url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
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https://mail.python.org/pipermail/python-list/2008-February/507476.html | 1,397,692,118,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00482-ip-10-147-4-33.ec2.internal.warc.gz | 816,217,513 | 1,963 | # Is there an easy way to sort a list by two criteria?
Duncan Booth duncan.booth at invalid.invalid
Sun Feb 10 06:44:37 CET 2008
thebjorn <BjornSteinarFjeldPettersen at gmail.com> wrote:
> I'm not sure which Python is default for Ubuntu 6.06, but assuming you
> can access a recent one (2.4), the list.sort() function takes a key
> argument (that seems to be rather sparsely documented in the tutorial
> and the docstring...). E.g.:
>
>>>> lst = [(1,2,4),(3,2,1),(2,2,2),(2,1,4),(2,4,1)]
>>>> lst.sort(key=lambda (a,b,c):(c,b))
>>>> lst
> [(3, 2, 1), (2, 4, 1), (2, 2, 2), (2, 1, 4), (1, 2, 4)]
>>>>
It may be simpler just to use the key argument multiple times (not
forgetting to specify the keys in reverse order, i.e. the most significant
comes last). So with this example, sorting by column 2 then column 1 and
ignoring column 0 can be done by:
>>> from operator import itemgetter
>>> lst = [(1,2,4),(3,2,1),(2,2,2),(2,1,4),(2,4,1)]
>>> lst.sort(key=itemgetter(1))
>>> lst.sort(key=itemgetter(2))
>>> lst
[(3, 2, 1), (2, 4, 1), (2, 2, 2), (2, 1, 4), (1, 2, 4)]
or even:
>>> from operator import itemgetter
>>> lst = [(1,2,4),(3,2,1),(2,2,2),(2,1,4),(2,4,1)]
>>> for keycolumn in reversed([2,1]):
lst.sort(key=itemgetter(keycolumn))
>>> lst
[(3, 2, 1), (2, 4, 1), (2, 2, 2), (2, 1, 4), (1, 2, 4)]
The important point here is to remember that the sort is stable (so you can
do multiple sorts without disrupting earlier results).
More information about the Python-list mailing list | 540 | 1,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2014-15 | latest | en | 0.716951 |
https://www.sanfoundry.com/highway-engineering-questions-answers-flexible-pavement-design-method-1/ | 1,716,401,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00482.warc.gz | 832,415,311 | 20,799 | # Highway Engineering Questions and Answers – Flexible Pavement Design Method – 1
This set of Highway Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Flexible Pavement Design Method – 1”.
1. How many types of methods are there to design a flexible pavement?
a) One
b) Two
c) Three
d) Four
Explanation: There are three types of methods to design a flexible pavement they are empirical, semi empirical and theoretical method.
2. Empirical method is dependent on the strength of _____________
a) Soil
b) Sub base
c) Base
d) Surface
Explanation: The strength of the soil sub grade is used to design the flexible pavement, because the sub base and base are more stronger than the sub grade, so design depends on sub grade only.
3. The stress strain approach is used in _____________
a) Empirical method
b) Semi empirical method
c) Theoretical
d) CBR method
Explanation: The design of pavement using stress strain approach is used in semi empirical method, which depends both on theory and graph.
4. CBR is a _____________
a) Measure of soil strength
b) Flexible pavement design method
c) Rigid pavement design method
d) Measure of soil characteristics
Explanation: CBR is a method of designing flexible pavement by using the soil characteristics.
5. The design charts are prepared based on _____________
a) Climate
b) Past experience
c) Location
d) Traffic
Explanation: The design charts are prepared based on the past experience hence they are not reliable, as it changes according to location.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!
6. For which material the semi empirical method is useful?
a) Elastic
b) Plastic
c) Rigid
d) Semi rigid
Explanation: If the material is elastic, then it will obey hookes law, so after that limit it is not so useful.
7. The CBR method was developed by _____________
a) California division of highway
b) IRC
c) MORTH
d) NHAI
Explanation: In 1925, the California division of the highway designed the CBR method for the design of flexible pavement.
8. The soaking period in CBR sample is _____________
a) 2 days
b) 3 days
c) 4 days
d) 5 days
Explanation: The soaking period usually specified is four days which may not be sufficient for some highways.
9. What is the total thickness of the pavement?
a) Constant
c) Changes with sub base
d) Changes with base
Explanation: The total thickness of the pavement remains constant as it does not depend on the number of layers.
10. As per MORTH the specified compaction of density is _____________
a) 95%
b) 96%
c) 100%
d) 99%
Explanation: MORTH recommends a field density of 97% for heavy compaction, in some cases it recommended even 99 or 100% of lab density.
11. The top 500mm of soil sub grade should be compacted at _____________
a) OMC
b) MDD
c) Dry density
d) Saturated density
Explanation: The top 500mm layer in the soil sub grade should be compacted at OMC, to obtain M.D.D, only at M.D.D the soil gets its maximum strength.
12. For how much amount of rainfall soaking of specimen is not required?
a) 100 mm
b) 200 mm
c) 300 mm
d) 500 mm
Explanation: For a rainy area which is having an amount of rainfall more than 500 mm, then soaking of specimen is not required.
