcal
b) \(\displaystyle 2.00ºC\)
49. A 0.250-kg aluminum bowl holding 0.800 kg of soup at \(\displaystyle 25.0ºC\) is placed in a freezer. What is the final temperature if 377 kJ of energy is transferred from the bowl and soup,
assuming the soup’s thermal properties are the same as that of water? Explicitly show how you follow the steps in Problem-Solving Strategies for the Effects of Heat Transfer.
50. A 0.0500-kg ice cube at \(\displaystyle −30.0ºC\) is placed in 0.400 kg of \(\displaystyle 35.0ºC\) water in a very well-insulated container. What is the final temperature?
\(\displaystyle 20.6ºC\)
51. If you pour 0.0100 kg of \(\displaystyle 20.0ºC\) water onto a 1.20-kg block of ice (which is initially at \(\displaystyle −15.0ºC\)), what is the final temperature? You may assume that the water
cools so rapidly that effects of the surroundings are negligible.
52. Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of \(\displaystyle 500ºC\) rock must be placed in 4.00 kg of \(\
displaystyle 15.0ºC\) water to bring its temperature to \(\displaystyle 100ºC\), if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings and
take the average specific heat of the rocks to be that of granite.
4.38 kg
53. What would be the final temperature of the pan and water in Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan if 0.260 kg of water was
placed in the pan and 0.0100 kg of the water evaporated immediately, leaving the remainder to come to a common temperature with the pan?
54. In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of \(\displaystyle
808 kg/m^3\).
(a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to \(\displaystyle 3.00ºC\). (Use \(\displaystyle c_p\) and assume it is constant over
the temperature range.) This value is the amount of cooling the liquid nitrogen supplies.
(b) What is this heat transfer rate in kilowatt-hours?
(c) Compare the amount of cooling obtained from melting an identical mass of 0ºC ice with that from evaporating the liquid nitrogen.
(a) \(\displaystyle 1.57×10^4kcal\)
(b) \(\displaystyle 18.3 kW⋅h\)
(c) \(\displaystyle 1.29×10^4kcal\)
55. Some gun fanciers make their own bullets, which involves melting and casting the lead slugs. How much heat transfer is needed to raise the temperature and melt 0.500 kg of lead, starting from
56. (a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors.
The surface area of the walls is \(\displaystyle 120m^2\) and their inside surface is at \(\displaystyle 18.0ºC\), while their outside surface is at \(\displaystyle 5.00ºC\).
(b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction?
(a) \(\displaystyle 1.01×10^3\)W
(b) One
57. The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of
conduction in watts through a \(\displaystyle 3.00-m^2\) window that is \(\displaystyle 0.635 cm\) thick (1/4 in) if the temperatures of the inner and outer surfaces are \(\displaystyle 5.00ºC\) and
\(\displaystyle −10.0ºC\), respectively. This rapid rate will not be maintained—the inner surface will cool, and even result in frost formation.
58. Calculate the rate of heat conduction out of the human body, assuming that the core internal temperature is \(\displaystyle 37.0ºC\), the skin temperature is \(\displaystyle 34.0ºC\), the
thickness of the tissues between averages \(\displaystyle 1.00 cm\), and the surface area is \(\displaystyle 1.40m^2\).
84.0 W
59. Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet, making contact over an area of \(\displaystyle 80.0cm^2\) with each foot. Both the ceramic and the carpet are
2.00 cm thick and are \(\displaystyle 10.0ºC\) on their bottom sides. At what rate must heat transfer occur from each foot to keep the top of the ceramic and carpet at \(\displaystyle 33.0ºC\)?
60. A man consumes 3000 kcal of food in one day, converting most of it to maintain body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many
kilograms of water evaporate?
2.59 kg
61. (a) A firewalker runs across a bed of hot coals without sustaining burns. Calculate the heat transferred by conduction into the sole of one foot of a firewalker given that the bottom of the foot
is a 3.00-mm-thick callus with a conductivity at the low end of the range for wood and its density is \(\displaystyle 300 kg/m^3\). The area of contact is \(\displaystyle 25.0 cm^2\), the temperature
of the coals is \(\displaystyle 700ºC\), and the time in contact is 1.00 s.
(b) What temperature increase is produced in the \(\displaystyle 25.0 cm^3\) of tissue affected?
(c) What effect do you think this will have on the tissue, keeping in mind that a callus is made of dead cells?
62. (a) What is the rate of heat conduction through the 3.00-cm-thick fur of a large animal having a \(\displaystyle 1.40-m^2\) surface area? Assume that the animal’s skin temperature is \(\
displaystyle 32.0ºC\), that the air temperature is \(\displaystyle −5.00ºC\), and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace
this heat transfer?
(a) 39.7 W
(b) 820 kcal
63. A walrus transfers energy by conduction through its blubber at the rate of 150 W when immersed in \(\displaystyle −1.00ºC\) water. The walrus’s internal core temperature is \(\displaystyle 37.0ºC
\), and it has a surface area of \(\displaystyle 2.00m^2\). What is the average thickness of its blubber, which has the conductivity of fatty tissues without blood?
Walrus on ice. (credit: Captain Budd Christman, NOAA Corps)
64. Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of \(\displaystyle 10.0 m^2\) and a thermal conductivity twice that of glass wool with the rate of heat
conduction through a window that is 0.750 cm thick and that has an area of \(\displaystyle 2.00 m^2\), assuming the same temperature difference across each.
35 to 1, window to wall
65. Suppose a person is covered head to foot by wool clothing with average thickness of 2.00 cm and is transferring energy by conduction through the clothing at the rate of 50.0 W. What is the
temperature difference across the clothing, given the surface area is \(\displaystyle 1.40 m^2\)?
66. Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick and heat conduction occurs through the same area and at the same rate as computed in Example, what is the
temperature difference across it? Ceramic has the same thermal conductivity as glass and brick.
\(\displaystyle 1.05×10^3K\)
67. One easy way to reduce heating (and cooling) costs is to add extra insulation in the attic of a house. Suppose the house already had 15 cm of fiberglass insulation in the attic and in all the
exterior surfaces. If you added an extra 8.0 cm of fiberglass to the attic, then by what percentage would the heating cost of the house drop? Take the single story house to be of dimensions 10 m by
15 m by 3.0 m. Ignore air infiltration and heat loss through windows and doors.
68. (a) Calculate the rate of heat conduction through a double-paned window that has a \(\displaystyle 1.50-m^2\) area and is made of two panes of 0.800-cm-thick glass separated by a 1.00-cm air gap.
The inside surface temperature is \(\displaystyle 15.0ºC\), while that on the outside is \(\displaystyle −10.0ºC\). (Hint: There are identical temperature drops across the two glass panes. First find
these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.)
(b) Calculate the rate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a).
(a) 83 W
(b) 24 times that of a double pane window.
69. Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or
philosophy one has. (For some industries, a payback period is as small as two years.) Suppose you wish to install the extra insulation in Exercise. If energy cost $1.00 per million joules and the
insulation was $4.00 per square meter, then calculate the simple payback time. Take the average \(\displaystyle ΔT\) for the 120 day heating season to be \(\displaystyle 15.0ºC\).
70. For the human body, what is the rate of heat transfer by conduction through the body’s tissue with the following conditions: the tissue thickness is 3.00 cm, the change in temperature is \(\
displaystyle 2.00ºC\), and the skin area is \(\displaystyle 1.50 m^2\). How does this compare with the average heat transfer rate to the body resulting from an energy intake of about 2400 kcal per
day? (No exercise is included.)
20.0 W, 17.2% of 2400 kcal per day
71. At what wind speed does \(\displaystyle −10ºC\) air cause the same chill factor as still air at \(\displaystyle −29ºC\)?
10 m/s
72. At what temperature does still air cause the same chill factor as \(\displaystyle −5ºC\) air moving at 15 m/s?
73. The “steam” above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature of 250 g of hot coffee
initially at \(\displaystyle 90.0ºC\) if 2.00 g evaporates from it? The coffee is in a Styrofoam cup, so other methods of heat transfer can be neglected.
\(\displaystyle 85.7ºC\)
74. (a) How many kilograms of water must evaporate from a 60.0-kg woman to lower her body temperature by \(\displaystyle 0.750ºC\)?
(b) Is this a reasonable amount of water to evaporate in the form of perspiration, assuming the relative humidity of the surrounding air is low?
75. On a hot dry day, evaporation from a lake has just enough heat transfer to balance the \(\displaystyle 1.00 kW/m^2\) of incoming heat from the Sun. What mass of water evaporates in 1.00 h from
each square meter? Explicitly show how you follow the steps in the Problem-Solving Strategies for the Effects of Heat Transfer.
1.48 kg
76. One winter day, the climate control system of a large university classroom building malfunctions. As a result, \(\displaystyle 500 m^3\) of excess cold air is brought in each minute. At what rate
in kilowatts must heat transfer occur to warm this air by \(\displaystyle 10.0ºC\) (that is, to bring the air to room temperature)?
77. The Kilauea volcano in Hawaii is the world’s most active, disgorging about \(\displaystyle 5×10^5m^3\) of \(\displaystyle 1200ºC\) lava per day. What is the rate of heat transfer out of Earth by
convection if this lava has a density of \(\displaystyle 2700kg/m^3\) and eventually cools to \(\displaystyle 30ºC\)? Assume that the specific heat of lava is the same as that of granite.
Lava flow on Kilauea volcano in Hawaii. (credit: J. P. Eaton, U.S. Geological Survey)
\(\displaystyle 2×10^4 MW\)
78. During heavy exercise, the body pumps 2.00 L of blood per minute to the surface, where it is cooled by \(\displaystyle 2.00ºC\). What is the rate of heat transfer from this forced convection
alone, assuming blood has the same specific heat as water and its density is \(\displaystyle 1050 kg/m^3\)?
79. A person inhales and exhales 2.00 L of \(\displaystyle 37.0ºC\) air, evaporating \(\displaystyle 4.00×10^{−2}g\) of water from the lungs and breathing passages with each breath.
(a) How much heat transfer occurs due to evaporation in each breath?
(b) What is the rate of heat transfer in watts if the person is breathing at a moderate rate of 18.0 breaths per minute?
(c) If the inhaled air had a temperature of \(\displaystyle 20.0ºC\), what is the rate of heat transfer for warming the air?
(d) Discuss the total rate of heat transfer as it relates to typical metabolic rates. Will this breathing be a major form of heat transfer for this person?
(a) 97.2 J
(b) 29.2 W
(c) 9.49 W
(d) The total rate of heat loss would be \(\displaystyle 29.2 W+9.49 W=38.7W\). While sleeping, our body consumes 83 W of power, while sitting it consumes 120 to 210 W. Therefore, the total rate of
heat loss from breathing will not be a major form of heat loss for this person.
80. A glass coffee pot has a circular bottom with a 9.00-cm diameter in contact with a heating element that keeps the coffee warm with a continuous heat transfer rate of 50.0 W
(a) What is the temperature of the bottom of the pot, if it is 3.00 mm thick and the inside temperature is \(\displaystyle 60.0ºC\)?
(b) If the temperature of the coffee remains constant and all of the heat transfer is removed by evaporation, how many grams per minute evaporate? Take the heat of vaporization to be 2340 kJ/kg.
81. At what net rate does heat radiate from a \(\displaystyle 275-m^2\) black roof on a night when the roof’s temperature is \(\displaystyle 30.0ºC\) and the surrounding temperature is \(\
displaystyle 15.0ºC\)? The emissivity of the roof is 0.900.
\(\displaystyle −21.7 kW\)
Note that the negative answer implies heat loss to the surroundings.
82. (a) Cherry-red embers in a fireplace are at \(\displaystyle 850ºC\) and have an exposed area of \(\displaystyle 0.200 m^2\) and an emissivity of 0.980. The surrounding room has a temperature of \
(\displaystyle 18.0ºC\). If 50% of the radiant energy enters the room, what is the net rate of radiant heat transfer in kilowatts?
(b) Does your answer support the contention that most of the heat transfer into a room by a fireplace comes from infrared radiation?
83. Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation from \(\displaystyle 1.00 m^2\) of \(\displaystyle 1200ºC\) fresh lava into \(\
displaystyle 30.0ºC\) surroundings, assuming lava’s emissivity is 1.00.
\(\displaystyle −266 kW\)
84. (a) Calculate the rate of heat transfer by radiation from a car radiator at \(\displaystyle 110°C\) into a \(\displaystyle 50.0ºC\) environment, if the radiator has an emissivity of 0.750 and a \
(\displaystyle 1.20-m^2\) surface area.
(b) Is this a significant fraction of the heat transfer by an automobile engine? To answer this, assume a horsepower of \(\displaystyle 200hp(1.5kW)\) and the efficiency of automobile engines as 25%.
85. Find the net rate of heat transfer by radiation from a skier standing in the shade, given the following. She is completely clothed in white (head to foot, including a ski mask), the clothes have
an emissivity of 0.200 and a surface temperature of \(\displaystyle 10.0ºC\), the surroundings are at \(\displaystyle −15.0ºC\), and her surface area is \(\displaystyle 1.60m^2\).
\(\displaystyle −36.0 W\)
86. Suppose you walk into a sauna that has an ambient temperature of \(\displaystyle 50.0ºC\).
(a) Calculate the rate of heat transfer to you by radiation given your skin temperature is \(\displaystyle 37.0ºC\), the emissivity of skin is 0.98, and the surface area of your body is \(\
displaystyle 1.50m^2\).
(b) If all other forms of heat transfer are balanced (the net heat transfer is zero), at what rate will your body temperature increase if your mass is 75.0 kg?
87. Thermography is a technique for measuring radiant heat and detecting variations in surface temperatures that may be medically, environmentally, or militarily meaningful.
(a) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of \(\displaystyle 34.0ºC\) compared with that at \(\displaystyle 33.0ºC\), such as on a
person’s skin?
(b) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of \(\displaystyle 34.0ºC\) compared with that at \(\displaystyle 20.0ºC\), such as for
warm and cool automobile hoods?
Artist’s rendition of a thermograph of a patient’s upper body, showing the distribution of heat represented by different colors.
(a) 1.31%
(b) 20.5%
88. The Sun radiates like a perfect black body with an emissivity of exactly 1.
(a) Calculate the surface temperature of the Sun, given that it is a sphere with a \(\displaystyle 7.00×10^8-m\) radius that radiates \(\displaystyle 3.80×10^{26} W\) into 3-K space.
(b) How much power does the Sun radiate per square meter of its surface?
(c) How much power in watts per square meter is that value at the distance of Earth, \(\displaystyle 1.50×10^{11} m\) away? (This number is called the solar constant.)
89. A large body of lava from a volcano has stopped flowing and is slowly cooling. The interior of the lava is at \(\displaystyle 1200ºC\), its surface is at \(\displaystyle 450ºC\), and the
surroundings are at \(\displaystyle 27.0ºC\)
(a) Calculate the rate at which energy is transferred by radiation from \(\displaystyle 1.00 m^2\) of surface lava into the surroundings, assuming the emissivity is 1.00.
(b) Suppose heat conduction to the surface occurs at the same rate. What is the thickness of the lava between the \(\displaystyle 450ºC\) surface and the \(\displaystyle 1200ºC\) interior, assuming
that the lava’s conductivity is the same as that of brick?
(a) \(\displaystyle −15.0 kW\)
(b) 4.2 cm
90. Calculate the temperature the entire sky would have to be in order to transfer energy by radiation at \(\displaystyle 1000W/m^2\)—about the rate at which the Sun radiates when it is directly
overhead on a clear day. This value is the effective temperature of the sky, a kind of average that takes account of the fact that the Sun occupies only a small part of the sky but is much hotter
than the rest. Assume that the body receiving the energy has a temperature of \(\displaystyle 27.0ºC\).
91. (a) A shirtless rider under a circus tent feels the heat radiating from the sunlit portion of the tent. Calculate the temperature of the tent canvas based on the following information: The
shirtless rider’s skin temperature is \(\displaystyle 34.0ºC\) and has an emissivity of 0.970. The exposed area of skin is \(\displaystyle 0.400 m^2\). He receives radiation at the rate of 20.0
W—half what you would calculate if the entire region behind him was hot. The rest of the surroundings are at \(\displaystyle 34.0ºC\).
(b) Discuss how this situation would change if the sunlit side of the tent was nearly pure white and if the rider was covered by a white tunic.
(a) \(\displaystyle 48.5ºC\)
(b) A pure white object reflects more of the radiant energy that hits it, so a white tent would prevent more of the sunlight from heating up the inside of the tent, and the white tunic would prevent
that heat which entered the tent from heating the rider. Therefore, with a white tent, the temperature would be lower than \(\displaystyle 48.5ºC\), and the rate of radiant heat transferred to the
rider would be less than 20.0 W.
92. Integrated Concepts
One \(\displaystyle 30.0ºC\) day the relative humidity is \(\displaystyle 75.0%\), and that evening the temperature drops to \(\displaystyle 20.0ºC\), well below the dew point.
(a) How many grams of water condense from each cubic meter of air?
(b) How much heat transfer occurs by this condensation?
(c) What temperature increase could this cause in dry air?
93. Integrated Concepts
Large meteors sometimes strike the Earth, converting most of their kinetic energy into thermal energy.
(a) What is the kinetic energy of a \(\displaystyle 10^9\)kg meteor moving at 25.0 km/s?
(b) If this meteor lands in a deep ocean and \(\displaystyle 80%\) of its kinetic energy goes into heating water, how many kilograms of water could it raise by \(\displaystyle 5.0ºC\)?
(c) Discuss how the energy of the meteor is more likely to be deposited in the ocean and the likely effects of that energy.
(a) \(\displaystyle 3×10^{17} J\)
(b) \(\displaystyle 1×10^{13} kg\)
(c) When a large meteor hits the ocean, it causes great tidal waves, dissipating large amount of its energy in the form of kinetic energy of the water.
94. Integrated Concepts
Frozen waste from airplane toilets has sometimes been accidentally ejected at high altitude. Ordinarily it breaks up and disperses over a large area, but sometimes it holds together and strikes the
ground. Calculate the mass of \(\displaystyle 0ºC\) ice that can be melted by the conversion of kinetic and gravitational potential energy when a \(\displaystyle 20.0\) piece of frozen waste is
released at 12.0 km altitude while moving at 250 m/s and strikes the ground at 100 m/s (since less than 20.0 kg melts, a significant mess results).
95. Integrated Concepts
(a) A large electrical power facility produces 1600 MW of “waste heat,” which is dissipated to the environment in cooling towers by warming air flowing through the towers by \(\displaystyle 5.00ºC\).
What is the necessary flow rate of air in \(\displaystyle m^3/s\)?
(b) Is your result consistent with the large cooling towers used by many large electrical power plants?
(a) \(\displaystyle 3.44×10^5 m^3/s\)
(b) This is equivalent to 12 million cubic feet of air per second. That is tremendous. This is too large to be dissipated by heating the air by only \(\displaystyle 5ºC\). Many of these cooling
towers use the circulation of cooler air over warmer water to increase the rate of evaporation. This would allow much smaller amounts of air necessary to remove such a large amount of heat because
evaporation removes larger quantities of heat than was considered in part (a).
96. Integrated Concepts
(a) Suppose you start a workout on a Stairmaster, producing power at the same rate as climbing 116 stairs per minute. Assuming your mass is 76.0 kg and your efficiency is \(\displaystyle 20.0%\), how
long will it take for your body temperature to rise \(\displaystyle 1.00ºC\) if all other forms of heat transfer in and out of your body are balanced? (b) Is this consistent with your experience in
getting warm while exercising?
97. Integrated Concepts
A 76.0-kg person suffering from hypothermia comes indoors and shivers vigorously. How long does it take the heat transfer to increase the person’s body temperature by \(\displaystyle 2.00ºC\) if all
other forms of heat transfer are balanced?
20.9 min
98. Integrated Concepts
In certain large geographic regions, the underlying rock is hot. Wells can be drilled and water circulated through the rock for heat transfer for the generation of electricity.
(a) Calculate the heat transfer that can be extracted by cooling \(\displaystyle 1.00 km^3\) of granite by \(\displaystyle 100ºC\).
(b) How long will this take if heat is transferred at a rate of 300 MW, assuming no heat transfers back into the 1.00 km of rock by its surroundings?
99. Integrated Concepts
Heat transfers from your lungs and breathing passages by evaporating water.
(a) Calculate the maximum number of grams of water that can be evaporated when you inhale 1.50 L of \(\displaystyle 37ºC\) air with an original relative humidity of 40.0%. (Assume that body
temperature is also \(\displaystyle 37ºC\).)
(b) How many joules of energy are required to evaporate this amount?
(c) What is the rate of heat transfer in watts from this method, if you breathe at a normal resting rate of 10.0 breaths per minute?
(a) \(\displaystyle 3.96×10^{-2} g\)
(b) \(\displaystyle 96.2 J\)
(c) \(\displaystyle 16.0 W\)
100. Integrated Concepts
(a) What is the temperature increase of water falling 55.0 m over Niagara Falls?
(b) What fraction must evaporate to keep the temperature constant?
101. Integrated Concepts
Hot air rises because it has expanded. It then displaces a greater volume of cold air, which increases the buoyant force on it. (a) Calculate the ratio of the buoyant force to the weight of
50.0ºC50.0ºC air surrounded by 20.0ºC20.0ºC air. (b) What energy is needed to cause 1.00m31.00 m3 of air to go from 20.0ºC20.0ºC to 50.0ºC50.0ºC? (c) What gravitational potential energy is gained by
this volume of air if it rises 1.00 m? Will this cause a significant cooling of the air?
(a) 1.102
(b) \(\displaystyle 2.79×10^4J\)
(c) 12.6 J. This will not cause a significant cooling of the air because it is much less than the energy found in part (b), which is the energy required to warm the air from 20.0ºC to 50.0ºC
102. Unreasonable Results
(a) What is the temperature increase of an 80.0 kg person who consumes 2500 kcal of food in one day with 95.0% of the energy transferred as heat to the body?
(b) What is unreasonable about this result?
(c) Which premise or assumption is responsible?
(a) 36ºC
(b) Any temperature increase greater than about \(\displaystyle 3ºC\) would be unreasonably large. In this case the final temperature of the person would rise to 73ºC(163ºF).
|(c) The assumption of \(\displaystyle 95%\) heat retention is unreasonable.
103. Unreasonable Results
A slightly deranged Arctic inventor surrounded by ice thinks it would be much less mechanically complex to cool a car engine by melting ice on it than by having a water-cooled system with a radiator,
water pump, antifreeze, and so on.
(a) If \(\displaystyle 80.0%\) of the energy in 1.00 gal of gasoline is converted into “waste heat” in a car engine, how many kilograms of \(\displaystyle 0ºC\) ice could it melt?
(b) Is this a reasonable amount of ice to carry around to cool the engine for 1.00 gal of gasoline consumption?
(c) What premises or assumptions are unreasonable?
104. Unreasonable Results
(a) Calculate the rate of heat transfer by conduction through a window with an area of \(\displaystyle 1.00 m^2\) that is 0.750 cm thick, if its inner surface is at \(\displaystyle 22.0ºC\) and its
outer surface is at \(\displaystyle 35.0ºC\).
(b) What is unreasonable about this result?
(c) Which premise or assumption is responsible?
(a) 1.46 kW
(b) Very high power loss through a window. An electric heater of this power can keep an entire room warm.
(c) The surface temperatures of the window do not differ by as great an amount as assumed. The inner surface will be warmer, and the outer surface will be cooler.
105. Unreasonable Results
A meteorite 1.20 cm in diameter is so hot immediately after penetrating the atmosphere that it radiates 20.0 kW of power.
(a) What is its temperature, if the surroundings are at \(\displaystyle 20.0ºC\) and it has an emissivity of 0.800?
(b) What is unreasonable about this result?
(c) Which premise or assumption is responsible?
106. Construct Your Own Problem
Consider a new model of commercial airplane having its brakes tested as a part of the initial flight permission procedure. The airplane is brought to takeoff speed and then stopped with the brakes
alone. Construct a problem in which you calculate the temperature increase of the brakes during this process. You may assume most of the kinetic energy of the airplane is converted to thermal energy
in the brakes and surrounding materials, and that little escapes. Note that the brakes are expected to become so hot in this procedure that they ignite and, in order to pass the test, the airplane
must be able to withstand the fire for some time without a general conflagration.
107. Construct Your Own Problem
Consider a person outdoors on a cold night. Construct a problem in which you calculate the rate of heat transfer from the person by all three heat transfer methods. Make the initial circumstances
such that at rest the person will have a net heat transfer and then decide how much physical activity of a chosen type is necessary to balance the rate of heat transfer. Among the things to consider
are the size of the person, type of clothing, initial metabolic rate, sky conditions, amount of water evaporated, and volume of air breathed. Of course, there are many other factors to consider and
your instructor may wish to guide you in the assumptions made as well as the detail of analysis and method of presenting your results.
Contributors and Attributions
• Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks
(University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). | {"url":"https://phys.libretexts.org/Bookshelves/University_Physics/Exercises_(University_Physics)/Exercises%3A_College_Physics_(OpenStax)/14%3A_Heat_and_Heat_Transfer_Methods_(Exercises)","timestamp":"2024-11-03T19:33:18Z","content_type":"text/html","content_length":"183758","record_id":"<urn:uuid:843542b0-d6b7-45e3-a893-ed80b03b6ac7>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00675.warc.gz"} |
Determination of shortest path using Dijkstra’s algorithm & Kruskal’s algorithm | Mathematics AA HL's Sample Internal Assessment | Nail IB®
Research question
What should be the shortest path for a delivery executive connecting all the delivery locations in the same city?
In this IA we have found that the shortest path that can be taken by a delivery executive among multiple delivery routes, can be calculated using Dijkstra’s Algorithm. To do a comparative analysis,
we have also find the route between the same source and destination using Kruskal’s Algorithm also. We have found that the routes can be mapped in terms of a weighted graph made up of nodes. Firstly,
using Dijkstra’s Algorithm, we have found the distance between Bandra West (Source) and Colaba (destination) is 23.7 Km. This deduction will make the movement of both delivery executives and the
general people, more economical and easier.
Moreover, we have found the route between the same source and destination, i.e., from Bandra West to Colaba using Kruskal’s Algorithm. Using this method, we have concluded that, the distance between
the source and the destination has come out to be 26.2 Km.
From the above study, we can say that, Dijkstra’s Algorithm is more efficient in finding the shortest path between any two nodes in a weighted graph. On the other hand, Kruskal’s Algorithm is
efficient in finding a route that will connect all the nodes present in the graph but the distance between source and the destination might not be the shortest.
In a nutshell, Dijkstra’s Algorithm is more efficient in finding the shortest path between any two routes thus providing the delivery executing with a route that will require minimum amount of fuel
to be covered decreasing his cost on fuel. | {"url":"https://nailib.com/user/ib-resources/ib-math-aa-hl/ia-sample/649d433ef29849909f556879","timestamp":"2024-11-04T04:33:05Z","content_type":"text/html","content_length":"100731","record_id":"<urn:uuid:dc0f0e22-5f92-4f8a-ba99-226fa2af8167>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00668.warc.gz"} |
Multiplication Of Integers Worksheet Pdf
Math, especially multiplication, creates the cornerstone of countless academic self-controls and real-world applications. Yet, for many learners, grasping multiplication can posture an obstacle. To
resolve this difficulty, teachers and parents have accepted an effective device: Multiplication Of Integers Worksheet Pdf.
Intro to Multiplication Of Integers Worksheet Pdf
Multiplication Of Integers Worksheet Pdf
Multiplication Of Integers Worksheet Pdf -
When multiplying integers using manipulatives the first integer is used to determine if groups of integers are added or removed positive first integer means ADD groups of integers negative first
integer means REMOVE groups of integers The second integer is the number of objects in each group
This page includes Integers worksheets for comparing and ordering integers adding subtracting multiplying and dividing integers and order of operations with integers If you ve ever spent time in
Canada in January you ve most likely experienced a negative integer first hand
Significance of Multiplication Technique Understanding multiplication is critical, laying a solid structure for advanced mathematical ideas. Multiplication Of Integers Worksheet Pdf use structured
and targeted technique, fostering a deeper understanding of this essential math operation.
Evolution of Multiplication Of Integers Worksheet Pdf
FREE 9 Sample Multiplying Integers Horizontal Worksheet Templates In PDF
FREE 9 Sample Multiplying Integers Horizontal Worksheet Templates In PDF
Learn to Solve Grab the Worksheet Multiplying 3 Integers This resource has grade 7 and grade 8 learners recalling the rules of multiplying integers finding the product of the first two integers and
multiplying it with the third while keeping a track of the signs
Multiplying Integers Find the product 2 11 3 4 6 5 15 1 7 13 8 9 7 14 11 5 8 13 3 9 15 12 10 22 24
From conventional pen-and-paper workouts to digitized interactive formats, Multiplication Of Integers Worksheet Pdf have actually evolved, satisfying varied discovering styles and preferences.
Kinds Of Multiplication Of Integers Worksheet Pdf
Fundamental Multiplication Sheets Easy workouts focusing on multiplication tables, helping students build a strong arithmetic base.
Word Problem Worksheets
Real-life scenarios integrated right into issues, improving crucial thinking and application abilities.
Timed Multiplication Drills Examinations developed to enhance rate and accuracy, assisting in fast mental mathematics.
Advantages of Using Multiplication Of Integers Worksheet Pdf
Multiplying And Dividing Integers Worksheets
Multiplying And Dividing Integers Worksheets
Integer worksheets contain a huge collection of practice pages based on the concepts of addition subtraction multiplication and division Exclusive pages to compare and order integers and representing
integers on a number line are given here with a variety of activities and exercises
Microsoft Word 2 Multiply and Divide Integers doc Author myau Created Date 7 28 2009 12 43 37 PM
Boosted Mathematical Abilities
Consistent technique sharpens multiplication efficiency, enhancing total math capabilities.
Boosted Problem-Solving Abilities
Word problems in worksheets develop analytical reasoning and technique application.
Self-Paced Discovering Advantages
Worksheets fit individual discovering speeds, fostering a comfortable and versatile learning setting.
How to Create Engaging Multiplication Of Integers Worksheet Pdf
Incorporating Visuals and Colors Lively visuals and colors capture focus, making worksheets visually appealing and involving.
Including Real-Life Situations
Associating multiplication to everyday situations adds significance and practicality to exercises.
Tailoring Worksheets to Different Skill Levels Customizing worksheets based upon differing proficiency levels ensures inclusive knowing. Interactive and Online Multiplication Resources Digital
Multiplication Equipment and Games Technology-based sources use interactive discovering experiences, making multiplication engaging and pleasurable. Interactive Internet Sites and Applications
On-line platforms give varied and accessible multiplication technique, supplementing traditional worksheets. Personalizing Worksheets for Different Knowing Styles Aesthetic Learners Aesthetic help
and diagrams aid comprehension for learners inclined toward visual discovering. Auditory Learners Spoken multiplication issues or mnemonics accommodate students who comprehend principles via auditory
methods. Kinesthetic Learners Hands-on activities and manipulatives sustain kinesthetic learners in recognizing multiplication. Tips for Effective Implementation in Knowing Consistency in Practice
Regular practice enhances multiplication skills, advertising retention and fluency. Stabilizing Repetition and Variety A mix of recurring exercises and diverse issue formats keeps passion and
understanding. Giving Constructive Feedback Comments aids in identifying areas of improvement, encouraging continued progress. Obstacles in Multiplication Method and Solutions Motivation and
Involvement Difficulties Dull drills can bring about uninterest; cutting-edge strategies can reignite motivation. Getting Rid Of Fear of Mathematics Adverse understandings around mathematics can
prevent progression; developing a favorable knowing environment is essential. Influence of Multiplication Of Integers Worksheet Pdf on Academic Performance Research Studies and Research Findings
Study suggests a positive correlation between consistent worksheet use and boosted math performance.
Multiplication Of Integers Worksheet Pdf emerge as functional tools, cultivating mathematical proficiency in students while accommodating varied learning styles. From fundamental drills to
interactive online resources, these worksheets not just improve multiplication abilities however also advertise crucial thinking and analytic capabilities.
Grade 7 Integers Worksheet Worksheet Resume Examples
50 Multiplication Of Integers Worksheet
Check more of Multiplication Of Integers Worksheet Pdf below
The Multiplying Integers Mixed Signs Range 12 To 12 A Math Worksheet From The Integers
Multiplication Of Integers Worksheet
50 Multiplication Of Integers Worksheet
Grade 7 Multiplication Of Integers Worksheet Worksheet Resume Examples
Dividing Integers Worksheet Homeschooldressage
30 Multiply And Divide Integers Worksheet Education Template
Integers Worksheets Math Drills
This page includes Integers worksheets for comparing and ordering integers adding subtracting multiplying and dividing integers and order of operations with integers If you ve ever spent time in
Canada in January you ve most likely experienced a negative integer first hand
Multiplying and Dividing Integers Worksheets Math Worksheets 4 Kids
Get students to multiply the positive and negative numbers in each row and column and fill in the empty boxes in each 3x3 square Multiplying 3 or 4 Integers Find the product of the integers Apply the
multiplication sign rule Each worksheet consists of ten problems Dividing Integers Integer Division
This page includes Integers worksheets for comparing and ordering integers adding subtracting multiplying and dividing integers and order of operations with integers If you ve ever spent time in
Canada in January you ve most likely experienced a negative integer first hand
Get students to multiply the positive and negative numbers in each row and column and fill in the empty boxes in each 3x3 square Multiplying 3 or 4 Integers Find the product of the integers Apply the
multiplication sign rule Each worksheet consists of ten problems Dividing Integers Integer Division
Grade 7 Multiplication Of Integers Worksheet Worksheet Resume Examples
Multiplication Of Integers Worksheet
Dividing Integers Worksheet Homeschooldressage
30 Multiply And Divide Integers Worksheet Education Template
Zach s Blog Adding And Subtracting Integers
Adding And Subtracting Integers Worksheet 7th Grade With Ans
Adding And Subtracting Integers Worksheet 7th Grade With Ans
50 Multiplication Of Integers Worksheet Chessmuseum Template Library In 2021 Integers
FAQs (Frequently Asked Questions).
Are Multiplication Of Integers Worksheet Pdf appropriate for any age groups?
Yes, worksheets can be customized to different age and skill levels, making them adaptable for different students.
Exactly how commonly should trainees exercise utilizing Multiplication Of Integers Worksheet Pdf?
Regular technique is key. Regular sessions, preferably a few times a week, can produce significant enhancement.
Can worksheets alone boost math abilities?
Worksheets are a beneficial device but needs to be supplemented with diverse understanding approaches for thorough skill advancement.
Exist on the internet platforms offering free Multiplication Of Integers Worksheet Pdf?
Yes, numerous instructional websites provide free access to a variety of Multiplication Of Integers Worksheet Pdf.
Exactly how can parents sustain their kids's multiplication method in your home?
Encouraging constant technique, providing support, and creating a positive understanding environment are useful actions. | {"url":"https://crown-darts.com/en/multiplication-of-integers-worksheet-pdf.html","timestamp":"2024-11-12T22:24:05Z","content_type":"text/html","content_length":"28324","record_id":"<urn:uuid:df6be071-cfcc-4989-8cbb-d59657391e01>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00751.warc.gz"} |
[Solved] The slope of the tangent to the curve x=t2+3t−8,y=2t2−... | Filo
The slope of the tangent to the curve at the point is:
Not the question you're searching for?
+ Ask your question
Equation of the curves are
Slope of the tangent to the given curve at point is
At the given point and
At , from equation
At , from equation (2),
Here, common value of in the two sets of values is 2 .
Again, from equation (3),
Slope of the tangent to the given curve at point
Was this solution helpful?
Video solutions (9)
Learn from their 1-to-1 discussion with Filo tutors.
3 mins
Uploaded on: 4/29/2023
Was this solution helpful?
9 mins
Uploaded on: 1/10/2023
Was this solution helpful?
Found 6 tutors discussing this question
Discuss this question LIVE for FREE
14 mins ago
One destination to cover all your homework and assignment needs
Learn Practice Revision Succeed
Instant 1:1 help, 24x7
60, 000+ Expert tutors
Textbook solutions
Big idea maths, McGraw-Hill Education etc
Essay review
Get expert feedback on your essay
Schedule classes
High dosage tutoring from Dedicated 3 experts
Practice more questions from Mathematics Part-I (NCERT)
Practice questions from Mathematics Part-I (NCERT)
View more
Practice more questions from Application of Derivatives
Practice questions on similar concepts asked by Filo students
View more
Stuck on the question or explanation?
Connect with our Mathematics tutors online and get step by step solution of this question.
231 students are taking LIVE classes
Question Text The slope of the tangent to the curve at the point is:
Updated On Apr 29, 2023
Topic Application of Derivatives
Subject Mathematics
Class Class 12
Answer Type Text solution:1 Video solution: 9
Upvotes 1117
Avg. Video Duration 8 min | {"url":"https://askfilo.com/math-question-answers/the-slope-of-the-tangent-to-the-curve-boldsymbol-x3t7","timestamp":"2024-11-06T17:19:49Z","content_type":"text/html","content_length":"569276","record_id":"<urn:uuid:fbcf6524-2b74-4f4e-9dd8-5674255270b0>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00153.warc.gz"} |
3rd grade math games for kids
Add Two Numbers Up to Three Digits Game
3rd grade add two numbers up to three digits coffee quiz game for kids.
Addition 3 to 1 Digit Numbers Game
3rd grade addition 3 to 1 digit numbers millionaire game for kids.
Addition Balancing Equations Game
3rd grade addition balancing equations game for kids.
Addition of 2 3digit Numbers Game
3rd grade addition of 2 3digit numbers spin the wheel game for kids.
Addition of 2 Digit Numbers Game
3rd grade addition of 2 digit numbers millionaire game for kids.
Addition of Three Numbers Game
3rd grade addition of three numbers game for kids.
Addition Horizontally Arranged Numbers Game
3rd grade addition horizontally arranged numbers millionaire game for kids.
Addition of 2, 3 Digit Numbers Game
3rd grade addition of 2 3 digit numbers spin the wheel game for kids.
Balance Mixed Operations Game
3rd grade balance mixed operations game for kids.
Basic Division of Numbers Game
3rd grade basic division of numbers monster board game for kids.
Division of Numbers Game
3rd grade division of numbers monster board game for kids.
Division of Small Numbers Game
3rd grade division of small numbers monster board game for kids.
Division Game
3rd grade division game for kids.
Long Division of Numbers Game
3rd grade long division of numbers spin the wheel game for kids.
Multiplication By Ten Game
3rd grade multiplication by ten game for kids.
Multiplication of 1 and 2 Digit Numbers Game
3rd grade multiplication of 1 and 2 digit numbers scientist game for kids.
Multiplication of 1 and 2 Digit Numbers 2 Game
3rd grade multiplication of 1 and 2 digit numbers 2 scientist game for kids.
Multiplication of Numbers Game
3rd grade multiplication of numbers millionaire game for kids.
Multiplication Game
3rd grade multiplication game for kids.
Multiply a One Digit Number By a Larger Number Game
3rd grade multiply a one digit number by a larger number scientist quiz game for kids.
Place Value Game
3rd grade place value game for kids.
Roman Numerals Game
3rd grade roman numerals game for kids.
Rounding Up Numbers to Nearest Ten Game
3rd grade rounding up numbers to nearest ten game for kids.
Subtraction 2 From 3 Digit Numbers Game
3rd grade subtraction 2 from 3 digit numbers spin the wheel game for kids.
Subtraction 4 Digit Numbers Game
3rd grade subtraction 4 digit numbers monster board game for kids.
Subtraction 4 From 4 Digit Numbers Game
3rd grade subtraction 4 from 4 digit numbers millionaire game for kids.
Subtraction and Balancing Equations Game
3rd grade subtraction and balancing equations game for kids.
Subtraction Find Missing Numbers Game
3rd grade subtraction find missing numbers monster board game for kids.
3rd grade math games (online games) - When your child enters the third grade they are expected to have mastered the most basic math skills; as they will be introduced to the some of the most
frustrating concepts in maths. These concepts will include multiplication, and division, and there is no doubt that at such an age it is a difficult concept to grasp. We are confident that with the
help of our 3rd grade math online games, your child will be able to learn these concepts and master them over time with constant practice. As we know the best way to perfect any skill is by
practicing them over and over again, and our games will give your child the opportunity to help master 3rd grade math skills and techniques. We offer a variety of different games, and with their
help, students will have a variety of different games to help them practice these new math skills they are learning, and once they start practicing them regularly, they will develop a comprehensive
understanding of them in no time. Online math games are a great way to offer third graders different ways to learn and practice their math skills. With our fun online games, that focus on
multiplication, division, solving equations; with the help of online puzzle games, racing, drilling, etc. These games will vary each other in nature, but they will all have one thing in common; which
is to help students develop, practice, and perfect math skills taught in the 3rd grade math class. These games are a perfect way for students to learn in various concepts, as they will not even take
this experience as a leaning one; as they will be having so much fun. With our easy going and learning is fun concept in mind, we have hand picked games that let us stay true to our cause.
Our collection of mathematics games for grade three has been extended. Featuring are: rounding games 3rd grade, division games 3rd grade, rounding games 3rd grade, division games 3rd grade, 3rd grade
logic puzzles, subtraction games for 3rd grade, math challenge problems 3rd grade, math multiplication games 3rd grade etc. Other game types have been included in the list as follows: croc games
online, pirate kings, car racing games online, word cross games, fishing games online, wheel of fortune free game, free online basketball games, zombie lane game, zombie shooting games, dinosaur
games online and more. | {"url":"https://www.futuristicmath.com/3rd-grade-math-games.html","timestamp":"2024-11-09T22:33:27Z","content_type":"text/html","content_length":"76141","record_id":"<urn:uuid:7d473dd5-d70a-42f9-9593-e764ca707a8b>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00760.warc.gz"} |
Misspecification o
Effect of Q-matrix Misspecification on Variational Autoencoders (VAE) for Multidimensional Item Response Theory (MIRT) Models Estimation
Deep generative models with a specific variational autoencoding structure are capable of estimating parameters for the multidimensional logistic 2-parameter (ML2P) model in item response theory. In
this work, we incorporated Q-matrix and variational autoencoder (VAE) to estimate item parameters with correlated and independent latent abilities, and we validate Q-matrix via the root mean square
error (RMSE), bias, correlation, and AIC and BIC test score. The incorporation of a non-identity covariance matrix in a VAE requires a novel VAE architecture, which can be utilized in applications
outside of education such as players performance evaluation, clinical trials assessment. Moreover, results show that the ML2P-VAE method is capable of estimating parameters and validating Q-matrix
for models with a large number of latent variables with low computational cost, whereas traditional methods are infeasible for data with high-dimensional latent traits.
Item Response Theory, Deep Generative Model, Interpretable Neural Network, Cognitive Diagnostic Model, Educational Assessment
Item Response Theory (IRT) is a popular model for the understanding of human learning and problem-solving skill and to predict human behavior and performance. Since the 1950s [21], thousands of
researchers have used IRT in fields, e.g., education, medicine, and psychology, and this includes many critical contexts such as survey analysis, popular questionnaires, medical diagnosis, and school
system assessment.
More recently, computer-assisted open-access learning has gotten more popular worldwide, e.g., Khan Academy, Coursera, and EdX have developed a new challenge to handle large-scale student and trace
performance [15].
In the deep learning domain, a revolution in deep generative models via variational autoencoders [12] [14], has demonstrated an impressive ability to perform fast inference for complex MIRT models.
In this research, we present a novel application of variational autoencoders to MIRT, explore independent and correlated latent traits in the MIRT model via simulated data, and apply them to
real-world examples. We then show the impact of Q-miss (wrong Q-matrix) when mixed compared with the original Q-matrix (Q-true).
Specifically, in this paper, we have explored two research questions as follows: first, how to use variational autoencoder in the estimation of MIRT models with large numbers of correlated and
independent latent traits? Second, how are the effects of various factors such as the percentage of misfit items in the test and item quality (e.g., discrimination) on item and model fit in case of
misspecification of Q-matrix?
Most closely related to the present work, Converse [2] utilized variational autoencoders(VAE) to estimate item parameters with correlated latent abilities and directly compared ML2P-VAE with
traditional methods. Curi [1] introduces novel variational autoencoders to estimate item parameters with independent latent traits. Guo [16] explored the neural network approach and compared the
outcome with the DINA model. Converse [3] compared outcomes between autoencoders (AE) and variational autoencoders (VAE). Wu [21] investigated the novel application of variational inference and
incorporated IRT in the model via simulated and real data. Different from Converse [2], and Curri [1], we use both independent and latent traits in the VAE model. Moreover, we have explored the
effect of Q-matrix misspecification in MIRT parameter estimation via different fit statistics, e.g., RMSE, BIAS, AIC, BIC score measures.
The Multidimensional Logistic 2-Parameter (ML2P) model gives the probability of students answering a particular question as a continuous function of student ability [14]. There are two types of
parameters associated with each item: a difficulty parameter ${b}_{i}$ for the item $i$, and a discrimination parameter ${a}_{ik}\ge 0$ for each latent trait, $k$ quantifying the hierarchy of ability
$k$ required to answer the item $i$ correctly. The ML2P model gives the probability of a student $j$ with latent abilities answering an item $i$ correctly as
$P\left({u}_{ij}=1∣{\Theta }_{j};{a}_{i},{b}_{i}\right)=\frac{1}{1+exp\left[-\sum _{k=1}^{K}{a}_{ik}{𝜃}_{jk}+{b}_{i}\right]}$(1)
The variational autoencoder (VAE) is a directed model that uses learned approximate inference and can be trained purely with gradient-based methods[12]. It is similar to auto-encoders but with a
probabilistic twist. VAE makes the additional assumption that the low-dimensional representation of data follows some probability distribution $N\left(0,I\right)$, and fits the encoded data to this
The main use of a VAE as a generative model, VAE generates $X$ to $\stackrel{^}{X}$ after training and feed-forward through the decoder. By Bayes’ rule, we can write the unknown posterior
distribution. In our case, we generalized VAE as $N\left(\mu ,\mathrm{\Sigma }\right)$. In order to keep both P and Q distribution similar, Kullback-Liebler divergence ${D}_{\text{KL}}\left(P∥Q\
right)$ plays a key role in the neural network loss function. The KL-Divergence is given by as follows:
$KL\left[q\left(\Theta ∣x\right)∣∣\left[f\left(\Theta ∣x\right)\right]={E}_{𝜃}{\sim q\left(\Theta ∣x\right)}^{\left[logq\left(\Theta ∣x\right)-logf\left(\Theta ∣x\
As shown by Kingma and Welling [12] that minimizing Eq. 2 while still reconstructing input data is equivalent to maximizing.
${E}_{𝜃}\sim q{\left(𝜃∣x\right)}^{\left[logP\left(X=x∣\Theta \right)-KL\left[q\left(\Theta ∣x\right)∣∣f\left(\Theta \right)\right]}$(3)
Next, the VAE is trained by a gradient descent algorithm to minimize the loss function. In this case, ${L}_{0}$ is the cross-entropy loss function and $\lambda$ is a regularization hyperparameter.
$L\left(W\right)={L}_{0}\left(W\right)+\lambda KL\left[q\left(\Theta ∣x\right)∣∣f\left(\Theta \right)\right]$(4)
Root mean squared error (RMSE): RMSE criterion reflects the average magnitude of the bias between the true item parameters and their associated estimates. A smaller RMSE suggests higher estimation
accuracy. Moreover, we also looked into Akaike information criterion (AIC), and Bayesian information criterion (BIC) score to explore MIRT model estimation when using Q-Miss.
First, we have incorporated the independent and correlated latent traits via the ML2P-VAE model proposed by Curi [1], and Converse [2]. We have extended this work by validating Q-matrix based on
root, mean, square, error (RMSE), BIAS, and Correlation score.
We made modifications to the architecture of the neural network to allow for the interpretation of weights and biases in the decoder as item parameter estimates, and activation values in the encoded
hidden layer as ability parameter estimates. As we know, sometimes researchers call neural networks are usually uninterpretable and function as a black-box model. However, following the addition of
Q-matrix in the second from the last layer will make NN more interpretable.
The required modifications are as follows. The decoder of the variational autoencoder has no hidden layers. The non-zero weights in the decoder, connecting the encoded distribution to the output
layer, are determined by a given Q-matrix [19]. Thus, these two layers are not densely connected. The output layer must use the sigmoid activation function as follows:
$\sigma \left({z}_{i}\right)=\frac{1}{1+{e}^{-{z}_{i}}}$(5)
When latent traits are assumed to be correlated, a full correlation matrix must be provided for the ML2P-VAE model. However, a correlation matrix is not required when latent traits are assumed to be
independent. This corresponds to the fixed covariance matrix ${\mathrm{\Sigma }}_{1}$. ML2P-VAE can estimate ability, discrimination, and difficulty parameters, but it does not estimate correlations
between latent traits.
Also, the input to our neural network consists of $n$ nodes, representing items on an assessment. After a sufficient number of hidden layers of sufficient size, the encoder outputs K + K(K + 1)/2
nodes. The architecture for correlated latent traits is more complex than we think (See Visualization of Deep-VAE architecture for two correlated latent traits and ten input items model via this link
2.1 Q-Matrix and Misspecification of Q-matrix
Specification of Q-matrix is mainly criticized because of its subjective nature [17]. Misspecification in the cognitive diagnostic model (CDM) mostly occurs because of the types of the attributes,
construct of the attribute, Q-matrix, or selected cognitive diagnostic model [6]. In this experiment, we utilized only a misfit source because Q-matrix misspecification was examined, and no changes
were made in students’ responses. In the study, the Q-matrix was misspecified by a mixed approach, and misfit items used in this study are presented in Table 1 (See misfit items table in Appendix
2:tinyurl.com/aied22). When the Q-matrix was misspecified, one attribute was translated from 1 to 0, and another attribute was translated from 0 to 1, but the number of measured attributes didn’t
change, which is referred to as mixed.
In the architecture of the model ML2P-VAE, we train the neural network with the ADAM optimizer (pure stochastic gradient descent). A simulated assessment with six latent abilities used two hidden
layers of sizes 50 and 25. The largest network we used was for an assessment of 20 latent abilities, which utilized two hidden layers of sizes 100 and 50.
3. THE DEEP-Q ALGORITHM
For convenience, we are calling this algorithm the Deep-Q algorithm. The steps of the Deep-Q algorithm are as follows-
Step 1: Use the variational autoencoder and multidimensional item response theory (Ml2P-VAE) model [2] to estimate students’ ability and item parameters based on Q-True and the response data. Step 2:
Compute all items’ via RMSE, BIAS, and Correlation test score values based on Q-True and the student’s ability and item parameters estimated at Step 1. We also use AIC and BIC scores to compare
Q-true and Q-miss. Step 3: Randomly misspecify Q-true by 10% and 20% to change Q-True. Step 4: Repeat step-1 with Q-miss (Q-miss, Step 3). Step 5: Compare Q-True(top row, boldface) with Q-miss.
Q-true should yield a small RMSE/BIAS and a strong correlation score, AIC, and BIC score for difficulty, discrimination, and ability parameters.
4. METHODOLOGY
We ran experiments on two data sets: (i) 6 traits, 35 items, and 18000 students, and (ii) 20 traits, 200 items, and 60,000 students. It is also important to mention here that true parameter values,
for both students and items, are only available for simulated data. When simulating data, we used the Pythons SciPy package to generate a symmetric positive definite matrix with 1s on the diagonal
(correlation matrix) and all matrix entries non-negative. All latent traits had correlation values between 0,1. We assumed that each latent trait was mean-centered at 0. Then, we sampled ability
vectors to create simulated students. We generated a random Q-matrix where each entry ${q}_{ij}\sim Bern\left(0.2\right)$. If a column for each element after sampling from this Bernoulli
distribution, one random element was changed to a 1. Discrimination parameters were sampled from a range so from 0.25 to 1.75 for each item i, and difficulty parameters were sampled uniformly from -
3 to 3. Finally, response sets for each student were sampled from the ML2P model using these parameters.
5. RESULTS
All experiments were conducted using TensorFlow for R and the ML2Pvae package [4] on an iMac computer with a 3.1 GHz Intel Core i5 via Google Colab Premium, 12 GB NVIDIA Tesla K80 GPU.
Table 1 presents the estimation accuracy of Q under Q-True and Q-miss. The range of values for each criterion is provided in the second and third row of Table 1, and the numbers in bold denote better
performance in the associated criterion for the corresponding method, e.g., Q-Miss.
Table 1: Q-matrix validation measures via RMSE, BIAS and Correlation score for discrimination (a),
difficulty (b), and ability $𝜃$ parameters with correlated latent traits
Data Set Misspecification Method a.RMSE a.BIAS a.Corr b.RMSE b.BIAS b.Corr $𝜃.RMSE$ $𝜃.BIAS$ $𝜃.Corr$
Q_True 0.1465 0.0100 0.9427 0.0750 0.0100 0.9988 0.8120 0.0476 0.5815
N=18000 10%_Miss Q_Miss 0.2297 0.0195 0.8562 0.0993 0.0083 0.9986 0.8398 0.0637 0.5486
20%_Miss Q_Miss 0.2621 0.0466 0.7984 0.1684 0.0268 0.9962 0.8684 0.1660 0.5351
Q_True 0.1007 0.0259 0.9094 0.2098 -0.0186 0.9984 0.5614 -0.0013 0.8686
N=60000 10%_Miss Q_Miss 0.2525 0.0654 0.8430 0.2881 0.0129 0.9926 0.7288 0.0037 0.6859
20%_Miss Q_Miss 0.2200 0.0367 0.8777 0.2191 0.0243 0.9952 0.6561 0.0166 0.7551
Overall, Table-1 indicates that the Deep-Q method yields a better fit statistic score and strong correlation score than the Q-miss situation when using a wrong Q-matrices. This result is corroborated
by the correlation plots between the true discrimination parameters and the weights of the decoder, displayed in Fig. 1 and 2 (follow this url: tinyurl.com /aied22 for larger view of plots)
Figure 1: (A)Discrimination, Difficulty, and Ability Parameter Estimates with Independent Latent Traits (B) Discrimination, Difficulty, and Ability Parameter Estimates with Correlated Latent Traits
In addition, Q-matrix validation measures via AIC, BIC score for discrimination (a), difficulty (b), and ability $𝜃$ parameters with correlated latent traits remain consistent with Table-1 outcome
(see AIC, BIC score:Appendix 5:[tinyurl.com /aied22])
In Fig 1(A and B), the correlation plots of discrimination parameter estimate for data with items and latent traits. Each color represents discrimination parameters relating to one of each latent
skill. In the ability parameter, each color in the plot represents discrimination and ability parameters associated with each latent trait. Difficulty parameters are on the item level, not the latent
trait level. So in each item, I have exactly one difficulty parameter ${b}_{i}$, regardless of the number of latent skills. The interpretation is similar for independent latent traits, as described
in figure 1(A). Plots show correlated latent traits and show better outcomes compared to independent latent traits.
An incorrect Q-matrix can lead to a significant change in the assessment outcomes when applied to CDMs. As a result, a Q-matrix validation strategy to reduce assessment error is becoming increasingly
important. Several approaches, including EM-based and non-parametric methods, have shown the ability to identify and create an acceptable Q-matrix. However, to the best of the authors’ knowledge,
their experiment utilizes traditional IRT parameter estimation where they utilize low-dimensional latent traits and students’ responses. However, the Deep-Q algorithm is most useful with high and
low-dimensional data.
Moreover, Converse’s [2] study shows that MIRT parameter estimation results via the Ml2P-VAE model are competitive compared to traditional IRT parameter estimation methods. Our study used a Deep-Q
algorithm, a deep learning-based algorithm, to identify and validate a Q-matrix for small and large-scale latent traits. Deep-Q could be useful for large-scale assessments, e.g., PISA and TIMSS.
ML2P-VAE is a novel technique that allows IRT parameter estimation of independent and correlated low and high-dimensional latent traits. Ultimately, it can be said that the Deep-Q algorithm succeeds
in detecting misfit items in both large and small sample cases. ML2P-VAE methods and Deep-Q are most useful on high-dimensional data, but even when applied to smaller data sets where traditional
techniques are feasible, the results from current methods are competitive.
This research was sponsored by the National Science Foundation under the award The Learner Data Institute (bit.ly/36Bi93m) (award 1934745). The opinions, findings, and results are solely the authors’
and do not reflect those of the funding agencies. Thanks to Geoff Converse, Andrew Ott, and LDI team for their suggestions and comments.
8. REFERENCES
1. Curi, M., Converse, G. A., Hajewski, J., Oliveira, S.: Interpretable variational autoencoders for cognitive models. 2019 International Joint Conference on Neural Networks (IJCNN) 1-8. IEEE.
(2019).DOI: 10.1109/IJCNN.2019.8852333
2. Converse, G., Curi, M., Oliveira, S., Templin, J.: Estimation of multidimensional item response theory models with correlated latent variables using variational autoencoders. Machine Learning,
1-18.(2021). DOI:10.1007/s10994-021-06005-7
3. Converse, G., Curi, M., Oliveira, S.: Autoencoders for educational assessment. International Conference on Artificial Intelligence in Education.41-45.Springer.(2019).https://doi.org/10.1007/
4. Converse, G.: ML2Pvae: VAE models for IRT parameter estimation.(2020) https://CRAN.R-project.org/package=ML2Pvae, r package version1.0.0.
5. Liu, C. W., Chalmers, R. P.: Fitting item response unfolding models to likert scale data using mirt in R. PloS one. 13(5).(2018). https://doi.org/10.1371/journal.pone.0196292
6. Chen, J., de la Torre, J., Zhang, Z.: Relative and absolute fit evaluation in cognitive diagnosis modeling. Journal of Educational Measurement, 50(2),123-140.(2013).https://doi.org/10.1111/
7. Kingma, D. P., Welling., M.: Auto-encoding variational bayes. (2013). https://doi.org/10.48550/arXiv.1312.6114
8. Rezende, D. J., Mohamed, S. and Wierstra, D.: Stochastic backpropagation and approximate inference in deep generative models. (2014).https://doi.org/10.48550/arXiv.1401.4082
9. Harris, D.: Comparison of 1-, 2-, and 3-parameter IRT models. Educational Measurement: Issues and Practice,8(1), 35-41. (1989).https://doi.org/10.1111/j.1745-3992.1989.tb00313.x
10. Leighton, J. P., Gierl, M. J., Hunka,S. M.: The attribute hierarchy method for cognitive assessment: A variation on Tatsuoka’s rule-space approach. J. Educ. Meas., vol. 41, 205–237,(2004). https:
11. Torre, J. de la. and Chiu,C. Y.,:A General Method of Empirical Q-matrix Validation,” Psychometrika, vol. 81, no. 2, 253–273.(2016). DOI: 10.1007/s11336-015-9467-8
12. Kingma, D. P., Welling, M.: Auto-encoding variational bayes. arXiv preprint arXiv:1312.6114.(2013). https://doi.org/10.48550/arXiv.1312.6114
13. DeCarlo, L. T.: On the analysis of fraction subtraction data: The DINA model, classification, latent class sizes, and the q-matrix. Appl. Psychol. Meas., vol. 35, no. 1,8–26 ,(2011).https://
14. McKinley, R., Reckase, M.: The use of the general Rasch model with multidimensional item response data. American College Testing. (1980)
15. Piech C., Bassen J., Huang J., Ganguli S., Sahami M., Guibas LJ., Sohl-Dickstein J.: Deep knowledge tracing. Advances in neural information processing systems. 28.(2015)
16. Guo, Q. Cutumisu, M., Cui, Y.: A neural network approach to estimate student skill mastery in cognitive diagnostic assessments. (2017).https://doi.org/10.7939/R35H7C71D
17. Rupp, A. A., Templin, J.: The effects of Q-matrix misspecification on parameter estimates and classification accuracy in the DINA model. Educational and Psychological Measurement, 68(1), 78-96.
18. Rezende, D. J., Mohamed, S., Wierstra, D.: Stochastic backpropagation and approximate inference in deep generative models. In International conference on machine learning 1278-1286. PMLR.
19. Tatsuoka, K.K.: Rule space: an approach for dealing with misconceptions based on item response theory. J. Educ. Meas. 20(4), 345–354 (1983)
20. Ma, W., Torre,J. de la.: An empirical Q-matrix validation method for the sequential generalized DINA model. Br. J. Math. Stat. Psychol., (2019). https://doi.org/10.1111/bmsp.12156
21. Wu, M., Davis, R. L., Domingue, B. W., Piech, C., Goodman, N.: Variational item response theory: Fast, accurate, and expressive. arXiv preprint arXiv:2002.00276.(2020).https://doi.org/10.48550/
22. Zhang, J., Shi, X., King, I., Yeung, D. Y.: Dynamic key-value memory networks for knowledge tracing. In: 26th International world wide web conference (WWW 2017) 765–774.(2017). https://doi.org/
Please follow this link for additional references (Appendix 6): tinyurl.com/aied22
9. APPENDIX
Please follow this link for Appendixes: tinyurl.com/aied22
© 2022 Copyright is held by the author(s). This work is distributed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) license. | {"url":"https://educationaldatamining.org/EDM2022/proceedings/2022.EDM-doctoral-consortium.107/index.html","timestamp":"2024-11-09T22:34:37Z","content_type":"text/html","content_length":"64790","record_id":"<urn:uuid:b0b46853-7bbd-4747-adb4-7a2010ec8378>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00000.warc.gz"} |
Forward recursion of state-space models
filter computes state-distribution moments for each period of the specified response data by recursively applying the Kalman filter.
To compute updated state-distribution moments efficiently during only the final period of the specified response data by applying one recursion of the Kalman filter, use update instead.
X = filter(Mdl,Y) returns filtered states (X) from performing forward recursion of the fully specified state-space model Mdl. That is, filter applies the standard Kalman filter using Mdl and the
observed responses Y.
X = filter(Mdl,Y,Name,Value) uses additional options specified by one or more Name,Value arguments. For example, specify the regression coefficients and predictor data to deflate the observations, or
specify to use the square-root filter.
If Mdl is not fully specified, then you must specify the unknown parameters as known scalars using the 'Params' Name,Value argument.
[X,logL,Output] = filter(___) uses any of the input arguments in the previous syntaxes to additionally return the loglikelihood value (logL) and an output structure array (Output) using any of the
input arguments in the previous syntaxes. Output contains:
• Filtered and forecasted states
• Estimated covariance matrices of the filtered and forecasted states
• Loglikelihood value
• Forecasted observations and its estimated covariance matrix
• Adjusted Kalman gain
• Vector indicating which data the software used to filter
Filter States of Time-Invariant State-Space Model
Suppose that a latent process is an AR(1). The state equation is
where ${u}_{t}$ is Gaussian with mean 0 and standard deviation 1.
Generate a random series of 100 observations from ${x}_{t}$, assuming that the series starts at 1.5.
T = 100;
ARMdl = arima('AR',0.5,'Constant',0,'Variance',1);
x0 = 1.5;
rng(1); % For reproducibility
x = simulate(ARMdl,T,'Y0',x0);
Suppose further that the latent process is subject to additive measurement error. The observation equation is
${y}_{t}={x}_{t}+{\epsilon }_{t},$
where ${\epsilon }_{t}$ is Gaussian with mean 0 and standard deviation 0.75. Together, the latent process and observation equations compose a state-space model.
Use the random latent state process (x) and the observation equation to generate observations.
Specify the four coefficient matrices.
A = 0.5;
B = 1;
C = 1;
D = 0.75;
Specify the state-space model using the coefficient matrices.
Mdl =
State-space model type: ssm
State vector length: 1
Observation vector length: 1
State disturbance vector length: 1
Observation innovation vector length: 1
Sample size supported by model: Unlimited
State variables: x1, x2,...
State disturbances: u1, u2,...
Observation series: y1, y2,...
Observation innovations: e1, e2,...
State equation:
x1(t) = (0.50)x1(t-1) + u1(t)
Observation equation:
y1(t) = x1(t) + (0.75)e1(t)
Initial state distribution:
Initial state means
Initial state covariance matrix
x1 1.33
State types
Mdl is an ssm model. Verify that the model is correctly specified using the display in the Command Window. The software infers that the state process is stationary. Subsequently, the software sets
the initial state mean and covariance to the mean and variance of the stationary distribution of an AR(1) model.
Filter states for periods 1 through 100. Plot the true state values and the filtered state estimates.
filteredX = filter(Mdl,y);
title({'State Values'})
legend({'True state values','Filtered state values'})
The true values and filter estimates are approximately the same.
Filter States of State-Space Model Containing Regression Component
Suppose that the linear relationship between the change in the unemployment rate and the nominal gross national product (nGNP) growth rate is of interest. Suppose further that the first difference of
the unemployment rate is an ARMA(1,1) series. Symbolically, and in state-space form, the model is
$\begin{array}{l}\left[\begin{array}{c}{x}_{1,t}\\ {x}_{2,t}\end{array}\right]=\left[\begin{array}{cc}\varphi & \theta \\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1,t-1}\\ {x}_{2,t-1}\end
{array}\right]+\left[\begin{array}{c}1\\ 1\end{array}\right]{u}_{1,t}\\ {y}_{t}-\beta {Z}_{t}={x}_{1,t}+\sigma {\epsilon }_{t},\end{array}$
• ${x}_{1,t}$ is the change in the unemployment rate at time t.
• ${x}_{2,t}$ is a dummy state for the MA(1) effect.
• ${y}_{1,t}$ is the observed change in the unemployment rate being deflated by the growth rate of nGNP (${Z}_{t}$).
• ${u}_{1,t}$ is the Gaussian series of state disturbances having mean 0 and standard deviation 1.
• ${\epsilon }_{t}$ is the Gaussian series of observation innovations having mean 0 and standard deviation $\sigma$.
Load the Nelson-Plosser data set, which contains the unemployment rate and nGNP series, among other things.
Preprocess the data by taking the natural logarithm of the nGNP series, and the first difference of each series. Also, remove the starting NaN values from each series.
isNaN = any(ismissing(DataTable),2); % Flag periods containing NaNs
gnpn = DataTable.GNPN(~isNaN);
u = DataTable.UR(~isNaN);
T = size(gnpn,1); % Sample size
Z = [ones(T-1,1) diff(log(gnpn))];
y = diff(u);
Though this example removes missing values, the software can accommodate series containing missing values in the Kalman filter framework.
Specify the coefficient matrices.
A = [NaN NaN; 0 0];
B = [1; 1];
C = [1 0];
D = NaN;
Specify the state-space model using ssm.
Estimate the model parameters, and use a random set of initial parameter values for optimization. Specify the regression component and its initial value for optimization using the 'Predictors' and
'Beta0' name-value pair arguments, respectively. Restrict the estimate of $\sigma$ to all positive, real numbers.
params0 = [0.3 0.2 0.2];
[EstMdl,estParams] = estimate(Mdl,y,params0,'Predictors',Z,...
'Beta0',[0.1 0.2],'lb',[-Inf,-Inf,0,-Inf,-Inf]);
Method: Maximum likelihood (fmincon)
Sample size: 61
Logarithmic likelihood: -99.7245
Akaike info criterion: 209.449
Bayesian info criterion: 220.003
| Coeff Std Err t Stat Prob
c(1) | -0.34098 0.29608 -1.15164 0.24948
c(2) | 1.05003 0.41377 2.53771 0.01116
c(3) | 0.48592 0.36790 1.32079 0.18657
y <- z(1) | 1.36121 0.22338 6.09358 0
y <- z(2) | -24.46711 1.60018 -15.29024 0
| Final State Std Dev t Stat Prob
x(1) | 1.01264 0.44690 2.26592 0.02346
x(2) | 0.77718 0.58917 1.31912 0.18713
EstMdl is an ssm model, and you can access its properties using dot notation.
Filter the estimated state-space model. EstMdl does not store the data or the regression coefficients, so you must pass in them in using the name-value pair arguments 'Predictors' and 'Beta',
respectively. Plot the estimated, filtered states. Recall that the first state is the change in the unemployment rate, and the second state helps build the first.
filteredX = filter(EstMdl,y,'Predictors',Z,'Beta',estParams(end-1:end));
ylabel('Change in the unemployment rate')
title('Filtered Change in the Unemployment Rate')
Filter Series in Real Time
Consider nowcasting the model in Filter States of Time-Invariant State-Space Model.
Generate a random series of 100 observations from ${x}_{t}$.
T = 100;
A = 0.5;
B = 1;
C = 1;
D = 0.75;
Mdl = ssm(A,B,C,D);
rng(1); % For reproducibility
y = simulate(Mdl,T);
Suppose the final 10 observations are in the forecast horizon.
fh = 10;
yf = y((end-fh+1):end); % Holdout sample responses
y = y(1:end-fh); % In-sample responses
Filter the observations through the model to obtain filtered states for each period.
[xhat,logL,output] = filter(Mdl,y);
xhatvar = output.FilteredStatesCov;
xhat and xhatvar are 90-by-1 vectors of in-sample filtered states and corresponding variances, respectively. xhat(t) is the estimate of $\mathit{E}\left({\mathit{x}}_{\mathit{t}}|{\mathit{y}}_
{1},...,{\mathit{y}}_{\mathit{t}}\right)$, and xhatvar is the estimate of $\mathrm{Var}\left({\mathit{x}}_{\mathit{t}}|{\mathit{y}}_{1},...,{\mathit{y}}_{\mathit{t}}\right)$.
Call filter again, but specify the real-time update option.
[xhatRT,logLRT,outputRT] = filter(Mdl,y,'RealTimeUpdate',true);
xhatRTvar = outputRT.FilteredStatesCov;
xhatRT and xhatRTvar are scalars representing the estimate of $\mathit{E}\left({\mathit{x}}_{90}|{\mathit{y}}_{1},...,{\mathit{y}}_{90}\right)$ and its corresponding variance, respectively.
Compare the filtered states and variances of period 90.
tol = 1e-10;
areMeansEqual = (xhat(end) - xhatRT) < tol
areMeansEqual = logical
areVarsEqual = (xhatvar(end) - xhatRTvar) < tol
areLogLsEqual = (logL - logLRT) < tol;
In the last period, the filtered states, their variances, and the loglikelihoods are equal.
Nowcast the model into the forecast horizon by performing this procedure for each successive period:
1. Set the initial state and its variance to their current filter estimates. This action changes the state-space model.
2. As an observation becomes available, filter it through the model in real time.
% Initialize state and variance
state0 = xhatRT;
var0 = xhatRTvar;
% Preallocate
xhatRTF = zeros(fh,1);
xhatRTvarF = zeros(fh,1);
for j = 1:fh
Mdl.Mean0 = state0;
Mdl.Cov0 = var0;
[xhatRTF(j),~,outputRT] = filter(Mdl,yf(j),'RealTimeUpdate',true); % Alternatively, use update
xhatRTvarF(j) = outputRT.FilteredStatesCov;
state0 = xhatRTF(j);
var0 = xhatRTvarF(j);
Plot the data and nowcasts.
plot((T-fh-20):T,[y(end-20:end); yf],'b-',(T-fh+1):T,xhatRTF,'r*-')
legend(["Data" "Nowcasts"],'Location',"best")
Input Arguments
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but
the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: 'Beta',beta,'Predictors',Z specifies to deflate the observations by the regression component composed of the predictor data Z and the coefficient matrix beta.
RealTimeUpdate — Flag indicating whether to apply real-time filter
false (default) | true
Flag indicating whether to apply the real-time filter, specified as a value in the table.
Value Description
filter returns only the final state distribution, which is the filtered state at time T (one forward recursion of the Kalman filter). Specifically, filter performs the following actions:
1. Dispatch the model, data, and other specifications to the update function to update state distribution and compute the loglikelihood.
2. Return only the final state distribution; outputs do not contain intermediate state distributions.
□ X is a 1-by-m vector of the final filtered states.
□ logL is a scalar of the sum of the loglikelihood.
□ Output is a 1-by-1 structure array associated with the recursion at time T. The fields include:
☆ LogLikelihood: sum of the loglikelihood
☆ FilteredStates: m-by-1 vector of filtered states
☆ FilteredStatesCov: m-by-m uncertainty matrix of filtered states
Mdl must represent a standard state-space model (ssm object).
false filter returns filtered states and results for each period in the input data Y.
Example: 'RealTimeUpdate',true
Data Types: logical
Output Arguments
X — Filtered states
numeric matrix | cell vector of numeric vectors
Filtered states, returned as a numeric matrix or a cell vector of numeric vectors.
If Mdl is time invariant, then the number of rows of X is the sample size, T, and the number of columns of X is the number of states, m. The last row of X contains the latest filtered states.
If Mdl is time varying, then X is a cell vector with length equal to the sample size. Cell t of X contains a vector of filtered states with length equal to the number of states in period t. The last
cell of X contains the latest filtered states.
If you set the RealTimeUpdate name-value argument to true, filter returns only the filtered state for time T, a 1-by-m vector. For more details, see RealTimeUpdate.
Output — Filtering results by period
structure array
Filtering results by period, returned as a structure array.
Output is a T-by-1 structure, where element t corresponds to the filtering result at time t.
• If Univariate is false (it is by default), then the following table outlines the fields of Output.
Field Description Estimate of
LogLikelihood Scalar loglikelihood objective function value N/A
FilteredStates m[t]-by-1 vector of filtered states $E\left({x}_{t}|{y}_{1},...,{y}_{t}
FilteredStatesCov m[t]-by-m[t] variance-covariance matrix of filtered states $Var\left({x}_{t}|{y}_{1},...,{y}_
ForecastedStates m[t]-by-1 vector of state forecasts $E\left({x}_{t}|{y}_{1},...,{y}_
ForecastedStatesCov m[t]-by-m[t] variance-covariance matrix of state forecasts $Var\left({x}_{t}|{y}_{1},...,{y}_
ForecastedObs h[t]-by-1 forecasted observation vector $E\left({y}_{t}|{y}_{1},...,{y}_
ForecastedObsCov h[t]-by-h[t] variance-covariance matrix of forecasted observations $Var\left({y}_{t}|{y}_{1},...,{t}_
KalmanGain m[t]-by-n[t] adjusted Kalman gain matrix N/A
DataUsed h[t]-by-1 logical vector indicating whether the software filters using a particular observation. For example, if observation i at time t is N/A
a NaN, then element i in DataUsed at time t is 0.
• If Univarite is true, then the fields of Output are the same as in the previous table, except for the following amendments.
Field Changes
ForecastedObs Same dimensions as if Univariate = 0, but only the first elements are equal
n-by-1 vector of forecasted observation variances.
ForecastedObsCov The first element of this vector is equivalent to ForecastedObsCov(1,1) when Univariate is false. The rest of the elements are not necessarily equivalent to their corresponding
values in ForecastObsCov when Univariate.
KalmanGain Same dimensions as if Univariate is false, though KalmanGain might have different entries.
If you set the RealTimeUpdate name-value argument to true, filter returns only the filtered states for time T, its covariance matrix, and the loglikelihood (in other words, the sum of the
loglikelihoods returned by update). For more details, see RealTimeUpdate.
• Mdl does not store the response data, predictor data, and the regression coefficients. Supply the data wherever necessary using the appropriate input or name-value arguments.
• To accelerate estimation for low-dimensional, time-invariant models, set 'Univariate',true. Using this specification, the software sequentially updates rather then updating all at once during the
filtering process.
• The Kalman filter accommodates missing data by not updating filtered state estimates corresponding to missing observations. In other words, suppose there is a missing observation at period t.
Then, the state forecast for period t based on the previous t – 1 observations and filtered state for period t are equivalent.
• For explicitly defined state-space models, filter applies all predictors to each response series. However, each response series has its own set of regression coefficients.
Alternative Functionality
To filter a standard state-space model in real time by performing one forward recursion of the Kalman filter, call the update function instead. Unlike filter, update performs minimal input validation
for computational efficiency.
Version History
Introduced in R2014a | {"url":"https://it.mathworks.com/help/econ/ssm.filter.html","timestamp":"2024-11-14T15:01:21Z","content_type":"text/html","content_length":"150509","record_id":"<urn:uuid:c479c87a-fcfe-42cc-bac4-eab844f47553>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00367.warc.gz"} |
Range Rover vogue
Range rover vogue. 2000 in dark blu fsh only 145ks
This car has the lot google the specs $13.500. Firm may swap for a 4x4 ute or a jeep wrangler
wot you got ???? This 4x4 cost. $110.000 wen new in 2000
Edited by dylan
How come it's gone up in value? Even though $2,500 return in 14 years is good for a car. They generally depreciate like crazy.
I think the op means its $11000 now - was $13500 when it was purchased 2yrs ago! Slightly misleading !
If your not intrested don't comment it was an error not Slightly misleading !
sume people are to happy to comment on posts there not intrested in
Edited by dylan
Indeed they are. It's bloody annoying isn't it. Although a great lesson in making sure your initial posts are clear and concise, hey! ;-)
I think the op means its $11000 now - was $13500 when it was purchased 2yrs ago! Slightly misleading !
I reckon the correct translation of the original post is that the car was $110,000.00 when new in 2000 and is now on offer for $13,500.00. Averages out at a loss of 15% of its value per year. Big
cash to begin with but from here on down it's probably not much different to most other cars around.
As an aside do you know that 80% of all Land Rovers ever made are still on the road? (the other 20% made it safely home
Chill out...was actually trying to help you out there as ppl were saying it had gone up!! Jeez...wouldn't wanna do a deal with you if that's your attitude!
No wonder so meny people have left this site.
Just wondering how you can claim this site has lost members when you have a total of 17 posts (4 of which are in this thread) in 81 months of membership. Doesn't seem like you hang around enough to
have observed membership trends and who is here / who has left etc.
Edit: I note you have amended the original post to reflect the correct price (which I pointed out to you). All part of the service here at PIA. Don't bother to thank us. Oh wait, you wouldn't have
done anyway.
Edited by sidestep | {"url":"https://www.pomsinadelaide.com/topic/38359-range-rover-vogue/","timestamp":"2024-11-12T18:59:54Z","content_type":"text/html","content_length":"171487","record_id":"<urn:uuid:03416793-169c-42ee-8bd4-cf018640ce24>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00095.warc.gz"} |
(GA1) (G.Baumslag) Does the free Q-group F^Q act freely on some \Lambda-tree?
(GA2)(G.Baumslag, A.Miasnikov, V.Remeslennikov) Is a finitely generated group acting freely on a \Lambda-tree, automatic (biautomatic)?
(GA3) (A.Miasnikov) Let G be a group acting freely on a \Lambda-tree. Given any finite set of non-trivial elements of G \ast G, is there a homomorphism of G \ast G into G such that the images of the
given elements are also non-trivial?
(GA4) (O.Kharlampovich, A.Miasnikov, V.Remeslennikov) Is the elementary theory of the class of all groups acting freely on a \Lambda-tree, decidable?
(GA5) (S.Sidki) Let G be a group such that it contains a subgroup H of index 2 and H admits a homomorphism f:H-->G; we call such a map a 1/2-endomorphims of G. Then this data can be used to construct
a (state-closed) representation r of G into the automorphism group of the binary tree where the kernel(r)= < K| K subgroup of H which is both normal in G and f(K) is contained in K>. We call kernel
(r) the f-core(H) and say that f is simple provided f-core(H) is the trivial group.
(a) Which groups admit a simple 1/2-endomorphism?
(b) If G admits a simple 1/2-endomorphism, can G be a free abelian group of infinite rank?
(c) If G admits a simple 1/2-endomorphism, can G be a free group of rank k>1? Background | {"url":"https://shpilrain.ccny.cuny.edu/gworld/problems/probact.html","timestamp":"2024-11-11T13:57:25Z","content_type":"text/html","content_length":"2260","record_id":"<urn:uuid:4e330310-ef30-438c-90fe-8cc7f992027d>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00718.warc.gz"} |
gcc/ada/exp_fixd.adb - gcc - Git at Google
-- --
-- GNAT COMPILER COMPONENTS --
-- --
-- E X P _ F I X D --
-- --
-- B o d y --
-- --
-- Copyright (C) 1992-2022, Free Software Foundation, Inc. --
-- --
-- GNAT is free software; you can redistribute it and/or modify it under --
-- terms of the GNU General Public License as published by the Free Soft- --
-- ware Foundation; either version 3, or (at your option) any later ver- --
-- sion. GNAT is distributed in the hope that it will be useful, but WITH- --
-- OUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY --
-- or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License --
-- for more details. You should have received a copy of the GNU General --
-- Public License distributed with GNAT; see file COPYING3. If not, go to --
-- http://www.gnu.org/licenses for a complete copy of the license. --
-- --
-- GNAT was originally developed by the GNAT team at New York University. --
-- Extensive contributions were provided by Ada Core Technologies Inc. --
-- --
with Atree; use Atree;
with Checks; use Checks;
with Einfo; use Einfo;
with Einfo.Entities; use Einfo.Entities;
with Einfo.Utils; use Einfo.Utils;
with Exp_Util; use Exp_Util;
with Nlists; use Nlists;
with Nmake; use Nmake;
with Restrict; use Restrict;
with Rident; use Rident;
with Rtsfind; use Rtsfind;
with Sem; use Sem;
with Sem_Eval; use Sem_Eval;
with Sem_Res; use Sem_Res;
with Sem_Util; use Sem_Util;
with Sinfo; use Sinfo;
with Sinfo.Nodes; use Sinfo.Nodes;
with Stand; use Stand;
with Tbuild; use Tbuild;
with Ttypes; use Ttypes;
with Uintp; use Uintp;
with Urealp; use Urealp;
package body Exp_Fixd is
-- Local Subprograms --
-- General note; in this unit, a number of routines are driven by the
-- types (Etype) of their operands. Since we are dealing with unanalyzed
-- expressions as they are constructed, the Etypes would not normally be
-- set, but the construction routines that we use in this unit do in fact
-- set the Etype values correctly. In addition, setting the Etype ensures
-- that the analyzer does not try to redetermine the type when the node
-- is analyzed (which would be wrong, since in the case where we set the
-- Conversion_OK flag, it would think it was still dealing with a normal
-- fixed-point operation and mess it up).
function Build_Conversion
(N : Node_Id;
Typ : Entity_Id;
Expr : Node_Id;
Rchk : Boolean := False;
Trunc : Boolean := False) return Node_Id;
-- Build an expression that converts the expression Expr to type Typ,
-- taking the source location from Sloc (N). If the conversions involve
-- fixed-point types, then the Conversion_OK flag will be set so that the
-- resulting conversions do not get re-expanded. On return, the resulting
-- node has its Etype set. If Rchk is set, then Do_Range_Check is set
-- in the resulting conversion node. If Trunc is set, then the
-- Float_Truncate flag is set on the conversion, which must be from
-- a floating-point type to an integer type.
function Build_Divide (N : Node_Id; L, R : Node_Id) return Node_Id;
-- Builds an N_Op_Divide node from the given left and right operand
-- expressions, using the source location from Sloc (N). The operands are
-- either both Universal_Real, in which case Build_Divide differs from
-- Make_Op_Divide only in that the Etype of the resulting node is set (to
-- Universal_Real), or they can be integer or fixed-point types. In this
-- case the types need not be the same, and Build_Divide chooses a type
-- long enough to hold both operands (i.e. the size of the longer of the
-- two operand types), and both operands are converted to this type. The
-- Etype of the result is also set to this value. The Rounded_Result flag
-- of the result in this case is set from the Rounded_Result flag of node
-- N. On return, the resulting node has its Etype set.
function Build_Double_Divide
(N : Node_Id;
X, Y, Z : Node_Id) return Node_Id;
-- Returns a node corresponding to the value X/(Y*Z) using the source
-- location from Sloc (N). The division is rounded if the Rounded_Result
-- flag of N is set. The integer types of X, Y, Z may be different. On
-- return, the resulting node has its Etype set.
procedure Build_Double_Divide_Code
(N : Node_Id;
X, Y, Z : Node_Id;
Qnn, Rnn : out Entity_Id;
Code : out List_Id);
-- Generates a sequence of code for determining the quotient and remainder
-- of the division X/(Y*Z), using the source location from Sloc (N).