13. What is the maximum aggregate size in CBR method?
a) 20 mm
b) 30 mm
c) 40 mm
d) 50 mm
Explanation: The maximum size of aggregate should be 20 mm in the design of flexible pavement if it is more than 20 mm then above layers are not valid.
14. The critical stress is considered in _____________
a) Surface
b) Sub base layer
d) Both sub grade and surface
Explanation: Both horizontal stress and vertical stress are considered for flexible pavement and measured at these two layers.
15. The design factor not considered in CBR is _____________
a) Weather
b) Traffic
c) VDF
d) Growth rate
Explanation: The weather factor is not considered in CBR method as it keeps changing frequently, growth rate keeps increasing, VDF is a constant so climate doesn’t actually affect the CBR.
Sanfoundry Global Education & Learning Series – Highway Engineering.
To practice all areas of Highway Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 1,048 | 4,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-22 | latest | en | 0.88432 |
https://www.physicsforums.com/threads/binomial-theorem-problem-on-the-terms-of-an-expansion.606427/ | 1,508,485,975,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823839.40/warc/CC-MAIN-20171020063725-20171020083725-00698.warc.gz | 976,216,643 | 14,997 | # Binomial theorem problem on the terms of an expansion
1. May 16, 2012
### agnibho
1. The problem statement, all variables and given/known data
Find an approximation of (0.99)5 using the first three terms of its expansion.
2. The attempt at a solution
To get to the binomial theorem I divided 0.99 into
(0.99)5 = (1-0.01)5 = {1+(-0.01)}5
Then,
T1 = 5C0(1)5 = 1 x 1=1
T2 = 5C1(1)5-1(-0.01)1 = 5x1x -0.01= (-0.05)
T3 = 5C2(1)5-2(-0.01)2 = 10 x 1x 0.0001 = (0.001)
2. May 16, 2012
### Ray Vickson
Why do you say "adding them doesn't get me to the answer"? Of course, it does not get you to the exact value of 0.99^5, but that is not the issue. Just adding the terms you have fulfills all the requirements of the problem.
RGV
3. May 17, 2012
### agnibho
OK thanks for the help! Actually I had some confusion about that operation.
Um....I's thinking that will I get to the same answer if I had divided 0.99 into (0.9+0.09)?
4. May 17, 2012
### HallsofIvy
Staff Emeritus
You would get another approximation, not necessarily the same answer. You would have to calculate the fifth, fourth, and third powers of .9 which is harder than the same powers of 1!
Last edited: May 17, 2012 | 412 | 1,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-43 | longest | en | 0.940948 |
andreychurkin.ru | 1,723,008,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00244.warc.gz | 77,871,520 | 13,273 | Search
# Basin stability
What do you see in the picture? Actually, as it usually happens in science, you see nature. In this case – the basin stability of a one-node power system, where green nodes stand for the regimes that fall into the attractor of the synchronous state.
As part of the Numerical Methods course led by prof. Aslan Kasimov, I studied the synchronization models of power systems and implemented the basin stability, a prominent concept introduced by Peter Menck et al. Explicit Runge-Kutta methods were used to solve the initial value problem posed by the classical swing equation system. It was found that the accuracy and the time step of the methods may significantly influence the estimation of the system’s stability. Convergence analysis of the errors showed that it is preferable to use higher order Runge-Kutta methods to reach a better balance between accuracy and computational cost. The complete report may be found here:
http://materials.andreychurkin.ru/Churkin%20project%20v2.4%2004.04.2020.pdf
In short, there are two possible outcomes of an electromechanical dynamic process caused by a large perturbation in the system: synchronization (generator’s angular frequency settles to a constant value) or loss of synchrony (generator accelerates until reaching the non-synchronous limit cycle). The following figures show how these outcomes depend on initial conditions.
Synchronization after the initial regime:
Loss of synchronization after the initial regime:
There is a systematic way of analyzing system stability: running numerous simulations with different initial conditions and calculating the basin stability – the ratio of the regimes that lead to the synchronous state to all possible initial conditions in the considered range. So, what you see in the figure below is the result of 40 401 simulations, where the green points depict synchronous states, and red points – loss of synchronization. The stability index, in this case, reaches 38%.
The solutions from the previous simulations can also be represented as trajectories in the stability basin. For example, the regime “1s” leads to a synchronous state where the solution reaches the major green region and starts swinging towards its center. The “1ns” initial values make the generator accelerating up to point “2ns” and further up to the non-synchronous limit cycle.
What else can we see using the basin stability concept? Well, we can play with the model and double the transmission capacity. In this case, the basin will significantly increase, and the stability index would reach 54%:
To further analyze the convergence of the explicit multistage methods for the dynamic power grid model, we estimated the initial stability basin but using an algorithm with a coarse time step. The modified basin got significantly deformed, especially at the bottom of the figure:
The modified stability index changed to 41%. So, the method with a coarse time step underestimated some oscillations and mistakenly enlarged the stability region. In the report, we performed a complete convergence analysis of the Runge-Kutta methods and concluded that higher order methods are superior in terms of accuracy/cost balance.
Error estimation of the methods:
I am grateful to prof. Aslan Kasimov, who recommended me investigating the oscillator networks and shared the paper by P. Menck on basin stability. This was a hard but entertaining journey.