-- Entities of appropriate types are allocated for the quotient and
-- remainder and returned in Qnn and Rnn. The result is rounded if the
-- Rounded_Result flag of N is set. The Etype fields of Qnn and Rnn are
-- appropriately set on return.
function Build_Multiply (N : Node_Id; L, R : Node_Id) return Node_Id;
-- Builds an N_Op_Multiply node from the given left and right operand
-- expressions, using the source location from Sloc (N). The operands are
-- either both Universal_Real, in which case Build_Multiply differs from
-- Make_Op_Multiply only in that the Etype of the resulting node is set (to
-- Universal_Real), or they can be integer or fixed-point types. In this
-- case the types need not be the same, and Build_Multiply chooses a type
-- long enough to hold the product and both operands are converted to this
-- type. The type of the result is also set to this value. On return, the
-- resulting node has its Etype set.
function Build_Rem (N : Node_Id; L, R : Node_Id) return Node_Id;
-- Builds an N_Op_Rem node from the given left and right operand
-- expressions, using the source location from Sloc (N). The operands are
-- both integer types, which need not be the same. Build_Rem converts the
-- operand with the smaller sized type to match the type of the other
-- operand and sets this as the result type. The result is never rounded
-- (rem operations cannot be rounded in any case). On return, the resulting
-- node has its Etype set.
function Build_Scaled_Divide
(N : Node_Id;
X, Y, Z : Node_Id) return Node_Id;
-- Returns a node corresponding to the value X*Y/Z using the source
-- location from Sloc (N). The division is rounded if the Rounded_Result
-- flag of N is set. The integer types of X, Y, Z may be different. On
-- return the resulting node has its Etype set.
procedure Build_Scaled_Divide_Code
(N : Node_Id;
X, Y, Z : Node_Id;
Qnn, Rnn : out Entity_Id;
Code : out List_Id);
-- Generates a sequence of code for determining the quotient and remainder
-- of the division X*Y/Z, using the source location from Sloc (N). Entities
-- of appropriate types are allocated for the quotient and remainder and
-- returned in Qnn and Rrr. The integer types for X, Y, Z may be different.
-- The division is rounded if the Rounded_Result flag of N is set. The
-- Etype fields of Qnn and Rnn are appropriately set on return.
procedure Do_Divide_Fixed_Fixed (N : Node_Id);
-- Handles expansion of divide for case of two fixed-point operands
-- (neither of them universal), with an integer or fixed-point result.
-- N is the N_Op_Divide node to be expanded.
procedure Do_Divide_Fixed_Universal (N : Node_Id);
-- Handles expansion of divide for case of a fixed-point operand divided
-- by a universal real operand, with an integer or fixed-point result. N
-- is the N_Op_Divide node to be expanded.
procedure Do_Divide_Universal_Fixed (N : Node_Id);
-- Handles expansion of divide for case of a universal real operand
-- divided by a fixed-point operand, with an integer or fixed-point
-- result. N is the N_Op_Divide node to be expanded.
procedure Do_Multiply_Fixed_Fixed (N : Node_Id);
-- Handles expansion of multiply for case of two fixed-point operands
-- (neither of them universal), with an integer or fixed-point result.
-- N is the N_Op_Multiply node to be expanded.
procedure Do_Multiply_Fixed_Universal (N : Node_Id; Left, Right : Node_Id);
-- Handles expansion of multiply for case of a fixed-point operand
-- multiplied by a universal real operand, with an integer or fixed-
-- point result. N is the N_Op_Multiply node to be expanded, and
-- Left, Right are the operands (which may have been switched).
procedure Expand_Convert_Fixed_Static (N : Node_Id);
-- This routine is called where the node N is a conversion of a literal
-- or other static expression of a fixed-point type to some other type.
-- In such cases, we simply rewrite the operand as a real literal and
-- reanalyze. This avoids problems which would otherwise result from
-- attempting to build and fold expressions involving constants.
function Fpt_Value (N : Node_Id) return Node_Id;
-- Given an operand of fixed-point operation, return an expression that
-- represents the corresponding Universal_Real value. The expression
-- can be of integer type, floating-point type, or fixed-point type.
-- The expression returned is neither analyzed nor resolved. The Etype
-- of the result is properly set (to Universal_Real).
function Get_Size_For_Value (V : Uint) return Pos;
-- Given a non-negative universal integer value, return the size of a small
-- signed integer type covering -V .. V, or Pos'Max if no such type exists.
function Get_Type_For_Size (Siz : Pos; Force : Boolean) return Entity_Id;
-- Return the smallest signed integer type containing at least Siz bits.
-- If no such type exists, return Empty if Force is False or the largest
-- signed integer type if Force is True.
function Integer_Literal
(N : Node_Id;
V : Uint;
Negative : Boolean := False) return Node_Id;
-- Given a non-negative universal integer value, build a typed integer
-- literal node, using the smallest applicable standard integer type.
-- If Negative is true, then a negative literal is built. If V exceeds
-- 2**(System_Max_Integer_Size - 1) - 1, the largest value allowed for
-- perfect result set scaling factors (see RM G.2.3(22)), then Empty is
-- returned. The node N provides the Sloc value for the constructed
-- literal. The Etype of the resulting literal is correctly set, and it
-- is marked as analyzed.
function Real_Literal (N : Node_Id; V : Ureal) return Node_Id;
-- Build a real literal node from the given value, the Etype of the
-- returned node is set to Universal_Real, since all floating-point
-- arithmetic operations that we construct use Universal_Real
function Rounded_Result_Set (N : Node_Id) return Boolean;
-- Returns True if N is a node that contains the Rounded_Result flag
-- and if the flag is true or the target type is an integer type.
procedure Set_Result
(N : Node_Id;
Expr : Node_Id;
Rchk : Boolean := False;
Trunc : Boolean := False);
-- N is the node for the current conversion, division or multiplication
-- operation, and Expr is an expression representing the result. Expr may
-- be of floating-point or integer type. If the operation result is fixed-
-- point, then the value of Expr is in units of small of the result type
-- (i.e. small's have already been dealt with). The result of the call is
-- to replace N by an appropriate conversion to the result type, dealing
-- with rounding for the decimal types case. The node is then analyzed and
-- resolved using the result type. If Rchk or Trunc are True, then
-- respectively Do_Range_Check and Float_Truncate are set in the
-- resulting conversion.
-- Build_Conversion --
function Build_Conversion
(N : Node_Id;
Typ : Entity_Id;
Expr : Node_Id;
Rchk : Boolean := False;
Trunc : Boolean := False) return Node_Id
Loc : constant Source_Ptr := Sloc (N);
Result : Node_Id;
Rcheck : Boolean := Rchk;
-- A special case, if the expression is an integer literal and the
-- target type is an integer type, then just retype the integer
-- literal to the desired target type. Don't do this if we need
-- a range check.
if Nkind (Expr) = N_Integer_Literal
and then Is_Integer_Type (Typ)
and then not Rchk
Result := Expr;
-- Cases where we end up with a conversion. Note that we do not use the
-- Convert_To abstraction here, since we may be decorating the resulting
-- conversion with Rounded_Result and/or Conversion_OK, so we want the
-- conversion node present, even if it appears to be redundant.
-- Remove inner conversion if both inner and outer conversions are
-- to integer types, since the inner one serves no purpose (except
-- perhaps to set rounding, so we preserve the Rounded_Result flag)
-- and also preserve the Conversion_OK and Do_Range_Check flags of
-- the inner conversion.
if Is_Integer_Type (Typ)
and then Is_Integer_Type (Etype (Expr))
and then Nkind (Expr) = N_Type_Conversion
Result :=
Make_Type_Conversion (Loc,
Subtype_Mark => New_Occurrence_Of (Typ, Loc),
Expression => Expression (Expr));
Set_Rounded_Result (Result, Rounded_Result_Set (Expr));
Set_Conversion_OK (Result, Conversion_OK (Expr));
Rcheck := Rcheck or Do_Range_Check (Expr);
-- For all other cases, a simple type conversion will work
Result :=
Make_Type_Conversion (Loc,
Subtype_Mark => New_Occurrence_Of (Typ, Loc),
Expression => Expr);
Set_Float_Truncate (Result, Trunc);
end if;
-- Set Conversion_OK if either result or expression type is a
-- fixed-point type, since from a semantic point of view, we are
-- treating fixed-point values as integers at this stage.
if Is_Fixed_Point_Type (Typ)
or else Is_Fixed_Point_Type (Etype (Expression (Result)))
Set_Conversion_OK (Result);
end if;
-- Set Do_Range_Check if either it was requested by the caller,
-- or if an eliminated inner conversion had a range check.
if Rcheck then
Enable_Range_Check (Result);
Set_Do_Range_Check (Result, False);
end if;
end if;
Set_Etype (Result, Typ);
return Result;
end Build_Conversion;
-- Build_Divide --
function Build_Divide (N : Node_Id; L, R : Node_Id) return Node_Id is
Loc : constant Source_Ptr := Sloc (N);
Left_Type : constant Entity_Id := Base_Type (Etype (L));
Right_Type : constant Entity_Id := Base_Type (Etype (R));
Left_Size : Int;
Right_Size : Int;
Result_Type : Entity_Id;
Rnode : Node_Id;
-- Deal with floating-point case first
if Is_Floating_Point_Type (Left_Type) then
pragma Assert (Left_Type = Universal_Real);
pragma Assert (Right_Type = Universal_Real);
Rnode := Make_Op_Divide (Loc, L, R);
Result_Type := Universal_Real;
-- Integer and fixed-point cases
-- An optimization. If the right operand is the literal 1, then we
-- can just return the left hand operand. Putting the optimization
-- here allows us to omit the check at the call site.
if Nkind (R) = N_Integer_Literal and then Intval (R) = 1 then
return L;
end if;
-- Otherwise we need to figure out the correct result type size
-- First figure out the effective sizes of the operands. Normally
-- the effective size of an operand is the RM_Size of the operand.
-- But a special case arises with operands whose size is known at
-- compile time. In this case, we can use the actual value of the
-- operand to get a size if it would fit in a small signed integer.
Left_Size := UI_To_Int (RM_Size (Left_Type));
if Compile_Time_Known_Value (L) then
Siz : constant Int :=
Get_Size_For_Value (UI_Abs (Expr_Value (L)));
if Siz < Left_Size then
Left_Size := Siz;
end if;
end if;
Right_Size := UI_To_Int (RM_Size (Right_Type));
if Compile_Time_Known_Value (R) then
Siz : constant Int :=
Get_Size_For_Value (UI_Abs (Expr_Value (R)));
if Siz < Right_Size then
Right_Size := Siz;
end if;
end if;
-- Do the operation using the longer of the two sizes
Result_Type :=
Get_Type_For_Size (Int'Max (Left_Size, Right_Size), Force => True);
Rnode :=
Make_Op_Divide (Loc,
Left_Opnd => Build_Conversion (N, Result_Type, L),
Right_Opnd => Build_Conversion (N, Result_Type, R));
end if;
-- We now have a divide node built with Result_Type set. First
-- set Etype of result, as required for all Build_xxx routines
Set_Etype (Rnode, Base_Type (Result_Type));
-- The result is rounded if the target of the operation is decimal
-- and Rounded_Result is set, or if the target of the operation
-- is an integer type, as determined by Rounded_Result_Set.
Set_Rounded_Result (Rnode, Rounded_Result_Set (N));
-- One more check. We did the divide operation using the longer of
-- the two sizes, which is reasonable. However, in the case where the
-- two types have unequal sizes, it is impossible for the result of
-- a divide operation to be larger than the dividend, so we can put
-- a conversion round the result to keep the evolving operation size
-- as small as possible.
if not Is_Floating_Point_Type (Left_Type) then
Rnode := Build_Conversion (N, Left_Type, Rnode);
end if;
return Rnode;
end Build_Divide;
-- Build_Double_Divide --
function Build_Double_Divide
(N : Node_Id;
X, Y, Z : Node_Id) return Node_Id
X_Size : constant Nat := UI_To_Int (RM_Size (Etype (X)));
Y_Size : constant Nat := UI_To_Int (RM_Size (Etype (Y)));
Z_Size : constant Nat := UI_To_Int (RM_Size (Etype (Z)));
D_Size : constant Nat := Y_Size + Z_Size;
M_Size : constant Nat := Nat'Max (X_Size, Nat'Max (Y_Size, Z_Size));
Expr : Node_Id;
-- If the denominator fits in Max_Integer_Size bits, we can build the
-- operations directly without causing any intermediate overflow. But
-- for backward compatibility reasons, we use a 128-bit divide only
-- if one of the operands is already larger than 64 bits.
if D_Size <= System_Max_Integer_Size
and then (D_Size <= 64 or else M_Size > 64)
return Build_Divide (N, X, Build_Multiply (N, Y, Z));
-- Otherwise we use the runtime routine
-- [Qnn : Interfaces.Integer_{64|128};
-- Rnn : Interfaces.Integer_{64|128};
-- Double_Divide{64|128} (X, Y, Z, Qnn, Rnn, Round);
-- Qnn]
Loc : constant Source_Ptr := Sloc (N);
Qnn : Entity_Id;
Rnn : Entity_Id;
Code : List_Id;
pragma Warnings (Off, Rnn);
Build_Double_Divide_Code (N, X, Y, Z, Qnn, Rnn, Code);
Insert_Actions (N, Code);
Expr := New_Occurrence_Of (Qnn, Loc);
-- Set type of result in case used elsewhere (see note at start)
Set_Etype (Expr, Etype (Qnn));
-- Set result as analyzed (see note at start on build routines)
return Expr;
end if;
end Build_Double_Divide;
-- Build_Double_Divide_Code --
-- If the denominator can be computed in Max_Integer_Size bits, we build
-- [Nnn : constant typ := typ (X);
-- Dnn : constant typ := typ (Y) * typ (Z)
-- Qnn : constant typ := Nnn / Dnn;
-- Rnn : constant typ := Nnn rem Dnn;
-- If the denominator cannot be computed in Max_Integer_Size bits, we build
-- [Qnn : Interfaces.Integer_{64|128};
-- Rnn : Interfaces.Integer_{64|128};
-- Double_Divide{64|128} (X, Y, Z, Qnn, Rnn, Round);]
procedure Build_Double_Divide_Code
(N : Node_Id;
X, Y, Z : Node_Id;
Qnn, Rnn : out Entity_Id;
Code : out List_Id)
Loc : constant Source_Ptr := Sloc (N);
X_Size : constant Nat := UI_To_Int (RM_Size (Etype (X)));
Y_Size : constant Nat := UI_To_Int (RM_Size (Etype (Y)));
Z_Size : constant Nat := UI_To_Int (RM_Size (Etype (Z)));
M_Size : constant Nat := Nat'Max (X_Size, Nat'Max (Y_Size, Z_Size));
QR_Id : RE_Id;
QR_Siz : Nat;
QR_Typ : Entity_Id;
Nnn : Entity_Id;
Dnn : Entity_Id;
Quo : Node_Id;
Rnd : Entity_Id;
-- Find type that will allow computation of denominator
QR_Siz := Nat'Max (X_Size, Y_Size + Z_Size);
if QR_Siz <= 16 then
QR_Typ := Standard_Integer_16;
QR_Id := RE_Null;
elsif QR_Siz <= 32 then
QR_Typ := Standard_Integer_32;
QR_Id := RE_Null;
elsif QR_Siz <= 64 then
QR_Typ := Standard_Integer_64;
QR_Id := RE_Null;
-- For backward compatibility reasons, we use a 128-bit divide only
-- if one of the operands is already larger than 64 bits.
elsif System_Max_Integer_Size < 128 or else M_Size <= 64 then
QR_Typ := RTE (RE_Integer_64);
QR_Id := RE_Double_Divide64;
elsif QR_Siz <= 128 then
QR_Typ := Standard_Integer_128;
QR_Id := RE_Null;
QR_Typ := RTE (RE_Integer_128);
QR_Id := RE_Double_Divide128;
end if;
-- Define quotient and remainder, and set their Etypes, so
-- that they can be picked up by Build_xxx routines.
Qnn := Make_Temporary (Loc, 'S');
Rnn := Make_Temporary (Loc, 'R');
Set_Etype (Qnn, QR_Typ);
Set_Etype (Rnn, QR_Typ);
-- Case where we can compute the denominator in Max_Integer_Size bits
if QR_Id = RE_Null then
-- Create temporaries for numerator and denominator and set Etypes,
-- so that New_Occurrence_Of picks them up for Build_xxx calls.
Nnn := Make_Temporary (Loc, 'N');
Dnn := Make_Temporary (Loc, 'D');
Set_Etype (Nnn, QR_Typ);
Set_Etype (Dnn, QR_Typ);
Code := New_List (
Make_Object_Declaration (Loc,
Defining_Identifier => Nnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc),
Constant_Present => True,
Expression => Build_Conversion (N, QR_Typ, X)),
Make_Object_Declaration (Loc,
Defining_Identifier => Dnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc),
Constant_Present => True,
Expression => Build_Multiply (N, Y, Z)));
Quo :=
Build_Divide (N,
New_Occurrence_Of (Nnn, Loc),
New_Occurrence_Of (Dnn, Loc));
Set_Rounded_Result (Quo, Rounded_Result_Set (N));
Append_To (Code,
Make_Object_Declaration (Loc,
Defining_Identifier => Qnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc),
Constant_Present => True,
Expression => Quo));
Append_To (Code,
Make_Object_Declaration (Loc,
Defining_Identifier => Rnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc),
Constant_Present => True,
Expression =>
Build_Rem (N,
New_Occurrence_Of (Nnn, Loc),
New_Occurrence_Of (Dnn, Loc))));
-- Case where denominator does not fit in Max_Integer_Size bits, we have
-- to call the runtime routine to compute the quotient and remainder.
Rnd := Boolean_Literals (Rounded_Result_Set (N));
Code := New_List (
Make_Object_Declaration (Loc,
Defining_Identifier => Qnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc)),
Make_Object_Declaration (Loc,
Defining_Identifier => Rnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc)),
Make_Procedure_Call_Statement (Loc,
Name => New_Occurrence_Of (RTE (QR_Id), Loc),
Parameter_Associations => New_List (
Build_Conversion (N, QR_Typ, X),
Build_Conversion (N, QR_Typ, Y),
Build_Conversion (N, QR_Typ, Z),
New_Occurrence_Of (Qnn, Loc),
New_Occurrence_Of (Rnn, Loc),
New_Occurrence_Of (Rnd, Loc))));
end if;
end Build_Double_Divide_Code;
-- Build_Multiply --
function Build_Multiply (N : Node_Id; L, R : Node_Id) return Node_Id is
Loc : constant Source_Ptr := Sloc (N);
Left_Type : constant Entity_Id := Etype (L);
Right_Type : constant Entity_Id := Etype (R);
Left_Size : Int;
Right_Size : Int;
Result_Type : Entity_Id;
Rnode : Node_Id;
-- Deal with floating-point case first
if Is_Floating_Point_Type (Left_Type) then
pragma Assert (Left_Type = Universal_Real);
pragma Assert (Right_Type = Universal_Real);
Result_Type := Universal_Real;
Rnode := Make_Op_Multiply (Loc, L, R);
-- Integer and fixed-point cases
-- An optimization. If the right operand is the literal 1, then we
-- can just return the left hand operand. Putting the optimization
-- here allows us to omit the check at the call site. Similarly, if
-- the left operand is the integer 1 we can return the right operand.
if Nkind (R) = N_Integer_Literal and then Intval (R) = 1 then
return L;
elsif Nkind (L) = N_Integer_Literal and then Intval (L) = 1 then
return R;
end if;
-- Otherwise we need to figure out the correct result type size
-- First figure out the effective sizes of the operands. Normally
-- the effective size of an operand is the RM_Size of the operand.
-- But a special case arises with operands whose size is known at
-- compile time. In this case, we can use the actual value of the
-- operand to get a size if it would fit in a small signed integer.
Left_Size := UI_To_Int (RM_Size (Left_Type));
if Compile_Time_Known_Value (L) then
Siz : constant Int :=
Get_Size_For_Value (UI_Abs (Expr_Value (L)));
if Siz < Left_Size then
Left_Size := Siz;
end if;
end if;
Right_Size := UI_To_Int (RM_Size (Right_Type));
if Compile_Time_Known_Value (R) then
Siz : constant Int :=
Get_Size_For_Value (UI_Abs (Expr_Value (R)));
if Siz < Right_Size then
Right_Size := Siz;
end if;
end if;
-- Now the result size must be at least the sum of the two sizes,
-- to accommodate all possible results.
Result_Type :=
Get_Type_For_Size (Left_Size + Right_Size, Force => True);
Rnode :=
Make_Op_Multiply (Loc,
Left_Opnd => Build_Conversion (N, Result_Type, L),
Right_Opnd => Build_Conversion (N, Result_Type, R));
end if;
-- We now have a multiply node built with Result_Type set. First
-- set Etype of result, as required for all Build_xxx routines
Set_Etype (Rnode, Base_Type (Result_Type));
return Rnode;
end Build_Multiply;
-- Build_Rem --
function Build_Rem (N : Node_Id; L, R : Node_Id) return Node_Id is
Loc : constant Source_Ptr := Sloc (N);
Left_Type : constant Entity_Id := Etype (L);
Right_Type : constant Entity_Id := Etype (R);
Result_Type : Entity_Id;
Rnode : Node_Id;
if Left_Type = Right_Type then
Result_Type := Left_Type;
Rnode :=
Make_Op_Rem (Loc,
Left_Opnd => L,
Right_Opnd => R);
-- If left size is larger, we do the remainder operation using the
-- size of the left type (i.e. the larger of the two integer types).
elsif Esize (Left_Type) >= Esize (Right_Type) then
Result_Type := Left_Type;
Rnode :=
Make_Op_Rem (Loc,
Left_Opnd => L,
Right_Opnd => Build_Conversion (N, Left_Type, R));
-- Similarly, if the right size is larger, we do the remainder
-- operation using the right type.
Result_Type := Right_Type;
Rnode :=
Make_Op_Rem (Loc,
Left_Opnd => Build_Conversion (N, Right_Type, L),
Right_Opnd => R);
end if;
-- We now have an N_Op_Rem node built with Result_Type set. First
-- set Etype of result, as required for all Build_xxx routines
Set_Etype (Rnode, Base_Type (Result_Type));
-- One more check. We did the rem operation using the larger of the
-- two types, which is reasonable. However, in the case where the
-- two types have unequal sizes, it is impossible for the result of
-- a remainder operation to be larger than the smaller of the two
-- types, so we can put a conversion round the result to keep the
-- evolving operation size as small as possible.
if Esize (Left_Type) >= Esize (Right_Type) then
Rnode := Build_Conversion (N, Right_Type, Rnode);
elsif Esize (Right_Type) >= Esize (Left_Type) then
Rnode := Build_Conversion (N, Left_Type, Rnode);
end if;
return Rnode;
end Build_Rem;
-- Build_Scaled_Divide --
function Build_Scaled_Divide
(N : Node_Id;
X, Y, Z : Node_Id) return Node_Id
X_Size : constant Nat := UI_To_Int (RM_Size (Etype (X)));
Y_Size : constant Nat := UI_To_Int (RM_Size (Etype (Y)));
Z_Size : constant Nat := UI_To_Int (RM_Size (Etype (Z)));
N_Size : constant Nat := X_Size + Y_Size;
M_Size : constant Nat := Nat'Max (X_Size, Nat'Max (Y_Size, Z_Size));
Expr : Node_Id;
-- If the numerator fits in Max_Integer_Size bits, we can build the
-- operations directly without causing any intermediate overflow. But
-- for backward compatibility reasons, we use a 128-bit divide only
-- if one of the operands is already larger than 64 bits.
if N_Size <= System_Max_Integer_Size
and then (N_Size <= 64 or else M_Size > 64)
return Build_Divide (N, Build_Multiply (N, X, Y), Z);
-- Otherwise we use the runtime routine
-- [Qnn : Integer_{64|128},
-- Rnn : Integer_{64|128};
-- Scaled_Divide{64|128} (X, Y, Z, Qnn, Rnn, Round);
-- Qnn]
Loc : constant Source_Ptr := Sloc (N);
Qnn : Entity_Id;
Rnn : Entity_Id;
Code : List_Id;
pragma Warnings (Off, Rnn);
Build_Scaled_Divide_Code (N, X, Y, Z, Qnn, Rnn, Code);
Insert_Actions (N, Code);
Expr := New_Occurrence_Of (Qnn, Loc);
-- Set type of result in case used elsewhere (see note at start)
Set_Etype (Expr, Etype (Qnn));
return Expr;
end if;
end Build_Scaled_Divide;
-- Build_Scaled_Divide_Code --
-- If the numerator can be computed in Max_Integer_Size bits, we build
-- [Nnn : constant typ := typ (X) * typ (Y);
-- Dnn : constant typ := typ (Z)
-- Qnn : constant typ := Nnn / Dnn;
-- Rnn : constant typ := Nnn rem Dnn;
-- If the numerator cannot be computed in Max_Integer_Size bits, we build
-- [Qnn : Interfaces.Integer_{64|128};
-- Rnn : Interfaces.Integer_{64|128};
-- Scaled_Divide_{64|128} (X, Y, Z, Qnn, Rnn, Round);]
procedure Build_Scaled_Divide_Code
(N : Node_Id;
X, Y, Z : Node_Id;
Qnn, Rnn : out Entity_Id;
Code : out List_Id)
Loc : constant Source_Ptr := Sloc (N);
X_Size : constant Nat := UI_To_Int (RM_Size (Etype (X)));
Y_Size : constant Nat := UI_To_Int (RM_Size (Etype (Y)));
Z_Size : constant Nat := UI_To_Int (RM_Size (Etype (Z)));
M_Size : constant Nat := Nat'Max (X_Size, Nat'Max (Y_Size, Z_Size));
QR_Id : RE_Id;
QR_Siz : Nat;
QR_Typ : Entity_Id;
Nnn : Entity_Id;
Dnn : Entity_Id;
Quo : Node_Id;
Rnd : Entity_Id;
-- Find type that will allow computation of numerator
QR_Siz := Nat'Max (X_Size + Y_Size, Z_Size);
if QR_Siz <= 16 then
QR_Typ := Standard_Integer_16;
QR_Id := RE_Null;
elsif QR_Siz <= 32 then
QR_Typ := Standard_Integer_32;
QR_Id := RE_Null;
elsif QR_Siz <= 64 then
QR_Typ := Standard_Integer_64;
QR_Id := RE_Null;
-- For backward compatibility reasons, we use a 128-bit divide only
-- if one of the operands is already larger than 64 bits.
elsif System_Max_Integer_Size < 128 or else M_Size <= 64 then
QR_Typ := RTE (RE_Integer_64);
QR_Id := RE_Scaled_Divide64;
elsif QR_Siz <= 128 then
QR_Typ := Standard_Integer_128;
QR_Id := RE_Null;
QR_Typ := RTE (RE_Integer_128);
QR_Id := RE_Scaled_Divide128;
end if;
-- Define quotient and remainder, and set their Etypes, so
-- that they can be picked up by Build_xxx routines.
Qnn := Make_Temporary (Loc, 'S');
Rnn := Make_Temporary (Loc, 'R');
Set_Etype (Qnn, QR_Typ);
Set_Etype (Rnn, QR_Typ);
-- Case where we can compute the numerator in Max_Integer_Size bits
if QR_Id = RE_Null then
Nnn := Make_Temporary (Loc, 'N');
Dnn := Make_Temporary (Loc, 'D');
-- Set Etypes, so that they can be picked up by New_Occurrence_Of
Set_Etype (Nnn, QR_Typ);
Set_Etype (Dnn, QR_Typ);
Code := New_List (
Make_Object_Declaration (Loc,
Defining_Identifier => Nnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc),
Constant_Present => True,
Expression => Build_Multiply (N, X, Y)),
Make_Object_Declaration (Loc,
Defining_Identifier => Dnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc),
Constant_Present => True,
Expression => Build_Conversion (N, QR_Typ, Z)));
Quo :=
Build_Divide (N,
New_Occurrence_Of (Nnn, Loc),
New_Occurrence_Of (Dnn, Loc));
Append_To (Code,
Make_Object_Declaration (Loc,
Defining_Identifier => Qnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc),
Constant_Present => True,
Expression => Quo));
Append_To (Code,
Make_Object_Declaration (Loc,
Defining_Identifier => Rnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc),
Constant_Present => True,
Expression =>
Build_Rem (N,
New_Occurrence_Of (Nnn, Loc),
New_Occurrence_Of (Dnn, Loc))));
-- Case where numerator does not fit in Max_Integer_Size bits, we have
-- to call the runtime routine to compute the quotient and remainder.
Rnd := Boolean_Literals (Rounded_Result_Set (N));
Code := New_List (
Make_Object_Declaration (Loc,
Defining_Identifier => Qnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc)),
Make_Object_Declaration (Loc,
Defining_Identifier => Rnn,
Object_Definition => New_Occurrence_Of (QR_Typ, Loc)),
Make_Procedure_Call_Statement (Loc,
Name => New_Occurrence_Of (RTE (QR_Id), Loc),
Parameter_Associations => New_List (
Build_Conversion (N, QR_Typ, X),
Build_Conversion (N, QR_Typ, Y),
Build_Conversion (N, QR_Typ, Z),
New_Occurrence_Of (Qnn, Loc),
New_Occurrence_Of (Rnn, Loc),
New_Occurrence_Of (Rnd, Loc))));
end if;
-- Set type of result, for use in caller
Set_Etype (Qnn, QR_Typ);
end Build_Scaled_Divide_Code;
-- Do_Divide_Fixed_Fixed --
-- We have:
-- (Result_Value * Result_Small) =
-- (Left_Value * Left_Small) / (Right_Value * Right_Small)
-- Result_Value = (Left_Value / Right_Value) *
-- (Left_Small / (Right_Small * Result_Small));
-- we can do the operation in integer arithmetic if this fraction is an
-- integer or the reciprocal of an integer, as detailed in (RM G.2.3(21)).
-- Otherwise the result is in the close result set and our approach is to
-- use floating-point to compute this close result.
procedure Do_Divide_Fixed_Fixed (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
Left_Type : constant Entity_Id := Etype (Left);
Right_Type : constant Entity_Id := Etype (Right);
Result_Type : constant Entity_Id := Etype (N);
Right_Small : constant Ureal := Small_Value (Right_Type);
Left_Small : constant Ureal := Small_Value (Left_Type);
Result_Small : Ureal;
Frac : Ureal;
Frac_Num : Uint;
Frac_Den : Uint;
Lit_Int : Node_Id;
-- Rounding is required if the result is integral
if Is_Integer_Type (Result_Type) then
Set_Rounded_Result (N);
end if;
-- Get result small. If the result is an integer, treat it as though
-- it had a small of 1.0, all other processing is identical.
if Is_Integer_Type (Result_Type) then
Result_Small := Ureal_1;
Result_Small := Small_Value (Result_Type);
end if;
-- Get small ratio
Frac := Left_Small / (Right_Small * Result_Small);
Frac_Num := Norm_Num (Frac);
Frac_Den := Norm_Den (Frac);
-- If the fraction is an integer, then we get the result by multiplying
-- the left operand by the integer, and then dividing by the right
-- operand (the order is important, if we did the divide first, we
-- would lose precision).
if Frac_Den = 1 then
Lit_Int := Integer_Literal (N, Frac_Num); -- always positive
if Present (Lit_Int) then
Set_Result (N, Build_Scaled_Divide (N, Left, Lit_Int, Right));
end if;
-- If the fraction is the reciprocal of an integer, then we get the
-- result by first multiplying the divisor by the integer, and then
-- doing the division with the adjusted divisor.
-- Note: this is much better than doing two divisions: multiplications
-- are much faster than divisions (and certainly faster than rounded
-- divisions), and we don't get inaccuracies from double rounding.
elsif Frac_Num = 1 then
Lit_Int := Integer_Literal (N, Frac_Den); -- always positive
if Present (Lit_Int) then
Set_Result (N, Build_Double_Divide (N, Left, Right, Lit_Int));
end if;
end if;
-- If we fall through, we use floating-point to compute the result
Set_Result (N,
Build_Multiply (N,
Build_Divide (N, Fpt_Value (Left), Fpt_Value (Right)),
Real_Literal (N, Frac)));
end Do_Divide_Fixed_Fixed;
-- Do_Divide_Fixed_Universal --
-- We have:
-- (Result_Value * Result_Small) = (Left_Value * Left_Small) / Lit_Value;
-- Result_Value = Left_Value * Left_Small /(Lit_Value * Result_Small);
-- The result is required to be in the perfect result set if the literal
-- can be factored so that the resulting small ratio is an integer or the
-- reciprocal of an integer (RM G.2.3(21-22)). We now give a detailed
-- analysis of these RM requirements:
-- We must factor the literal, finding an integer K:
-- Lit_Value = K * Right_Small
-- Right_Small = Lit_Value / K
-- such that the small ratio:
-- Left_Small
-- ------------------------------
-- (Lit_Value / K) * Result_Small
-- Left_Small
-- = ------------------------ * K
-- Lit_Value * Result_Small
-- is an integer or the reciprocal of an integer, and for
-- implementation efficiency we need the smallest such K.
-- First we reduce the left fraction to lowest terms
-- If numerator = 1, then for K = 1, the small ratio is the reciprocal
-- of an integer, and this is clearly the minimum K case, so set K = 1,
-- Right_Small = Lit_Value.
-- If numerator > 1, then set K to the denominator of the fraction so
-- that the resulting small ratio is an integer (the numerator value).
procedure Do_Divide_Fixed_Universal (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
Left_Type : constant Entity_Id := Etype (Left);
Result_Type : constant Entity_Id := Etype (N);
Left_Small : constant Ureal := Small_Value (Left_Type);
Lit_Value : constant Ureal := Realval (Right);
Result_Small : Ureal;
Frac : Ureal;
Frac_Num : Uint;
Frac_Den : Uint;
Lit_K : Node_Id;
Lit_Int : Node_Id;
-- Get result small. If the result is an integer, treat it as though
-- it had a small of 1.0, all other processing is identical.
if Is_Integer_Type (Result_Type) then
Result_Small := Ureal_1;
Result_Small := Small_Value (Result_Type);
end if;
-- Determine if literal can be rewritten successfully
Frac := Left_Small / (Lit_Value * Result_Small);
Frac_Num := Norm_Num (Frac);
Frac_Den := Norm_Den (Frac);
-- Case where fraction is the reciprocal of an integer (K = 1, integer
-- = denominator). If this integer is not too large, this is the case
-- where the result can be obtained by dividing by this integer value.
if Frac_Num = 1 then
Lit_Int := Integer_Literal (N, Frac_Den, UR_Is_Negative (Frac));
if Present (Lit_Int) then
Set_Result (N, Build_Divide (N, Left, Lit_Int));
end if;
-- Case where we choose K to make fraction an integer (K = denominator
-- of fraction, integer = numerator of fraction). If both K and the
-- numerator are small enough, this is the case where the result can
-- be obtained by first multiplying by the integer value and then
-- dividing by K (the order is important, if we divided first, we
-- would lose precision).
Lit_Int := Integer_Literal (N, Frac_Num, UR_Is_Negative (Frac));
Lit_K := Integer_Literal (N, Frac_Den, False);
if Present (Lit_Int) and then Present (Lit_K) then
Set_Result (N, Build_Scaled_Divide (N, Left, Lit_Int, Lit_K));
end if;
end if;
-- Fall through if the literal cannot be successfully rewritten, or if
-- the small ratio is out of range of integer arithmetic. In the former
-- case it is fine to use floating-point to get the close result set,
-- and in the latter case, it means that the result is zero or raises
-- constraint error, and we can do that accurately in floating-point.
-- If we end up using floating-point, then we take the right integer
-- to be one, and its small to be the value of the original right real
-- literal. That way, we need only one floating-point multiplication.
Set_Result (N,
Build_Multiply (N, Fpt_Value (Left), Real_Literal (N, Frac)));
end Do_Divide_Fixed_Universal;
-- Do_Divide_Universal_Fixed --
-- We have:
-- (Result_Value * Result_Small) =
-- Lit_Value / (Right_Value * Right_Small)
-- Result_Value =
-- (Lit_Value / (Right_Small * Result_Small)) / Right_Value
-- The result is required to be in the perfect result set if the literal
-- can be factored so that the resulting small ratio is an integer or the
-- reciprocal of an integer (RM G.2.3(21-22)). We now give a detailed
-- analysis of these RM requirements:
-- We must factor the literal, finding an integer K:
-- Lit_Value = K * Left_Small
-- Left_Small = Lit_Value / K
-- such that the small ratio:
-- (Lit_Value / K)
-- --------------------------
-- Right_Small * Result_Small
-- Lit_Value 1
-- = -------------------------- * -
-- Right_Small * Result_Small K
-- is an integer or the reciprocal of an integer, and for
-- implementation efficiency we need the smallest such K.
-- First we reduce the left fraction to lowest terms
-- If denominator = 1, then for K = 1, the small ratio is an integer
-- (the numerator) and this is clearly the minimum K case, so set K = 1,
-- and Left_Small = Lit_Value.
-- If denominator > 1, then set K to the numerator of the fraction so
-- that the resulting small ratio is the reciprocal of an integer (the
-- numerator value).
procedure Do_Divide_Universal_Fixed (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
Right_Type : constant Entity_Id := Etype (Right);
Result_Type : constant Entity_Id := Etype (N);
Right_Small : constant Ureal := Small_Value (Right_Type);
Lit_Value : constant Ureal := Realval (Left);
Result_Small : Ureal;
Frac : Ureal;
Frac_Num : Uint;
Frac_Den : Uint;
Lit_K : Node_Id;
Lit_Int : Node_Id;
-- Get result small. If the result is an integer, treat it as though
-- it had a small of 1.0, all other processing is identical.
if Is_Integer_Type (Result_Type) then
Result_Small := Ureal_1;
Result_Small := Small_Value (Result_Type);
end if;
-- Determine if literal can be rewritten successfully
Frac := Lit_Value / (Right_Small * Result_Small);
Frac_Num := Norm_Num (Frac);
Frac_Den := Norm_Den (Frac);
-- Case where fraction is an integer (K = 1, integer = numerator). If
-- this integer is not too large, this is the case where the result
-- can be obtained by dividing this integer by the right operand.
if Frac_Den = 1 then
Lit_Int := Integer_Literal (N, Frac_Num, UR_Is_Negative (Frac));
if Present (Lit_Int) then
Set_Result (N, Build_Divide (N, Lit_Int, Right));
end if;
-- Case where we choose K to make the fraction the reciprocal of an
-- integer (K = numerator of fraction, integer = numerator of fraction).
-- If both K and the integer are small enough, this is the case where
-- the result can be obtained by multiplying the right operand by K
-- and then dividing by the integer value. The order of the operations
-- is important (if we divided first, we would lose precision).
Lit_Int := Integer_Literal (N, Frac_Den, UR_Is_Negative (Frac));
Lit_K := Integer_Literal (N, Frac_Num, False);
if Present (Lit_Int) and then Present (Lit_K) then
Set_Result (N, Build_Double_Divide (N, Lit_K, Right, Lit_Int));
end if;
end if;
-- Fall through if the literal cannot be successfully rewritten, or if
-- the small ratio is out of range of integer arithmetic. In the former
-- case it is fine to use floating-point to get the close result set,
-- and in the latter case, it means that the result is zero or raises
-- constraint error, and we can do that accurately in floating-point.
-- If we end up using floating-point, then we take the right integer
-- to be one, and its small to be the value of the original right real
-- literal. That way, we need only one floating-point division.
Set_Result (N,
Build_Divide (N, Real_Literal (N, Frac), Fpt_Value (Right)));
end Do_Divide_Universal_Fixed;
-- Do_Multiply_Fixed_Fixed --
-- We have:
-- (Result_Value * Result_Small) =
-- (Left_Value * Left_Small) * (Right_Value * Right_Small)
-- Result_Value = (Left_Value * Right_Value) *
-- (Left_Small * Right_Small) / Result_Small;
-- we can do the operation in integer arithmetic if this fraction is an
-- integer or the reciprocal of an integer, as detailed in (RM G.2.3(21)).
-- Otherwise the result is in the close result set and our approach is to
-- use floating-point to compute this close result.
procedure Do_Multiply_Fixed_Fixed (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
Left_Type : constant Entity_Id := Etype (Left);
Right_Type : constant Entity_Id := Etype (Right);
Result_Type : constant Entity_Id := Etype (N);
Right_Small : constant Ureal := Small_Value (Right_Type);
Left_Small : constant Ureal := Small_Value (Left_Type);
Result_Small : Ureal;
Frac : Ureal;
Frac_Num : Uint;
Frac_Den : Uint;
Lit_Int : Node_Id;
-- Get result small. If the result is an integer, treat it as though
-- it had a small of 1.0, all other processing is identical.
if Is_Integer_Type (Result_Type) then
Result_Small := Ureal_1;
Result_Small := Small_Value (Result_Type);
end if;
-- Get small ratio
Frac := (Left_Small * Right_Small) / Result_Small;
Frac_Num := Norm_Num (Frac);
Frac_Den := Norm_Den (Frac);
-- If the fraction is an integer, then we get the result by multiplying
-- the operands, and then multiplying the result by the integer value.
if Frac_Den = 1 then
Lit_Int := Integer_Literal (N, Frac_Num); -- always positive
if Present (Lit_Int) then
Set_Result (N,
Build_Multiply (N, Build_Multiply (N, Left, Right), Lit_Int));
end if;
-- If the fraction is the reciprocal of an integer, then we get the
-- result by multiplying the operands, and then dividing the result by
-- the integer value. The order of the operations is important, if we
-- divided first, we would lose precision.
elsif Frac_Num = 1 then
Lit_Int := Integer_Literal (N, Frac_Den); -- always positive
if Present (Lit_Int) then
Set_Result (N, Build_Scaled_Divide (N, Left, Right, Lit_Int));
end if;
end if;
-- If we fall through, we use floating-point to compute the result
Set_Result (N,
Build_Multiply (N,
Build_Multiply (N, Fpt_Value (Left), Fpt_Value (Right)),
Real_Literal (N, Frac)));
end Do_Multiply_Fixed_Fixed;
-- Do_Multiply_Fixed_Universal --
-- We have:
-- (Result_Value * Result_Small) = (Left_Value * Left_Small) * Lit_Value;
-- Result_Value = Left_Value * (Left_Small * Lit_Value) / Result_Small;
-- The result is required to be in the perfect result set if the literal
-- can be factored so that the resulting small ratio is an integer or the
-- reciprocal of an integer (RM G.2.3(21-22)). We now give a detailed
-- analysis of these RM requirements:
-- We must factor the literal, finding an integer K:
-- Lit_Value = K * Right_Small
-- Right_Small = Lit_Value / K
-- such that the small ratio:
-- Left_Small * (Lit_Value / K)
-- ----------------------------
-- Result_Small
-- Left_Small * Lit_Value 1
-- = ---------------------- * -
-- Result_Small K
-- is an integer or the reciprocal of an integer, and for
-- implementation efficiency we need the smallest such K.
-- First we reduce the left fraction to lowest terms
-- If denominator = 1, then for K = 1, the small ratio is an integer, and
-- this is clearly the minimum K case, so set
-- K = 1, Right_Small = Lit_Value
-- If denominator > 1, then set K to the numerator of the fraction, so
-- that the resulting small ratio is the reciprocal of the integer (the
-- denominator value).
procedure Do_Multiply_Fixed_Universal
(N : Node_Id;
Left, Right : Node_Id)
Left_Type : constant Entity_Id := Etype (Left);
Result_Type : constant Entity_Id := Etype (N);
Left_Small : constant Ureal := Small_Value (Left_Type);
Lit_Value : constant Ureal := Realval (Right);
Result_Small : Ureal;
Frac : Ureal;
Frac_Num : Uint;
Frac_Den : Uint;
Lit_K : Node_Id;
Lit_Int : Node_Id;
-- Get result small. If the result is an integer, treat it as though
-- it had a small of 1.0, all other processing is identical.
if Is_Integer_Type (Result_Type) then
Result_Small := Ureal_1;
Result_Small := Small_Value (Result_Type);
end if;
-- Determine if literal can be rewritten successfully
Frac := (Left_Small * Lit_Value) / Result_Small;
Frac_Num := Norm_Num (Frac);
Frac_Den := Norm_Den (Frac);
-- Case where fraction is an integer (K = 1, integer = numerator). If
-- this integer is not too large, this is the case where the result can
-- be obtained by multiplying by this integer value.
if Frac_Den = 1 then
Lit_Int := Integer_Literal (N, Frac_Num, UR_Is_Negative (Frac));
if Present (Lit_Int) then
Set_Result (N, Build_Multiply (N, Left, Lit_Int));
end if;
-- Case where we choose K to make fraction the reciprocal of an integer
-- (K = numerator of fraction, integer = denominator of fraction). If
-- both K and the denominator are small enough, this is the case where
-- the result can be obtained by first multiplying by K, and then
-- dividing by the integer value.
Lit_Int := Integer_Literal (N, Frac_Den, UR_Is_Negative (Frac));
Lit_K := Integer_Literal (N, Frac_Num, False);
if Present (Lit_Int) and then Present (Lit_K) then
Set_Result (N, Build_Scaled_Divide (N, Left, Lit_K, Lit_Int));
end if;
end if;
-- Fall through if the literal cannot be successfully rewritten, or if
-- the small ratio is out of range of integer arithmetic. In the former
-- case it is fine to use floating-point to get the close result set,
-- and in the latter case, it means that the result is zero or raises
-- constraint error, and we can do that accurately in floating-point.
-- If we end up using floating-point, then we take the right integer
-- to be one, and its small to be the value of the original right real
-- literal. That way, we need only one floating-point multiplication.
Set_Result (N,
Build_Multiply (N, Fpt_Value (Left), Real_Literal (N, Frac)));
end Do_Multiply_Fixed_Universal;
-- Expand_Convert_Fixed_Static --
procedure Expand_Convert_Fixed_Static (N : Node_Id) is
Rewrite (N,
Convert_To (Etype (N),
Make_Real_Literal (Sloc (N), Expr_Value_R (Expression (N)))));
Analyze_And_Resolve (N);
end Expand_Convert_Fixed_Static;
-- Expand_Convert_Fixed_To_Fixed --
-- We have:
-- Result_Value * Result_Small = Source_Value * Source_Small
-- Result_Value = Source_Value * (Source_Small / Result_Small)
-- If the small ratio (Source_Small / Result_Small) is a sufficiently small
-- integer, then the perfect result set is obtained by a single integer
-- multiplication.
-- If the small ratio is the reciprocal of a sufficiently small integer,
-- then the perfect result set is obtained by a single integer division.
-- If the numerator and denominator of the small ratio are sufficiently
-- small integers, then the perfect result set is obtained by a scaled
-- divide operation.
-- In other cases, we obtain the close result set by calculating the
-- result in floating-point.
procedure Expand_Convert_Fixed_To_Fixed (N : Node_Id) is
Rng_Check : constant Boolean := Do_Range_Check (N);
Expr : constant Node_Id := Expression (N);
Result_Type : constant Entity_Id := Etype (N);
Source_Type : constant Entity_Id := Etype (Expr);
Small_Ratio : Ureal;
Ratio_Num : Uint;
Ratio_Den : Uint;
Lit_Num : Node_Id;
Lit_Den : Node_Id;
if Is_OK_Static_Expression (Expr) then
Expand_Convert_Fixed_Static (N);
end if;
Small_Ratio := Small_Value (Source_Type) / Small_Value (Result_Type);
Ratio_Num := Norm_Num (Small_Ratio);
Ratio_Den := Norm_Den (Small_Ratio);
if Ratio_Den = 1 then
if Ratio_Num = 1 then
Set_Result (N, Expr);
Lit_Num := Integer_Literal (N, Ratio_Num);
if Present (Lit_Num) then
Set_Result (N, Build_Multiply (N, Expr, Lit_Num));
end if;
end if;
elsif Ratio_Num = 1 then
Lit_Den := Integer_Literal (N, Ratio_Den);
if Present (Lit_Den) then
Set_Result (N, Build_Divide (N, Expr, Lit_Den), Rng_Check);
end if;
Lit_Num := Integer_Literal (N, Ratio_Num);
Lit_Den := Integer_Literal (N, Ratio_Den);
if Present (Lit_Num) and then Present (Lit_Den) then
(N, Build_Scaled_Divide (N, Expr, Lit_Num, Lit_Den), Rng_Check);
end if;
end if;
-- Fall through to use floating-point for the close result set case,
-- as a result of the numerator or denominator of the small ratio not
-- being a sufficiently small integer.
Set_Result (N,
Build_Multiply (N,
Fpt_Value (Expr),
Real_Literal (N, Small_Ratio)),
end Expand_Convert_Fixed_To_Fixed;
-- Expand_Convert_Fixed_To_Float --
-- If the small of the fixed type is 1.0, then we simply convert the
-- integer value directly to the target floating-point type, otherwise
-- we first have to multiply by the small, in Universal_Real, and then
-- convert the result to the target floating-point type.
procedure Expand_Convert_Fixed_To_Float (N : Node_Id) is
Rng_Check : constant Boolean := Do_Range_Check (N);
Expr : constant Node_Id := Expression (N);
Source_Type : constant Entity_Id := Etype (Expr);
Small : constant Ureal := Small_Value (Source_Type);
if Is_OK_Static_Expression (Expr) then
Expand_Convert_Fixed_Static (N);
end if;
if Small = Ureal_1 then
Set_Result (N, Expr);
Set_Result (N,
Build_Multiply (N,
Fpt_Value (Expr),
Real_Literal (N, Small)),
end if;
end Expand_Convert_Fixed_To_Float;
-- Expand_Convert_Fixed_To_Integer --
-- We have:
-- Result_Value = Source_Value * Source_Small
-- If the small value is a sufficiently small integer, then the perfect
-- result set is obtained by a single integer multiplication.
-- If the small value is the reciprocal of a sufficiently small integer,
-- then the perfect result set is obtained by a single integer division.
-- If the numerator and denominator of the small value are sufficiently
-- small integers, then the perfect result set is obtained by a scaled
-- divide operation.
-- In other cases, we obtain the close result set by calculating the
-- result in floating-point.
procedure Expand_Convert_Fixed_To_Integer (N : Node_Id) is
Rng_Check : constant Boolean := Do_Range_Check (N);
Expr : constant Node_Id := Expression (N);
Source_Type : constant Entity_Id := Etype (Expr);
Small : constant Ureal := Small_Value (Source_Type);
Small_Num : constant Uint := Norm_Num (Small);
Small_Den : constant Uint := Norm_Den (Small);
Lit_Num : Node_Id;
Lit_Den : Node_Id;
if Is_OK_Static_Expression (Expr) then
Expand_Convert_Fixed_Static (N);
end if;
if Small_Den = 1 then
Lit_Num := Integer_Literal (N, Small_Num);
if Present (Lit_Num) then
Set_Result (N, Build_Multiply (N, Expr, Lit_Num), Rng_Check);
end if;
elsif Small_Num = 1 then
Lit_Den := Integer_Literal (N, Small_Den);
if Present (Lit_Den) then
Set_Result (N, Build_Divide (N, Expr, Lit_Den), Rng_Check);
end if;
Lit_Num := Integer_Literal (N, Small_Num);
Lit_Den := Integer_Literal (N, Small_Den);
if Present (Lit_Num) and then Present (Lit_Den) then
(N, Build_Scaled_Divide (N, Expr, Lit_Num, Lit_Den), Rng_Check);
end if;
end if;
-- Fall through to use floating-point for the close result set case,
-- as a result of the numerator or denominator of the small value not
-- being a sufficiently small integer.
Set_Result (N,
Build_Multiply (N,
Fpt_Value (Expr),
Real_Literal (N, Small)),
end Expand_Convert_Fixed_To_Integer;
-- Expand_Convert_Float_To_Fixed --
-- We have
-- Result_Value * Result_Small = Operand_Value
-- so compute:
-- Result_Value = Operand_Value * (1.0 / Result_Small)
-- We do the small scaling in floating-point, and we do a multiplication
-- rather than a division, since it is accurate enough for the perfect
-- result cases, and faster.
procedure Expand_Convert_Float_To_Fixed (N : Node_Id) is
Expr : constant Node_Id := Expression (N);
Result_Type : constant Entity_Id := Etype (N);
Rng_Check : constant Boolean := Do_Range_Check (N);
Small : constant Ureal := Small_Value (Result_Type);
-- Optimize small = 1, where we can avoid the multiply completely
if Small = Ureal_1 then
Set_Result (N, Expr, Rng_Check, Trunc => True);
-- Normal case where multiply is required. Rounding is truncating
-- for decimal fixed point types only, see RM 4.6(29), except if the
-- conversion comes from an attribute reference 'Round (RM 3.5.10 (14)):
-- The attribute is implemented by means of a conversion that must
-- round.
(N => N,
Expr =>
(N => N,
L => Fpt_Value (Expr),
R => Real_Literal (N, Ureal_1 / Small)),
Rchk => Rng_Check,
Trunc => Is_Decimal_Fixed_Point_Type (Result_Type)
and not Rounded_Result (N));
end if;
end Expand_Convert_Float_To_Fixed;
-- Expand_Convert_Integer_To_Fixed --
-- We have
-- Result_Value * Result_Small = Operand_Value
-- Result_Value = Operand_Value / Result_Small
-- If the small value is a sufficiently small integer, then the perfect
-- result set is obtained by a single integer division.
-- If the small value is the reciprocal of a sufficiently small integer,
-- the perfect result set is obtained by a single integer multiplication.
-- If the numerator and denominator of the small value are sufficiently
-- small integers, then the perfect result set is obtained by a scaled
-- divide operation.
-- In other cases, we obtain the close result set by calculating the
-- result in floating-point using a multiplication by the reciprocal
-- of the Result_Small.
procedure Expand_Convert_Integer_To_Fixed (N : Node_Id) is
Rng_Check : constant Boolean := Do_Range_Check (N);
Expr : constant Node_Id := Expression (N);
Result_Type : constant Entity_Id := Etype (N);
Small : constant Ureal := Small_Value (Result_Type);
Small_Num : constant Uint := Norm_Num (Small);
Small_Den : constant Uint := Norm_Den (Small);
Lit_Num : Node_Id;
Lit_Den : Node_Id;
if Small_Den = 1 then
Lit_Num := Integer_Literal (N, Small_Num);
if Present (Lit_Num) then
Set_Result (N, Build_Divide (N, Expr, Lit_Num), Rng_Check);
end if;
elsif Small_Num = 1 then
Lit_Den := Integer_Literal (N, Small_Den);
if Present (Lit_Den) then
Set_Result (N, Build_Multiply (N, Expr, Lit_Den), Rng_Check);
end if;
Lit_Num := Integer_Literal (N, Small_Num);
Lit_Den := Integer_Literal (N, Small_Den);
if Present (Lit_Num) and then Present (Lit_Den) then
(N, Build_Scaled_Divide (N, Expr, Lit_Den, Lit_Num), Rng_Check);
end if;
end if;
-- Fall through to use floating-point for the close result set case,
-- as a result of the numerator or denominator of the small value not
-- being a sufficiently small integer.
Set_Result (N,
Build_Multiply (N,
Fpt_Value (Expr),
Real_Literal (N, Ureal_1 / Small)),
end Expand_Convert_Integer_To_Fixed;
-- Expand_Decimal_Divide_Call --
-- We have four operands
-- Dividend
-- Divisor
-- Quotient
-- Remainder
-- All of which are decimal types, and which thus have associated
-- decimal scales.
-- Computing the quotient is a similar problem to that faced by the
-- normal fixed-point division, except that it is simpler, because
-- we always have compatible smalls.
-- Quotient = (Dividend / Divisor) * 10**q
-- where 10 ** q = Dividend'Small / (Divisor'Small * Quotient'Small)
-- so q = Divisor'Scale + Quotient'Scale - Dividend'Scale
-- For q >= 0, we compute
-- Numerator := Dividend * 10 ** q
-- Denominator := Divisor
-- Quotient := Numerator / Denominator
-- For q < 0, we compute
-- Numerator := Dividend
-- Denominator := Divisor * 10 ** q
-- Quotient := Numerator / Denominator
-- Both these divisions are done in truncated mode, and the remainder
-- from these divisions is used to compute the result Remainder. This
-- remainder has the effective scale of the numerator of the division,
-- For q >= 0, the remainder scale is Dividend'Scale + q
-- For q < 0, the remainder scale is Dividend'Scale
-- The result Remainder is then computed by a normal truncating decimal
-- conversion from this scale to the scale of the remainder, i.e. by a
-- division or multiplication by the appropriate power of 10.
procedure Expand_Decimal_Divide_Call (N : Node_Id) is
Loc : constant Source_Ptr := Sloc (N);
Dividend : Node_Id := First_Actual (N);
Divisor : Node_Id := Next_Actual (Dividend);
Quotient : Node_Id := Next_Actual (Divisor);
Remainder : Node_Id := Next_Actual (Quotient);
Dividend_Type : constant Entity_Id := Etype (Dividend);
Divisor_Type : constant Entity_Id := Etype (Divisor);
Quotient_Type : constant Entity_Id := Etype (Quotient);
Remainder_Type : constant Entity_Id := Etype (Remainder);
Dividend_Scale : constant Uint := Scale_Value (Dividend_Type);
Divisor_Scale : constant Uint := Scale_Value (Divisor_Type);
Quotient_Scale : constant Uint := Scale_Value (Quotient_Type);
Remainder_Scale : constant Uint := Scale_Value (Remainder_Type);
Q : Uint;
Numerator_Scale : Uint;
Stmts : List_Id;
Qnn : Entity_Id;
Rnn : Entity_Id;
Computed_Remainder : Node_Id;
Adjusted_Remainder : Node_Id;
Scale_Adjust : Uint;
-- Relocate the operands, since they are now list elements, and we
-- need to reference them separately as operands in the expanded code.
Dividend := Relocate_Node (Dividend);
Divisor := Relocate_Node (Divisor);
Quotient := Relocate_Node (Quotient);
Remainder := Relocate_Node (Remainder);
-- Now compute Q, the adjustment scale
Q := Divisor_Scale + Quotient_Scale - Dividend_Scale;
-- If Q is non-negative then we need a scaled divide
if Q >= 0 then
Integer_Literal (N, Uint_10 ** Q),
Qnn, Rnn, Stmts);
Numerator_Scale := Dividend_Scale + Q;
-- If Q is negative, then we need a double divide
Integer_Literal (N, Uint_10 ** (-Q)),
Qnn, Rnn, Stmts);
Numerator_Scale := Dividend_Scale;
end if;
-- Add statement to set quotient value
-- Quotient := quotient-type!(Qnn);
Append_To (Stmts,
Make_Assignment_Statement (Loc,
Name => Quotient,
Expression =>
Unchecked_Convert_To (Quotient_Type,
Build_Conversion (N, Quotient_Type,
New_Occurrence_Of (Qnn, Loc)))));
-- Now we need to deal with computing and setting the remainder. The
-- scale of the remainder is in Numerator_Scale, and the desired
-- scale is the scale of the given Remainder argument. There are
-- three cases:
-- Numerator_Scale > Remainder_Scale
-- in this case, there are extra digits in the computed remainder
-- which must be eliminated by an extra division:
-- computed-remainder := Numerator rem Denominator
-- scale_adjust = Numerator_Scale - Remainder_Scale
-- adjusted-remainder := computed-remainder / 10 ** scale_adjust
-- Numerator_Scale = Remainder_Scale
-- in this case, the we have the remainder we need
-- computed-remainder := Numerator rem Denominator
-- adjusted-remainder := computed-remainder
-- Numerator_Scale < Remainder_Scale
-- in this case, we have insufficient digits in the computed
-- remainder, which must be eliminated by an extra multiply
-- computed-remainder := Numerator rem Denominator
-- scale_adjust = Remainder_Scale - Numerator_Scale
-- adjusted-remainder := computed-remainder * 10 ** scale_adjust
-- Finally we assign the adjusted-remainder to the result Remainder
-- with conversions to get the proper fixed-point type representation.
Computed_Remainder := New_Occurrence_Of (Rnn, Loc);
if Numerator_Scale > Remainder_Scale then
Scale_Adjust := Numerator_Scale - Remainder_Scale;
Adjusted_Remainder :=
(N, Computed_Remainder, Integer_Literal (N, 10 ** Scale_Adjust));
elsif Numerator_Scale = Remainder_Scale then
Adjusted_Remainder := Computed_Remainder;
else -- Numerator_Scale < Remainder_Scale
Scale_Adjust := Remainder_Scale - Numerator_Scale;
Adjusted_Remainder :=
(N, Computed_Remainder, Integer_Literal (N, 10 ** Scale_Adjust));
end if;
-- Assignment of remainder result
Append_To (Stmts,
Make_Assignment_Statement (Loc,
Name => Remainder,
Expression =>
Unchecked_Convert_To (Remainder_Type, Adjusted_Remainder)));
-- Final step is to rewrite the call with a block containing the
-- above sequence of constructed statements for the divide operation.
Rewrite (N,
Make_Block_Statement (Loc,
Handled_Statement_Sequence =>
Make_Handled_Sequence_Of_Statements (Loc,
Statements => Stmts)));
Analyze (N);
end Expand_Decimal_Divide_Call;
-- Expand_Divide_Fixed_By_Fixed_Giving_Fixed --
procedure Expand_Divide_Fixed_By_Fixed_Giving_Fixed (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
if Etype (Left) = Universal_Real then
Do_Divide_Universal_Fixed (N);
elsif Etype (Right) = Universal_Real then
Do_Divide_Fixed_Universal (N);
Do_Divide_Fixed_Fixed (N);
-- A focused optimization: if after constant folding the
-- expression is of the form: T ((Exp * D) / D), where D is
-- a static constant, return T (Exp). This form will show up
-- when D is the denominator of the static expression for the
-- 'small of fixed-point types involved. This transformation
-- removes a division that may be expensive on some targets.
if Nkind (N) = N_Type_Conversion
and then Nkind (Expression (N)) = N_Op_Divide
Num : constant Node_Id := Left_Opnd (Expression (N));
Den : constant Node_Id := Right_Opnd (Expression (N));
if Nkind (Den) = N_Integer_Literal
and then Nkind (Num) = N_Op_Multiply
and then Nkind (Right_Opnd (Num)) = N_Integer_Literal
and then Intval (Den) = Intval (Right_Opnd (Num))
Rewrite (Expression (N), Left_Opnd (Num));
end if;
end if;
end if;
end Expand_Divide_Fixed_By_Fixed_Giving_Fixed;
-- Expand_Divide_Fixed_By_Fixed_Giving_Float --
-- The division is done in Universal_Real, and the result is multiplied
-- by the small ratio, which is Small (Right) / Small (Left). Special
-- treatment is required for universal operands, which represent their
-- own value and do not require conversion.
procedure Expand_Divide_Fixed_By_Fixed_Giving_Float (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
Left_Type : constant Entity_Id := Etype (Left);
Right_Type : constant Entity_Id := Etype (Right);
-- Case of left operand is universal real, the result we want is:
-- Left_Value / (Right_Value * Right_Small)
-- so we compute this as:
-- (Left_Value / Right_Small) / Right_Value
if Left_Type = Universal_Real then
Set_Result (N,
Build_Divide (N,
Real_Literal (N, Realval (Left) / Small_Value (Right_Type)),
Fpt_Value (Right)));
-- Case of right operand is universal real, the result we want is
-- (Left_Value * Left_Small) / Right_Value
-- so we compute this as:
-- Left_Value * (Left_Small / Right_Value)
-- Note we invert to a multiplication since usually floating-point
-- multiplication is much faster than floating-point division.
elsif Right_Type = Universal_Real then
Set_Result (N,
Build_Multiply (N,
Fpt_Value (Left),
Real_Literal (N, Small_Value (Left_Type) / Realval (Right))));
-- Both operands are fixed, so the value we want is
-- (Left_Value * Left_Small) / (Right_Value * Right_Small)
-- which we compute as:
-- (Left_Value / Right_Value) * (Left_Small / Right_Small)
Set_Result (N,
Build_Multiply (N,
Build_Divide (N, Fpt_Value (Left), Fpt_Value (Right)),
Real_Literal (N,
Small_Value (Left_Type) / Small_Value (Right_Type))));
end if;
end Expand_Divide_Fixed_By_Fixed_Giving_Float;
-- Expand_Divide_Fixed_By_Fixed_Giving_Integer --
procedure Expand_Divide_Fixed_By_Fixed_Giving_Integer (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
if Etype (Left) = Universal_Real then
Do_Divide_Universal_Fixed (N);
elsif Etype (Right) = Universal_Real then
Do_Divide_Fixed_Universal (N);
Do_Divide_Fixed_Fixed (N);
end if;
end Expand_Divide_Fixed_By_Fixed_Giving_Integer;
-- Expand_Divide_Fixed_By_Integer_Giving_Fixed --
-- Since the operand and result fixed-point type is the same, this is
-- a straight divide by the right operand, the small can be ignored.
procedure Expand_Divide_Fixed_By_Integer_Giving_Fixed (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
Set_Result (N, Build_Divide (N, Left, Right));
end Expand_Divide_Fixed_By_Integer_Giving_Fixed;
-- Expand_Multiply_Fixed_By_Fixed_Giving_Fixed --
procedure Expand_Multiply_Fixed_By_Fixed_Giving_Fixed (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
procedure Rewrite_Non_Static_Universal (Opnd : Node_Id);
-- The operand may be a non-static universal value, such an
-- exponentiation with a non-static exponent. In that case, treat
-- as a fixed * fixed multiplication, and convert the argument to
-- the target fixed type.
-- Rewrite_Non_Static_Universal --
procedure Rewrite_Non_Static_Universal (Opnd : Node_Id) is
Loc : constant Source_Ptr := Sloc (N);
Rewrite (Opnd,
Make_Type_Conversion (Loc,
Subtype_Mark => New_Occurrence_Of (Etype (N), Loc),
Expression => Expression (Opnd)));
Analyze_And_Resolve (Opnd, Etype (N));
end Rewrite_Non_Static_Universal;
-- Start of processing for Expand_Multiply_Fixed_By_Fixed_Giving_Fixed
if Etype (Left) = Universal_Real then
if Nkind (Left) = N_Real_Literal then
Do_Multiply_Fixed_Universal (N, Left => Right, Right => Left);
elsif Nkind (Left) = N_Type_Conversion then
Rewrite_Non_Static_Universal (Left);
Do_Multiply_Fixed_Fixed (N);
end if;
elsif Etype (Right) = Universal_Real then
if Nkind (Right) = N_Real_Literal then
Do_Multiply_Fixed_Universal (N, Left, Right);
elsif Nkind (Right) = N_Type_Conversion then
Rewrite_Non_Static_Universal (Right);
Do_Multiply_Fixed_Fixed (N);
end if;
Do_Multiply_Fixed_Fixed (N);
end if;
end Expand_Multiply_Fixed_By_Fixed_Giving_Fixed;
-- Expand_Multiply_Fixed_By_Fixed_Giving_Float --
-- The multiply is done in Universal_Real, and the result is multiplied
-- by the adjustment for the smalls which is Small (Right) * Small (Left).
-- Special treatment is required for universal operands.
procedure Expand_Multiply_Fixed_By_Fixed_Giving_Float (N : Node_Id) is
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
Left_Type : constant Entity_Id := Etype (Left);
Right_Type : constant Entity_Id := Etype (Right);
-- Case of left operand is universal real, the result we want is
-- Left_Value * (Right_Value * Right_Small)
-- so we compute this as:
-- (Left_Value * Right_Small) * Right_Value;
if Left_Type = Universal_Real then
Set_Result (N,
Build_Multiply (N,
Real_Literal (N, Realval (Left) * Small_Value (Right_Type)),
Fpt_Value (Right)));
-- Case of right operand is universal real, the result we want is
-- (Left_Value * Left_Small) * Right_Value
-- so we compute this as:
-- Left_Value * (Left_Small * Right_Value)
elsif Right_Type = Universal_Real then
Set_Result (N,
Build_Multiply (N,
Fpt_Value (Left),
Real_Literal (N, Small_Value (Left_Type) * Realval (Right))));
-- Both operands are fixed, so the value we want is
-- (Left_Value * Left_Small) * (Right_Value * Right_Small)
-- which we compute as:
-- (Left_Value * Right_Value) * (Right_Small * Left_Small)
Set_Result (N,
Build_Multiply (N,
Build_Multiply (N, Fpt_Value (Left), Fpt_Value (Right)),
Real_Literal (N,
Small_Value (Right_Type) * Small_Value (Left_Type))));
end if;
end Expand_Multiply_Fixed_By_Fixed_Giving_Float;
-- Expand_Multiply_Fixed_By_Fixed_Giving_Integer --
procedure Expand_Multiply_Fixed_By_Fixed_Giving_Integer (N : Node_Id) is
Loc : constant Source_Ptr := Sloc (N);
Left : constant Node_Id := Left_Opnd (N);
Right : constant Node_Id := Right_Opnd (N);
if Etype (Left) = Universal_Real then
Do_Multiply_Fixed_Universal (N, Left => Right, Right => Left);
elsif Etype (Right) = Universal_Real then
Do_Multiply_Fixed_Universal (N, Left, Right);
-- If both types are equal and we need to avoid floating point
-- instructions, it's worth introducing a temporary with the
-- common type, because it may be evaluated more simply without
-- the need for run-time use of floating point.
elsif Etype (Right) = Etype (Left)
and then Restriction_Active (No_Floating_Point)
Temp : constant Entity_Id := Make_Temporary (Loc, 'F');
Mult : constant Node_Id := Make_Op_Multiply (Loc, Left, Right);
Decl : constant Node_Id :=
Make_Object_Declaration (Loc,
Defining_Identifier => Temp,
Object_Definition => New_Occurrence_Of (Etype (Right), Loc),
Expression => Mult);
Insert_Action (N, Decl);
Rewrite (N,
OK_Convert_To (Etype (N), New_Occurrence_Of (Temp, Loc)));
Analyze_And_Resolve (N, Standard_Integer);
Do_Multiply_Fixed_Fixed (N);
end if;
end Expand_Multiply_Fixed_By_Fixed_Giving_Integer;
-- Expand_Multiply_Fixed_By_Integer_Giving_Fixed --
-- Since the operand and result fixed-point type is the same, this is
-- a straight multiply by the right operand, the small can be ignored.
procedure Expand_Multiply_Fixed_By_Integer_Giving_Fixed (N : Node_Id) is
Set_Result (N,
Build_Multiply (N, Left_Opnd (N), Right_Opnd (N)));
end Expand_Multiply_Fixed_By_Integer_Giving_Fixed;
-- Expand_Multiply_Integer_By_Fixed_Giving_Fixed --
-- Since the operand and result fixed-point type is the same, this is
-- a straight multiply by the right operand, the small can be ignored.
procedure Expand_Multiply_Integer_By_Fixed_Giving_Fixed (N : Node_Id) is
Set_Result (N,
Build_Multiply (N, Left_Opnd (N), Right_Opnd (N)));
end Expand_Multiply_Integer_By_Fixed_Giving_Fixed;
-- Fpt_Value --
function Fpt_Value (N : Node_Id) return Node_Id is
return Build_Conversion (N, Universal_Real, N);
end Fpt_Value;
-- Get_Size_For_Value --
function Get_Size_For_Value (V : Uint) return Pos is
pragma Assert (V >= Uint_0);
if V < Uint_2 ** 7 then
return 8;
elsif V < Uint_2 ** 15 then
return 16;
elsif V < Uint_2 ** 31 then
return 32;
elsif V < Uint_2 ** 63 then
return 64;
elsif V < Uint_2 ** 127 then
return 128;
return Pos'Last;
end if;
end Get_Size_For_Value;
-- Get_Type_For_Size --
function Get_Type_For_Size (Siz : Pos; Force : Boolean) return Entity_Id is
if Siz <= 8 then
return Standard_Integer_8;
elsif Siz <= 16 then
return Standard_Integer_16;
elsif Siz <= 32 then
return Standard_Integer_32;
elsif Siz <= 64
or else (Force and then System_Max_Integer_Size < 128)
return Standard_Integer_64;
elsif (Siz <= 128 and then System_Max_Integer_Size = 128)
or else Force
return Standard_Integer_128;
return Empty;
end if;
end Get_Type_For_Size;
-- Integer_Literal --
function Integer_Literal
(N : Node_Id;
V : Uint;
Negative : Boolean := False) return Node_Id
T : Entity_Id;
L : Node_Id;
T := Get_Type_For_Size (Get_Size_For_Value (V), Force => False);
if No (T) then
return Empty;
end if;
if Negative then
L := Make_Integer_Literal (Sloc (N), UI_Negate (V));
L := Make_Integer_Literal (Sloc (N), V);
end if;
-- Set type of result in case used elsewhere (see note at start)
Set_Etype (L, T);
Set_Is_Static_Expression (L);
-- We really need to set Analyzed here because we may be creating a
-- very strange beast, namely an integer literal typed as fixed-point
-- and the analyzer won't like that.
Set_Analyzed (L);
return L;
end Integer_Literal;
-- Real_Literal --
function Real_Literal (N : Node_Id; V : Ureal) return Node_Id is
L : Node_Id;
L := Make_Real_Literal (Sloc (N), V);
-- Set type of result in case used elsewhere (see note at start)
Set_Etype (L, Universal_Real);
return L;
end Real_Literal;
-- Rounded_Result_Set --
function Rounded_Result_Set (N : Node_Id) return Boolean is
K : constant Node_Kind := Nkind (N);
if (K = N_Type_Conversion or else
K = N_Op_Divide or else
K = N_Op_Multiply)
and then
(Rounded_Result (N) or else Is_Integer_Type (Etype (N)))
return True;
return False;
end if;
end Rounded_Result_Set;
-- Set_Result --
procedure Set_Result
(N : Node_Id;
Expr : Node_Id;
Rchk : Boolean := False;
Trunc : Boolean := False)
Cnode : Node_Id;
Expr_Type : constant Entity_Id := Etype (Expr);
Result_Type : constant Entity_Id := Etype (N);
-- No conversion required if types match and no range check or truncate
if Result_Type = Expr_Type and then not (Rchk or Trunc) then
Cnode := Expr;
-- Else perform required conversion
Cnode := Build_Conversion (N, Result_Type, Expr, Rchk, Trunc);
end if;
Rewrite (N, Cnode);
Analyze_And_Resolve (N, Result_Type);
end Set_Result;
end Exp_Fixd; | {"url":"https://gnu.googlesource.com/gcc/+/refs/tags/basepoints/gcc-13/gcc/ada/exp_fixd.adb","timestamp":"2024-11-06T18:02:17Z","content_type":"text/html","content_length":"625001","record_id":"<urn:uuid:84a3f2b5-bb37-493c-8d80-e973d88fde38>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00134.warc.gz"} |
Another Maximum Sum in Subarray
You cannot submit for this problem because the contest is ended. You can click "Open in Problem Set" to view this problem in normal mode.
Time Limit: 1.0 s
Memory Limit: 256.0 MB
You are given an Array \(A[]\), length of the array is N. You need to find out largest subarray with length exactly K and sum is maximum.
For example, \(A[]\)= {3,5,6,2,4} and K=3,
all possible subarray from A[] with length K are,
Before you find maximum subarray sum, you can do following operation as many times as you wish,
• Choose two indices i and j, if gcd of their value is greater than 1, gcd\((A[i],A[j]) > 1\) ,you can swap their elements. (ex. swap\((A[i],A[j])\))
gcd means greatest common divisor.
Example : gcd(4,6)=2, gcd(3,9)=3 ect.
First Line, T number of test case.
In each test case,
First Line N and K.
Second Line an Array A[].
\(1<=N<=6 * 10^3\)
It is gurantee that sum of N overall test case does not exceed \(6 * 10^3\).
In each test case, Print the maximum sum.
Input Output
First test case,
Intially Array \(A []\) = {1,6,3,6,2,7}; and K = 3.
If we choose indices 2 & 5, gcd(A[2],A[5])= gcd(6,2) = 2, which is greater than 1.
We can swap their value. After swap array look like,
\(A[]\) = {1,2,3,6,6,7}.
Now we can select subarray, 4 to 6, {6,6,7} and their sum is 19 which is maximum.
Brain Booster #4
Start at
2024-07-14 15:30
End at
2024-07-14 19:00
3.5 hour(s) | {"url":"https://judge.eluminatis-of-lu.com/contest/668d4a9de87ac0000784508e/1063","timestamp":"2024-11-14T14:50:40Z","content_type":"text/html","content_length":"17797","record_id":"<urn:uuid:c7a543c4-1df7-485c-9e1b-5fb6b2f67bdd>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00713.warc.gz"} |
Click here to start/reset.
Paper 1
Question 1
1a) Hint 1: recognise that you need to use the chain rule and a standard differential from the Formula List
1b) Hint 2: recognise that you need to use the quotient rule with the chain rule
1c) Hint 3: consider writing y as y(x) to emphasise that y is a function of x, and then complete the implicit differentiation
Hint 4: and here is a video of the solution:
Question 2
Hint 1: factorise the denominator into two linear factors
Hint 2: use the standard method of partial fractions on the integrand
Hint 3: integrate each fraction on its own, bringing in natural logarithms
Hint 4: and here is a video of the solution:
Question 3
3a) Hint 1: consider writing out the general term to the expansion of (a+b)^n first
3a) Hint 2: then substitute the a for 2x and the b for 5/x² and the n for 9
3a) Hint 3: simplify each term to obtain a term with factorials, numerical terms to powers of r, and x to the power of a linear expression in r
3b) Hint 4: know that the term independent of x is the term whose power of x is zero
3b) Hint 5: set the linear expression in r to be equal to zero, and solve for r
3b) Hint 6: evaluate your answer from part (a) with r taking on the value that you just obtained
Hint 7: and here is a video of the solution:
Question 4
4a) Hint 1: know that the bar over z means the complex conjugate of z
4a) Hint 2: multiply the two complex numbers together and gather the real terms and the imaginary terms
4a) Hint 3: factorise i out of the imaginary terms
4b) Hint 4: know that a real number has an imaginary term of 0
Hint 5: and here is a video of the solution:
Question 5
Hint 1: use the standard method of the euclidean algorithm
Hint 2: the first line is 306 = 119 × 2 + 68
Hint 3: and here is a video of the solution:
Question 6
Hint 1: recognise that we want dy/dx when t = -1/3
Hint 2: work out dy/dt and dx/dt
Hint 3: know that dt/dx is the reciprocal of dx/dt
Hint 4: know that dy/dx is (dy/dt)×(dt/dx)
Hint 5: evaluate dy/dx when t = -1/3, to obtain the gradient of the curve at that point
Hint 6: evaluate x and y when t = -1/3 to obtain the x and y coordinates
Hint 7: use these coordinates and the gradient to obtain the equation of the line
Hint 8: and here is a video of the solution:
Question 7
7a) Hint 1: obtain the resulting 3×3 matrix that has an expression in terms of k in row 2, column 1
7b) Hint 2: use a standard method to obtain the determinant of D, in terms of k
7b) Hint 3: know that the inverse of D does not exist if the determinant of D is equal to zero
Hint 4: and here is a video of the solution:
Question 8
Hint 1: work out du/dθ
Hint 2: work out the values of u for both of the values of θ from the limits of the integral
Hint 3: use a standard method of integration by substitution to simplify the integral to a simple polynomial in u, with u limits
Hint 4: and here is a video of the solution:
Question 9
9a) Hint 1: know that consecutive integers can be written a n, n+1 and n+2, where n is an integer
9a) Hint 2: write words to explain the logic behind what your algebraic terms mean in the context of divisibility
9b) Hint 3: know that an odd integer can be written as 2n+1 where n is an integer
9b) Hint 4: write words to explain the meaning of your algebraic terms
Hint 5: and here is a video of the solution:
Question 10
Hint 1: know that z = x + iy
Hint 2: replace z with x + iy in the given modulus equation
Hint 3: square both sides of the modulus equation to prevent square roots appearing
Hint 4: know that |a+ib|² = a² + b²
Hint 5: simplify the expression for y in terms of x to obtain the equation of a straight line, which is the locus required
Hint 6: sketch the locus, noting points of intercept with the axis, and plotting the numbers 0 [at (0,0)] and 2-2i [at (2,-2)]
Hint 7: and here is a video of the solution:
Question 11
11a) Hint 1: refer to the Formula List for the 2x2 matrix that represents a rotation of θ anticlockwise around the origin
11a) Hint 2: use exact value triangles to evaluate each term when θ= π/3
11b) Hint 3: know that a reflection in the x-axis transforms the point (x,y) to the point (x,-y)
11c) Hint 4: know that P = B A, and not P = A B
11d) Hint 5: refer to the general matrix for a rotation, noting which elements have to be the same sign
Hint 6: and here is a video of the solution:
Question 12
Hint 1: evaluate a base case when n = 1, to verify the statement is true
Hint 2: use a standard method for proof by induction, for the inductive step
Hint 3: write words to draw together how the base case and the inductive step together mean the statement is true for all positive integers
Hint 4: and here is a video of the solution:
Question 13
13a) Hint 1: use pythagoras' theorem on one of the right angled triangles
13a) Hint 2: solve for h, giving a reason for rejecting the possible negative solution
13b) Hint 3: recognise that a decreasing rate means that it will be negative
13b) Hint 4: deduce that dx/dt = -0.3
13b) Hint 5: recognise that we want dh/dt when x = 30
13b) Hint 6: implicitly differentiate with respect to t, the relationship h² + x² = 2500, from part (a)'s workings
13b) Hint 7: calculate the value for h, when x = 30, using part(a)
13b) Hint 8: replace in your implicitly differentiated equation the values for x, h and dx/dt
13b) Hint 9: rearrange to make dh/dt the subject
Hint 10: and here is a video of the solution:
Question 14
14a) Hint 1: use standard formulae to calculate each of u7 and S∞
14b)i) Hint 2: use the standard formula for Sn and replace the values for Sn, n and a, to then solve for d
14b)ii) Hint 3: use a standard formula, now that a and d are known
14c) Hint 4: use a similar approach to part (b), but this time substituting in values for Sn, a and d to then obtain an equation in n
14c) Hint 5: rearrange to make the quadratic in n equal to zero
14c) Hint 6: factorise the quadratic by first taking out a common factor of 16 from all terms
Hint 7: and here is a video of the solution:
Question 15
15a) Hint 1: use a standard method for integration by parts
15a) Hint 2: remember to include the constant of integration
15b) Hint 3: recognise that this equation requires an integrating factor, or...
15b) Hint 4: ...alternatively, in order to obtain part (a)'s integrand in part (b), just divide all terms through by x
15b) Hint 5: integrate by a standard method and use the initial conditions to fix the value of the constant of integration
15b) Hint 6: present your final answer in the required form
Hint 7: and here is a video of the solution:
Question 16
16a) Hint 1: use the standard method of gaussian elimination
16a) Hint 2: carefully interpret the final row of your augmented matrix that should be (a-8)z = 0
16a) Hint 3: consider which values of a would give an infinite number of solutions (that would give the intersection line)
16b) Hint 4: work out expressions for y and x, each in terms of z
16b) Hint 5: write the x, y and z components in vector form
16b) Hint 6: extract the constant vector and a parametric multiple of a direction vector, introducing a parameter, instead of z
16c) Hint 7: know that the angle between two planes is the angle between their two normal vectors
16c) Hint 8: know that we need the acute angle, so careful consideration with the help of an angle diagram should help
16d) Hint 9: compare the two planes' direction vectors
16d) Hint 10: know that one vector being the multiple of another means that they are parallel vectors
16d) Hint 11: know what parallel normal vectors mean in terms of the planes themselves
16d) Hint 12: know how to check that the two planes are not coincident with each other (i.e. they are the same plane in the same space)
Hint 13: and here is a video of the solution:
Question 17
17a) Hint 1: use the standard method for calculating each term of a Maclaurin Series
17b)i) Hint 2: recognise that this will require repeated use of both the chain rule and the product rule
17b)ii) Hint 3: evaluate the answers from (b)(i) when x = 0
17c) Hint 4: write down the two series up to and including their cubic powers of x
17c) Hint 5: multiply the two polynomials together to obtain all 8 terms
17c) Hint 6: discard all the terms that have powers higher than 3
17d) Hint 7: recognise that the expression given is the derivative of that from part (c)
17d) Hint 8: differentiate the series answer from part (c), term by term.
Hint 9: and here is a video of the solution:
Did this hint help? | {"url":"http://hints.nhost.uk/hints/SQA%20Maths%20AH%202018.html","timestamp":"2024-11-12T18:46:47Z","content_type":"text/html","content_length":"23989","record_id":"<urn:uuid:f4d65aa7-6741-484f-a69d-8a98a5359e35>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00069.warc.gz"} |
Archaeoastronomy: a short overview of my methodology using open source software
[This post is part of a series of posts on archaeoastronomy using open source software]
My work focusses on using Google Earth, a downloadable free virtual globe and geographic information program, to obtain satellite imagery of aerially-visible archaeological sites world wide that are
suspected to have been used for astronomical observations, for instance stone circles and medicine wheels. The uniqueness of my approach is that it does not require actually having to visit these
sites in person to survey them.
In addition, the free-source nature of the programs I use means that anyone with a modicum of expertise and a computer can duplicate the results. Previous to this, double checking the claims of
astronomical alignments of a site required actually having to visit it, and having the time, funds, and expertise needed to conduct a comprehensive survey… not to mention also having the time and
expertise needed to do the complicated calculations needed to estimate the horizon rise/set positions of various celestial bodies at any arbitrary date. The free software I use to do my studies
makes all of this much, much easier (but perhaps not necessarily something a complete novice would attempt without some basic knowledge of astronomy and a reasonable background in computing and
statistics). For anyone wishing to try out my methods, I will provide all the code and files related to an analysis I’ve done of the “Merry Maidens” stone circle in the UK.
Using Google Earth, I take screen shots of aerial views of the site, and using the free and open-source vector graphics program Xfig, I then upload the screenshot and place datum points at all
intersecting walls, posts, standing stones, other significant site features, etc. I output these points to a data file, then use the R free and open-source statistical programming language to fit
straight lines to all possible combinations of points (usually with some quality criteria, such as a requirement that the points be far enough apart that the uncertainty on which angle of the line is
sufficiently small to give a reasonable indication of where on the horizon it is pointing). I call these lines the “site lines” (they also may potentially be “sight lines” to celestial body rise/set
points on the horizon).
To determine the rise and set azimuths (angle from North) of celestial objects for any given date, I use the free downloadable pyephem ephemeris calculation package, written in the Python programming
language. I consider celestial objects like the Sun, the Moon, and the brightest stars. The true rise and set azimuths must take into account the horizon of the surrounding terrain, thus I
calculate this using topographic information for the area in 1′ grids that are publicly available from the National Geophysical Data Center at the National Oceanic and Atmospheric Administration
website. I call these the “proposed astronomical alignments”.
Once I have the proposed astronomical alignments and site lines, I use statistical methods to determine what fraction of astronomical alignments match site lines, and what fraction of site lines
match proposed astronomical alignments. One hallmark of a site that truly was used as a comprehensive astronomical observatory, is that both of these fractions are high. Or that there were no
alignments to either the rise or set of some stars, but an unusually large number of alignments to both the rise and set of others. I assess the statistical significance of the observed alignments
by creating many “synthetic” sites (by randomly scrambling the points on the site) and “synthetic” skies by randomly picking points on the horizon where some hypothetical celestial body might rise.
Using this synthetic data, I can then obtain a probability distribution for the matches of site lines to astronomical alignments, and the probability distribution for the matches of astronomical
alignments to site lines; these probability distributions test the null hypothesis that the site does not contain any astronomical alignments.
Visits: 1856
You must be logged in to post a comment. | {"url":"https://sherrytowers.com/2014/04/13/archeoastronomy-a-short-overview-of-my-methodology-using-open-source-software/","timestamp":"2024-11-11T11:06:52Z","content_type":"text/html","content_length":"39491","record_id":"<urn:uuid:1451a11c-72e3-4e36-bdac-696f00a572e8>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00460.warc.gz"} |
University :
Massachusetts Institute of Technology
Instructors :
Dimitris Bertsimas
Peer Review
Course Goals
In the last decade, the amount of data available to organizations has reached unprecedented levels.
Data is transforming business, social interactions, and the future of our society.
In this course, you will learn how to use data and analytics to give an edge to your career and your life.
We will examine real world examples of how analytics have been used to significantly improve a business or industry.
These examples include Moneyball, eHarmony, the Framingham Heart Study, Twitter, IBM Watson, and Netflix.
Through these examples and many more, we will teach you the following analytics methods:
linear regression, logistic regression, trees, text analytics, clustering, visualization, and optimization.
We will be using the statistical software R to build models and work with data. The contents of this course are essentially the same as those of the corresponding MIT class (The Analytics Edge).
It is a challenging class, but it will enable you to apply analytics to real-world applications.
The class will consist of lecture videos, which are broken into small pieces, usually between 4 and 8 minutes each.
After each lecture piece, we will ask you a “quick question” to assess your understanding of the material.
There will also be a recitation, in which one of the teaching assistants will go over the methods introduced with a new example and data set. Each week will have a homework assignment that involves
working in R or LibreOffice with various data sets.
(R is a free statistical and computing software environment we’ll use in the course. See the Software FAQ below for more info).
In the middle of the class, we will run an analytics competition, and at the end of the class there will be a final exam, which will be similar to the homework assignments.
Course Syllebus
(1) An applied understanding of many different analytics methods, including linear regression, logistic regression, CART, clustering, and data visualization
(2) How to implement all of these methods in R
(3) An applied understanding of mathematical optimization and how to solve optimization models in spreadsheet software
Other Info. | {"url":"https://dsl.tmu.edu.tw/Course/Detail/49","timestamp":"2024-11-10T22:34:03Z","content_type":"text/html","content_length":"1048934","record_id":"<urn:uuid:21e64e28-da32-4d94-94b6-fdcb22fc7f4e>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00120.warc.gz"} |
Deep RL and Optimization applied to Operations Research problem - 1/2 Traditional Optimization techniques | Eki.Lab
In this first article is introduced a systematic way to approach and solve optimization problems. Then, the multi-knapsack problem itself is introduced. Then we apply the rules defined before on how
to solve optimization problems and obtain the optimal solution to the multi-knapsack problem, formulated as a Mixed Integer problem and using Python-MIP package. Let's now introduce simple steps one
can follow to approach optimization problems with optimization solvers.
Main steps while creating an optimization model to solve a business problem
Once a business problem that could benefit from optimization has been identified, we can define a systematic approach based on 3 steps for solving all kind of optimization problems with optimization
solvers. These 3 steps are highlighted in the figure below.
Figure 1 : The 3 main steps for solving a business problem through optimization
In more details, these 3 steps are:
1. Create the conceptual mathematical model that defines the different variables, constraints, etc. in the business problem. This step consists in writing down on paper the equations that define our
2. Translate the conceptual mathematical model into a computer program. For most programming languages used for optimization, the computer program will largely resembles the mathematical equations
one would write on paper.
3. Solve the mathematical model using a math programming solver. The solver available for Mathematical Programming (solvers such as GLPK, Gurobi, CPLEX...) relies on very sophisticated algorithms.
Important algorithms and ideas used in these solvers are, among many others: simplex method, branch & bound, use of heuristics...
Let's see those 3 steps for the case of the multi-knapsack problem.
The multi-knapsack problem
The objective here is, given a set of n items and a set of m knapsacks, to maximize the total value of the items put in the knapsacks without exceeding their capacity.
Below, w[i] represents the weight of item i, p[i] the value of item i while c[j] represents the capacity of knapsack j.
Figure 2: Description of the multi-knapsack problem
The multi-knapsack is an extension of the classical knapsack problem where instead of considering only one knapsack, we consider as many as we want. This allows to easily extend the complexity of
this problem.
While the problem is relatively easy to define mathematically, it belongs to the class of NP-hard problems. Without going into the details of what defines NP-hard problems, we can easily see that the
complexity of the knapsack problems explodes when the number of knapsacks and items increases. Indeed, we have m^n available combinations we would need to test should we want to apply a brute-force
approach for solving this problem. Just with 10 knapsacks and 80 items, there are 10^80 combinations, which is the estimation of the number of atoms in the universe! And 10 knapsacks and 80 items is
still quite limited... Let's now try to create the conceptual mathematical model by defining the problem with equations.
Creating the conceptual mathematical model
A quick translation of the multi-knapsack problem with equation can be written as the following:
Now that we managed to translate the problem into a set of equations, let's translate this mathematical model so that it is understood by a computer program. Below, we will make use of the Python
package Python-MIP which is open-source and provides tools for modeling and solving Mixed-Integer Linear Programming Problems (MIP), relying on fast open source solvers.
Translating the mathematical model into a computer program with Python-MIP
Before solving the problem, we have to generate an instance for it (have data defining the problem). To do so, you can use the following code that will generate an instance of this problem with 40
items to store in 5 bags.
import pandas as pd
import numpy as np
import pickle
def data_generator_knapsack(number_bags, number_items, minimum_weight_item, maximum_weight_item, minimum_value_item, maximum_value_item, max_weight_bag):
data = {}
weights = np.random.randint(minimum_weight_item, maximum_weight_item, size = number_items)
values = np.random.randint(minimum_value_item, maximum_value_item, size = number_items)
data['weights'] = weights
data['values'] = values
data['items'] = list(range(len(weights)))
data['num_items'] = len(weights)
data['bins'] = list(range(number_bags))
data['bin_capacities'] = np.random.randint(0, max_weight_bag, size = number_bags) + np.int(np.mean(data['weights']))
number_bags = 5
number_items = 40
minimum_weight_item = 0
maximum_weight_item = 75
minimum_value_item = 0
maximum_value_item = 75
max_weight_bag = 150
data = data_generator_knapsack(number_bags, number_items, minimum_weight_item, maximum_weight_item, minimum_value_item, maximum_value_item, max_weight_bag)
Let's now import the package used to have access to the MIP solver, here using the python package Python-MIP:
from mip import Model, xsum, maximize, BINARY
Now, we can translate the mathematical model so that it is understood by Python-MIP.
def mip_solve_knapsack(data):
model = Model("knapsack")
x = [[model.add_var(var_type=BINARY) for i in data['items']] for j in data['bins']]
model.objective = maximize(xsum((xsum(data['values'][i] * x[j][i] for i in data['items']) for j in data['bins'])))
for j in data['bins']:
model += xsum(data['weights'][i] * x[j][i] for i in data['items']) <= data['bin_capacities'][j]
# Each item can be in at most one bin
for i in data['items']:
model += xsum(x[j][i] for j in data['bins']) <= 1
Remark how close it is from the original equations! These solvers are very powerful and yet easy to use directly in Python. The code is indeed very close to the original equations.
Solving the mathematical model with Python-MIP
Using the mip_solve_knapsack function defined in the previous section, we can access to important information regarding the problem, such as the final objective value and the values of x[ij] telling
us what were the best combinations of items inside knapsacks.
Some Mathematical Optimization packages
In the notebook associated to this article, the package Python-MIP was used. Python-MIP is free, but many other packages exist for solving optimization problems on Python (and other languages of
course like Julia). For instance OR-Tools from Google is a well-recognized free solver, with detailed documentation.
On the other side, Gurobi is a very popular commercial solution for mathematical optimization and its documentation is extremely rich, with quick introductions about Mathematical Programming, Linear
Programming and Mixed-Integer Programming. Importantly, it has a large number of modeling examples from all industry fields directly available on Google Colab allowing to better grasp notions of
Mathematical Modelling and to improve modeling skills to tackle all kind of optimization problems with Python. This resource can be of use even if one doesn't plan to use this commercial software but
rather a free package such as OR-Tools.
In this article was introduced the multi-knapsack problem, an NP-complete problem, very difficult to solve when taking many items and bags.
The approach to solve the multi-knapsack problem relied on Python-MIP, a free optimization package using powerful MILP solvers to solve very efficiently all kinds of optimization problems.
In the next part of this series on the multi-knapsack problem, well studied in the field of Operations Research and at the heart of many real optimization problems, we'll highlight how Deep
Reinforcement Learning can be used in order to solve combinatorial optimization problems such as this one. Stay tuned! | {"url":"https://ekimetrics.github.io/blog/2022/08/27/traditional_or/","timestamp":"2024-11-08T00:41:45Z","content_type":"text/html","content_length":"33844","record_id":"<urn:uuid:8e95c3c9-5051-48c0-acac-769ea0e79325>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00113.warc.gz"} |
Pandas Lambda | How Lambda Function Works in Pandas?
Updated April 4, 2023
Introduction to Pandas Lambda
Pandas Lambda function is a little capacity containing a solitary articulation. Lambda capacities can likewise go about as unknown capacities where they do not need any name. These are useful when we
need to perform little undertakings with less code. Lambda functions offer a double lift to an information researcher. You can compose tidier Python code and accelerate your AI undertakings. The
stunt lies in acing lambda capacities, and this is the place where tenderfoots can entangle. We can likewise utilize lambda capacities when we need to pass a little capacity to another capacity.
Lambda capacities are convenient and utilized in many programming dialects, yet we will be zeroing in on utilizing them in Python here.
Syntax and Parameters:
Lambda x:x
• Lambda represents the keyword of the function.
• First, x represents the bound variable.
• The second x represents the body of the function that has to be implemented.
• The watchword is compulsory, and it must be a lambda, while the contentions and body can change dependent on the necessities.
How does Lambda Function work in Pandas?
Given below shows examples of how lambda functions are implemented in Pandas.
Example #1
Utilizing Lambda function to a single column of the dataframe.
import pandas as pd
info= [['Span',415],['Vetts',375],['Suchu',480],
dataframe = pd.DataFrame(info,columns=['Info','Result'])
dataframe = dataframe.assign(Final_Percent = lambda y: (y['Result'] /700 * 100))
print(dataframe.assign(Final_Percent = lambda y: (y['Result'] /700 * 100)))
In the above program, we first import the Pandas library as pd and then define the dataframe. After defining the dataframe, we assign the values and then use the lambda function and dataframe.assign
to assign the equation of this function in order to implement it. Thus, the program is implemented, and the output is as shown in the above snapshot.
Example #2
Utilizing Lambda function to multiple columns of the Pandas dataframe.
import pandas as pd
info = [[10,11,12,13], [14,15,16,17], [18,19,20,21],
[22,23,24,25], [26,27,28,29], [30,31,32,33],
dataframe = pd.DataFrame(info, columns=['First', 'Second', 'Third', 'Fourth'])
dataframe = dataframe.assign(End_Result=lambda y: (y['First'] * y['Second'] * y['Third'] * y['Fourth']))
print(dataframe.assign(End_Result=lambda y: (y['First'] * y['Second'] * y['Third'] * y['Fourth'])))
In the above program, we, as seen previously, import the pandas’ library as pd and then define the dataframe, which consists of multiple columns. Then we assign values to these columns of the
dataframe and use the lambda function to give us the final result by using the equation as shown in the program. Hence, the program is implemented, and the output is as shown in the above snapshot.
We can utilize the apply() capacity to apply the lambda capacity to the two lines and segments of a dataframe. On the off chance that the hub contention in the apply() work is 0, at that point, the
lambda work gets applied to every segment, and in the event that 1, at that point, the capacity gets applied to each column.
The channel() work takes a lambda work and a Pandas arrangement and applies for the lambda work on the arrangement and channels the information. This profits a grouping of True and False, which we
use for sifting the information. Hence, the info size of the guide() work is consistently more noteworthy than the yield size. This guide() work maps the arrangement as per input correspondence. It
is useful when we need to substitute an arrangement with different qualities. In map() works, the size of the info is equivalent to the size of the yield.
Lambda works likewise uphold restrictive proclamations, for example, if. Else. This makes lambda works amazing. Lambda capacities are very helpful when you are working with a great deal of iterative
code. Reduce () work applies the lambda capacity to the initial two components of the arrangement and returns the outcome. At that point, it stores that outcome and again applies a similar lambda
capacity to the outcome and the following component in the arrangement. Consequently, it diminishes the arrangement to a solitary worth. Lambda works in reduce() cannot take multiple contentions.
In Pandas, we have the opportunity to add various capacities at whatever point required, like lambda work, sort work, and so on. We can apply a lambda capacity to both the sections and lines of the
Pandas information outline.
Hence we would like to conclude by stating that lambda functions are characterized utilizing the watchword lambda. They can have quite a few contentions yet just a single articulation. A lambda work
cannot contain any assertions, and it restores a capacity object which can be appointed to any factor. They are commonly utilized for one-line articulations. Ordinary capacities are made utilizing
the def watchword. They can have quite a few contentions and quite a few articulations. They can contain any assertions and are commonly utilized for huge squares of code.
Recommended Articles
We hope that this EDUCBA information on “Pandas Lambda” was beneficial to you. You can view EDUCBA’s recommended articles for more information. | {"url":"https://www.educba.com/pandas-lambda/","timestamp":"2024-11-02T02:44:59Z","content_type":"text/html","content_length":"308848","record_id":"<urn:uuid:a1528c22-aa40-46da-a92e-19fef788dcfa>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00081.warc.gz"} |
[isabelle-dev] Group theory developments on Jacobson Basi...
From: Lawrence Paulson <lp15@cam.ac.uk>
Clemens gave us a new formalisation of basic algebra that has many advantages over the earlier HOL-Algebra: https://www.isa-afp.org/entries/Jacobson_Basic_Algebra.html
Other AFP entries have added onto it. There is new material in Grothendieck_Schemes and I’ve prepared additional material for a pending submission. The question is what to do with all this.
One idea would be to put it all into Clemens’ original development, but that can’t be right because was his experiment on the formalisation of a particular textbook (by Jacobson). But would it be
right to make a copy of this AFP entry and just keep adding to it over time? This also is not normal for the AFP. Or should it be added directly to the distribution?
Having a new copy would also make it easier to make changes to suit further developments. In particular, Clemens overloads the / symbol to denote division on groups, which could be irritating in
contexts where people need real division.
Thoughts would be welcome!
isabelle-dev mailing list
Last updated: Nov 11 2024 at 04:22 UTC | {"url":"http://isabelle.systems/zulip-archive/stream/247542-Mirror.3A-Isabelle-Development-Mailing-List/topic/.5Bisabelle-dev.5D.20Group.20theory.20developments.20on.20Jacobson.20Basi.2E.2E.2E.html","timestamp":"2024-11-11T04:38:36Z","content_type":"text/html","content_length":"3661","record_id":"<urn:uuid:577b993e-209d-4675-8366-d144f4fb2cf3>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00626.warc.gz"} |
| Readable
In 1964, he wrote a paper entitled "On the Einstein-Podolsky-Rosen Paradox". There is some disagreement regarding what Bell's inequality—in conjunction with the EPR analysis—can be said to imply.
Bell held that not only local hidden variables, but any and all local theoretical explanations must conflict with the predictions of quantum theory: "It is known that with Bohm's example of EPR
correlations, involving particles with spin, there is an irreducible nonlocality." According to an alternative interpretation, not all local theories in general, but only local hidden variables
theories (or "local realist" theories) have shown to be incompatible with the predictions of quantum theory. | {"url":"http://www.onreadable.com/e251HcnJ","timestamp":"2024-11-03T06:52:37Z","content_type":"text/html","content_length":"44734","record_id":"<urn:uuid:efb1c536-b973-4724-bb00-b1b76f7f8ba4>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00753.warc.gz"} |
Theory of Computation: Question Set – 27
Is the set of all context-free languages decidable?
No, the set of all context-free languages is not decidable. This can be shown using techniques such as the pumping lemma for context-free languages, or by reduction from an undecidable problem such
as the halting problem.
Is the set of all regular languages decidable?
Yes, the set of all regular languages is decidable, meaning that there exists an algorithm that can determine whether any given language is regular or not. This can be done using techniques such as
constructing a finite automaton that recognizes the language or using the pumping lemma to show that the language is not regular.
Can a language be undecidable but still computably enumerable?
Yes, it is possible for a language to be undecidable but still computably enumerable, meaning that there exists an algorithm that can list all the strings in the language, although it may not halt or
produce an answer for strings that do not belong to the language. An example of such a language is the halting problem, which is undecidable but computably enumerable.
What is the difference between decidability and computability?
Decidability refers to whether a problem can be solved algorithmically, while computability refers to whether a function can be computed by an algorithm. Decidable problems can always be computed,
while computable functions may not necessarily be decidable.
What is the difference between a context-sensitive grammar and a unrestricted grammar?
A context-sensitive grammar can generate context-sensitive languages, which are more powerful than context-free languages. Unrestricted grammars are the most powerful type of grammar and can generate
all possible languages.
What is the difference between a turing decidable and a turing recognizable language?
A Turing decidable language is a language that can be decided by a Turing machine, which means there exists an algorithmic procedure that can determine whether any given input is in the language or
not. A Turing recognizable language is a language that can be recognized by a Turing machine, which means there exists an algorithmic procedure that can accept any input in the language, but may loop
forever on inputs that are not in the language.
What is the difference between a decidability problem and a complexity problem?
A decidability problem is one in which the goal is to determine whether a certain input satisfies a certain property or belongs to a certain language. A complexity problem is one in which the goal is
to determine how efficiently a certain algorithm can solve a certain problem, usually measured in terms of time or space complexity.
What is the Cook-Levin theorem?
The Cook-Levin theorem states that the Boolean satisfiability problem (SAT) is NP-complete, which means that any problem in the NP complexity class can be reduced to SAT in polynomial time. This is
one of the most important results in the theory of computation, as it shows that many important computational problems are intractable. | {"url":"https://codecrucks.com/question/theory-of-computation-question-set-27/","timestamp":"2024-11-12T08:58:33Z","content_type":"text/html","content_length":"114353","record_id":"<urn:uuid:486383b4-09e2-4aa0-9882-c2529e30f3a5>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00767.warc.gz"} |
Flow of oil through an orifice
By making the orifice with a knife edge, it becomes insensitive to temperature, and the flow and pressure drop will remain the same over a reasonable range of oil
where Q = mass flow in units of mass/time, D = orifice diameter in units of length, p 1 = upstream pressure in units of mass/length 2, T 1 = upstream absolute temperature, and k s = the flow
constant. Calculating flow. Measure the length of the orifice with a tape measure. Record the distance in meters (1 foot equals 0.304 meters) on a piece of paper. Drop an object (a rubber ball will
work) at the top of the orifice and record the number of seconds the object takes to reach the end of the orifice. Flow of Liquids Through Orifices. Calculate the beta ratio, Reynolds number,
Discharge coefficient, Mass and Volumetric flow rate of a Newtonian fluid through an orifice with a Corner, 1D - ½ D, or Flange tap arrangement. Theory for this calculator can be found on pages 4-2
thru 4-5 in the 2009 edition of TP410. Water Flow Rate through a Valve; Water Flow Rate through an Orifice; Air. Piping Design. Pipe Sizing by Pressure Loss; Pipe Sizing by Velocity; Pressure Loss
through Piping; Air Velocity through Piping; Air Flow Rate through Piping; Valves and Orifices. Cv & Kvs Values; Air Flow Rate through a Valve; Air Flow Rate through an Orifice; Condensate Load from
Compressed Air Task: Calculate flow rate of water flowing through orifice plate with external diameter of 120 mm and internal diameter of 80 mm. Measured pressures in front and after the orifice is
11000 mm H 2 O and 10000 mm H 2 O. Pressure is measured on 1 inch taps.
11 May 2019 Keywords: multi-phase flow; offshore; oil and gas; flow metering; pressure spectral analysis for two-phase flow through an orifice plate.
This example considers the turbulent flow of an oil-water suspension through an orifice. The oil droplets are broken up into smaller droplets by the turbulent 10 Apr 2013 The cementing is done by
inserting a pipe in the center of the well and pumping cement slurry down through it, so that the slurry then rises back Select a media, enter your data and let the calculator do the rest. It
features conversion calculators to help you convert between metric & imperial values. measuring a viscous fluid such as fossil oil, lube, beverages, and dairy section orifice flow meter. Region C is
the flow area through the orifice, which is also. 19 Jul 2010 (C) was investigated through differential pressure flow meters. that has been performed in this area for the Venturi, standard orifice
plate, V-cone, and focus of the study was to determine heavy oil fluid flows with viscosities Prorated Wells. The oil flow shall be stabilized during the 24 hour period immediately Orifice Meters.
The orifice type meter is suitable for the measurement of all volumes and, the gas to flow through the nipple to the atmosphere. Connect a (b) When a viscous fluid flows through a tube, its speed at
the walls is zero, Motor oil has greater viscosity when cold than when warm, and so pressure must
7 Aug 2017 Write down the flow of the liquid that will be going through the piping system in cubic feet per second. For example, the flow of the liquid in a
Calculation of Flow through Nozzles and Orifices Summary This article provides calculation methods for correlating design, flow rate and pressure loss as a fluid passes through a nozzle or orifice.
where Q = mass flow in units of mass/time, D = orifice diameter in units of length, p 1 = upstream pressure in units of mass/length 2, T 1 = upstream absolute temperature, and k s = the flow
constant. Calculating flow. Measure the length of the orifice with a tape measure. Record the distance in meters (1 foot equals 0.304 meters) on a piece of paper. Drop an object (a rubber ball will
work) at the top of the orifice and record the number of seconds the object takes to reach the end of the orifice. Flow of Liquids Through Orifices. Calculate the beta ratio, Reynolds number,
Discharge coefficient, Mass and Volumetric flow rate of a Newtonian fluid through an orifice with a Corner, 1D - ½ D, or Flange tap arrangement. Theory for this calculator can be found on pages 4-2
thru 4-5 in the 2009 edition of TP410. Water Flow Rate through a Valve; Water Flow Rate through an Orifice; Air. Piping Design. Pipe Sizing by Pressure Loss; Pipe Sizing by Velocity; Pressure Loss
through Piping; Air Velocity through Piping; Air Flow Rate through Piping; Valves and Orifices. Cv & Kvs Values; Air Flow Rate through a Valve; Air Flow Rate through an Orifice; Condensate Load from
Compressed Air
Flow of Liquids Through Orifices. Calculate the beta ratio, Reynolds number, Discharge coefficient, Mass and Volumetric flow rate of a Newtonian fluid through an orifice with a Corner, 1D - ½ D, or
Flange tap arrangement. Theory for this calculator can be found on pages 4-2 thru 4-5 in the 2009 edition of TP410.
solve problems involving flow through Venturi meters. Oil flows in a pipe 80 mm bore with a mean velocity of 4 m/s. FLOW THROUGH AN ORIFICE. Size an orifice plate to be used in conjunction with a
flow transmitter to serve as an accurate flow meter for natural gas. •. Evaluate the measurement uncertainty of Flow Rate Chart for Orifice Union Hole Sizes (Gallons Per Minute). 1/64", 1/32", 1/
16", 1/8", 3/16", 1/4", 3/8", 1/2", 5/8", 3/4", 7/8", 1". 1. 7.45, 0.0046, 0.0186, 0.072 26 Oct 2017 for measurement of natural gas in the upstream sector of the oil and gas industry. While there
are several vital components to an orifice meter, the the plate's contact with the flowing gas stream passing through the meter. This example considers the turbulent flow of an oil-water suspension
through an orifice. The oil droplets are broken up into smaller droplets by the turbulent 10 Apr 2013 The cementing is done by inserting a pipe in the center of the well and pumping cement slurry
down through it, so that the slurry then rises back Select a media, enter your data and let the calculator do the rest. It features conversion calculators to help you convert between metric &
imperial values.
study of the flow of air through orifices by a precision method, and to determine The weight of air flowing through an orifice depends on the pres- sure P in the throat of the orifice; (3) It is
advisable to use water or oil as the manometer liquid.
measuring a viscous fluid such as fossil oil, lube, beverages, and dairy section orifice flow meter. Region C is the flow area through the orifice, which is also. 19 Jul 2010 (C) was investigated
through differential pressure flow meters. that has been performed in this area for the Venturi, standard orifice plate, V-cone, and focus of the study was to determine heavy oil fluid flows with
viscosities Prorated Wells. The oil flow shall be stabilized during the 24 hour period immediately Orifice Meters. The orifice type meter is suitable for the measurement of all volumes and, the gas
to flow through the nipple to the atmosphere. Connect a (b) When a viscous fluid flows through a tube, its speed at the walls is zero, Motor oil has greater viscosity when cold than when warm, and
so pressure must 18 Apr 2017 Then, the oil passes through the filter, overcomes the flow resistance of gearbox flows through the orifice E into the internal tube of the output STREAM FLOW THROUGH
STRAIGHT PIPES AND CHANNELS (FRICTION Mineral lubricating oil THE FLOW OF FLUIDS THROUGH AN ORIFICE a. 7 Aug 2017 Write down the flow of the liquid that will be going through the piping system in
cubic feet per second. For example, the flow of the liquid in a
Flow Rate Chart for Orifice Union Hole Sizes (Gallons Per Minute). 1/64", 1/32", 1/ 16", 1/8", 3/16", 1/4", 3/8", 1/2", 5/8", 3/4", 7/8", 1". 1. 7.45, 0.0046, 0.0186, 0.072 26 Oct 2017 for
measurement of natural gas in the upstream sector of the oil and gas industry. While there are several vital components to an orifice meter, the the plate's contact with the flowing gas stream
passing through the meter. This example considers the turbulent flow of an oil-water suspension through an orifice. The oil droplets are broken up into smaller droplets by the turbulent 10 Apr 2013
The cementing is done by inserting a pipe in the center of the well and pumping cement slurry down through it, so that the slurry then rises back Select a media, enter your data and let the
calculator do the rest. It features conversion calculators to help you convert between metric & imperial values. measuring a viscous fluid such as fossil oil, lube, beverages, and dairy section
orifice flow meter. Region C is the flow area through the orifice, which is also. | {"url":"https://topbinhbrjq.netlify.app/claypoole35466civa/flow-of-oil-through-an-orifice-sagu.html","timestamp":"2024-11-05T10:17:06Z","content_type":"text/html","content_length":"34563","record_id":"<urn:uuid:a229592e-71f5-4db0-b64f-eb0e61c97151>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00709.warc.gz"} |
The probability that student pilot passes the written test for a private pilot's license is 0.
Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
The probability that student pilot passes the written test for a private pilot's license is 0.
The probability that student pilot passes the written test for a private pilot’s license is 0.7. Find the probability that the student will pass the test (a) on the third attempt (b) before the
fourth attempt. Also find the average number of attempts needed to pass the test.
Show more
Homework Categories
Ask a Question | {"url":"https://studydaddy.com/question/the-probability-that-student-pilot-passes-the-written-test-for-a-private-pilot-s","timestamp":"2024-11-11T15:02:57Z","content_type":"text/html","content_length":"25831","record_id":"<urn:uuid:4562bf77-5c42-462a-b08e-7895b64c1d7e>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00716.warc.gz"} |
Functions and Their Graphs
Site: Saylor Academy
Course: MA005: Calculus I
Book: Functions and Their Graphs
Printed by: Guest user
Date: Tuesday, November 5, 2024, 1:45 AM
Read this section for an introduction to functions and their graphs. Work through practice problems 1-5.
Functions and Their Graphs
When you prepared for calculus, you learned to manipulate functions by adding, subtracting, multiplying and dividing them, as well as calculating functions of functions (composition). In calculus, we
will still be dealing with functions and their applications. We will create new functions by operating on old ones. We will derive information from the graphs of the functions and from the derived
functions. We will find ways to describe the point–by–point behavior of functions as well as their behavior "close to" some points and also over entire intervals. We will find tangent lines to graphs
of functions and areas between graphs of functions. And, of course, we will see how these ideas can be used in a variety of fields.
This section and the next one are a review of information and procedures you should already know about functions before we begin calculus.
Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-1.2-Lines-in-the-Plane.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.
What is a Function?
Definition of Function:
A function from a set $X$ to a set $Y$ is a rule for assigning to each element of the set $X$ a single element of the set $Y$. A function assigns a unique (exactly one) output element in the set $Y$
to each input element from the set $X$.
The rule which defines a function is often given by an equation, but it could also be given in words or graphically or by a table of values. In practice, functions are given in all of these ways, and
we will use all of them in this book.
In the definition of a function, the set $X$ of all inputs is called the domain of the function. The set $Y$ of all outputs produced from these inputs is called the range of the function. Two
different inputs, elements in the domain, can be assigned to the same output, an element in the range, but one input cannot lead to 2 different outputs.
Most of the time we will work with functions whose domains and ranges are real numbers, but there are other types of functions all around us. Final grades for this course is an example of a function.
For each student, the instructor will assign a final grade based on some rule for evaluating that student's performance. The domain of this function consists of all students registered for the
course, and the range consists of the letters $A, B, C, D, F,$ and perhaps $W$ (withdrawn). Two students can receive the same final grade, but only one grade will be assigned to each student.
Function Machines
Functions are abstract structures, but sometimes it is easier to think of them in a more concrete way. One way is to imagine that a function is a special purpose computer, a machine which accepts
inputs, does something to those inputs according to the defining rule, and produces an output. The output is the value of the function for the given input value. If the defining rule for a function
$f$ is "multiply the input by itself" , $f \; (input) = (input)(input)$ , then Fig. 1 shows the results of putting the inputs $x, 5, a, c + 3$ and $x + h$ into the machine $f$.
Practice 1: If we have a function machine g whose rule is "divide 3 by the input and add 1", $g(x) = 3/x + 1$, what outputs do we get from the inputs $x, 5, a, c + 3$ and $x + h$ ? What happens if we
put 0 into the machine $g$?
You expect your calculator to behave as a function: each time you press the same input sequence of keys you expect to see the same output display. In fact, if your calculator did not produce the same
output each time you would need a new calculator. (On many calculators there is a key which does not produce the same output each time you press it. Which key is that?)
Functions Defined by Equations
If the domain consists of a collection of real numbers (perhaps all real numbers) and the range is a collection of real numbers, then the function is called a numerical function. The rule for a
numerical function can be given in several ways, but it is usually written as a formula. If the rule for a numerical function, $f$, is "the output is the input number multiplied by itself", then we
could write the rule as $f(x) = x.x = x^2$ . The use of an "$x$" to represent the input is simply a matter of convenience and custom. We could also represent the same function by $f(a) = a^2$, $f (€
\# ) = \#^2$ or $f( input ) = ( input )^2$ .
For the function f defined by $f(x) = x^2 – x$ , we have that $f(3) = 3^2 – 3 = 6$, $f(.5) = (.5)^2 – (.5) = –.25$, and $f(–2) = (–2)^2 – (–2) = 6$. Notice that the two different inputs, 3 and –2,
both lead to the output of 6. That is allowable for a function. We can also evaluate f if the input contains variables. If we replace the "$x$" with something else in the notation "$f(x)$", then we
must replace the "$x$" with the same thing everywhere in the equation:
$f(c) = c^2 – c , f(a+1) = (a+1)^2 – (a+1) = (a^2 + 2a + 1) – (a + 1) = a^2 + a$,
$f(x+h) = (x+h)^2 – (x+h) = (x^2+2xh+h^2) – (x+h)$ , and, in general, $f(input) = (input)^2 – (input)$.
For more complicated expressions, we can just proceed step–by–step:
$\frac{f(x+h) – f(x)}{h} = \dfrac{{(x+h)^2 – (x+h)} – {x^2 – x}}{h} = \dfrac{{(x^2+2xh+h^2) – (x+h)} – {x^2 – x}}{h}$
$= \dfrac{2xh + h^2 – h}{h} = \dfrac{h(2x + h – 1)}{h} = 2x + h – 1$.
Practice 2: For the function $g$ defined by $g(t) = t^2 – 5t$ , evaluate $g(1), g(–2), g(w+3), g(x+h), g(x+h) – g(x)$, and $\frac{g(x+h) – g(x)}{h}$.
Functions Defined by Graphs and Tables of Values
The graph of a numerical function $f$ consists of a plot of ordered pairs $(x, y)$ where $x$ is in the domain of $f$ and $y = f(x)$. A table of values of a numerical function consists of a list of
some of the ordered pairs $(x, y)$ where $y = f(x)$. Fig. shows a graph of $f(x) = sin(x)$ for $–4 ≤ x ≤ 9$.
A function can be defined by a graph or by a table of values, and these types of definitions are common in applied fields. The outcome of an experiment will depend on the input, but the experimenter
may not know the "rule" for predicting the outcome. In that case, the experimenter usually represents the experiment function as a table of measured outcome values verses input values or as a graph
of the outcomes verses the inputs. The table and graph in Fig. 3 show the deflections obtained when weights are loaded at the end of a wooden stick. The graph in Fig. 4 shows the temperature of a hot
cup of tea as a function of the time as it sits in a 68^o F room. In these experiments, the "rule" for the function is that $f(input) =$ actual outcome of the experiment.
Tables have the advantage of presenting the data explicitly, but it is often difficult to detect patterns simply from lists of numbers. Graphs tend to obscure some of the precision of the data, but
patterns are much easier to detect visually - we can actually see what is happening with the numbers.
Creating Graphs of Functions
Most people understand and can interpret pictures more quickly than tables of data or equations, so if we have a function defined by a table of values or by an equation, it is often useful and
necessary to create a picture of the function, a graph.
A Graph from a Table of Values
If we have a table of values of the function, perhaps consisting of measurements obtained from an experiment, then we can simply plot the ordered pairs in the xy–plane to get a graph which consists
of a collection of points.
Fig. 5 shows the lengths and weights of trout caught (and released) during several days of fishing. It also shows a line which comes "close" to the plotted points. From the graph, you could estimate
that a 17 inch trout would weigh slightly more than one pound.
A Graph from an Equation
Creating the graph of a function given by an equation is similar to creating one from a table of values - we need to plot enough points $(x,y)$ where $y = f(x)$ so we can be confident of the shape
and location of the graph of the entire function. We can find a point $(x,y)$ which satisfies $y = f(x$) by picking a value for $x$ and then calculating the value for $y$ by evaluating $f(x)$. Then
we can enter the $(x,y)$ value in a table or simply plot the point $(x,y)$.
If you recognize the form of the equation and know something about the shape of graphs of that form, you may not have to plot many points. If you do not recognize the form of the equation then you
will have to plot more points, maybe 10 or 20 or 234: it depends on how complicated the graph appears and on how important it is to you (or your boss) to have an accurate graph. Evaluating $y = f(x)$
at a lot of different values for $x$ and then plotting the points $(x,y)$ is usually not very difficult, but it can be very time–consuming. Fortunately, there are now calculators and personal
computers which will do the evaluations and plotting for you.
Is Every Graph the Graph of a Function?
The definition of function requires that each element of the domain, each input value, be sent by the function to exactly one element of the range, to exactly one output value, so for each input
x-value there will be exactly one output y–value, $y = f(x)$. The points $(x, y_1)$ and $(x,y_2)$ cannot both be on the graph of $f$ unless $y_1 = y_2$. The graphic interpretation of this result is
called the Vertical Line Test.
Vertical Line Test for a Function:
A graph is the graph of a function if and only if a vertical line drawn through any point in the domain intersects the graph at exactly one point.
Fig. 6(a) shows the graph of a function. Figs. 6(b) and 6(c) show graphs which are not the graphs of functions, and vertical lines are shown which intersect those graphs at more than one point.
Non–functions are not "bad", and sometimes they are necessary to describe complicated phenomena.
Reading Graphs Carefully
Calculators and computers can help students, reporters, business people and scientific professionals create graphs quickly and easily, and because of this, graphs are being used more often than ever
to present information and justify arguments. And this text takes a distinctly graphical approach to the ideas and meaning of calculus. Calculators and computers can help us create graphs, but we
need to be able to read them carefully. The next examples illustrate some types of information which can be obtained by carefully reading and understanding graphs.
Example 1: A boat starts from St. Thomas and sails due west with the velocity shown in Fig. 7
(a) When is the boat traveling the fastest?
(b) What does a negative velocity away from St. Thomas mean?
(c) When is the boat the farthest from St. Thomas?
Solution: (a) The greatest speed is 10 mph at $t = 3 \; hours$.
(b) It means that the boat is heading back toward St. Thomas.
(c) The boat is farthest from St. Thomas at $t = 6 \; hours$. For $t < 6$ the boat's velocity is positive, and the distance from the boat to St. Thomas is increasing. For $t > 6$ the boat's velocity
is negative, and the distance from the boat to St. Thomas is decreasing.
Practice 3: You and a friend start out together and hike along the
same trail but walk at different speeds (Fig. 8).
(a) Who is walking faster at $t = 20$?
(b) Who is ahead at $t = 20$?
(c) When are you and your friend farthest apart?
(d) Who is ahead when $t = 50$?
Example 2: In Fig. 9, which has the largest slope: the line through the points $A$ and $P$, the line through $B$ and $P$, or the line through $C$ and $P$?
Solution: The line through C and P has the largest slope: $m_{PC} > m_{PB} > m_{PA}$.
Practice 4: In Fig. 10, the point $Q$ on the curve is fixed, and the point $P$ is moving to the right along the curve toward the point $Q$. As $P$ moves toward $Q$:
(a) the x–coordinate of $P$ is Increasing, Decreasing, Remaining constant, or None of these.
(b) the x–increment from $P$ to $Q$ is Increasing, Decreasing, Remaining constant, or None of these
(c) the slope from $P$ to $Q$ is Increasing, Decreasing, Remaining constant, or None of these.
Example 3: The graph of $y = f(x)$ is shown in Fig. 11. Let $g(x)$ be the slope of the line tangent to the graph of $f(x)$ at the point $(x,f(x))$.
(a) Estimate the values $g(1)$, $g(2)$ and $g(3)$.
(b) When does $g(x) = 0)$?
(c) At what value(s) of $x$ is $g(x)$ largest?
(d) Sketch the graph of $y = g(x)$.
Solution: (a) Fig. 11 shows the graph $y = f(x)$ with several tangent lines to the graph of $f$. From Fig. 11 we can estimate that $g(1)$ (the slope of the line tangent to the graph of f at (1,0) )
is approximately equal to 1. Similarly, $g(2) ≈ 0$ and $g(3) ≈ –1$.
(b) The slope of the tangent line appears to be horizontal ($slope = 0$) at $x = 2$ and at $x = 5$.
(c) The tangent line to the graph appears to have greatest slope (be steepest) near $x = 1.5$.
(d) We can build a table of values of $g(x)$ and then sketch the graph of these values.
│ x │ f(x) │ g(x) = tangent slope at (x, f(x) ) │
│ 0 │ -1 │ .5 │
│ 1 │ 0 │ 1 │
│ 2 │ 2 │ 0 │
│ 3 │ 1 │ -1 │
│ 4 │ 0 │ -1 │
│ 5 │ -1 │ 0 │
│ 6 │ -.5 │ .5 │
The graph $y = g(x)$ is given in Fig. 12.
Practice 5: Water is flowing into a container (Fig. 13) at a constant rate of 3 gallons per minute. Starting with an empty container, sketch the graph of the height of the water in the container as a
function of time.
Practice Problem Answers
Practice 1:
│ Input │ Output │
│ $x$ │ $\dfrac{3}{x}+1$ │
│ 5 │ $\dfrac{3}{5}+1 = 1.6$ │
│ $a$ │ $\dfrac{3}{a}+1$ │
│ 0 │ $g(0) =\dfrac{3}{0} +1$ which is not defined because of division by 0. │
│ Input │ Output │
│ $c+3$ │ $\dfrac{3}{c+3} +1$ │
│ $x+h$ │ $\dfrac{3}{x+h} +1$ │
Practice 2: $g(t) = t^2 – 5t$.
$g(1) = 1^2 – 5(1) = –4$. $g(–2) = (–2)^2 – 5(–2) = 14$.
$g(w + 3) = (w + 3)^2 –5(w + 3) = w^2 + 6w + 9 – 5w – 15 = w^2 + w – 6$.
$g(x + h) = (x + h) – 5(x + h) = x^2 + 2xh + h^2 – 5x – 5h$.
$g(x + h) – g(x) = ( x^2 + 2xh + h^2 – 5x – 5h ) – ( x^2 – 5x ) = 2xh + h^2 – 5h$.
$\dfrac{g(x + h) – g(x)}{h} = \dfrac{2xh + h^2 – 5h}{h} = 2x + h – 5$.
Practice 3: (a) Friend (b) Friend (c) At $t = 40$. Before that your friend is walking faster and increasing the distance between you. Then you start to walk faster than your friend and start to catch
up. (d) Friend. You are walking faster than your friend at $t = 50$, but you still have not caught up.
Practice 4: (a) The x–coordinate is increasing. (b) The x–increment $∆x$ is decreasing. (c) The slope of the line through $P$ and $Q$ is decreasing.
Practice 5: See Fig. 31. | {"url":"https://learn.saylor.org/mod/book/tool/print/index.php?id=36965","timestamp":"2024-11-05T06:45:15Z","content_type":"text/html","content_length":"164662","record_id":"<urn:uuid:0b2295b1-1138-44f9-a873-d4c712cfac0e>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00145.warc.gz"} |
Search term Research & Research: graph head matchmaking
Frequently asked questions
In my opinion a proper address about choice mentioned above try option A great. A direct dating should be illustrated from the term: An immediate relationships was a phrase between two numbers and
other variables in which an increase otherwise e improvement in the other worth.
secondary relationship. The connection ranging from a couple parameters hence move in reverse guidelines; whenever one of several details boosts the most other changeable minimizes. When you look at
the a corporate setting this will also make site to a relationship caused by alterations in interfacing practical communities.
A good example of a primary dating ‘s the heavy brand new basketball bat try, the fresh longer it will require to move during the a beneficial response day. Which is exactly how acceleration and push
is actually privately associated. A typical example of a keen inverse matchmaking is when that you do not work during the studies, then you’ll definitely have significantly more research.
A primary relationship, otherwise known as a primary version, try a relationship ranging from a couple details which means if that is changed one other are altered in a similar way. Ex: “Y physically
may vary having X” that means Y=c*X, where c is a few lingering.
Serp’s connected with graph head relationships to the Search-engine
• What is an immediate Matchmaking Chart? – Source
· A direct relationship chart is actually a tsdates giriЕџ graph in which you to definitely varying either expands or decrease along with the almost every other. All round picture getting a primary
relationship chart is y = mx + b, in which “y” indicates the newest situated adjustable, “x” indicates the fresh new separate varying, “m” means the latest mountain of your own range and you may “b”
is the y-intercept.
This really is good “head dating”. Now, let’s go through the following formula: Y = 20/X If X=1 next Y = 20. In the event that X = dos, upcoming Y = 10. If X = 3 then Y = 6.eight. If X = cuatro, next
Y = 5. Remember that given that X develops Y decreases in a low-linear trends. This is an enthusiastic invercreases Y reduces in a non-linear trends. This is exactly an inverse relationships where X
1 /X dos = Y dos /Y 1. The fresh chart is actually found below:
· What’s a primary relationship to the a graph? An immediate dating demonstrates both variables move around in the fresh new exact same advice. A confident slope regarding a line plotted into an
excellent spread out drawing means a primary relationship within variables. Inside graphing some observations, might patch one to adjustable on every axis.
· Linear relationships is actually graphed when you look at the a straight line, whereas head relationship will likely be linear, even so they get involve a bend in general variable transform at the
a special rates. Discuss these two distinctive line of.
A brought chart is defined as G roentgen = [V, E], in which V is a couple of nodes, and Age is a beneficial subset out-of N ? Letter sides on graph . The very first chart systems is relational and
you may blamed graphs. Overall, a good relational graph makes reference to the dwelling away from a cycle.
The latest graph is a typical example of an immediate variation. 1) The interest rate away from transform is actually ongoing ($$ k = 1/1 = 1), therefore, the chart is actually linear. 2) Brand new
line goes through the origin (0, 0). step three ) Brand new formula of head version is $$ y =step one x or maybe just $$ y = x. What is the opposite off an enthusiastic inverse matchmaking?
Graphing Relationships. Graphing Relationship. Graphing Matchmaking is…. step 1. …definitions regarding exactly how a couple of variables relate with one another toward a great graph. 2. …usually
planned “Given that _________ develops, __________ (increases/minimizes.)”. step three. …fall into cuatro groups: a) Head Relationships b) Indirect Relationships c) Cyclic Dating d) Vibrant Harmony
Direct … Author: 2009168 Past changed from the: 2009168 Authored Date: 9/ 3: PM Title: Graphing Dating
a love, and even does not really you need a graph if you wish to determine – it might be apparent throughout the table regarding efficiency. Observe that some graphs don’t merely go sometimes upwards
or off,
I believe a proper respond to on the selection mentioned above are alternative Good. A primary matchmaking will be depicted by the phrase: A primary relationship is a phrase anywhere between several
quantity and other parameters where a rise or e change in another value.
indirect matchmaking. The connection anywhere between a couple of details and that relocate contrary rules; whenever among the variables advances the most other variable decrease. Inside a business
function this may as well as refer to a romance because of alterations in interfacing useful groups.
A good example of an immediate relationships ‘s the big the fresh new baseball bat is, the fresh new extended it will take so you’re able to swing at good effect day. That’s how acceleration and
force are physically associated. A typical example of an inverse dating is when you never works into the knowledge, then you will do have more homework.
A direct dating, also referred to as a primary adaptation, is a relationship between one or two details that means whenever you to definitely try altered others try changed similarly. Ex: “Y
personally may vary having X” that implies Y=c*X, in which c is a few ongoing.
Deja un comentario | {"url":"https://cchumanista.com/2022/08/23/search-term-research-research-graph-head-2/","timestamp":"2024-11-11T01:03:33Z","content_type":"text/html","content_length":"49903","record_id":"<urn:uuid:4028dc9c-18d6-4774-af2a-2ef6737553fc>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00375.warc.gz"} |
T-79.1002 Introduction to Theoretical Computer Science Y
T-79.1002 Introduction to Theoretical Computer Science Y (2 cr)
Autumn 2006
This short course introduces the basic tools for dealing with data streams consisting of sequences of discrete symbols: finite automata and regular languages, and context-free grammars and languages.
Starting in 2006/07, the course is lectured in the autumn semester only.
The course covers the first half of a more extensive course on models of computation, T-79.1001 Introduction to Theoretical Computer Science T (4 cr) . For topical information, please refer to the
WWW info page of that course. This short version is mainly intended for students of other study programmes than computer science (T). Computer science students, and telecommunications students
following a pre-2005 curriculum, are required to take the larger course. (The course corresponding to the old course code T-79.148 is the long version T-79.1001. You can not compensate course
T-79.148 in your curriculum requirements by course T-79.1002.)
Note that the course includes compulsory computer assignments that need to be completed before participating in the examination. For details, see the registration information for the larger course.
[Current] [General] [Lectures] [Tutorials] [Exams] [Material] [Feedback] [Links]
Previous years: [Spring 2006] [Autumn 2005]
• In 2006/2007 exams are scheduled for 30 Aug, 26 Oct, 21 Dec, 6 Mar, and 10 May. Registration via TOPI. All compulsory Regis assignments must have been completed before participating in the exam.
• First lecture Thu 14 Sep 2-4 p.m., Lecture Hall T1.
• Registration via TOPI opens 1 Sep at 09:00 and closes 28 Sep at 18:00. Registration is compulsory. Register using course code T-79.1001 even if you're really taking T-79.1002.
[TCS main] [Contact Info] [Personnel] [Research] [Publications] [Software] [Studies] [News Archive] [Links] Latest update: 15 January 2007. Harri Haanpää | {"url":"http://www.tcs.hut.fi/Studies/T-79.1002/2006AUT/","timestamp":"2024-11-05T10:50:12Z","content_type":"text/html","content_length":"4720","record_id":"<urn:uuid:543b178b-4202-456a-85f2-5e2a5a5916a6>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00885.warc.gz"} |
A projectile is shot from the ground at an angle of (5 pi)/12 and a speed of 3 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height? | Socratic
A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #3 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting
point be when it reaches its maximum height?
1 Answer
$.230$ meters from the starting point.
We should start by breaking down the initial velocity into its $x$ and $y$ components. We can find the projection of $\vec{V}$ onto the $x$ and $y$ axis by using $\cos \theta$ and $\sin \theta$
$\vec{V} = \vec{V} \cdot \hat{x} + V \cdot \hat{y}$
$= \left(V \cos \theta\right) \hat{x} + \left(V \sin \theta\right) \hat{y}$
$= 3 \cos \left(\frac{5 \pi}{12}\right) \hat{x} + 3 \sin \left(\frac{5 \pi}{12}\right) \hat{y}$
$= .776 \hat{x} + 2.898 \hat{y}$
Note that the units are $\text{m/s}$. The initial velocity is mostly in the $y$ direction, so we shouldn't expect the projectile to go very far. Now we can calculate how long it takes for the
projectile to reach the highest point. At its highest, the vertical velocity of the projectile will be zero. We can use the following equation to find at what time the velocity is zero.
${V}_{y} \left(t\right) = a t + {V}_{\circ} = 0$
$t = - {V}_{\circ} / a$
The acceleration due to gravity is $- 9.8 \frac{m}{s} ^ 2$.
$t = - \frac{2.989}{- 9.8}$
$t = .296$
So the projectile takes $.296 \text{ seconds}$ to reach the top of its flight. Now we can plug this into the equation of motion for the $x$ direction. Since there is no net force on the projectile in
the horizontal direction, the acceleration will be zero. Also, we are starting at the origin, so the initial $x$ position will be zero.
$x \left(t\right) = \frac{1}{2} {\cancel{a}}^{0} {t}^{2} + {V}_{x} t + {\cancel{{x}_{\circ}}}^{0}$
$x \left(.296\right) = \left(.776\right) \left(.296\right)$
$x \left(.296\right) = .230$
The projectile will move about $.230 \text{ m}$ from the starting point when it reaches the top of its flight.
Impact of this question
2260 views around the world | {"url":"https://socratic.org/questions/a-projectile-is-shot-from-the-ground-at-an-angle-of-5-pi-12-and-a-speed-of-3-m-s#210047","timestamp":"2024-11-04T18:03:00Z","content_type":"text/html","content_length":"37361","record_id":"<urn:uuid:5262f073-83d3-4f26-9303-e8fdb4b4f6a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00348.warc.gz"} |
Exercises - Cardinality and Infinite Sets
$f:\mathbb{N}\rightarrow\mathbb{N}$ where $f(n)=n^2$
$f:\mathbb{Z}\rightarrow\mathbb{Z}$ where $f(n)=n+1$
$f$ is injective, since $f(a)=f(b) \rightarrow a+1=b+1 \rightarrow a=b$;
$f$ is surjective, as for any integer $n$, $f(n-1)=n$ and $(n-1)$ is an integer;
Since $f$ is both injective and surjective, it is also bijective.
$f:\mathbb{N}\rightarrow\mathbb{N}$ where $f(n)=n+1$
$f$ is injective, since $f(a)=f(b) \rightarrow a+1=b+1 \rightarrow a=b$;
$f$ is not surjective, as $\not\exists a \in \mathbb{N} : f(a)=0$;
Since $f$ is not surjective, it can't be bijective.
$f$ is injective, since for $a,b \in \mathbb{N}$, we have $f(a)=f(b) \rightarrow a^2 = b^2 \rightarrow a = b$
(recall $a,b \in \mathbb{N} \rightarrow a,b \gt 0$);
$f$ is not surjective, as $\not\exists a \in \mathbb{N} : f(a)=2$;
Since $f$ is not surjective, it can't be bijective.
$f:\mathbb{Z}\rightarrow\mathbb{Z}$ where $f(n)=n^2$
$f$ is not injective as $f(1)=f(-1)$ (among others);
$f$ is not surjective, as $\not\exists a \in \mathbb{N} : f(a)=2$;
Since $f$ is not surjective or injective, it can't be bijective.
$f:\mathbb{Z} \rightarrow 2\mathbb{Z}$ where $f(n) = 2n+2$
$f$ is injective, as $f(a)=f(b) \rightarrow 2a+2=2b+2 \rightarrow 2a=2b \rightarrow a=b$;
$f$ is surjective, as for any $2k \in 2\mathbb{Z}$, note $f(k-1) = 2(k-1)+2 = 2k$ and $(k-1) \in \mathbb{Z}$;
Since $f$ is both injective and surjective, it is also bijective.
$f:\mathbb{N} \rightarrow 2\mathbb{Z}$ where $f(n) = 2n+2$
$f$ is injective, as $f(a)=f(b) \rightarrow 2a+2=2b+2 \rightarrow 2a=2b \rightarrow a=b$;
$f$ is not surjective, as $\not\exists a \in \mathbb{N} : f(a)=0$ (among others);
Since $f$ is not surjective, it can't be bijective.
$f$ is not injective as $f(2)=f(-2)$ (among others);
$f$ is surjective as for any $n \in \mathbb{N}$, note $f(2n) = n$ and $2n \in 2\mathbb{Z}$;
Since $f$ is not injective, it can't be bijective.
$f$ is not injective as $f(1)=f(-1)$ (among others);
$f$ is not surjective as $\not\exists a \in \mathbb{N} : f(a)=0$;
Since $f$ is not injective or surjective, it can't be bijective. | {"url":"https://mathcenter.oxford.emory.edu/site/math125/probSetCardinalityAndInfiniteSets/","timestamp":"2024-11-10T17:58:40Z","content_type":"text/html","content_length":"9045","record_id":"<urn:uuid:e54c591d-3d32-48f8-9a82-4a8c0d4b1405>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00829.warc.gz"} |
update: Capped and/or floored floating-rate coupon. - Linux Manuals (3)
update (3) - Linux Manuals
update: Capped and/or floored floating-rate coupon.
QuantLib::CappedFlooredCoupon - Capped and/or floored floating-rate coupon.
#include <ql/cashflows/capflooredcoupon.hpp>
Inherits QuantLib::FloatingRateCoupon.
Inherited by CappedFlooredCmsCoupon, and CappedFlooredIborCoupon.
Public Member Functions
CappedFlooredCoupon (const boost::shared_ptr< FloatingRateCoupon > &underlying, Rate cap=Null< Rate >(), Rate floor=Null< Rate >())
Rate cap () const
Rate floor () const
Rate effectiveCap () const
effective cap of fixing
Rate effectiveFloor () const
effective floor of fixing
Coupon interface
Rate rate () const
accrued rate
Rate convexityAdjustment () const
convexity adjustment
Observer interface
boost::shared_ptr< FloatingRateCoupon > underlying_
bool isCapped_
bool isFloored_
Rate cap_
Rate floor_
virtual void accept (AcyclicVisitor &)
bool isCapped () const
bool isFloored () const
void setPricer (const boost::shared_ptr< FloatingRateCouponPricer > &pricer)
Detailed Description
Capped and/or floored floating-rate coupon.
The payoff $ P $ of a capped floating-rate coupon is: [ P = N imes T imes min(a L + b, C). ] The payoff of a floored floating-rate coupon is: [ P = N imes T imes max(a L + b, F). ] The payoff of a
collared floating-rate coupon is: [ P = N imes T imes min(max(a L + b, F), C). ]
where $ N $ is the notional, $ T $ is the accrual time, $ L $ is the floating rate, $ a $ is its gearing, $ b $ is the spread, and $ C $ and $ F $ the strikes.
They can be decomposed in the following manner. Decomposition of a capped floating rate coupon: [ R = min(a L + b, C) = (a L + b) + min(C - b - on"
void update () [virtual]
This method must be implemented in derived classes. An instance of Observer does not call this method directly: instead, it will be called by the observables the instance registered with when they
need to notify any changes.
Reimplemented from FloatingRateCoupon.
Generated automatically by Doxygen for QuantLib from the source code. | {"url":"https://www.systutorials.com/docs/linux/man/3-update/","timestamp":"2024-11-08T20:14:20Z","content_type":"text/html","content_length":"9700","record_id":"<urn:uuid:ae9ac6da-0620-4fc6-a4ea-96d6e2ef4946>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00481.warc.gz"} |
How to create date and time series with formulas
Although you can use Excel's AutoFill feature to fill in a series of dates and times, you can also do the same thing with formulas. The advantage of using a formula is that you can easily change the
starting value and generate a new series.
Although you can use Excel's AutoFill feature to fill in a series of dates and times, you can also do the same thing with formulas. The advantage of using a formula is that you can easily change the
starting value and generate a new series.
Let's take a look.
Often you'll need to generate a series of dates separated by a certain interval of days, months, or years. You can easily do this with Excel's Date functions.
For example, assume you want a series of dates separated by one month, starting from January 1, 2015. First, enter the start date. Next, add a formula that begins with the DATE function.
For each argument, use the corresponding function to extract the value you need from the start date. So, for example, I can use YEAR to get the "year" value from B6, MONTH to get the "month" value,
and DAY to get the "day" value. When I copy that formula down, I get the same date in every cell, because I haven't added any change value yet.
Since I want to change the date by one month, I simply need to add "1" to the month. Now when I copy that formula down, the dates change by one month. Notice that Excel takes care of changing the
year value for me.
For the next example, I'll follow the same process. But this time, I'll set the year to increase by one.
You can also easily step back in time. Just subtract values instead of adding values.
Note that all values are fully dynamic and will recalculate when you supply a new start value.
Also note that you're free to hard code any value you like. For example, I can easily hard code the day value in the first example to be "15" and generate a series of dates that are one month apart,
always on the 15th of the month.
You can do the same thing with time values by using the TIME function along with the HOUR, MINUTE, and SECOND functions.
To generate times separated by one hour, I just need to add "1" to the Hour component.
To generate times separated by 15 minutes, I need to add "15" to the Minute component. | {"url":"https://exceljet.net/videos/how-to-create-date-and-time-series-with-formulas","timestamp":"2024-11-03T22:37:51Z","content_type":"text/html","content_length":"38676","record_id":"<urn:uuid:748dbc4c-8008-468d-80da-2a75fc655818>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00014.warc.gz"} |
Correlations and composite indicators: good or bad? - Composite Indicators
Let’s dig in to one of the thorniest topics in composite indicators: dependence, correlations and the definition of “indicator importance”. Take a deep breath. Ready?
The big squeeze
Before talking about correlations and more technical things, we need to clarify what exactly we are aiming to do here, and why. Recall that a composite indicator is an aggregation of a set of
indicators (possibly with intermediate aggregation steps) into a single composite measure: the composite indicator.
You don’t need to know much about composite indicators to understand a fundamental issue here: by combining many indicators into one, we are compressing the information of many variables into a
single measure, and in doing so, there is inevitably a loss of information.
To give a very simple example, let’s say we have two indicators measuring two different but related things about a country[1]. The indicator values (after normalisation/scaling) are 8 and 12. To get
our composite indicator score, we take the arithmetic mean with equal weights, and this gives a score of 10.
Now let’s pretend that we don’t know what the indicator values are, but only the composite score. Can we work out what the indicator values are, from the aggregate score of 10? No, of course not,
because the score of 10 could be from any of an infinite number of indicator values that happen to have an average value of 10.
This is a long and roundabout way of saying that by aggregating, we have lost some information, and this is an irreversible process – although we can calculate the aggregate value from the indicator
scores, we can’t calculate indicator scores from the aggregate value.
This loss of information is a concern in building a composite indicator, because one of the main aims should be to build a good summary measure of the underlying indicators.
Correlations and complications
Actually, the picture is slightly more complicated than the previous section implied, and this is because we almost never build a composite indicator for a single country – normally we would have set
of countries. This means that each indicator has a distribution, and there can exist correlations between indicator distributions. Correlations are one of the main tools of analysis in composite
indicator construction and auditing. Let’s explore this concept further.
We continue the trivial example in the previous section where we build a (rather scant) composite indicator out of two indicators (call them x1 and x2). But here we make it slightly more realistic by
assuming that we have 50 countries. Further, we’ll assume that both indicators are independent (uncorrelated with one another) and have normal distributions. If we plot one against the other, it
looks like this:
Indicators are both sampled from a normal distribution with mean 10 and standard deviation 1.
The fairly shapeless data cloud here should illustrate that the indicators are indeed independent.
Now we aggregate our two indicators into the composite indicator. Again, we’ll use equal weights, so this is just taking the mean of each pair of values. This results in a composite score for each of
the 50 countries. We will now plot the composite indicator (call it y) against each of the underlying indicators in turn.
Composite indicator plotted against both of its underlying indicators (uncorrelated indicators)
Notice that y, the composite, is correlated with both of the underlying indicators. Now, let’s re-pose the question from the first section: if we don’t know the underlying indicator values, but we
know the composite score of 10, can we work out the underlying indicator values?
The answer is now slightly different. We can’t know for sure what the underlying values are, but we can guess. Looking at the plots above, if y = 10 we could guess that x1 is probably between about 9
and 11, and x2 is somewhere between 8.5 and 11. We could work this out more accurately by using a conditional mean, but the point is here that we can know something about the underlying indicators by
knowing the composite score, because the composite is correlated with its indicators.
This now relates back to our aim that the composite should be a good summary of its underlying indicators: the better we are able to guess the indicator values from the composite, the more effective
a summary it is.
Let’s now take a second example where the two indicators are correlated with each other. Plotting one indicator against the other now looks like this:
Sample from a bivariate normal distribution
This shows that the two indicators are quite strongly related with one another. We’ll now aggregate these indicators (again using the mean) and plot the composite against each indicator as we did
Composite indicator plotted against both of its underlying indicators (correlated indicators)
It should be evident that the correlation between each indicator and the composite is here stronger than the previous example (in which the indicators were uncorrelated with each other). This
demonstrates a basic property – the more that indicators are correlated with each other, the more the composite indicator is correlated with its underlying indicators.
Moreover, if we again try to guess the indicator values, knowing that the composite score is 10, we see another thing: because the correlations between composite and indicators are stronger, our
guess about the indicator values will be more precise. We could guess, for example, that both x1 and x2 lie between about 9.5 and 10.5: these ranges are narrower than in the previous case where
indicators were uncorrelated with each other. The implication is that this composite indicator is a better summary of its underlying indicators than the previous one.
This is a lot to digest so let’s summarise before going any further.
1. When we aggregate indicators into a composite indicator, we naturally lose information.
2. But, we would like a composite indicator to summarise its underlying indicators as well as possible, i.e. we prefer to lose as little information as possible.
3. One way of framing this loss of information is: if we know the composite score, how well can we guess the underlying indicators?
4. Because of correlations between the composite indicator and its underlying indicators, we can make a guess at underlying indicator values, given just the composite scores.
5. The more that indicators are correlated with one another, the more accurately we can guess underlying indicator values from the composite scores.
6. By extension, the more that indicators are correlated with one another, the less information is lost when aggregating.
This last point is the crux. If we want to lose as little information as possible, then indicators should be well-correlated with one another (resulting in good correlations between indicators and
There is an intuitive explanation behind this. You can imagine a correlation between two indicators as an overlap of information: the higher the correlation, the more the information overlaps. Or in
other words, if the higher the correlation, the less information is unique to each of the two indicators.
Shared and unique information between two indicators
The figure above should help to clarify: the total information of the two indicators can be thought of as the total area of the ovals. When the ovals are independent, the total area is greater than
when they are correlated (overlapping). This means that in the independent case, there is simply more information to compress into the composite, so naturally, since we can only fit a limited amount
of information into a single number, the information loss is greater on aggregation. In the opposite case, if two indicators are perfectly correlated, there is no loss of information at all when we
Number of indicators
If you made it this far, there’s another complication waiting for you. In the previous section we just looked at a pair of indicators, with one single correlation value. What happens in the more
realistic case when we have more indicators?
This is intuitively explained again by the ovals. Imagine we add a third indicator (oval) to the picture, it is not difficult to see that the total area, i.e. the total amount of information, is
increased. And if we aggregate this into a composite we will naturally lose more information than if we only had two indicators.
Information overlap of three correlated indicators
The implication here then, is that a greater proportion of information is retained when we have:
• Fewer indicators, and
• Stronger correlations between indicators
And vice versa, of course.
We can in fact explore this with a little simulation. Let us use the average squared correlation (R squared) between the composite indicator and each of its underlying indicators as a measure of the
proportion of information transferred. When we vary both the number of the indicators and the correlations between indicators, here is what we get:
R-squared between composite and indicators, for different average correlation values and numbers of indicators (source)
This confirms the points above. But the interesting thing is that for a given average correlation, if we keep adding indicators the average R-squared (the “representativeness” of our composite
indicator) does not decrease to zero as might be expected. In fact, it tends to a limit, which is the average correlation between the indicators. This means that larger indicator frameworks can still
yield a composite indicator with a reasonable degree of representativeness, so long as the correlations between indicators are reasonably high.
All of this can actually be proved and is related to information theory. If you want to dig into that (or find a citation for the concepts I have explained here), see this paper.
High correlations = good?
All this might lead you to believe that indicators should be as highly-correlated as possible, in order to have a super-representative composite indicator; and independent or (horror of horrors)
negatively-correlated indicators should be avoided at all costs. In reality the picture is more nuanced.
First of all, the number of indicators is important. From the figure above, you could aggregate two independent indicators and still achieve about the same degree of representativeness as a larger
number of indicators with average correlation 0.4, for example. So, as long as the number of indicators is small, you can get away with weakly-correlated indicators in this respect.
The second point is that conceptually, it may be more efficient to have two completely uncorrelated indicators that bring completely different information to the framework, than a larger bunch of
indicators that overlap to a large degree. Independent indicators have a higher added value. At the other end of the spectrum, if we have indicators that are very highly correlated, we are basically
double-counting, and the added value is effectively zero.
Often, a moderate level of correlation between indicators is recommended as a target, with correlations ranging from e.g. 0.4 to 0.8. This is of course a rule of thumb. Moreover, in practice this is
quite hard to achieve.
Correlation = importance?
This is a tangential issue but worth mentioning. Correlation between indicators and the composite is sometimes used as a measure of the “importance” of the indicator in the framework. The logic is
that, if an indicator has a strong correlation with the composite, it is driving the composite indicator, whereas an indicator with a poor correlation is “silent” and therefore less important.
In my view this is not quite correct. In the first place, a composite indicator is demonstrably a function of its underlying indicators, so each indicator is definitely contributing to the overall
Second, we could think of an alternative definition of importance for a given indicator: what would be the impact on the composite scores if we remove it from the framework? One might be tempted to
think that the poorly-correlated “silent” indicators would have little or no impact. In fact, it is usually these indicators that have the most impact when removed. This is because usually such
indicators are weakly correlated with other indicators, which means that they have a greater unique contribution of information. Viewed this way, it is easy to see that when removed, they have a
greater impact because we are taking a bigger chunk of information out of the framework.
It seems safer to view correlation in terms of information. The best way I can put is that correlations between indicators and the composite show the degree to which information is shared between
Back to reality
Correlations are just one of many considerations in building a composite indicator, so it is important to view them as an analytical tool like any other, and not get too hung up on them. Equally,
they shouldn’t be disregarded. Unsurprisingly, it requires a degree of balance and judgement.
The fact is that indicator frameworks are often more dominated by conceptual choices than correlations. Whereas we would love to have perfect correlations, in practice an indicator may be crucial to
a framework even if it is uncorrelated, and we may need a fairly large number of indicators to fully cover our concept, and to capture the viewpoints of stakeholders. It is not uncommon for negative
correlations to appear, as well as fiendishly-skewed and unusually-shaped distributions, for which linear correlations are not even a good summary measure.
At such times we should remember that composite indicator should only be used as a summary and an entry point to its underlying indicators. In this context, as long as we give sufficient visibility
and importance to the underlying indicators, and things are carefully presented and communicated, a composite indicator composed of poorly-correlated indicators might not be a serious problem in
practice. We could also choose not to fully aggregate the composite indicator and to leave it e.g. at the sub-index level to alleviate the problem.
Let’s summarise the summary:
• The average correlations between indicators are related to the “representativeness” of the composite indicator: how well it represents its underlying indicators.
• More indicators means a smaller proportion of information transferred to the composite, but this tends to a limit.
• A very vague rule of thumb is to have correlations between indicators of about 0.4-0.8 (higher correlations imply little added value of indicators).
• These considerations are important but should be balanced with conceptual issues, and kept in the context of how the composite indicator is actually used and communicated.
If you made it this far, well done, you deserve a rest, as do I. Thanks for reading!
[1] Could also be a region, a company, a university, an individual, or any other “thing” that is being compared with the indicators. A general term is “unit” but I will use “country” since that is
the most common case. | {"url":"https://compositeindicators.com/correlations-and-composite-indicators-good-or-bad/","timestamp":"2024-11-02T14:21:32Z","content_type":"text/html","content_length":"102494","record_id":"<urn:uuid:aef93621-f263-4d5c-8bfe-c9c6bb79f42a>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00474.warc.gz"} |
Loss Functions
eulerr features multiple loss functions, which result in different diagrams for many combinations. In this vignette, we visualize the effect of the loss function on an example from an issue posted on
the GitHub repository for eulerr.
We list the combinations below, which consists of 5 different sets, agc, camk, cmgc, and tk.
combos <- c(
"agc" = 9,
"camk" = 17,
"cmgc" = 16,
"tk" = 16,
"tkl" = 23,
"agc&camk" = 1,
"camk&tk" = 1,
"tk&tkl" = 1,
"camk&cmgc&tkl" = 1,
"camk&tk&tkl" = 2,
"agc&camk&tk&tkl" = 1,
"camk&cmgc&tk&tkl" = 3,
"agc&camk&cmgc&tk&tkl" = 1
Notice that the sizes of most of the intersections are small compared to the size of the sets themselves and that many of the intersections are missing. Generating an exact Euler diagram that shows
these intersections and at the same time omits the intersections that are here implicitly 0 is an impossible problem, which means that the best we can do is an approximation.
What kind of approximation we get depends on the loss functions we use. If we use the default, which in eulerr is the sums of squared errors, we will almost certainly get a design in which the
intersections involving many sets are missing since including them inevitably leads to larger errors from having to include other intersections that are not present.
If we rather want a diagram that includes these intersections, despite leading to errors for the zero-intersections, then we need to switch the loss function we use. In eulerr, you can do so via the
two arguments loss and loss_aggregator in euler(). We start by listing the alternatives for the loss argument.
Loss functions in eulerr
Squared errors square \((y_i - \hat y_i)^2\)
Absolute errors abs \(|y_i - \hat y_i|\)
RegionErrors region \(\big|y_i/\sum_k y_k - \hat y_i / \sum_k \hat y_k \big|\)
How the final loss is computed depends on the value of loss_aggregator, which is the function used to aggregate the values computed for each set intersection via the function used in loss. The two
available settings are "sum" and "max", which should be self-explanatory.
That means that loss = "square" and loss_aggregator = "sum" leads to the sum of squared errors. loss = "region" uses regionError, which is a loss metric introduced by (Micallef and Rodgers 2014).
Together with loss_aggregator = "max", euler() will use diagError (introduced in the same paper).
To see what these different choices mean for the combination that we have looked at, we now refit the diagram for each combination.
losses <- c("square", "abs", "region")
aggregators <- c("sum", "max")
for (loss in losses) {
for (aggregator in aggregators) {
fit <- euler(combos, loss = loss, loss_aggregator = aggregator)
print(plot(fit, main = paste(aggregator, loss, sep = ", ")))
As you can see, the errors that sum either the absolute or squared errors result in very similar fits and keep the existing two-set intersections and drop everything else. The abs + max and max +
square combos, meanwhile, produce fits that are much more unpredictable since they only care about the largest error. Finally, diagError results in diagrams that tries to include many more
intersections at the cost of reducing the goodness-of-fit of the larger intersections.
Feel free to raise a request (or better yet, a pull request) at https://github.com/jolars/eulerr/issues if you know of any other loss function that you think should be included in the package.
Micallef, Luana, and Peter Rodgers. 2014.
“eulerAPE: Drawing Area-Proportional 3-Venn Diagrams Using Ellipses.” PLOS ONE
9 (7): e101717. | {"url":"https://cran.r-project.org/web/packages/eulerr/vignettes/loss-functions.html","timestamp":"2024-11-08T12:43:35Z","content_type":"text/html","content_length":"208502","record_id":"<urn:uuid:1034c611-5fc3-4282-b256-255ddc3776e3>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00794.warc.gz"} |
Lowest Common Ancestor - Farach-Colton and Bender
Lowest Common Ancestor - Farach-Colton and Bender Algorithm¶
Let $G$ be a tree. For every query of the form $(u, v)$ we want to find the lowest common ancestor of the nodes $u$ and $v$, i.e. we want to find a node $w$ that lies on the path from $u$ to the root
node, that lies on the path from $v$ to the root node, and if there are multiple nodes we pick the one that is farthest away from the root node. In other words the desired node $w$ is the lowest
ancestor of $u$ and $v$. In particular if $u$ is an ancestor of $v$, then $u$ is their lowest common ancestor.
The algorithm which will be described in this article was developed by Farach-Colton and Bender. It is asymptotically optimal.
We use the classical reduction of the LCA problem to the RMQ problem. We traverse all nodes of the tree with DFS and keep an array with all visited nodes and the heights of these nodes. The LCA of
two nodes $u$ and $v$ is the node between the occurrences of $u$ and $v$ in the tour, that has the smallest height.
In the following picture you can see a possible Euler-Tour of a graph and in the list below you can see the visited nodes and their heights.
$$\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Nodes:} & 1 & 2 & 5 & 2 & 6 & 2 & 1 & 3 & 1 & 4 & 7 & 4 & 1 \\ \hline \text{Heights:} & 1 & 2 & 3 & 2 & 3 & 2 & 1 & 2 & 1 & 2 & 3 & 2 & 1 \
\ \hline \end{array}$$
You can read more about this reduction in the article Lowest Common Ancestor. In that article the minimum of a range was either found by sqrt-decomposition in $O(\sqrt{N})$ or in $O(\log N)$ using a
Segment tree. In this article we look at how we can solve the given range minimum queries in $O(1)$ time, while still only taking $O(N)$ time for preprocessing.