Andrey Churkin (Андрей Чуркин) 2020
## One Comment
• ### Dr. Dre
When you have brilliant mind you can afford journey even during pandemy of COVID-19. Nice work. | 720 | 3,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.912959 |
https://pdfcoffee.com/aerodynamics-mock-board-2-pdf-free.html | 1,656,922,854,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00279.warc.gz | 492,559,835 | 12,456 | # Aerodynamics Mock Board 2
##### Citation preview
Feati University Helios St. Sta. Cruz Metro Manila Feati Aeronautical Engineering Board Review Program Computational Aerodynamics Mock Board Exam Directions: Write the letter of the correct answer on the space provided before the number. 1. Consider an airplane flying at a pressure altitude of 33,500 ft and a density altitude of 32,000 ft. Calculate the outside air temperature. a. 378 R c. 368 R b. 380 R d. 370 R 2. An F-15 supersonic fighter aircraft is in rapid climb. At the instant it passes through a standard altitude of 25,000 ft, its time rate of change of altitude is 500 ft/s, which by definition is the rate of climb. Corresponding to this rate of climb at 25,000 ft is a time rate of change of ambient pressure. Calculate this rate of change of pressure in units of lb/ (ft2) (s). a. -16.16 lb/ft2 s c. -20.25 lb/ft2 s 2 b. -17.17 lb/ft s d. -19.88 lb/ft2 s 3. An airplane is flying at a velocity of 130 mi/h at a standard altitude of 5000 ft. At a point on the wing, the pressure is 1750.0 lb/ft2. Calculate the velocity at that point, assuming incompressible flow. a. 220.5 fps c. 216.8 fps b. 218.9 fps d. 212.8 fps 4. A supersonic transport is flying at a velocity of 1500 mi/h at a standard altitude of 50,000 ft. The temperature at a point in the flow over the wing is 793.32R. Calculate the flow velocity at that point. a. 6.9 fps c. 7.9 fps b. 6.3 fps d. 7.2 fps 5. Consider a low speed subsonic wind tunnel with a nozzle contraction ratio of 1:20. One side of the mercury manometer is connected to the settling chamber, and the other side to the test section. The pressure and temperature in the test section are 1 atm and 300 K, respectively. What is the height difference between the two columns of mercury when the test section velocity is 80 m/s? a. 3.8 cm c. 2.8 cm b. 3.2 cm d. 2.0 cm 6. A pitot tube is mounted in the test section of a low speed subsonic wind tunnel. The flow in the test section has a velocity, static pressure, and temperature of 150 mi/h, 1 atm, and 70 F, respectively. Calculate the pressure measured by the pitot tube. a. 2172 lb/ft2 c. 2072 lb/ft2 2 b. 3172 lb/ft d. 3072 lb/ft2 7. The altimeter on a low speed airplane reads 2 km. The airspeed indicator reads 50 m/s. If the outside air temperature is 280 K, what is the true velocity of the airplane? a. 56 mps c. 50 mps b. 54 mps d. 52 mps 8. A pitot tube is mounted in the test section of a high speed subsonic wind tunnel. The pressure and temperature of the airflow are 1 atm and 270 K, respectively. If the flow velocity is 250 m/s, what is the pressure measured by the pitot tube? a. 1.53 x 105 N/m2 c. 1.48 x 105 N/m2 5 2 b. 1.55 x 10 N/m d. 1.42 x 105 N/m2 9. Consider the flow of air through a supersonic nozzle. The reservoir pressure and temperature are 5 atm and 500 K, respectively. If the Mach number at the nozzle exit is 3, calculate the exit pressure, temperature, and density. a. 1.67 x 104 N/m2, 190 K, 0.550 kg/m3 c. 1.45 x 104 N/m2 , 181.6 K, 0.252 kg/m3 4 2 3 b. 2.09 x 10 N/m , 208.7 K, 0.673 kg/m d. 1.37 x 104 N/m2 , 178.6 K, 0.267 kg/m3 10. Consider an airfoil mounted in a low speed subsonic wind tunnel. The flow velocity in the test section is 100 ft/s, and the conditions are standard sea level. If the pressure at a point on the airfoil is 2012 lb/ft2, what is the pressure coefficient? a. 1.18 c. 1.28 b. -1.18 d. -1.28 11. Consider the Northtrop F-5 fighter airplane, which has a wing area of 170 ft2. The wing is generating 18,000 lb of lift. For a flight velocity of 250 mi/h at standard sea level, calculate the lift coefficient. a. 0.6626 c. 0.6526 b. 0.5662 d. 0.4662 12. Consider an airfoil in a freestream with a velocity of 50 m/s at standard sea level conditions. At a point on the airfoil, the pressure is 9.5 x 104 N/m2. What is the pressure coefficient at this point? a. -3.91 c. -3.51 b. -3.71 d. -3.81