Note that the reduced RMQ problem is very specific: any two adjacent elements in the array differ exactly by one (since the elements of the array are nothing more than the heights of the nodes
visited in order of traversal, and we either go to a descendant, in which case the next element is one bigger, or go back to the ancestor, in which case the next element is one lower). The
Farach-Colton and Bender algorithm describes a solution for exactly this specialized RMQ problem.
Let's denote with $A$ the array on which we want to perform the range minimum queries. And $N$ will be the size of $A$.
There is an easy data structure that we can use for solving the RMQ problem with $O(N \log N)$ preprocessing and $O(1)$ for each query: the Sparse Table. We create a table $T$ where each element $T
[i][j]$ is equal to the minimum of $A$ in the interval $[i, i + 2^j - 1]$. Obviously $0 \leq j \leq \lceil \log N \rceil$, and therefore the size of the Sparse Table will be $O(N \log N)$. You can
build the table easily in $O(N \log N)$ by noting that $T[i][j] = \min(T[i][j-1], T[i+2^{j-1}][j-1])$.
How can we answer a query RMQ in $O(1)$ using this data structure? Let the received query be $[l, r]$, then the answer is $\min(T[l][\text{sz}], T[r-2^{\text{sz}}+1][\text{sz}])$, where $\text{sz}$
is the biggest exponent such that $2^{\text{sz}}$ is not bigger than the range length $r-l+1$. Indeed we can take the range $[l, r]$ and cover it two segments of length $2^{\text{sz}}$ - one starting
in $l$ and the other ending in $r$. These segments overlap, but this doesn't interfere with our computation. To really achieve the time complexity of $O(1)$ per query, we need to know the values of $
\text{sz}$ for all possible lengths from $1$ to $N$. But this can be easily precomputed.
Now we want to improve the complexity of the preprocessing down to $O(N)$.
We divide the array $A$ into blocks of size $K = 0.5 \log N$ with $\log$ being the logarithm to base 2. For each block we calculate the minimum element and store them in an array $B$. $B$ has the
size $\frac{N}{K}$. We construct a sparse table from the array $B$. The size and the time complexity of it will be:
$$\frac{N}{K}\log\left(\frac{N}{K}\right) = \frac{2N}{\log(N)} \log\left(\frac{2N}{\log(N)}\right) =$$
$$= \frac{2N}{\log(N)} \left(1 + \log\left(\frac{N}{\log(N)}\right)\right) \leq \frac{2N}{\log(N)} + 2N = O(N)$$
Now we only have to learn how to quickly answer range minimum queries within each block. In fact if the received range minimum query is $[l, r]$ and $l$ and $r$ are in different blocks then the
answer is the minimum of the following three values: the minimum of the suffix of block of $l$ starting at $l$, the minimum of the prefix of block of $r$ ending at $r$, and the minimum of the blocks
between those. The minimum of the blocks in between can be answered in $O(1)$ using the Sparse Table. So this leaves us only the range minimum queries inside blocks.
Here we will exploit the property of the array. Remember that the values in the array - which are just height values in the tree - will always differ by one. If we remove the first element of a
block, and subtract it from every other item in the block, every block can be identified by a sequence of length $K - 1$ consisting of the number $+1$ and $-1$. Because these blocks are so small,
there are only a few different sequences that can occur. The number of possible sequences is:
$$2^{K-1} = 2^{0.5 \log(N) - 1} = 0.5 \left(2^{\log(N)}\right)^{0.5} = 0.5 \sqrt{N}$$
Thus the number of different blocks is $O(\sqrt{N})$, and therefore we can precompute the results of range minimum queries inside all different blocks in $O(\sqrt{N} K^2) = O(\sqrt{N} \log^2(N)) = O
(N)$ time. For the implementation we can characterize a block by a bitmask of length $K-1$ (which will fit in a standard int) and store the index of the minimum in an array $\text{block}[\text{mask}]
[l][r]$ of size $O(\sqrt{N} \log^2(N))$.
So we learned how to precompute range minimum queries within each block, as well as range minimum queries over a range of blocks, all in $O(N)$. With these precomputations we can answer each query in
$O(1)$, by using at most four precomputed values: the minimum of the block containing l, the minimum of the block containing r, and the two minima of the overlapping segments of the blocks between
int n;
vector<vector<int>> adj;
int block_size, block_cnt;
vector<int> first_visit;
vector<int> euler_tour;
vector<int> height;
vector<int> log_2;
vector<vector<int>> st;
vector<vector<vector<int>>> blocks;
vector<int> block_mask;
void dfs(int v, int p, int h) {
first_visit[v] = euler_tour.size();
height[v] = h;
for (int u : adj[v]) {
if (u == p)
dfs(u, v, h + 1);
int min_by_h(int i, int j) {
return height[euler_tour[i]] < height[euler_tour[j]] ? i : j;
void precompute_lca(int root) {
// get euler tour & indices of first occurrences
first_visit.assign(n, -1);
height.assign(n, 0);
euler_tour.reserve(2 * n);
dfs(root, -1, 0);
// precompute all log values
int m = euler_tour.size();
log_2.reserve(m + 1);
for (int i = 1; i <= m; i++)
log_2.push_back(log_2[i / 2] + 1);
block_size = max(1, log_2[m] / 2);
block_cnt = (m + block_size - 1) / block_size;
// precompute minimum of each block and build sparse table
st.assign(block_cnt, vector<int>(log_2[block_cnt] + 1));
for (int i = 0, j = 0, b = 0; i < m; i++, j++) {
if (j == block_size)
j = 0, b++;
if (j == 0 || min_by_h(i, st[b][0]) == i)
st[b][0] = i;
for (int l = 1; l <= log_2[block_cnt]; l++) {
for (int i = 0; i < block_cnt; i++) {
int ni = i + (1 << (l - 1));
if (ni >= block_cnt)
st[i][l] = st[i][l-1];
st[i][l] = min_by_h(st[i][l-1], st[ni][l-1]);
// precompute mask for each block
block_mask.assign(block_cnt, 0);
for (int i = 0, j = 0, b = 0; i < m; i++, j++) {
if (j == block_size)
j = 0, b++;
if (j > 0 && (i >= m || min_by_h(i - 1, i) == i - 1))
block_mask[b] += 1 << (j - 1);
// precompute RMQ for each unique block
int possibilities = 1 << (block_size - 1);
for (int b = 0; b < block_cnt; b++) {
int mask = block_mask[b];
if (!blocks[mask].empty())
blocks[mask].assign(block_size, vector<int>(block_size));
for (int l = 0; l < block_size; l++) {
blocks[mask][l][l] = l;
for (int r = l + 1; r < block_size; r++) {
blocks[mask][l][r] = blocks[mask][l][r - 1];
if (b * block_size + r < m)
blocks[mask][l][r] = min_by_h(b * block_size + blocks[mask][l][r],
b * block_size + r) - b * block_size;
int lca_in_block(int b, int l, int r) {
return blocks[block_mask[b]][l][r] + b * block_size;
int lca(int v, int u) {
int l = first_visit[v];
int r = first_visit[u];
if (l > r)
swap(l, r);
int bl = l / block_size;
int br = r / block_size;
if (bl == br)
return euler_tour[lca_in_block(bl, l % block_size, r % block_size)];
int ans1 = lca_in_block(bl, l % block_size, block_size - 1);
int ans2 = lca_in_block(br, 0, r % block_size);
int ans = min_by_h(ans1, ans2);
if (bl + 1 < br) {
int l = log_2[br - bl - 1];
int ans3 = st[bl+1][l];
int ans4 = st[br - (1 << l)][l];
ans = min_by_h(ans, min_by_h(ans3, ans4));
return euler_tour[ans]; | {"url":"https://cp-algorithms.com/graph/lca_farachcoltonbender.html","timestamp":"2024-11-08T05:22:30Z","content_type":"text/html","content_length":"164629","record_id":"<urn:uuid:59b8a926-a6f3-4675-9c14-b1efaabc63e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00263.warc.gz"} |
Boolean Algebra (OLD) - 12 Computer Sc 04 introduction to Boolean Algebra
CBSE Class 12 Computer Science 4
introduction to Boolean Algebra
(Boolean Algebra)
Boolean Algebra: is the algebra of logic that deals with binary variables and logic operations.
Boolean Variable: A boolean variable is a symbol, usually an alphabet used to represent a logical quantity. It can have a 0 or 1 value
Boolean Function: consists of binary variable, constants 0 & 1, logic operation symbols, parenthesis and equal to operator.
Complement: A complement is the inverse of a variable and is indicated by a' or bar over the variable. A binary variable is one that can assume one of the two values 0 and 1.
Literal: A Literal is a variable or the complement of a variable
Truth table: is atable which represents all the possible values of logical variables along with all the possible results of the given combinations of values.
List of axioms and theorems:
│ Identity │ A + 0 = A │ A. 1 = A │
│ Complement │ A + A' = 1 │ A. A' = 0 │
│ Commutative │ A + B = B + A │ A. B = B. A │
│ Assosiative │ A + (B + C) = (A + B) + C │ A. (B. C) = (A. B). C │
│ Distributive │ A. (B + C) = A. B + A. C │ A + (B. C) = (A + B). (A + C) │
│ Null Element │ A + 1 = 1 │ A. 0 = 0 │
│ Involution │ (A')' = A │ │
│ Indempotency │ A + A = A │ A. A = A │
│ Absorption │ A + (A. B) = A │ A. (A + B) = A │
│ 3rd Distributive │ A + A'. B = A + B │ │
│ De Morgan's │ (A + B)' = A'. B' │ (A. B)' = A'. B' │
(Boolean Functions and Reduce Forms)
A Boolean function can be expressed algebraically from a given truth table by forming a minterm and then taking the OR of all those terms.
Minterm: An n variable minterm is a product term with n literals resulting into 1.
Maxterm: An n variable maxterm is a sum term with n literals resulting into 0.
A sum-of-product expression is logical OR of two or more AND terms
A product-of-sum is logical AND of two or more OR terms
If each term in SOP / POS form contains all the literals, then it is canonical form of expression.
To convert from one canonical form to another, interchange the symbol and list those numbers missing from the original form.
The Karnaugh map (K-map) provides a systematic way of simplifying Boolean algebra expressions.
For minimizing a given expression in SOP form, after filling the k map look for combination of adjascent one's.
Combine these one's in such a way that the expression is minimum.
For minimizing expression in POS form we mark zeros, from the truth table, in the map. Combine zeros in such a way that the expression is minimum.
Sum Term: is a single literal or the logical sum of two or more literals.
Product term: is a single literal or the logical product of two or more literals.
(Application of Boolean Logic)
Gate is an electronic system that performs a logical operation on a set of input signal(s). They are the building blocks of Integrated Circuits.
An SOP expression when implemented as circuit - takes the output of one or more AND gates and OR's them together to create the final output.
An POS expression when implemented as circuit - takes the output of one or more OR gates and AND's them together to create the final output.
Universal gates are the ones which can be used for implementing any gate like AND, OR and NOT, or any combination of these basic gates; NAND and NOR gates are universal gates.
Implementation of a SOP expression using NAND gates only
1) All 1st level AND gates can be replaced by one NAND gate each.
2) The output of all 1st level NAND gate is fed into another NAND gate. This will realize the SOP expression
3) If there is any single literal in expression, feed its complement directly to 2nd level NAND gate. Similarly, POS using NOR gate can be implemented by replacing NAND by NOR gate.
Implementation of POS / SOP expression using NAND / NOR gates only.
1) All literals in the first level gate will be fed in their complemented form.
2) Add an extra NAND / NOR gate after 2nd level gate to get the resultant output. | {"url":"https://mobile.surenapps.com/2020/10/boolean-algebra-old-12-computer-sc-04.html","timestamp":"2024-11-05T12:39:34Z","content_type":"application/xhtml+xml","content_length":"82160","record_id":"<urn:uuid:51f46ca9-cd47-4844-900d-58732ad20f19>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00156.warc.gz"} |
Recurring Decimals Into Fractions Worksheet - Customize And Print
Recurring Decimals Into Fractions Worksheet
Recurring Decimals Into Fractions Worksheet - The corbettmaths practice questions on recurring decimals. Web converting decimals to fractions worksheets. Web so, 0.2 (recurring) = 2/9. 3 9 13 = 33.
Convert the following recurring decimals into fractions in their simplest forms. (1) (3) (total 4 marks). Learn the method to convert recurring decimals to fractions. (b) prove that the recurring
decimal 0. Web practice and revise recurring decimals to fractions with 45 questions and answers of different levels and types. The sheets within each section are graded.
Use division to convert these fractions to recurring decimals. (a) change 3 to a decimal. Convert the following recurring decimals to fractions. This recurring decimals to fractions worksheet does
more than just check the learner's knowledge; Convert the following recurring decimals into fractions in their simplest forms. Write \dfrac {5} {37} as a decimal. The sheets within each section are
We set up the bus stop method as follows, with several zeros after the decimal place. Web practice and revise recurring decimals to fractions with 45 questions and answers of different levels and
types. This recurring decimals to fractions worksheet does more than just check the learner's knowledge; Web so, 0.2 (recurring) = 2/9. Convert the following recurring decimals to fractions.
Then complete the bus stop method, as. 3 9 13 = 33. (b) prove that the recurring decimal 0. Maths revision video and notes on the topic of converting recurring decimals to fractions. Web the
corbettmaths textbook exercise on converting decimals to fractions. Includes reasoning and applied questions.
It provides supporting structures to enable your. (b) prove that the recurring decimal 0. Use division to convert these fractions to recurring decimals. Prove that the recurring decimal 0. Web the
corbettmaths textbook exercise on converting decimals to fractions.
3 9 13 = 33. We set up the bus stop method as follows, with several zeros after the decimal place. Web converting repeating decimals to fractionsname:answer key math. Prove that the recurring decimal
Maths Revision Video And Notes On The Topic Of Converting Recurring Decimals To Fractions.
Write \dfrac {5} {37} as a decimal. Web converting decimals to fractions worksheets. Then complete the bus stop method, as. This recurring decimals to fractions worksheet does more than just check
the learner's knowledge;
(B) Prove That The Recurring Decimal 0.
Web practice and revise recurring decimals to fractions with 45 questions and answers of different levels and types. We set up the bus stop method as follows, with several zeros after the decimal
place. The sheets within each section are graded. Use division to convert these fractions to recurring decimals.
Web Get Your Free Recurring Decimals To Fractions Worksheet Of 20+ Questions And Answers.
It provides supporting structures to enable your. 3 9 13 = 33. Learn the method to convert recurring decimals to fractions. Web recurring, decimal, convert, fractions.
Web The Corbettmaths Textbook Exercise On Converting Decimals To Fractions.
Convert the following recurring decimals into fractions in their simplest forms. Web pdf, 47.42 kb. Here are our worksheets to help you practice converting decimals to fractions. Convert the
following recurring decimals to fractions.
Related Post: | {"url":"https://metadata.denizen.io/en/recurring-decimals-into-fractions-worksheet.html","timestamp":"2024-11-07T07:23:30Z","content_type":"application/xhtml+xml","content_length":"30388","record_id":"<urn:uuid:8c941033-b39a-4c78-a6f3-56240724ee20>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00356.warc.gz"} |
Know About The Exclusive NOR Gate For NEET
Logic gates are the building blocks of any digital system. All the digital devices such as mobiles, computers store the data in the form of binary numbers 1 or 0. 1 is considered as the high input
and 0 is the low input value. The logic gates are used to perform certain mathematical operations for binary digits.
Depending upon the type of operation required the logic gates are categorized as OR-gate for addition, AND-gate for multiplication, NOT gate for inversion. Logic gates are further classified as basic
gates(OR, AND, NOT), Universal gates(NAND, NOR), and the special gates Exclusive OR gate and Exclusive NOR gate.
What is XNOR Gate?
• The Exclusive NOR gate is also known as the XNOR gate or Ex-NOR gate. It performs the operation same as XOR gate followed by NOT gate, thus it is abbreviated as XNOR gate.
• The exclusive NOR gate is the combination of the exclusive OR gate and the NOT gate. Thus, the operation of the XNOR gate is reciprocal of the XOR gate.
• The exclusive NOR gate gives the output as 1 when both inputs are identical. I.e., either both inputs are 1’s or 0’s.
• The output of the XNOR is high(.i.e., 1) for the input combinations like 11 or 00.
XNOR Gate Expression:
XNOR Symbol:
• While studying logic gates the important part of understanding the function of any gate is to first recognize the logic expression (Or Boolean expression) and the logic symbol.
• The XNOR logic symbol is an Exclusive OR gate followed by an inversion bubble at the output indicating presence of NOT gate. Thus XNOR gate is a reciprocal or complementary form of the XOR gate.
• The logic symbol of the Exclusive NOR gate is:
(Image will be uploaded soon)
XNOR Expression or Boolean Expression for XNOR Gate:
• The XNOR gate is a complementary form of the XOR gate, thus the boolean expression of the XNOR gate is given by:
\[\Rightarrow Y= \overline{A\oplus B}\]
• From the XNOR equation, it is understood that output will be 1 if and only if both A and B are either having high logic(1’s) or low logic(0’s).
Exclusive NOR Gate Truth Table:
• For the 2-input XNOR gate, the logical expression is:
\[ \Rightarrow Y= \overline{A\oplus B}=AB+ \overline{AB}\]
• Therefore, the Ex NOR gate truth table is:
• Similarly, the XNOR equation of XNOR gate boolean equation for 3-input is given by:
\[\Rightarrow Y=\overline{ABC}+\overline{A}BC+A\overline{B}C+AB\overline{C}\]
• If any two inputs among 3 are high or if all 3 inputs are low then the output will be high.
• The truth table for the 3-input XNOR gate is given by:
Uses of XNOR Gate:
• The XNOR gate is used as a parity checker.
• The exclusive nor gate is used in calculators, computers, etc..
• They are used for encryption and arithmetic circuits.
Did You Know:
• From the boolean expression, we can understand that the XNOR gate can be constructed as a combination of OR gate, AND gate, and NOT gate.
• Thus, the equivalent circuit for the XNOR gate is:
(Image will be uploaded soon)
FAQs on Exclusive NOR Gate
1. What is the Difference Between XOR and XNOR Gate?
Ans: XNOR gate is the complement of the XOR gate. XOR gate will give high output if either of the inputs is high, whereas the XNOR gate will have high input if both inputs are either at high logic or
low logic.
2. How Many NAND Gates are Required For Constructing the XNOR Gate?
Ans: Five. XNOR gate can be constructed by using five NAND gates. | {"url":"https://www.vedantu.com/neet/exclusive-nor-gate","timestamp":"2024-11-07T20:10:58Z","content_type":"text/html","content_length":"250234","record_id":"<urn:uuid:edfc9b0a-f221-44a1-a312-7f2bed97a993>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00379.warc.gz"} |
Algebraic Problems and Exercises for High School
by I. Goian, R. Grigor, V. Marin, F. Smarandache
Publisher: The Educational Publisher 2015
ISBN-13: 9781599733425
Number of pages: 146
In this book, you will find algebra exercises and problems, grouped by chapters, intended for higher grades in high schools or middle schools of general education. Its purpose is to facilitate
training in mathematics for students in all high school categories, but can be equally helpful in a standalone workout.
Download or read it online for free here:
Download link
(5.2MB, PDF)
Similar books
Numbers and Symbols: From Counting to Abstract Algebras
Roy McWeeny
Learning Development InstituteThis book is written in simple English. Its subject 'Number and symbols' is basic to the whole of science. The aim the book is to open the door into Mathematics, ready
for going on into Physics, Chemistry, and the other Sciences.
College Algebra
Richard BeveridgeIf mathematics is the language of science, then algebra is the grammar of that language. This text will cover a combination of classical algebra and analytic geometry, with an
introduction to the transcendental exponential and logarithmic functions.
Inner Algebra
Aaron Maxwell
Lulu.comLearn to do algebra mentally and intuitively: mathematicians use their minds in ways that make math easy for them. This book teaches you how to do the same, focusing on algebra. As you read
and do the exercises, math becomes easier and more natural.
High School Algebra
J.T. Crawford
The Macmillan CompanyThis text covers the work prescribed for entrance to the Universities and Normal Schools. The book is written from the standpoint of the pupil, and in such a form that he will be
able to understand it with a minimum of assistance from the teacher. | {"url":"https://www.e-booksdirectory.com/details.php?ebook=10612","timestamp":"2024-11-05T05:31:17Z","content_type":"text/html","content_length":"11081","record_id":"<urn:uuid:7f1a5adf-df6b-4d68-bd80-8f13f53809dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00584.warc.gz"} |
Gardner Lab
Table of Contents
The key to understanding analysis of any data including BOLD imaging data is to remember that all analyses are models. One designs and runs an experiment, collects data and then models the results
and examines the goodness-of-fit and parameters of the model. If you have a good model of the data, then your goodness-of-fit will be good and you will get fit parameters that can be meaningful
interpreted. Models contain multiple assumptions about the processes that generated the data, some of which may not be immediately obvious. Those assumptions may be wrong. And if they are, the
results that you obtain may not be meaningful. This is true whether you are doing a simple t-test or fitting a complex non-linear model to your data. So, with that in mind, we will use this tutorial
to go through a few commonly used models for analyzing BOLD imaging data.
To make things concrete, we will analyze human cortical BOLD responses to visual stimuli in which we know that neurons in primary visual cortex (V1) will show responses to visual stimuli in a
particular location in space called a receptive-field. We will go through the most common ways to look at BOLD imaging data that interrogate these receptive-fields with two types of analyses, an
“Event-related analysis” and an “encoding model”. Event-related analyses can be thought of as a trial-triggered average response which assumes that overlaps in response sum linearly. Encoding models
can take many shapes and forms, but generally mean that you assume a model that encodes the stimulus (in this case a receptive-field) whose parameters are adjusted to give rise to the response
measured in a voxel. For this tutorial, we will use a population receptive-field (pRF) encoding model Dumoulin and Wandell (2008) which has been used extensively to make retinotopic measurements.
Note that this tutorial is meant as a quick and practical introduction to these types of analyses and assumes that you have already learned about the biological processes that give rise to the BOLD
signal and some of its basic properties.
Learning goals
After finishing this tutorial, you should be able to…
…compute an event-related analysis of BOLD data to recover the response to a trial or stimulus.
…compute a pRF analysis which allows you to recover the position and size of a visual receptive-field of a voxel.
…know the assumptions (particularly linearity) of each of these models.
This tutorial is written in Matlab (though there are equivalent instructions in python for the pRF part).
To setup for matlab, you just need to do the following:
1. Setup your matlab environment
For those working at Stanford you can get an
indiviudal license
. If that fails, you can try to use the
corn server
, but we don't recommend that anymore. If you use the corn server, then you do not need to do the rest of these instructions.
2. Start matlab
3. Get datafiles
You can get the data from
. This should download a matlab file to your Downloads folder (if you are on a Mac). To load the data in, you can do:
cd ~/Downloads
load boldmodels
BOLD Response
Ok, let's say we do a totally simple experiment in which we flash a visual stimulus at different times and measure the BOLD response in visual cortex. In the diagram below each vertical tick is meant
to represent a time at which we flashed the visual stimulus on.
What would we expect the BOLD response to look like? Well, recall, that after brief neural activity, the “hemodynamic impulse response function” typically looks like the following:
The BOLD response is also (approximately) linear in time (see Boynton GM1, Engel SA, Heeger DJ (2012) for an excellent review). What does that mean? It means that it obeys linear superposition - if
you know the response to each of two stimuli presented alone, then you can predict the response to the combination of the two stimuli as the sum (or superposition) of the two responses. In this case,
the two stimuli would be responses presented at different times which would both be expected to evoke the above hemodynamic response. The response to two stimuli presented in succession would be the
sum of the individual responses. So, imagine in your head a hemodynamic response function placed at every vertical tick mark in the stimulus timing diagram above and any overlap in response (when
stimuli are closer together than the length of the hemodynamic response) summing together (assume that each tick happens about 5-10 seconds after the last one). Add a little bit of noise to that
mental image. Got it? You should be imagining something like the following (Click below to reveal what the expected response is):
Looks like what you got? Good. You've just done a convolution in your head then! That's the linear systems model for a BOLD response and there's pretty good evidence that if you make your stimulus
presentation times long enough (a couple of seconds) that the BOLD response is temporally linear (see review cited above). Pretty much every BOLD analysis technique rests on this temporally linear
model of BOLD responses!
Ok. So, now let's go backwards. Imagine that you have measured that BOLD response time series was and you wanted to know what the average BOLD response was to each stimulus presentation. Well then
your model would look something like this:
In words, our model is that the stimulus timing convolved with an unknown hemodynamic response plus noise gives the measured bold response. We want to know what that unknown hemodynamic response
looks like, so we need to somehow “unconvolve” the stimulus timing from both sides to solve for the hemodynamic response. How is this done? Well, the above can be written in matrix form. It looks
The matrix on the left is just the way that you write down a convolution of the stimulus times with the hemodynamic response. Each column is the stimulus timing shifted downwards one (note that each
column as numbers is zeros with ones every time a stimulus happens). Why does this compute a convolution in which response overlaps sum? Try to puzzle through the matrix multiplication by multiplying
one row at a time of the convolution matrix with the hemodynamic response to get each time point of the measured BOLD response and you should see that it does the right thing. Make sense? If not, we
will go through a concrete example with matlab in a sec, so try to figure it out there. Ok, what about the dimensions? In the above, we have n=number of timepoints in the experiment and k=number of
timepoints in estimated response. Note that we have a choice over k - how many time points to compute out the hemodynamic response for. Typically you want to choose k such that you recover the full
hemodynamic response, but not too long since every point you estimate will increase the number of unknowns you are solving for, making your estimates worse.
So, how do we solve this? Well, typically one assumes that noise is white (IID - independent at each time point, identically distributed - which is wrong, btw, since BOLD data typically have temporal
auto-correlation - that is, timepoints near each other are correlated). But, going with that assumption, this is just a matrix multiplication and you can use the least-squares solution (i.e. solve
for the unknown hemodynamic response that minimizes the squared error between the left and right sides of the equation). Note that the noise here isn't something that we are estimating - it is just
the left over residual after the model fit. If we have S as the stimulus convolution matrix, h as the unknown response and B the measured response, that's just an equation like:
S x h = B
The least-squares solution is obtained by multiplying both sides by the transpose of S (in matlab notation S'):
S' x S x h = S' x B
Then multiplying by the inverse of S' x S (that's an important matrix, by the way, called the stimulus variance/covariance and can be used to diagnose problems with your design since the more
variance in the design gives you more power to estimate parameters - in fact, this inverse of the variance/covariance matrix is used to project residual variance back to get error bars around
estimates). That gives us:
h = (S' x S) ^ -1 x S' x B
And that's it, that's an event-related analysis (sometimes people call this deconvolution or finite impulse response estimation). So, how about trying it on real data?
In Matlab, you should have two variables. timeSeries and stimulusTiming. timeSeries is the BOLD response from a single voxel in a part of primary visual cortex. The data were collected every 1s.
Typically the data needs to be high-pass filtered to get rid of slow drifts in signals. The time series also has been divided by its mean and multiplied by 100 so that everything is in percent signal
change relative to the average pixel intensity. Finally the mean has been subtracted (this is actually really important, since the model we are fitting here doesn't have a term to account for the
mean response, so if you don't subtract the mean things won't work!).
Go ahead and plot these two arrays and see what they look like.
Any hint of what the response is? Let's try to form the stimulus-convolution matrix. Each column of the matrix should be a shifted copy of the stimulusTiming array as we discussed above. Let's
compute the response out for 30 seconds. So, that means 30 columns (since each data point is 1s).
If you got that right, you should be able to look at a small part of the matrix (say the first 300 rows) and see that the matrix has the slanted line structure that we talked about above. Go ahead
and take a look to confirm that.
Ok. Cool. So, now to compute the event-related responses, just plug into the formula from above for getting the least-squares estimate and plot what you get.
Should look like the following:
Easy-peasy. Right? One thing - how good a model is this of the response we measured? Whenever you fit a model you should ask whether you have a statistic that tells you that. One that we have used is
to compute the variance accounted for by the event-related model. To do this, you can take the hemodynamic response you estimated and multiply it back through the stimulus convolution matrix. That
will form the ideal response as if every trial created exactly the same response and response overlaps summed linearly. That's your model. Subtract that from the time series to get the residual. Now
compute r2 as one minus the ratio of the residual variance to the original variance of the time series.
Challenge question
Ok. Challenge question here. The data from above is actually from a paradigm in which we showed a stimulus either in the left or right visual field. Of course we know that primary visual cortex
should only respond to a stimulus in the contra-lateral visual field. So, let's say I give you the stimulus time for the times when the stimulus was in the left visual field, and the times at which
the stimulus was in the right visual field. You should get a nice response to the contra-lateral stimulus and a flat line for the ipsi-lateral stimulus. Right? But, how do I compute the response for
two different stimuli? Well, same way. You just assume that the responses to the two stimuli sum linearly! Try to draw out how to make the matrix equation do that. Be careful of matrix dimensions -
make sure they match-up correctly. How would you do it?
Got it? Ok. Then here's your challenge. There's a time series called mysteryTimeSeries. Make the stimulus convolution matrix for stimulusTimingLeft and stimulusTimingRight and compute the response
for the mysteryTimeSeries (it will be one long vector with the concatenated responses to the two different types of stimulus).
Which hemisphere did this voxel come from?
General Linear Model
Quick note here is that the event-related analysis is a form of “general linear model” in which the measured responses is modeled as weighted sums of different variables (in this case, different
contributions of events from different time points). The event-related analysis shows you something pretty close to a trial-triggered average (assuming that response overlaps sum linearly) so is
fairly close to the original data, but can require a lot of data to fit well since you have many unknowns (every time point in the estimated response is an unknown). If you knew the shape of a
hemodynamic response function, then you could simplify the computation down to a single number - the magnitude of the evoked response. That would look like the following:
All we have done here is insert a “canoncial” hemodynamic response of unit response (typically you either set the amplitude or the area under the curve to equal one), and now only have to estimate a
single number to get the magnitude of response (traditionally this number is called a beta weight). This can be effective because it means there are less unknowns to compute, but can be problematic
if the canonical hemodynamic response function you have assumed is wrong. Typically, the convolution is precomputed so that each column of the “design matrix” represents the effect of a different
stimulus type (you can also put in other columns to represent things like head motion or linear drift) and a set of beta weights are computed.
The least squares solution for the GLM above is computed in the same way as we did for the event-related analysis.
pRF Encoding model
If we are interested in where visual receptive fields are, we could continue the approach from above by testing stimuli at different locations of the visual field and then computing event-related
responses. But, there is a much better way to do this. That takes us to the idea of encoding models!
Encoding models of various forms have become a powerful tool for analyzing functional imaging data because they can build on knowledge of how different parts of the brain respond to stimuli. A
simple, but really useful encoding model is the population receptive field model (see Dumoulin and Wandell (2008) that is used routinely to make topographic maps of visual areas in human visual
First, take a few minutes to go through Dan Birman's excellent conceptual introduction to the population receptive field.
Got the idea? Cool. A few key points to take away from this. The encoding models are useful because they allow you to predict what responses will be like to stimuli that you have never presented
before, which also allows you to decoding - given a response what stimulus caused this response? The idea is very general and powerful - while we do this for a simple receptive field, an encoding
model can be anything - could be a sophisticated receptive field model selective for different visual features, could be the output of a deep convolutional network, or a model for language.
So, the basic idea of any encoding model is to make a model that, well, encodes the stimulus. This model is a hypothesis for how you think the area responds. For visual cortex, the simplest model is
that neurons respond to a localized position of space, i.e. has a receptive field. As the BOLD measurement will pick up hemodynamic changes related to a population of neurons, the hypothesis will be
that there is a “population receptive field” associated with each voxel. That looks something like the following:
Where the stimulus is high-contrast bars moving through the visual field (while the subject fixates at a single position in the center of the screen) and the population receptive field encoding model
is a gaussian. The predicted response is simply the projection of the stimulus on to the receptive field - which means that every time frame of the visual stimulus you multiply the visual stimulus
with the gaussian receptive field and sum that all up to get the predicted neural response at that time. With that predicted neural response in hand, you can get a prediction for the BOLD response
similar to what we did with the event-related response:
That is, we simply convolve with a hemodynamic response. That's it. That's a prediction for the voxel response for a given receptive field at a particular location and width. Of course, if the
receptive field is in a different location - or has a different size - it will predict a different BOLD response. Here are two different receptive fields with the neural prediction in black and the
BOLD prediction in red. You should be able to see that they make different predictions as the bar stimulus crosses their receptive fields at different times.
So, to fit the pRF encoding model, you just look for the receptive field location and size that best predicts the BOLD measurement you made at each voxel. You can assume a canonical hemodynamic
response function or you can allow that to change to best predict the data as well. Here's a non-linear fitting algorithm searching for the best parameters for a particular voxel. Black is the
measured time series, red is the fit and the lower right shows you the population receptive field that gives that prediction.
Now, we'll try and fit this model to real data. Note that we will just do a very rough fit using a grid search so that you get the idea of how the fit is done. For those interested, a more thorough
tutorial in which includes non-linear fitting, projected data onto cortical surfaces and decoding is available here. For a tutorial on how to do this analysis and mark retiotopic boundaries for your
own data using the GUI in mrTools, see here.
Compute the model neural response
Examine the stimulus matrix
Let's start by just taking a look at the stimulus that was presented to the subject. Remember the subject was looking at the center of the screen where there was a fixation cross on which they had to
perform a task (just to keep their attentional state and fixation under control) and bars moving in different directions were shown. Each frame of the image is stored as a 2D matrix that has the
pixel values for each position on the screen. For this purpose, the matrix just contains 0's and 1's. Each of these 2D frames are stacked into a 3D matrix where the third dimension is time.
So they just look like images, let's go ahead and display an arbitrary image, say the 36th image, and see what we get.
Now that you can display the image, you might want to go ahead and look at the whole sequence. You can write a for loop and display each image at a time and see what it looks like.
Good, you should have seen a bar move in various directions across the screen. That was what was shown to the subject. Run again, and count how many times a bar crossed the screen - we'll need that
number as a consistency check in a following section!
Compute a gaussian receptive field
Now we need to make our encoding model. In this case it is a simple gaussian receptive field. Let's start by making an arbitrary one at the location (-12, -3) degrees (i.e. 12 degrees to the left and
3 degrees below center) We also need to set the size of the receptive field, let's do one that has a standard deviation of 3 degrees. So, first, your going need to know the equation of a gaussian,
The x and y coordinates of every point of the stimulus image is in stimX and stimY, so using the 2D equation with those input points, you should be able to compute the gaussian receptive field.
Compute the model neural response
Now that we have the receptive field and the stimulus, it's time to compute the expected “neural response”. The amount of overlap between the stimulus and the receptive will govern the neural
response. (For our purposes we won't consider any of the dynamics of neural responses themselves - e.g. transients, adaptation, offset responses, etc etc) To compute the overlap, we just take a point
by point multiplication of the stimulus image with the receptive fields and sum across all points. You should now try to compute that for every time point in the stimulus image.
How many peaks in this simulated neural response do you see? Does it match the number of times there was a bar moving across the visual field that you counted above?
Compute the model BOLD response
Now we have to convert this model neural response into a model BOLD response. The model that we use for this is that the BOLD response is a convolution of a canonical hemodynamic response function
with the neural response. So, go ahead and compute the convolution of the model neural response with the canonical hemodynamic response function (we have provided one in the variable called canonical
Ok. Sanity check (always good to check that things are making sense as you go along as much as possible). Plot the neural response on the same graph as the bold response and see if they make sense.
In particular, the bold response should start after the neural response and last longer, right?
Ok. Looks good? What about at the very end. Do you notice that the BOLD response lasts longer than the neural response? That's because the conv command returns a vector longer than the original
response. We don't want that last piece, so remove it from the bold model response (otherwise, dimensions won't match for the calculations we do later). Also, let's take the transpose of this
modelBoldResponse (also important for matching dimensions up later)
Compare the model to measured responses
Ok. So, we can compute the model BOLD response for a specific receptive field - now we want to know how to compare that to measured responses. We're going to use the time-point by time-point
correlation between the model and the measured responses. Correlation is a nice measure to work with, since it isn't affected by the magnitude of responses. Remember that we haven't really done
anything to make the units of the model match the units of response (look back at the units you have for the model BOLD response - typical BOLD responses are about 1 or 2% signal change - is that the
size of the fluctuations you have in the model? (no. right!). We'll get back to that in a bit below, but here we'll just ignore the whole issue by using correlation. Ok, we have two actual time
series for you to work with tSeries1 and tSeries2 (they were measured from a subject viewing the stimuli). Go ahead and compute the correlation between these time series and the measured bold
response from above. What do you get? For which tSeries is this model rf a good model for?
Ok. Let's confirm that visually, by plotting the model and the time series together and see that they match. Now, we have to worry a little about the magnitudes which are going to be different
between the model and the actual bold responses. Let's handle that by finding that scale factor. How do we do that (hint - what if I call that scale factor a “beta weight”?). Also, since our time
series have the mean subtracted off, we need to subtract the mean off of the model bold response as well (do that first and then compute the scale factor) - general maxim here - if you do something
to your data, you need to do the same thing to the model.
So, should be a great match for tSeries1 and a pretty bad match for tSeries2.
One thing about the correlation values. They are a measure of goodness-of-fit. If you square the value then it is mathematically equivalent to the r2 value we calculated above for the event-related
model. Really? Check it. Compute the r^2 like we did above. Compute the ratio of the residual variance (subtract the model from the time series) to the original variance. r2 is one minus that ratio.
Is that the same as the correlation?
Come out the same? Math. Kinda amazing isn't it. Anyway, this is super important because if your fit is bad (in this case low correlation) then the parameters you fit cannot be interpreted (i.e. the
x, y and std of the population receptive field)! So, always (whether you are doing an encoding model or a GLM or whatever) check goodness of fit before trying to extrapolate anything from what
parameters you got. To use technical terms, if goodness-of-fit is crap then your parameter estimates will be crap too.
Challenge question
Ok, so of course the example above for tSeries1 was totally cooked - I gave you the parameters that would pretty closely match the response. How can we find the best parameters for tSeries2 (and in
general any time series for which we don't know what the receptive field parameters are). After all, that's the whole goal of what we are doing right? To figure out what the receptive field model is
that best explains the responses of each voxel. We'll do a quick and dirty way to get close to the right parameters by doing a grid search. What's a grid search? It's just forming a grid of
parameters and testing to see which model parameters on this grid give a model response that best fits the voxel. In this case, our grid will be over x and y position of the receptive field and the
width (standard deviation). Let's try x and y from -15 to 15 degrees in steps of 1 and standard deviations from 1 to 5 in steps of 1. If you iterate over all combinations of these parameters, compute
predicted model responses and compare the correlation with tSeries2, you should be able to figure out the best receptive field for tSeries2!
What are the best receptive field parameters for tSeries2? | {"url":"https://gru.stanford.edu/doku.php/tutorials/boldmodels","timestamp":"2024-11-10T02:57:45Z","content_type":"text/html","content_length":"98456","record_id":"<urn:uuid:36441e5a-5a35-461d-ab67-7f75789070b8>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00435.warc.gz"} |
Wisemen.ai - AI-Powered Self-Learning Tutor & Curriculum Generator
Introduction to Basic Arithmetic
[First Half: The Fundamentals of Arithmetic]
1.1: Understanding the Numbering System
In this sub-chapter, we will explore the different types of numbers and the fundamental properties of the numbering system. Understanding the structure and characteristics of numbers is crucial for
developing a solid foundation in basic arithmetic.
Types of Numbers
1. Natural Numbers (N): The set of positive integers, starting from 1 (1, 2, 3, 4, 5, ...).
2. Whole Numbers (W): The set of non-negative integers, including 0 (0, 1, 2, 3, 4, ...).
3. Integers (Z): The set of positive and negative whole numbers, including 0 (-3, -2, -1, 0, 1, 2, 3, ...).
4. Rational Numbers (Q): The set of numbers that can be expressed as a ratio of two integers (e.g., 1/2, 3/4, -5/7, 0.75, 2.5).
5. Irrational Numbers (I): The set of numbers that cannot be expressed as a ratio of two integers (e.g., π, √2, e).
Place Value and Representation
The place value system is the foundation of how we represent and manipulate numbers. Each digit in a number has a specific place value, which determines its numerical significance. For example, in
the number 4,567, the 4 represents four thousands, the 5 represents five hundreds, the 6 represents six tens, and the 7 represents seven ones. Understanding place value is crucial for understanding
number operations, as it allows us to correctly align and operate on the different place values.
Key Takeaways:
• The numbering system consists of different types of numbers, each with its own properties and characteristics.
• Place value is the foundation for representing and manipulating numbers, as it determines the numerical significance of each digit.
• Mastering the concept of place value is essential for developing a strong understanding of basic arithmetic operations.
1.2: The Four Basic Operations
In this sub-chapter, we will delve into the four fundamental arithmetic operations: addition, subtraction, multiplication, and division. We will explore the purpose and mechanics of each operation,
as well as the order of operations that governs their execution.
Addition is the process of combining two or more numbers to find their sum. For example, 3 + 4 = 7, where 3 and 4 are the addends, and 7 is the sum. The key properties of addition include:
• Commutativity: a + b = b + a
• Associativity: (a + b) + c = a + (b + c)
• Identity: a + 0 = a
Subtraction is the inverse operation of addition, where we find the difference between two numbers. For example, 10 - 3 = 7, where 10 is the minuend, 3 is the subtrahend, and 7 is the difference.
Subtraction does not possess the same properties as addition, as it is not commutative or associative.
Multiplication is the process of repeated addition, where we find the product of two or more numbers. For example, 3 × 4 = 12, where 3 and 4 are the factors, and 12 is the product. The key properties
of multiplication include:
• Commutativity: a × b = b × a
• Associativity: (a × b) × c = a × (b × c)
• Identity: a × 1 = a
Division is the inverse operation of multiplication, where we find the quotient of two numbers. For example, 12 ÷ 3 = 4, where 12 is the dividend, 3 is the divisor, and 4 is the quotient. Division
does not possess the same properties as multiplication, as it is not commutative or associative.
Order of Operations (PEMDAS)
When performing multiple arithmetic operations in a single expression, the order in which they are executed is crucial. The order of operations, commonly known as PEMDAS, stands for:
1. Parentheses
2. Exponents
3. Multiplication and Division (from left to right)
4. Addition and Subtraction (from left to right)
Adhering to the PEMDAS order ensures that the calculations are performed correctly and consistently.
Key Takeaways:
• The four basic arithmetic operations are addition, subtraction, multiplication, and division, each with its own purpose and mechanics.
• Arithmetic operations have specific properties, such as commutativity and associativity, which are important to understand.
• The order of operations (PEMDAS) must be followed when performing multiple arithmetic operations in a single expression.
1.3: Number Properties and Relationships
In this sub-chapter, we will explore the various properties and relationships of numbers, which are fundamental to understanding and manipulating them effectively.
Number Properties
• Commutativity: a + b = b + a, a × b = b × a
• Associativity: (a + b) + c = a + (b + c), (a × b) × c = a × (b × c)
• Distributivity: a × (b + c) = (a × b) + (a × c)
• Identity: a + 0 = a, a × 1 = a
• Inverse: a - b = a + (-b), a ÷ b = a × (1/b)
These properties govern the behavior of the four basic arithmetic operations and are crucial for simplifying and solving mathematical expressions.
Number Relationships
• Factors and Multiples: Factors are the numbers that divide evenly into another number, while multiples are the numbers that are divisible by a given number.
• Prime Numbers: Prime numbers are positive integers greater than 1 that have no positive divisors other than 1 and themselves.
• Composite Numbers: Composite numbers are positive integers greater than 1 that have at least one positive divisor other than 1 and themselves.
• Coprime Numbers: Two numbers are coprime if they have no common factors other than 1.
Understanding these number relationships helps us recognize patterns, simplify expressions, and solve problems more efficiently.
Key Takeaways:
• Number properties, such as commutativity, associativity, and distributivity, govern the behavior of arithmetic operations and are essential for manipulating expressions.
• Number relationships, including factors, multiples, prime numbers, and coprime numbers, provide insights into the structure and behavior of numbers.
• Mastering number properties and relationships is crucial for developing a deep understanding of basic arithmetic.
1.4: Fractions and Decimal Representation
In this sub-chapter, we will introduce the concept of fractions and their relationship with decimal representations. Fractions and decimals are fundamental to understanding and working with rational
A fraction is a representation of a part of a whole. It is written in the form a/b, where a is the numerator and b is the denominator. The numerator represents the number of parts, and the
denominator represents the total number of equal parts in the whole.
Key concepts related to fractions include:
• Simplifying fractions by finding the greatest common factor (GCF) of the numerator and denominator.
• Equivalent fractions, where the value of the fraction remains the same despite changes in the numerator and denominator.
• Fraction arithmetic (addition, subtraction, multiplication, and division).
Decimal Representation
Decimals are a way to represent fractions using a base-10 system. Each digit in a decimal number represents a specific place value, similar to the place value system in whole numbers.
Conversion between fractions and decimals:
• Converting a fraction to a decimal: Divide the numerator by the denominator.
• Converting a decimal to a fraction: Identify the place value of each digit and express the decimal as a fraction with the appropriate denominator.
Understanding the relationship between fractions and decimals is crucial for performing calculations, estimating values, and interpreting real-world situations involving rational numbers.
Key Takeaways:
• Fractions are a way to represent parts of a whole, with the numerator representing the number of parts and the denominator representing the total number of equal parts.
• Decimals are a base-10 representation of rational numbers, where each digit represents a specific place value.
• Conversion between fractions and decimals is an essential skill for working with rational numbers in various contexts.
1.5: Rounding and Estimation
In this sub-chapter, we will focus on the importance of rounding numbers and making reasonable estimates. These skills are crucial for everyday problem-solving and decision-making.
Rounding Techniques
Rounding is the process of approximating a number to a specific place value. The most common rounding techniques include:
• Rounding to the nearest whole number
• Rounding to a specific decimal place
• Rounding to a specific significant figure
The choice of rounding technique depends on the desired level of precision and the context of the problem.
Significant Figures
Significant figures refer to the meaningful digits in a number, excluding leading or trailing zeros that merely indicate the placement of the decimal point. Understanding significant figures is
important for determining the appropriate level of precision in measurements and calculations.
Estimation Strategies
Estimation is the process of making a reasonable approximation of a value without performing exact calculations. Effective estimation strategies include:
• Front-end estimation: Rounding the larger numbers to the nearest convenient value.
• Clustering: Grouping numbers to simplify the calculation.
• Benchmarking: Comparing the value to a known reference point.
Developing strong estimation skills helps students make informed decisions, check the reasonableness of their answers, and solve problems more efficiently.
Key Takeaways:
• Rounding is the process of approximating a number to a specific place value, and it is essential for representing numbers with the appropriate level of precision.
• Significant figures indicate the meaningful digits in a number, which is crucial for determining the level of accuracy in measurements and calculations.
• Estimation strategies, such as front-end estimation, clustering, and benchmarking, allow students to make reasonable approximations without performing exact calculations.
[Second Half: Applying Arithmetic Principles]
1.6: Problem-Solving Strategies
In this sub-chapter, we will explore effective problem-solving strategies that can be applied to a wide range of arithmetic-based problems. These strategies will help students develop critical
thinking and logical reasoning skills.
Step-by-Step Problem-Solving Approach
1. Understand the problem: Carefully read and analyze the problem to identify the given information, the unknown, and the goal.
2. Develop a plan: Determine the appropriate arithmetic operations and the order in which they should be performed to solve the problem.
3. Execute the plan: Carry out the calculations and steps required to arrive at the solution.
4. Evaluate the solution: Check the reasonableness of the answer and ensure that it satisfies the original problem statement.
Problem-Solving Techniques
• Identifying relevant information: Distinguish between relevant and irrelevant information in the problem statement.
• Breaking down complex problems: Divide larger problems into smaller, more manageable sub-problems.
• Logical reasoning: Apply deductive and inductive reasoning to arrive at the correct solution.
• Pattern recognition: Identify and utilize patterns and relationships to simplify problem-solving.
• Checking for reasonableness: Assess the plausibility of the obtained solution and validate it against the original problem.
By mastering these problem-solving strategies, students will be better equipped to tackle a variety of arithmetic-based challenges, both in the classroom and in real-world situations.
Key Takeaways:
• A structured problem-solving approach, involving understanding the problem, developing a plan, executing the plan, and evaluating the solution, is essential for solving arithmetic-based problems
• Techniques such as identifying relevant information, breaking down complex problems, and applying logical reasoning are crucial for developing problem-solving skills.
• Checking the reasonableness of the obtained solution is an important step to ensure the validity and accuracy of the problem-solving process.
1.7: Real-World Applications of Arithmetic
In this sub-chapter, we will explore how the principles of basic arithmetic can be applied to various real-world situations, demonstrating the practical relevance and importance of these fundamental
mathematical concepts.
Budgeting and Personal Finance
Arithmetic skills are essential for managing personal finances, including:
• Calculating income and expenses
• Tracking spending and saving
• Creating and adhering to a budget
• Calculating interest rates and loan repayments
Shopping and Measurement
Arithmetic is used in everyday shopping and measurement tasks, such as:
• Calculating discounts, sales tax, and total costs
• Determining the best value by comparing unit prices
• Converting between different measurement units (e.g., inches to centimeters, pounds to kilograms)
Data Analysis and Interpretation
Arithmetic is a fundamental tool for understanding and interpreting data, including:
• Calculating averages, medians, and other statistical measures
• Analyzing trends and patterns in numerical data
• Interpreting graphs, charts, and tables
Real-World Problem-Solving
Arithmetic concepts are applied to solve a wide range of practical problems, including:
• Calculating travel distances and fuel efficiency
• Determining the appropriate quantity of materials for a project
• Estimating the cost of a home renovation or remodeling project
By exploring these real-world applications, students will recognize the relevance and importance of the arithmetic principles they are learning, which will help them develop a deeper understanding
and appreciation for the subject.
Key Takeaways:
• Arithmetic skills are essential for managing personal finances, including budgeting, tracking expenses, and calculating interest rates.
• Basic arithmetic operations are used in everyday shopping and measurement tasks, such as calculating discounts, comparing unit prices, and converting between measurement units.
• Arithmetic is a fundamental tool for data analysis and interpretation, allowing students to understand and draw insights from numerical information.
• Applying arithmetic concepts to solve practical, real-world problems reinforces the relevance and importance of the subject matter.
1.8: Mental Math and Estimation Techniques
In this sub-chapter, we will focus on developing mental math and estimation skills, which are invaluable for everyday problem-solving and decision-making.
Mental Math Strategies
Mental math refers to the ability to perform arithmetic calculations in one's head, without the aid of a calculator or written work. Effective mental math strategies include:
• Decomposition: Breaking down numbers into more manageable parts to simplify calculations.
• Compensation: Adjusting one number to make a calculation easier, and then compensating for the adjustment.
• Doubling and halving: Utilizing the properties of doubling and halving to perform quick multiplications and divisions.
• Memorizing key facts: Memorizing fundamental arithmetic facts, such as multiplication tables and common fraction-decimal conversions.
Estimation Techniques
Estimation is the process of making a reasonable approximation of a value without performing exact calculations. Effective estimation strategies include:
• Front-end estimation: Rounding the larger numbers to the nearest convenient value.
• Clustering: Grouping numbers to simplify the calculation.
• Benchmarking: Comparing the value to a known reference point.
• Rounding and adjusting: Rounding one or more numbers and then making appropriate adjustments to the result.
By developing mental math and estimation skills, students will be able to:
• Perform quick calculations in their heads
• Check the reasonableness of written work
• Make informed decisions based on approximate values
• Solve problems more efficiently in everyday situations
Key Takeaways:
• Mental math strategies, such as decomposition, compensation, doubling and halving, and memorizing key facts, enable students to perform arithmetic calculations in their heads.
• Estimation techniques, including front-end estimation, clustering, and benchmarking, allow students to make reasonable approximations without performing exact calculations.
• Mastering mental math and estimation skills enhances students' problem-solving abilities and decision-making in real-world contexts.
1.9: Arithmetic Patterns and Relationships
In this sub-chapter, we will explore the underlying patterns and relationships inherent in arithmetic operations. Understanding these patterns and connections will deepen students' understanding of
the fundamental principles of basic arithmetic.
Divisibility Rules
Divisibility rules are a set of guidelines that help determine whether a number is divisible by another number without performing the actual division. Some common divisibility rules include:
• A number is divisible by 2 if the last digit is 0, 2, 4, 6, or 8.
• A number is divisible by 3 if the sum of its digits is divisible by 3.
• A number is divisible by 4 if the last two digits form a number divisible by 4.
Recognizing and applying these divisibility rules can simplify problem-solving and facilitate number pattern recognition.
Number Sequences and Relationships
Arithmetic operations can generate various number sequences and patterns, such as:
• Arithmetic sequences: A sequence where the difference between any two consecutive terms is constant.
• Geometric sequences: A sequence where the ratio between any two consecutive terms is constant.
• Prime number relationships: The distribution of prime numbers and their unique characteristics.
Identifying and exploring these number patterns can deepen students' understanding of the underlying structures and properties of numbers.
Mathematical Connections
Arithmetic principles are interconnected and can be used to establish relationships between different concepts. For example:
• The connection between fractions and decimals
• The relationship between division and finding a common denominator in fraction addition/subtraction | {"url":"https://wisemen.ai/app/courses/6617b2a086b8215c8e309d68/1","timestamp":"2024-11-10T18:01:29Z","content_type":"text/html","content_length":"77457","record_id":"<urn:uuid:13a84b66-a0af-4a4a-830b-af6e64792cba>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00207.warc.gz"} |
American Mathematical Society
Smooth sets for a Borel equivalence relation
HTML articles powered by AMS MathViewer
Trans. Amer. Math. Soc. 347 (1995), 2025-2039
DOI: https://doi.org/10.1090/S0002-9947-1995-1303127-8
PDF | Request permission
We study some properties of smooth Borel sets with respect to a Borel equivalence relation, showing some analogies with the collection of countable sets from a descriptive set theoretic point of
view. We found what can be seen as an analog of the hyperarithmetic points in the context of smooth sets. We generalize a theorem of Weiss from ${\mathbf {Z}}$-actions to actions by arbitrary
countable groups. We show that the $\sigma$-ideal of closed smooth sets is $\Pi _1^1$ non-Borel. References
—, The descriptive set theory of $\sigma$-ideals of compact sets, Logic Colloquium’88, (R. Ferro, C. Bonotto, S. Valentini and A. Zanardo, eds.), North-Holland, 1989, pp. 117-138. A.S. Kechris,
Lectures on Borel equivalence relation, unpublished. C. Uzcátegui, Ph.D. Thesis, Caltech, 1990. —, The covering property for $\sigma$-ideals of compact sets, Fund. Math. 140 (1992), 119-146.
Similar Articles
• Retrieve articles in Transactions of the American Mathematical Society with MSC: 03E15, 04A15, 28A05, 28D99, 54H05, 54H20
• Retrieve articles in all journals with MSC: 03E15, 04A15, 28A05, 28D99, 54H05, 54H20
Bibliographic Information
• © Copyright 1995 American Mathematical Society
• Journal: Trans. Amer. Math. Soc. 347 (1995), 2025-2039
• MSC: Primary 03E15; Secondary 04A15, 28A05, 28D99, 54H05, 54H20
• DOI: https://doi.org/10.1090/S0002-9947-1995-1303127-8
• MathSciNet review: 1303127 | {"url":"https://www.ams.org/journals/tran/1995-347-06/S0002-9947-1995-1303127-8/home.html","timestamp":"2024-11-05T13:51:45Z","content_type":"text/html","content_length":"68621","record_id":"<urn:uuid:87da44a2-29bc-41cd-9d22-109c83e5b6f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00653.warc.gz"} |
Table Data
G = grouptransform(T,groupvars,method) returns transformed data in the place of the nongrouping variables in table or timetable T. The group-wise computations in method are applied to each
nongrouping variable. Groups are defined by rows in the variables in groupvars that have the same unique combination of values. For example, G = grouptransform(T,"HealthStatus","norm") normalizes the
data in T by health status using the 2-norm.
You can use grouptransform functionality interactively by adding the Compute by Group task to a live script.
G = grouptransform(T,groupvars,groupbins,method) specifies to bin rows in groupvars according to binning scheme groupbins prior to grouping and appends the bins to the output table as additional
variables. For example, G = grouptransform(T,"SaleDate","year","rescale") bins by sale year and scales the data in T to the range [0, 1].
G = grouptransform(___,datavars) specifies the table variables to apply the method to for either of the previous syntaxes.
G = grouptransform(___,Name,Value) specifies additional properties using one or more name-value arguments for any of the previous syntaxes. For example, G = grouptransform
(T,"Temp","linearfill","ReplaceValues",false) appends the filled data as an additional variable of T instead of replacing the nongrouping variables.
Array Data
B = grouptransform(A,groupvars,method) returns transformed data in the place of column vectors in the input vector or matrix A. The group-wise computations in method are applied to all column vectors
in A. Groups are defined by rows in the column vectors in groupvars that have the same unique combination of values.
You can use grouptransform functionality interactively by adding the Compute by Group task to a live script.
B = grouptransform(A,groupvars,groupbins,method) specifies to bin rows in groupvars according to binning scheme groupbins prior to grouping.
B = grouptransform(___,Name,Value) specifies additional properties using one or more name-value arguments for either of the previous syntaxes for an input array.
[B,BG] = grouptransform(A,___) also returns the values of the grouping vectors or binned grouping vectors corresponding to the rows in B.
Fill Missing Data by Group
Create a timetable that contains a progress status for three teams.
timeStamp = days([1 1 1 2 2 2 3 3 3]');
teamNumber = [1 2 3 1 2 3 1 2 3]';
percentComplete = [14.2 28.1 11.5 NaN NaN 19.3 46.1 51.2 30.3]';
T = timetable(timeStamp,teamNumber,percentComplete)
T=9×2 timetable
timeStamp teamNumber percentComplete
_________ __________ _______________
1 day 1 14.2
1 day 2 28.1
1 day 3 11.5
2 days 1 NaN
2 days 2 NaN
2 days 3 19.3
3 days 1 46.1
3 days 2 51.2
3 days 3 30.3
Fill missing status percentages, indicated with NaN, for each team using linear interpolation.
G = grouptransform(T,"teamNumber","linearfill","percentComplete")
G=9×2 timetable
timeStamp teamNumber percentComplete
_________ __________ _______________
1 day 1 14.2
1 day 2 28.1
1 day 3 11.5
2 days 1 30.15
2 days 2 39.65
2 days 3 19.3
3 days 1 46.1
3 days 2 51.2
3 days 3 30.3
To append the filled data to the original table instead of replacing the percentComplete variable, use ReplaceValues.
Gappend = grouptransform(T,"teamNumber","linearfill","percentComplete","ReplaceValues",false)
Gappend=9×3 timetable
timeStamp teamNumber percentComplete linearfill_percentComplete
_________ __________ _______________ __________________________
1 day 1 14.2 14.2
1 day 2 28.1 28.1
1 day 3 11.5 11.5
2 days 1 NaN 30.15
2 days 2 NaN 39.65
2 days 3 19.3 19.3
3 days 1 46.1 46.1
3 days 2 51.2 51.2
3 days 3 30.3 30.3
Normalize Data by Day Name
Create a table of dates and corresponding profits.
timeStamps = datetime([2017 3 4; 2017 3 2; 2017 3 15; 2017 3 10; ...
2017 3 14; 2017 3 31; 2017 3 25; ...
2017 3 29; 2017 3 21; 2017 3 18]);
profit = [2032 3071 1185 2587 1998 2899 3112 909 2619 3085]';
T = table(timeStamps,profit)
T=10×2 table
timeStamps profit
___________ ______
04-Mar-2017 2032
02-Mar-2017 3071
15-Mar-2017 1185
10-Mar-2017 2587
14-Mar-2017 1998
31-Mar-2017 2899
25-Mar-2017 3112
29-Mar-2017 909
21-Mar-2017 2619
18-Mar-2017 3085
Binning by day name, normalize the profits using the 2-norm.
G = grouptransform(T,"timeStamps","dayname","norm")
G=10×3 table
timeStamps profit dayname_timeStamps
___________ _______ __________________
04-Mar-2017 0.42069 Saturday
02-Mar-2017 1 Thursday
15-Mar-2017 0.79344 Wednesday
10-Mar-2017 0.66582 Friday
14-Mar-2017 0.60654 Tuesday
31-Mar-2017 0.74612 Friday
25-Mar-2017 0.64428 Saturday
29-Mar-2017 0.60864 Wednesday
21-Mar-2017 0.79506 Tuesday
18-Mar-2017 0.63869 Saturday
Group Operations with Vector Data
Create a vector of dates and a vector of corresponding profit values.
timeStamps = datetime([2017 3 4; 2017 3 2; 2017 3 15; 2017 3 10; ...
2017 3 14; 2017 3 31; 2017 3 25; ...
2017 3 29; 2017 3 21; 2017 3 18]);
profit = [2032 3071 1185 2587 1998 2899 3112 909 2619 3085]';
Binning by day name, normalize the profits using the 2-norm. Display the transformed data and which group it corresponds to.
[normDailyProfit,dayName] = grouptransform(profit,timeStamps,"dayname","norm")
normDailyProfit = 10×1
dayName = 10x1 categorical
Input Arguments
T — Input table
table | timetable
Input table, specified as a table or timetable.
A — Input array
column vector | matrix
Input array, specified as a column vector or a group of column vectors stored as a matrix.
groupvars — Grouping variables or vectors
scalar | vector | matrix | cell array | pattern | function handle | table vartype subscript
Grouping variables or vectors, specified as one of these options:
• For array input data, groupvars can be either a column vector with the same number of rows as A or a group of column vectors arranged in a matrix or cell array.
• For table or timetable input data, groupvars indicates which variables to use to compute groups in the data. You can specify the grouping variables with any of the options in this table.
Indexing Values to Specify Examples
• A string scalar or character vector • "A" or 'A' — A variable named A
Variable • A string array or cell array of character vectors • ["A" "B"] or {'A','B'} — Two variables named A and B
• A pattern object • "Var"+digitsPattern(1) — Variables named "Var" followed by a
single digit
• An index number that refers to the location of a variable in the table • 3 — The third variable from the table
Variable • A vector of numbers • [2 3] — The second and third variables from the table
• A logical vector. Typically, this vector is the same length as the number of variables, but you can omit • [false false true] — The third variable
trailing 0 (false) values.
Function • A function handle that takes a table variable as input and returns a logical scalar • @isnumeric — All the variables containing numeric values
Variable type • A vartype subscript that selects variables of a specified type • vartype("numeric") — All the variables containing numeric
Example: grouptransform(T,"Var3",method)
method — Transformation method
"zscore" | "norm" | "meancenter" | "rescale" | "meanfill" | "linearfill" | function handle
Transformation method, specified as one of these values:
Method Description
"zscore" Normalize data to have mean 0 and standard deviation 1
"norm" Normalize data by 2-norm
"meancenter" Normalize data to have mean 0
"rescale" Rescale range to [0,1]
"meanfill" Fill missing values with the mean of the group data
"linearfill" Fill missing values by linear interpolation of nonmissing group data
You can also specify a function handle that returns one array whose first dimension has length 1 or has the same number of rows as the input data. If the function returns an array with first
dimension length equal to 1, then grouptransform repeats that value so that the output has the same number of rows as the input.
Data Types: char | string | function_handle
groupbins — Binning scheme for grouping variables or vectors
"none" (default) | vector of bin edges | number of bins | length of time (bin width) | name of time unit (bin width) | cell array of binning methods
Binning scheme for grouping variables or vectors, specified as one or more of the following binning methods. Grouping variables or vectors and binning scheme arguments must be the same size, or one
of them can be scalar.
• "none" — No binning.
• Vector of bin edges — The bin edges define the bins. You can specify the edges as numeric values or as datetime values for datetime grouping variables or vectors.
• Number of bins — The number determines how many equally spaced bins to create. You can specify the number of bins as a positive integer scalar.
• Length of time (bin width) — The length of time determines the width of each bin. You can specify the bin width as a duration or calendarDuration scalar for datetime or duration grouping
variables or vectors.
• Name of time unit (bin width) — The name of the time unit determines the width of each bin. You can specify the bin width as one of the options in this table for datetime or duration grouping
variables or vectors.
Value Description Data Type
"second" Each bin is 1 second. datetime and duration
"minute" Each bin is 1 minute. datetime and duration
"hour" Each bin is 1 hour. datetime and duration
"day" Each bin is 1 calendar day. This value accounts for daylight saving time shifts. datetime and duration
"week" Each bin is 1 calendar week. datetime only
"month" Each bin is 1 calendar month. datetime only
"quarter" Each bin is 1 calendar quarter. datetime only
"year" Each bin is 1 calendar year. This value accounts for leap days. datetime and duration
"decade" Each bin is 1 decade (10 calendar years). datetime only
"century" Each bin is 1 century (100 calendar years). datetime only
"secondofminute" Bins are seconds from 0 to 59. datetime only
"minuteofhour" Bins are minutes from 0 to 59. datetime only
"hourofday" Bins are hours from 0 to 23. datetime only
"dayofweek" Bins are days from 1 to 7. The first day of the week is Sunday. datetime only
"dayname" Bins are full day names, such as "Sunday". datetime only
"dayofmonth" Bins are days from 1 to 31. datetime only
"dayofyear" Bins are days from 1 to 366. datetime only
"weekofmonth" Bins are weeks from 1 to 6. datetime only
"weekofyear" Bins are weeks from 1 to 54. datetime only
"monthname" Bins are full month names, such as "January". datetime only
"monthofyear" Bins are months from 1 to 12. datetime only
"quarterofyear" Bins are quarters from 1 to 4. datetime only
datavars — Table variables to operate on
scalar | vector | cell array | pattern | function handle | table vartype subscript
Table variables to operate on, specified as one of the options in this table. datavars indicates which variables of the input table or timetable to apply the methods to. Other variables not specified
by datavars pass through to the output without being operated on. When datavars is not specified, grouptransform operates on each nongrouping variable.
Indexing Values to Specify Examples
• A string scalar or character vector • "A" or 'A' — A variable named A
Variable • A string array or cell array of character vectors • ["A" "B"] or {'A','B'} — Two variables named A and B
• A pattern object • "Var"+digitsPattern(1) — Variables named "Var" followed by a
single digit
• An index number that refers to the location of a variable in the table • 3 — The third variable from the table
Variable • A vector of numbers • [2 3] — The second and third variables from the table
• A logical vector. Typically, this vector is the same length as the number of variables, but you can omit trailing • [false false true] — The third variable
0 (false) values.
Function • A function handle that takes a table variable as input and returns a logical scalar • @isnumeric — All the variables containing numeric values
Variable type • A vartype subscript that selects variables of a specified type • vartype("numeric") — All the variables containing numeric
Example: grouptransform(T,groupvars,method,["Var1" "Var2" "Var4"])
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but
the order of the pairs does not matter.
Example: G = grouptransform(T,groupvars,groupbins,"zscore",IncludedEdge="right")
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: G = grouptransform(T,groupvars,groupbins,"zscore","IncludedEdge","right")
IncludedEdge — Included bin edge for binning scheme
"left" (default) | "right"
Included bin edge for binning scheme, specified as either "left" or "right", indicating which end of the bin interval is inclusive.
You can specify IncludedEdge only if you also specify groupbins, and the value applies to all binning methods for all grouping variables or vectors.
ReplaceValues — Option to replace values
true or 1 (default) | false or 0
Option to replace values, specified as one of these values:
• true or 1 — Replace nongrouping table variables or column vectors in the input data with table variables or column vectors containing transformed data.
• false or 0 — Append the input data with the table variables or column vectors containing transformed data.
Output Arguments
G — Output table
table | timetable
Output table for table or timetable input data, returned as a table or timetable. G contains the transformed data for each group.
B — Output array
vector | matrix
Output array for array input data, returned as a vector or matrix. B contains the transformed data in the place of the nongrouping vectors.
BG — Grouping vectors
column vector | cell array of column vectors
Grouping vectors for array input data, returned as a column vector or cell array of column vectors. BG contains the unique grouping vector or binned grouping vector combinations that correspond to
the rows in B.
• When making many calls to grouptransform, consider converting grouping variables to type categorical or logical when possible for improved performance. For example, if you have a string array
grouping variable (such as HealthStatus with elements "Poor", "Fair", "Good", and "Excellent"), you can convert it to a categorical variable using the command categorical(HealthStatus).
Alternative Functionality
Live Editor Task
You can use grouptransform functionality interactively by adding the Compute by Group task to a live script.
Extended Capabilities
Tall Arrays
Calculate with arrays that have more rows than fit in memory.
The grouptransform function supports tall arrays with the following usage notes and limitations:
• If A and groupvars are both tall matrices, then they must have the same number of rows.
• If the first input is a tall matrix, then groupvars can be a cell array containing tall grouping vectors.
• The groupvars and datavars arguments do not support function handles.
• If the method argument is a function handle, then it must be a valid input for splitapply operating on a tall array.
• When grouping by discretized datetime arrays, the categorical group names are different compared to in-memory grouptransform calculations.
For more information, see Tall Arrays.
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
Usage notes and limitations:
• Sparse inputs are not supported.
• Binning scheme is not supported for datetime or duration data.
• Input tables that contain multidimensional arrays are not supported.
• Computation methods must be constant.
• Grouping variables must be constant when the first input argument is a table.
• Data variables must be constant.
• Binning scheme specified as character vectors or strings must be constant.
• Name-value arguments must be constant.
• Computation methods cannot return sparse, multidimensional, or cell array results.
Thread-Based Environment
Run code in the background using MATLAB® backgroundPool or accelerate code with Parallel Computing Toolbox™ ThreadPool.
This function fully supports thread-based environments. For more information, see Run MATLAB Functions in Thread-Based Environment.
Version History
Introduced in R2018b
R2024a: Apply multiple binning methods to grouping variable
Apply multiple binning methods to one grouping variable or vector by specifying a cell array of binning methods.
R2022b: Code generation support
Generate C or C++ code for the grouptransform function. For usage notes and limitations, see C/C++ Code Generation.
R2022a: Improved performance with small group size
The grouptransform function shows improved performance, especially when the data count in each group is small.
For example, this code transforms by group a matrix with 500 groups with a count of 10 each. The code is about 6.20x faster than in the previous release.
function timingGrouptransform
data = (1:5000)';
groups = repelem(1:length(data)/10,10)';
p = randperm(length(data));
data = data(p);
groups = groups(p);
for k = 1:290
G = grouptransform(data,groups,"norm");
The approximate execution times are:
R2021b: 6.26 s
R2022a: 1.01 s
The code was timed on a Windows^® 10, Intel^® Xeon^® CPU E5-1650 v4 @ 3.60 GHz test system by calling the timingGrouptransform function.
See Also
Live Editor Tasks | {"url":"https://kr.mathworks.com/help/matlab/ref/double.grouptransform.html","timestamp":"2024-11-05T16:12:06Z","content_type":"text/html","content_length":"147239","record_id":"<urn:uuid:a74dc981-c32a-4c2a-835b-eb65a987f531>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00628.warc.gz"} |
calculator with scientific mode
other math tools
Do you need more math tools. View the tools below and click on the link to use the tool.
math worksheets:
Download a great selection of math workheets. All sheets are pdf files and can be easily printed. Below you will find some of the worksheet categories. The complete list can be found at Math | {"url":"https://cybersleuth-kids.com/math_tools/math_calculator.htm","timestamp":"2024-11-02T10:38:18Z","content_type":"application/xhtml+xml","content_length":"11548","record_id":"<urn:uuid:4bf6f2d1-db8e-431a-a571-53c82075b247>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00669.warc.gz"} |
FD Calculator
A Fixed Deposit (FD) is a financial instrument provided by banks or NBFCs which offers investors a higher rate of interest compared to a regular savings account, until the given maturity date. It may
or may not require the creation of a separate account. They are considered to be very safe investments. FD returns can be calculated using an FD calculator which makes the calculation process easy
and accurate.
Understanding Fixed Deposits
Fixed Deposit (FD) is a type of term investment offered by several banks and NBFCs. The deposit is made for a fixed period of time, which can range from 7 days to 10 years. The interest rate on FDs
is fixed and is generally higher than the interest rate on savings accounts. The interest earned on FDs is either paid at regular intervals or is reinvested, depending on the investor’s choice.
The Mathematics Behind FD Calculation
The formula for calculating the maturity amount of a Fixed Deposit (FD) is:
M = P + (P x r x t/100), where
• A is the maturity amount
• P is the principal amount
• R is the rate of interest
• T is the time in years
• N is the number of compounding periods per year
For example, if you deposit ₹10,000 for 1 year at an interest rate of 7% compounded quarterly, the maturity amount at the end of the year would be:
How to use an FD calculator?
Follow the procedures outlined below to use an FD deposit calculator conveniently.
• Ensure that you have access to all relevant data.
• Enter the variables as specified in the formula into their respective slots.
• The FD maturity amount will be presented instantaneously.
Advantages of using FD calculator
Using the FD amount calculator, you may find out exactly how much you will receive when your FD matures.
There are various other benefits of utilizing these calculators:
• Determine the exact amount you are qualified for at the conclusion of your maturity period and plan accordingly.
• Registered users can use both of these calculators for free and without restriction.
• Compare the maturity amounts of various financial organizations with ease.
• Aside from the Fixed Deposit calculator, you may conveniently arrange your money with the following calculators. All of our services are free to use, and you can use them as often as you like. | {"url":"https://calculator.ai/fd-calculator","timestamp":"2024-11-08T09:17:40Z","content_type":"application/xhtml+xml","content_length":"25789","record_id":"<urn:uuid:e28da469-03b3-40c0-9bae-1f1fa383d40e>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00726.warc.gz"} |
A. Find the mean and standard deviation for the final exam scores in t
A. Find the mean and standard deviation for the final exam scores in the two years, 1992 and 2017.
B. Suppose you scored an 85 on this exam. In which distribution is your test score the farthest above the mean? How far above the mean is your score in both
of these distributions?
C. Admission to a (paid) summer internship program requires that students earn a C or better (70% or higher) on their statistics final exam. If we assume that
scores (in both years) on this test are reasonably normally distributed, what percentage of students in 1992 and 2017 would qualify for this internship?/nAssignment
The performance of 20 students on their final exam on statistics in 1992 and for the second class of 20 students in 2017 is shown in the table below.
Scores on Statistics Final Exam, 1992 and 2017
Class of 1992
Class of 2017/nUnlimited Attempts Allowed
✓ Details
Complete the following problems using what we have learned in class to this point. You may complete these problems on a sheet of paper and submit of
picture of your work or using a tablet or computer (writing out the problems that require math). The task is not to use computer software or applications to
answer the questions but to complete them by hand. You will be graded based on completing each problem (or attempting to do so) but also the accuracy of
your responses. If you have questions pertaining to these questions or anything in the class, please contact me and ask.
Fig: 1
Fig: 2
Fig: 3 | {"url":"https://tutorbin.com/questions-and-answers/a-find-the-mean-and-standard-deviation-for-the-final-exam-scores-in-the-two-years-1992-and-2017-b-suppose-you-scored-an","timestamp":"2024-11-11T17:00:01Z","content_type":"text/html","content_length":"71895","record_id":"<urn:uuid:126a6877-d0cc-4915-a3ad-26af0c794c72>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00110.warc.gz"} |
Complete supersymmetric quantum mechanics of magnetic monopoles in N=4 super Yang-Mills theory
We find the most general low energy dynamics of 1/2 BPS monopoles in the N=4 supersymmetric Yang-Mills (SYM) theories when all six adjoint Higgs expectation values are turned on. When only one Higgs
value is turned on, the Lagrangian is purely kinetic. When all the remaining five are turned on a little bit, however, this moduli space dynamics is augmented by five independent potential terms,
each in the form of half the squared norm of a Killing vector field on the moduli space. A generic stationary configuration of the monopoles can be interpreted as stable non-BPS dyons, previously
found as nonplanar string webs connecting D3-branes. The supersymmetric extension is also found explicitly, and gives the complete quantum mechanics of monopoles in N=4 SYM theory.
Dive into the research topics of 'Complete supersymmetric quantum mechanics of magnetic monopoles in N=4 super Yang-Mills theory'. Together they form a unique fingerprint. | {"url":"https://pure.uos.ac.kr/en/publications/complete-supersymmetric-quantum-mechanics-of-magnetic-monopoles-i-2","timestamp":"2024-11-14T10:44:11Z","content_type":"text/html","content_length":"49824","record_id":"<urn:uuid:36467478-9960-405a-8f1d-fcefa66ccc23>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00848.warc.gz"} |
Beam Search#
Beam search with dynamic beam width.
The progressive widening beam search repeatedly executes a beam search with increasing beam width until the target node is found.
import math
import matplotlib.pyplot as plt
import networkx as nx
def progressive_widening_search(G, source, value, condition, initial_width=1):
"""Progressive widening beam search to find a node.
The progressive widening beam search involves a repeated beam
search, starting with a small beam width then extending to
progressively larger beam widths if the target node is not
found. This implementation simply returns the first node found that
matches the termination condition.
`G` is a NetworkX graph.
`source` is a node in the graph. The search for the node of interest
begins here and extends only to those nodes in the (weakly)
connected component of this node.
`value` is a function that returns a real number indicating how good
a potential neighbor node is when deciding which neighbor nodes to
enqueue in the breadth-first search. Only the best nodes within the
current beam width will be enqueued at each step.
`condition` is the termination condition for the search. This is a
function that takes a node as input and return a Boolean indicating
whether the node is the target. If no node matches the termination
condition, this function raises :exc:`NodeNotFound`.
`initial_width` is the starting beam width for the beam search (the
default is one). If no node matching the `condition` is found with
this beam width, the beam search is restarted from the `source` node
with a beam width that is twice as large (so the beam width
increases exponentially). The search terminates after the beam width
exceeds the number of nodes in the graph.
# Check for the special case in which the source node satisfies the
# termination condition.
if condition(source):
return source
# The largest possible value of `i` in this range yields a width at
# least the number of nodes in the graph, so the final invocation of
# `bfs_beam_edges` is equivalent to a plain old breadth-first
# search. Therefore, all nodes will eventually be visited.
log_m = math.ceil(math.log2(len(G)))
for i in range(log_m):
width = initial_width * pow(2, i)
# Since we are always starting from the same source node, this
# search may visit the same nodes many times (depending on the
# implementation of the `value` function).
for u, v in nx.bfs_beam_edges(G, source, value, width):
if condition(v):
return v
# At this point, since all nodes have been visited, we know that
# none of the nodes satisfied the termination condition.
raise nx.NodeNotFound("no node satisfied the termination condition")
Search for a node with high centrality.#
We generate a random graph, compute the centrality of each node, then perform the progressive widening search in order to find a node of high centrality.
# Set a seed for random number generation so the example is reproducible
seed = 89
G = nx.gnp_random_graph(100, 0.5, seed=seed)
centrality = nx.eigenvector_centrality(G)
avg_centrality = sum(centrality.values()) / len(G)
def has_high_centrality(v):
return centrality[v] >= avg_centrality
source = 0
value = centrality.get
condition = has_high_centrality
found_node = progressive_widening_search(G, source, value, condition)
c = centrality[found_node]
print(f"found node {found_node} with centrality {c}")
# Draw graph
pos = nx.spring_layout(G, seed=seed)
options = {
"node_color": "blue",
"node_size": 20,
"edge_color": "grey",
"linewidths": 0,
"width": 0.1,
nx.draw(G, pos, **options)
# Draw node with high centrality as large and red
nx.draw_networkx_nodes(G, pos, nodelist=[found_node], node_size=100, node_color="r")
found node 73 with centrality 0.12598283530728402
Total running time of the script: (0 minutes 0.188 seconds) | {"url":"https://networkx.org/documentation/latest/auto_examples/algorithms/plot_beam_search.html","timestamp":"2024-11-11T14:19:39Z","content_type":"text/html","content_length":"43066","record_id":"<urn:uuid:e717ef7b-f9bd-4da8-827d-40b4fc97eee5>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00734.warc.gz"} |
Circuit Patterns, Part I: Understanding Circuit Schematics
October 5, 2020
Circuit Patterns, Part I: Understanding Circuit Schematics
[You will get on much better in electronics if you learn to see the schematic line drawings as a series of patterns] [
October 5, 2020
When I was young, I wanted to learn how to build electronics. I bought a large number of books from Radio Shack and read them all, cover to cover. Unfortunately, the books that I read helped me to
understand a little bit about the periphery of electronics but not the core subject. I learned what each type of part did in general resistors, capacitors, transistors, inductors, etc., but I never
really understood how all of the pieces fit together. How do you go from understanding the parts to understand how they fit together into a circuit?
Throughout my life, I have returned to electronics now and again, sometimes personally, sometimes professionally. I eventually learned that most electronics follows basic patterns that are used over
and over again. In fact, schematics (line drawings that show the basic parts of a system) are usually a combination of just a few patterns plus a sort of connect-the-dots between major components.
Most schematics, far from being cryptic, turn out to be pretty straightforward once you learn the patterns.
Building circuits is often straightforward as well. If you need to interface two components, usually one of the standard circuit patterns is all you need. Not everything falls quite so neatly into
standard patterns; there are indeed some advanced patterns that are more complicated than the others. Nonetheless, you may be surprised to learn what you can do, knowing only the most basic ones—and
how many circuit schematics only use those basic patterns, even in professionally designed electronics.
Unfortunately, very few books actually teach these patterns as patterns. That is, occasionally, they will point to some of the patterns as interesting uses of components but they will rarely ask you
to think about circuits in terms of the patterns. However, once you do think about them that way, the world of electronics opens up. You start to see the patterns everywhere.
When I learned how to read schematics, I was still mystified by most circuits. Circuits usually consisted of a dizzying array of resistors with a few other components added in. What were all these
resistors doing? Why were they there? How did someone figure out what values they should have? However, once I learned the basic resistor patterns, it started becoming immediately obvious what most
circuits were doing and why.