13. The wing area of the Lockheed F-104 straight wing supersonic fighter is approximately 210 ft2. If the airplane weighs 16,000 lbs and is flying in level flight at Mach 2.2 at a standard altitude of 36,000 ft, estimate the wave drag on the wings. a. 366 lbs c. 326 lbs b. 386 lbs d. 376 lbs 14. The Cessna Cardinal, a single engine light plane, has a wing with an area of 16.2 m2 and an aspect ratio of 7.31. Assume the span efficiency factor is 0.62. If the airplane is flying at standard sea level conditions with a velocity of 251 km/h, what is the induced drag when the total weight is 9800 N? a. 149.5 N c. 139.5 N b. 159.5 N d. 129.5 N 15. Consider a light, single engine airplane such as the Piper Super Cub. If the maximum gross weight of the airplane is 7780 N, the wing area is 16.6 m2, and the maximum lift coefficient is 2.1 with flaps down, calculate the stalling speed at sea level. a. 64.5 km/h c. 75.3 km/h b. 32.8 km/h d. 68.7 km/h 16. Consider an airplane patterned after the twin engine Beechcraft Queen Air executive transport. The airplane weight is 38,220 N, wing area is 27.3 m2, aspect ratio is 7.5. Oswald efficiency factor is 0.9, and parasite drag coefficient CD,o = 0.03. Calculate the thrust required to fly at a velocity of 350 km/h at a.) Standard sea level and b.) An altitude of 4.5 km. a. TR (at sea level) = 5179 N, TR (at 4.5km) = 3711 N c. TR (at sea level) = 5200 N, TR (at 4.5 km) = 3800 N b. TR (at sea level) = 5600 N, TR (at 4.5 km) = 4000 N d. TR (at sea level) = 4179 N, TR (at 4.5 km) = 3611 N 17. The maximum lift to drag ratio of the World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine fails, how far can it glide in terms of distance measured along the ground? a. 629 miles c. 700 miles b. 719 miles d. 729 miles 18. Consider an airplane with a parasite drag coefficient of 0.025, an aspect ratio of 6.72, and an Oswald efficiency factor of 0.9. Calculate the value of (L/D) max a. 13.78 c. 14.78 b. 13.55 d. 14.55 19. A supersonic nozzle is also a convergent-divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the reservoir of the nozzle, the pressure and temperature are 10 atm and 300 K, respectively. At the nozzle exit, the pressure is 1 atm. Calculate the temperature and density of the flow at the exit. Assume the flow is isentropic and, of course, compressible. a. TE = 155 K, ρE = 2.26 kg/m3 c. TE = 150 K, ρE = 2.20 kg/m3 3 b. TE = 145 K, ρE = 1.26 kg/m d. TE = 153 K, ρE = 2.22 kg/m3 20. The altimeter on a low speed Piper Aztec reads 8000 ft. A pitot tube mounted on the wing tip measures a pressure of 1650 lb/ft2. If the outside air temperature is 500 R, what is the true velocity of the airplane? a. 300 fps c. 290 fps b. 295 fps d. 292 fps 21. Find R.N. for an airplane wing 4 ft chord, moving at 150 mph. Air is 40 C; barometer 21 in Hg a. 4,500,000 c. 4,300,000 b. 3,400,000 d. 5,300,000 22. A horizontal pipe, 1 ft in diameter, tapers gradually to 8 in. in diameter. If the flow is 500 cu ft of water per minute, what is the difference between the pressures at the two sections? a. 3.08 psi c. 4.08 psi b. 3.01 psi d. 4.01 psi 23. Alcohol (spec. Grav = 0.80) is flowing through a horizontal pipe, which is 10 in. in diameter, with a velocity of 40 fps. At a smaller section of the pipe, there is 6 psi less pressure. Asssuming that the flow is smooth, what is the diameter there? a. 0.85 ft c. 0.45 ft b. 0.73 ft d. 0.80 ft 24. A stream of air 72 sq ft in cross section is moving horizontally at a speed of 100 mph. What force is required to deflect it downward 10 degrees without loss in speed? a. 600 lb c. 655 lb b. 632 lb d. 642 lb 25. An airplane weighs 7,000 lb, the wing area is 460 sq ft, the cl of the wing is 0.85 and the cd is 0.053. What power is required by the wing at sea level? a. 97.4 hp c. 95.9 hp b. 96.9 hp d. 98.7 hp 26. A cylinder 3 ft in diameter and 8 ft long is rotating at 150 rpm in air stream of 50 mph. What is the total lift? a. 310 lb c. 415 lb b. 410 lb d. 315 lb 27. At an airspeed of 95 mph at sea level, what is the induced drag of a monoplane weighing 4,700 lb and having a wing span of 52 ft? a. 115.8 lb c. 115.9 lb b. 112.6 lb d. 113.5 lb
28. A Fairchild monoplane weighs 2,550 lb; its wing span is 36 ft, 4in. At 5000 ft altitude, what is the induced drag at an airspeed of 122 mph? a. 67.9 lb c. 47.8 lb b. 55.9 lb d. 40.9 lb 29. An airplane is flying at 480 mph at an altitude of 30,000 ft. What is the critical velocity? a. 659 mph c. 780 mph b. 649 mph d. 770 mph 30. An airplane is flying at 500 knots in air at -50 F. What is the critical velocity? a. 574 knots c. 500 knots b. 600 knots d. 692 knots 31. For an airplane flying at 400 knots at 25,000 ft altitude, find the critical value of the pressure coefficient. a. -0.934 c. 0.915 b. -0.884 d. 0.873 32. For an airplane flying at 450 knots in air at -35 F, find the critical value of the pressure coefficient a. -0.765 c. -0.583 b. -0.632 d. -0.683 33. The service ceiling of an airplane is 15,000 ft. Its rate of climb at sea level is 890 fpm. What is the absolute ceiling? a. 14,900 ft c. 17,500 ft b. 16,900 ft d. 15,900 ft 34. An airplane weighs 4,200 lb. What is the excess horsepower at service ceiling? a. 13.9 hp c. 10.8 hp b. 12.7 hp d. 11.7 hp 35. The service ceiling of an airplane is 21,300 ft. The rate of climb at sea level is 950 fpm. What is the rate of climb at 13,000 ft altitude? a. 521 fpm c. 431 fpm b. 473 fpm d. 545 fpm 36. At sea level, an airplane weighing 5,200 lb has 135 excess horsepower. Its absolute ceiling is 19,000 ft. How long will it take to climb from sea level to 7,000 ft? a. 10.2 min c. 11.46 min b. 11.2 min d. 9.50 min 37. A Luscombe trainer climbs 900 fpm at sea level. How long does it take to climb to its service ceiling, 15,000 ft? a. 41.2 min c. 43.4 min b. 45.6 min d. 42.9 min 38. An airplane takes 7 min 30 sec to reach 8000 ft altitude. In that same time interval (i.e., 15 min from sea level) it reaches 13,600 ft altitude. What is the ceiling? a. 25,400 ft c. 26,667 ft b. 26,467 ft d. 27,100 ft 39. In still air, an airplane can maintain an 8 degrees angle of glide with airspeed of 96 mph. At that same airspeed, what is the true angle of glide when gliding in the same direction as a 20 mph wind? a. 5.9 deg c. 6.0 deg b. 6.6 deg d. 5.0 deg 40. An airplane weighs 3,000 lb and has a landing speed of 50 mph. What is the landing speed with 500 lb additional load? a. 50 mph c. 60 mph b. 53 mph d. 64 mph 41. A basic training plane weighing 3,530 lb has a landing speed of 58.5 mph. What is the landing speed after 350 lb of fuel have been burned? a. 65.9 mph c. 53.8 mph b. 60.5 mph d. 55.5 mph 42. An airplane with 300 sq ft of wing area has a landing speed of 40 mph. If wing area is reduced to 250 sq ft, what is the landing speed? a. 43.8 mph c. 44.9 mph b. 42.5 mph d. 42.9 mph 43. An airplane is making an endurance flight and is therefore flying constantly at the angle of minimum power required at which angle CL = 0.715, and CDt = 0.060. If the engine burns 0.5 lb per hp per hour, if the propeller efficiency is 83% and if the wing area is 255 sq ft, how long may flight be maintained if the airplane takes off with 6,000 lb gross weight of which 2,000 lb are fuel? a. 27.5 hr c. 28.4 hr b. 26.9 hr d. 29.4 hr
44. An airplane weighs 4,000 lb and takes off with 80 gal of fuel. It has a Clark Y wing of 216 sq ft area and has 3.8 sq ft equivalent flat plate area of parasite. It has a 180 hp engine. The maximum efficiency of the propeller is 78 %. Assume fuel consumption to be 0.55 lb per bhp per hr and maximum velocity to be 135 mph. Find the range. a. 770 miles c. 780 miles b. 760 miles d. 750 miles 45. A plane of 3,800 lb gross weight is turning at 175 mph with an angle of bank of 50 deg. Calculate the centrifugal force. a. 4529 lb c. 4429 lb b. 4625 lb d. 4629 lb 46. An airplane is making a 40 deg. banked turn of 565 ft radius. What should be the airspeed? a. 83.2 mph c. 80.3 mph b. 84.2 mph d. 81.9 mph 47. An airplane is making a turn of 1/8 mile radius at a speed of 225 mph. What is the load factor? a. 5.0 c. 5.5 b. 5.9 d. 5.2 48. An airplane is flying horizontally at 250 knots. Three seconds later, it is flying at the same speed but an angle of climb of 20 deg. What is the acceleration? a. 1.53 g c. 1.48 g b. 1.75 g d. 1.68 g
Prepared By: Engr. Kenneth Rene Ian M. Talag Aerodynamics Reviewer
Answer Key: 1. A 2. B 3. C 4. B 5. C 6. A 7. A 8. C 9. D 10. B 11. A 12. A 13. A 14. C 15. D 16. A 17. D 18. A 19. A 20. D
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
B A B D A A B C B A A C B B C A A C B B
41 42 43 44 45 46 47 48
D A D A A B D A | 4,240 | 12,602 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2022-27 | latest | en | 0.841071 |
https://www.primidi.com/force | 1,611,574,344,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565541.79/warc/CC-MAIN-20210125092143-20210125122143-00592.warc.gz | 917,740,310 | 4,756 | # Force
In physics, a force is any influence that causes an object to undergo a certain change, either concerning its movement, direction, or geometrical construction. It is measured with the SI unit of newtons and represented by the symbol F. In other words, a force is that which can cause an object with mass to change its velocity (which includes to begin moving from a state of rest), i.e., to accelerate, or which can cause a flexible object to deform. Force can also be described by intuitive concepts such as a push or pull. A force has both magnitude and direction, making it a vector quantity.
The original form of Newton's second law states that the net force acting upon an object is equal to the rate at which its momentum changes with time. If the mass of the object is constant, this law implies that the acceleration of an object is directly proportional to the net force acting on the object, is in the direction of the net force, and is inversely proportional the mass of the object. As a formula, this is expressed as:
where the arrows imply a vector quantity possessing both magnitude and direction.