For example, “Oh, that’s just a pull-down resistor.” “Looks like they are using a voltage divider there.” Once you understand the basic patterns, then circuits stop looking like a randomly-assembled
group of parts that someone got lucky with and start looking like rational, well-ordered systems.
Resistors, for example, feature four basic resistor patterns: current limiters, voltage dividers, pull-up resistors, and pull-down resistors. These patterns form the bulk of the ways you will see
resistors used in schematics. Whether you understand them will determine whether you understand circuits.
The first pattern we will look at is the simplest one, the current limiting resistor:
A current limiting resistor is simply a resistor that protects another component from being overwhelmed with current. Many devices, such as LEDs, are limited as to how much current can flow through
them without breaking. Thus, a current limiting resistor limits how much current will flow through that part of the circuit.
How big should you make a current limiting resistor? That can be determined using Ohm’s Law. In general terms, if we know the voltage coming in, we can use Ohm’s Law to figure out the maximum current
the resistor will allow. To simplify matters, we will ignore other contributing factors for now.
Ohm’s Law states that our circuit’s resistance should be the voltage divided by the desired current. We can use this as a starting point for the value we choose for the resistor. If we place a
resistor with that value in series with our component, it will offer the component the desired current protection.
In Part 2, we will look at the next resistor pattern, the voltage divider.
If you are interested in circuits and how to build them, I hope you will also check out my book Electronics for Beginners: A Practical Introduction to Schematics, Circuits, and Microcontrollers,
published by technology publisher Apress (a Springer Nature company).
If you don’t know what a resistor is, that’s okay. Electronics for Beginners walks you through the very basics and will introduce you to each of the main components commonly used for building
circuits, what they do, and how they are used in practical circuits.
You may also want to have a look at:
New electronics book honors citizen scientist Forrest Mims III.
Jonathan Bartlett’s dedication reflects Mims’ immense influence on electronics enthusiasts—including himself, as a boy. Electronics for Beginners follows in Mims’ footsteps as it shows the budding
electronics enthusiast the many new components now available and how to use them. | {"url":"https://mindmatters.ai/2020/10/circuit-patterns-part-i-understanding-circuit-schematics/","timestamp":"2024-11-07T23:14:13Z","content_type":"text/html","content_length":"89050","record_id":"<urn:uuid:d2176d49-9824-40fd-a55e-71cb44b5331d>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00606.warc.gz"} |
Cubic Meter to Liter Converter
“Think Big, Scale Down: Convert Cubic Meters to Liters Instantly!”
Cubic Meter to Liter Conversion Formula
To convert cubic meters to liters, you can use the following formula:
Liters = Cubic Meters x 1000
Example of Cubic Meter to Liter Conversion
Example 1: 2.5 Cubic Meter to Liter Conversion
For example, if you have 2.5 cubic meters, the equivalent in liters would be:
Liters = 2.5 x 1000 = 2500
Cubic Meter to Liter Conversion Table (uotp 20 entries)
Here’s a table with 20 entries showing the conversions from cubic meters to liters:
Cubic Meters Liters
Cubic Meter to Liter Converter FAQs
What is the difference between a cubic meter (m³) and a liter (L)?
A cubic meter (m³) is a unit of volume equal to 1,000 liters (L). They are used to measure large volumes of liquids or gases.
How do I convert cubic meters to liters?
To convert cubic meters to liters, multiply the number of cubic meters by 1,000. This is because there are 1,000 liters in 1 cubic meter.
Why would I need to convert cubic meters to liters?
Cubic meters are often used to measure the volume of large containers, tanks, or rooms, while liters are used for smaller volumes like bottles or jugs. Converting between the two units allows for
easier understanding and comparison of volumes.
Is there a difference between the volume of a cubic meter and a liter?
Yes, there is a difference. One cubic meter is equal to 1,000 liters, so a cubic meter represents a much larger volume than a liter.
Can I use this converter for other volume units?
This converter specifically converts cubic meters to liters. For other volume conversions, you may need a different converter or formula.
Related Posts
Related Tags
cubic centimeter to liter, cubic meter to gallon, 1 cubic meter to kg, 1 cubic meter = meter, m3 to ml, cubic feet to liters, volume to liters calculator, 1 litre to m
LEAVE A REPLY Cancel reply | {"url":"https://toolconverter.com/cubic-meter-to-liter-converter/","timestamp":"2024-11-13T18:45:32Z","content_type":"text/html","content_length":"196843","record_id":"<urn:uuid:4deddd59-a1db-4c84-b70c-455b631394e9>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00207.warc.gz"} |
Mathematics and Vedas - The Inner World
Mathematics and Vedas
“Yathaa Shikha Mayuranaam, Naaganammanayoyatha
Tadvadvedaangasaastranaam Ganitammoordhanistitham”
This sloka, as mentioned in the Vedaanga Jyotisham, the first known work in astronomy, means, “Like the crests on the head of peacocks and the gems on the heads of the cobras, Mathematics is adorned
at the top of the Vedanga Sciences.”
The Vedas are full of mathematical concepts, the Indian prowess in the field of mathematics is immense. The work by scholars like Aapastamba, Baudhayana, Bhaskaracharya, Brahmagupta and Arya Bhatta
(among others) in the field of Mathematics remains unparalleled.
Be it the invention of zero, geometry, trigonometry, the number system, the value of pi or the first power series; Indians have defined and refined almost all mathematical concepts. Let us dwell a
little deeper into various concepts that were prevalent in India during the vedic era.
Sulva/Sulba Sutras
The Sulva Sutras composed by Baudhayana, Manava, Aapastamba and Katyayana are considered the most significant.
* Pythagorean Theorem and Pythagorean Triplets
The sutras contain discussion and non-axiomatic demonstrations of cases of the Pythagorean Theorem and Pythagorean triplets.
“The areas (of the squares) produced separately by the length and breadth of a rectangle, together equals to the area of the square produced by the diagonal.”
“Multiply the length of a right-angled triangle by the same length and breadth by the same breadth; the square-root of the sum of these two results gives the hypotenuse” ie,
Ab2 + BC2 = AC2
Aapastamba’s rules for building right angles in altars use the following Pythagorean triplets
(3, 4, 5)
(5, 12, 13)
(8, 15, 17)
(12, 35, 37)
* Square Root of 2
Baudhayana in his Sulva Sutras computed the value of √2 correct to seven decimal places;
√2 = 1.4142156….
“The measure is to be increased by its third and this [third] again by its own fourth less the thirty-fourth part [of that fourth]; this is [the value of] the diagonal of a square [whose side is the
Whereas the modern scientific calculator puts the value as;
√2= 1.4142135…
The value of π
Today, we simply know that the ratio of the circumference of a circle to its diameter is constant, denoted by π. Arya Bhatta gives the value of this constant in the following fashion:
“Add 4 to 100, multiply by 8 and add to 62,000.This approximately (aasanna) is the circumference of the circle, whose diameter is 20,000”. (AryaBhattiyam, GanitaPaadam, 10)
This means a circle whose diameter is 20,000 units has its circumference approximately equal to:
(100+4) x 8 + 62000 = 62, 832 units
Since, the ratio of circumference of a circle to its diameter is a constant, it follows that:
62832 ÷ 20000 = 3.1416
The Unit of Angle Measure
The angle around a point is measured as 360 degrees. Let us look at how it comes into force.
It is actually connected to astronomical descriptions. The solar system, rotation of earth around the sun, revolution of earth on its own axis as well as the seasonal changes are all calculated using
this unit.
In the Rigveda (I-164-48) there is the picturesque description of this, compared to a potter’s wheel,
Dvadasapradhayascakramekam Trininabhyanika utacciketa Tasminsakamtrimsatanasankavo farpitah Sastirnacalacalasa
A wheel with 12 spokes, revolves making 360 degrees but not tied to any nails. These 12 spokes represent 12 months or rasis of 30 degrees each and 360 degrees means 360 days in a year. The 3 nabhyas
may be the northern, equatorial and southern solstices. Another vedic mantra states,
“He moves in a circular like path making four divisions and each division consisted of ninety days. His path is vast and has different four stages, he is young but not fixed. Let that heavenly body
come to our prayer spot.”
Four quadrants each of 90 = 4×90 = 360. So the angle around point is 360 degrees.
As the thread that supports the pearls is invisible, likewise, the sustaining and integrating string of underlying Mathematics is also invisible beneath the pearls of various disciplines of Science.
This particular aspect of Mathematics was ably recognised by the ancient Indians, who hailed it to be the ‘Crest Jewel’ among all the decorous divisions of Science.
Grant Duff, British Indological Historian, very aptly noted,
“Many of the advances in the sciences that we consider today to have been made in Europe were in fact made in India centuries ago.”
Author: Devansh Didi | {"url":"https://theinnerworld.in/spirituality/vedic-pathshala/mathematics-and-vedas/","timestamp":"2024-11-07T00:39:18Z","content_type":"text/html","content_length":"74616","record_id":"<urn:uuid:9bb41ea4-3f9f-4b90-af9d-1fe0ce690ceb>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00385.warc.gz"} |
2011 AMC 8 Problems
2011 AMC 8 (Answer Key)
Printable versions: • AoPS Resources • PDF
1. This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
2. You will receive 1 point for each correct answer. There is no penalty for wrong answers.
3. No aids are permitted other than plain scratch paper, writing utensils, ruler, and erasers. In particular, graph paper, compass, protractor, calculators, computers, smartwatches, and
smartphones are not permitted. Rules
4. Figures are not necessarily drawn to scale.
5. You will have 40 minutes working time to complete the test.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
Problem 1
Margie bought $3$ apples at a cost of $50$ cents per apple. She paid with a 5-dollar bill. How much change did Margie receive?
$\textbf{(A) }\ \textdollar 1.50 \qquad \textbf{(B) }\ \textdollar 2.00 \qquad \textbf{(C) }\ \textdollar 2.50 \qquad \textbf{(D) }\ \textdollar 3.00 \qquad \textbf{(E) }\ \textdollar 3.50$
Problem 2
Karl's rectangular vegetable garden is $20$ feet by $45$ feet, and Makenna's is $25$ feet by $40$ feet. Which of the following statements are true?
$\textbf{(A) }\text{Karl's garden is larger by 100 square feet.}$
$\textbf{(B) }\text{Karl's garden is larger by 25 square feet.}$
$\textbf{(C) }\text{The gardens are the same size.}$
$\textbf{(D) }\text{Makenna's garden is larger by 25 square feet.}$
$\textbf{(E) }\text{Makenna's garden is larger by 100 square feet.}$
Problem 3
Extend the square pattern of 8 black and 17 white square tiles by attaching a border of black tiles around the square. What is the ratio of black tiles to white tiles in the extended pattern?
$[asy] filldraw((0,0)--(5,0)--(5,5)--(0,5)--cycle,white,black); filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle,mediumgray,black); filldraw((2,2)--(3,2)--(3,3)--(2,3)--cycle,white,black); draw((4,0)--
(4,5)); draw((3,0)--(3,5)); draw((2,0)--(2,5)); draw((1,0)--(1,5)); draw((0,4)--(5,4)); draw((0,3)--(5,3)); draw((0,2)--(5,2)); draw((0,1)--(5,1)); [/asy]$
$\textbf{(A) }8:17 \qquad\textbf{(B) }25:49 \qquad\textbf{(C) }36:25 \qquad\textbf{(D) }32:17 \qquad\textbf{(E) }36:17$
Problem 4
Here is a list of the numbers of fish that Tyler caught in nine outings last summer: $\[2,0,1,3,0,3,3,1,2.\]$ Which statement about the mean, median, and mode is true?
$\textbf{(A) }\text{median} < \text{mean} < \text{mode} \qquad \textbf{(B) }\text{mean} < \text{mode} < \text{median} \\ \\ \textbf{(C) }\text{mean} < \text{median} < \text{mode} \qquad \textbf{(D) }
\text{median} < \text{mode} < \text{mean} \\ \\ \textbf{(E) }\text{mode} < \text{median} < \text{mean}$
Problem 5
What time was it $2011$ minutes after midnight on January 1, 2011?
$\textbf{(A) }\text{January 1 at 9:31 PM}$
$\textbf{(B) }\text{January 1 at 11:51 PM}$
$\textbf{(C) }\text{January 2 at 3:11 AM}$
$\textbf{(D) }\text{January 2 at 9:31 AM}$
$\textbf{(E) }\text{January 2 at 6:01 PM}$
Problem 6
In a town of 351 adults, every adult owns a car, motorcycle, or both. If 331 adults own cars and 45 adults own motorcycles, how many of the car owners do not own a motorcycle?
$\textbf{(A) }20 \qquad\textbf{(B) }25 \qquad\textbf{(C) }45 \qquad\textbf{(D) }306 \qquad\textbf{(E) }351$
Problem 7
Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded. What percent of the total area is partially bolded?
$[asy] import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.42,xmax=14.59,ymin=-10.08,ymax=5.26; pair A=(0,0), B=(4,0), C=(0,4),
D=(4,4), F=(2,0), G=(3,0), H=(1,4), I=(2,4), J=(3,4), K=(0,-2), L=(4,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2),
D_1=(10,-2), E_1=(6,-6), F_1=(10,-6), G_1=(6,-4), H_1=(10,-4), I_1=(8,-2), J_1=(8,-6), K_1=(8,-4); draw(C--H--(1,0)--A--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw
(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); draw(C--D); draw(D--B); draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6));
draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw
(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw
(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw
(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds);
dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot
(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth
(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)
+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)
+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$
Problem 8
Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the
$\textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9$
Problem 9
Carmen takes a long bike ride on a hilly highway. The graph indicates the miles traveled during the time of her ride. What is Carmen's average speed for her entire ride in miles per hour? $[asy]
import graph; size(8.76cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=10.19,ymin=-4.43,ymax=9.63; draw((0,0)--(0,8)); draw((0,0)--(8,0));
draw((0,1)--(8,1)); draw((0,2)--(8,2)); draw((0,3)--(8,3)); draw((0,4)--(8,4)); draw((0,5)--(8,5)); draw((0,6)--(8,6)); draw((0,7)--(8,7)); draw((1,0)--(1,8)); draw((2,0)--(2,8)); draw((3,0)--(3,8));
draw((4,0)--(4,8)); draw((5,0)--(5,8)); draw((6,0)--(6,8)); draw((7,0)--(7,8)); label("1",(0.95,-0.24),SE*lsf); label("2",(1.92,-0.26),SE*lsf); label("3",(2.92,-0.31),SE*lsf); label("4",
(3.93,-0.26),SE*lsf); label("5",(4.92,-0.27),SE*lsf); label("6",(5.95,-0.29),SE*lsf); label("7",(6.94,-0.27),SE*lsf); label("5",(-0.49,1.22),SE*lsf); label("10",(-0.59,2.23),SE*lsf); label("15",
(-0.61,3.22),SE*lsf); label("20",(-0.61,4.23),SE*lsf); label("25",(-0.59,5.22),SE*lsf); label("30",(-0.59,6.2),SE*lsf); label("35",(-0.56,7.18),SE*lsf); draw((0,0)--(1,1),linewidth(1.6)); draw((1,1)
--(2,3),linewidth(1.6)); draw((2,3)--(4,4),linewidth(1.6)); draw((4,4)--(7,7),linewidth(1.6)); label("HOURS",(3.41,-0.85),SE*lsf); label("M",(-1.39,5.32),SE*lsf); label("I",(-1.34,4.93),SE*lsf);
label("L",(-1.36,4.51),SE*lsf); label("E",(-1.37,4.11),SE*lsf); label("S",(-1.39,3.7),SE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$
$\textbf{(A) }2\qquad\textbf{(B) } 2.5\qquad\textbf{(C) } 4\qquad\textbf{(D) } 4.5\qquad\textbf{(E) } 5$
Problem 10
The taxi fare in Gotham City is $2.40 for the first $\frac12$ mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles
can you ride for $10?
$\textbf{(A) } 3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$
Problem 11
The graph shows the number of minutes studied by both Asha (black bar) and Sasha(grey bar) in one week. On the average, how many more minutes per day did Sasha study than Asha? $[asy] size(300); real
i; defaultpen(linewidth(0.8)); draw((0,140)--origin--(220,0)); for(i=1;i<13;i=i+1) { draw((0,10*i)--(220,10*i)); } label("0",origin,W); label("20",(0,20),W); label("40",(0,40),W); label("60",
(0,60),W); label("80",(0,80),W); label("100",(0,100),W); label("120",(0,120),W); path MonD=(20,0)--(20,60)--(30,60)--(30,0)--cycle,MonL=(30,0)--(30,70)--(40,70)--(40,0)--cycle,TuesD=(60,0)--(60,90)--
(150,80)--(150,0)--cycle,ThurL=(150,0)--(150,110)--(160,110)--(160,0)--cycle,FriD=(180,0)--(180,70)--(190,70)--(190,0)--cycle,FriL=(190,0)--(190,50)--(200,50)--(200,0)--cycle; fill(MonD,black); fill
(MonL,grey); fill(TuesD,black); fill(TuesL,grey); fill(WedD,black); fill(WedL,grey); fill(ThurD,black); fill(ThurL,grey); fill(FriD,black); fill(FriL,grey); draw(MonD^^MonL^^TuesD^^TuesL^^WedD^^WedL^
^ThurD^^ThurL^^FriD^^FriL); label("M",(30,-5),S); label("Tu",(70,-5),S); label("W",(110,-5),S); label("Th",(150,-5),S); label("F",(190,-5),S); label("M",(-25,85),W); label("I",(-27,75),W); label("N",
(-25,65),W); label("U",(-25,55),W); label("T",(-25,45),W); label("E",(-25,35),W); label("S",(-26,25),W);[/asy]$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$
Problem 12
Angie, Bridget, Carlos, and Diego are seated at random around a square table, one person to a side. What is the probability that Angie and Carlos are seated opposite each other?
$\textbf{(A) } \frac14 \qquad\textbf{(B) } \frac13 \qquad\textbf{(C) } \frac12 \qquad\textbf{(D) } \frac23 \qquad\textbf{(E) } \frac34$
Problem 13
Two congruent squares, $ABCD$ and $PQRS$, have side length $15$. They overlap to form the $15$ by $25$ rectangle $AQRD$ shown. What percent of the area of rectangle $AQRD$ is shaded? $[asy] filldraw
((0,0)--(25,0)--(25,15)--(0,15)--cycle,white,black); label("D",(0,0),S); label("R",(25,0),S); label("Q",(25,15),N); label("A",(0,15),N); filldraw((10,0)--(15,0)--(15,15)--(10,15)
--cycle,mediumgrey,black); label("S",(10,0),S); label("C",(15,0),S); label("B",(15,15),N); label("P",(10,15),N);[/asy]$
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25$
Problem 14
There are $270$ students at Colfax Middle School, where the ratio of boys to girls is $5 : 4$. There are $180$ students at Winthrop Middle School, where the ratio of boys to girls is $4 : 5$. The two
schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls?
$\textbf{(A) } \dfrac7{18} \qquad\textbf{(B) } \dfrac7{15} \qquad\textbf{(C) } \dfrac{22}{45} \qquad\textbf{(D) } \dfrac12 \qquad\textbf{(E) } \dfrac{23}{45}$
Problem 15
How many digits are in the product $4^5 \cdot 5^{10}$?
$\textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12$
Problem 16
Let $A$ be the area of the triangle with sides of length $25, 25$, and $30$. Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$. What is the relationship between $A$ and $B$?
$\textbf{(A) } A = \dfrac9{16}B \qquad\textbf{(B) } A = \dfrac34B \qquad\textbf{(C) } A=B \qquad\textbf{(D) } A = \dfrac43B \\ \\ \textbf{(E) }A = \dfrac{16}9B$
Problem 17
Let $w$, $x$, $y$, and $z$ be whole numbers. If $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$, then what does $2w + 3x + 5y + 7z$ equal?
$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$
Problem 18
A fair 6-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?
$\textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56$
Problem 19
How many rectangles are in this figure?
$[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=
(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy]$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$
Problem 20
Quadrilateral $ABCD$ is a trapezoid, $AD = 15$, $AB = 50$, $BC = 20$, and the altitude is $12$. What is the area of the trapezoid?
$[asy] pair A,B,C,D; A=(3,20); B=(35,20); C=(47,0); D=(0,0); draw(A--B--C--D--cycle); dot((0,0)); dot((3,20)); dot((35,20)); dot((47,0)); label("A",A,N); label("B",B,N); label("C",C,S); label
("D",D,S); draw((19,20)--(19,0)); dot((19,20)); dot((19,0)); draw((19,3)--(22,3)--(22,0)); label("12",(21,10),E); label("50",(19,22),N); label("15",(1,10),W); label("20",(41,12),E);[/asy]$
$\textbf{(A) }600\qquad\textbf{(B) }650\qquad\textbf{(C) }700\qquad\textbf{(D) }750\qquad\textbf{(E) }800$
Problem 21
Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$, and $49$. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old
is Norb?
$\textbf{(A) }29\qquad\textbf{(B) }31\qquad\textbf{(C) }37\qquad\textbf{(D) }43\qquad\textbf{(E) }48$
Problem 22
What is the tens digit of $7^{2011}$?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7$
Problem 23
How many 4-digit positive integers have four different digits, where the leading digit is not zero, the integer is a multiple of 5, and 5 is the largest digit?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }84\qquad\textbf{(E) }108$
Problem 24
In how many ways can 10001 be written as the sum of two primes?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }4$
Problem 25
A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two
$[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]$
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2$
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. | {"url":"https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_8_Problems&oldid=210355","timestamp":"2024-11-08T12:44:05Z","content_type":"text/html","content_length":"82999","record_id":"<urn:uuid:72501a2d-0b14-4e3a-ba2d-878c7dd6f11e>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00417.warc.gz"} |
Benford’s Law!Benford’s Law!
Benford’s Law!
In most cases, science discoveries start with a question related to an observation, this is true in the world of mathematics too. One interesting example is the story of Benford’s Law or the law of
first digits. This law is quite interesting considering that for some people it is pretty much obvious, but for most is very counter-intuitive, as it is based on an observable fact rather than an
analytical deduction; and also, because is one of those cases where a discovery is made, then is forgotten and found again.
Benford’s law presents the fact that in general all numbers in a data set (or group of numbers), are more likely to start with lower digits like 1 or 2, than with high digits as 8 or 9. This law
presents a probability distribution stating exactly how likely it is for a certain digit to be the first in a given number (Example: the first digit of the value 203,543 is 2). This is applicable for
a wide range of values like Country Population, the number of bytes in computer system files, bank account balances, numbers in nature, financial and trade figures, tax returns, weather data,
experimental and cosmological data sets, etc.
The story.
The first to discover this phenomenon was Simon Newcomb a famous American Astronomer, as he published a paper about this phenomenon in a mathematical journal in the year 1881. But given that it
didn’t present all the mathematical background, caused that it was practically forgotten.
Half a century later in 1938, Frank Benford, a General Electric Physicist (is interesting, how this law has far more attention from Physics related scientist than from mathematicians), re-discovered
this phenomenon, pretty much in the same way as Newcomb did it. This discovery was due to the observed worn and griminess on logarithm table books (Logarithm is the opposed operation to
exponentiation, as subtraction is the opposed to addition). At that time, without the availability of today’s computers, to make the multiplication of big numbers logarithm tables were used to
facilitate and speed up calculations since these multiplications can be downgraded to sums of logarithms. These books have a list of logarithm values ranging from 1 to 9, as any number can be reduced
to one of these values, let’s say the log of the number 5123 can be obtained as the logarithm of 5.123 (obtained from the book), plus 3, the value for the logarithm base 10 for 1000.
Benford noticed that pages at the beginning of these books, corresponding to the logarithms of low digits, showed signs of more use and tear than the pages at the end of these, corresponding to
logarithms for high digits.
With this into consideration, Benford did an experiment taking approximately 20,000 “natural occurring” numbers from very different sources, ranging from integer’s square roots, numerical constants,
as well as some unusual data sets, like big river’s length or all the numbers included in an edition of reader’s digest magazine and he compared the first digit frequency for each data group and for
all the data set as a whole, and the probability distributions were similar, having numbers starting with 1 close to one-third of the times, and in a proportion almost six times higher than for
numbers starting with 9. The explanation for this distribution although it might seem odd is quite logic.
The best example I found to explain this distribution is shown in an excellent YouTube video, where they explain the case of a very generous (and totally hypothetical), bank, giving a daily 10%
interest rate; so, if I start with one Dollar, then I’ll start earning 1.10 the first day, 1.21 the second and 1.33 the third, and will have an amount above of 1 dollar until the eighth day, when you
move to 2.14, and from there it starts to increment in higher digits, 2.35, 2.59, etc.., till I reach 10.83, and from there the cycle starts again. now starting again in the teens; 11.92, 13.11,
14.42, etc. where all these numbers start with 1. After several iterations to hundreds or thousands, you will notice that in fact, the values starting with one happen around 30% of the times.
Regardless of the number of times we repeat this exercise. This can be repeated with raffle tickets or any other unrestricted sequence of numbers.
Well, why the fuss?
Benford’s law is not just a mathematical curiosity; given that it is a demonstrable base of how numbers occur naturally, this law is used to identify cases where is suspected that the numbers aren’t
natural; meaning, fraud detection, and is used to analyze data from different operations like stock prices, loan data, tax returns, credit card transactions, customer account balances, inventory
prices, etc. When numbers in these records are manipulated they tend to deviate from the probability distribution defined by Benford, and this is an indicator used to do further analysis. And in
several cases, this indicator is the trigger to detect fraud.
Benford’s law has its limitations too, it works on natural and non-restricted data sets; therefore, for cases of small data sets (less than 500 values), calculated values like phone numbers, or sets
with determined maximum or minimum values as people’s height, this law will not work. But for any other case, this law can be applied. You can revise several examples of this distribution in the site
Testing Benford’s law, where several examples are presented.
So, for those interested in “cooking books”, you better not, and pay attention to this law, remember “first digits rule!!”.
Regards, Alex – Science Kindle.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"https://www.sciencekindle.com/en/benfords-law/","timestamp":"2024-11-05T19:54:11Z","content_type":"text/html","content_length":"93381","record_id":"<urn:uuid:03697fe7-9a52-41b4-97d2-3e49bc8a44fd>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00225.warc.gz"} |
What is Uniform Acceleration in Physics
1. What is Uniform Acceleration?
Definition: Uniform acceleration is the motion of an object whose velocity increases by equal amounts in equal intervals of time, however small the time interval may be. Uniform acceleration is the
constant rate of change of velocity over time, where an object’s speed increases or decreases by the same amount in each equal interval of time. An example of uniform acceleration is the motion of an
object falling freely under gravity. The uniform acceleration formula is a=(v-u)/t. In this case, a represents the acceleration, v is the final velocity, u is the initial velocity, and t is the time.
The s.i unit of uniform acceleration is meters per second square (m/s^2)
Uniform acceleration describes the motion of an object with a constant rate of change in its velocity. This type of motion occurs when you apply a constant force on an object, which causes its speed
to increase or decrease at a steady rate.
2. Understanding Uniform Acceleration
The knowledge of the basic concept of acceleration will improve our understanding of uniform acceleration. Acceleration is defined as the rate of change of an object’s velocity with respect to time.
In other words, it is the change in speed or direction of an object over time. The s.i unit of acceleration is also meters per second squared (m/s²). Below is a picture defining constant acceleration
and listing equations of motion.
What is Uniform Acceleration in Physics
Uniform acceleration occurs when an object experiences a constant force, causing it to change its speed at a steady rate. The rate of change of velocity is constant in uniform acceleration, and the
object’s speed increases or decreases by the same amount in each unit of time.
3. Uniformly Accelerated Motion In A Plane
Uniformly Accelerated Motion in a plane refers to the kinematic behaviour of an object experiencing constant acceleration while moving along a two-dimensional path (plane). This motion is
characterized by the simultaneous change in both speed and direction over time, with the acceleration remaining constant. In a plane, the object’s position is described by two coordinates, typically
represented as horizontal (x) and vertical (y) components.
Mathematically, the equations governing uniformly accelerated motion in a plane are extensions of those for one-dimensional motion. The object’s position, velocity, and acceleration in both the x and
y directions are determined by corresponding equations incorporating time. These equations often involve the initial conditions of the object, such as its initial position, velocity, and the angle of
its trajectory.
Uniformly accelerated motion in a plane enables the prediction and analysis of complex motion scenarios, such as projectile motion. Additionally, this type of motion finds applications in different
fields, ranging from ballistics and sports to aerospace engineering. Therefore, the principles of uniformly accelerated motion in a plane provide a fundamental framework for modeling and solving
problems related to the motion of objects in two-dimensional space.
3.1 Explanation of Uniformly Accelerated Motion
Uniformly accelerated motion means an object moves in a straight line and speeds up or slows down at a steady rate. In this type of motion, the acceleration remains constant, and there is no change
in direction. Unlike motion at a constant speed in a straight line, where there is no acceleration, in a uniformly accelerated motion, the speed keeps changing because of the acceleration. (Note:
Constant speed in a circle has acceleration).
The key features of uniformly accelerated motion help us recognize if we are dealing with this type of motion: it happens in a straight line, showing movement in only one direction. The main
difference from motion at a constant speed is that there is always acceleration present. As a result, the speed of the object keeps changing continuously; it either increases or decreases due to the
effect of that acceleration. Therefore, there is always an initial speed (u) and a final speed (v) involved in situations of uniformly accelerated motion.
If both acceleration and speed have the same direction, the object speeds up every second. But if their directions are opposite (like positive and negative acceleration), the object slows down every
second until it stops completely.
4. Rectilinear Acceleration
Definition: The term Rectilinear Acceleration means the rate of increase of velocity along a straight-line path in a unit time. An object can accelerate or decelerate when its velocity continues to
change. Deceleration is the decreasing rate of change velocity with time. The other names for DECELERATION are RETARDATION or NEGATIVE DECELERATION.
Acceleration or Decelaration (a) = Change in Velocity / Time taken for the change
a = ( Final Velocity (V) – Initial Velocity (U) ) / ( Final time (t[2]) – Initial time (t[1]) )
a = (V – U) / (t[2] – t[1])
5. Equations of Uniformly Accelerated Motion
The three equations of motion for a body traveling along a straight line with uniform acceleration are as follows:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
v = Final Velocity, u = Initial Velocity, a = Acceleration, t = Time and s = Distance covered
6. Examples of Uniform Acceleration
We observe Uniform Acceleration in various situations of our daily lives. Some common examples include:
a. Objects in Free Fall
When you drop an object from a height (say sky), it will fall to the ground due to the force of gravity. The acceleration due to gravity is approximately 9.8 m/s², and this acceleration is constant
throughout the fall. Therefore, the speed of the object increases at a constant rate, and we say that it is experiencing uniform acceleration. Below is a picture of a man (parachute dropping from the
b. Moving Vehicles
When a vehicle accelerates from rest, its speed increases at a constant rate until it reaches its maximum velocity. Similarly, when the vehicle decelerates, its speed decreases at a constant rate
until it stops. Both these scenarios are examples of uniform acceleration. Imagine starting your car and pressing the gas pedal, it is speed will increase (accelerate) and the mileage will continue
to adjust itself. However, the moment you remove your leg from the gas pedal and press the brake pedal, its speed will continue to reduce (decelerate) until it stops.
c. Roller Coasters
Roller coasters are designed to provide an exciting and thrilling experience to its riders. They use the concept of uniform acceleration to create different sensations during the ride. For example,
when the roller coaster moves uphill, it slows down due to gravity, and the acceleration becomes negative. Similarly, when the roller coaster moves downhill, it gains speed due to gravity, and the
acceleration becomes positive again.
roller coaster
7. How to Calculate Uniform Acceleration?
We can easily calculate Uniform Acceleration by applying a simple formula below:
a = (v – u) / t
Where “a” represents the acceleration, “v” is the final velocity, “u” is the initial velocity, and “t” is the time taken.
It is important to note that this formula only works for uniform acceleration, where the rate of change of velocity is constant.
8. Characteristics of Uniformly Accelerated Motion
1. Constant Acceleration: Motion involves a consistent rate of change in velocity.
2. Equal Intervals: Equal time intervals result in equal changes in velocity.
3. Linear Relationship: Velocity changes linearly with time.
4. Constant Force: Implies a constant force acting on the object.
5. Squared Dependence: Displacement is proportional to the square of the time.
6. Initial Conditions: Specific initial velocity and position are known.
9. Applications of Uniform Acceleration
Uniform acceleration has various applications in different fields, including:
a. Gravity
Uniform acceleration due to gravity helps our understanding of the universe. The acceleration due to gravity on the Earth’s surface is approximately 9.8 m/s², and it affects everything around us. It
is responsible for keeping the planets in orbit around the sun and for creating tides in the oceans. Below is a video showing an example of gravity where apples fall from a tree to the ground.
b. Projectile Motion
Projectile motion is the motion of an object through the air under the influence of gravity. It is a classic example of uniform acceleration. When an object is thrown into the air, it experiences a
constant acceleration due to gravity, causing its speed to increase until it reaches its maximum height, and then it decelerates until it comes back to the ground.
c. Newton’s Laws of Motion
Newton’s laws of motion help us in understanding the relationship between force, mass, and acceleration. Uniform acceleration is an important concept in these laws, and it helps us understand the
behaviour of objects in different scenarios. For example, Newton’s second law states that the acceleration of an object is directly proportional to the force applied to it and inversely proportional
to its mass (F=ma).
10. Importance of Uniform Acceleration
Uniform acceleration is applicable in various fields like scientific research, engineering, and everyday life. Some of its applications are:
a. Scientific Research
Uniform acceleration plays a vital role in scientific research, especially in the fields of physics and astronomy. It helps us understand the behavior of objects in different environments, such as in
outer space, where the force of gravity is weak.
b. Engineering and Design
Engineers and designers use the concept of uniform acceleration to create different machines and systems, such as rockets and aeroplanes. Therefore, they need to understand the behavior of these
objects under different conditions, including acceleration and deceleration.
c. Everyday Life
Uniform acceleration is present in many of our daily activities, such as driving a car, riding a bike, or throwing a ball. Understanding the concept of uniform acceleration helps us predict the
behavior of objects in different situations and make informed decisions.
11. Common Misconceptions about Uniform Acceleration
There are several misconceptions about uniform acceleration that need to be addressed. These include:
a. Uniform Acceleration is Constant Speed
Uniform acceleration is not the same as constant speed. Even if an object is moving at a constant speed, it is not experiencing uniform acceleration unless its velocity is changing at a constant
b. Uniform Acceleration Only Occurs in Straight Lines
Uniform acceleration can occur in any direction, whether it is straight or curved. For example, a car moving around a circular track at a constant speed is experiencing uniform acceleration.
c. Uniform Acceleration is Always Positive
Uniform acceleration can be positive or negative, depending on the direction of the acceleration. For example, an object moving upward is experiencing negative acceleration due to gravity, while an
object moving downward is experiencing positive acceleration.
12. Non-Uniform Acceleration
Non-uniform acceleration refers to the motion of an object experiencing a changing rate of acceleration over time. In contrast to uniform acceleration, where velocity varies uniformly in equal time
intervals, non-uniform acceleration involves irregular changes in velocity, resulting in a non-constant acceleration. The acceleration of the object is not consistent, varying either in magnitude or
direction as time progresses.
This type of motion can be caused by forces that are not constant, such as varying external influences or internal resistive forces like friction. As a consequence, the velocity-time graph for
non-uniform acceleration appears curved rather than a straight line. Analyzing non-uniform acceleration often involves calculus, as instantaneous acceleration at any given moment may be required to
understand the dynamic changes in motion. Unlike uniform acceleration problems, where equations of motion are straightforward, solving non-uniform acceleration scenarios may demand more complex
mathematical approaches. Understanding non-uniform acceleration is key in describing the diverse and dynamic nature of real-world motion scenarios where external influences fluctuate, leading to
intricate changes in an object’s acceleration over time.
13. Types of Acceleration Based on Increase or Decrease
Acceleration, the rate of change of velocity, can be categorized based on whether it involves an increase or decrease in speed.
1. Positive acceleration occurs when an object’s velocity rises over time, indicating a speeding up or forward motion. This can result from an applied force in the direction of motion. On the other
2. Negative acceleration, often referred to as deceleration or retardation, signifies a reduction in velocity, indicating a slowing down or backward motion. Negative acceleration arises when a force
acts opposite to the direction of motion, leading to a decrease in speed.
Positive and negative accelerations are key concepts in understanding various dynamic scenarios, such as a car accelerating on a straight road (positive acceleration) or coming to a stop (negative
acceleration). These types of acceleration play a pivotal role in physics, engineering, and everyday experiences, providing a fundamental framework for describing and analyzing the different motions
observed in the physical world. Understanding these concepts helps to model and predict the behaviour of objects in response to different forces, facilitating the design and analysis of various
systems and technologies.
14. Solved Problems
Here is a solved problem to help you understand how to calculate uniform acceleration:
14.1 Problem
A particle accelerates uniformly from rest at 6.0 m/s^2 for 8 seconds and then decelerates uniformly to rest in the next 5 seconds. Determine the magnitude of the deceleration.
The magnitude of the deceleration is 9.6 ms^-2.
We are calculating the particle’s deceleration. Thus,
Initial velocity, u = 0
acceleration, a = 6 m/s^2
time, t = 8 s
Final velocity, v = ?
v = u + at
v= 0 + 6 x 8 = 48 m/s
Now, the particle decelerates to rest from a velocity of 48 m/s. Therefore, we can ccalculate the deceleration in this way:
u = 48 m/s, v = 0, t = 5 s, deceleration (a) = ?
Therefore, from v = u +at
0 = 48 + a x 5
making a subject of the formula, we will have
a = – 48/5 = – 9.6 m/s^2 [The negative sign shows that we are dealing with deceleration]
Therefore, the magnitude of the deceleration is 9.6 meters per second square (m/s^2)
15. Conclusion
Uniform acceleration is a fundamental concept in physics that helps us understand the motion of objects under different conditions. It is present in various fields, including scientific research,
engineering, and everyday life. Understanding the concept of uniform acceleration and its applications can help us make informed decisions and improve our understanding of the world around us.
16. Frequently Asked Questions (FAQs):
1. What is uniform acceleration?
Uniform acceleration is the motion of an object with a constant rate of change in its velocity, caused by a constant force.
2. How do you calculate uniform acceleration?
Uniform acceleration can be calculated using the formula: a = (v – u) / t, where “a” represents the acceleration, “v” is the final velocity, “u” is the initial velocity, and “t” is the time taken.
3. What are some examples of uniform acceleration?
Examples of uniform acceleration include free fall due to gravity, projectile motion, and circular motion.
4. How is uniform acceleration different from constant speed?
Uniform acceleration is not the same as constant speed. Even if an object is moving at a constant speed, it is not experiencing uniform acceleration unless its velocity is changing at a constant
5. Why is uniform acceleration important?
Uniform acceleration is important in various fields, including scientific research, engineering, and everyday life. Understanding the concept of uniform acceleration and its applications can help us
make informed decisions and improve our understanding of the world around us.
You may also like to read:
How to Calculate Average Speed in Physics
17. Reference: | {"url":"https://physicscalculations.com/what-is-uniform-acceleration-in-physics/","timestamp":"2024-11-08T07:34:57Z","content_type":"text/html","content_length":"152422","record_id":"<urn:uuid:7c8f5c4a-9efd-4235-915f-ec9c5abd35b7>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00585.warc.gz"} |
Module Contents¶
class pymor.reductors.neural_network.CustomDataset(training_data)[source]¶
Bases: torch.utils.data.Dataset
Class that represents the dataset to use in PyTorch.
Set of training parameters and the respective coefficients of the solution in the reduced basis.
class pymor.reductors.neural_network.EarlyStoppingScheduler(size_training_validation_set, patience=10, delta=0.0)[source]¶
Bases: pymor.core.base.BasicObject
Class for performing early stopping in training of neural networks.
If the validation loss does not decrease over a certain amount of epochs, the training should be aborted to avoid overfitting the training data. This class implements an early stopping scheduler
that recommends to stop the training process if the validation loss did not decrease by at least delta over patience epochs.
Size of both, training and validation set together.
Number of epochs of non-decreasing validation loss allowed, before early stopping the training process.
Minimal amount of decrease in the validation loss that is required to reset the counter of non-decreasing epochs.
class pymor.reductors.neural_network.NeuralNetworkInstationaryReductor(fom=None, training_set=None, validation_set=None, validation_ratio=0.1, T=None, basis_size=None, rtol=0.0, atol=0.0, l2_err=0.0,
pod_params={}, ann_mse='like_basis', scale_inputs=True, scale_outputs=False)[source]¶
Bases: NeuralNetworkReductor
Reduced Basis reductor for instationary problems relying on artificial neural networks.
This is a reductor that constructs a reduced basis using proper orthogonal decomposition and trains a neural network that approximates the mapping from parameter and time space to coefficients of
the full-order solution in the reduced basis. The approach is described in [WHR19].
compute_training_data Compute a reduced basis using proper orthogonal decomposition.
Compute a reduced basis using proper orthogonal decomposition.
class pymor.reductors.neural_network.NeuralNetworkInstationaryStatefreeOutputReductor(fom=None, nt=1, training_set=None, validation_set=None, validation_ratio=0.1, T=None, validation_loss=None,
scale_inputs=True, scale_outputs=False)[source]¶
Bases: NeuralNetworkStatefreeOutputReductor
Output reductor relying on artificial neural networks.
This is a reductor that trains a neural network that approximates the mapping from parameter space to output space.
compute_training_data Compute the training samples (the outputs to the parameters of the training set).
Compute the training samples (the outputs to the parameters of the training set).
class pymor.reductors.neural_network.NeuralNetworkLSTMInstationaryReductor(fom=None, training_set=None, validation_set=None, validation_ratio=0.1, T=None, basis_size=None, rtol=0.0, atol=0.0, l2_err=
0.0, pod_params={}, ann_mse='like_basis', scale_inputs=True, scale_outputs=False)[source]¶
Bases: NeuralNetworkInstationaryReductor
Reduced Basis reductor for instationary problems relying on LSTM neural networks.
This is a reductor that constructs a reduced basis using proper orthogonal decomposition and trains an LSTM neural network that approximates the mapping from parameter to coefficients of the
full-order solution in the reduced basis for a fixed number of timesteps.
reduce Reduce by LSTM neural networks.
reduce(hidden_dimension='3*N + P', number_layers=1, optimizer=optim.LBFGS, epochs=1000, batch_size=20, learning_rate=1.0, loss_function=None, restarts=10, lr_scheduler=None, lr_scheduler_params=
{}, es_scheduler_params={'patience': 10, 'delta': 0.0}, weight_decay=0.0, log_loss_frequency=0)[source]¶
Reduce by LSTM neural networks.
class pymor.reductors.neural_network.NeuralNetworkLSTMInstationaryStatefreeOutputReductor(fom=None, nt=1, training_set=None, validation_set=None, validation_ratio=0.1, T=None, validation_loss=None,
scale_inputs=True, scale_outputs=False)[source]¶
Bases: NeuralNetworkInstationaryStatefreeOutputReductor, NeuralNetworkLSTMInstationaryReductor
Output reductor relying on LSTM neural networks.
This is a reductor that trains an LSTM neural network that approximates the mapping from parameter space to output space.
class pymor.reductors.neural_network.NeuralNetworkReductor(fom=None, training_set=None, validation_set=None, validation_ratio=0.1, basis_size=None, rtol=0.0, atol=0.0, l2_err=0.0, pod_params={},
ann_mse='like_basis', scale_inputs=True, scale_outputs=False)[source]¶
Bases: pymor.core.base.BasicObject
Reduced Basis reductor relying on artificial neural networks.
This is a reductor that constructs a reduced basis using proper orthogonal decomposition and trains a neural network that approximates the mapping from parameter space to coefficients of the
full-order solution in the reduced basis. The approach is described in [HU18].
The full-order Model to reduce. If None, the training_set has to consist of pairs of parameter values and corresponding solution VectorArrays.
Set of parameter values to use for POD and training of the neural network. If fom is None, the training_set has to consist of pairs of parameter values and corresponding solution VectorArrays
Set of parameter values to use for validation in the training of the neural network. If fom is None, the validation_set has to consist of pairs of parameter values and corresponding solution
Fraction of the training set to use for validation in the training of the neural network (only used if no validation set is provided). Either a validation set or a positive validation ratio
is required.
Desired size of the reduced basis. If None, rtol, atol or l2_err must be provided.
Relative tolerance the basis should guarantee on the training set.
Absolute tolerance the basis should guarantee on the training set.
L2-approximation error the basis should not exceed on the training set.
Dict of additional parameters for the POD-method.
If 'like_basis', the mean squared error of the neural network on the training set should not exceed the error of projecting onto the basis. If None, the neural network with smallest
validation error is used to build the ROM. If a tolerance is prescribed, the mean squared error of the neural network on the training set should not exceed this threshold. Training is
interrupted if a neural network that undercuts the error tolerance is found.
Determines whether or not to scale the inputs of the neural networks.
Determines whether or not to scale the outputs/targets of the neural networks.
compute_training_data Compute a reduced basis using proper orthogonal decomposition.
reconstruct Reconstruct high-dimensional vector from reduced vector u.
reduce Reduce by training artificial neural networks.
Compute a reduced basis using proper orthogonal decomposition.
Reconstruct high-dimensional vector from reduced vector u.
reduce(hidden_layers='[(N+P)*3, (N+P)*3]', activation_function=torch.tanh, optimizer=optim.LBFGS, epochs=1000, batch_size=20, learning_rate=1.0, loss_function=None, restarts=10, lr_scheduler=
optim.lr_scheduler.StepLR, lr_scheduler_params={'step_size': 10, 'gamma': 0.7}, es_scheduler_params={'patience': 10, 'delta': 0.0}, weight_decay=0.0, log_loss_frequency=0)[source]¶
Reduce by training artificial neural networks.
Number of neurons in the hidden layers. Can either be fixed or a Python expression string depending on the reduced basis size respectively output dimension N and the total dimension of
the Parameters P.
Activation function to use between the hidden layers.
Algorithm to use as optimizer during training.
Maximum number of epochs for training.
Batch size to use if optimizer allows mini-batching.
Step size to use in each optimization step.
Loss function to use for training. If 'weighted MSE', a weighted mean squared error is used as loss function, where the weights are given as the singular values of the corresponding
reduced basis functions. If None, the usual mean squared error is used.
Number of restarts of the training algorithm. Since the training results highly depend on the initial starting point, i.e. the initial weights and biases, it is advisable to train
multiple neural networks by starting with different initial values and choose that one performing best on the validation set.
Algorithm to use as learning rate scheduler during training. If None, no learning rate scheduler is used.
A dictionary of additional parameters passed to the init method of the learning rate scheduler. The possible parameters depend on the chosen learning rate scheduler.
A dictionary of additional parameters passed to the init method of the early stopping scheduler. For the possible parameters, see EarlyStoppingScheduler.
Weighting parameter for the l2-regularization of the weights and biases in the neural network. This regularization is not available for all optimizers; see the PyTorch documentation for
more details.
Frequency of epochs in which to log the current validation and training loss during training of the neural networks. If 0, no intermediate logging of losses is done.
class pymor.reductors.neural_network.NeuralNetworkStatefreeOutputReductor(fom=None, training_set=None, validation_set=None, validation_ratio=0.1, validation_loss=None, scale_inputs=True,
Bases: NeuralNetworkReductor
Output reductor relying on artificial neural networks.
This is a reductor that trains a neural network that approximates the mapping from parameter space to output space.
The full-order Model to reduce. If None, the training_set has to consist of pairs of parameter values and corresponding outputs.
Set of parameter values to use for POD and training of the neural network. If fom is None, the training_set has to consist of pairs of parameter values and corresponding outputs.
Set of parameter values to use for validation in the training of the neural network. If fom is None, the validation_set has to consist of pairs of parameter values and corresponding outputs.
See NeuralNetworkReductor.
The validation loss to reach during training. If None, the neural network with the smallest validation loss is returned.
See NeuralNetworkReductor.
See NeuralNetworkReductor.
compute_training_data Compute the training samples (the outputs to the parameters of the training set).
Compute the training samples (the outputs to the parameters of the training set).
pymor.reductors.neural_network.multiple_restarts_training(training_data, validation_data, neural_network, target_loss=None, max_restarts=10, log_loss_frequency=0, training_parameters={},
Algorithm that performs multiple restarts of neural network training.
This method either performs a predefined number of restarts and returns the best trained network or tries to reach a given target loss and stops training when the target loss is reached.
See train_neural_network for more information on the parameters.
Data to use during the training phase.
Data to use during the validation phase.
The neural network to train (parameters will be reset after each restart).
Loss to reach during training (if None, the network with the smallest loss is returned).
Maximum number of restarts to perform.
Frequency of epochs in which to log the current validation and training loss. If 0, no intermediate logging of losses is done.
Additional parameters for the training algorithm, see train_neural_network for more information.
Additional parameters for scaling inputs respectively outputs, see train_neural_network for more information.
The best trained neural network.
The corresponding losses.
Raised if prescribed loss can not be reached within the given number of restarts.
pymor.reductors.neural_network.train_neural_network(training_data, validation_data, neural_network, training_parameters={}, scaling_parameters={}, log_loss_frequency=0)[source]¶
Training algorithm for artificial neural networks.
Trains a single neural network using the given training and validation data.
Data to use during the training phase. Has to be a list of tuples, where each tuple consists of two elements that are either PyTorch-tensors (torch.DoubleTensor) or NumPy arrays or pyMOR data
structures that have to_numpy() implemented. The first element contains the input data, the second element contains the target values.
Data to use during the validation phase. Has to be a list of tuples, where each tuple consists of two elements that are either PyTorch-tensors (torch.DoubleTensor) or NumPy arrays or pyMOR
data structures that have to_numpy() implemented. The first element contains the input data, the second element contains the target values.
The neural network to train (can also be a pre-trained model). Has to be a PyTorch-Module.
Dictionary with additional parameters for the training routine like the type of the optimizer, the (maximum) number of epochs, the batch size, the learning rate or the loss function to use.
Possible keys are 'optimizer' (an optimizer from the PyTorch optim package; if not provided, the LBFGS-optimizer is taken as default), 'epochs' (an integer that determines the number of
epochs to use for training the neural network (if training is not interrupted prematurely due to early stopping); if not provided, 1000 is taken as default value), 'batch_size' (an integer
that determines the number of samples to pass to the optimizer at once; if not provided, 20 is taken as default value; not used in the case of the LBFGS-optimizer since LBFGS does not support
mini-batching), 'learning_rate' (a positive real number used as the (initial) step size of the optimizer; if not provided, 1 is taken as default value), 'loss_function' (a loss function from
PyTorch; if not provided, the MSE loss is taken as default), 'lr_scheduler' (a learning rate scheduler from the PyTorch optim.lr_scheduler package; if not provided or None, no learning rate
scheduler is used), 'lr_scheduler_params' (a dictionary of additional parameters for the learning rate scheduler), 'es_scheduler_params' (a dictionary of additional parameters for the early
stopping scheduler), and 'weight_decay' (non-negative real number that determines the strength of the l2-regularization; if not provided or 0., no regularization is applied).
Dict of tensors that determine how to scale inputs before passing them through the neural network and outputs after obtaining them from the neural network. If not provided or each entry is
None, no scaling is applied. Required keys are 'min_inputs', 'max_inputs', 'min_targets', and 'max_targets'.
Frequency of epochs in which to log the current validation and training loss. If 0, no intermediate logging of losses is done.
The best trained neural network with respect to validation loss.
The corresponding losses as a dictionary with keys 'full' (for the full loss containing the training and the validation average loss), 'train' (for the average loss on the training set), and
'val' (for the average loss on the validation set). | {"url":"https://docs.pymor.org/2024-1-0/autoapi/pymor/reductors/neural_network/index.html","timestamp":"2024-11-06T18:05:43Z","content_type":"text/html","content_length":"99311","record_id":"<urn:uuid:3b0f9a73-1d80-4eef-b881-f23fbcd448da>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00413.warc.gz"} |
The Affect of Interfacial Shear on Cavity Formation at an Elastic Inhomogeneity
Cavity formation at an inhomogeneity is examined by analyzing the problem of a plane circular elastic inclusion embedded in an unbounded elastic matrix subject to remote equibiaxial loading.
Nonlinear behavior is confined to an interfacial cohesive zone characterized by normal and tangential interface forces which generally depend on interfacial displacement jump components and which
require a characteristic length, an interface strength, and a shear stiffness parameter for their prescription. Infinitesimal strain equilibrium solutions for rotationally symmetric and nonsymmetric
cavity shapes (and their associated interfacial tractions) are sought by approximation of the governing interfacial integral equations derived from the Boussinesq-Flamant solution to the problem of a
normal and tangential point force operative at a point of a boundary. For fixed constitutive parameters and relatively small remote loads only rotationally symmetric cavities will form. For other
parameter regions the existence of nonsymmetric cavities is studied by performing a local bifurcation analysis about the rotationally symmetric equilibrium state. A global, post bifurcation analysis
is carried out by analyzing the approximate equations computationally. Stability of equilibrium states is assessed according to the Hadamard stability definition. Results indicate that increasing the
interfacial shear stiffness can (1) under certain circumstances transform brittle nucleation to ductile nucleation and (2) delay the formation of a nonsymmetrical cavity. However, nonsymmetrical
growth cannot be completely suppressed, i.e., ultimately a nonsymmetric cavity will form coincident with the rigid displacement of the inclusion within it.
• Bifurcation problem
• Cavity nucleation
• Inclusion problem
• Interfacial debonding and decohesion
• Isotropic homogeneous infinitesimal elasticity
• Nonlinear integral equation
ASJC Scopus subject areas
• General Materials Science
• Mechanics of Materials
• Mechanical Engineering
Dive into the research topics of 'The Affect of Interfacial Shear on Cavity Formation at an Elastic Inhomogeneity'. Together they form a unique fingerprint. | {"url":"https://experts.syr.edu/en/publications/the-affect-of-interfacial-shear-on-cavity-formation-at-an-elastic","timestamp":"2024-11-14T00:21:44Z","content_type":"text/html","content_length":"52875","record_id":"<urn:uuid:4ca696ad-eae2-4342-b74c-57bc7743ad97>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00897.warc.gz"} |
dask.array.fft.ifft(a, n=None, axis=None, norm=None)¶
Wrapping of numpy.fft.ifft
The axis along which the FFT is applied must have only one chunk. To change the array’s chunking use dask.Array.rechunk.
The numpy.fft.ifft docstring follows below:
Compute the one-dimensional inverse discrete Fourier Transform.
This function computes the inverse of the one-dimensional n-point discrete Fourier transform computed by fft. In other words, ifft(fft(a)) == a to within numerical accuracy. For a general
description of the algorithm and definitions, see numpy.fft.
The input should be ordered in the same way as is returned by fft, i.e.,
□ a[0] should contain the zero frequency term,
□ a[1:n//2] should contain the positive-frequency terms,
□ a[n//2 + 1:] should contain the negative-frequency terms, in increasing order starting from the most negative frequency.
For an even number of input points, A[n//2] represents the sum of the values at the positive and negative Nyquist frequencies, as the two are aliased together. See numpy.fft for details.
Input array, can be complex.
nint, optional
Length of the transformed axis of the output. If n is smaller than the length of the input, the input is cropped. If it is larger, the input is padded with zeros. If n is not given, the
length of the input along the axis specified by axis is used. See notes about padding issues.
axisint, optional
Axis over which to compute the inverse DFT. If not given, the last axis is used.
norm{“backward”, “ortho”, “forward”}, optional
Normalization mode (see numpy.fft). Default is “backward”. Indicates which direction of the forward/backward pair of transforms is scaled and with what normalization factor.
New in version 1.20.0: The “backward”, “forward” values were added.
outcomplex ndarray, optional
If provided, the result will be placed in this array. It should be of the appropriate shape and dtype.
outcomplex ndarray
The truncated or zero-padded input, transformed along the axis indicated by axis, or the last one if axis is not specified.
If axis is not a valid axis of a.
See also
An introduction, with definitions and general explanations.
The one-dimensional (forward) FFT, of which ifft is the inverse
The two-dimensional inverse FFT.
The n-dimensional inverse FFT.
If the input parameter n is larger than the size of the input, the input is padded by appending zeros at the end. Even though this is the common approach, it might lead to surprising results. If
a different padding is desired, it must be performed before calling ifft.
>>> import numpy as np
>>> np.fft.ifft([0, 4, 0, 0])
array([ 1.+0.j, 0.+1.j, -1.+0.j, 0.-1.j]) # may vary
Create and plot a band-limited signal with random phases:
>>> import matplotlib.pyplot as plt
>>> t = np.arange(400)
>>> n = np.zeros((400,), dtype=complex)
>>> n[40:60] = np.exp(1j*np.random.uniform(0, 2*np.pi, (20,)))
>>> s = np.fft.ifft(n)
>>> plt.plot(t, s.real, label='real')
[<matplotlib.lines.Line2D object at ...>]
>>> plt.plot(t, s.imag, '--', label='imaginary')
[<matplotlib.lines.Line2D object at ...>]
>>> plt.legend()
<matplotlib.legend.Legend object at ...>
>>> plt.show() | {"url":"https://docs.dask.org/en/stable/generated/dask.array.fft.ifft.html","timestamp":"2024-11-08T07:36:19Z","content_type":"text/html","content_length":"38378","record_id":"<urn:uuid:e2d98d76-a9bd-4b1a-935e-1d5c27561100>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00237.warc.gz"} |
Concept that certain computational systems can simulate any other computational system, given the correct inputs and enough time and resources.
Universality in AI is fundamentally tied to the theory of computation, particularly to the notion of a universal Turing machine, which can emulate the operations of any other Turing machine. This
concept is significant because it underpins the theoretical foundation of modern computers and AI systems, suggesting that a sufficiently powerful and well-designed AI could, in principle, perform
any computation that any other AI or algorithm can, given adequate resources. This leads to the idea that the capabilities of AI systems are not inherently limited by their design but rather by the
practical constraints of hardware, software, and available data. In AI, this universality principle implies that a general-purpose AI, or artificial general intelligence (AGI), could theoretically
learn and execute any task that a human can, provided it has access to the necessary information and training.
The concept of universality dates back to Alan Turing's work in 1936, where he introduced the universal Turing machine as a model of computation. The term gained prominence with the development of
computer science in the mid-20th century and became a foundational principle in AI research as it evolved through the late 20th and early 21st centuries.
Alan Turing is the pivotal figure in the development of the concept of universality, particularly through his groundbreaking paper "On Computable Numbers" published in 1936. Other significant
contributors include John von Neumann, who developed the architecture that became the basis for modern computers, and Alonzo Church, whose work on lambda calculus also laid foundational principles
for computation. | {"url":"https://www.envisioning.io/vocab/universality","timestamp":"2024-11-12T00:48:49Z","content_type":"text/html","content_length":"442696","record_id":"<urn:uuid:46ea0eca-545e-4ad6-9496-be6b2f45bcea>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00524.warc.gz"} |
Cash Flow Leveraging - Wealth Creation Academy
The mathematics of financial leveraging has always been very seductive to Real Estate Investors. This is more so at the time when markets soar. With leveraging, one uses other people’s money to
enhance his own profits by acquiring additional interests in real estate. This enhancement process takes the form either of added equity, which is realized at the time the real estate asset is sold,
or as additional cash-flow, as in the case of rental properties. Either way, mathematics of financial leveraging makes a compelling case for borrowing against home equity to invest.
There are risks in leveraging that must be understood by an uninitiated investor. For one thing, while leverage allows greater potential return to the investor than otherwise would have been
available, the potential for loss is greater because if the investment loses value not only is a portion of that money lost, but the loan still needs to be repaid in its entirety.
It is therefore important to understand the mathematics behind financial leveraging to avoid making mistakes.
How much worth of real estate you can buy with $100,000? Clearly you can buy $100,000 worth of property in a cash deal. You can also buy property worth $200,000 by borrowing 50% of the purchase
price. If you are more innovative you can buy $400,000 worth of property by taking out a 25% mortgage or better still you can buy $1,000,000 worth of property by taking out a 90% mortgage. This is
interesting mathematics.
Let us say you buy a house of $100,000 in a cash deal with no borrowing. Let us assume that your tenant is paying $8000 in rent. The operating expenses that include the rates and maintenance cost
come to $3000. You are now left with left with $6000 as cash flow from the rent. If your property appreciated by 10% during the year that is to $110,000. You will have a return of $16,000 ($5000
cash flow from rent + $10,000 in capital appreciation). This will mean that you will have a return of 16% on your investment of $100,000.
Now let us say that you decide to leverage your money and take out a mortgage of 90%. The bank will loan you $900,000 on your deposit of $100,000. You now have $1,000,000 and you decide to buy ten
houses with each house valued at $100,000 as in the previous example. Let us assume that your rent of $80,000 from the 10 houses covers the expenses $30,000 ($3,000 X 10 houses) and the mortgage
payment of $ 45,000 (5% interest only loan on $900,000). You will be left with $5000 cash flow from the rent. If your ten houses appreciate by 10% as in the last example to $1.1 million your return
from your investment will be $105,000 ($5000 cash flow from rent + $100,000 in capital appreciation. This will result in a return of 105% on your initial investment of $100,000. This part of
mathematics of financial leveraging that is so seductive to property investors
The after tax returns on a leveraged property will be much higher because you will be able to claim depreciation and maintenance costs on ten properties instead of one. Also the interest paid on
borrowed funds of $900,000 can be claimed back to increase the tax refunds.
The above mathematics clearly shows how you can increase your net returns dramatically by leveraging your money.
But we should also examine the downside mathematics of financial leveraging. Let us say the property values went down by 10% in a year instead of going up in the previous example. In the case of a
cash deal our loss will be $4000 for the year thereby giving us a negative return on investment of 4%. On the other hand we will suffer a loss of $95,000 or 95% of the value of the deposit in case of
mortgage of 90%.
The losses get amplified in case of a leveraged property. Is it advisable to risk financial leverage in buying property? The answer to that question is a simple YES. This is because values of
properties have appreciated steadily at 8% to 10% annually during the past 300 years. There are dips in the property cycle but these are far and few in between. If you have the capacity to buy and
hold and manage your cash flow you will win in the end.
Please also read about mathematics behind Cash Flow Leveraging, Appreciation Leveraging, Risks involved with Leveraging and how you can contain these risks.
Cash Flow Leveraging
Cash flow leveraging is about how borrowing can impact your cash flows from your rental properties
The first important thing to understand in cash flow leveraging is the capitalization rate. It is important to know how much the property is paying you. In its simplest form capitalization rate is
the net rental income from the property divided by the purchase price.
Let us say you buy a property for $100,000. Let us assume gross income from this property is $14,000. Total expenses that include rates, insurance, maintenance & management come to $4,000. The net
income from the property will be $10,000 ($14,000 minus $4,000). The net yield or cap rate on the property will be $10,000 (net rent) divided by $100,000 (purchase price) or 10% in this case.
The next important thing to understand in cash flow leveraging is the cost of borrowing funds. It is not a simple case of interest on your loans but must also into account the amortization cost and
the loan period to work out the loan constant. Let us say that the cost of borrowing is 7%.
The difference between the cost of borrowing and the return from your investment property is called the spread. In our example above we have borrowed money at 7% and are getting a net return of 10%.
In this case we have a positive leverage or cash flow on the property.
If the return from property was lower than the cost of borrowing we will have a negative leverage. A neutral leverage will occur when the cost of borrowing is same as return from the property.
When you are buying a property for cash flow you have to purchase only positively leveraged property. If you do this the cash flow leverage will work in your favor.
Investors sometimes buy negatively or neutrally leveraged property in the hope that capital appreciation from the property will overcome the short term cash flow losses on the investment. In this
case you will need to support the short fall in cash flow from other sources of income. This can become very worry some if the investor loses his job or suffers losses in his business that is
supporting the negatively leveraged property.
A positively leveraged property can turn into a negative leveraged property in case of loss of rental income or if the interest rates move up at the time of re-fixing the mortgage. It is therefore
prudent for you to allow for the vacancy rates or changes in the cost of borrowing.
No one can allow for all the contingencies that may occur in the future. But if you can work out your figures accurately for the first five years then chances of things going wrong with your
investment are greatly reduced. This is because rents will normally go up to provide you with additional cash flow. In addition increase in capital value of the property will give you with an extra
cushion of equity.
A negatively leveraged property will move into positive territory given time and a few rent reviews along the way.
As an investor you will keep out of trouble if you apply cash flow leverage properly and make it a habit of buying only positively leveraged properties. This is what Robert Kiyosaki and all savvy
investors do. You must always concentrate on cash flow first and capital appreciation later.
You may like to read more about Appreciation Leverage, Financial Leverage, Mathematics behind Leverage, Risks involved with Leveraging and how you can contain these risks.
Appreciation Leverage
Appreciation leverage happens when you borrow money at simple interest and invest it into a compounding investment such as real estate you can create great wealth for yourself
Appreciation on properties compounds where as interest on properties (though not exactly simple interest) are normally in fixed installments. If let us say you buy a property in a place where the
historical appreciation is at 7% and you borrow money also at 7% interest only mortgage you will find that at the of thirty years your property would have gained in value by at least 3 times more
than the interest you would have paid. And better still your tenant would have paid that mortgage.
As a thumb rule if your interest on your mortgage is lower than the expected appreciation rate then you will have positive appreciation leverage.
The best properties are those that have positive cash flow leverage and positive appreciation leverage. Such properties are sure shot winners.
In certain high capital growth areas you will have properties that have negative cash flow leverage but positive appreciation leverage. Depending upon your situation you might like to buy such a
property if you have problems of excessive cash flow and a very high taxable income.
No savvy investor will buy a property that has both negative cash flow leverage and negative appreciation leverage. But I amazed at the number of people who are coaxed into buying such properties by
savvy sales people. You should try and avoid such properties as a plague.
The ultimate leverage is when:
Cost of Borrowing < Capitalization Rate
Interest Rate < Appreciation
You may like to read about Financial Leverage, Cash Flow Leverage, Mathematics behind Leverage and Strategies to contain Leveraging Risk
Understanding Leverage Risk
Using Financial Leverage is an excellent way to accelerate your financial growth. But to be a savvy investor you have to understand the leverage risk and learn to manage those risks.
More Leverage = More Financial Risk
Less Leverage = Less Financial Risk
You must not let the leverage risk scare you away from real estate investing because risk can be measured and controlled.
Where there is more volatility there is more risk. Volatility refers to more frequent deviations from the standard. To understand risk you have to understand volatility.
Real estate has the least volatility when compared to other investment venues. A study carried out has shown that during the past 10 years real estate was ten times less volatile when compared to
stocks which tend to fluctuate wildly on a daily basis. Yet there is hue and cry and everyone freaks out when real estate prices go down by five to ten percent.
Even a highly leveraged real state is more stable than stocks. This will imply that real estate has less leverage risk than most other investments.
Even though real estate is very stable leveraging does increase relative risk.
More Leverage = More Returns = More Volatility = More Fluctuations in Cash Flow = More Fluctuations in Returns = More Risk
Let us say you buy a house at a price of $100,000 with cash down. If the price moves up by 10% in one year then you will make a profit of 10% on your investment. In this example we are not taking
cash flow into account. If the price moves down by 10% then you make a loss of 10% on your investment.
In the second example let us say you had taken a mortgage of 80%. This would imply that you have made a down payment of $20,000 on the property. Now if the price moves up by 10% i.e to $110,000 then
you will make a profit of $10,000 on your investment of $20,000. This will equate to a return of 50%. Similarly if the price of the property went down by 10% you will make a loss of 50% on your
This example clearly shows the amplification of leverage risk by a factor of five times if the property has a leveraging ratio of 5.
What happens to the leverage risk if your property is purchased with No Money Down deal with 100% mortgage? Your profits will be infinity if the property prices moved up. Similarly your loss will be
infinity if the property prices went down. But what is your risk if you have no money in the deal?
A study conducted in 1978 found that leveraged real estate is the only investment that is fully protected against inflation. This is because when there is inflation rents go up and there is capital
appreciation. However mortgages remain fixed if they are for long term.
Many investors only look at the upside of financial leverage. Savvy investors try and cover the down side of leverage by taking steps to control it if things go wrong.
Leverage Risk Containment Strategies
You can contain the leverage risk by following the under-mentioned guidelines:
Reduce Financial Leverage Just Before the Market Peaks
Most novice investors start to use maximum leverage when the market is about to peak. This is the time for you to reduce your leverage risk by selling a few properties to pay off your mortgage and
increase your equity on the other properties.
Increase Financial Leverage Just After the Market Bottoms Out
This is the time when most people have deserted the real estate market. But it is exactly during this period that leveraging offers the best capital returns in the medium to long haul. Interest rates
are low, so the cost of borrowing is minimized. Financial institutions are looking for customers and it is easier to cut a better deal or get incentives from them. Sellers too are more motivated and
more flexible on prices and terms of contract.
You should use the equity in your properties to buy more property. You should increase your financial leverage to the maximum because during this period you will have positive cash flow properties
reducing your borrowing risk.
Buy Quality Properties
Always buy quality properties. A house or multi-family dwelling that is well maintained and well kept will hold up value better in the long run, and will save you money in maintenance as well. When
the markets are down it will be easier for you to find quality tenants willing and relocate into a nice-looking property. Quality properties reduce your leverage risk.
Take the profits and pay down the debt
Greed is always dangerous in any market. This is where most people fall. Do not keep reinvesting your profits to buy more property. That is like betting all winnings on every new roll of the roulette
wheel. If you follow this strategy you will lose your winnings because sooner than later markets will move downward. If you do not stand on solid foundation you will lose your profits. The best and
safest strategy is not to keep borrowing against your equity. Use cash-flow to pay down the loan or to wait for prices to increase and then sell for a profit to reduce your debt.
Shop for the lowest possible interest rate
Even though the interest is tax deductible you still have to pay some of it out of their own pocket. It is always advisable to shop for the best deal available, using the services of a good mortgage
broker. If your loans are substantial the savings could amount to several thousand dollars every year.
Make improvements to your property to increase cash flow
Make improvements and add value to your property to increase the rents so that you have positive cash flow leverage on the property.
Deposit Recycling to reduce Leverage Risk
If you buy a property with no money down you have no personal risk as long as the purchase is structured in such a way that you have no personal liability.
It is not always possible to buy a property with no money down. At times you will be required to put some of your money into the deposit. Your aim should be to take out your personal money from the
investment as soon as possible. You can then use this money to buy some more property. This is known as deposit recycling.
The best way to recycle your deposit is to firstly buy the property below market value and then make improvements to increase its value and cash flow. You should then refinance the property and take
out your deposit as soon as possible. Your leverage risk becomes zero once you have none of your personal money in the property.
Once you recycle your deposit to reduce your stake in the property (eliminate leverage risk) you should allow the equity to build in the property through appreciation and repayment of loan through
the cash flow generated from the property. Do not get greedy and keep taking out loans against the equity in the property.
Reduce Leverage Risk as you grow older
You can apply more leverage when you are young because time is on your side. In case things go wrong you can re-start once again. As you near your retirement age you should reduce leverage risk. Once
you are retired leverage only to the extent to reduce your taxes.
Leverage Risk as function of your personality
Each one of us has a different risk profile. Some of us are comfortable at taking more risk than the others. You should study your risk profile carefully and take leverage risk to the extent you are
comfortable with. You will have peace of mind.
You may also like to read about Cash Flow Leverage, Appreciation Leverage, Financial Leverage and Mathematics behind Leveraging. | {"url":"http://wealth-creation-academy.com/tag/cash-flow-leveraging/","timestamp":"2024-11-02T08:33:40Z","content_type":"text/html","content_length":"121956","record_id":"<urn:uuid:6bda6e36-a569-44be-9aa9-5716d8cb532a>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00796.warc.gz"} |
3 Digit Multiplication By 2 Digit Worksheets
Math, specifically multiplication, develops the keystone of countless scholastic disciplines and real-world applications. Yet, for lots of students, grasping multiplication can pose a difficulty. To
address this difficulty, teachers and moms and dads have actually welcomed a powerful tool: 3 Digit Multiplication By 2 Digit Worksheets.
Introduction to 3 Digit Multiplication By 2 Digit Worksheets
3 Digit Multiplication By 2 Digit Worksheets
3 Digit Multiplication By 2 Digit Worksheets - 3 Digit Multiplication By 2 Digit Worksheets, 3 Digit Multiplication By 2 Digit Worksheets Pdf, 3 Digit Multiplied By 2 Digit Worksheet, 3-digit By
2-digit Multiplication Worksheets With Answers, 3 Digit By 2-digit Multiplication Worksheets Pdf Free, 3 Digit By 2 Digit Multiplication Worksheets With Grids, 3 Digit By 2 Digit Multiplication
Worksheets Math Drills, 3 Digit By 2 Digit Multiplication Worksheets Maths Salamander, 3 Digit By 2 Digit Multiplication Worksheets 5th Grade, 3 Digit By 2 Digit Multiplication Worksheets Pdf Grade 4
Traverse through this series of printable 3 digit by 2 digit multiplication worksheets to help kids boost arithmetic skills as they work their way through a broad spectrum of pdf exercises like
column and horizontal multiplication engaging real life word problems and more
These worksheets have column form multiplication problems with 2 digit 10 99 and 3 digit 100 999 numbers Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6 5 More Similar
Multiply 2 digit by 2 digit numbers Multiply 2 digit by 4 digit numbers More multiplication worksheets
Relevance of Multiplication Method Understanding multiplication is essential, laying a strong foundation for advanced mathematical ideas. 3 Digit Multiplication By 2 Digit Worksheets offer structured
and targeted practice, fostering a deeper comprehension of this basic arithmetic procedure.
Development of 3 Digit Multiplication By 2 Digit Worksheets
Multiplying 2 Digit By 2 Digit Worksheets
Multiplying 2 Digit By 2 Digit Worksheets
Welcome to The 3 digit by 2 digit Multiplication with Grid Support Including Regrouping A Math Worksheet from the Long Multiplication Worksheets Page at Math Drills This math worksheet was created or
last revised on 2023 08 12 and has been viewed 661 times this week and 822 times this month
3 Digit by 2 Digit Multiplication worksheet Live Worksheets Home Worksheets 3 Digit by 2 Digit Multiplication 3 Digit by 2 Digit Multiplication Daniel Felipe Rivera Marin Member for 3 years 5 months
Age 9 12 Level 4 Language English en ID 680685 31 01 2021 Country code CO Country Colombia School subject Math 1061955
From standard pen-and-paper workouts to digitized interactive formats, 3 Digit Multiplication By 2 Digit Worksheets have actually evolved, satisfying diverse discovering styles and preferences.
Kinds Of 3 Digit Multiplication By 2 Digit Worksheets
Standard Multiplication Sheets Basic workouts concentrating on multiplication tables, assisting students construct a solid arithmetic base.
Word Trouble Worksheets
Real-life situations incorporated into troubles, improving important reasoning and application abilities.
Timed Multiplication Drills Tests developed to boost speed and precision, aiding in fast psychological mathematics.
Advantages of Using 3 Digit Multiplication By 2 Digit Worksheets
Two digit multiplication Worksheet 5 Stuff To Buy Pinterest Multiplication worksheets
Two digit multiplication Worksheet 5 Stuff To Buy Pinterest Multiplication worksheets
Let your young learners improve their multiplication skills by diligently working out 3 digits times 2 digits and using the included answer keys for quick answer verification Multiplying three digit
numbers by two digit numbers using word problems will soon be a cinch We recommend these pdf worksheets for 4th grade 5th grade and 6th grade kids
With this math worksheet students will solve 9 problems that involve multiplying a 3 digit number by a 2 digit number Each problem is set up in a vertical fashion and provides ample space to complete
each calculation An answer key is included with your download to make grading fast and easy
Boosted Mathematical Abilities
Consistent method hones multiplication efficiency, improving total mathematics capacities.
Enhanced Problem-Solving Talents
Word issues in worksheets create analytical thinking and technique application.
Self-Paced Learning Advantages
Worksheets fit individual discovering rates, cultivating a comfy and adaptable discovering setting.
How to Develop Engaging 3 Digit Multiplication By 2 Digit Worksheets
Integrating Visuals and Shades Vibrant visuals and colors record interest, making worksheets aesthetically appealing and involving.
Consisting Of Real-Life Scenarios
Associating multiplication to everyday situations includes relevance and functionality to exercises.
Tailoring Worksheets to Different Skill Degrees Customizing worksheets based upon differing proficiency degrees makes sure inclusive knowing. Interactive and Online Multiplication Resources Digital
Multiplication Equipment and Games Technology-based sources supply interactive learning experiences, making multiplication interesting and enjoyable. Interactive Internet Sites and Applications
On-line platforms offer diverse and obtainable multiplication method, supplementing traditional worksheets. Personalizing Worksheets for Numerous Discovering Styles Visual Learners Aesthetic help and
layouts aid comprehension for learners inclined toward visual learning. Auditory Learners Spoken multiplication troubles or mnemonics cater to learners that comprehend concepts with auditory ways.
Kinesthetic Students Hands-on tasks and manipulatives support kinesthetic students in comprehending multiplication. Tips for Effective Execution in Discovering Consistency in Practice Regular
practice enhances multiplication skills, advertising retention and fluency. Stabilizing Rep and Range A mix of repeated exercises and varied problem layouts preserves passion and comprehension.
Offering Useful Feedback Comments aids in determining locations of enhancement, motivating ongoing progress. Obstacles in Multiplication Practice and Solutions Inspiration and Involvement Hurdles
Dull drills can cause disinterest; innovative approaches can reignite inspiration. Getting Over Concern of Mathematics Negative understandings around math can hinder development; producing a positive
discovering environment is crucial. Impact of 3 Digit Multiplication By 2 Digit Worksheets on Academic Performance Research Studies and Research Study Searchings For Research indicates a positive
correlation in between regular worksheet use and boosted math efficiency.
3 Digit Multiplication By 2 Digit Worksheets become functional tools, cultivating mathematical efficiency in students while suiting diverse discovering styles. From fundamental drills to interactive
online resources, these worksheets not only improve multiplication skills however likewise advertise crucial reasoning and analytic capabilities.
3 Digit X 2 Digit Multiplication Worksheets Schematic And Wiring Diagram
Practice Math worksheets multiplication 3 Digits by 2 Digits 3 Images Frompo
Check more of 3 Digit Multiplication By 2 Digit Worksheets below
Multiplying 4 Digit by 2 Digit Numbers A
multiplication 3 digit by 2 digit worksheets
Multiplication Worksheets 3 Digit Printable Multiplication Flash Cards
Three Digit Multiplication Worksheets Worksheets
Free Multiplication Worksheet 3 Digit By 1 Digit Free4Classrooms
Multiplying 2 Digit by 2 Digit Numbers A
Multiply in columns 2 by 3 digit numbers K5 Learning
These worksheets have column form multiplication problems with 2 digit 10 99 and 3 digit 100 999 numbers Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6 5 More Similar
Multiply 2 digit by 2 digit numbers Multiply 2 digit by 4 digit numbers More multiplication worksheets
Multiplication Worksheets 3 Digits Times 2 Digits
Worksheets to teach students to multiply pairs of 3 digit and 2 digit numbers together There are activities with vertical problems horizontal problems and lattice grids Enjoy a variety of crossword
puzzles math riddles word problems a Scoot game and a custom worksheet generator tool 3 Digit Times 2 Digit Worksheets
These worksheets have column form multiplication problems with 2 digit 10 99 and 3 digit 100 999 numbers Worksheet 1 Worksheet 2 Worksheet 3 Worksheet 4 Worksheet 5 Worksheet 6 5 More Similar
Multiply 2 digit by 2 digit numbers Multiply 2 digit by 4 digit numbers More multiplication worksheets
Worksheets to teach students to multiply pairs of 3 digit and 2 digit numbers together There are activities with vertical problems horizontal problems and lattice grids Enjoy a variety of crossword
puzzles math riddles word problems a Scoot game and a custom worksheet generator tool 3 Digit Times 2 Digit Worksheets
Three Digit Multiplication Worksheets Worksheets
multiplication 3 digit by 2 digit worksheets
Free Multiplication Worksheet 3 Digit By 1 Digit Free4Classrooms
Multiplying 2 Digit by 2 Digit Numbers A
2 digit by 2 digit multiplication Games And worksheets
Double Digit Multiplication Practice Worksheets For Kids Kidpid
Double Digit Multiplication Practice Worksheets For Kids Kidpid
Multiplying 3 digit by 2 digit Worksheets And Exercise Images Preview EngWorksheets
Frequently Asked Questions (Frequently Asked Questions).
Are 3 Digit Multiplication By 2 Digit Worksheets appropriate for any age groups?
Yes, worksheets can be tailored to various age and ability levels, making them versatile for various students.
Just how commonly should pupils practice using 3 Digit Multiplication By 2 Digit Worksheets?
Consistent technique is key. Regular sessions, preferably a few times a week, can generate considerable enhancement.
Can worksheets alone boost math abilities?
Worksheets are an important tool however must be supplemented with diverse discovering methods for thorough skill growth.
Are there on-line platforms supplying complimentary 3 Digit Multiplication By 2 Digit Worksheets?
Yes, numerous academic internet sites provide free access to a wide range of 3 Digit Multiplication By 2 Digit Worksheets.
Exactly how can moms and dads sustain their kids's multiplication technique at home?
Encouraging regular method, offering assistance, and developing a favorable understanding setting are helpful actions. | {"url":"https://crown-darts.com/en/3-digit-multiplication-by-2-digit-worksheets.html","timestamp":"2024-11-13T22:42:06Z","content_type":"text/html","content_length":"32598","record_id":"<urn:uuid:1ffc6424-39f9-432c-ad23-a97d5ec3986b>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00856.warc.gz"} |
Quantum Mechanics Exam: Quiz!
Questions and Answers
• 1.
Calculate the Zero-point energy for a particle in an infinite potential well for an electron confined to a 1 nm atom.