Related concepts to force include: thrust, which increases the velocity of an object; drag, which decreases the velocity of an object; and torque which produces changes in rotational speed of an object. Forces which do not act uniformly on all parts of a body will also cause mechanical stresses, a technical term for influences which cause deformation of matter. While mechanical stress can remain embedded in a solid object, gradually deforming it, mechanical stress in a fluid determines changes in its pressure and volume.
### Other articles related to "force":
Meteorology - History - Research Into Cyclones and Air Flow
... with air being deflected by the Coriolis force to create the prevailing westerly winds ... extent of the large scale interaction of pressure gradient force and deflecting force that in the end causes air masses to move along isobars was understood ... By 1912, this deflecting force was named the Coriolis effect ...
Knightmare - Life Force
... The life force was a combined clock and progress meter used to track the energy status of the dungeoneer (the main contestant) ... would often tell the team "You're wasting Life Force"), taking "damage" through being attacked by monsters or obstacles, taking the wrong route or making bad decisions ... In the first five series, the life force was a computer animated image of an adventurer wearing a helmet ...
Force - Units of Measurement
... The SI unit of force is the newton (symbol N), which is the force required to accelerate a one kilogram mass at a rate of one meter per second squared, or kg·m·s−2 ... The corresponding CGS unit is the dyne, the force required to accelerate a one gram mass by one centimeter per second squared, or g·cm·s−2 ... foot-pound-second English unit of force is the pound-force (lbf), defined as the force exerted by gravity on a pound-mass in the standard gravitational field of 9.80665 m·s−2 ...
... featuring the words Royal Australian Air Force, beneath which scroll work displays the Latin motto (shared with the Royal Air Force) Per Ardua Ad Astra ...
Lift (force)
... A fluid flowing past the surface of a body exerts surface force on it ... Lift is the component of this force that is perpendicular to the oncoming flow direction ... It contrasts with the drag force, which is the component of the surface force parallel to the flow direction ...
### Famous quotes containing the word force:
The body, what is it, Father, but a sign
To love the force that grows us, to give back
What in Thy palm is senselessness and mud?
Karl Shapiro (b. 1913)
Men have two ways of righting their wrongs, by force and by the ballot. Both are denied to women, one by nature, the other by man.
Ida A. Harper 1851–1931, U.S. women’s magazine contributor. Fireman’s Magazine, repr. In The Woman’s Magazine, pp. 423-5 (May 1887)
Awareness of the stars and their light pervades the Koran, which reflects the brightness of the heavenly bodies in many verses. The blossoming of mathematics and astronomy was a natural consequence of this awareness. Understanding the cosmos and the movements of the stars means understanding the marvels created by Allah. There would be no persecuted Galileo in Islam, because Islam, unlike Christianity, did not force people to believe in a “fixed” heaven.
Fatima Mernissi, Moroccan sociologist. Islam and Democracy, ch. 9, Addison-Wesley Publishing Co. (Trans. 1992) | 981 | 4,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-04 | longest | en | 0.94722 |
https://cboard.cprogramming.com/c-programming/152418-short-sorting-problem.html | 1,508,483,888,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823839.40/warc/CC-MAIN-20171020063725-20171020083725-00397.warc.gz | 638,473,799 | 12,648 | 1. Short Sorting Problem
I am having trouble debugging my segmentation fault error in this short code!
Code:
```void radixsort(unsigned long array[]){
int arraysize = sizeof(array)/sizeof(*array);
unsigned long move[arraysize];
while(index<4){
for(i=0;i<arraysize;i++){
val=array[i]>>(index*group);
}
for(i=1;i<256;i++){
map[i]=map[i-1]+count[i-1];
}
for(i=0;i<arraysize;i++){
val=array[i]>>(index*group);
}
index++;
}
array = move;
}```
2. You don't initialize "count" and "map".
Bye, Andreas
3. This may be a dumb question but is there an easy way to initialize without using a simple for loop and setting everything to 0?
4. Code:
```int main(void)
{
int array[5] = { 1,1,1,1,1 };
return 0;
}```
I think you see while the loops are so beloved by many (including me)
5. Originally Posted by workisnotfun
This may be a dumb question but is there an easy way to initialize without using a simple for loop and setting everything to 0?
If you want to set every element to 0 try
Code:
`int count[256] = { 0 }, map[256] = { 0 };`
By default, all missing elements in an initializer list are set to 0 by the compiler.
Bye, Andreas
6. Thanks for the tip, the seg fault's fixed. On another note, is there something wrong with my sorting logic or is the parameter suppose to be a pointer to an array? When I print the values of my array before and after calling this function, the values are in the same place, unmoved.
Code:
``` unsigned long j[size];
argv++;
for(i=0;i<size;i++){
j[i]=atoi(*argv++);
}
`int count[256] = {}, map[256] = {};` | 425 | 1,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-43 | latest | en | 0.716783 |
https://www.enotes.com/homework-help/verify-4-3i-3-4i-2-1-2i-203633 | 1,632,009,641,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056578.5/warc/CC-MAIN-20210918214805-20210919004805-00406.warc.gz | 789,078,182 | 17,150 | # Verify if (4+3i)/(3-4i)-(2+i)/(1-2i)
let z= (4+3i)/(3-4i) - (2+i)/(1-2i)
First lest us simplify:
==> z= (4+3i)(3+4i)/(3-4i)(3+4i) - (2+i)(1+2i)/(1-2i)(1+2i)
==> z= (12+25i-12)/(3^2 +4^2) - (2+3i -2)/(1^2 + 2^2)
==> z= 25i/25 - 3i/3
==> z= i -i = 0
==> z= 0 + 0i
Approved by eNotes Editorial Team | 175 | 306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-39 | latest | en | 0.418115 |
https://www.physicsforums.com/threads/conservation-of-energy.97808/ | 1,508,753,977,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825889.47/warc/CC-MAIN-20171023092524-20171023112524-00577.warc.gz | 976,752,421 | 19,917 | # Conservation of Energy?
1. Nov 1, 2005
### PhiJ
By reletivity,
K.E.=mc^2-m0c^2
For light, m0=0, so K.E.=mc^2
We know that photons have gravitational potential energy, but that all of the photon's mass-energy is kinetic energy, so doesn't that mean that gravitational potential energy is not really energy, and thus the mass energy of the universe is not constant?
There must be something wrong with that argument, but I can't see it.
Last edited: Nov 1, 2005
2. Nov 1, 2005
### Garth
Photons do not have gravitational potential energy - that concept is foreign to GR, which conserves energy-momentum not, in general, energy .
For a photon $$E = h\nu$$
and energy is a frame dependent quantity, a particle's energy is measured as
$$E = -p_{\alpha}U^{\alpha}$$
where $$p_{\alpha}$$ is the particle's 4-momentum and $$U^{\alpha}$$ is the observer's 4-velocity.
Garth
Last edited: Nov 1, 2005
3. Nov 1, 2005
### PhiJ
Oh! I knew it would be obvious!
4. Nov 3, 2005
### pmb_phy
I can't see how you came to that conclusion since you used an expression from SR to make a conclusion in GR. E.g. In the weak field limit the total energy for a partilce in a gravitational field is the rest energy + kinetic energy + potential energy. If the rest energy is zero then what is left is kinetic energy + potential energy. You can see this derivation on my web site here
http://www.geocities.com/physics_world/gr/red_shift.htm
The old definition of potential energy in my opinion should have been replaced by a new one which can be stated as "Energy of a particle by virtue of position only." There are complications with that since its bound to be misused.
Pete
Last edited: Nov 3, 2005
5. Nov 3, 2005
### Rebel
I have another energy related question:
If the universe is expanding in a way explained at least in part by a posittive cosmological constant, even then one assumes the energy of the universe to be constant? or is it explained using other therms? I was wondering about it cause there are some hyphotesis where universe is accelerating in some stages and in others is decelerating, some using tachyons to explain the dark energy. I believe this should be a though work in relationing cosmology with thermodynamics, e.g. how fundamental are the laws of thermodynamics compared with others.
6. Nov 3, 2005
### pervect
Staff Emeritus
Take a look at the sci.physics.faq on Energy & Gr.http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html" [Broken]
The textbooks I've seen, such as MTW and Wald, make a point of saying that the universe doesn't have a well-defined energy.
Last edited by a moderator: May 2, 2017
7. Nov 3, 2005
### Rebel
T
Thank you pervect. Actually im reading the MTW book, but it will take me a lot more of time to finish it. Regards.
8. Nov 3, 2005
### pervect
Staff Emeritus
LOL, I'll bet it will take just a little bit of time to read all 1200+ pages.
Since you have MTW, you might want to check out pg 457 "Mass and angular momentum of a closed universe" in MTW, also pg 705 has some useful notes on the topic.
9. Nov 5, 2005
### PhiJ
OK, there seems to be a disagreement.
Can anyone back pmb phy or Garth up, and pmb phy, if light does have GPE, won't this then add to it's mass and then slow it down?
10. Nov 5, 2005
### Garth
Where's the disagreement? (So far)
That is a good reason why it doesn't have GPE! As I said GPE is a classical concept that does not carry through to GR. In GR gravitational red-shift is the observation of a time dilation effect between the bottom and top of a gravitational 'pit'.
The null-geodsics of consecutive pulses of light diverge as they traverse curved space-time and are received further apart in time at the top of the pit than they were transmitted at the bottom.
If you do have access to Misner Thorne and Wheeler - Gravitation - you need to read 7.3 page 187 - 189 carefully. They follow an argument of Schild which uses the diagram fig 7.1 to prove that if space-time were flat (SR) the null-geodesics of succesive pulses of light ascending a gravitaiton 'pit' will not diverge and red shift will not be observed.
To understand what really is going on you have to draw the diagram on a curved surface, the inside surface of the Schwarzschild 'funnel', to make the null-geodesics diverge.
I hope this helps.
Garth
11. Nov 5, 2005
### pervect
Staff Emeritus
Other than my previously referenced FAQ article,
http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
you'll have to read some textbooks if you want more info. MTW's gravitation is good, though somewhat dated. Wald's "General Relativity" has one of the better modern treatments of energy in GR, Be prepared for some heavy reading, though - especially in Wald, MTW has an informal style that might allow someone to get some understanding without the math, Wald's approach is rather difficult.