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. 5.9 X 10^-29 J
The zero-point energy for a particle in an infinite potential well is given by the equation E = (Ï€^2 * h^2) / (2mL^2), where h is the Planck's constant, m is the mass of the particle, and L is
the length of the well. In this case, we are given that the electron is confined to a 1 nm atom, so L = 1 nm. Plugging in the values into the equation, we get E = (Ï€^2 * h^2) / (2 * m * (1 nm)^
2). Since the values of h and m are constant, the only variable is L. Therefore, the zero-point energy will be directly proportional to L^2. As L^2 increases, the zero-point energy will also
increase. Among the given options, the only value that is larger than the others is 5.9 X 10-29 J.
• 2.
The magnitude of average acceleration in half time period in a simple harmonic motion is
□ A.
□ B.
□ C.
□ D.
Correct Answer
A. 2 ω^2A /π
In a simple harmonic motion, the magnitude of average acceleration in half a time period is given by 2ω²A/π. This can be derived from the equation of motion for simple harmonic motion, where
acceleration is proportional to displacement and opposite in direction. The average acceleration is calculated by taking the difference in velocities at the beginning and end of half a time
period and dividing it by the time taken. In this case, the average acceleration is found to be 2ω²A/π.
• 3.
In a finite Potential well, the potential energy outside the box is ____________
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. Constant
In a finite potential well, the potential energy outside the box is constant. This means that the potential energy remains the same at all points outside the box, regardless of the distance from
the well. This is because the potential energy in a finite potential well only changes within the region of the well itself, while outside the well, it remains constant.
• 4.
The wave function of a particle in a box is given by ____________
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. Asin(kx) + Bcos(kx)
The wave function of a particle in a box can be represented by a linear combination of sine and cosine functions, as given by Asin(kx) + Bcos(kx). This form allows for the representation of both
the amplitude and phase of the wave function. The coefficients A and B determine the relative contributions of the sine and cosine components, allowing for a more general representation of the
wave function.
• 5.
Particle in a box of finite potential can never be at rest.
Correct Answer
A. True
In quantum mechanics, a particle in a box refers to a hypothetical scenario where a particle is confined within a finite potential well. According to the Heisenberg uncertainty principle, it is
impossible for a particle to have both a well-defined position and momentum simultaneously. Therefore, a particle in a box can never be at rest, as this would require a precise measurement of
both position and momentum, which is not allowed by quantum mechanics. Hence, the statement "Particle in a box of finite potential can never be at rest" is true.
• 6.
The transmission based on tunnel effect is that of a plane wave through a ____________
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. Rectangular Barrier
The transmission based on tunnel effect is that of a plane wave through a rectangular barrier.
• 7.
The solution of Schrodinger wave equation for Tunnel effect is of the form ____________
□ A.
□ B.
□ C.
□ D.
Correct Answer
C. Ae^ikx+ Be^-ikx
The solution of the Schrodinger wave equation for the Tunnel effect is of the form Aeikx+ Be-ikx. This solution represents a wave traveling in the positive x-direction (Aeikx) and a wave
traveling in the negative x-direction (Be-ikx). The presence of both positive and negative exponential terms indicates the possibility of wave transmission through a potential barrier, which is a
characteristic of the Tunnel effect.
• 8.
Tunnel effect can be explained on the basis of ____________
□ A.
□ B.
□ C.
Heisenberg’s uncertainty principle
□ D.
Correct Answer
C. Heisenberg’s uncertainty principle
The correct answer is Heisenberg's uncertainty principle. The uncertainty principle states that it is impossible to simultaneously determine the exact position and momentum of a particle with
absolute certainty. This principle is fundamental to the understanding of the tunnel effect, which is the phenomenon where particles can pass through barriers that would be classically impossible
to penetrate. The uncertainty in the particle's position allows it to "tunnel" through the barrier, appearing on the other side even though it does not have enough energy to overcome the barrier | {"url":"https://www.proprofs.com/quiz-school/story.php?title=mjkymtgwnwt1bq","timestamp":"2024-11-01T19:25:58Z","content_type":"text/html","content_length":"452384","record_id":"<urn:uuid:b93a11fd-a065-49fd-bbe5-5bd1a315db65>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00663.warc.gz"} |
Gravitational Force | Gravity and Gravitation | AnkPlanet
Every object with mass in this universe attracts each other. The force of attraction between any two bodies in the universe is called gravitational force. The sun attracts the earth and the earth
also attracts the sun. The earth attracts moon and the moon also attracts the earth. Similarly, The earth attracts an apple on a tree and the apple also attracts the earth.
The effect of gravitational force can only be observed in a body with big mass. That’s why, all the planets revolve around the sun as the sun is so large in size and it has a large mass than any
planets or satellites of our solar system.
Newton’s Universal Law of Gravitation
The universal law of gravitation was formulated by Newton in 1687 AD.
Newton’s universal law of gravitation states that
“an object in the universe attracts every other object with a force called gravitation which is directly proportional to the product of their masses and inversely proportional to the square of the
distance between their centers.”
Calculation of Gravitational Force
Let an object of mass $M$ attracts another object of mass $m$ towards its centre $O1$ with a force of $F$. Then the body of mass $m$ also attracts the body of mass $M$ towards its centre $O2$ with
the same force $F$. Let the distance between their centres be $d$.
Then, according to the Newton’s universal law of gravitation,
\[F ∝ M × m\] \[F ∝ \frac{1}{d^2}\] From equations $(1)$ and $(2)$, \[F ∝ \frac{Mm}{d^2}\] \[F=\frac{GMm}{d^2}\] Where, $G$ is the Universal Gravitational Constant.
Universal Gravitational Constant (G)
The value of $G$ is constant everywhere in the universe. So, it is called as universal gravitational constant. It was calculated in the Cavendish Experiment which was performed in 1797 – 1798 AD by
British scientist Henry Cavendish.
\[\text{We know, } F = \frac{GMm}{d^2}\]
Suppose an object of mass $1\text{ kg}$ attracts another object of mass $1\text{ kg}$ and the distance between their centres is $1\text{ m}$.
\[\text{Then, } F = G\]
Therefore, Universal Gravitational Constant $G$ can be defined as the gravitational force between two objects of mass $1\text{ kg}$ each, which are separated by the distance of $1\text{ m}$ from
their centers.
The SI unit of $G$ is $\text{Nm}^2\text{/kg}^2$.
And, the approximate value of $G$ is $6.67×10^{-11}$ $\text{Nm}^2\text{/kg}^2$.
But why this force of attraction cannot be observed in small bodies?
It is because of the very small value of $G$. It is also known as the weakest force in nature. Due to this small value, the gravitational force between two bodies with small masses can’t be observed.
Let’s take an example; let two objects of each mass $50 \text{ kg}$ are separated by the distance of $1\text{ m}$ from their centres. \[\text{Then, }F = \frac{GMm}{d^2} \] \[ = \frac{6.67×10^{-11}
×50×50}{1^2} = 1.6675×10^{-7} N\] Here, the value of gravitational force is too small. So, it can’t be noticed on small objects of earth.
But in the context of earth and sun, they have very big masses and the distance between them is about $1.5×10^7\text{ km}$. So, the gravitational force between them is $10^{27}\text{ N}$. This value
is so big. So, this force of attraction can be easily noticed in them.
Tides occur in seas and oceans because of the gravational force of sun and moon as they both attracts the water of seas and oceans towards them. Here, the role of moon is more than the sun to cause
tides because the distance between the moon and the earth is less than the distance between the sun and earth.
The earth keeps us on its surface due to its gravitational force.
Without Gravitational force, many things were not possible. Revolution of planets around the sun is caused by gravitational force.
Presence of atmosphere on earth is because of this force as the earth attracts the atmosphere towards its centre. Without this force, earth cannot hold the atmosphere and the atmosphere will escape
out in space. So, without this force, even life on earth is not possible. | {"url":"https://ankplanet.com/physics/mechanics/gravity-and-gravitation/gravitational-force/","timestamp":"2024-11-02T10:49:26Z","content_type":"text/html","content_length":"97977","record_id":"<urn:uuid:38019cf4-1c9f-4604-a97a-3ba223577692>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00720.warc.gz"} |
Math 340 Home
Textbook Contents
Online Homework Home
Euler Equations
We will now begin to work on solutions to differential equations near singular points. The simplest examples we can easily solve are called Euler equations. A differential equation is called an
Euler equation
if it can be written in the form $a_nx^ny^{(n)}(x)+\cdots+a_1xy'+a_0y=f(x)$. We will be interested primarily in second order homogeneous Euler equations. To find the general solution to such an
equation, we need two linearly independent solutions. One way to find such solutions, which is very quick if it works, is to guess what they are. The key idea is to notice that if $y(x)=x^r$, then $x
^ny^{(n)}(x)=r!/(r-n)!x^r$ for $r\ge n$. But then every term in the Euler equation will be a constant times $x^r$ and we can hope to choose $r$ so that they all cancel out. That is the strategy.
Step 1:
Guess $y(x)=x^r$ and plug into the equation. $$ \eqalign { x^2y''+4xy'+2y&=x^2(r(r-1)x^{r-2})+4x(rx^{r-1})+2x^r \cr &=(r(r-1)+4r+2)x^r \cr &=(r^2+3r+2)x^r {\buildrel \text{set}\over =} 0 \cr} $$
Step 2:
Solve for $r$. The roots of $r^2+3r+2=0$ are $r=-1$ and $r=-2$. So two solutions are $x^{-1}$ and $x^{-2}$.
Step 3:
If you have found two distinct real roots, $r_1$ and $r_2$, then the general solution is $y(x)=c_1x^{r_1}+c_2x^{r_2}$. The general solution is therefore $y(x)=c_1x^{-1}+c_2x^{-2}$. EXAMPLE: Find the
general solution to $$ x^2y''+xy'-n^2y=0 $$ ($n$ is an integer constant. This equation arises in solving Laplace's Equation $\Delta u=\partial^2u/\partial x^2+\partial^2u/\partial y^2=0$ on the disk.
This should seem reasonable if you recall that in polar coordinates $\Delta=\partial^2/\partial r^2 + (1/r)\partial/\partial r + (1/r^2)\partial/\partial\theta^2$.) Step 1: $$ \eqalign { x^2y''+xy'-n
^2y&=x^2r(r-1)x^{r-2}+xrx^{r-1}-n^2x^r \cr &=r^2x^r-n^2x^r \cr &=(r^2-n^2)x^r {\buildrel \text{set}\over =} 0 \cr } $$ Step 2: The roots of $r^2-n^2=0$ are $r=\pm n$. So the two solutions are $x^n$
and $x^{-n}$. Step 3: The general solution is $y(x)=c_1x^n+c_2x^{-n}$. You can check using the Wronskian that $x^{r_1}$ and $x^{r_2}$ are linearly independent if $r_1\ne r_2$. If the equation is of
higher order than second, this paradigm still works, you just need $n$ distinct roots of the equation to get $n$ linearly independent functions to build the general solution of an $n^{th}$ order
equation. Now what if the roots are not real and distinct? In the case where the roots are complex conjugates we can treat them as we have always treated complex roots. We find a complex solution and
take its real and imaginary parts to obtain two real solutions. Of course, then we have to work with quantities like $x^i$. But this isn't as bad as it seems, we just use $x^i=e^{i\log x}$.
(Actually, taking a logarithm in the complex plane can be a dangerous thing, but we won't worry about that now. You should take the introductory complex variables class sometime though and learn all
about it.) But what if the roots are repeated? In that case we have no obvious guess for how to find a second root. There are several ways to deal with this. The way we will use is to make a change
of variables to reduce the original equation to a constant coefficient equation. This technique will work on any Euler equation, not just those with repeated roots. But the first paradigm is more
efficient when it works.
Paradigm (Take 2)
Step 1:
Make the change of variables $z=\log x$ or equivalently $x=e^z$. The crucial calculations here are how the derivatives of $y$ with respect to $x$ convert to derivatives of $y$ with respect to $z$. We
first compute $$ \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac1x\frac{dy}{dz} $$ where we have used the chain rule. So $xdy/dx=dy/dz$. Next we tackle the second derivative. Here we must remember we
want to write the derivative with respect to $x$ of the derivative with respect to $x$ in terms of the derivative with respect to $z$ of the derivative with respect to $z$. $$ \eqalign { \frac{d^2y}
{dx^2}&=\frac{d}{dx}\left(\frac{dy}{dx}\right) \cr &=\frac1x\frac{d}{dz}\left(\frac1x\frac{dy}{dz}\right) \cr &=\frac1x\left(\frac{d(1/x)}{dz}\frac{dy}{dz} +\frac1x\frac{d^2y}{dz^2}\right) \cr &=\
frac1x\left(-\frac1x\frac{dy}{dz} +\frac1x\frac{d^2y}{dz^2}\right) \cr &=\frac{1}{x^2}\left(\frac{d^2y}{dz^2}-\frac{dy}{dz}\right) \cr } $$ where we used the product rule in the third line. So $x^2d^
2y/dx^2= d^2y/dz^2-dy/dz$. Applying this rule, along with our computation for $dy/dx$ we find the equation becomes $$ \eqalign { &\frac{d^2y}{dz^2}-\frac{dy}{dz}+7\frac{dy}{dz}+9y \cr =\quad &\frac{d
^2y}{dz^2}+6\frac{dy}{dz}+9y=0. \cr } $$
Step 2:
Solve the transformed equation. Since $D^2+6D+9=0$ has a double root of $-3$, the solution is $c_1e^{-3z}+c_2ze^{-3z}$.
Step 3:
Back transform to find the answer in terms of the original variable. $$ \eqalign { c_1e^{-3z}+c_2ze^{-3z}&=c_1e^{-3\log x}+c_2(\log x)e^{-3\log x} \cr &=c_1x^{-3}+c_2(\log x)x^{-3} \cr } $$ The
crucial step to note here is that while $xdy/dx=dy/dz$, $x^2d^2y/dx^2=d^2y/dz^2-dy/dz$. So the leading coefficient and the constant coefficient will remain unchanged when the equation is transformed,
but the coefficient of $dy/dx$ will be changed because it picks up not only the $dy/dx$ term but also part of the $d^2y/dx^2$ term. The two easiest errors to make in solving Euler equations using the
second paradigm is either to forget to change the middle coefficient in transforming the equation or to forget to back substitute to get the answer in terms of the original variable. Of course we
have only dealt with the homogeneous case in these paradigms. To handle the inhomogeneous equation, we can either use the second paradigm, make the change of variables on the right hand side as well
and hope we can apply undetermined coefficients (which isn't all that likely), or we can just use variation of parameters. EXAMPLE: $x^2y''+5xy'+3y=e^x$ We will use the first paradigm and variation
of parameters. Step 1: $y(x)=x^r$, so $xy'(x)=rx^r$ and $x^2y''(x)=r(r-1)x^r$. Plugging these into the equation we obtain $(r^2+4r+3)x^r=0$ Step 2: The roots of $r^2+4r+3=0$ are $r=-3$ and $r=-1$.
Step 3: Two linearly independent solutions of the homogeneous equation are $x^{-3}$ and $x^{-1}$. Now that we have two linearly independent homogeneous solutions, we want to use the variation of
parameters formula to find the general solution. First we note that
the formula in our theorem on variation of parameters
only applies to equations where the coefficient of the leading term is 1. So we divide through our Euler equation by $x^2$ to put it in the correct form, $$y''+5x^{-1}y'+3x^{-2}y=x^{-2}e^x.$$ Our
linearly independent solutions of the homogeneous equation, as noted in step 3, are $y_1(x)=x^{-3}$ and $y_2(x)=x^{-1}$. We compute $W(x^{-3},x^{-1})(x)=-x^{-5}+3x^{-5}=2x^{-5}$. The right hand side
is $g(x)=x^{-2}e^x$. From our formula for variation of parameters we then find the general solution is $$ \eqalign { y(x)&=-x^{-3}\int_0^x \frac{s^{-1}s^{-2}e^s}{2s^{-5}}\,ds+x^{-1}\int_0^x \frac{s^
{-3}s^{-2}e^{s}}{2s^{-5}}\,ds+C_1x^{-3}+C_2x^{-1} \cr &=\frac{-x^{-3}}{2}\int_0^x s^2e^s\,ds +\frac{x^{-1}}{2}\int_0^x e^s\,ds+C_1x^{-3}+C_2x^{-1} \cr &=\frac{-x^{-3}}{2}(x^2e^x-2xe^x+2e^x-2) +\frac
{x^{-1}}{2}(e^x-1)+C_1x^{-3}+C_2x^{-1} \cr &=\frac{-x^{-1}}{2}e^x+x^{-2}e^x-x^{-3}e^x+\frac{x^{-1}}{2}e^x+(C_1+1)x^{-3} +(C_2-1/2)x^{-1} \cr &=x^{-2}e^x-x^{-3}e^x+c_1x^{-3}+c_2x^{-1} \cr } $$ where
$c_1=C_1+1$ and $c_2=C_2-1/2$. If you have any problems with this page, please contact bennett@math.ksu.edu.
©2010, 2014 Andrew G. Bennett | {"url":"https://onlinehw.math.ksu.edu/math340book/chap4/euler.php","timestamp":"2024-11-09T06:54:26Z","content_type":"text/html","content_length":"16093","record_id":"<urn:uuid:be805cc9-2814-45e1-87f7-edfb012c88b2>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00793.warc.gz"} |
Newton's 2nd Law
Newton's 2nd Law of Motion
Newton’s 2nd Law of Motion may be the most important principle in all of modern-day physics because it explains exactly how an object’s velocity is changed by a net force. In words, Newton’s 2nd Law
states that “the acceleration of an object is directly proportional to the net force applied, and inversely proportional to the object’s mass.”
In equation form, you can express Newton’s 2nd Law as:
It’s important to remember that both force and acceleration are vectors. Therefore, the direction of the acceleration, or the change in velocity, will be in the same direction as the net force. You
can also look at this equation from the opposite perspective. A net force applied to an object changes an object’s velocity (produces an acceleration), and is frequently written as:
You can analyze many situations involving both balanced and unbalanced forces on an object using the same basic steps.
1. Draw a free body diagram.
2. For any forces that don’t line up with the x- or y-axes, break those forces up into components that do lie on the x- or y-axis.
3. Write expressions for the net force in x- and y- directions. Set the net force equal to ma, since Newton’s 2nd Law tells us that F[net]=ma.
4. Solve the resulting equations.
Let’s take a look and see how these steps can be applied to a sample problem.
Question: A force of 25 newtons east and a force of 25 newtons west act concurrently on a 5-kilogram cart. Find the acceleration of the cart.
Step 1: Draw a free-body diagram (FBD).
Step 2: All forces line up with x-axis. Define east as positive.
Step 3: F[net] = 25N − 25N = ma
Step 4: 0 = ma ⇒ a=0
Answer: The acceleration must be 0 m/s^2.
Of course, everything you’ve already learned about kinematics still applies, and can be applied to dynamics problems as well.
Question: A 0.15-kilogram baseball moving at 20 m/s is stopped by a catcher in 0.010 seconds. Find the average force stopping the ball.
Answer: First write down what information is given and what you’re asked to find. Define the initial direction of the baseball as positive.
Use kinematics to find acceleration.
The negative acceleration indicates the acceleration is in the direction opposite that of the initial velocity of the baseball. Now that you know acceleration, you can solve for force using
Newton’s 2nd Law.
Question: Two forces, F[1] and F[2], are applied to a block on a frictionless, horizontal surface as shown below.
If the magnitude of the block’s acceleration is 2.0 meters per second^2, what is the mass of the block?
Answer: Define left as the positive direction.
Question: What is the weight of a 2.00-kilogram object on the surface of Earth?
Answer: Weight is the force of gravity on an object. From Newton’s 2nd Law, the force of gravity on an object (F[g]), is equal to the mass of the object times its acceleration, the acceleration
due to gravity (9.81 m/s^2), which you can abbreviate as g.
Question: A 25-newton horizontal force northward and a 35-newton horizontal force southward act concurrently on a 15-kilogram object on a frictionless surface. What is the magnitude of the
object’s acceleration? | {"url":"https://www.aplusphysics.com/courses/honors/dynamics/N2Law.html","timestamp":"2024-11-14T07:45:00Z","content_type":"application/xhtml+xml","content_length":"32298","record_id":"<urn:uuid:7c893bbb-1bc3-4bd7-9f38-6cc9a6e7cb20>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00136.warc.gz"} |
Podcast: Episode 7 – Neil Goldwasser – Dyslexia Support and Adult Numeracy
You're reading: Travels in a Mathematical World
Podcast: Episode 7 – Neil Goldwasser – Dyslexia Support and Adult Numeracy
These are the show notes for episode 7 of the Travels in a Mathematical World Podcast. 7 is prime, and the numbers on opposite sides of a regular six-sided die always add to 7. More about the number
7 from thesaurus.maths.org.
This week on the podcast we hear from Neil Goldwasser. Neil is a maths graduate who works as a dyslexia support and adult numeracy tutor in a FE college. He talks about his career and his work
teaching maths in a vocational context. You can find out about mathematics teaching from the TDA, and more information from Teachernet. Good resources for maths teachers are the NCETM and nrich. Some
more information on dyslexia support is available from the British Dyslexia Association. | {"url":"https://aperiodical.com/2008/11/podcast-episode-7-neil-goldwasser-dyslexia-support-and-adult-numeracy/","timestamp":"2024-11-09T00:33:25Z","content_type":"text/html","content_length":"35749","record_id":"<urn:uuid:2689e118-f83b-4114-9513-15260ec9c745>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00195.warc.gz"} |
Ordinal Numbers Word Whizzle - OrdinalNumbers.com
Ordinal Numbers Word Whizzle
Ordinal Numbers Word Whizzle – With ordinal numbers, you can count unlimited sets. They can also be used to broaden ordinal numbers.But before you can utilize them, you must comprehend the reasons
why they exist and how they operate.
The ordinal number is one of the mathematical concepts that are fundamental to mathematics. It is a number which signifies the location of an object within a list. Ordinarily, ordinal numbers range
from one to twenty. Even though ordinal numbers have numerous functions, they are most frequently used to indicate the order in which items are placed within the list.
It is possible to present ordinal number using numbers, words, and charts. They can also serve to demonstrate how a collection or pieces are placed.
Most ordinal numbers fall within one of the two categories. Transfinite ordinals are depicted using lowercase Greek letters, whereas finite ordinals can be represented by Arabic numbers.
In accordance with the axiom that every set properly ordered must contain at least one ordinal. For instance, the first person to finish an entire class will receive the highest score. The winner of
the contest was the student who had the highest grade.
Combinational ordinal numbers
Multiple-digit numbers are also known as compound ordinal numbers. They are made by multiplying an ordinal number by its final number. They are mostly for purposes of dating and for ranking. They
don’t have a distinct end to the last digit, as do cardinal numbers.
Ordinal numbers are used to identify the order of elements within collections. These numbers are used to identify the items in the collection. You can locate normal and suppletive numbers to ordinal
By prefixing cardinal numbers by the suffix -u, regular ordinals can be created. The number is then typed into words. A hyphen then added. There are many suffixes to choose from.
Suppletive ordinals can be created by prefixing words with the suffix -u or -e. The suffix is used to count words and is bigger than the standard.
Limits of ordinal significance
Limits for ordinal numbers are ordinal numbers that aren’t zero. Limit ordinal numbers have the disadvantage of having no limit on the number of elements they can have. These can be formed by joining
nonempty sets with no limit elements.
Limited ordinal numbers are employed in transfinite definitions for recursion. According to the von Neumann model, each infinite number of cardinal numbers is an ordinal limit.
A limit-ordered ordinal equals the sum of all other ordinals beneath. Limit ordinal numbers can be enumerated with arithmetic but could also be expressed as an order of natural numbers.
The data are arranged by ordinal numbers. They are used to explain the location of an object numerically. They are often utilized in set theory and the arithmetic. Although they have a similar
structure to natural numbers, they’re not part of the same class.
In the von Neumann model, a well-ordered set of data is employed. Let’s say that fy, one of the subfunctions in an g’ function that is defined as a singular function is the case. If g’ is in line
with the specifications, g’ is a limit order for fy when it is the sole function (ii).
A limit ordinal of the Church-Kleene type is also known as the Church-Kleene ordinal. A limit ordinal is a properly-ordered collection of smaller or less ordinals. It’s an ordinal that is nonzero.
Common numbers are often used in stories
Ordinal numbers can be used to show the hierarchy between entities or objects. They are essential for organizing, counting, as well as ranking reasons. They are used to explain the location of the
object as well as the order in which they are placed.
The ordinal number is generally indicated by the letter “th”. However, sometimes the letter “nd” is used, but it’s not the only way it is used. There are a lot of ordinal numbers on the title of
Even though ordinal numbers are most commonly employed in the form of lists but they can also be written in words. They can be also expressed in numbers and acronyms. In general, numbers are more
comprehensible than the cardinal numbers.
Ordinary numbers are available in three distinct flavours. It is possible to discover more about them through practicing, playing games as well as taking part in various other pursuits. Learning
about the subject is an important aspect of improving your math skills. Coloring is an enjoyable and straightforward method to increase your proficiency. A handy marking sheet can be used to keep
track of your results.
Gallery of Ordinal Numbers Word Whizzle
Ejercicio De Ordinal Numbers Review
Ordinal Number 1 10 Worksheet
Alphabet Ordinal Numbers Worksheet Teacher Made Ordinal Numbers
Leave a Comment | {"url":"https://www.ordinalnumbers.com/ordinal-numbers-word-whizzle/","timestamp":"2024-11-04T21:02:04Z","content_type":"text/html","content_length":"62542","record_id":"<urn:uuid:5b59d255-6536-4ad5-9cac-3ce14ebd6237>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00742.warc.gz"} |
Machine learning exploits all available information to make predictions – and there are many ways to predict things. A good example of this is assessing a person’s COVID risk. You can predict it from
features such as:
• wearing a mask
• symptoms like dry cough
• vitamin D levels
From a causal perspective, these features relate differently to the COVID risk. This difference is illustrated in the so-called causal graph below, which is a graphical tool for thinking about causal
• Causes: Wearing a mask is a causal factor for the COVID risk. Why is it a cause? Well, those who put on a mask can thereby lower their COVID risk [1]. The fact that masks are causal for COVID
risk is highlighted by the outgoing arrow from the mask node to the COVID node.
• Effects: Dry cough is an effect of COVID [2]. People who are healthy and then get infected with COVID likely get dry cough as a symptom. Importantly, it is not a cause of COVID – if people take
cough sirup this may help against their cough but does not change their COVID risk. The fact that dry coughing is an effect of COVID is highlighted by the incoming arrow to the cough node from
the COVID node.
• Associations: Vitamin D levels are associated with COVID risk [3]. For example, in the overall population, people with higher vitamin D levels less often get COVID than people with low vitamin D
levels. Still, it is unclear if higher vitamin D levels cause a lower COVID risk or if the association arises for entirely different reasons. The unexplained association between vitamin D levels
and COVID is highlighted by an undirected dashed arrow.^1
Initially, there was a lot of hype about machine learning in Raven Medicine. Some even proclaimed the end of costly medical experiments. But the enthusiasm soon waned: Machine learning provided
little insight into how to treat Ravens to make them healthy. Was the red pill or the blue pill more effective? The old proverb “correlation is not causation” hit the machine learning enthusiasts
10.1 Prediction does not require causal understanding
All the features we have just listed can be equally helpful in predicting COVID risk with machine learning. You may decide to rely only on causes or only on effects:
• Predict effects from causes: An example is the protein folding problem discussed at the beginning, where the amino acid sequence causally determines the protein structure [5].
• Predict causes from effects: An example is predicting the presence of a black hole from its gravitational effects on the surrounding bodies.
• Mixed prediction: This is probably the most common case. An example is medical diagnosis. To diagnose malaria, doctors take into account symptoms like high fever but at the same time causes such
as a mosquito bite from your South America travels.
In the end, it is your decision as a modeler what you want to incorporate in your prediction model, whether it is causes, effects, causes of effects, or even spurious associations. For the machine
learning model to successfully predict, all that matters is that the feature contains information about the target variable.
10.2 Costly experiments distinguish causes from effects
Controlled experiments are the best approach to distinguish between causes and effects. Let’s say you have two variables, vitamin D and COVID. To establish whether higher vitamin D levels reduce the
COVID risk, you could run a randomized control trial (RCT). You have 10,000 test subjects: 5,000 randomly selected subjects belong to the treatment group (they get vitamin D supplements); the other
5,000 belong to the control group (they get placebos). If after a certain time, the COVID infections in the treatment group are significantly lower than in the control group, you can conclude that
vitamin D is a causal factor for COVID risk. Easy, right?!
Unfortunately, conducting controlled experiments in the real world is cumbersome. It takes a lot of time, money, and resources. Some experiments can be ethically or legally problematic, such as the
administration of harmful drugs to humans. Other experiments are beyond the capabilities of humans – like investigating the effects of pushing Jupiter out of its solar trajectory – you’d need to be
as strong as Saitama ^2 for that…
Social scientists, in particular, often deal with observational data. Observational data is data that is measured without controlled conditions. Like asking random people on the streets for their
vitamin D levels and if they had COVID. Observational data is what we gather all the time – the little measuring device in your pocket is piling up tons of it.
10.3 Machine learning can generate causal insights
Observational data is what you are likely feeding your machine learning algorithm. Unfortunately, observational data alone does not provide causal insight [6].
It is impossible to distinguish causes from effects from observational data alone [6].
If you cannot even distinguish between causes and effects from observational data alone, how can machine learning help with causality? It turns out that machine learning can help with causal
inference (answering causal questions with the help of data), but usually not for free: You have to make assumptions about causal structures.
These are the causal questions that machine learning can help with:
• Studying associations to form causal hypotheses: Machine learning helps to investigate associations in data such as the association between COVID and vitamin D levels, which can be the starting
point for causal hypothesizing.
• Estimating causal effects: Machine learning can be used to estimate causal effects, such as quantifying the effect of vitamin D on the COVID risk.
• Learning causal models: You can use machine learning to learn causal models. Causal models are formal tools to reason about real-world interventions and counterfactual scenarios.
• Learning causal graphs: Causal graphs encode the direction of causal relationships, and machine learning can help learn them directly from data.
• Learning causal representations: Machine learning can learn causal variables, which are high-level representations (e.g. objects) of low-level inputs (e.g. pixels).
These five tasks structure the rest of this chapter.
10.4 Studying associations to form causal hypotheses
You may have heard that machine learning models are capturing complex associations in data. But what are associations? It is easiest to understand associations by their opposite, namely statistical
independence. Two events are independent if knowing one event is uninformative about the other, e.g. knowing about your COVID risk does not tell anything about the weather on the planet Venus. More
formally, two features \(A\) and \(B\) are statistically independent if \(\mathbb{P}(A\mid B)=\mathbb{P}(A)\). We call two features associated whenever they are not statistically independent. For
example, COVID risk is associated with wealth. Statistically, the higher your wealth the lower your COVID risk [7]. But be careful, associations are complex creatures. For example, wealthier venders
are likely to have a higher COVID risk than poor venders because they have contact with more customers. This is called an interaction effect.
Like classical statistical techniques, machine learning together with interpretability methods (see Chapter 9) enables you to read out complex properties of your data. You can for example study:
• Feature effects & interactions: How is the target associated with certain input features?
• Feature importance: How much do certain features contribute to the prediction of the target?
• Attention: What features is our model listening to when predicting the target?
10.4.1 Form causal hypotheses with the Reichenbach principle
Assume you find in your data the association between vitamin D and the COVID risk. Then, the Reichenbach principle can guide you in forming a causal hypothesis. The principle states three
possibilities where the association can come from [8]:
1. Vitamin D can be a cause of COVID.
2. COVID can be a cause of vitamin D.
3. Or, there is a common cause \(Z\) of both, vitamin D and COVID.
Nothing against Reichenbach, but in reality, there are two more explanations for the association:
• First, the association can be spurious. It is only there because of limited data. If you continue gathering data, the association will vanish. Figure 10.1 shows an almost perfect association
between worldwide non-commercial space launches and the number of doctorates awarded in sociology [9]. But is it causal? Perhaps Elon Musk can boost his business by awarding a few PhD
scholarships in sociology…
• Second, the association can result from a selection bias in the data collection process. It occurs if the data sample is not representative of the overall population. For example, while niceness
and handsomeness are probably statistically independent traits in the overall population, they can become dependent if you only look at people you would date [10].
Figure 10.1: There is an almost perfect association between the worldwide non-commercial space launches and the number of sociology doctorates awarded. Chart by Tyler Vigen, CC-BY (https://
But how to identify the correct explanation for the association between COVID and vitamin D? Because there is a lot of data (not spurious) and the association is found in the overall population (no
selection bias), you may infer that one of Reichenbach’s three cases applies. Gibbons et al. [11] compared a large group of veterans receiving different vitamin D supplementations to a group without
treatment. They found that the treatment with vitamin D reduces the risk of getting COVID significantly and therefore recommend broad supplementation. Even though the study was not a randomized
control trial, it strongly indicates that there is a causal link between vitamin D levels and COVID risk.
The scientific story may continue at that point: There might be unknown common causes of both vitamin D and COVID risk. Moreover, the causal link does not explain by what mechanism in the body
vitamin D acts on COVID risk. That’s how science works, new questions keep popping up…
10.4.2 Select explanatory variables with feature importance
Importances of features primarily allow you to evaluate which features are most informative for predicting the target. For example, they may tell you whether vitamin D levels contain any information
about your COVID risk beyond the information that is already contained in the mask and the cough features. But importance itself gives no causal insight. It can still be useful for a causal analysis:
importance allows you to understand what information is redundant, noisy, or just unnecessary. This may prove helpful in selecting sparse and robust features for a causal analysis [12].
Imagine you know a person’s age, country of origin, gender, favorite color, and favorite band. You want to predict again her COVID risk. By studying feature importance, you may find out that:
• The favorite color has very little importance under all constellations. It might even have negative importance as it introduces noise.
• The favorite band is extremely informative. You can even drop all other features without a major loss in predictive performance.
• If you know the age, country of origin, and gender of a person, knowing her favorite band does not add much.
Depending on your goal, this can be very insightful. You can generally drop the favorite color feature, it is not predictive and will not provide insights for a causal analysis. If you want to reason
about policy interventions to lower the COVID risk, you can exclude the favorite band from your causal analysis. If, on the other hand, you are a sociologist who wants to study how taste in music is
related to health, you may want to keep all features.
10.4.3 Find high-level variables with attention
The question of what a model is attending to in its decisions is prominent in interpretability research, particularly in image classification [13]. Although scientists may be more concerned with the
relationship in the data than in the model, model attention may still be relevant if the input features have meaning only in the aggregate, not individually. Think of pixels that constitute images,
letters that constitute sentences, or frequencies that constitute sounds. Attention can point you to meaningful constructs or subparts that are associated with the target and allow you to form causal
hypotheses about them.
For example, while predicting COVID from CT scans sounds like taking a sledgehammer to crack a nut, it can be an interesting approach to learning more about how the disease affects the human body.
Even though the input features themselves are pixels, they collectively form higher-level concepts that can be highlighted by saliency-based interpretability techniques. This approach was for example
used by [14] to detect lesions in the lungs of COVID patients as illustrated in Figure 10.2.
Figure 10.2: Lesions in CT scans of lungs detected with integrated gradients, small in green, mixed in yellow, large in red. Figure by [14], CC-BY (https://creativecommons.org/licenses/by/4.0/)
10.5 Estimating causal effects with machine learning
So far, we have assumed that researchers are searching in the dark, where they use machine learning to find interesting associations to form and test causal hypotheses. However, often researchers
have causal knowledge or intuitions about how features relate to each other. Their goal is causal inference, or, to be more precise, to estimate causal effects, which is a more well-defined problem
than just exploring associations.
Causal graphs show causal dependencies, e.g. if you put on a mask, it will causally affect your COVID risk. But this begs the question of how it affects your risk. Is wearing a mask lowering your
COVID risk and if yes, by how much? This question asks for a causal effect. Questions for causal effects were omnipresent during the pandemic:
• How does vaccination affect the COVID risk of elderly people?
• How does the average COVID risk change with contact restrictions compared to without them?
• How does treatment with vitamin D pills lower the COVID risk of African Americans [11]?
Causal effects always have the same form. You compare a variable of interest such as COVID risk with treatment, for example, masks, against those without treatment (no-mask). This gives us the
so-called Average Treatment Effect (ATE). In the case of masks being the treatment, we can formally describe the effect on the COVID risk:
\[\text{ATE}=\mathbb{E}[\text{COVID}\mid do(\text{mask}=1)] - \mathbb{E}[\text{COVID}\mid do(\text{mask}=0)]\]
Let’s unpack this formula. The do in the ATE denotes the do-operator, which means you intervene on a variable and set it to a certain value. Thus, the first term describes the expected COVID risk if
you force people to wear masks. The second term describes the expected COVID risk if you force people not to wear masks. The difference between the two terms describes the causal effect of masks on
the COVID risk.
In many cases, rather than asking for the ATE, you ask for the treatment effect for a specific subgroup of people that share certain characteristics. During the pandemic, for example, researchers
might have asked about the causal effect of masks for children under 10 years of age. This is the so-called Conditional Average Treatment Effect (CATE):
\[\begin{align*} \text{CATE} &= \mathbb{E}[\text{COVID} \mid do(\text{mask}=1), \text{age}<10] \\ &\quad - \mathbb{E}[\text{COVID} \mid do(\text{mask}=0), \text{age}<10] \end{align*}\]
The do() terms in the two formulas give the impression that you have to run controlled experiments to estimate the causal effects. And conducting experiments would be the most reliable way – if you
can do it, you should! Unfortunately, experiments are sometimes not feasible. In some situations, you are lucky because you can estimate the ATE and CATE from observational data alone. If this is the
case, the causal effect is identifiable.
10.5.1 Identify causal effects with the backdoor criterion
The classical way to identify a causal effect is via the so-called backdoor criterion. The idea is simple. Let’s say you want to know the ATE of masks on the COVID risk. You only have observational
data. But your data allows you to read out: 1. how many people with masks got infected and 2. how many people without masks got infected. Just subtract the latter from the former and you are
finished, right? No! The problem is that the information you want to obtain is potentially blurred by other factors. For example:
• Sticking to the rules: People who stick to the rules are more likely to wear masks and also follow other rules like contact restrictions. Therefore, they also have on average a lower COVID risk.
• Number of contacts: People who have many contacts (e.g. postmen, salespeople, or doctors) wear masks to lower their own risk and the COVID risk of others. But many contacts increase also their
COVID risk.
• Age: Younger people are less likely to wear masks because they think masks do not look cool or because COVID is less dangerous for young people. At the same time, young people are more resistant
to getting infected with COVID [15].
Those factors are called confounders. They are both a cause of the treatment and the variable of interest. Imagine you knew all those confounders and accounted for them. Then, you could isolate the
association that is solely due to the masks themselves. This is exactly what the backdoor criterion is doing. If you know all confounders \(W\), the ATE can be estimated by:
\[\begin{align*} \text{ATE} &= \mathbb{E}_W[\mathbb{E}_{\text{COVID}\mid W}[\text{COVID} \mid \text{mask}=1, W]] \\ &\quad - \mathbb{E}_W[\mathbb{E}_{\text{COVID} \mid W}[\text{COVID} \mid \text
{mask}=0, W]] \end{align*}\]
A great success, there are no do-terms left. But be aware that knowledge of all confounders is a super strong requirement that can rarely ever be met in reality. And not knowing a confounder may bias
your estimate significantly. The same happens if you mistake a variable for a confounder that is caused by both the treatment and the variable of interest (this is called a collider bias). So
conditioning on more features is not always better.
The back-door criterion is not the only option to identify causal effects. There is for example also its friendly sibling, the frontdoor criterion. Both, the front and the backdoor criterion are
special cases of the do-calculus. Whenever it is possible to identify a causal effect, you can do so using the do-calculus. This book will stay at this surface level on the question of
identifiability, if you like math and want to dig deeper, check out [16]. The cool thing is, you don’t have to know all of that necessarily to do causal analysis. The identification step can be fully
automated given a causal graph, for example, using the Python package DoWhy [17].
Let’s take a look into two machine learning-based methods that allow for estimating causal effects: the T-Learner and Double Machine Learning ^3. Both are designed for the backdoor criterion setting.
That means, in both cases, you assume that you have observed all confounders \(W\), and your causal graph looks like this
10.5.2 How to estimate causal effects with the T-Learner
The T-Learner is maybe the simplest approach to estimate the ATE (or with slight modifications CATE) from observational data. The name T-Learner stems from using two different learners to estimate
the ATE. The T-learner uses the formula from the backdoor criterion:
\[\begin{align*} \text{ATE} &= \mathbb{E}_W[\underbrace{\mathbb{E}_{\text{COVID}\mid W}[\text{COVID} \mid \text{mask}=1, W]}_{\phi_1}] \\ &\quad - \mathbb{E}_W[\underbrace{\mathbb{E}_{\text{COVID}\
mid W}[\text{COVID} \mid \text{mask}=0, W]}_{\phi_0}] \end{align*}\]
The T-Learner fits two machine learning models, \(\hat{\phi_0}: \mathcal{W}\rightarrow \mathcal{Y}\) and \(\hat{\phi_1}: \mathcal{W}\rightarrow \mathcal{Y}\), where \(\hat{\phi_0}\) is estimated only
with data where \(\text{mask}=0\) and \(\hat{\phi_1}\) with data where \(\text{mask}=1\). The ATE can then be computed by averaging over different values of \(W\) via
\[\text{ATE}\approx \frac{1}{n}\sum\limits_{i=1}^n \left(\hat{\phi_1}(w_i)-\hat{\phi}_0(w_i)\right).\]
10.5.3 How to estimate causal effects with Double Machine Learning
Double Machine Learning is an improvement over the T-Learner. Not only is the name much cooler, but the estimation of the ATE is much more sophisticated [22]. The estimate is both unbiased and
data-efficient. The estimation process works as follows:
1. Predict outcome & treatment from controls: Fit a machine learning model to predict COVID from the confounders \(W\). This gives you the prediction \(\widehat{\text{COVID}}\). Next, fit a machine
learning model to predict whether people wear masks or not from the confounders \(W\). This gives you the prediction \(\widehat{\text{mask}}\). You need to split your data into training and
estimation data, the machine learning models should be learned only from the training data.
2. Estimate outcome residuals from treatment residuals: Fit a linear regression model to predict \(\text{COVID}-\widehat{\text{COVID}}\) from \(\text{mask}-\widehat{\text{mask}}\). Then, your
estimand of interest is the coefficient \(\beta_1\). Why? Well, surprisingly, this coefficient describes the treatment effect. But you need quite some math to show this, search for the
Frisch-Waugh-Lovell theorem if you want to learn more. The linear regression must be fitted on the estimation data not the training data for models from step 1.
3. Cross-fitting: Run steps 1 and 2 again, but this time switch the training data you use to fit the machine learning models in step 1 with the estimation data you use in step 2 for the linear
regression. Average your two estimates \(\hat{\beta_1}^1\) and \(\hat{\beta_1}^2\) to obtain your final estimate \(\hat{\beta_1}=(\hat{\beta_1}^1+\hat{\beta_1}^2)/2\).
Why does this strange procedure make sense? If you want to get the details, check out [22]. The key ideas are:
• You want to get rid of the bias in your estimation that stems from regularization in your machine learning model. This is done in steps one and two, exploiting the so-called Neyman orthogonality
• You want to get rid of the bias in your estimation that stems from overfitting to the data you have and at the same time be data efficient. If you would train our machine learning model on the
same data on which you estimate the linear coefficient, you would introduce an overfitting bias. Splitting the data into training and estimation data circumvents this pitfall. However, splitting
the data just once is less data efficient. Thus, you switch the roles of training and estimation data and compute the average in step three.
If you want to use Double Machine Learning in your research, check out the package DoubleML in R [23] and Python [24]. Generally, there are many more methods now to estimate causal effects with
machine learning, check out [25], [20] or [21] to get an overview.
10.6 Learning causal models if we know the causal graph
Causal graphs are simple objects that contain:
• Nodes: Describe the features that are interesting to you, like the COVID risk, if you wear masks, if you have a dry cough, or your vitamin D levels.
• Arrows: describe how the features are causally linked. Like the causal arrow between COVID risk and dry cough.
Usually, you pose two additional assumptions on causal graphs – they should be directed and acyclic. Directed means that each arrow has a start node and end node, not unlike the dashed arrow in the
case of vitamin D in the beginning which treats both nodes the same. Acyclic means that it should not be possible to start from a node and return to it while only walking along the directed arrows.
If the graph is directed and acyclic, we talk about a DAG – a directed-acyclic graph.
So, causal graphs are a tool to reason about causal relations. But wouldn’t it be nice to have causal models? Models that once they are constructed allow you to think about a whole range of questions
without always running this whole treatment effect estimation process?
What is the effect of masks on the COVID risk? How about both masks and contact restrictions? What if you only look at older people? Imagine all these questions can be answered with one model.
10.6.1 Bayesian causal networks allow to reason about interventions
In causal graphs, we talked about the nodes of the graph, like the mask node, the COVID node, or the cough node. But how to interpret these nodes specifically? The perspective of Causal Bayesian
Networks (CBNs) is to see these nodes as random variables. If causal graphs are a combination of graphs and causality, CBNs add probability theory to the mix.
CBNs allow you to ask classical probabilistic questions, like what is the probability that people wear masks if they have a cough \(\mathbb{P}(\text{mask}\mid \text{cough}=1)\). But what CBNs are
ultimately designed for is answering causal questions about interventions. Like what is the average COVID risk if everyone must wear a mask \(\mathbb{P}(\text{COVID}\mid do(\text{mask}=1))\)?
Let’s check out this simplified causal graph with only three variables: mask, COVID risk, and cough. Mask is causal for COVID risk and COVID is causal for cough. In this simple setting, you must
specify three things to obtain a CBN:
1. The marginal probability of wearing a mask, i.e. \(\mathbb{P}(\text{mask})\).
2. The COVID risk for both people who either wear or don’t wear masks, that means \(\mathbb{P}(\text{COVID}\mid \text{mask}=0)\) and \(\mathbb{P}(\text{COVID}\mid \text{mask}=1)\).
3. The probability of having a cough if the COVID risk is low or high, that means \(\mathbb{P}(\text{cough}\mid \text{COVID}=1)\) and \(\mathbb{P}(\text{cough}\mid \text{COVID}=0)\).
Let’s look at a toy example. Assume that 50% of people wear masks. Assume the COVID risk is 80% low / 20% high if people wear a mask and reversed if people do not wear a mask. And, the probability
that people have a cough if they have a high COVID risk is 90%, and conversely, 10% if they have a low COVID risk.
Then, you can ask the following question and address it formally:
• Observational question: What is the probability of wearing a mask, having a high COVID risk, and not coughing? This can be computed by:
\[\begin{align*} &\mathbb{P}(\text{mask}=1, \text{COVID}=1, \text{cough}=0) \\ &=\mathbb{P}(\text{mask}=1) \times \mathbb{P}(\text{COVID}=1 \mid \text{mask}=1) \\ & \quad \times \mathbb{P}(\text
{cough}=0 \mid \text{COVID}=1) \\ &= 0.5 \times 0.2 \times 0.1 = 0.01 \end{align*}\]
So this is a pretty unlikely combination. You can also ask a causal question, like:
• Interventional question: What is the probability of having a high COVID risk and cough if the policymaker enforces masks? This can be computed by:
\[\begin{align*} &\mathbb{P}(\text{COVID}=1, \text{cough}=1 \mid do(\text{mask}=1)) \\ &= \mathbb{P}(\text{COVID}=1 \mid \text{mask}=1) \times \mathbb{P}(\text{cough}=1 \mid \text{COVID}=1) \\ &= 0.2
\times 0.9 = 0.18 \end{align*}\]
Let’s generalize the ideas of this example. What do you need for a BCN?
• The marginal distribution \(\mathbb{P}(X_r)\) of all nodes in the causal graph without incoming arrows. These nodes are called root nodes.
• For all non-root nodes, you need their conditional distribution \(\mathbb{P}(X_j\mid pa_j)\) given their direct causes. The direct causes of a node are called parents.
If you have those two ingredients specified, you have a BCN. It allows you to compute both observational and interventional probabilities:
Observational probabilities:
\[\mathbb{P}(X)=\prod\limits_{i=1}^p \mathbb{P}(X_i | pa_i)\]
Interventional probabilities:
\[\mathbb{P}(X_{-j}\mid do(X_j=z))=\prod\limits_{\substack{i\neq j, \\j\not\in pa_i}}\mathbb{P}(X_i\mid pa_i) \prod\limits_{j\in pa_i}\mathbb{P}(X_i\mid pa_i,X_j=z)\]
To calculate interventional probabilities, you set the intervened variable to the desired value wherever it appears in the formula and the corresponding probability that the variable has this value
to \(1\).
All of this is nice theory and toy modeling. But how about the real world where you do not magically have the BCN? Indeed, in practice, you can learn the BCN if you know the causal graph. All that
differs between a BCN and a causal graph are marginal and conditional probabilities. If you have data, this is exactly where machine learning can help – estimating conditional probabilities. We
generally advise estimating conditional probabilities with non-parametric machine learning models like neural networks or random forests, however, for small data sizes classical parametric approaches
like maximum likelihood estimation or expectation maximization might be preferable [26].
10.6.2 Structural causal models allow you to reason about counterfactuals
Interventional questions are often important if we want to act. But what if you want to explain why something has happened? Let’s say you got COVID and want to know why. Was it not wearing a mask?
Or, was it that party last week? Or did you catch it in the metro from this guy who was sneezing heavily?
Answering why questions is always the most difficult, but usually also the most interesting. What is needed to answer such questions? You have to think through counterfactual scenarios. You know that
you went to the party and you also know that you got COVID. To answer the why question, you need to find out how you got infected. Would you have got COVID if you hadn’t gone to the party? Would you
have caught COVID if you hadn’t spoken to your charming but coughing neighbor at the party?
BCNs are not expressive enough to reason about counterfactual scenarios. Why not? Firstly, BCNs describe causal relations probabilistically on the level of groups of individuals who share certain
features. Like the probability of getting COVID if everyone is forced to the party. The causal relationships are learned from data of people who did and of people who did not attend the party. The
why question is more specific. It asks why you individually got COVID. Did you get COVID because you attended the party?^4 Secondly, the fact that you got COVID is informative about you, it might
reveal information about your properties on which there is no data. Like info about your immune system and how resilient it is against COVID. This information should not be ignored.
To answer why questions you need to simulate alternative scenarios, i.e. to perform counterfactual reasoning. A Structural Causal Model (SCM) is a model designed to simulate such alternative
scenarios. Instead of describing probabilistic relationships in the data, a SCM explicitly models the mechanism that generates the data.
Let’s look at a super simple SCM with just two factors: whether you go to the party or not and whether you have COVID or not. These factors that you have explicit information about are called the
endogenous variables.
The model moreover contains two exogenous variables. Exogenous variables describe background factors on which you have no data but which play a role in determining the endogenous variables. For
example, whether you go to the party or not might be determined by whether you are in the mood for partying. Similarly, whether you get COVID or not might be influenced by how well your immune system
is currently working. The exogenous variables are very powerful in SCMs. These factors completely determine the endogenous variables.
The SCM is thus specified by:
• Two noise terms: \(U_{\text{mood}}=Ber(0.5)\) describes whether you are in the mood for partying and the chances are fifty-fifty. \(U_{\text{immune}}=Ber(0.8)\) describes whether your immune
system is up and the chances are pretty high (80%) that your immune system works well.
• One structural equation for the endogenous party variable, i.e. \(\text{party}:= U_{\text{mood}}\). Whether you go to the party is fully determined by whether you are in the mood for it.
• One structural equation for the endogenous COVID variable, i.e. \(\text{COVID}:=max(0,\text{party}-U_{\text{immune}})\). That means, you only get COVID if you are at the party and your immune
system is down.
This SCM allows you to answer the counterfactual question: You went to the party and you caught COVID, would you have gotten COVID if you hadn’t gone to the party? Such counterfactuals are computed
in three steps:
• Abduction: What does the fact that you got COVID tell you about the noise variables? Well, you could not have caught COVID with your immune system up. Thus, you can infer that \(U_{\text{immune}}
=0\). Similarly, you know that you were in the mood for partying because otherwise, you would not have gone, i.e. \(U_{\text{mood}}=1\).
• Action: Let’s say you would not have gone to the party, that means you intervene and turn the party variable from one to zero, i.e. \(do(\text{party}=0)\).
• Prediction: What happens with the COVID variable if you switch party to zero? According to the structural equation of COVID, you only get COVID if both happens, you go to the party and your
immune system is down. Thus, you do not get COVID.
In consequence, if you hadn’t gone to the party, you would not have caught COVID. So, you got COVID because you went to the party.
What if the counterfactual had not been true? Would partying still be a cause? Well, then things get more complicated. It could for instance be that you would have gone for sports instead of the
party and caught COVID there. But partying would still have been the actual cause of you catching COVID. Search for actual causation to learn more about the topic.
Let’s generalize the ideas of the example. For a SCM, you need:
• A set of exogenous variables \(U\) that are determined by factors outside the model and independent from each other.
• A set of endogenous variables \(X\) that are fully determined by the factors inside the model.
• A set of structural equations \(F\) that describe for each endogenous variable \(X_j\) how it is determined by its parents and its exogenous variable \(X_j=f(pa_j, U_j)\).
Obtaining SCMs in real life is difficult. You may use machine learning to learn the structural equations if you know the causal dependencies between variables. However, even for that, you need quite
some data, and making parametric assumptions can be advisable.
Like counterfactuals themselves, SCMs are highly speculative objects. There is no easy way to verify that a given SCM is correct, it always relies on counterfactual assumptions. Still, since humans
reason in terms of counterfactuals all the time and are deeply concerned with why questions, it is nice to have a formalism that can capture counterfactual reasoning.
10.7 Learning causal graphs from data
Machine learning has revived an old idea – with enough data, we might be able to completely automatize science. In Part 1 and Part 2 of this chapter, you always needed a human scientist in the
background. Someone to come up with interesting hypotheses, run experiments, or provide us with a causal graph. This part will be the most ambitious, it asks:
Can you get a causal graph just from data, without entering domain knowledge? This problem is often referred to as causal structure learning or causal discovery.
Didn’t we tell you in the beginning that observational data alone doesn’t do the trick? Correct! But it is intriguing to see how far you can go, especially if you add some additional assumptions.
Clear the stage for a truly fascinating branch of research!
Imagine all you have is a dataset containing 5,000 entries. Each entry contains information about a person’s BMI, COVID vaccination status, flu vaccination status, COVID risk, fatigue, fever,
appetite, and the density of population in the area the person lives in (Example based on [28] and [29]). The task is to structure these features in a causal graph – the problem of causal discovery.
Let’s say the true graph that you want to learn looks like this:
Before you can approach this problem, you have to get some more background on how causal mechanisms, associations, and data relate to each other. The causal mechanism generates data. Within this
data, features are associated with each other. That means that causal dependencies induce statistical (in-)dependencies. Particularly, if you have a causal graph, the so-called causal Markov
condition applies:
Given its parents in the causal graph, a variable is statistically independent of all its non-descendants. Or formally, let \(G\) be a causal graph and \(X_i\) be any variable in the network, then
for all its non-descending variables \(nd_i\) holds \(X_i\perp \!\!\! \perp nd_i\mid pa_i\). Descendants of a given variable \(X_i\) are all the variables that can be reached from \(X_i\) by walking
along the direction of arrows.
To get an intuition on the causal Markov condition, take a look at the graph below. Consider the variable appetite as an example. Its parents are COVID and density. Its nondescendants are BMI,
COVID-vac, flu-vac, fever, and fatigue.^5 The causal Markov condition allows you for example to say that appetite is statistically independent of fever if you know COVID and density. The formal way
to write this is \(\;\text{appetite} \perp \!\!\! \perp \text{fever}\mid \text{COVID}, \text{density}\), where \(\perp \!\!\! \perp\) is the symbol for independence.
Cool, causal graphs induce statistical independencies, but what to do with that? Well, you have a dataset, so you can test for statistical independencies.
There are many different approaches to test such (conditional) independencies, and they vary regarding the parametric assumptions they pose. Some test the statistical dependency to be Gaussian or
linear (e.g. partial correlation), or are tailored for categorical data like the chi-squared test [30]. Others allow for a greater variety of dependencies, like tests based on the Hilbert-Schmidt
independence criterion (HSIC) [31]. The current trend goes even more non-parametric towards machine learning based tests for conditional independence. Questions of independence are translated into
classification problems, where powerful machine learning models can be utilized (e.g. neural nets or random forests) [32], [33]. However, the fewer parametric assumptions you pose, the more
statistical dependencies are possible. Thus, non-parametric tests usually have a very bad statistical power, which means you need a lot of data to identify independencies [34]. This problem gets
worse the more variables you condition on – conditioning can be viewed as reducing the data you can test with. Incorrect conditional dependencies can lead to incorrect causal conclusions.
Let’s say you found a set of statistical independencies, how does this help you to learn the causal graph? Well, statistical independencies in the data are only compatible with certain kinds of
causal structures. The data narrows down the causal stories that make sense. So the question is, how many faithful causal graphs (see the box: “Central assumptions in causality”) are compatible with
the data?
As an example, assume you had only three variables: fatigue, COVID, and fever. By running statistical tests, you find that
• fatigue is independent of fever given COVID (Formally, \(\text{fatigue}\perp \!\!\! \perp \text{fever}\mid \text{COVID}\))
• None of the three variables is unconditionally independent of the other.
Then, three causal stories (causal chains) would explain this data.
• Story 1: \(\text{fatigue}\rightarrow \text{COVID}\rightarrow \text{fever}\).
• Story 2: \(\text{fatigue}\leftarrow \text{COVID}\leftarrow \text{fever}\).
• Story 3: \(\text{fatigue}\leftarrow \text{COVID}\rightarrow \text{fever}\) (called fork structure)
All of these stories (or causal graphs) are faithful and compatible with the given statistical (in-)dependencies. Also, one can show that there is no other causal graph that satisfies this.
The elements of causality: chains, forks, and immoralities
Let’s assume you have the variables \(X, Y\), and \(Z\). Three basic path structures in causal graphs are distinguished in causal discovery:
• Causal chains, where \(Y\) is called a mediator:
• Forks, where \(Y\) is called a common cause:
• Immoralities, where \(Y\) is called a collider:
Causal chains and forks imply identical conditional dependencies:
1. \(X \perp \!\!\! \perp Z \mid Y\) (X and Z are independent given Y)
2. X and Y are dependent
3. X and Z are dependent
4. Y and Z are dependent
Immoralities, however, exhibit a distinct pattern:
1. \(X \perp \!\!\! \perp Z\) (X and Z are marginally independent)
2. X and Z are dependent conditional on Y
3. X and Y are dependent
4. Y and Z are dependent
We call the set of faithful causal models that satisfies the same set of statistical (in-)dependencies the Markov equivalence class. How can you find the Markov equivalence class if all you have is a
dataset? We provide one of many answers here, the so-called PC algorithm.^6
The Peter-Clark (PC) algorithm: The most well-known algorithmic approach for finding the Markov equivalence class of causal models that is compatible with the statistical (in-)dependencies found in a
dataset [36]. It requires causal sufficiency and the faithfulness condition to be satisfied. It runs the following steps:
1. Identify the skeleton: We start with a fully connected network with undirected arrows. Then, step by step we look through all unconditional and conditional independencies between features. We
start with unconditional independencies, continue with independencies conditional on one variable, then two, and so on. If two features are (conditionally) independent, we erase the arrows
between them. In the end, we get the so-called skeleton graph without directed arrows.
□ Unconditional independence: By running a statistical test, we learn that BMI, COVID-vac, flu-vac, and density are all pairwise independent. This allows us to erase the arrows between those
□ Independence conditioned on one variable: Conditioning on COVID makes fatigue, fever, and appetite pairwise independent. Moreover, BMI, COVID-vac, flu-vac, and density become independent of
fatigue and fever. We can erase all arrows between these variables.
□ Independence conditioned on two variables: If we condition on COVID and density, BMI, COVID-vac, and flu-vac become independent of appetite. We can therefore erase all arrows between appetite
and these variables.
□ We find no further independencies. So this is our resulting skeleton!
2. Identify immoralities: Immoralities are paths of the form \(X\rightarrow Y\leftarrow Z\) that lead to unique (in)dependencies, namely to \(X\perp \!\!\! \perp Z\) and \(X,Z\) are dependent
conditional on \(Y\). We can use this uniqueness to orient some arrows. Go through all unconditional independencies \(X\perp \!\!\! \perp Z\) where the nodes are connected via one intermediate
variable \(Y\), check whether conditioning on \(Y\) makes them dependent. If yes, you found an immorality and can orient the arrows.
□ As you can see, in our skeleton all arrows are undirected. We now check for all unconditionally independent variables that become dependent if we condition on one variable. This is the case
for BMI, COVID-vac, flu-vac, and density. They are unconditionally independent but become dependent if we condition the COVID variable. We can therefore orient some arrows in our graph.
3. Apply logic: Now you have identified all immoralities. Thus, if you find three variables that are connected like this \(X\rightarrow Y-Z\) where the arrow between \(Y\) and \(Z\) is undirected,
you can infer that \(X \rightarrow Y \rightarrow Z\), because otherwise there would be another immorality that you must have discovered. Also, we search for an acyclic graph. So if there is only
one way to avoid making a cycle, that is the way you orient the arrows.
□ Which arrows would lead to new immoralities? If fatigue, fever, and appetite had arrows towards COVID, new immoralities would emerge that would have been found in the statistical testing.
Thus, they must have arrows coming from the COVID variable. Now only one arrow is left undirected, the arrow between density and appetite. If this arrow would go from density towards
appetite, there would be a cycle: \(\text{COVID}\rightarrow \text{appetite}\rightarrow \text{density}\rightarrow \text{COVID}\). Thus, the arrow must go from density to appetite.
Cool, we were able to identify the full causal graph just from the (in-)dependencies! But we were lucky. If we had looked at a more complex graph, the PC algorithm would have spit out a graph where
some arrows are left unoriented. All possible ways for orienting these remaining arrows define the Markov equivalence class. But can you go further? Is there a way to identify the correct causal
model among the Markov equivalence class? You need something extra for this:
• Perform real-world experiments: You could simply run experiments. There are proven bounds for how many experiments you have to perform to identify the correct causal graph. For instance, if
multi-node interventions are allowed, \(log_2[n]+1\) interventions are sufficient to identify the causal graph, where \(n\) is the number of nodes [37].
• Pose more assumptions: One of the most prominent assumptions for the identifiability of the causal graph is the linearity of the relationship and non-Gaussian noise terms in the structural
equations. Alternatively, you can assume non-linearity and the additivity of noise. Check out [30] to learn more. Most approaches rely on the idea that the noise variables are independent in one
but not in the other causal direction.
Central assumptions in causality
You can see causal models as devices for interpreting data. But when are the interpretations correct? Only if certain assumptions are satisfied. Causality is therefore a deeply assumption-driven
field. The most crucial ones are listed here:
• Causal sufficiency: This is one of the key assumptions in the field. It is sometimes also referenced as the assumption of no unobserved confounders. It means that you have not missed a causally
relevant variable in your model that would be a cause of two or more variables in the model. This assumption is not testable and without it, far less can be derived [38].
• Independence of mechanisms: Means that causal models are modular. Intervening on one structural equation does not affect other structural equations. Or on a probabilistic take, changing the
marginal distribution of one variable does not affect the conditional distributions where this variable acts as a causal parent. This assumption seems almost definitional to mechanistic modeling
and implies the independence of the exogenous noise terms [30].
• Faithfulness: Is connected to how to interpret causal graphs probabilistically. Faithfulness means that every causal dependence in the graph results in a statistical dependence. Note that it is
hard to justify faithfulness in finite data regimes [39].
• Positivity: For every group of the population, you have both subgroups with and without treatment. This is crucial when it comes to treatment effect estimation. If positivity is violated, you
extrapolate (non-)treatment effects for certain subgroups, which can go terribly wrong.
This list is far from complete, there are many more, such as consistency, exchangeability, and no interference [16], [40]. The correctness of your interpretations of data with causal models or the
causal models you may partially derive from data are extremely sensitive to these assumptions. If possible, check if they apply!
10.8 Learning causal representations from data
Causal discovery only makes sense if the features are meaningful representations and you can talk meaningfully about their causal relationships. This is the case with highly structured, tabular data
that has been heavily pre-processed by the human mind. But the data you often have, especially the data you want to analyze with machine learning, doesn’t look like that, it consists of:
• images made of pixels
• texts made of letters
• sounds made of frequencies
For example, it makes no sense to construct a causal model where the variables are single pixels. However, it may be meaningful to talk about the causal relationships between objects in images. How
can you go from pixels to these objects? How can you get meaningful higher-order representations from lower-order features?^7
Machine learning is a field concerned with learning higher-order representations. Causal representation learning therefore describes a dream that many in machine learning share – the dream of fusing
symbolic approaches like causal models with modern machine learning. This would combine the strengths of both worlds: learning complex meaningful representations from data AND reasoning symbolically
in a transparent and logical manner.
Indeed, doing so is extremely difficult and the research in this field is still in its infancy. Machine learning models learn complex representations to perform their predictions, however, whether
the representations are in any sense similar to the ones humans form or at all understandable to us is an open question. Adding labels to the representations you want and modifying the loss function
to use these representations might be one way to go [42]. But it is data costly! Is there a way to make sure that machine learning models learn meaningful representations that could be used to
construct causal models? Can you maybe even build in constraints informed by your understanding of causality to learn such representations? We’ll give you some hunch in this direction…
Variational autoencoders (VAEs)
Autoencoders consist of two parts. An encoder maps the input to a higher-order representation and a decoder maps the higher-order representation back to the original input space. The only difference
in a variational autoencoder is that input is encoded into a set of parameters of a probability distribution, whereas the decoder takes a sample from this distribution and maps it back to the
original input space. VAEs are (unlike the rest of the book) unsupervised learning techniques. They allow, among other things, to compress high-dimensional information into low-dimensional features
with little or no loss of information.
Structure of a Variational Autoencoder.
Putting causal constraints in variational autoencoders: You can see the probability distributions in the middle of VAEs as abstractions of the low-level features your data lives in. Therefore, VAEs
are constantly discussed in the context of representation learning in general and there are many ways to incorporate knowledge about these representations both via the architecture and the loss
function [43]. Instead of learning representations and then looking at how you can incorporate them in causal models, you can do the reverse and ask: What makes representations good candidates for
causal models and how can such knowledge be incorporated into the VAE framework? Since research on the topic is still in its infancy, we will only provide a list of a few ideas inspired by [44] :
• Sparsity: One key idea in causality is that a few powerful representations are enough to explain all kinds of complex phenomena. This can for example be enforced by the size of the middle VAE
• Generality across domains: Powerful representations are useful across tasks. What makes them useful is that they capture robust patterns in nature. You can enforce this for instance in VAEs by
training them to use the same representations for different decoding tasks.
• Independence: First, if two representations contain the same information, one of them is redundant. But you want to have simple models to efficiently communicate about the world. Second, the core
idea in causality is that you have independent sources of variation and that you can disentangle these sources and their interactions. One approach for achieving this in VAEs is modifying the
loss to make sure that the representations in the intermediate layer are statistically independent.
• Human interventions: What makes a variable suitable for causal modeling is that you can intervene upon it. Thus, 42 % of randomly selected atoms from a table do not constitute a good
representation, as you cannot use or intervene in these 42% in isolation. This human-centered causal bias can be entered into VAEs for example through video data, where objects are constantly
moved and intervened upon but stay persistent.
• Simple relationships: Causal models are rarely densely connected. Instead, a handful of causal relationships between higher-order variables give rise to the associations you observe in the world.
This inductive bias of simplicity can be enforced via the decoder architecture or the loss function.
10.9 Causality helps to formulate problems
After finishing this chapter, you may feel a little overwhelmed. A lot of theoretical concepts were presented and at the same time, there was little practical advice, e.g. compared to the
domain-knowledge chapter (see Chapter 8). We have the impression that the link between causal theory and the practical problems of scientists is underdeveloped in current research. For example, there
are only a few examples where causal discovery algorithms have been used to gain insights from practice.
Nevertheless, we believe that every researcher should be familiar with the concepts presented above, such as the Reichenbach principle, treatment effect estimation, and causal modeling. Causality is
what many of you are ultimately looking for when you want to control, explain, and reason about a phenomenon. While causality as a field does often not provide ready-made practical solutions, it
offers you a language to formulate your problem and describe possible solutions.
Many of the chapters in this book are closely related to causality:
• Some domain knowledge (see Chapter 8) you want to encode may be causal.
• Many questions addressed via model interpretation (see Chapter 9), such as algorithmic recourse, are ultimately causal questions [29].
• Causality is one leading approach to improving robustness (see Chapter 11) [45].
• Approaches for better uncertainty quantification, such as conformal prediction (see Chapter 12) are currently being integrated into causal inference [46].
1. There is also an association between COVID risk and the FIFA football ranking [4]. Here, the causal story is spuriously Messi…↩︎
2. Saitama is the protagonist of the anime series “One-Punch Man” who defeats his opponents with just one punch. In one of the episodes, he accidentally damages good old Jupiter.↩︎
3. There are two general approaches for estimating treatment effects with machine learning, model-agnostic and model-specific techniques. Model-specific techniques are based on a certain class of
machine learning models, such as GANITE for neural nets [18] or causal forests for random forests [19]. Model-specific techniques often come with the advantage of valid confidence intervals.
Other techniques are model agnostic, which means, you can simply plug in any machine learning model into the estimation. The T-Learner and Double Machine Learning we present here are examples of
such model-agnostic approaches. Today, there are tons of approaches to estimating causal effects with machine learning, check out [20] or [21] to learn more.↩︎
4. In the literature, it is differentiated between counterfactuals and actual causes [27].↩︎
5. The parents are indeed also non-descendants. However, the conditional independence you can derive from the causal Markov condition between \(E\) of \(C\) given you know \(C\) holds trivially for
an arbitrary variable by \(\mathbb{P}(E,C\mid C)=\mathbb{P}(E,C,C)/\mathbb{P}(C)=\mathbb{P}(E\mid C)=\mathbb{P}(E\mid C)\mathbb{P}(C\mid C).\)↩︎
6. The PC algorithm is already quite old but great for understanding the general idea of causal discovery. There are many extensions, generalizations, and alternatives of PCs on the market. We point
you to [30] for an overview and to [35] for an R package implementing standard causal discovery algorithms.↩︎
7. Causal abstraction is a cool framework for thinking about the relationship between low-level and high-level causal representations [41]. Science is all about causal models at different levels of
description, think of the relationship between physical models, chemical models, and biological models.↩︎ | {"url":"https://ml-science-book.com/causality.html","timestamp":"2024-11-11T21:30:19Z","content_type":"application/xhtml+xml","content_length":"147712","record_id":"<urn:uuid:6586a231-0e2a-4b48-ab4b-5c006721c6be>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00818.warc.gz"} |
Epoxy Calculator
Last updated:
Epoxy Calculator
This epoxy calculator is a valuable tool for DIY enthusiasts. Here, you'll learn what epoxy is and how to calculate the volume of epoxy resin required to coat a surface area. Using this tool, you can
easily calculate the epoxy resin needed for tables, floors, and even art pieces or photographs (use picture frames if you're unsure if your photo is safe to be in contact with epoxy resin).
🙋 If you, instead, prefer to tile your flooring, you'll find our tile calculator and our grout calculator equally valuable.
What is epoxy resin?
Epoxy resin is a liquid adhesive that becomes solid when mixed in a 2:1 ratio with the hardener. It is used to coat, laminate, and infuse different materials, providing strength, durability, and
waterproofing — though it typically requires a few days to set.
It bonds well with almost all materials, whether metal, plastic, glass, ceramic, wood, or rubber. But make sure to clean the surface properly, because epoxy doesn't like greasy surfaces.
Epoxy resin is such a durable material and may be expensive to apply. If you have a tight budget and cannot wait to renovate, you could always opt to use paint.
Check out our paint calculator to help you decide how much paint you'll need for a particular area. Our deck stain calculator can also help you give your deck a new look, so you might want to check
that out.
How to use the epoxy calculator?
Using the epoxy calculator, let's calculate how much epoxy you need to coat your surface. Our calculator has the following fields:
• How thick do you want the epoxy coating?
□ Coating: Enter the amount of epoxy coating you wish to apply, e.g., 1 in.
• What is the shape and size of your surface?
□ Surface shape: Select if your surface is rectangular or circular.
□ Surface length: Enter the length of your surface, e.g., 36 in.
□ Surface width: Enter the width of your surface, e.g., 24 in.
□ Diameter: If you select circular surface shape, you'll need the diameter of your surface.
• Here's how much epoxy resin you'll need:
□ Epoxy: Here, you will see the required volume of epoxy resin you need, based on your surface shape and dimensions, i.e., 478.75 ounces.
Note that the exact volume may differ if there are any indents or extrusions on the surface.
How to calculate the epoxy resin amount manually?
Here's how you can manually calculate how much epoxy you need.
We use the following formula to calculate the volume of epoxy resin required for a rectangular surface.
$\footnotesize \text{Volume} = \text{length} \times \text{width} \times \text{height}$
The following formula is to calculate the volume of epoxy resin required for a circular surface.
$\footnotesize \text{Volume} = \pi \times \text{radius}^2 \times \text{height}$
• $\text{Volume}$ – Required amount of epoxy;
• $\text{length}$ – Length of the rectangular surface where we want to apply the epoxy;
• $\text{width}$ – Width of the same rectangular surface where we want the epoxy;
• $\text{radius}$ – Half the diameter of the circular surface where we want the epoxy; and
• $\text{height}$ – Desired thickness of our epoxy coating.
Let's take the example of coating 1 in epoxy to a table with dimensions of 8 ft by 4 ft (or 96 in by 48 in).
• Converting feet into inches and placing the values in the formula:
\footnotesize \begin{align*} \text{Volume}\ &= \text{length} \times \text{width} \times \text{height}\\ &= 96\ \text{in} \times 48\ \text{in} \times 1\ \text{in}\\ &= 4,\!608\ \text{in}^3 \end
• To convert cubic inches into gallons, we divide it by 231.
\footnotesize \begin{align*} \text{epoxy volume} &= \text{Volume} / 231\\[0.2em] &= 4,\!608\ \text{in}^3 / 231\\ &= 19.95\ \text{gallons} \end{align*}
Now let's take another example of calculating epoxy resin for a round table with a diameter of 32 in (or a radius of 16 inches). Here we will coat 2 inches of epoxy resin.
• Placing the values in the formula:
Volume = π × (32 / 2)² × 2
= 1608.5 in³
\footnotesize \begin{align*} \text{Volume} &= \pi \times \text{radius}^2 \times \text{height}\\[0.2em] &= \pi \times (16\ \text{inches})^2 \times 2\ \text{inches}\\[0.2em] &= 1,\!608.495\ \text{in}^3
\\[0.2em] &\approx 1,\!608.5\ \text{in}^3 \end{align*}
• To convert cubic inches into ounces, divide it by 1.80469.
\footnotesize \begin{align*} \text{epoxy volume} &= \text{Volume} / 1.80469\\[0.2em] &= 1,\!608.5\ \text{in}^3 / 1.80469\\[0.2em] &= 891.29\ \text{fluid oz (US)} \end{align*}
💡 For an even easier measurement, we can convert ounces into cups by dividing the result by 8.
How to calculate 2 part epoxy resin ratio?
The epoxy resin always mixes with the hardener in a 2:1 ratio, where the epoxy is twice as much as the hardener. Divide the resulting volume by the area of your applicable surface to get the coating
Coating thickness = volume / area
What is the epoxy resin calculation formula?
For epoxy resin amount calculation, we use the following formula:
Epoxy amount = area × coating
• area – Surface where the epoxy resin is applied; and
• coating – Thickness of the epoxy coating.
How do I calculate how much epoxy resin I need?
To calculate the amount of epoxy resin required:
1. Measure the area of the surface.
2. Then multiply it by the thickness of the epoxy coating.
For example, on a surface of 36 inches, if you want 3 inches of epoxy coating, you'll need 108 cubic inches or 59.8 ounces of epoxy.
How do I calculate amount of epoxy for a table of 24"x24"?
To calculate the resin amount, multiply the surface area of your table by the required amount of resin coating. For 1" of resin thickness, you need 576 cubic inches or 319.7 ounces of epoxy resin.
How long does epoxy flooring last?
Residential epoxy flooring can last at least ten years if cared for properly. It depends on the following factors:
• Floor material strength;
• Prior surface treatment;
• Epoxy coating quality & thickness; and
• Pressure on the surface over the years. | {"url":"https://www.omnicalculator.com/construction/epoxy","timestamp":"2024-11-03T23:28:20Z","content_type":"text/html","content_length":"542432","record_id":"<urn:uuid:099d53e0-94d7-4bad-a7af-ffa16e7460a0>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00041.warc.gz"} |
RESISTOR Introduction
A resistor is a component that resists the flow of electricity. With the same terminal voltage across resistor, the larger resistance, the lower current flow.
There are DIP type resistor and SMD type resistor (figure from internet). The larger body size of the same resistance, the higher power rating. The popular power rating of DIP type resistor is 1/8W,1
/4W,1/2W,1W etc. and the popular size of SMD type resistor is 0201,0402,0603,0805,1206,1210 etc.
The behavior of a resistor is dictated by the relationship specified by Ohm’s law : I (current)=V (voltage) / R (resistance).
Ohm’s law states that the voltage (V) across a resistor is proportional to the current (I), where the constant of proportionality is the resistance (R)
Unit : K = 1000, m=1/1000(=0.001)
1K Ohm resistor = 1000 Ohm
Example :
10V is applied to a 1K ohm resistor, the current flow though this resistor is 10 volt / 1000 ohm = 0.01A = 10mA. | {"url":"https://drm.com.tw/2019/02/resistor-introduction/","timestamp":"2024-11-10T05:36:44Z","content_type":"text/html","content_length":"33702","record_id":"<urn:uuid:d73c41eb-5cb3-4b70-b502-955cbf5de4ee>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00177.warc.gz"} |
Greta Malaspina
We develop and analyze stochastic inexact Gauss-Newton methods for nonlinear least-squares problems and inexact Newton methods for nonlinear systems of equations. Random models are formed using
suitable sampling strategies for the matrices involved in the deterministic models. The analysis of the expected number of iterations needed in the worst case to achieve a desired level … Read more
Splitted Levenberg-Marquardt Method for Large-Scale Sparse Problems
We consider large-scale nonlinear least squares problems with sparse residuals, each of them depending on a small number of variables. A decoupling procedure which results in a splitting of the
original problems into a sequence of independent problems of smaller sizes is proposed and analysed. The smaller size problems are modified in a way that … Read more | {"url":"https://optimization-online.org/author/malaspinag/","timestamp":"2024-11-04T05:48:33Z","content_type":"text/html","content_length":"85764","record_id":"<urn:uuid:c825fbdd-035a-4047-97fb-2cade0ca48f9>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00088.warc.gz"} |
Natural logarithm
The natural logarithm of a number is its logarithm to the base of the mathematical constant e, where e is an irrational and transcendental number approximately equal to 2.718281828459. The natural
logarithm of x is generally written as ln x, log[e] x, or sometimes, if the base e is implicit, simply log x.^[1] Parentheses are sometimes added for clarity, giving ln(x), log[e](x) or log(x). This
is done in particular when the argument to the logarithm is not a single symbol, to prevent ambiguity.
The natural logarithm of x is the power to which e would have to be raised to equal x. For example, ln(7.5) is 2.0149..., because e^2.0149... = 7.5. The natural log of e itself, ln(e), is 1, because
e^1 = e, while the natural logarithm of 1, ln(1), is 0, since e^0 = 1.
The natural logarithm can be defined for any positive real number a as the area under the curve y = 1/x from 1 to a (the area being taken as negative when a < 1). The simplicity of this definition,
which is matched in many other formulas involving the natural logarithm, leads to the term "natural". The definition of the natural logarithm can be extended to give logarithm values for negative
numbers and for all non-zero complex numbers, although this leads to a multi-valued function: see Complex logarithm.
The natural logarithm function, if considered as a real-valued function of a real variable, is the inverse function of the exponential function, leading to the identities:
Like all logarithms, the natural logarithm maps multiplication into addition:
Thus, the logarithm function is a group isomorphism from positive real numbers under multiplication to the group of real numbers under addition, represented as a function:
Logarithms can be defined to any positive base other than 1, not only e. However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and are usually defined in
terms of the latter. For instance, the binary logarithm is the natural logarithm divided by ln(2), the natural logarithm of 2. Logarithms are useful for solving equations in which the unknown appears
as the exponent of some other quantity. For example, logarithms are used to solve for the half-life, decay constant, or unknown time in exponential decay problems. They are important in many branches
of mathematics and the sciences and are used in finance to solve problems involving compound interest.
By Lindemann–Weierstrass theorem, the natural logarithm of any positive algebraic number other than 1 is a transcendental number.
Natural logarithm
Indefinite Integral
The concept of the natural logarithm was worked out by Gregoire de Saint-Vincent and Alphonse Antonio de Sarasa before 1649.^[2] Their work involved quadrature of the hyperbola xy = 1 by
determination of the area of hyperbolic sectors. Their solution generated the requisite "hyperbolic logarithm" function having properties now associated with the natural logarithm.
An early mention of the natural logarithm was by Nicholas Mercator in his work Logarithmotechnia published in 1668,^[3] although the mathematics teacher John Speidell had already in 1619 compiled a
table of what in fact were effectively natural logarithms.^[4]
Notational conventions
The notations "ln x" and "log[e] x" both refer unambiguously to the natural logarithm of x. "log x" without an explicit base may also refer to the natural logarithm. This usage is common in
mathematics and some scientific contexts as well as in many programming languages.^[5] In some other contexts, however, "log x" can be used to denote the common (base 10) logarithm.
Historically, the notations "l." and "l" were in use at least since the 1730s,^[6]^[7] and until at least the 1840s,^[8] then "log."^[9] or "log",^[10] at least since the 1790s. Finally, in the
twentieth century, the notations "Log"^[11] and "logh"^[12] are attested.
Origin of the term natural logarithm
The graph of the natural logarithm function shown earlier on the right side of the page enables one to glean some of the basic characteristics that logarithms to any base have in common. Chief among
them are: the logarithm of the number one is zero; and the logarithm as x approaches zero from the right is negative infinity.
What makes natural logarithms unique is to be found at the single point where all logarithms are zero, namely the logarithm of the number one. At that specific point the "slope" of the curve of the
graph of the natural logarithm is also precisely one. Logarithms to a higher base than e, such as those to the base 10, exhibit a slope at that point less than one, while logarithms to a lower base
than e, such as those to the base 2, exhibit a slope at that point greater than one. While the methods for computing the "value" of e are fascinating from various mathematical perspectives, they all
can be thought of as resulting from the pursuit of this condition.
Another way of conceptualizing this is to realize that, for any numeric value close to the number one, the natural logarithm can be computed by subtracting the number one from the numeric value. For
example, the natural logarithm of 1.01 is 0.01 to an accuracy better than 5 parts per thousand. With similar accuracy one can assert that the natural logarithm of 0.99 is negative 0.01. The accuracy
of this concept increases as one approaches the number one ever more closely, and reaches completeness of accuracy precisely there. To the same extent that the number one itself is a number common to
all systems of counting, so also the natural logarithm is independent of all systems of counting. In the English language the term adopted to encapsulate this concept is the word "natural".
Initially, it might seem that since the common numbering system is base 10, this base would be more "natural" than base e. But mathematically, the number 10 is not particularly significant. Its use
culturally—as the basis for many societies’ numbering systems—likely arises from humans’ typical number of fingers.^[13] Other cultures have based their counting systems on such choices as 5, 8, 12,
20, and 60.^[14]^[15]^[16]
log[e] is a "natural" log because it automatically springs from, and appears so often in, mathematics. For example, consider the problem of differentiating a logarithmic function:^[17]
If the base b equals e, then the derivative is simply 1/x, and at x = 1 this derivative equals 1. Another sense in which the base-e-logarithm is the most natural is that it can be defined quite
easily in terms of a simple integral or Taylor series and this is not true of other logarithms.
Further senses of this naturalness make no use of calculus. As an example, there are a number of simple series involving the natural logarithm. Pietro Mengoli and Nicholas Mercator called it
logarithmus naturalis a few decades before Newton and Leibniz developed calculus.^[18]
Formally, ln(a) may be defined as the area under the hyperbola 1/x. This is the integral,
This function is a logarithm because it satisfies the fundamental property of a logarithm:
This can be demonstrated by splitting the integral that defines ln(ab) into two parts and then making the variable substitution x = ta in the second part, as follows:
In elementary terms, this is simply scaling by 1/a in the horizontal direction and by a in the vertical direction. Area does not change under this transformation, but the region between a and ab is
reconfigured. Because the function a/(ax) is equal to the function 1/x, the resulting area is precisely ln(b).
The number e is defined as the unique real number a such that ln(a) = 1.
Alternatively, if the exponential function has been defined first, say by using an infinite series, the natural logarithm may be defined as its inverse function, i.e., ln is that function such that
exp(ln(x)) = x. Since the range of the exponential function on real arguments is all positive real numbers and since the exponential function is strictly increasing, this is well-defined for all
positive x.
1. The statement is true for , and we now show that for all , which completes the proof by the fundamental theorem of calculus. Hence, we want to show that
(Note that we have not yet proved that this statement is true.) If this is true, then by multiplying the middle statement by the positive quantity and subtracting we would obtain
This statement is trivially true for since the left hand side is negative or zero. For it is still true since both factors on the left are less than 1 (recall that ). Thus this last statement is
true and by repeating our steps in reverse order we find that for all . This completes the proof.
Derivative, Taylor series
The derivative of the natural logarithm is given by
Proof 1 by the first part of the fundamental theorem of calculus:
Proof 2 from this video:
This leads to the Taylor series for ln(1 + x) around 0; also known as the Mercator series
(Leonhard Euler^[19] nevertheless boldly applied this series to x= -1, in order to show that the harmonic series equals the (natural) logarithm of 1/(1-1), that is the logarithm of infinity.
Nowadays, more formally but perhaps less vividly, we prove that the harmonic series truncated at N is close to the logarithm of N, when N is large.)
At right is a picture of ln(1 + x) and some of its Taylor polynomials around 0. These approximations converge to the function only in the region −1 < x ≤ 1; outside of this region the higher-degree
Taylor polynomials are worse approximations for the function.
Substituting x − 1 for x, we obtain an alternative form for ln(x) itself, namely
By using the Euler transform on the Mercator series, one obtains the following, which is valid for any x with absolute value greater than 1:
This series is similar to a BBP-type formula.
Also note that is its own inverse function, so to yield the natural logarithm of a certain number y, simply put in for x.
The natural logarithm in integration
The natural logarithm allows simple integration of functions of the form g(x) = f '(x)/f(x): an antiderivative of g(x) is given by ln(|f(x)|). This is the case because of the chain rule and the
following fact:
In other words,
Here is an example in the case of g(x) = tan(x):
Letting f(x) = cos(x) and f'(x)= – sin(x):
where C is an arbitrary constant of integration.
The natural logarithm can be integrated using integration by parts:
Numerical value
To calculate the numerical value of the natural logarithm of a number, the Taylor series expansion can be rewritten as:
To obtain a better rate of convergence, the following identity can be used.
provided that y = (x−1)/(x+1) and Re(x) > 0.
For ln(x) where x > 1, the closer the value of x is to 1, the faster the rate of convergence. The identities associated with the logarithm can be leveraged to exploit this:
Such techniques were used before calculators, by referring to numerical tables and performing manipulations such as those above.
Natural logarithm of 10
The natural logarithm of 10, which has the decimal expansion 2.30258509...,^[21] plays a role for example in the computation of natural logarithms of numbers represented in scientific notation, as a
mantissa multiplied by a power of 10:
This means that one can effectively calculate the logarithms of numbers with very large or very small magnitude using the logarithms of a relatively small set of decimals in the range .
High precision
To compute the natural logarithm with many digits of precision, the Taylor series approach is not efficient since the convergence is slow. If x is near 1, an alternative is to use Newton's method to
invert the exponential function, whose series converges more quickly. For an optimal function, the iteration simplifies to
which has cubic convergence to ln(x).
Another alternative for extremely high precision calculation is the formula^[22] ^[23]
where M denotes the arithmetic-geometric mean of 1 and 4/s, and
with m chosen so that p bits of precision is attained. (For most purposes, the value of 8 for m is sufficient.) In fact, if this method is used, Newton inversion of the natural logarithm may
conversely be used to calculate the exponential function efficiently. (The constants ln 2 and π can be pre-computed to the desired precision using any of several known quickly converging series.)
Based on a proposal by William Kahan and first implemented in the Hewlett-Packard HP-41C calculator in 1979, some calculators, computer algebra systems and programming languages (for example C99^[24]
) provide a special natural logarithm plus 1 function alternatively named LN1+X, LN(1+X), lnp1(x),^[25]^[26] ln1p(x) or log1p(x)^[24] to give more accurate results for values of x close to zero
compared to using ln(x+1) directly.^[24]^[25]^[26] This function is implemented using a different internal algorithm to avoid an intermediate result near 1, thereby allowing both the argument and the
result to be near zero.^[25]^[26] Similar inverse functions named expm1(x),^[24] expm(x)^[25]^[26] or exp1m(x) exist as well.^[nb 1]
Computational complexity
The computational complexity of computing the natural logarithm (using the arithmetic-geometric mean) is O(M(n) ln n). Here n is the number of digits of precision at which the natural logarithm is to
be evaluated and M(n) is the computational complexity of multiplying two n-digit numbers.
Continued fractions
While no simple continued fractions are available, several generalized continued fractions are, including:
These continued fractions—particularly the last—converge rapidly for values close to 1. However, the natural logarithms of much larger numbers can easily be computed by repeatedly adding those of
smaller numbers, with similarly rapid convergence.
For example, since 2 = 1.25^3 × 1.024, the natural logarithm of 2 can be computed as:
Furthermore, since 10 = 1.25^10 × 1.024^3, even the natural logarithm of 10 similarly can be computed as:
Complex logarithms
The exponential function can be extended to a function which gives a complex number as e^x for any arbitrary complex number x; simply use the infinite series with x complex. This exponential function
can be inverted to form a complex logarithm that exhibits most of the properties of the ordinary logarithm. There are two difficulties involved: no x has e^x = 0; and it turns out that e^2πi = 1 = e^
0. Since the multiplicative property still works for the complex exponential function, e^z = e^z+2πki, for all complex z and integers k.
So the logarithm cannot be defined for the whole complex plane, and even then it is multi-valued – any complex logarithm can be changed into an "equivalent" logarithm by adding any integer multiple
of 2πi at will. The complex logarithm can only be single-valued on the cut plane. For example, ln(i) = πi/2 or 5πi/2 or -3πi/2, etc.; and although i^4 = 1, 4 log(i) can be defined as 2πi, or 10πi or
−6πi, and so on.
• Plots of the natural logarithm function on the complex plane (principal branch)
Superposition of the previous 3 graphs
See also
This article is issued from
- version of the 11/15/2016. The text is available under the
Creative Commons Attribution/Share Alike
but additional terms may apply for the media files. | {"url":"https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Natural_logarithm.html","timestamp":"2024-11-10T11:26:01Z","content_type":"text/html","content_length":"104655","record_id":"<urn:uuid:7ecbcfc0-9342-4420-bd0e-6401a52ef725>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00022.warc.gz"} |
Finding Age
Question: Pooja and Esha met each other after long time. In the course of their conversation, Pooja asked Esha her age. Esha replied, "If you reverse my age, you will get my husband's age. He is of
course older than me. Also, the difference between our age is 1/11th of the sum of our age."
Can you help out Pooja in finding Esha's age?
Esha's age is 45 years.
Assume that Esha's age is 10X+Y years. Hence, her hunsbands age is (10Y + X) years.
It is given that difference between their age is 1/11th of the sum of their age. Hence,
[(10Y + X) - (10X + Y)] = (1/11)[(10Y + X) + (10X + Y)]
(9Y - 9X) = (1/11)(11X + 11Y)
9Y - 9X = X + Y
8Y = 10X
4Y = 5X
Hence, the possible values are X=4, Y=5 and Esha's age is 45 years. | {"url":"http://www.tutioncentral.com/2012/02/finding-age.html","timestamp":"2024-11-14T15:25:08Z","content_type":"application/xhtml+xml","content_length":"118367","record_id":"<urn:uuid:3b527979-41ae-4060-ac26-8926a6b4d072>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00295.warc.gz"} |
OFUS ÅRSBOK om Färdtjänst 2000-2001 Epub-nedladdning
18 Nästa hem idéer i 2021 hem inredning, inredning, interiör
Use an initial guess for . F. in eq. (3.3.4) and use the resulting value for – N. i. in eq. (3.3.5). This will give you a new value of .
Axelsson, T. Midtvedt, A.-C. Bylund-Fellenius, The Role of Intestinal be used for Ren 709 and P3 grading. We also offer ARRI laser scanning; a method by which digital image files are transferred to
35 mm film. Deformationsegenskaper hos slagna betongpålar Bengt Fellenius & Torsten The pile specimens were loaded axially in the same way as the piles in the fieltl.
F. does not change.
J. Sjöstrand: मुफ़्त में डाउनलोड. ई-बुक
tan(– )} + W. i. cos( ) tan(– ) i=1 n. W. i. sin( i) (3.1.6) Modified Bishop method .
Basics of Foundation Design - Bengt Fellenius - Häftad
The forces between the slices are neglected and each slice is considered to be an independent column of soil of unit thickness. The Fellenius (or Swedish ) Solution. It is assumed that for each slice
the resultant of the interslice forces is zero.
Göteborgsområdet. Daniel Salomons Daniel Salomons-bild Björn BreidegardYvonne ErikssonKerstin FelleniusBodil JönssonKenneth HolmqvistSven Improving Quality in Higher Education by using Living Lab
Mall konkursansokan
Also, no limit was imposed on the shaft resistance calculation for the Takesue method since the … D-Geo Stability is a slope stability package for soft soils. Previous releases of D-Geo Stability
were named MStab. Deltares systems tools follow best practice, and are developed according to modern software standards; therefore they are user-friendly and easy to learn. However, instead of doing
that, they placed great emphasis on the factor of safety of 1.14 that they obtained using the Fellenius Method with total unit weights, which was only 57 percent of their “most correct” solution. The
details of their programming are unknown and we have not been able to reproduce that number, but that is not critical. In the method of slices, also called OMS or the Fellenius method, the sliding
mass above the failure surface is divided into a number of slices. The forces acting on each slice are obtained by considering the mechanical (force and moment) equilibrium for the slices.
Sökning: "Kerstin Fellenius". Hittade 1 avhandling innehållade orden Kerstin Fellenius. 1. Reading acquisition in pupils with visual impairments in mainstream KR Massarsch, C Wersäll, BH Fellenius.
Proceedings of the Computer Methods and Recent Advances in Geomechanics: Proceedings of the …, 2015. 5, 2015. 154 följare, 25 följer, 11951 pins – Se vad Elisabeth Fellenius (elisabethfellen) hittade
på Pinterest, platsen för världens bästa idéer.
Ulcus kolit
N' = Wcosa — ul. Hence the factor of safety in terms of effective stress(Equation 3) is given by: The geotechnical engineer frequently uses limit equilibrium methods of analysis when studying slope
stability problems, for example, Ordinary or Fellenius’ method (sometimes referred to as the Swedish circle method or the conventional method), Simplified Bishop method, Spencer’s method, Janbu’s
simplified method, Janbu’s rigorous method, Morgenstern-Price method, or unified solution scheme [1–3]. In order to reduce the influence of the assumptions made in limit equilibrium methods In the
current practice, to determine the safety factor of a slope with two-dimensional circular potential failure surface, one of the searching methods for the critical slip surface is Genetic Algorithm
(GA), while the method to calculate the slope safety factor is Fellenius slicesmethod.HoweverGAneedstobevalidatedwithmorenumerictests,whileFellenius slicesmethodisjustanapprox- imatemethodlike
niteelementmethod. ispaperproposedanewmethodtodeterminetheminimumslopesafetyfactorwhich Based on the method of infinite article points to the theoretical basis of the loess slope stability analysis
is even the Fellenius article point method. Limit equilibrium methods : Fellenius; Bishop; Janbu; Spencer; G.L.E.; Sarma. Seismic analysis: Pseudostatic method ; Simplified dynamic method by
empirical f
Stability Analysis of Finite Slopes Using the Swedish Circle Method: The simplest and most basic limit equilibrium method is known as the Fellenius Method of Slices.The Fellenius Method assumes that
the moment arm is the same for both the driving and resisting forces, i.e, that these forces are collinear. This method was developed by the Swedish geotechnical pioneer Wolmar Fellenius (1876-1957).
Swedish circle method for analyzing the slope stability: This method was first introduced by Fellenius (1926) for analysis of slope stability of C-φ soil. In this method, the soil mass above the
assumed slip circle is divided into a number of vertical slices of equal width. The forces between the slices are neglected and each slice is considered to be an independent column of soil of unit
Stefan lofven svetsare
Frida Kvennefelt FILMCAFE.SE
Fellenius method : 원형파괴단면, 모멘트 평형만 고려. 절편력 합력이 절편 바닥에 평행하게 작용한다고 가정(사실상 절편력 영향 고려하지 않는 것임). 쉽지만 오차가 커서 실무에 거의 쓰이지 않음. 2010-05-05
3 4. Dzielimy obszar ograniczony kraw ędzi ą skarpy i wycinkiem koła na 10 -20 pionowych pasków (mog ą by ć ró Ŝnej szeroko ści, The Fellenius (Swedish) Method Fellenius assumed that the resultant of
the inter-slice forces is zero, then N’ = W cos α–ul Hence the factor of safety in terms of effective stress is given by: The components W cosαand W sinαcan be determined graphically while angle a
can be calculated or measured Stability Analysis Methods (General): There are several available methods that can be used to perform a circular arc stability analysis for an approach embankment over
soft ground. The simplest basic method is known as the Normal or Ordinary Method of Slices, also known as Fellenius’ method or the Swedish circle method of analysis. methods of slope stability using
limit equilibrium, finite element, and other methods are frequently reported (Duncan and Wright 2005).
Harga lampu belakang avanza
J. Sjöstrand: मुफ़्त में डाउनलोड. ई-बुक
Continue iterating between these equations until . | {"url":"https://londzmoixp.netlify.app/50022/57547.html","timestamp":"2024-11-14T10:05:10Z","content_type":"text/html","content_length":"17104","record_id":"<urn:uuid:1ddcf671-c5ef-4b54-801d-7732f5bc6b97>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00484.warc.gz"} |
STDEVP Function
Get standard deviation of population
Return value
Estimated standard deviation
• number1 - First number or reference in the sample.
• number2 - [optional] Second number or reference.
How to use
The STDEVP function calculates the standard deviation for an entire population. Standard deviation is a measure of how much variance there is in a set of numbers compared to the average (mean) of the
numbers. The STDEVP function is meant to estimate standard deviation for an entire population. If data represents a sample, use the STDEV function.
Note: STDEVP has been replaced with a newer function called STDEV.P, which has identical behavior. Although STDEVP still exists for backwards compatibility, Microsoft recommends that people use the
newer STDEV.P function instead.
Standard Deviation functions in Excel
The table below summarizes the standard deviation functions provided by Excel.
• STDEVP calculates standard deviation using the "n" method, ignoring logical values and text.
• STDEVP assumes your data is the entire population. When your data is a sample set only, calculate standard deviation using the STDEV function (or its more current replacement, the STDEV.S
• Numbers are supplied as arguments. They can be supplied as actual numbers, ranges, arrays, or references that contain numbers.
• The STDEVP function ignores logical values and text. If you want to include logical values and/or numbers as text in a reference, use the STDEVA function. | {"url":"https://exceljet.net/functions/stdevp-function","timestamp":"2024-11-09T09:30:55Z","content_type":"text/html","content_length":"48471","record_id":"<urn:uuid:5eea2b1a-d78c-429c-801d-8b0432b04846>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00651.warc.gz"} |
2005 AMC 12A Problems/Problem 19
A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled? $(\mathrm
{A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$
Solution 1
We find the number of numbers with a $4$ and subtract from $2005$. Quick counting tells us that there are $200$ numbers with a 4 in the hundreds place, $200$ numbers with a 4 in the tens place, and
$201$ numbers with a 4 in the units place (counting $2004$). Now we apply the Principle of Inclusion-Exclusion. There are $20$ numbers with a 4 in the hundreds and in the tens, and $20$ for both the
other two intersections. The intersection of all three sets is just $2$. So we get:
$2005-(200+200+201-20-20-20+2) = 1462 \Longrightarrow \mathrm{(B)}$
Solution 2
Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$; only, in base $9$, the number $9$ is not counted. Since $4$ is skipped, the symbol $5$
represents $4$ miles of travel, and we have traveled $2004_9$ miles. By basic conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}$
Solution 3
Since any numbers containing one or more $4$s were skipped, we need only to find the numbers that don't contain a $4$ at all. First we consider $1$ - $1999$. Single digits are simply four digit
numbers with a zero in all but the ones place (this concept applies to double and triple digits numbers as well). From $1$ - $1999$, we have $2$ possibilities for the thousands place, and $9$
possibilities for the hundreds, tens, and ones places. This is $2 \cdot 9 \cdot 9 \cdot 9-1$ possibilities (because $0000$ doesn't count) or $1457$ numbers. From $2000$ - $2005$ there are $6$
numbers, $5$ of which don't contain a $4$. Therefore the total is $1457 + 5$, or $1462$$\Rightarrow$$\boxed{\text{B}}$.
Solution 4
We seek to find the amount of numbers that contain at least one $4,$ and subtract this number from $2005.$
We can simply apply casework to this problem.
The amount of numbers with at least one $4$ that are one or two digit numbers are $4,14,24,34,40-49,54,\cdots,94$ which gives $19$ numbers.
The amount of three digit numbers with at least one $4$ is $8*19+100=252.$
The amount of four digit numbers with at least one $4$ is $252+1+19=272$
This, our answer is $2005-19-252-272=1462,$ or $\boxed{B}.$
Solution 5(Super fast)
This is very analogous to base $9$. But, in base $9$, we don't have a $9$. So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 =
1463$. $1463 - 1 = 1462$
Therefore, our answer is $\boxed {1462}$
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. | {"url":"https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_19&oldid=193410","timestamp":"2024-11-04T01:23:40Z","content_type":"text/html","content_length":"54836","record_id":"<urn:uuid:4178ee57-49d2-4ed6-afea-08ae0f8caae2>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00535.warc.gz"} |
Computational Complexity
Expander graphs informally are graphs that given any subset S that is not too large, the set of vertices connected to S contains a large number of vertices outside of S. There are many constructions
and applications for expander graphs leading to entire
on the subject.
The adjacency matrix A of a graph G of n vertices is an n×n matrix such that a[i,j] is 1 if there is an edge between vertices i and j and 0 otherwise. Noga Alon noticed that a graph that has a large
gap between the first and second eigenvalue of the adjacency matrix will be a good expander.
We can use ε-biased sets to get expanders. Let S be a ε-biased set for F^m for F the field of 2 elements. Consider the graph G consisting of 2^m vertices labelled with the elements of F^m and an edge
from x to y if y=x+s or x=y+s. This kind of graph G is known as a Cayley graph.
By looking at the eigenvalues the adjacency matrix A of G we can show G is an expander. The eigenvectors v are just the vectors corresponding to the functions g in L described earlier. For any vector
a we have
(Ag)(a) = Σ[s in S] g(a+s) = g(a) Σ[s in S] g(s)
since g(a+s) = g(a)g(s). Let g(S) = Σ
[s in S]
g(s). We now have that
Ag = g(S) g.
So g is an eigenvector with eigenvalue g(S). If g is the constant one function then g(S)=|S|. Since S is an ε-biased set, g(S)≤ε|S| for every other g, so the second eigenvalue is much smaller than
the largest eigenvalue and G must be an expander.
The June 2003 SIGACT News is out. Aduri Pavan wrote this months Complexity Theory Column on "Comparison of Reductions and Completeness Notions".
As I have mentioned before in this weblog, I heartily encourage joining SIGACT, the ACM Special Interest Group on Algorithms and Computation Theory. You get the SIGACT News, discounts on conferences
and as I discovered last night from home, you apparently get online access to the STOC proceedings. Not to mention supporting the theory community. All this for the low price of $18 ($9 for
What about the ACM itself? I have been an ACM member since graduate school since I feel it is important to support the main computer science organization. But for the additional $96 ($42 for
students) there are no real significant benefits over joining SIGACT alone.
ε-biased sets are an interesting concept that I have seen recently in a few papers but never seemed to have a clear description. At FCRC Eli Ben-Sasson gave me a good explanation and I will try to
recreate it here.
Let F be the field of 2 elements 0 and 1 with addition and multiplication done modulo 2. Fix a dimension m. Let L be the set of functions g mapping elements of F^m to {-1,1} with the property that g
(x+y)=g(x)g(y). Here x+y represents addition done coordinate-wise modulo 2. One example of a g in L is g(x[1],x[2],x[3])=(-1)^x[1] (-1)^x[3].
There is the trivial function g in L that always maps to 1. For every non-trivial g in L exactly half of the elements in F^m map to 1 and the others to -1. If one picks a reasonably large subset S of
F^m at random then high probability, g will map about half the elements to 1 and the rest to -1. In other words the expected value of g(x) for x uniformly chosen in S is smaller than some small value
ε. If this is true we say S is ε-biased for g.
An ε-biased set is a set S such that for all nontrivial g in L, S is ε-biased for g. Formally this means that
Σ[x in S] g(x) ≤ ε|S|.
Not only do reasonable size ε-biased sets exists but they can be found efficiently. Naor and Naor found the first efficiently constructible ε-biased sets of size polynomial in m and 1/ε.
One can extend the notion of ε-biased sets to fields F of p elements for arbitrary prime p. L would now be the set of functions g mapping elements of F^m to the complex pth roots of unity, e^2π(j/p)i
for 0≤j≤p-1 again with the property that g(x+y)=g(x)g(y). Various constructions have created generalized ε-biased sets of size polynomial in m, 1/ε and log p.
For applications let me quote from the recent STOC paper by Ben-Sasson, Sudan, Vadhan and Wigderson that used ε-biased sets to get efficient low-degree tests and smaller probabilistically checkable
proofs. You can get more information and references from that paper.
Since the introduction of explicit ε-biased sets, the set and diversity of applications of these objects grew quickly, establishing their fundamental role in theoretical computer science. The
settings where ε-biased sets are used include: the direct derandomization of algorithms such as fast verification of matrix multiplication and communication protocols for equality; the construction
of almost k-wise independent random variables, which in turn have many applications; inapproximability results for quadratic equation over GF(2); learning theory; explicit constructions of Ramsey
graphs; and elementary constructions of Cayley expanders.
old post on conference covers, Pekka Orponen left a comment saying that Alvy Ray Smith originally designed the FOCS cover back in 1973, when the conference was SWAT: Switching and Automata Theory.
Here is Smith's story about the cover where he states the hands are his.
Smith had three papers in SWAT before moving into computer graphics and co-founding companies such as Pixar. So next time you are looking up a FOCS paper take a look at the hands of a one-time
theorist who made a difference.
After the FCRC meetings I attended were concluded, I headed up to UCSD for the celebration of Walter Savitch for his sixtieth birthday and upcoming retirement. He gained his fame in complexity for
Savitch's Theorem that shows "P=NP" for space.
I learned quite a bit at the meeting. Walt Savitch was Steve Cook's first student, his only student while Cook was at Berkeley in his pre-Toronto pre-"SAT is NP-complete" days. Also as Cook said,
Savitch is the only student he has had with a theorem named after him. That theorem made up a good part of Savitch's Ph.D. thesis. At the celebration Cook gave an overview on propositional proof
After coming to UCSD, Savitch did some work on computational linguistics and one of the leaders of the field, Aravind Joshi, gave a talk on combining trees to keep the structure when parsing
Savitch is probably best known now in computer science for his textbooks in introductory programming that likley many of you have used.
Congrats Walt on a fine career and here's hoping retirement doesn't slow you down.
The list of accepted papers for FOCS 2003 has been posted. As usual, many (but not all) of the papers are available on the authors' web sites.
As promised I added links to the papers in the post on the STOC business meeting. Let me say some more words on the winner of the Gödel prize
Valiant developed the concept of PAC (Probably Approximably Correct) learning as roughly where a learner sees a small number of labelled examples from a distribution and with high confidence will
generate a hypothesis that with high probability will correctly label instances drawn from the same distribution.
A strong learner has confidence close to 100%; a weak learner has confidence only slightly better than 50%. Schapire, using a technique called boosting, showed how to convert a weak learner to a
strong learner. This is a wonderful theoretical result but the algorithm had problems that made it difficult to implement.
In their Gödel prize winning paper, A decision-theoretic generalization of on-line learning and an application to boosting, Freund and Schapire develop the adaboost algorithm that solves many of
these issues and has become a staple of the theoretical and practical machine learning community.
Boosting has its own web site where you can find much more information about the algorithms and applications.
Alonzo Church was born a hundred years ago today in Washington, DC. Church is best known for the λ-calculus, a simple method for expressing and applying functions that has the same computational
power as Turing machines.
With Rosser in 1936, he showed that λ-expressions that reduce to an irreducible normal form have a unique normal form. In that same year he showed the impossibility of decided whether such a normal
form existed.
Church's thesis, which he states as a definition: "An effectively calculable function of the positive integers is a λ-definable function of the positive integers."
Again in 1936, Kleene and Church showed that computing normal forms have the equivalent power of the recursive functions of Turing machines. And thus the Church-Turing thesis was born: Everything
computable is computable by a Turing machine.
The λ-calculus also set the stage for many of the functional programming languages like lisp and scheme.
Alonzo Church passed away on August 11, 1995 in Ohio.
I have mixed feelings about the Federated Computing Research Conference. It is a good idea to get many different areas of computer science together. I do get to see many people I haven't seen in
years who went into non-theoretical areas of CS.
On the other hand 2200 participants made the place quite crowded and it seemed to take away from the informal atmosphere of most theory conference. Since STOC and Electronic Commerce had nearly a
complete overlap I jumped back and forth between talks never really feeling fully part of either conference.
For the first time the Complexity conference was not part of FCRC because 2003 is a Europe year for Complexity. In an informal poll I took of STOC people interested in complexity most liked having
both conferences at the same place but would rather that happen in isolation, like last year in Montreal, rather than as part of the much larger FCRC meeting.
In what seems to be a trend in CS conferences, wireless internet was made available at the conference site. As you walked around you would pass many people sitting on chairs and on the ground hunched
over their laptops disconnected from the conference and connected into another world. Seemed a bit depressing but I too found the net hard to resist--it is always tempting to simply open my laptop
and connect, checking email and posting to this weblog.
Update (9/3/03): STOC 2004 CFP
Last night was the business meeting for STOC. This used to be a raucous affair with major battles over issues like whether to have parallel sessions, but now in its old age is more for distributing
information like
• Awards: Godel Prize for best paper published in a journal in last seven years: Yoav Freund and Rob Schapire for their adaboost algorithm.
Knuth Prize for lifetime research activity: Miklos Ajtai.
STOC Best Paper: Jointly to Impagliazzo and Kabanets for "Derandomizing Polynomial Identity Tests Means Proving Circuit Lower Bounds" (you read it here first) and Oded Regev for "New Lattice
Based Cryptographic Constructions".
The Danny Lewin Best Student Paper Award: Thomas Hayes for "Randomly Coloring Graphs of Girth at Least Five" Thomas Hayes is from my once and future department at U. Chicago.
• Future Conferences: FOCS 2003 in Boston, STOC 2004 in Chicago and FOCS 2004 in Rome, its first time outside North America. On Friday, a day after agreeing to return to Chicago, Janos Simon asked
me to give the presentation on STOC 2004, which is always fun. Laszlo Babai, another member of the Chicago faculty, will be PC chair of that meeting.
• Registration numbers, NSF report, progress on scanning in old papers and redoing submission software. I'm not going to do a full business meeting report--someone else does that and will
eventually appear in SIGACT news.
This week I'm at the Federated Computing Research Conference (FCRC), a combination of thirty conferences and workshops with 2200 participants. I'm here for the theory conference (STOC) and Electronic
Last night, Adleman, Rivest and Shamir gave their Turing award lecture, each giving twenty minutes of an hour long talk. Their basic them on how cryptology has changed in the last 25 years:
1. Cryptography is now done publicly rather than in secret. This has led to researchers building on each others ideas to create better and better encryption schemes and protocols. But also it has
allowed more people to attack these protocols and weed out the bad ones.
2. Cryptography has moved from art to science. Now we have protocols based on mathematical ideas like number theory instead of just creating seemingly complexity functions.
Adi Shamir made other interesting comments like that perfect cryptography is impossible, though very good cryptography can be had at a modest cost. Most attacks on practical implementations of
cryptographic protocols work on the implementation as opposed to the protocol.
The lectures were taped and may show up on-line someday. I'll let you know if I find them there--definitely recommended viewing.
A personal note: I have accepted an offer to return to the computer science department of the University of Chicago starting this fall. As NEC Labs has been moving its focus away from basic research,
it is time for me to go back to an academic life.
I plan to keep this weblog going in Chicago as long as I have things to say.
Previous Lesson
In this lesson we will prove the Immerman-Szelepcsényi Theorem.
Theorem (Immerman-Szelepcsényi): For reasonable s(n)≥ log n, NSPACE(s(n))=co-NSPACE(s(n)).
Let M be a nondeterministic machine using s(n) space. We will create a nondeterministic machine N such that for all inputs x, N(x) accepts if and only if M(x) rejects.
Fix an input x and let s=s(|x|). The total number of configurations of M(x) can be at most c^s for some constant c. Let t=c^s. We can also bound the running time of M(x) by t because any computation
path of length more than t must repeat a configuration and thus could be shortened.
Let I be the initial configuration of M(x). Let m be the number of possible configurations reachable from I on some nondeterministic path. Suppose we knew the value of m. We now show how N(x) can
correctly determine that M(x) does not accept.
Let r=0
For all nonaccepting configurations C of M(x)
Try to guess a computation path from I to C
If found let r=r+1
If r=m then accept o.w. reject
If M(x) accepts then there is some accepting configuration reachable from I so there must be less than m non-accepting configurations reachable from I so N(x) cannot accept. If M(x) rejects then
there is no accepting configurations reachable from I so N(x) on some nondeterministic path will find all m nonaccepting paths and accept. The total space is at most O(s) since we are looking only at
one configuration at a time.
Of course we cannot assume that we know m. To get m we use an idea called inductive counting. Let m[i] be the number of configurations reachable from I in at most i steps. We have m[0]=1 and m[t]=m.
We show how to compute m[i+1] from m[i]. Then starting at m[0] we compute m[1] then m[2] all the way up to m[t]=m and then run the algorithm above.
Here is the algorithm to nondeterministically compute m[i+1] from m[i].
Let m[i+1]=0
For all configurations C
Let b=0, r=0
For all configurations D
Guess a path from I to D in at most i steps
If found
Let r=r+1
If D=C or D goes to C in 1 step
Let b=1
If r<m[i] halt and reject
Let m[i+1]=m[i+1]+b
The test that r<m[i] guarantees that we have looked at all of the configurations D reachable from I in i steps. If we pass the test each time then we will have correctly computed b to be equal to 1
if C is reachable from I in at most i+1 steps and b equals 0 otherwise.
We are only remembering a constant number of configurations and variables so again the space is bounded by O(s). Since we only need to remember m[i] to get m[i+1] we can run the whole algorithm in
space O(s). | {"url":"https://blog.computationalcomplexity.org/2003/06/?m=0","timestamp":"2024-11-08T10:41:39Z","content_type":"application/xhtml+xml","content_length":"232591","record_id":"<urn:uuid:174cabbf-4525-407a-8012-67ff6c7337ff>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00593.warc.gz"} |
Hydropneumatic Tank Sizing - HVAC/R & Solar
Hydropneumatic tanks are primarily used in a domestic water system for draw down purposes when the pressure booster system is off on no-flow shutdown (NFSD). The NFSD circuitry turns the lead pump
off when there is no demand on the system. While the system is off in this condition, the hydropneumatic tank will satisfy small demands on the system. Without the tank, the booster would restart
upon the slightest call for flow such as a single toilet being flushed or even a minute leak in the piping system.
Expansion Tank
Hydropneumatic tank sizing is dependent on two factors:
1. Length of time you wish the pumps to remain off in a no-flow situation.
2. The tank location in relation to the pressure booster.
Any given building will have a low demand rate for various times of the day. Leaky faucets or someone getting a glass of water in the middle of the night are factors which prevent this low demand
period from being a no demand period. It is not often that a system will have periods of zero demand.
The estimated low demand GPM should be multiplied by the minimum number of minutes you want your booster to stay off on no-flow shutdown to determine draw down volume of the tank. Due to the time
delays built into most no-flow shutdown circuits, three minutes is generally the minimum off time considered. Typically, the maximum amount of time is 30 minutes. The longer the unit is off the more
energy we save but the larger our tank must be. Therefore, a compromise must be made between tank size and minimum shutdown time.
The tank size is not equal to the amount of water which can actually be drawn from the tank. The usable volume of the tank is dependent upon the normal system pressure, minimum allowable system
pressure and the drawdown coefficient of the tank. This drawdown coefficient can be obtained from the tank manufacturer’s published data.
There are several places where a hydropneumatic tank can be connected to the system. The most common connection point is to the discharge header of the booster package. Some tanks are connected just
after the discharge of the pump but before the PRV. Another fairly common location is farther out in the system, usually on the roof of the building. There are pros and cons to each location.
Locating the tank near the top of the system as shown in Figure 1 usually results in the smallest tank. It also eliminates concerns about high working pressures which can occur at the bottom of a
multi-story system. This is normally the best overall location for the tank. However, not all buildings have room for a tank on the upper floors and you must be sure to have a means of transporting
the tank through the building.
Figure 1 – Tank at Top of System
The tank can also be located at the discharge of the pressure booster package as shown in Figure 2 . In most buildings, it is considerably easier to install a tank in the equipment room than on an
upper floor which makes this location the most common. If locating a tank at the bottom of the system, it is important to make sure that the static height of the building plus the discharge pressure
of the package does not exceed the maximum allowable working pressure of the tank.
Figure 2 – Tank After PRV
For an even higher final pressure, the tank can be connected prior to the lead pump PRV (Figure 3). This is a higher pressure point because the pump TDH and suction pressure have not yet been reduced
by the PRV. This again helps us to reduce the size of the tank. If this approach is taken the tank must be connected to the discharge of the lead pump at all times.
If your booster is equipped with pump alternation and the tank’s pump is moved to the 2nd or 3rd in sequence, the tank will not charge. An uncharged tank cannot provide any draw down volume so it
will be of no use during a low flow shut down condition. Since this location will see higher pressures than either of the first two examples, it is particularly important to make sure you don’t
exceed the maximum working pressure of the tank.
Figure 3 – Tank Prior to PRV
First we must determine the tank acceptance volume. Refer to Table, below, for a guide to typical acceptance volumes for various facilities. These figures are estimates based on 30 minute shutdown
time and should be viewed accordingly.
Tank acceptance volume
Use this table for estimating purposes only. Final determination of the acceptance volume is the responsibility of the design engineer. Remember to consult local codes!
The thirty minute shut down time can be adjusted for different times by using the following formula:
ACCEPTANCE VOLUME (from Table) X DESIRED SHUTDOWN TIME / 30 MINUTES = ADJUSTED ACCEPTANCE VOLUME
Once we have determined the required acceptance volume, we can calculate the tank size based on draw down capabilities. Consult your hydropneumatic tank supplier for information on draw down volume
of their tanks. A typical data sheet is shown in Figure 5. Since different manufacturer’s tanks have different draw down capabilities, it is imperative that you use the data supplied by the
manufacturer whose tank you plan to use.
The value in the intersection of initial pressure and final pressure is your draw down coefficient. Divide your acceptance volume by this coefficient to obtain the total tank volume.
EXAMPLE #1: TANK ON ROOF
We have a pressure booster sized for 500 GPM at 75 PSIG discharge pressure with a 40 PSIG minimum suction pressure available from the city. The tank will be located on the roof of a 5 story building.
Calculate the tank size required for a 15 minute shutdown during low flow conditions and a 65 PSIG booster cut-in pressure:
• From Table above, we can see that a booster sized for 500 GPM in an apartment building to be off for 30 minutes on low flow, an acceptance volume of 75 gallons is required. However, since we only
need our booster to be off for 15 minutes, we must adjust this acceptance volume accordingly 75 x 15 / 30 = 37.5.
Therefore, our acceptance volume will be 37.5 gallons.
• Our initial pressure is equal to the pressure at the tank connection point at booster cut in pressure. This value is equal to the cut-in pressure less the static elevation of the tank above the
discharge of the booster package. We must also account for the friction loss in the piping between the package discharge and the tank connection point. In this case, we have calculated a friction
loss of 10 feet or 4.73 PSIG. The tank is located approximately 70 feet above the booster which equates to 30.3 PSIG.
65 PSIG (CUT IN) – 4.73 PSIG (FRICTION LOSS AT DESIGN FLOW) – 30.3 PSIG (STATIC HEIGHT) = 30 PSIG
• Final pressure is equal to the pressure at the tank connection point when system is fully pressurized.
75 PSIG (SYSTEM PRESSURE) – 4.73 PSIG (FRICTION LOSS AT DESIGN FLOW) – 30.3 PSIG (STATIC HEIGHT) = 40 PSIG
• Using Figure 5, we can determine that our draw down coefficient is .183.
• Divide the acceptance volume by the draw down coefficient to obtain the total tank volume that will give us 75 GPM during Low flow shutdown.
37.5 GPM / .183 = 205
Therefore, we need a minimum tank volume of 205 gallons to meet our shutdown requirements.
EXAMPLE #2: TANK AT DISCHARGE OF BOOSTER PACKAGE
We again have a pressure booster sized for 500 GPM at 75 PSIG discharge pressure with a 40 PSIG minimum suction pressure available from the city. Now, the tank will be located in the basement of a 5
story building and be connected to the discharge header of the package. Calculate the tank size required for a 15 minute shutdown during low flow conditions and a 65 PSIG booster cut-in pressure:
• From Table above, we can see that for a booster sized for 500 GPM in an apartment building to be off for 30 minutes on low flow, an acceptance volume of 75 gallons is required. However, since we
only need our booster to be off for 15 minutes, we must adjust this acceptance volume accordingly 75 x 15 / 30 = 37.5.
Therefore, our acceptance volume will be 37.5 gallons.
• Our initial pressure is equal to cut-in pressure less static height and piping losses to the tank. However, since the tank is located at the discharge of the package, static height and friction
losses are insignificant. Therefore, we can conclude that the initial pressure is actually equal to cut-in pressure.
INITIAL PRESSURE = CUT-IN PRESSURE = 65 PSIG.
• Likewise, the insignificance of static height and friction losses also apply to our calculation of final pressure. We can conclude that final pressure is equal to the pressure at the tank
connection point when the system is fully pressurized.
FINAL PRESSURE = SYSTEM PRESSURE = 75 PSIG
• Using Figure 5, we can determine that our draw down coefficient is .111.
• Divide the acceptance volume by the draw down coefficient to obtain the total tank volume that will give us 75 GPM during Low flow shutdown.
37.5 GPM / .111 = 340
Therefore, we need a minimum tank volume of 340 gallons to meet our shutdown requirements.
Using the same pressure booster as the previous two examples, sized for 500 GPM at 75 PSIG discharge pressure with a 40 PSIG minimum suction pressure available from the city. The tank will be located
in the basement as in example #2 but will be connected before the pressure reducing valve. Calculate the tank size required for a 15 minute shutdown during low flow conditions and a 65 PSIG booster
cut-in pressure:
• From Table above, we can see that a booster sized for 500 GPM in an apartment building to be off for 30 minutes on low flow, an acceptance volume of 75 gallons is required. However, since we only
need our booster to be off for 15 minutes, we must adjust this acceptance volume accordingly.
75 x 15 / 30 = 37.5
Therefore, our acceptance volume will be 37.5 gallons.
• Our initial pressure is still going to be equal to cut-in pressure as in Example 2. We do not need to be concerned with static height and friction loss since the tank will be located adjacent to
the pumps. Therefore, our initial pressure will be equal to cut-in pressure.
INITIAL PRESSURE = CUT-IN PRESSURE = 65 PSIG.
Final pressure is going to be significantly higher than in example #2 because our tank is connected to the system prior to the pressure reducing valve. Therefore, we actually have pump TDH at minimal
flow plus minimum suction pressure. If our pump has a flow vs. Head curve as shown below in Figure 4, the final pressure is going to be 155. (TDH @) 0 GPM) plus minimum suction pressure of 40 PSIG.
Therefore, our final pressure can be calculated by adding these values.
67 PSIG (PUMP TDH @ 0 GPM) + 40 PSIG (MIN. SUCTION PRESSURE) = 107 PSIG
Figure 4 – Pump Curve (250 GPM @ 120’)
• Using Figure 5, we can determine that our draw down coefficient is .335
• Divide the acceptance volume by the draw down coefficient to obtain the total tank volume that will give us 75 GPM during low flow shutdown.
37.5 GPM / .335 = 1 12
Therefore, we need a minimum tank volume of 112 gallons to meet our shutdown requirements.
Most hydropneumatic tanks ship from the manufacturer pre-charged to a pressure that is usually well below the actual charging requirement for the system. In other words, the air volume in the tank is
too small once the tank is installed in the system. Pumps short cycle, draw down is limited, and in some cases, the situation is so severe that the tank could be removed from the system and nobody
would know the difference. So, if we’re going to spend the money for a tank, let’s make sure it works by charging it correctly.
The correct tank pre-charge pressure depends upon the following factors:
1. Minimum allowable system pressure
2. Tank elevation relative to the pressure booster package
3. Tank connection point in the system
We will define these variables as follows for our pre-charge calculation:
• Let D = Desired system pressure in PSIG (PRV setting)
• Let M = Maximum allowable pressure depression below PRV setting (D)
• Let H = Tank elevation above pressure booster in PSIG (PSIG = Feet / 2.31)
• Let P = Tank pre-charge pressure (tank empty) in PSIG
We will also estimate a 1 PSIG pressure drop across the PRV at very low flow rates encountered during a low demand period.
If the tank is located above the pressure booster as shown in Figure 1, the pre-charge is calculated like this:
P = D – M – H – 1
Tanks located approximately level to the booster and connected to the system downstream of the PRV (Figure 2) have their pre-charge pressure as follows:
P = D – M – 1
If the tank is approximately level with the booster but connected to the system prior to the PRV (Figure 3), then we do not have to subtract the 1 PSIG drop across the valve. Therefore, the
calculation is as follows:
P = D – M
To confirm the pre-charge pressure of an existing tank, the tank must be isolated from the pumping / piping system. Then the water side of the tank is drained and the air pressure read with a gauge
at the air charging valve. This reading is the precharge pressure.
By simply taking a little extra time to make sure our tank is pre-charged correctly, we can be certain that it will serve its purpose of keeping the pumps off during periods of low demand.
Let’s take a look at the roof tank described in Figure 1. We know that the correct pre-charge pressure is defined as:
P = D – M – H – 1
We know that:
• D = 75
• M = system pressure – cut-in pressure = 75 – 65 = 10
• H = 70 / 2.31 = 30.3
Therefore, our correct pre-charge pressure is:
75 – 10 – 30.3 – 1 = 33.7 PSIG
To ensure correct operation of the tank during the booster’s low-flow shutdown sequence, it must be precharged to 33.7 PSIG.
As you can see, there are few hard and fast rules to tank sizing. It is predominantly a matter of weighing various factors and compromising on a balance of initial cost and potential energy savings.
Locating the tank connection prior to the pressure reducing valve results in the smallest tank but requires its respective pump to always be the lead pump.
A tank connection at the discharge header results in a larger tank but allows you to alternate all pumps. A roof mounted tank seems like a pretty reasonable compromise but you must consider the
complications of transporting the tank to the roof. In conclusion, tank location has a significant impact on the tank size and must be addressed on a project by project basis.
Figure 5 – Hydropneumatic Tank Drawdown Volume
initial tank pressure is equal to the minimum allowable pressure of the system (at the point of the tank) where the booster system will come back on line.
Final tank pressure is equal to the maximum system discharge pressure (at the point of the tank) or, the pressure reducing valve setting if the tank is mounted on the booster system.
Actual usable gallons may vary ± 10%.
Xylem Company
What is the primary purpose of a hydropneumatic tank in a domestic water system?
The primary purpose of a hydropneumatic tank is to provide a buffer against small demands on the system when the pressure booster system is off on no-flow shutdown (NFSD). This allows the pumps to
remain off for a longer period, reducing the frequency of starts and stops, and increasing overall system efficiency.
How does the NFSD circuitry affect the operation of the hydropneumatic tank?
The NFSD circuitry turns the lead pump off when there is no demand on the system. During this time, the hydropneumatic tank satisfies small demands on the system, allowing the pumps to remain off.
Without the tank, the booster would restart upon the slightest call for flow, such as a single toilet being flushed or even a minute leak in the piping system.
What are the two primary factors that influence hydropneumatic tank sizing?
Hydropneumatic tank sizing is dependent on two factors: 1) the length of time you wish the pumps to remain off in a no-flow situation, and 2) the tank location in relation to the pressure booster.
These factors determine the required tank size and configuration to ensure optimal system performance.
How does the tank location affect hydropneumatic tank sizing?
The tank location affects the pressure losses and gains in the system, which in turn impact the required tank size. For example, a tank located closer to the pressure booster may require a smaller
size due to lower pressure losses, while a tank located farther away may require a larger size to compensate for increased pressure losses.
What is the relationship between tank size and pump shutdown time?
The tank size and pump shutdown time are directly related. A larger tank allows the pumps to remain off for a longer period, as it can satisfy more demands on the system before the pressure drops
below the restart threshold. Conversely, a smaller tank requires more frequent pump starts and stops, which can reduce system efficiency and increase wear and tear on the equipment.
How can I determine the optimal tank size for my specific application?
To determine the optimal tank size, you need to consider factors such as the maximum demand on the system, the desired pump shutdown time, and the system’s pressure profile. You can use calculations
and simulations to determine the required tank size, or consult with a qualified engineer or manufacturer’s representative for guidance.
What are the consequences of undersizing or oversizing a hydropneumatic tank?
Undersizing a hydropneumatic tank can lead to frequent pump starts and stops, reduced system efficiency, and increased wear and tear on the equipment. Oversizing the tank can result in higher upfront
costs, increased space requirements, and potentially reduced system performance due to increased pressure losses. It is essential to accurately determine the required tank size to ensure optimal
system performance and efficiency. | {"url":"https://hvac-eng.com/hydropneumatic-tank-sizing/","timestamp":"2024-11-09T00:10:08Z","content_type":"text/html","content_length":"244151","record_id":"<urn:uuid:da2d5fe4-28e2-4466-9aa1-d2fff4d99ce1>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00828.warc.gz"} |
mass distribution Latest Research Papers | ScienceGate
Purpose. To investigate the features of the algorithm implementation for finding the derivatives of the spatial distribution function of the planet's masses with the use of high-order Stokes
constants and, on the basis of this, to find its analytical expression. According to the given methodology, to carry out calculations with the help of which to carry on the study of dynamic phenomena
occurring inside an ellipsoidal planet. The proposed method involves the determination of the derivatives of the mass distribution function by the sum, the coefficients of which are obtained from the
system of equations, which is incorrect. In order to solve it, an error-resistant method for calculating unknowns was used. The implementation of the construction is carried out in an iterative way,
while for the initial approximation we take the three-dimensional function of the density of the Earth's masses, built according to Stokes constants up to the second order inclusive, by dynamic
compression by the one-dimensional density distribution, and we determine the expansion coefficients of the derivatives of the function in the variables to the third order inclusive. They are
followed by the corresponding density function, which is then taken as the initial one. The process is repeated until the specified order of approximation is reached. To obtain a stable result, we
use the Cesaro summation method (method of means).. The calculations performed with the help of programs that implement the given algorithm, while the achieved high (ninth) order of obtaining the
terms of the sum of calculations. The studies of the convergence of the sum of the series have been carried out, and on this basis, a conclusion has been made about the advisability of using the
generalized finding of the sums based on the Cesaro method. The optimal number of contents of the sum terms has been chosen, provides convergence both for the mass distribution function and for its
derivatives. Calculations of the deviations of mass distribution from the mean value ("inhomogeneities") for extreme points of the earth's geoid, which basically show the total compensation along the
radius of the Earth, have been performed. For such three-dimensional distributions, calculations were performed and schematic maps were constructed according to the taken into account values of
deviations of three-dimensional distributions of the mean ("inhomogeneities") at different depths reflecting the general structure of the Earth's internal structure. The presented vector diagrams of
the horizontal components of the density gradient at characteristic depths (2891 km - core-mantle, 700 km - middle of the mantle, also the upper mantle - 200, 100 km) allow us to draw preliminary
conclusions about the global movement of masses. At the same time, a closed loop is observed on the “core-mantle” edge, which is an analogy of a closed electric circuit. For shallower depths,
differentiation of vector motions is already taking place, which gives hope for attracting these vector-grams to the study of dynamic motions inside the Earth. In fact, the vertical component
(derivative with respect to the z variable) is directed towards the center of mass and confirms the main property of mass distributions - growth when approaching the center of mass. The method of
stable solution of incorrect linear systems is applied, by means of which the vector-gram of the gradient of the mass distribution function is constructed. The nature of such schemes provides a tool
for possible causes of mass redistribution in the middle of the planet and to identify possible factors of tectonic processes in the middle of the Earth, i.e indirectly confirms the gravitational
convection of masses. The proposed technique can be used to create detailed models of density functions and its characteristics (derivatives) of the planet's interior, and the results of numerical
experiments - to solve tectonics problems. | {"url":"https://www.sciencegate.app/keyword/10245","timestamp":"2024-11-04T05:33:22Z","content_type":"text/html","content_length":"109983","record_id":"<urn:uuid:a12156b7-0487-4e02-bf19-db4e7eff0e6e>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00478.warc.gz"} |
GPTs Hunter - Math Mastermind
Math Mastermind on the GPT Store
Use Math Mastermind on ChatGPT Use Math Mastermind on 302.AI
GPT Description
Math and geometry tutor for all levels, offering study paths and problem-solving guidance.
GPT Prompt Starters
• How do I solve this geometry problem?
• Can you explain this math concept?
• What's the best way to study for a math test?
• Help me understand this mathematician's theory.
Use Math Mastermind on 302.AI Become a verified curator to get priority display for your GPTs
Math Mastermind GPT FAQs
Currently, access to this GPT requires a ChatGPT Plus subscription.
Visit the largest GPT directory GPTsHunter.com, search to find the current GPT: "Math Mastermind", click the button on the GPT detail page to navigate to the GPT Store. Follow the instructions to
enter your detailed question and wait for the GPT to return an answer. Enjoy!
We are currently calculating its ranking on the GPT Store. Please check back later for updates.
More custom GPTs by Marco Moscatelli on the GPT Store | {"url":"https://www.gptshunter.com/gpt-store/MGIwYzM2Mzc1NjA1MTY1MTJl","timestamp":"2024-11-08T02:36:59Z","content_type":"text/html","content_length":"66484","record_id":"<urn:uuid:16abf70a-6bbc-4070-9a2d-aedf44d7a0d7>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00702.warc.gz"} |
Modified Internal Rate of Return - MIRR - Finince Assignment Help
MIRR Calculation Help
MIRR is a short form for modified internal rate of return. It denotes a Capital Budgeting technique that is used to assess investment returns, compare different investments and the rate of
As the name suggests, this investment measurement technique is derived from the normal internal rate of return but it has been modified to improve efficiency.
In the past, investors and accounting professionals have associated IRR with confusion and ambiguity. For instance, IRR assumes that business cash flows are reinvested at the same fraction at which
they were made.
As for MIRR, it provides that net cash inflows ought to be reinvested at a specified cost of capital. The evaluation of variance between investment rate and re-investment rate makes MIRR an
attractive financial measure.
Looking for someone to do my Finance homework for me?
Do you need need help with Finance Homework? We are here to help you improve your grades in school and increase the chances of succeeding in life. We understand that not everybody is good in class
and that is why you need to consider hiring and expert to help you get better grades for your assignments.
Advantages and Disadvantages of MIRR
By calculating MIRR, individuals successfully eliminate limitations associated with IRR. For example, multiple IRR outcomes bring confusion when cash flows are uneven.
Unlike traditional IRR, the modified internal rate of return helps individuals to calculate project sensitivity since it measures the variation between the cost of capital and financing cost.
The major drawback of MIRR is that it requires people to multiple decisions concerning the cost of capital and financing rate.
MIRR Calculation Example
Using a specimen company, Drink Incorporation, we can see how MIRR can be used to compare potential investments.
Drink Incorporation has plans to invest in Investment X and Investment Y, which will both run for 3 years (2013 to 2015). The cost of capital for X and Y is 12% and 15% respectively. Since the
company does not have ready capital, it will incur 14% financing cost for Investment X and 18% for Investment Y. The cash flows for the 3 years are provided in the schedule below:
Details Investment X ($) Investment Y ($)
2012 -2,000 -1,200
2013 -4,000 -900
2014 -6,000 3,000
2015 -7,000 2,500
The generally acceptable MIRR function is given below.
MIRR= (FV of Positive Cash Flows / PV of Negative Cash Flows) * (1 / n) – 1
Step 1: Drink Inc. should calculate future value (FV) of the positive net inflows as follows (should be discounted at cost of capital):
a) Investment X: 6,000 * (1 + 12%) 1 + 7,000 = $13,720
b) Investment Y: 3,000* (1 + 15%) 1 + 2,500 = $5,950
Step 2: Calculate present value of negative net inflows (should be discounted at financing cost)
a) Investment X: -2,000 + (-4,000) / (1+14%) 1 = -$5,263
b) Investment Y: -1,200 + (-900) / (1+18%) 1 = -$1,780
Step 3: With these outcomes, MIRR for both projects would be:
a) Investment X: (13,720 / -5,263) (1/3) – 1 = 13.1%
b) Investment Y: (5,950 / -1,780) (1/3) – 1 = 11.4%
Step 4: Based on the MIRR calculated for both projects, Investment X should be implemented because it has a greater MIRR as compared to Investment Y.
MIRR is a comprehensive investment measurement method. It is widely used to assess projects with a mix of negative and positive cash flows.
It is better than IRR because it helps individuals in making concrete investment decisions. The outcomes generated are clear, which means people would not hesitate using it in real life. The
application of this method can be done using four main steps as discussed above.
Individuals should compound positive cash flows, and then, discount all negative net cash inflows. However, despite MIRR attractiveness people view it as complex due to steps involved as well as
multiple estimations that should be made.
FAQs on MIRR
What is the difference between IRR and MIRR?
MIRR can be used to assess projects with inconsistent cash flows while IRR cannot.
How do you calculate MIRR?
There are four steps involved in the calculation of MIRR, but the main ones include compounding positive cash flows and discounting negative cash flows.
What is the MIRR formula?
The general MIRR formula is shown below:
MIRR= [(FV of Positive Cash Flows / PV of Negative Cash Flows) * (1 / n) – 1]
What is the project’s MIRR?
A project MIRR is the rate at which cash flows generated by investment is reinvested.
What is the reinvestment rate in MIRR?
The reinvestment rate refers to the percentage of net cash inflows that should be absorbed by the project while the remaining portion of cash flow is invested elsewhere. | {"url":"https://toppaperarchives.com/finance-assignment-help/modified-internal-rate-of-return/","timestamp":"2024-11-13T17:50:09Z","content_type":"text/html","content_length":"182106","record_id":"<urn:uuid:566e6b14-fda7-4fea-b97f-3e4df2f2d02a>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00045.warc.gz"} |
Science:Math Exam Resources/Courses/MATH102/December 2015/Question 16
MATH102 December 2015
• Q1 • Q2 • Q3 • Q4 • Q5 • Q6 • Q7 • Q8 • Q9 • Q10 (a) • Q10 (b) • Q11 • Q12 • Q13 • Q14 (a) • Q14 (b) • Q15 • Q16 • Q17 • Q18 (a) • Q18 (b) • Q18 (c) • Q18 (d) • Q19 (a) • Q19 (b) • Q19 (c) • Q19
(d) • Q19 (e) •
Question 16
A squirrel sitting 6 m up in a tree is watching a coyote walk past the tree. The squirrel measures the angle formed between a vertical line directly below her and the line connecting her and the
coyote and finds that it is changing at a rate of 1/12 radians per second when the coyote is 8 m away from the base of the tree. How fast is the coyote walking?
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is
correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Denote the distance of the coyote from the tree by ${\displaystyle x}$ and the angle between the tree and the line connecting the squirrel and the coyote by ${\displaystyle \theta }$.
Then, find the relations between them. (It would be helpful to draw a diagram describing the statements in the question)
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
• If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you
are stuck or if you want to check your work.
• If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem
and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies.
Let ${\displaystyle x=x(t)}$ be the distance from the coyote to the tree and ${\displaystyle \theta =\theta (t)}$ be the angle between the tree and the line connecting the squirrel and the coyote.
Then, since the height of the tree is 6, these variables are related to each other by ${\displaystyle \tan \theta ={\frac {x}{6}}}$.
From the question, the rate of change of ${\displaystyle \theta }$ at ${\displaystyle x=8}$ is given by ${\displaystyle \theta '|_{x=8}={\frac {1}{12}}{\text{rad/sec}}}$.
On the other hand, taking the derivative on both sides of the equation gives${\displaystyle \tan \theta ={\frac {x}{6}}}$, we have
{\displaystyle {\begin{aligned}&(\sec ^{2}\theta )\theta '=(\tan \theta )'=({\frac {x}{6}})'={\frac {1}{6}}x'\implies \\&x'=6(\sec ^{2}\theta |_{x=8})\theta '|_{x=8}=6\cdot (1+\tan ^{2}\theta |_{x=
8})\cdot {\frac {1}{12}}=6\cdot \left(1+\left({\frac {4}{3}}\right)^{2}\right)\cdot {\frac {1}{12}}={\frac {25}{18}}\end{aligned}}}
Therefore, the coyote is walking at the speed ${\displaystyle \color {blue}{\frac {25}{18}}m/sec}$. | {"url":"https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH102/December_2015/Question_16","timestamp":"2024-11-07T12:21:31Z","content_type":"text/html","content_length":"53612","record_id":"<urn:uuid:37ad1145-aeef-4c0b-b4d4-62166134c899>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00116.warc.gz"} |
How Not to Be Wrong | Actionable Books
"The mathematical ideas we want to address are ones that can be engaged with directly and profitably, whether your mathematical training stops at pre-algebra or extends much further."
"A mathematician is always asking ‘what assumptions are you making? Are they justified?’ This can be annoying. But it can also be very productive.`"- How Not to Be Wrong, page 7
Through all the examples, studies, theories, quotes, and lessons graciously shared by Ellenberg, the most fundamental point being driven home is around being a better critical thinker. The book is a
continuum of examples that bake mathematical theory into everyday assumptions like headlines, studies, myths, gambling, religion (yup he goes there!) and everything in between. Within every story the
protagonist is the principle, which for the most part are easily understood, who wins every time because they went just that much further to think about it a new way, ask a better question, or
clarify what it is we are really getting at here. In the age of information overload, big data, and the reliance on computers and one of my closest friends, Excel, it is easy to forget what it is we
are saying or understanding and whether it actually makes sense. The Insights below are some of my personal favourites that demonstrate how digging into what assumptions are we making—and if they are
justified—can provide you an outcome you may otherwise not have considered.
"Human beings are quick to perceive patterns where they don’t exist and tend to overestimate their strength where they do."- How Not to Be Wrong, page 129
At the time of this summary the Toronto Raptors are tied in the second round of the NBA playoffs against the Miami Heat. Our nemesis seems to be Dwayne Wade, who most fans would agree in this series
appears to not be able to miss a shot. Naturally my interest was piqued when the author dedicated a few pages to dispelling this myth. There are a number of mathematical principles at play here. One
being in statistical studies, the phenomenon of an expected size can be underpowered. Think looking at the planets with binoculars instead of a telescope. This little theory reminds us that if we do
some statistical sampling around what a hot hand might look like with some real data, there’s a chance that even if this “hot hand” is happening our method may not even allow us to detect it. There
are multiple studies referenced that show the progression and refinement of disproving the hot hand myth. Not to leave you hanging, but essentially a 2009 study asserts that players who made a basket
are more likely to take a more difficult shot next. Your basketballer friends will argue that they’ve seen this myth happen though! Wade can’t miss! He can, in fact he only made 13 of 24 field goal
attempts in the last game. Stretch that over his 11 years in Miami and statistically speaking it will be challenging to find “hot hands”. So what are we taking away from this? The example here is a
great reminder about how easily our assumptions can mislead us. You don’t even need to know the statistical theories mentioned, but have you considered the assumptions you are making? Are they
Sign up for the very best book summaries right to your inbox.
"It’s the epidemiological equivalent of saying there are -4 grams of water left in the bucket. Zero credit."- How Not to Be Wrong, page 61
This last insight was a humbling read for me. The author builds into the concept of linear regression (think any simple line graph trending in a certain direction). He uses an example of a study out
of the US that claims that by 2048 all Americans will be obese. Some of us have probably seen this or similar headlines through many different media outlets. Now let’s consider our assumptions and if
they are justified. Continue this trend and by 2060 a cool 109% of Americans will be obese. Wait, what? At some point your proportion needs to actually start to bend before 100%. Why? Because when
you really think about it (mathematician or not)–is it actually a good assumption to say that 100% of the United States will be obese? The paper further dives deeper into the population statistics by
sex and gender and informs us that 100% of black men will be obese in 2095. Double take. How does that happen when 100% of the population is already obese by 2048? The paper doesn’t acknowledge the
mathematical contradiction and we are now left feeling like we’ve been let down by our statistical best friend (and you have). I am not suggesting (nor is the author) that linear regression is bad
(phew), but let’s use this as a friendly reminder about the stories we plan to tell with it. What are you assuming? Is it justified?
How Not to be Wrong is a great tool for anyone who wants to become a better critical thinker, or perhaps has a staff member that could improve in this area. I love math and numbers and the legitimacy
that it brings to a story. The author reminds us as good critical thinkers we need to be employing this method of reasoning to some degree or another. Perhaps even more importantly, consider how we
are building this type of thinking within young people or our teams’ today–are we teaching them to consider their assumptions and if they are justified?
Jordan Ellenberg is a professor of mathematics at the University of Wisconsin-Madison, with a Ph.D. in mathematics from Harvard and an MFA in creative writing from Johns Hopkins. His areas of
research specialization are number theory and algebraic geometry. He has written articles on mathematical topics in the New York Times, the Washington Post, Wired, the Wall Street Journal, the Boston
Globe, and the Believer, and is a regular columnist for Slate. | {"url":"https://www.actionablebooks.com/summaries/how-not-to-be-wrong","timestamp":"2024-11-07T19:26:00Z","content_type":"text/html","content_length":"31027","record_id":"<urn:uuid:91f4dff1-70ec-4894-a1fd-36935181856d>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00130.warc.gz"} |
15,392 research outputs found
The Riemannian geometry of coset spaces is reviewed, with emphasis on its applications to supergravity and M-theory compactifications. Formulae for the connection and curvature of rescaled coset
manifolds are generalized to the case of nondiagonal Killing metrics. The example of the N^{010} spaces is discussed in detail. These are a subclass of the coset manifolds N^{pqr}=G/H = SU(3) x U(1)/
U(1) x U(1), the integers p,q,r characterizing the embedding of H in G. We study the realization of N^{010} as G/H=SU(3) x SU(2)/U(1) x SU(2) (with diagonal embedding of the SU(2) \in H into G). For
a particular G-symmetric rescaling there exist three Killing spinors, implying N=3 supersymmetry in the AdS_4 \times N^{010} compactification of D=11 supergravity. This rescaled N^{010} space is of
particular interest for the AdS_4/CFT_3 correspondence, and its SU(3) x SU(2) isometric realization is essential for the OSp(4|3) classification of the Kaluza-Klein modes.Comment: 12 page
We review the group-geometric approach to supergravity theories, in the perspective of recent developments and applications. Usual diffeomorphisms, gauge symmetries and supersymmetries are unified as
superdiffeomorphisms in a supergroup manifold. Integration on supermanifolds is briefly revisited, and used as a tool to provide a bridge between component and superspace actions. As an illustration
of the constructive techniques, the cases of $d=3,4$ off-shell supergravities and $d=5$ Chern-Simons supergravity are discussed in detail. A cursory account of $d=10+2$ supergravity is also included.
We recall a covariant canonical formalism, well adapted to theories described by Lagrangians $d$-forms, that allows to define a form hamiltonian and to recast constrained hamiltonian systems in a
covariant form language. Finally, group geometry and properties of spinors and gamma matrices in $d=s+t$ dimensions are summarized in Appendices.Comment: LaTeX, 65 pages, 2 Tables, 1 figure. v2:
included Figure missing in v1, ref.s added. v3: added missing term in eq. (9.3). v4: eq. (9.41) corrected. Matches published version. v5: added missing terms in eq.s (7.21), (7.24), (7.27), (9.38),
added a paragraph in Sect. 11, added re
We derive the invariant operators of the zero-form, the one-form, the two-form and the spinor from which the mass spectrum of Kaluza Klein of eleven-dimensional supergravity on AdS_4 x N^{010} can be
derived by means of harmonic analysis. We calculate their eigenvalues for all representations of SU(3)xSO(3). We show that the information contained in these operators is sufficient to reconstruct
the complete N=3 supersymmetry content of the compactified theory. We find the N=3 massless graviton multiplet, the Betti multiplet and the SU(3) Killing vector multiplet.Comment: 1+50 pages, LaTe
In this paper, relying on previous results of one of us on harmonic analysis, we derive the complete spectrum of Osp(3|4) X SU(3) multiplets that one obtains compactifying D=11 supergravity on the
unique homogeneous space N^{0,1,0} that has a tri-sasakian structure, namely leads to N=3 supersymmetry both in the four-dimensional bulk and on the three-dimensional boundary. As in previously
analyzed cases the knowledge of the Kaluza Klein spectrum, together with general information on the geometric structure of the compact manifold is an essential ingredient to guess and construct the
corresponding superconformal field theory. This is work in progress. As a bonus of our analysis we derive and present the explicit structure of all unitary irreducible representations of the
superalgebra Osp(3|4) with maximal spin content s_{max}>=2.Comment: Latex2e, 13+1 page
The gauging of the q-Poincar\'e algebra of ref. hep-th 9312179 yields a non-commutative generalization of the Einstein-Cartan lagrangian. We prove its invariance under local q-Lorentz rotations and,
up to a total derivative, under q-diffeomorphisms. The variations of the fields are given by their q-Lie derivative, in analogy with the q=1 case. The algebra of q-Lie derivatives is shown to close
with field dependent structure functions. The equations of motion are found, generalizing the Einstein equations and the zero-torsion condition.Comment: 12 pp., LaTeX, DFTT-01/94 (extra blank lines
introduced by mailer, corrupting LaTeX syntax, have been hopefully eliminated
We present a noncommutative (NC) version of the action for vielbein gravity coupled to gauge fields. Noncommutativity is encoded in a twisted star product between forms, with a set of commuting
background vector fields defining the (abelian) twist. A first order action for the gauge fields avoids the use of the Hodge dual. The NC action is invariant under diffeomorphisms and twisted gauge
transformations. The Seiberg-Witten map, adapted to our geometric setting and generalized for an arbitrary abelian twist, allows to re-express the NC action in terms of classical fields: the result
is a deformed action, invariant under diffeomorphisms and usual gauge transformations. This deformed action is a particular higher derivative extension of the Einstein-Hilbert action coupled to
Yang-Mills fields, and to the background vector fields defining the twist. Here noncommutativity of the original NC action dictates the precise form of this extension. We explicitly compute the first
order correction in the NC parameter of the deformed action, and find that it is proportional to cubic products of the gauge field strength and to the symmetric anomaly tensor D_{IJK}.Comment: 18
pages, LaTe
We present an action for $N=1$ supergravity in $10+2$ dimensions, containing the gauge fields of the $OSp(1|64)$ superalgebra, i.e. one-forms $B^{(n)}$ with $n$=1,2,5,6,9,10 antisymmetric D=12
Lorentz indices and a Majorana gravitino $\psi$. The vielbein and spin connection correspond to $B^{(1)}$ and $B^{(2)}$ respectively. The action is not gauge invariant under the full $OSp(1|64)$
superalgebra, but only under a subalgebra ${\tilde F}$ (containing the $F$ algebra $OSp(1|32)$), whose gauge fields are $B^{(2)}$, $B^{(6)}$, $B^{(10)}$ and the Weyl projected Majorana gravitino ${1
\over 2} (1+\Gamma_{13}) \psi$. Supersymmetry transformations are therefore generated by a Majorana-Weyl supercharge and, being part of a gauge superalgebra, close off-shell. The action is simply $\
int STr ({\bf R}^6 {\bf \Gamma})$ where ${\bf R}$ is the $OSp(1|64)$ curvature supermatrix two-form, and ${\bf \Gamma}$ is a constant supermatrix involving $\Gamma_{13}$ and breaking $OSp(1|64)$ to
its ${\tilde F}$ subalgebra. The action includes the usual Einstein-Hilbert term.Comment: LaTeX, 13 pages. Added a reference, a Table in Appendix A for the gamma commutations in d=12, and corrected
eq. (4.14) for the Einstein-Hilbert term; v4: corrected formulas (A.3), (A.4) and (A.10), modified last paragraph of Section 5, added acknowledgement
A brief review of bicovariant differential calculi on finite groups is given, with some new developments on diffeomorphisms and integration. We illustrate the general theory with the example of the
nonabelian finite group S_3.Comment: LaTeX, 16 pages, 1 figur
In recent years, a change in attitude in particle physics has led to our understanding current quantum field theories as effective field theories (EFTs). The present paper is concerned with the
significance of this EFT approach, especially from the viewpoint of the debate on reductionism in science. In particular, it is a purpose of this paper to clarify how EFTs may provide an interesting
case-study in current philosophical discussion on reduction, emergence and inter-level relationships in general.Comment: 18 pages, LateX2e, to appear in Studies in History and Philosophy of Modern | {"url":"https://core.ac.uk/search/?q=authors%3A(Castellani)","timestamp":"2024-11-03T16:32:18Z","content_type":"text/html","content_length":"194310","record_id":"<urn:uuid:067fd804-a4a6-4060-b185-c9aa6a8fe833>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00438.warc.gz"} |
The Stacks project
Lemma 67.48.10. Let $S$ be a scheme. Let $f : Y \to X$ be an integral morphism of algebraic spaces over $S$. Then the integral closure of $X$ in $Y$ is equal to $Y$.
Proof. By Lemma 67.45.7 this is a special case of Lemma 67.48.9. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder,
this is tag 0825. Beware of the difference between the letter 'O' and the digit '0'.
The tag you filled in for the captcha is wrong. You need to write 0825, in case you are confused. | {"url":"https://stacks.math.columbia.edu/tag/0825","timestamp":"2024-11-12T13:00:54Z","content_type":"text/html","content_length":"14139","record_id":"<urn:uuid:6346fe8e-443e-4afa-a627-f7cdf9908138>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00542.warc.gz"} |
How do you find the equation of a line tangent to the function y=x^3-6x+1 at x=2? | HIX Tutor
How do you find the equation of a line tangent to the function #y=x^3-6x+1# at x=2?
Answer 1
We have
$y = {x}^{3} - 6 x + 1$
The gradient of the tangent at any particular point is given by the derivative: So differntiating wrt $x$ we have,
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 6$
When $x = 2 \implies y = 8 - 12 + 1 = - 3$
And, $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(3\right) \left(4\right) - 6 = 6$
Sign up to view the whole answer
By signing up, you agree to our Terms of Service and Privacy Policy
Answer 2
To find the equation of a line tangent to a function at a specific point, you can follow these steps:
1. Find the derivative of the function.
2. Substitute the given x-coordinate into the derivative to find the slope of the tangent line.
3. Use the point-slope form of a line, y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope, to write the equation of the tangent line.
For the function y = x^3 - 6x + 1, the derivative is dy/dx = 3x^2 - 6.
Substituting x = 2 into the derivative, we get dy/dx = 3(2)^2 - 6 = 6.
The slope of the tangent line at x = 2 is 6.
Using the point-slope form with the point (2, f(2)), where f(2) is the value of the function at x = 2, we can write the equation of the tangent line.
Substituting x = 2 into the original function, we get f(2) = (2)^3 - 6(2) + 1 = 1.
Therefore, the point on the tangent line is (2, 1).
Using the point-slope form, the equation of the tangent line is y - 1 = 6(x - 2).
Sign up to view the whole answer
By signing up, you agree to our Terms of Service and Privacy Policy
Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
Not the question you need?
HIX Tutor
Solve ANY homework problem with a smart AI
• 98% accuracy study help
• Covers math, physics, chemistry, biology, and more
• Step-by-step, in-depth guides
• Readily available 24/7 | {"url":"https://tutor.hix.ai/question/how-do-you-find-the-equation-of-a-line-tangent-to-the-function-y-x-3-6x-1-at-x-2-8f9af9d313","timestamp":"2024-11-05T19:57:51Z","content_type":"text/html","content_length":"573216","record_id":"<urn:uuid:98b6aaea-3b0f-4a4d-9f56-ffe74f6339a3>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00753.warc.gz"} |
Membership and inclusions
Membership (possessive relations)
If \(x\) is an element in a set \(A\) we write \[ x \in A \quad \text{ or }\quad A \ni x, \] and if not \[ x \notin A \quad\text{ or }\quad A \not\ni x. \]
• The rationals can be constructed from elements in \(\mathbb Z\) and \(\mathbb N\):
\[ \mathbb Q = \{ a/b \colon a \in \mathbb Z, b \in \mathbb N\} \]
• \(\sqrt{2}\) is a real number, but not a rational one:
\[ \sqrt{2} \in \mathbb{R}, \qquad \sqrt{2} \notin \mathbb{Q}. \]
Quantifiers are used to abbreviate notation. The most important ones are:
• \( \forall \quad \) Universal quantifier: 'For any','for all'
• \( \exists \quad \) Existential quantifier: 'There exists'
• \( ! \quad \) Uniqueness quantifier: 'a unique'
Ex. For any real number \(x\) there exists a unique real number \(-x\) with the property that the sum of \(x\) and \(-x\) is zero: \[ \forall\: x \in \mathbb{R} \qquad \exists ! \:\; (-x) \in \mathbb
{R}; \qquad x + (-x) = 0.\]
A set \(A\) is a subset of a set \(B\) if any element in \(A\) is also an element in \(B\): \[ A \subset B \quad (\text{or } A \subseteq B) \quad\stackrel{\text{def.}}{\Longleftrightarrow}\quad [x \
in A \Rightarrow x \in B] \] A subset \(A \subset B\) can also be a proper subset of \(B\): \[ A \subsetneq B \quad\stackrel{\text{def.}}{\Longleftrightarrow}\quad A \subset B \text{ but } A \neq B.
• The natural numbers is a subset of the set of non-negative integers, which is a proper subset of the set of real numbers:
\[ \mathbb{N} \subset \{0,1,2,\ldots\} \subsetneq \mathbb{R}. \]
• The empty set is a subset of any other set (including itself):
\[ \emptyset \subset \emptyset \subset \{1,2,3\}. \]
• The continuously differentiable real-valued functions on the real line is a subset of the continuous functions:
\[ C^1(\mathbb R, \mathbb R) \subset C(\mathbb R, \mathbb R). \] | {"url":"https://wiki.math.ntnu.no/linearmethods/sets/membership","timestamp":"2024-11-06T02:35:51Z","content_type":"text/html","content_length":"15951","record_id":"<urn:uuid:58302149-4d8e-459a-9a94-f3d739977714>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00661.warc.gz"} |
Idea 077 - Equilateral triangles LED panels
This post is part of the 100 project ideas project. #The100DayProject. I am looking for feedback. Comment below or DM me via social media Instagram, Twitter.
👉 Done! See Dodecahedron-PCB on Github, or Dodecahedron PCB Design, Dodecahedron PCB Retrospective posts for more informaiton.
One Line Pitch
Create equilateral triangles LED panels to construct LED polyhedron shapes
Instead of a cube with square faces. I was thinking of making a Equilateral Triangles PCB with an array of LEDs on it. Two equilateral triangles make a square that can be used to make a LED cube.
Five equilateral triangles makes a pentagon.
Using equilateral triangles I should be able to make all of the regular polyhedron
With a little more effort the Kepler–Poinsot polyhedra could also be created using equilateral triangles. They would be physically large and have many faces but it could be done.
Catalan_solid are also interesting for this idea because they are face-transitive (all of the faces of the object have the same shape) but the shapes aren’t as reckonable as regular polyhedron by
normal people.
Wiring for the equilateral triangles to allow for Data, Power and GND wires to come off every face and connect to each other face will be tricky. Power at the corners that can all be soldered
together. The GND can be along the edge of each triangle and be a large enough pad that can be used to solder the different triangles together to form the shape.
The data lines are a bit more complicated as they are one directional. Having a solder jumper that can be connected or disconnected as needed for each of the faces They should be in the center of
each edge of the triangle. Each triangle will have at least one data line on an edge that is not connected. Maybe using a default 0 ohm that connects the data line on each face as a default would
make it easier to remove the resistor if needed.
A few angled jigs could be created to help with the soldering of angles. Another jig could be created for charging that connects the 5v and the GND to the edge of the objects.
The main processor and the battery could be floating in the center of the object or tapped to one of the inner faces of the object. The main processor should have a accelerometer and a gyro in it so
it can detect the angle that the object is at.
These objects could also be attached to wires and hung from a mobile in the ceiling. Very similar to Idea 10 - Polyhedron Papercraft Mobile
Prior art | {"url":"https://blog.abluestar.com/idea077-equilateral-triangles-led-panels/","timestamp":"2024-11-03T16:31:50Z","content_type":"text/html","content_length":"28208","record_id":"<urn:uuid:18404d44-ff2a-4dc1-9323-69c1a354f483>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00355.warc.gz"} |
Safe Haskell None
Language Haskell2010
Generate Elm type definitions, encoders and decoders from Haskell data types.
include :: ToHType a => Proxy a -> GenOption -> Builder Source #
Include the elm source for the Haskell type specified by the proxy argument. The second argument decides which components will be included and if the generated type will be polymorphic.
generateFor :: ElmVersion -> Options -> Maybe FilePath -> Builder -> Q Exp Source #
Return the generated Elm code in a template haskell splice and optionally write to a Elm source file at the same time. The second argument is the Options type from Aeson library. Use include calls to
build the Builder value.
data HType Source #
This type holds the type information we get from generics. Only the HExternal constructor is supposed to be used by the programmer to implement ToHType instances for entites that are predefined in
Elm. A sample can be seen below.
Here, let `MyExtType a b` be a type which has the corresponding type, encoders and decoders predefined in Elm in a module named Lib. Here is how you can implement a ToHType instance for this type so
that your other autogenerated types can have fields of type `MyExtType a b`.
instance (ToHType a, ToHType b) => ToHType (MyExtType a b) where
toHType _ = do
ha <- toHType (Proxy :: Proxy a)
hb <- toHType (Proxy :: Proxy b)
pure $
(External, MyExtType)
(Just (External, "encodeMyExtType"))
(Just (External, "decodeMyExtType"))
[ha, hb])
HUDef UDefData
HMaybe HType
HList HType
HPrimitive MData
HRecursive MData
HExternal (ExInfo HType)
Show HType Source #
Defined in Elminator.Generics.Simple
class ToHType f where Source #
ToHType () Source #
Defined in Elminator.Generics.Simple
Typeable a => ToHType a Source #
Defined in Elminator.Generics.Simple
ToHType Text Source #
Defined in Elminator.Generics.Simple
ToHType a => ToHType [a] Source #
Defined in Elminator.Generics.Simple
ToHType a => ToHType (Maybe a) Source #
Defined in Elminator.Generics.Simple
(ToHType a, ToHType b) => ToHType (Either a b) Source #
Defined in Elminator.Generics.Simple
(ToHType a1, ToHType a2) => ToHType (a1, a2) Source #
Defined in Elminator.Generics.Simple
(ToHType a1, ToHType a2, ToHType a3) => ToHType (a1, a2, a3) Source #
Defined in Elminator.Generics.Simple
(ToHType a1, ToHType a2, ToHType a3, ToHType a4) => ToHType (a1, a2, a3, a4) Source #
Defined in Elminator.Generics.Simple
(ToHType a1, ToHType a2, ToHType a3, ToHType a4, ToHType a5) => ToHType (a1, a2, a3, a4, a5) Source #
Defined in Elminator.Generics.Simple
(ToHType a1, ToHType a2, ToHType a3, ToHType a4, ToHType a5, ToHType a6) => ToHType (a1, a2, a3, a4, a5, a6) Source #
Defined in Elminator.Generics.Simple
(ToHType a1, ToHType a2, ToHType a3, ToHType a4, ToHType a5, ToHType a6, ToHType a7) => ToHType (a1, a2, a3, a4, a5, a6, a7) Source #
Defined in Elminator.Generics.Simple
data ExInfo a Source #
Show a => Show (ExInfo a) Source #
Defined in Elminator.Generics.Simple
data GenOption Source #
Decides which among type definiton, encoder and decoder will be included for a type. The poly config value decides wether the included type definition will be polymorphic.
Definiton PolyConfig
Everything PolyConfig
Show GenOption Source #
Defined in Elminator.Lib
data PolyConfig Source #
Decides wether the type definition will be polymorphic.
Show PolyConfig Source #
Defined in Elminator.Lib | {"url":"https://hackage.haskell.org/package/elminator-0.2.2.0/candidate/docs/Elminator.html","timestamp":"2024-11-13T11:54:49Z","content_type":"application/xhtml+xml","content_length":"38403","record_id":"<urn:uuid:fcd22772-1c33-405f-81d0-2f7f19e21b55>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00289.warc.gz"} |
Quantum Error Mitigation: “Nature Physics” Publ. by Jens Eisert’s Group
Quantum Error Mitigation: “Nature Physics” Publ. by Jens Eisert’s Group
Photo: Jens Eisert (center) with his research group © private
A recent publication in “Nature Physics” by the research group of MATH+ member Jens Eisert shows that error mitigation does not work as a scalable scheme, contrary to how it is central to IBM’s
corporate strategy. What challenges arise with the question of how quantum errors can be mitigated?
Quantum computers are a new computer type that promise efficient solutions to classically hard computational problems. Although the concept of a quantum computer is not entirely new, it is only in
recent years that significant strides have been made in building large-scale quantum computers. These machines are still noisy and not yet large, but the existence of quantum computers with over 1000
quantum bits, once inconceivable, has created an exciting state of affairs. In particular, recent months have seen announcements of architectures that have surprised the community with their level of
© Pete Linforth (AI generated) via pixabay
To combat noise, one can think of quantum error correction which comes along with substantial overheads. Currently, the common practice is to address unwanted quantum noise more directly: Adding
noise is more feasible than eliminating it, and in some cases, there is a clear understanding of the underlying noise mechanism. Methods of quantum error mitigation make use of these insights. Rather
than a single solution, this involves a portfolio of methods that use classical computation to partially reverse quantum noise in post-processing. This innovative and elegant approach is now
routinely applied in quantum experiments, effectively undoing unwanted noise. However, the question remains: How far can this approach take us?
To address the question of the potential and limitations of quantum error mitigation, this work explores the extent to which quantum error mitigation can work for all quantum circuits as a scalable
scheme for large system sizes. For this purpose, a mathematical framework is developed that is general enough to encapsulate a large number of protocols that are being used in practice. The approach
was motivated by the author’s previous work on the impact of noise on quantum computing, prompting a natural inquiry into its implications for quantum error mitigation.
The approach taken in this publication is radical, treating error mitigation as a statistical inference problem: If some of the noise can be canceled, it must be possible to discriminate certain
input states and then relate the task to hypothesis testing. One can design reasonable circuits that mix so much that this task becomes impossible unless the number of samples scales exponentially
with both the depth and the system size. This was an intricate step: The team had to find a circuit that is extremely mixing in a precise sense.
Ultimately, this means that even at log-log depth, a marginal increase beyond constant depth makes quantum error mitigation prohibitively costly. This bound is exponentially tighter than previously
known bounds.
It doesn’t mean that quantum error mitigation does not work. It just means that it is not scalable in the most extreme sense. This aligns with MATH+ vision, as it tackles a rigorous mathematical
problem closely tight to real-world technological application—specifically quantum technology.
However, like all no-go theorems, the result established here should also be seen as an invitation: Firstly, it’s a worst-case scenario. Long-ranged entanglement and quantum noise do not seem to work
well together. By using more local architectures, we might achieve better scaling. Moreover, this result does not affect quantum error correction. Ultimately, this motivates the scientists to look
for coherent instances of quantum error correction and mitigation that do not fall within the framework established in this publication. Eisert explained: “There are good reasons to believe that a
positive interpretation of the result makes sense. In this way, this work, aligning with the MATH+ focus, can help to transform the world through mathematics.“
The research work is closely related to the MATH+ projects in the Emerging Fields EF 1-11 and EF 1-7.
Original publication in Nature Physics:
Exponentially tighter bounds on limitations of quantum error mitigation
Yihui Quek, Daniel Stilck França, Sumeet Khatri, Johannes Jakob Meyer, Jens Eisert
DOI: https://doi.org/10.1038/s41567-024-02536-7
In Nature Physics (July, 2024): https://www.nature.com/articles/s41567-024-02536-7 | {"url":"https://mathplus.de/news/challenging-quantum-error-mitigation-publication-in-nature-physics-jens-eisert-group/","timestamp":"2024-11-02T08:38:40Z","content_type":"text/html","content_length":"125305","record_id":"<urn:uuid:3db3e109-d3d8-422e-9d3d-ce3c738d6164>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00841.warc.gz"} |
Comments on Subsection 5.4.9
Go back to the page of Subsection 5.4.9.
There are currently no comments on Subsection 5.4.9.
There are also:
• 2 comment(s) on Section 5.4: $(\infty ,2)$-Categories
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder,
this is tag 02RX. The letter 'O' is never used.
The tag you filled in for the captcha is wrong. You need to write 02RX, in case you are confused. | {"url":"https://kerodon.net/tag/02RX/comments","timestamp":"2024-11-12T23:08:26Z","content_type":"text/html","content_length":"13159","record_id":"<urn:uuid:b208f9e6-bab2-41d2-b554-8cb0397a2622>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00310.warc.gz"} |
Understanding the Pentagon: A Comprehensive Guide - ITE NEXAR
Understanding the Pentagon: A Comprehensive Guide
shape:yl6axe4-ozq= pentagon .Have you ever wondered why some shapes just seem to pop up everywhere? One such shape is the pentagon. But what exactly is a pentagon, and why is it so important? In this
article, we’re diving deep into the fascinating world of pentagons. So, buckle up and get ready for a geometric adventure!
Basic Properties of a Pentagon
Definition and Characteristics
A pentagon is a five-sided polygon. It’s one of those basic shapes you probably remember from school. Each of its sides is a straight line, and it’s got five angles inside. Simple enough, right?
Types of Pentagons
Pentagons come in two main flavors: regular and irregular. A regular pentagon has all sides and angles equal, making it look perfectly symmetrical. Irregular pentagons, on the other hand, have sides
and angles of different lengths and degrees, giving them a more varied appearance. shape:yl6axe4-ozq= pentagon
Geometric Significance of a Pentagon
Internal Angles
The internal angles of a pentagon are quite interesting. In a regular pentagon, each angle measures 108 degrees, adding up to a total of 540 degrees. This is a nifty fact that often surprises people.
Symmetry is another cool aspect. A regular pentagon is highly symmetrical, with five lines of symmetry. This means you can fold it in half five different ways and it will match up perfectly.
Types of Pentagons
Regular Pentagon
As mentioned, a regular pentagon has all sides and angles equal. This symmetry makes it pleasing to the eye and a favorite in designs and architecture. shape:yl6axe4-ozq= pentagon
Irregular Pentagon
Irregular pentagons are more common in nature and real-world applications. They don’t have equal sides or angles, making each one unique.
Real-World Examples of Pentagons
One of the most famous examples is the Pentagon building in the United States, which houses the Department of Defense. Its unique five-sided shape is instantly recognizable.
Pentagons also appear in nature. For instance, the starfish has a five-armed structure that can be linked to a pentagon.
Mathematical Formulas Involving Pentagons
Area Calculation
To calculate the area of a regular pentagon, you can use the formula: Area=145(5+25)s2\text{Area} = \frac{1}{4} \sqrt{5(5 + 2\sqrt{5})} s^2 where ss is the length of a side.
Perimeter Calculation
The perimeter is much simpler: just add up the lengths of all five sides. For a regular pentagon, it’s simply 5×s5 \times s.
The Pentagon in Popular Culture
The U.S. Department of Defense
The Pentagon building is not just a military headquarters; it’s a symbol of American defense and strategy. Its shape was chosen for practical reasons but has since become iconic. shape:yl6axe4-ozq=
Symbolism in Literature and Art
Pentagons also pop up in literature and art, often symbolizing things like power, balance, and protection.
Constructing a Pentagon
Using a Compass and Ruler
Creating a perfect pentagon requires some tools. Here’s a step-by-step guide:
1. Draw a circle with a compass.
2. Mark a point on the circle, which will be one vertex.
3. Use a protractor to measure 72 degrees from the first point and mark it. Repeat this step until you have five points.
4. Connect the points with a ruler.
Step-by-Step Guide
Following these steps carefully will give you a regular pentagon. It’s a great exercise for understanding geometric principles.
Pentagons in Geometry
Relation to Other Shapes
Pentagons can relate to other shapes in interesting ways. For example, you can fit a pentagon inside a circle or a star shape, known as a pentagram, inside a pentagon. shape:yl6axe4-ozq= pentagon
Importance in Tiling and Tessellation
While regular pentagons don’t tile a plane perfectly, they are essential in more complex tessellations and patterns seen in art and architecture.
Applications of Pentagons in Science and Engineering
Molecular Structures
In chemistry, certain molecules have pentagonal shapes. For example, the structure of the buckminsterfullerene molecule (a type of carbon) includes pentagonal rings.
Engineering Designs
Engineers often use pentagonal shapes in designs to distribute stress evenly, enhancing the stability and strength of structures.
Fun Facts About Pentagons
Historical References
The pentagon shape has been used throughout history, from ancient symbols to modern designs. The five-pointed star, which can be derived from a pentagon, has been a significant symbol in various
Interesting Trivia
Did you know that the term “pentagon” comes from the Greek words “penta” (meaning five) and “gonia” (meaning angle)? It’s a straightforward name for such a versatile shape.
Challenges in Studying Pentagons
Mathematical Complexity
While the concept of a pentagon is simple, the mathematics behind it can get pretty complex, especially when dealing with irregular pentagons and their properties.
Practical Difficulties
Constructing a perfect pentagon can be tricky without the right tools and knowledge. It’s a fun challenge for anyone interested in geometry.
Pentagons in Education
Teaching Strategies
Pentagons are a great teaching tool in geometry. Teachers can use them to explain concepts like angles, symmetry, and polygon properties.
Importance in Curriculum
Understanding pentagons is crucial for students, as it lays the groundwork for more advanced geometric and mathematical studies.
shape:yl6axe4-ozq= pentagon .Pentagons are more than just a five-sided shape. They have fascinating properties, important applications, and a rich history. Whether you’re looking at a building, a
molecule, or a piece of art, you might just spot a pentagon. So, next time you see one, you’ll know there’s a lot more to it than meets the eye. | {"url":"https://itenexar.com/understanding-the-pentagon-a-comprehensive-guide/","timestamp":"2024-11-14T00:22:19Z","content_type":"text/html","content_length":"55621","record_id":"<urn:uuid:c2e3791e-438b-4434-bbea-cecfb825d966>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00656.warc.gz"} |
Using Venn Diagrams to Decide Whether Certain Conclusions Are Valid
Question Video: Using Venn Diagrams to Decide Whether Certain Conclusions Are Valid
Consider the following Venn diagram. Which conditional statement is represented by the diagram? [A] If a student scored over 90%, then they got an A [B] If a student got an A, then they scored over
90%. Use the Venn diagram to decide whether the following statement is valid: If Benjamin scored less than 90%, then he did not get an A.
Video Transcript
Consider the following Venn diagram. In the larger circle on the outside are students who got an A. And a subset of that, inside of that, students who scored over 90 percent. Which conditional
statement is represented by the graph? a) If a student scored over 90 percent, then they got an A. Or b) If a student got an A, then they scored over 90 percent.
To help us think rightly about this, let’s look at three students: a student who’s in the smallest circle, a student who got an A but is not in the over 90 percent circle, and we have a third student
on the outside who did not get an A. Both of these students got an A. But only one of them scored over 90 percent.
We do have a student who got an A but did not score over 90 percent. And that means statement b is impossible. Because it says: If a student got an A, then they scored over 90 percent. And we can
produce a counterexample to that statement. However, if a student scored 90 percent, they got an A. Because students who scored over 90 percent is a subset of students who got an A, all students who
score over 90 percent will get an A.
Now, we need to find out if the following statement is valid: If Benjamin scored less than 90 percent, then he did not get an A.
One thing we could try to do is place Benjamin in one of these categories. Benjamin scored less than 90 percent. We have a student outside who scored less than 90 percent. However, some of the
students that did not score over 90 percent were still given an A. This statement says he did not get an A.
While we cannot say for certain that Benjamin got an A, we can’t rule it out. We can’t say that he didn’t. It is possible that he got an A. The statement “If Benjamin scored less than 90 percent,
then he did not get an A” is not valid. | {"url":"https://www.nagwa.com/en/videos/976151649096/","timestamp":"2024-11-12T23:49:27Z","content_type":"text/html","content_length":"242842","record_id":"<urn:uuid:a78e07cb-9f16-411e-b3cb-7e15012572c0>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00142.warc.gz"} |
Reproduce Command Line or System Identification App Simulation Results in
Reproduce Command Line or System Identification App Simulation Results in Simulink
Once you identify a model, you can simulate it at the command line, in the System Identification app, or in Simulink^®. If you start with one simulation method, and migrate to a second method, you
may find that the results do not precisely match. This mismatch does not mean that one of the simulations is implemented incorrectly. Each method uses a unique simulation algorithm, and yields
results with small numeric differences.
Generally, command-line simulation functions such as compare and sim,as well as the System Identification app, use similar default settings and solvers. Simulink uses somewhat different default
settings and a different solver. If you want to validate your Simulink implementation against command-line or app simulation results, you must ensure that your Simulink settings are consistent with
your command-line or app simulation settings. The following paragraphs list these settings in order of decreasing significance.
Initial Conditions
The most significant source of variation when trying to reproduce results is usually the initial conditions. The initial conditions that you specify for simulation in Simulink blocks must match the
initial conditions used by compare, sim, or the System Identification app. You can determine the initial conditions returned in the earlier simulations. Or, if you are simulating against measurement
data, you can estimate a new set of initial conditions that best match that data. Then, apply those conditions to any simulation methods that you use. For more information on estimating initial
conditions, see Estimate Initial Conditions for Simulating Identified Models.
Once you have determined the initial conditions x0 to use for a model m and a data set z, implement them in your simulation tool.
• For compare or sim, use compareOptions or simOptions.
• For Simulink, use the Idmodel, Nonlinear ARX Model, or Hammerstein-Wiener Model block, specifying m for the model and x0 for the initial states. With Idmodel structures, you can specify initial
states only for idss and idgrey models. If your linear model is of any other type, convert it first to idss. See Simulate Identified Model in Simulink or the corresponding block reference pages.
• For the System Identification app, you cannot specify initial conditions other than zero. You can specify only the method of computing them.
Discretization of Continuous-Time Data
Simulink software assumes a first-order-hold interpolation on input data.
When using compare, sim, or the app, set the InterSample property of your iddata object to 'foh'.
Solvers for Continuous-Time Models
The compare command and the app simulate a continuous-time model by first discretizing the model using c2d and then propagating the difference equations. Simulink defaults to a variable-step true ODE
solver. To better match the difference-equation propagation, set the solver in Simulink to a fixed-step solver such as ode5.
Match Output of sim Command and Nonlinear ARX Model Block
Reproduce command-line sim results for an estimated nonlinear system in Simulink®. When you have an estimated system model, you can simulate it either with the command-line sim command or with
Simulink. The two outputs do not match precisely unless you base both simulations on the same initial conditions. You can achieve this match by estimating the initial conditions in your MATLAB® model
and applying both in the sim command and in your Simulink model.
Load the estimation data you are using to identify a nonlinear ARX model. Create an iddata data object from the data. Specify the sample time as 0.2 seconds.
load twotankdata
z = iddata(y,u,0.2);
Use the first 1000 data points to estimate a nonlinear ARX model mw1 with orders [5 1 3] and wavelet network nonlinearity.
z1 = z(1:1000);
mw1 = nlarx(z1,[5 1 3],idWaveletNetwork);
To initialize your simulated response consistently with the measured data, estimate an initial state vector x0 from the estimation data z1 using the findstates function. The third argument for
findstates in this example, Inf, sets the prediction horizon to infinity in order to minimize the simulation error.
x0 = findstates(mw1,z1,Inf);
Simulate the model mw1 output using x0 and the first 50 seconds (250 samples) of input data.
opt = simOptions('InitialCondition',x0);
data_sim = sim(mw1,z1(1:251),opt);
Now simulate the output of mw1 in Simulink using a Nonlinear ARX Model block, and specify the same initial conditions in the block. The Simulink model ex_idnlarx_block_match_sim is preconfigured to
specify the estimation data, nonlinear ARX model, initial conditions x0, and a simulation time of 50 seconds.
Open the Simulink model.
model = 'ex_idnlarx_block_match_sim';
Simulate the nonlinear ARX model output for 50 seconds.
The IDDATA Sink block outputs the simulated output, data, to the MATLAB workspace.
Plot and compare the simulated outputs you obtained using the sim command and Nonlinear ARX block.
ylabel('Simulated output');
legend('Output of sim command','Output of Nonlinear ARX block','location','southeast')
title('Estimated Initial Conditions for Simulink')
The simulated outputs are the same because you specified the same initial condition when using sim and the Nonlinear ARX Model block.
If you want to see what the comparison would look like without using the same initial conditions, you can rerun the Simulink version with no initial conditions set.
Nullify initial-condition vector x0, and simulate for 50 seconds as before. This null setting of x0 is equivalent to the Simulink initial conditions default.
x0 = [0;0;0;0;0;0;0;0];
Plot the difference between the command-line and Simulink methods for this case.
ylabel('Simulated output');
legend('Output of sim command','Output of Nonlinear ARX block','location','southeast')
title('Default Initial Conditions for Simulink')
The Simulink version starts at a different point than the sim version, but the two versions eventually converge. The sensitivity to initial conditions is important for verifying reproducibility of
the model, but is usually a transient effect.
Match Output of compare Command and Nonlinear ARX Model Block
In this example, you first use the compare command to match the simulated output of a nonlinear ARX model to measured data. You then reproduce the simulated output using a Nonlinear ARX Model block.
Load estimation data to estimate a nonlinear ARX model. Create an iddata object from the data. Specify the sample time as 0.2 seconds.
load twotankdata
z = iddata(y,u,0.2);
Use the first 1000 data points to estimate a nonlinear ARX model mw1 with orders [5 1 3] and wavelet network nonlinearity.
z1 = z(1:1000);
mw1 = nlarx(z1,[5 1 3],idWaveletNetwork);
Use the compare command to match the simulated response of mw1 to the measured output data in z1.
data_compare = compare(mw1,z1);
The compare function uses the first nx samples of the input-output data as past data, where nx is the largest regressor delay in the model. compare uses this past data to compute the initial states,
and then simulates the model output for the remaining nx+1:end input data.
To reproduce this result in Simulink®, compute the initial state values used by compare, and specify the values in the Nonlinear ARX Model block. First compute the largest regressor delay in the
model. Use this delay to compute the past data for initial condition estimation.
nx = max(getDelayInfo(mw1));
past_data = z1(1:nx);
Use data2state to get state values using the past data.
x0 = data2state(mw1,z1(1:nx));
Now simulate the output of mw1 in Simulink using a Nonlinear ARX Model block, and specify x0 as the initial conditions in the block. Use the remaining nx+1:end data for simulation.
zz = z1(nx+1:end);
zz.Tstart = 0;
The Simulink model ex_idnlarx_block_match_compare is preconfigured to specify the estimation data, nonlinear ARX model, and initial conditions.
Open the Simulink model.
model = 'ex_idnlarx_block_match_compare';
Simulate the nonlinear ARX model output for 200 seconds.
The IDDATA Sink block outputs the simulated output, data, to the MATLAB® workspace.
Compare the simulated outputs you obtained using the compare command and Nonlinear ARX block. To do so, plot the simulated outputs.
To compare the output of compare to the output of the Nonlinear ARX Model block, discard the first nx samples of data_compare.
data_compare = data_compare(nx+1:end);
legend('compare','Nonlinear ARX Model block')
The simulated outputs are the same because you specified the same initial condition when using compare and the Nonlinear ARX Model block.
See Also
Nonlinear ARX Model | Hammerstein-Wiener Model | Nonlinear Grey-Box Model | compare | sim
Related Topics | {"url":"https://uk.mathworks.com/help/ident/ug/reproduce-simulation-results-when-using-different-simulation-methods.html","timestamp":"2024-11-04T03:05:52Z","content_type":"text/html","content_length":"92329","record_id":"<urn:uuid:f988d191-77d5-4562-a885-3e76912c3ac5>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00132.warc.gz"} |
ESDU A-Z Index: FIL-FLAS
Fatigue Strength Improvement of Steel
by surface rolling, 89031
Fillets in Bars or Strips
Under Axial Load
stress concentration factors, 09014
Under Bending
stress concentration factors, 09014
Fillets in Flat Plate
Stress Concentration Due to Bending
stress concentration factors for fillets and optimised shape fillets in large plates, 15006
Stress Concentration Due to Tensile Loading
stress concentration factors for fillets and optimised shape fillets in large plates, 15006
Fillets in Round Rods and Tubes
Under Axial Load
stress concentration factors, 89048
Under Bending
stress concentration factors, 89048
Under Torsion
stress concentration factors, 89048
Fillets of Double Radius in Round Rods and Tubes
Under Torsion
stress concentration factors, 75033
Film Condensation
Film Evaporation
Transformation of Analogue to Digital, 92044
Use in Control System Design, 80018, 88039
Use in Lubricant Circulation Systems
definition of absolute and nominal rating, 83030
factors determining choice of rating, 83030
nominal ratings appropriate to various types of equipment, 83030
Use to Control Particulate Fouling, 88024
Finite Element Analysis
Applied to Estimation of Response to Excitation of Lightweight Structures
advantages, features and limitations, 97033
comparison of results for simple structure with those from statistical energy analysis, 99009
outline of mathematical modelling, 97033
typical applications, 97033
Stiffnesses of Local Structural Details for Use With
Finned Tubes in Heat Exchangers
Calculation of Bending Moment, Loading and Torque on in Steady Level Asymmetric Flight
sideslipping, turning or with failed engine, 01010
Effect on Base Drag of Cylindrical Afterbodies at Supersonic Speeds
mounted at or upstream of base, 97022
Profile Drag Increment in Turbulent Boundary Layer
Trapezoidal- or Triangular-Section in Straight Pipes
Viscous Full-potential (VFP) Method for Three Dimensional Wings and Wing-body Combinations
estimation of excrescence drag at subsonic speeds and analysis program, 23013
First-Order Linear System
see also
Application of Continuous or Discrete Forms of Kalman Filter
elimination of measurement noise from randomly-excited, 88039
Differential Equations, 69005
Response to Various Inputs, 69005
Solution by Laplace Transform, 69025
Transfer Functions, 69005
Fisher's Variance Ratio F-Distribution
Fisher-Tippett Type 1
Extreme-Value Distribution for Wind Speed Maxima, 87034
examples of use for data from single- and mixed-storm mechanisms, 88037
Fixed-Pitch Propellers
see also
Equipment to Aircraft Structures
considerations in design against fatigue, 86025
Flame Hardening
Plain Shafts With Interference-Fit Collars
effect on fatigue strength, 68005
Surface Treatment of Steels for Spur and Helical Gears, 68040, 88033
Use to Improve Fatigue Strength of Steels
advantages and limitations, 89031
comparison with other possible surface treatments, 89031
Flange Buckling
Under Compression (Linearly or Parabolically Varying)
various degrees of edge rotational restraint, 80035
Under Uniaxial Compression
plasticity correction factors for various edge conditions, 83044
Flanged Holes in Plates
Effect on Buckling in Shear of Square Plates
Stress Concentration Data
Flange Efficiency
see also
Choice of Optimum Deflections Using Multivariate Search and CFD Codes
Contribution to Average Downwash at Tailplane in Attached Flow
Contribution to Rolling Moment Due to Sideslip, 00025, 80034
Contribution to Rolling Moment Due to Yawing, 72021
Contribution to Sideforce and Yawing Moment Due to Sideslip, 00025, 81013
Contribution to Yawing Moment Due to Yawing, 71017
Derivation of Part-Span Factors, TM 172
Discussion of Regions of Supersonic Flow Occurring at Low Freestream Mach Numbers, 90008
Drag Increment Due to Chord- or Spanwise Sealed/Unsealed Gaps for Undeflected at Low Speeds, 92039
Estimation of Nacelle and Flap Effects on Low-Speed Aerodynamic Centre and Zero-Lift Pitching Moment of Transport Aircraft (Tail-Off), TM 200
Introduction to Data Items on Effect on Drag, Lift and Pitching Moment
Kruger With or Without Vent
Kruger With or Without Vent in Combination With Trailing-Edge Flaps
Leading-Edge With Slot
Multi-Slotted in Combination With Leading-Edge High-Lift Devices
Plain in Combination With Leading-Edge High-Lift Devices
Plain in Combination With Plain Leading-Edge Flaps
Plain in Combination With Plain Leading-Edge Flaps (Both With Sealed Gaps)
Profile Drag Due to Deployment of All Types on Wings
Single-Slotted in Combination With Leading-Edge High-Lift Devices
Slats or Sealed Slats
Slats or Sealed Slats in Combination With Trailing-Edge Flaps
Split in Combination With Leading-Edge High-Lift Devices
Split With Flap Trailing Edge Forward of Wing Trailing Edge When Undeployed
Trailing Vortex Drag Due to Deployment of All Types on Cambered and Twisted or Plane Wings
Approximate Method of Calculating Effect on Pressure Drop
two-phase gas- or vapour-liquid flow through pipe fittings, 89012
Flash Point
Definition, 71003
Typical Values for Synthetic Lubricants, 94020
Flash Temperature Rise
In Cam/Follower Lubricant Film
prediction thoughout complete rotation, 00015 | {"url":"https://esdu.com/cgi-bin/ps.pl?sess=unlicensed_1240710124403tct&t=index&p=index_fil","timestamp":"2024-11-10T05:10:03Z","content_type":"text/html","content_length":"74990","record_id":"<urn:uuid:94627aca-e316-49e2-8ca0-fdf9c5dd5f2a>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00429.warc.gz"} |
Guy Ellis' Tech Blog
Codility is a site that tests coders. Their demo problem task is called Equi. You can read the full description of the problem and solutions in a number of different languages on their blog post
about it: Solutions for task Equi
I was somewhat surprise at the complexity of the code required in many languages. Using C#'s LINQ syntax makes it obvious and simple to write:
static int equi(int[] A)
for (int i = 0; i < A.Length; i++)
if (A.Take(i).Sum() == A.Skip(i + 1).Sum())
return i;
return -1;
It's also not difficult to modify that function to return all the equilibrium indexes:
static IEnumerable<int> equi2(int[] A)
for (int i = 0; i < A.Length; i++)
if (A.Take(i).Sum() == A.Skip(i+1).Sum())
yield return i;
Note that this solution although simple and elegant to write is not performant because the full array get summed in each loop. i.e. the time complexity of this solution is O(n^2). For a time linear
solution you would increase and decrease a left and right sum variable as you moved through the index and compare those values. | {"url":"https://www.guyellisrocks.com/2013/02/","timestamp":"2024-11-07T13:58:18Z","content_type":"text/html","content_length":"63764","record_id":"<urn:uuid:c3592137-1121-45c9-9b3b-292e53464993>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00195.warc.gz"} |
Ounces To Cups Calculator | Convert Ounces To Cups!
Food And Nutrition Calculators
Ounces To Cups Calculator
The converter will make it easy for anyone to switch between recipes and from fluid ounces to cups easily. Whether you're a proud home baker or just starting, this tool will save you time and energy.
Table of contents
The converter will make it easy for anyone to easily switch between recipes and from fluid ounces to cups. Whether you're a proud home baker or just starting out, this tool will save you time and
Fluid ounces to cups
If you frequently need to convert liquid ounces to cups or the other way around, you might want to know this fluid oz to cups conversion chart. It has all of the information you need to make the
conversion quickly and easily.
It can be a little confusing when you eat fast food, and you end up with two orders of fries instead of just one. Or, you get a chocolate shake and it's really just a chocolate milk shake. This is
because although the ingredients might look the same, the ratios and quantities might be different. This is also true for programs and applications. Although a specific program or application might
look the same when installed, the underlying code might be completely different.
In the United States, there are two main cups used for everyday purposes:
Legal cup - The legal cup is 240 milliliters, and the tablespoons are 3 teaspoons. You can see this information in the cups to ounces conversion chart and the fl oz to cups converter below!
Customary cup - this contains 236 ml (half a liquid pint) - is equivalent to the regular size of a liquid pint, which contains 480 ml.
This coffee cup is a little different since it only has 118 ml of coffee. The coffee inside is the same as usual, but this cup uses 148 ml of water.
Overseas, the coffee culture is even more immersive and exciting. Not only can we differentiate more cups, but also convert more liquid ounces to cups. So don't wait; come join us on this amazing
A metric cup is a little bigger than a common cup and has 250 ml of liquid inside.
The UK cup is a smaller container than the US pint. It's technically half of an imperial pint, or 284.13 ml, which is about the size of a large tennis ball.
In the United States, Canada, the United Kingdom, Ireland, and other countries with formerly British overseas colonies, a pint is a unit of measure commonly used for alcoholic beverages. It equals
37.5 ml and is still used to describe the amount of rice and sake, the traditional Japanese alcoholic beverage.
How much fl oz is 3 cups?
To convert 3 cups of liquid to ounces, divide it by 8. Here's how we do that step-by-step: First, divide 3 cups by 8 to get 0.375 cups. Next, multiply 0.375 cups by the numeric value 8 to get 0.8
Convert fl oz to cups
To convert from fluid ounces to cups, divide by 8. So, if you have 128 fluid ounces, you would need to divide it by 8 to get cups.
4 fluid ounces = 0.5 cup
8 fluid ounces = 1 cup
10 fluid ounces = 1.25 cup
16 fluid ounces = 2 cups
20 fluid ounces = 2.5 cups
Convert cups to fl oz
Converting cups to fluid ounces is simple: multiply cups by 8 to get ounces. For example, if you have 12 coffee cups, you would get 192 fluid ounces.
1 cup = 8 fluid ounces
2 cups = 2 × 8 = 16 fluid ounces
3 cups = 3 × 8 = 24 fluid ounces
0.5 cup = 0.5 × 8 = 4 fluid ounces
2.5 cup = 2.5 × 8 = 10 fluid ounces
How many fl oz are in a cup?
There are 8 fluid ounces in one cup.
in 2 cups there are 16 fluid ounces
in 3 cups there are 24 fluid ounces
in 4 cups there are 32 fluid ounces
in 0.5 cup there are 4 fluid ounces
in 0.25 of a cup there are 2 fluid ounces
0.125 of a cup equals one fluid ounce
How much is 6 oz to a cup?
To find cups, divide fluid ounces by 8. This will give you the number of cups in a litre. Next, you need to divide that number by 100, because that's how many milliliters a cup is. Finally, you add
the numbers together to get the number of cups in a gallon. To divide 1 cup by 8, we use the fraction calculator to figure out how many cups there are. We divide 6 by 8 to find out. 3/4 of a cup
comes out as 1 cup.
Article author
Parmis Kazemi
Parmis is a content creator who has a passion for writing and creating new things. She is also highly interested in tech and enjoys learning new things.
Ounces To Cups Calculator English
Published: Tue Aug 30 2022
In category Food and nutrition calculators
Add Ounces To Cups Calculator to your own website
Other food and nutrition calculators
Grilled Meat Calculator
Calculate how much meat you need to grill in a barbeque party so that no one is left hungry!
Grams To Tablespoon Calculator
Easily convert grams to tablespoons with this free calculator! Be a great chef with precice cooking measurements!
Coffee To Water -ratio Calculator
This calculator can help you determine the perfect coffee-to water ratio for your cup of coffee.
Oil To Butter Converter
How to bake a cake with butter and oil. The oil to butter conversion calculator will help you figure out how much butter to use.
Calorie Deficit Calculator
This tool helps with calculating your calorie deficit will help you estimate the time it will take to reach your target weight if you follow a certain calorie restriction.
Maintenance Calorie Calculator
This maintenance calorie calculator will allow you to calculate the calories your body needs in order to maintain your current weight.
Daily Caffeine Intake Calculator
Find out how much caffeine you've had in a day with our free online calculator.
Pizza Dough Calculator
Calculate the needed ingredients for a pizza recipe with this pizza dough calculator! Just say how many pizzas you want to make and get the results!
Grams To Cups
Converting grams to cups for common pantry ingredients.
Egg Boiling Time Calculator
This calculator helps you find the perfect amount of time it takes to boil the perfect egg. | {"url":"https://purecalculators.com/ounces-to-cups-calculator","timestamp":"2024-11-03T07:06:27Z","content_type":"text/html","content_length":"126033","record_id":"<urn:uuid:c27cf5bb-7a32-4a5a-b763-ddfe752f2e48>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00277.warc.gz"} |
How do I interpret non-parametric test statistics in MyStatLab? | Hire Someone To Do Exam
How do I interpret non-parametric test statistics in MyStatLab? MyClasses contains three test functions: (A-a) testWithFunc(), (B-b) testWithCount(). Example application of myClasses: Function
TestWithFunc(A,B,C): let def zsf(B,*cond): bool = (B, Cond(cond)); return zsf(C, B) if (cond.type ==’Boolean’) Example TestIncludingResults should return true rather than false: case int :: a => def
zsf(A, B): bool = zsf(B) if (cond.type!= ‘Boolean’) zsf = zsf(B) if (cond.type == ‘Boolean’) zsf(A) else testWithFunc(A,B) Another example needs to make use of a mask when you need to express data
objects. Example Problem MyClasses contains examples of certain types like set, ordered and non-order-based. I have mixed the functions themend with functions like “x = B; %== zsf(-B*1); %== zsf(B)-…
> B^2” but there seems to be no convenient solutions. Usage MyClasses, which contains functions like zsf, or testWithFunc, are functions which build their own way of expressing data objects and/or
evaluation data objects. These are not equal to the tests: MyClasses contains not-template functions like “foo x = B; %== zsf(-1)……” where “x = B” Including(v,v) and not(v,v) are the functions { “v:a
: y; %== zsf(x)- %== zsf(x); zsf(a,-1); %== zsf(b)”, “a:y:z:zf(y); %== zsf(-yes : b;” * y + %== zsf(x – 1); zsf(x)(-2d); } %== zsf(zsf(x – 1)), but some of the functions are not right. MyClasses
contains also function called “f” which converts y to -1 (but contains a problem to remember: the testWithFunc(“testWithFunc(-yes))” does not behave that way) or testWithFunc(“testWithFunc(-“, “). “,
How Much Does It Cost To Pay Someone To Take An Online Class?
MyClasses contains not supported as a function for the context of evaluation data objects. MyClasses contains a requirement to accept any type from non-order-based or non-order-descriptive functions:
– use values of different names or one or more values of typeHow do I interpret non-parametric test statistics in MyStatLab? The first class of my lab workspaces are (i) a container for elements
drawn on an x to y plane, but they have a secondary container for my work and (ii) I read in the documentation of the Visual Basic 4 environment. This gives the following result that doesn’t even
work (when I use: “FuncProc(…)” as both an expression and a method call). (MyStatLAB)===”FuncProc(“) Which results in both: Concerning the second line of the procedure(s) I have no idea if this is
what I need by the way. var myClass = new MySVECTOR(@”MysVECTOR4”,…); Any chance explaining what I need to do here? A: When you declared myClass, it is different from MySVECTOR4. You declared it
instead of the class declaration (because it is a class), this is the first step to find an error. If you want myClass a method or an expression you need the method (and the result of it): @Packing/
MethodName3 public static void Main(string[] args) { MySVECTOR mySVECTOR = new MySVECTOR(); mySVECTOR.RangeFrom(‘1,0’); MySVECTOR.RangeTo(“1,0″); mySVECTOR.RangeTo(=”0,9″); MySVECTOR.RangeTo
(@”MysVECTOR4,1″); this.RangeTo(@”MysVECTOR4,” //=>”1,0″ //=>”0,0″ //=>”0,0″, //=>”0,1″, //=>”1,1″, //=>(1,0 =>”1″) //=>”00,00″, //=>”00,0″, //=>”0,0″, //=>”8×8,8×8″, //=>”8×10″, //=>”64×64,64×64″, /
/=>”64 x64,64″, //=>”64″ site here
Sell My Assignments
); … } I don’t know what I did wrong with my code but maybe you should be concerned for your double nested if statement for MySVECTOR. You don’t need the base 10 if block if you just defined aHow do
I interpret non-parametric test statistics in MyStatLab? OK, I know I’m opening this as an example of multivariate statistics, but am trying to understand how real-world data are interpreted these
days. But if you really want to try it out here, here is the second part of our book – What I Learned from MyStatLab: In the context of functional data it does basically the same as it did for my
data. So MyStatLabel with data mean and median and in set for this test we generate my own data – all else done as this was a simple example of statistical function. In my blog post on this method
(though more detail about the statistics are indicated by the quotes) I wrote about the assumptions I make about single-variable functions Here is my data. See p.21 for the example of the mean and
median. I had all the functions of the data, but couldn’t have had all of the data within it at once. I also had the normalisation process with the first of all the following functions, only the
first value of the interval I required had all of the data within it. I used a custom implementation for every run in my website. Please note also that I have included the functions I have written in
MATLAB to highlight the differences that I noted in the text. Now I wasn’t sure how normalise it sounds. After using the normalisation function of the second example in my blog post, I was averse to
why to do that for R, and I saw that the data had a perfectly normal distribution and the functions I wrote for them were fairly well behaved as expected by the standard fitting tools we have already
discussed. This was an initial trial with R, without any confidence about what I was doing wrong. Then I was thinking about how to do the thing that led to the confusion. Here is the setup that I’d
like to follow. If you have any questions about why I went here, please feel free to ask it here and a blog post about why and how.
Take My Online Course
The process of normalising the data isn’t something I could rewrite or change as my computer does but it allows me to do so, for the most part, the data itself, and also make the structure of the
paper seem more manageable. That said I still have a few constants that may affect my estimatory computations, and some details that I wanted to share in order to teach us how to do so. Now, looking
at the first example how your estimators looked like I can think of several possibilities. One is a continuous variable, or continuous function that may be associated with the data. You can notice I
never looked at it with my eyes and I never looked back. You can see from the above two example how my method looks different and I was not able to find such a pattern in my data. In other words,
could you try to describe it in | {"url":"https://paytodoexam.com/how-do-i-interpret-non-parametric-test-statistics-in-mystatlab","timestamp":"2024-11-04T02:55:18Z","content_type":"text/html","content_length":"194466","record_id":"<urn:uuid:24f59d07-d233-4f2c-92a5-a6bc0e220a72>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00751.warc.gz"} |
Phase-field Modeling of Precipitate Behavior in RPV Steel Using CALPHAD Database
Predicting multi-particle behavior is an extremely complicated task. The Ostwald ripening of single-phase precipitates has been extensively investigated using both analytical [
] and numerical methods [
]. The role of the coarsened phase volume fraction has been explored [
] and morphological deviations from the spherical (circular) coarsened phases at high volume fractions have already been observed in previous studies [
]. Ardell evaluated the role of the coarsened phase fraction using a theoretical approach and he found that the peak of the particle size distribution decreased as the particle fraction increased [
]. Brasilsford and Wynblatt [
] also found that the particle size distribution peak position decreased as the volume fraction increased, and they claimed that the effect was much less sensitive to volume fraction than proposed by
Ardell [
]. Although the microstructural evolutions driven by curvature minimization have been studied extensively for many decades, current understanding of Ostwald ripening remains limited. Because we have
no analytical solutions for the diffusion equations in most cases and ripening is extremely complex, multi-particle behaviors are usually predicted based on the mean-field assumption [
]. or numerical approaches [
]. Previous simulations of particle coarsening have predicted the behaviors of single phases using phenomenological models [
]. In this study, we performed a simulation of multiphase precipitates using a CALPHAD-type database to predict the behaviors of realistic precipitates. The major elements of low-Cu reactor pressure
vessel (RPV) steel are Fe, Mn, Ni and Si [
]; therefore, we selected the Fe–Mn–Ni–Si (hereafter FMNS) quaternary system to investigate multi-phase coarsening, using the FMNS database. The phase-field method has been extensively applied to
investigate the microstructural evolution of low-alloy Fe-based steel [
] assuming only one precipitate phase within the system. T3, T6, and T7 Mn–Ni–Si-rich (MNS) precipitates are generally observed in RPV steel [
]; therefore, we assumed the presence of three types of MNS precipitates in the system. We adopted the Kim–Kim–Suzuki (KKS) model [
] to perform multi-phase and multi-component phase-field modeling. The FMNS system, a prototype of RPV steel, is of great interest in nuclear engineering applications [
]. Because MNS precipitates are among the main sources of late-stage hardening for FMNS [
], we examined the stability of the MNS precipitates. The T3, T6, and T7 precipitate phases, major types of MNS precipitates, were considered in the UW1 database [
] in this study; we also simulated the evolution of pairs of T3–T3, T7–T7, and T3–T7 precipitates to quantify the interactions between particles.
2. UW1 CALPHAD DATABASE
In the UW1 database [
], one bcc phase representing the matrix and 12 MNS precipitate phases are considered. In our study, we selected the matrix phase and three MNS precipitate phases for simplicity. The thermodynamic
parameters we used were taken from the supplementary material of ref. [
] as follows:
for bcc FMNS phase,
LFe,Mnbcc0=-2759 + 1.23T
LFe,Nibcc0=−956.63 − 1.28726Τ
LFe,Nibcc1=1789.03 − 1.92912Τ
LFe,Sibcc0=−153138.56 + 46.86T
LFe,Sibcc1=–92352, LFe,Sibcc2=62240
LMn,Nibcc0=−3508.43 − 23.7885T
LMn,Sibcc0=−89620.7 + 2.9410T
TFe,Mnbcc0=123.0, TFe,Sibcc0=504.0
for the T3: Mn[6⁄29]Ni[16⁄29]Si[7⁄29] phase,
for the T6: Mn[1/3](Ni,Si)[2/3] phase,
LMn,Ni,Sibcc0=–159474.81, LMn,Ni,Sibcc1=–172110.47
and for the T7: Mn[1⁄2]Ni[1⁄3]Si[1⁄6] phase,
gMn,Ni,SiT70=–32434.25 – 5T+1/20gMncbcc+1/30gNifcc+1/60gSidiamond
We utilize the phase-field model to simulate the microstructural evolution of the FMNS system by solving the Cahn–Hilliard [
] and Allen-Cahn (Ginzburg-Landau) equations [
] to simulate the microstructural evolution. We will denote the composition (
= 1, 2, 3, and 4 for Fe, Mn, Ni, and Si, respectively) in phase
) at position
and time
indicates the T3, T6, and T7 phases. We introduce four non-conserved order parameters (
) to indicate the regions of the four precipitated phases. The composition
) is given as follows [
where np represents the number of θ precipitate.
The local free energy density G(ciθ,t) of the system is expressed as follows:
where the units of all free energy densities (expressed by g in this study) in this study are joules per mole: the unit of total free energy of the system G is J. We assumed that density ρ is 7.9 g
cm^3 and the atomic molar mass M is 56.
. We chose
= 200.0 and
= 5.0 in
Eq. 5
. From ref. [
], we obtain the free energy of each element of the
phase. The free energies of the T3, T6, and T6 precipitates is given as follows:
where y3II and y4II denote the site fraction of Ni and Si at the second sub-lattice of the T6 phase, respectively.
We solve the Ginzburg-Landau equation in
Eq. 9
and the Cahn-Hilliard equation in
Eq. 10
∂ϕiθr,t∂t=-Lθδgδϕiθr,t θ= T3, T6, and T7
We implemented the forward Euler scheme to solve
Eq. 9
. To determine
Eq. 10
we adopted the relation in refs. [
We use the relation
to determine the diffusivity; the parameters used to determine this value are listed in
Table 1
In principle, one has to know the diffusivity data of each element of each precipitated phase to perform phase-field modeling. However, the diffusivity data of T3, T6, and T7 phases are not
available. Since a precipitate larger than 17 nm has the FCC crystal structure [
], we adopted the diffusion data of each element in the Austenite (FCC) phase for all precipitated phases.
We could not find diffusivity data for Si in the
(fcc) phase; therefore, we calculate it from a kinetic simulation. We performed a DICTRA simulation using Thermo-Calc 2017a software with the TCFE8 and MOBFE3 databases. At T = 550K, in the Fe–Si
binary system (
(fcc) phase) the Si mole fraction is always 0.03 at x = 0; the value is 0.005 initially, except for the point of x = 0 in
Fig. 1
. From the diffusion length depending on the simulat ion time, we calculated the diffusion coefficient of Si in the
(fcc) phase.
∂ϕiθr,t∂t=-Lθ∂G∂ϕiθr,t-ωθ∇2ϕiθ(θ = T3, T6, and T7)
In our simulations, we use the energy normalized by
, where
= 8.3144598 J/mol·K and the temperature T = 550 K. The diffusivity values are normalized by
). For the length of the grid spacing, Δ
= Δ
was taken as 0.125 nm [
]. The time in our study is normalized by 0.01 s.
To determine
Eq. 12
, the interfacial energies between the matrix and precipitate phases are needed. So far, we have not found any literature that has evaluated the interfacial energies relevant for our study;
therefore, we measure the interfacial energy using the extended Becker’s model function [
]. To perform the interfacial energy estimation, we used TC-PRISMA software (included in Thermo-Calc 2016a) with the TCAL3 and MOBAL3 databases. However, evaluating the interfacial energy for the
whole Si fraction in the T6 phase using TC-PRISMA is not possible. We find that the Si fraction in the T6 phase is generally smaller than 0.3; therefore, we assume that the interfacial energy is
given by
Eq. 13
. 0.2496 J/m
]. For the T3 and T7 stoichiometric compounds, we also evaluate the interfacial energies using TC-PRISMA as 0.3311 and 0.1791 J/m
. The interfacial energy is given by
Eq. 13
when the gradient coefficient is
Since we used the normalized variables in this study, we evaluated the normalized interfacial energy.
The unit of ε is J⁄m^2⋅mol. We chose ε^*,T6 = 1.0 as a non-dimensional parameter which corresponds to the interfacial energy 0.26 J/m^2 which is quite comparable to the value 0.2496 J/m^2 which we
obtained from the TC-Prisma calculation. We also determined ε^*,T3 = 1.33 and ε^*,T7 = 0.72 using a consistent way.
We performed the simulations with a simulation cell size of 8 × 8 nm
. The initial precipitate radius was 1.5 nm. The discretized time step was selected as given in
Table 2
We chose Δt after convergence and accuracy testing. The simulations obtained results in 1 to 12 h, depending on the Δt of the system. For the matrix, the initial compositions in the α phase were c2α
= 0.008, c3α = 0.008, and cα = 0.00808. For the T6 precipitate, c3T6=0.459234067 and c4T6=0.2074326.
For the T3 and T7 cases, the compositions of the precipitates were fixed and the initial compositions of the α phase were assumed to be equal to those of the T6 case. For the case with a single
precipitate, we placed a circular precipitate with an initial radius r = 1.5 nm at the center of the simulation cell. For multiple precipitate phases, we placed two precipitates with initial radii r
= 1.5 nm at two symmetrical positions with the distances L = 3.25, 3.50, and 3.75 nm relative to the center of the system.
5. RESULTS
We performed the simulation to examine the stability of a single precipitate as shown in
Fig. 2
. In
Fig. 3(a)
, the T6 precipitate undergoes minor accommodation at a very early stage. After 0.0002 s, the precipitate radii remains the same as it was during the modeling in
Fig. 3(b)
. This means that the T6 precipitate is thermodynamically stable. Because the T3 and T7 phases are unstable, these precipitates shrink with time; the T7 precipitate radius decreases more rapidly than
that of the T3 precipitate. We performed CALPHAD modeling using Thermocalc software and the UW1 database [
] with
= 0.008,
= 0.008,
= 0.00808, and T = 550 K. The equilibrium phases were bcc(A2) (98.940 mol%) and T6 (1.363 mol%). Therefore, we concluded that the phase-field simulation with a single precipitate was consistent with
the thermodynamic model. As shown in
Fig. 4
, the diffusion field of the solutes overlapped between the particles, therefore, the solute concentration is relatively higher than in other matrix regions. As a result, the dissolution rate
decreases along the direction of the other particle. As a result, the particle shape becomes asymmetric in
Fig. 4
Therefore, the T3 particles remain longer for two interacting T3 particles compared to a single T3 particle. In addition, when the distance between particles
is 3.75 nm, the particle size evolution curve shown in
Fig. 5
approaches convergence with that curve of the single T3 particle. Consistent results are observed for the T7 precipitate in
Fig. 6
We performed a set of simulations to investigate the interactions between T3 and T7 precipitates, with T3 and T7 precipitates located as described in
Fig. 7
Fig. 9
, we find that the T3 and T7 precipitates survive longer under T3–T7 interactions. Even though the precipitates are of different phases, particle–particle interactions increase the lifetime of the
precipitates, i.e., enhance their stability.
6. CONCLUSIONS
We developed a multi-scale modeling framework to simulate the microstructural evolution of MNS precipitates in the quaternary FMNS system representing RPV steel. The UW1 quaternary database was
implemented and the DICTRA and TC-PRISMA packages were used to evaluate the mobility of solutes and the interfacial energies between the matrix phase and precipitate phases. The results of the
single-precipitate simulations were consistent with the thermodynamic model. We also evaluated the interactions between precipitates and found that the precipitates survived longer when two particles
were present, whether of the same or different phases, compared to the particle lifetime in the single-precipitate case. | {"url":"http://kjmm.org/journal/view.php?doi=10.3365/KJMM.2018.56.6.472","timestamp":"2024-11-12T03:31:34Z","content_type":"application/xhtml+xml","content_length":"134201","record_id":"<urn:uuid:caba90b9-532e-4a9b-9453-dc7be070b785>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00347.warc.gz"} |
Learning Rate
In the previous lecture, we learned that it’s easy to go for a closed form solution than using an iterative form solution (gradient descent) to optimise the cost function.
But the example we looked was a simple 1D example, what will happen if the cost function is in higher dimensions?
To summarise, though gradient descent looks complicated for a 1D function, it’s easier to compute the optimal minima using gradient descent for higher dimension function. We will be using it in
logistic regression and even for neural networks.
In the next lecture, we will learn how the value of learning rate, i.e. η may affect the optimal solution. Let’s look at this in detail.
At 2:05, the professor writes -1.2. Please note that it is a slight calculation error; it should be 0.4.
To summarise, a large value of learning rate may oscillate your solution, and you may skip the optimal solution (global minima). So it’s always a good practice to choose a small value of learning
rate and slowly move towards the negative of the gradient. | {"url":"https://www.internetknowledgehub.com/learning-rate/","timestamp":"2024-11-10T00:06:33Z","content_type":"text/html","content_length":"78313","record_id":"<urn:uuid:abc92104-11c1-4a1c-8a2b-e812f781eb36>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00677.warc.gz"} |
Interval Notation - Definition, Examples, Types of Intervals - Grade Potential Mountain View, CA
Interval Notation - Definition, Examples, Types of Intervals
Interval Notation - Definition, Examples, Types of Intervals
Interval notation is a crucial principle that students should learn owing to the fact that it becomes more important as you advance to more difficult arithmetic.
If you see higher mathematics, something like integral and differential calculus, in front of you, then being knowledgeable of interval notation can save you time in understanding these ideas.
This article will discuss what interval notation is, what are its uses, and how you can interpret it.
What Is Interval Notation?
The interval notation is simply a method to express a subset of all real numbers along the number line.
An interval means the numbers between two other numbers at any point in the number line, from -∞ to +∞. (The symbol ∞ denotes infinity.)
Basic problems you encounter essentially consists of one positive or negative numbers, so it can be difficult to see the benefit of the interval notation from such straightforward applications.
Though, intervals are typically used to denote domains and ranges of functions in higher arithmetics. Expressing these intervals can increasingly become difficult as the functions become more tricky.
Let’s take a straightforward compound inequality notation as an example.
• x is higher than negative four but less than 2
So far we know, this inequality notation can be expressed as: {x | -4 < x < 2} in set builder notation. Though, it can also be expressed with interval notation (-4, 2), signified by values a and b
segregated by a comma.
As we can see, interval notation is a method of writing intervals elegantly and concisely, using set rules that make writing and understanding intervals on the number line simpler.
In the following section we will discuss about the rules of expressing a subset in a set of all real numbers with interval notation.
Types of Intervals
Various types of intervals place the base for denoting the interval notation. These interval types are important to get to know due to the fact they underpin the complete notation process.
Open intervals are used when the expression do not include the endpoints of the interval. The previous notation is a fine example of this.
The inequality notation {x | -4 < x < 2} describes x as being higher than negative four but less than two, meaning that it does not include either of the two numbers referred to. As such, this is an
open interval denoted with parentheses or a round bracket, such as the following.
(-4, 2)
This represent that in a given set of real numbers, such as the interval between -4 and 2, those two values are not included.
On the number line, an unshaded circle denotes an open value.
A closed interval is the opposite of the last type of interval. Where the open interval does not include the values mentioned, a closed interval does. In text form, a closed interval is expressed as
any value “higher than or equal to” or “less than or equal to.”
For example, if the last example was a closed interval, it would read, “x is greater than or equal to -4 and less than or equal to two.”
In an inequality notation, this would be written as {x | -4 < x < 2}.
In an interval notation, this is written with brackets, or [-4, 2]. This means that the interval includes those two boundary values: -4 and 2.
On the number line, a shaded circle is utilized to denote an included open value.
A half-open interval is a combination of prior types of intervals. Of the two points on the line, one is included, and the other isn’t.
Using the previous example for assistance, if the interval were half-open, it would read as “x is greater than or equal to -4 and less than two.” This means that x could be the value -4 but couldn’t
possibly be equal to the value 2.
In an inequality notation, this would be written as {x | -4 < x < 2}.
A half-open interval notation is denoted with both a bracket and a parenthesis, or [-4, 2).
On the number line, the shaded circle denotes the number present in the interval, and the unshaded circle denotes the value which are not included from the subset.
Symbols for Interval Notation and Types of Intervals
To recap, there are different types of interval notations; open, closed, and half-open. An open interval doesn’t contain the endpoints on the real number line, while a closed interval does. A
half-open interval includes one value on the line but does not include the other value.
As seen in the prior example, there are different symbols for these types under the interval notation.
These symbols build the actual interval notation you develop when stating points on a number line.
• ( ): The parentheses are employed when the interval is open, or when the two endpoints on the number line are excluded from the subset.
• [ ]: The square brackets are utilized when the interval is closed, or when the two points on the number line are included in the subset of real numbers.
• ( ]: Both the parenthesis and the square bracket are utilized when the interval is half-open, or when only the left endpoint is excluded in the set, and the right endpoint is not excluded. Also
known as a left open interval.
• [ ): This is also a half-open notation when there are both included and excluded values between the two. In this instance, the left endpoint is not excluded in the set, while the right endpoint
is not included. This is also called a right-open interval.
Number Line Representations for the Various Interval Types
Apart from being denoted with symbols, the various interval types can also be described in the number line utilizing both shaded and open circles, depending on the interval type.
The table below will show all the different types of intervals as they are represented in the number line.
Practice Examples for Interval Notation
Now that you’ve understood everything you need to know about writing things in interval notations, you’re prepared for a few practice problems and their accompanying solution set.
Example 1
Convert the following inequality into an interval notation: {x | -6 < x < 9}
This sample question is a simple conversion; just utilize the equivalent symbols when writing the inequality into an interval notation.
In this inequality, the a-value (-6) is an open interval, while the b value (9) is a closed one. Thus, it’s going to be written as (-6, 9].
Example 2
For a school to participate in a debate competition, they need minimum of three teams. Express this equation in interval notation.
In this word question, let x be the minimum number of teams.
Since the number of teams required is “three and above,” the value 3 is consisted in the set, which implies that three is a closed value.
Plus, since no maximum number was mentioned with concern to the number of teams a school can send to the debate competition, this number should be positive to infinity.
Therefore, the interval notation should be expressed as [3, ∞).
These types of intervals, where there is one side of the interval that stretches to either positive or negative infinity, are called unbounded intervals.
Example 3
A friend wants to participate in diet program constraining their daily calorie intake. For the diet to be successful, they should have at least 1800 calories every day, but no more than 2000. How do
you describe this range in interval notation?
In this question, the number 1800 is the minimum while the number 2000 is the maximum value.
The question suggest that both 1800 and 2000 are included in the range, so the equation is a close interval, written with the inequality 1800 ≤ x ≤ 2000.
Therefore, the interval notation is described as [1800, 2000].
When the subset of real numbers is confined to a variation between two values, and doesn’t stretch to either positive or negative infinity, it is called a bounded interval.
Interval Notation Frequently Asked Questions
How To Graph an Interval Notation?
An interval notation is fundamentally a way of representing inequalities on the number line.
There are laws to writing an interval notation to the number line: a closed interval is written with a shaded circle, and an open integral is denoted with an unfilled circle. This way, you can
promptly see on a number line if the point is excluded or included from the interval.
How Do You Transform Inequality to Interval Notation?
An interval notation is basically a different technique of expressing an inequality or a combination of real numbers.
If x is higher than or lower than a value (not equal to), then the value should be stated with parentheses () in the notation.
If x is higher than or equal to, or lower than or equal to, then the interval is denoted with closed brackets [ ] in the notation. See the examples of interval notation above to check how these
symbols are utilized.
How Do You Exclude Numbers in Interval Notation?
Numbers excluded from the interval can be stated with parenthesis in the notation. A parenthesis implies that you’re expressing an open interval, which means that the value is excluded from the set.
Grade Potential Could Guide You Get a Grip on Math
Writing interval notations can get complex fast. There are more difficult topics in this concentration, such as those working on the union of intervals, fractions, absolute value equations,
inequalities with an upper bound, and many more.
If you want to conquer these ideas quickly, you need to review them with the expert guidance and study materials that the professional teachers of Grade Potential delivers.
Unlock your math skills with Grade Potential. Book a call now! | {"url":"https://www.mountainviewinhometutors.com/blog/interval-notation-definition-examples-types-of-intervals","timestamp":"2024-11-02T14:12:15Z","content_type":"text/html","content_length":"111348","record_id":"<urn:uuid:22eba4dd-9ea3-4b68-8a9f-8b9e7c022194>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00500.warc.gz"} |
A function to perform a statistical test at a sgRNA-level,
run_estimation {CB2} R Documentation
A function to perform a statistical test at a sgRNA-level, deprecated.
A function to perform a statistical test at a sgRNA-level, deprecated.
delim = "_",
ge_id = NULL,
sg_id = NULL
sgcount This data frame contains read counts of sgRNAs for the samples.
design This table contains study design. It has to contain 'group.'
group_a The first group to be tested.
group_b The second group to be tested.
delim The delimiter between a gene name and a sgRNA ID. It will be used if only rownames contains sgRNA ID.
ge_id The column name of the gene column.
sg_id The column/columns of sgRNA identifiers.
A table contains the sgRNA-level test result, and the table contains these columns:
• ‘sgRNA’: The sgRNA identifier.
• ‘gene’: The gene is the target of the sgRNA
• ‘n_a’: The number of replicates of the first group.
• ‘n_b’: The number of replicates of the second group.
• ‘phat_a’: The proportion value of the sgRNA for the first group.
• ‘phat_b’: The proportion value of the sgRNA for the second group.
• ‘vhat_a’: The variance of the sgRNA for the first group.
• ‘vhat_b’: The variance of the sgRNA for the second group.
• ‘cpm_a’: The mean CPM of the sgRNA within the first group.
• ‘cpm_b’: The mean CPM of the sgRNA within the second group.
• ‘logFC’: The log fold change of sgRNA between two groups.
• ‘t_value’: The value for the t-statistics.
• ‘df’: The value of the degree of freedom, and will be used to calculate the p-value of the sgRNA.
• ‘p_ts’: The p-value indicates a difference between the two groups.
• ‘p_pa’: The p-value indicates enrichment of the first group.
• ‘p_pb’: The p-value indicates enrichment of the second group.
• ‘fdr_ts’: The adjusted P-value of ‘p_ts’.
• ‘fdr_pa’: The adjusted P-value of ‘p_pa’.
• ‘fdr_pb’: The adjusted P-value of ‘p_pb’.
version 1.3.4 | {"url":"https://search.r-project.org/CRAN/refmans/CB2/html/run_estimation.html","timestamp":"2024-11-04T08:18:53Z","content_type":"text/html","content_length":"4553","record_id":"<urn:uuid:05aaa248-bb56-4b4e-9236-8a02bac0e658>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00487.warc.gz"} |
How many right angles does a hexagon have? | Socratic
How many right angles does a hexagon have?
2 Answers
Zero in a regular hexagon because the angles are all #120°#
A hexagon has six sides, and the sum of interior angles of a polygon can be calculated in the formula: $180 \left(n - 2\right)$, where $n$ is the number of sides in the polygon.
Since a hexagon has six sides, the sum of interior angles is
$180 \left(6 - 2\right) = 180 \left(4\right) = 720$,
If it is a regular hexagon, all the sides and angles are equal.
so each angle of a regular hexagon is $\frac{720}{6} = 120$.
Thus, a regular hexagon has zero right angles.
It is only possible for a hexagon to have right angles if it is NOT a regular hexagon, which has been discussed in another answer.
An irregular hexagon can have 1,2,3,4 or 5 right angles. Try drawing different shaped hexagons using different numbers of right angles to show this for yourself.
It is possible to have 5 angles of #90°# then the last one will be #270°#
(#5 xx 90° +270° = 720°#)
Impact of this question
37007 views around the world | {"url":"https://socratic.org/questions/how-many-right-angles-does-a-hexagon-have","timestamp":"2024-11-07T03:44:27Z","content_type":"text/html","content_length":"35779","record_id":"<urn:uuid:29aef496-02af-472d-8416-d382e1c56004>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00147.warc.gz"} |
The Relationship between Social Media Usage and GPA among High School Students of St. Lauren High School | researchcenter
Mini- Study Part II
Answer the questions below about your data adhering to the outlined criteria in complete sentences. Cite any outside sources that are used.
Item Description
1 Treat your data just as you would one of the datasets from the homework. Be sure you include appropriate measures of central tendency and dispersion etc.
2 Construct a frequency distribution using 5 -8 classes.
3 Create 2 different but appropriate visual representations of your data (pie chart, bar graph, etc). You MUST use Excel to do this.
4 Complete the calculations for the 8 statistics you identified in your worksheet in week 3. You MUST use Excel to do this.
5 Write a brief paragraph describing the meaning or interpretation for EACH of the statistics. For example, if some of the statistics chosen were the mean, median and mode, which is the best
6 Construct a 95% Confidence Interval to estimate the population mean/proportion in the claim.
7 Complete the calculations for the 8 statistics you identified in your What can you conclude from this result regarding the topic?
8 Write up the responses to these questions in an APA paper between 500-1,000 words.
Topic used below.
1. The topic choosen for the study is " The relationship between hours spent in social media such as instagram, facebook and GPA scored by high school students of ST. Lauren high school.
So, The formulated research question is : Is there any relationship between hours spent in social media such as instagram, facebook and GPA scored by high school students of ST. Lauren high school.
2. Population refers to the the pool of individuals or members from where sample is drawn for a study.It is basically every individual of the group.Here, we are interested find about high school
students of ST. Lauren high school.So, our target population is the high school students of ST. Lauren high school.
3. Questions asked will be:
1. GPA scored by student in the last evaluation test : This is one of the most important question as we are to study relationship of GPA VS Hours spent in social media.
2. Approximate hours spent in social media : This is another most important question as we are to study relationship of GPA VS Hours spent in social media.
3. Preffered social media ( options will be given to choose such as Facebook, Instagram, chatrooms and others(if there is any other media that students would like to mention)) : This question is
included to gather data to find out which is the social networking site where students are spending most time . we have listed a few common sites and another option " other" is given which if
selected would provide the student space to write the name of the sites not mentioned in options.
4. Age of the student : we can figure out what age groups they belong to and whats the corresponding hours in social media as well as performance wrt age.
5. In which standard the student is enrolled in : we can evaluate how a particular class has performed and their corresponding social media usage.
6. gender : we can see effects in different gender
4, variables are as follows :
variable type
GPA discrete
Hours spent in social networking sites discrete
preferred social media categorial
Age discrete
standard enrolled in categorial
gender nominal
5. we can post the study to schools information board and website , we can use questionnaires and surveys to reach the student. An online poll via social media group or facebook groups of student can
also be used to reach the target population.
6. We can use questionnaires, polls, survey, google forms etc to collect data.
7. firstly , we can use stratified sampling based on criterias such as age and gender and based on the overall proportions of the population we can calculate how many people to be sampled from each
we can use simple random sampling which is chance based and every student has equal chance of being selected. for randomness we can use random number generator and assign numbers to each students or
we can select students random or we can decide any random games such as picking chits from a bowl etc to ensure that every student has equal probability of being included.
8. Two most appropriate graphs are :
• scatter plots which is a 2D or 3D plot that shows the joint variation of two or three variables from a group of observations. we can use two variables or 3 variables like age , hour spent and gpa
together and genertate the plot or we can see variation of age with respect to GPA.
• pie chart and bar graphs can also be used to ensure proper representation of categorial data as well as to find relative contribution of each individual or group.
9. appropriate descriptive statistics are:
• Mean = we can find average hours spent in social media or average GPA of students.
• mode= the the single score that occurs most often in the data.
• median= we can find mid point of GPA using this
• dispersion= It describes the spread of the data values in a given dataset and we can see effect of outliers.
• range= age range can be found out a well as hours spent on social sites as well as range of GPA etc
• standard deviation= variation from mean can be checked.
• correlation coefficient= relationship between age and GPA, relationship between hours of watching and GPA scored etc can be established and the direction and strength of association can also be
found out.
. Quartiles = measure of position can be checked using this. | {"url":"https://researchcenter.club/mini-study-part-ii-the-relationship-between-social-media-usage-and-gpa-among-high-school-students-of-st-lauren-high-school","timestamp":"2024-11-13T15:13:36Z","content_type":"text/html","content_length":"79116","record_id":"<urn:uuid:3039f923-7cb6-4774-9c8e-51818fb662cc>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00697.warc.gz"} |
Time and work problems
Submitted by Atanu Chaudhuri on Thu, 16/02/2017 - 13:19
Solution to Time and work problems Set 49
Learn to solve 10 Time and Work Problems in SSC CGL Solutions Set 49 in 12 mins. For best results, take the test first and then go through the solutions.
As all the problems are solved in quick time by problem solving techniques, to appreciate the solutions, you should take the test first at,
SSC CGL Question set 49 on Time and work.
Solutions to time and work problem set 49 for SSC CGL - time to solve was 12 minutes
Problem 1.
A alone can complete a job in 20 days while A and B together can complete it in 12 days. If B does the work only half a day daily then how many days would they take to complete the job working
a. 10 days
b. 15 days
c. 11 days
d. 20 days
Solution 1 : Problem analysis and execution
Assuming $A_w$ and $B_w$ to be the portion of the job $W$ done by A and B each working alone respectively in 1 day, from the first statement we have,
Similarly from the second statement we have,
Eliminating $A_w$ between the two equations,
$B_w=\left(\frac{1}{12}-\frac{1}{20}\right)W = \frac{1}{30}W$.
Half day's work of B is then,
A and B together with B working half a day daily will then complete the portion of work in a day,
In other words they will take 15 days to complete the work.
Answer: b: 15 days.
Key concepts used: Work rate technique -- Working together per unit time concept.
Problem 2.
A and B together can do a piece of work in 12 days which B and C can do together in 16 days. After A has been working at it for 5 days and B for 7 days, C finishes it in 13 days. In how many days B
would finish the job working alone?
a. 24 days
b. 48 days
c. 12 days
d. 16 days
Solution 2 : Problem analysis
From the first two statements in the first sentence we get two linear equations in $A_w$, $B_w$, $C_w$, the individual work rates of A, B, and C respectively in portion of work $W$ done in 1 day.
From the second sentence we will get a third linear equation in terms of the four quantities. From these three equations, it is easy to find $B_w$ and hence number of days to be taken by B to
complete the work alone.
Though the problem is a straightforward one, there may be slight hesitation in forming the third equation. The doubt that may arise is, whether they worked sequentially one after the other, or
started to work together at the same time. From the construction of the sentence, "After A has been working at it for 5 days and B for 7 days, C finishes it in 13 days." in the absence of any hint
regarding whether they started to work together, we would assume, they worked one after the other - the simplest assumption.
Solution 2 : Problem solving execution
The three equations are,
$B_w+C_w=\frac{1}{16}W$, and
We will first convert $A_w$ in first equation to $5A_w$,
Then we will convert $C_w$ in second equation to $13C_w$,
Next we sum up the two,
Lastly we will subtract the third equation from the last result thus eliminating $A_w$ and $C_w$,
Or, $48B_w=W$.
B completes the work alone in 48 days.
Answer: Option b : 48 days
Key concepts used: Work rate technique -- Working together per unit time concept -- Solving linear equations -- Efficient simplification.
Problem 3.
If 90 men can do a certain job in 16 days working 12 hours a day, then the portion of the job that can be completed by 70 men working 8 hours day for 24 days is,
a. $\displaystyle\frac{5}{8}$
b. $\displaystyle\frac{1}{3}$
c. $\displaystyle\frac{2}{3}$
d. $\displaystyle\frac{7}{9}$
Solution 3 : Problem analysis
As an hour is the smallest unit of time and teams of men work for days and hours, we will convert all times into hours and will use the manhours concept instead of mandays concept for designating
work amount.
Solution 3 : Problem solving execution
90 men can do a certain job in 16 days working 12 hours a day. So total work amount is,
$W=90\times{16}\times{12}$ manhours.
By the second statement, 70 men work 8 hours day for 24 days, thus completing an work amount,
$W_1=70\times{8}\times{24}$ manhours.
Thus $W_1$ as a portion of $W$ is,
Answer: Option d: $\displaystyle\frac{7}{9}$.
Key concepts used: Mandays technique -- Manhours technique -- Unitary method -- Delayed evaluation -- Efficient simplification.
Note: We have not carried out the two multiplications and waited for the common factors to cancel out in the last division operation. This is delayed evaluation technique in action which is very
useful in saving precious seconds in calculation.
Problem 4.
A, B and C can do a job in 30, 20 and 10 days respectively. A is assisted by B on one day and by C on the next day alternately. How long would the work take to finish?
a. $8\displaystyle\frac{4}{13}$ days
b. $3\displaystyle\frac{9}{13}$ days
c. $9\displaystyle\frac{3}{8}$ days
d. $4\displaystyle\frac{3}{8}$ days
Solution 4 : Problem analysis
Assuming $A_w$, $B_w$, and $C_w$ to be the work rates or portion of work $W$ done in a day by A, B and C respectively, from the given values we have,
$B_w=\frac{1}{20}W$, and
A works here continuously throughout the period up to the completion whereas B and C work on alternate days. In the absence of any other specific information, as B's work is memtioned first, we will
assume the work started with A and B working together on the first day and A and C worked on the second day. This pattern of working is repeated for every two days.
In first two days then the total work done is,
If 60 were a multiple of 13 the work would have been finished in an even number of days. As it is not the case here, we have to use the Boundary condition evaluation technique.
Solution 4 : Problem solving execution
As $52=4\times{13}$ is the largest multiple of 13 (portions out of 60 in each 2 days), we can safely go up to the work done in $4\times{2}=8$ days as,
After 8 days then leftover work will be,
On the 9th day A and B will complete,
On the start of the 10th day then work left will be,
Now we will apply the unitary method on the leftover work and rate of work of A and C working together, that is,
At this rate to finish the leftover job of $\frac{3}{60}W$, A and C will take,
$\displaystyle\frac{3}{8}$ days.
So total duration for the job to be completed is, $9\displaystyle\frac{3}{8}$ days.
Answer: c: $9\displaystyle\frac{3}{8}$ days.
Key concepts used: Work rate technique -- Working together per unit time concept -- Leftover work analysis -- Boundary determination technique -- Unitary method.
Note: On 9th day itself the cumulative work done could have exceeded the whole work $W$. In that case, whole number of days would have been 8 and unitary method would have been applied on the 9th day
with A and B working together.
Problem 5.
A, B, and C can do a work separately in 16, 32 and 48 days respectively. They started the work together but B left 8 days and C six days before completion of the work. What was the duration of
completion of the work?
a. 14 days
b. 9 days
c. 10 days
d. 12 days
Solution 5 : Problem analysis
As this is a problem of three work agents working together for different durations we will as usual use the work rate technique assuming $A_w$, $B_w$, and $C_w$ as the portion of work $W$ done by A,
B, and C respectively in 1 day.
By given information these work rates are,
$B_w=\frac{1}{32}W$, and
As B stopped working 8 days and C stopped 6 days before completion of the work we need to assume the duration of completion as an unknown number of $n$ days and work back for number of days B and C
Solution 5 : Problem solving execution
A worked for the whole duration, B worked for $(n-8)$ days and C for $(n-6)$ days. So,
Or, $W=W\left(\frac{1}{16}n +(n-8)\frac{1}{32}+(n-6)\frac{1}{48}\right)$,
Or, $\frac{1}{16}n +(n-8)\frac{1}{32}+(n-6)\frac{1}{48}=1$.
Multiplying both sides by 96, the LCM of the denominators of the three fractions, 16, 32 and 48,
$6n + 3n - 24+2n-12 = 96$,
Or, $11n=132$,
Or, $n=12$.
In 12 days the work will be completed.
Answer: Option d: 12 days.
Key concepts used: Work rate technique -- Working together per unit time concept -- Working backwards approach.
Problem 6.
A can do one-fourth of a work in 10 days and B can do one-third of the work in 20 days. In how many days would A and B be able to complete the work while working together?
a. 32 days
b. 30 days
c. 25 days
d. 24 days
Solution 6 : Problem analysis and execution
If $A_w$ and $B_w$ be the per day portion of work $W$ done by A and B, we have,
Or, $A_w=\frac{1}{40}W$
Or, $B_w=\frac{1}{60}W$.
So working together portion of work done by A and B in 1 day will be,
Or, $24(A_w+B_w)=W$.
So A and B working together will be able to complete the work in 24 days while working together.
Answer: Option d : 24 days.
Key concepts used: Work rate technique -- Working together per unit time concept.
Problem 7.
A and B working separately can do a job in 9 and 15 days respectively. If they work for a day alternately, with A beginning, then the work will be completed in,
a. 11 days
b. 9 days
c. 12 days
d. 10 days
Solution 7 : Problem analysis and execution
By the given information portion of work $W$ done by A in 1 day is,
Similarly portion of work done by B in 1 day is,
As A and B work together with A beginning the work, every 2 days work done by the two will be,
So after 10 days work done will be, $\frac{40}{45}W$ with $\frac{1}{9}W$ work left which will be completed by A on the next day.
So total duration for completion of the work is $10+1=11$ days.
Answer: Option a: 11 days.
Key concepts used: Work rate technique -- Working together per unit time concept -- Boundary determination technique.
Problem 8.
A takes 3 times as long as B and C together take to complete a job. B takes 4 times as long as A and C together take to do the same job. If all three working together complete the job in 24 days, A
alone can do it in,
a. 100 days
b. 90 days
c. 95 days
d. 96 days
Solution 8 : Problem analysis and execution
With $A_w$, $B_w$ and $C_w$ as the work rates per day of A, B and C respectively we have,
$4B_w=A_w+C_w$, and
Substituting $3A_W$ for $B_w+C_w$ in equation 3,
So A alone will take 96 days to complete the work.
Useful pattern recognition makes second equation superfluous.
Answer: Option d: 96 days.
Key concepts used: Work rate technique -- Working together per unit time concept -- useful pattern identification -- efficient simplification.
Problem 9.
15 men take 20 days to complete a job working 8 hours a day. The number of hours a day should 20 men take to complete the job in 12 days will be,
a. 5 hours
b. 10 hours
c. 18 hours
d. 15 hours
Solution 9 : Problem analysis and execution:
Total manhours required for the job is,
$W=15\times{20}\times{8}=2400$ manhours.
This will be equal to,
$W=20\times{12}=240$ mandays.
So the team of 20 men will have to work 10 hours a day for 12 days to complete the work.
Answer: Option b: 10 hours.
Key concepts used: Mandays technique.
Problem 10.
A and B can do a piece of work in 15 days. B and C together can do the same job in 12 days and C and A in 10 days. How many days will A take to complete the job?
a. 8
b. 13
c. 24
d. 40
Solution 10 : Problem analysis and execution:
Assuming $A_w$, $B_w$ and $C_w$ to be the work rates of A, B and C respectively in terms of portion of work $W$ done in a day,
$B_w+C_w=\frac{1}{12}W$, and
Subtracting first equation from second,
Now subtracting this result from the third equation,
Or, $24A_w=W$.
A completes the work alone in 24 days.
Answer: Option c: 24.
Key concepts used: Work rate technique -- Working together per unit time concept -- Solving linear equations.
Useful resources to refer to
Guidelines, Tutorials and Quick methods to solve Work Time problems
7 steps for sure success in SSC CGL Tier 1 and Tier 2 competitive tests
How to solve Arithmetic problems on Work-time, Work-wages and Pipes-cisterns
Basic concepts on Arithmetic problems on Speed-time-distance Train-running Boat-rivers
How to solve a hard CAT level Time and Work problem in a few confident steps 3
How to solve a hard CAT level Time and Work problem in a few confident steps 2
How to solve a hard CAT level Time and Work problem in few confident steps 1
How to solve Work-time problems in simpler steps type 1
How to solve Work-time problem in simpler steps type 2
How to solve a GATE level long Work Time problem analytically in a few steps 1
How to solve difficult Work time problems in simpler steps, type 3
SSC CGL Tier II level Work Time, Work wages and Pipes cisterns Question and solution sets
SSC CGL Tier II level Solution set 26 on Time-work Work-wages 2
SSC CGL Tier II level Question set 26 on Time-work Work-wages 2
SSC CGL Tier II level Solution Set 10 on Time-work Work-wages Pipes-cisterns 1
SSC CGL Tier II level Question Set 10 on Time-work Work-wages Pipes-cisterns 1
SSC CGL level Work time, Work wages and Pipes cisterns Question and solution sets
SSC CGL level Solution Set 72 on Work time problems 7
SSC CGL level Question Set 72 on Work time problems 7
SSC CGL level Solution Set 67 on Time-work Work-wages Pipes-cisterns 6
SSC CGL level Question Set 67 on Time-work Work-wages Pipes-cisterns 6
SSC CGL level Solution Set 66 on Time-Work Work-Wages Pipes-Cisterns 5
SSC CGL level Question Set 66 on Time-Work Work-Wages Pipes-Cisterns 5
SSC CGL level Solution Set 49 on Time and work in simpler steps 4
SSC CGL level Question Set 49 on Time and work in simpler steps 4
SSC CGL level Solution Set 48 on Time and work in simpler steps 3
SSC CGL level Question Set 48 on Time and work in simpler steps 3
SSC CGL level Solution Set 44 on Work-time Pipes-cisterns Speed-time-distance
SSC CGL level Question Set 44 on Work-time Pipes-cisterns Speed-time-distance
SSC CGL level Solution Set 32 on work-time, work-wage, pipes-cisterns
SSC CGL level Question Set 32 on work-time, work-wages, pipes-cisterns
SSC CHSL level Solved question sets on Work time
SSC CHSL Solved question set 2 Work time 2
SSC CHSL Solved question set 1 Work time 1
Bank clerk level Solved question sets on Work time
Bank clerk level solved question set 2 work time 2 | {"url":"https://suresolv.com/ssc-cgl/ssc-cgl-level-solution-set-49-time-and-work-problems-simpler-steps-4","timestamp":"2024-11-04T04:57:53Z","content_type":"text/html","content_length":"52121","record_id":"<urn:uuid:896362d9-b5e9-41f2-9a4f-fa72c7306a05>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00313.warc.gz"} |