You might also look at the quote from Steve Carlip that I posted in another thread
https://www.physicsforums.com/showpost.php?p=810720&postcount=4
Here's a link to some of the honors Steve Carlip has won
http://www.physics.ucdavis.edu/Text/Carlip.html#Honors
While this doesn't necessarily mean he can't be wrong, he's definitely a heavy hitter in the relativity world, and one of the few such who makes some "outreach" efforts to talk about GR.
12. Nov 6, 2005
### PhiJ
pmb phy said:
so he's saying that there is GPE in GR, and your saying their isn't.
Unless I'm just misinterpreting or generally being stupid...
13. Nov 6, 2005
### pmb_phy
The gravitational potential energy of position will not add to the photons rest mass and make it non-zero. It will, however, add to its "relativistic" mass and alter it, yes. So in that sense it does change its mass, just not its proper mass. It is also a matter of observation/prediction etc. the coordinate speed of light in a gravitational field changes.
Note: The increase in relativistic mass is not the reason for the slowing of light in a g-field. The speed of anything in a g-field is independant of its mass. In this case the inertial mass of light changes along with its passive gravitational mass so as to cancel out.
Pete
Last edited: Nov 6, 2005
14. Nov 6, 2005
### pmb_phy
The arguement of Schild's was argued to be wrong in the American Journal of Physics. I disagree with Schild myself and have explained myself in that link I gave above as I recall.
Pete
15. Nov 6, 2005
### pervect
Staff Emeritus
You appear to be conflating invariant mass with "relativistic mass".
"Slowing it down" refers to the fact that only objects with an invariant mass of zero can travel at the speed of light. Photons always have a zero invariant mass in both GR and SR as I will explain below.
"Adding to its mass" refers to the "relativitic mass", which in the case of a photon is just another name for its energy.
In SR, invariant mass is given by E^2 - px^2 - py^2 - pz^2, and is always a constant, zero for a photon. E is the energy, and px, py, and pz are the x, y, and z components of the momentum.
In GR invariant mass is given by a different formula:
g_00 E^2 - g_11 px^2 - g_22 py^2 - g_33 pz^2
in a diagonal metric (not the most general possible, but easy to write down and understand). For completness, in the most general case
$$m = g_ij P^i P^j$$
where P^i is the energy-momentum 4-vector (P^0 being the energy).
As is the case in SR, in GR the invariant mass of a photon is always zero.
Note that an effective gravitational potential can exist in very simple cases in GR (such as a photon or particle falling into a black hole). However, the general existence of such an effective potential is not guaranteed, it exists only when the system is static.
Note also the use of the word "effective".
16. Nov 6, 2005
### pmb_phy
To be exact that should read
$$m^2 = g_{\alpha\beta} P^{\alpha}P^{\beta}$$
You forgot the square on m pervect (I like to use greek letters for indices which range from 0->3).
Pete
17. Nov 8, 2005
### PhiJ
Arite, I seem to be sort of understanding you, and it looks like the rest I will understand after I read my differential geometry + GR text.
Does the space-time contain the GPE then?
btw. How do you do LaTeX images on these forums?
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Week 6 Assignment Appendix E
# Week 6 Assignment Appendix E - Axia College Material...
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Axia College Material Appendix E Fueling Up Motorists often complain about rising gas prices. Some motorists purchase fuel-efficient vehicles and participate in trip reduction plans, such as carpooling and using alternative transportation. Other drivers try to drive only when necessary. Application Practice Answer the following questions. Use Equation Editor to write mathematical expressions and equations. First, save this file to your hard drive by selecting Save As from the File menu. Click the white space below each question to maintain proper formatting. 1. Imagine you are at a gas station filling your tank with gas. The function C ( g ) represents the cost C of filling up the gas tank with g gallons. Given the equation ) ( 03 . 3 ) ( g g C = a. What does the number 3.03 represent? It represents the cost of a gallon of gas. b. Find C (2). ( 29 ) 2 ( 03 . 3 2 = C = ( 29 06 . 6 2 = C c. Find C (9). ( 29 ( 29 ( 29 27 . 27 9 9 03 . 3 9 = = = C C d. For the average motorist, name one value for g that would be inappropriate for this function’s purpose. Explain why you chose that number.
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### MEAP Preparation - Grade 7 Mathematics1.98 Expressing One Quantity as a Percentage of Another Quantity
Directions: Answer the following qustions. Also write at least ten examples of your own.
Q 1: What percent of 20 dollars is 4 dollars?40%20%50% Q 2: What percent of 4 1/2 gallons is 3/4 gallons?18 2/3%16 2/3%16 3/2% Q 3: What percent of 9 miles is 6 miles?56 2/3%66 2/3%64 2/3% Q 4: What percent of 70 gallons is 50 gallons?75 3/7%72 3/7%71 3/7% Q 5: What percent of 8 pounds is 2.5 pounds?31 1/4%35 1/4%40 1/4% Q 6: What percent of 40 dollars is 10 dollars?30%20%25% Q 7: What percent of 18 dozens is 6 dozens?33 2/3%33 1/3%33 1/4% Q 8: What percent of 30 miles is 15 miles?100%50%200% Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 350 | 983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-51 | longest | en | 0.857385 |
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# Relative Valuation
Aswath Damodaran
Aswath Damodaran 1
Why relative valuation?
If you think Im crazy, you should see the guy who lives across the hall
Jerry Seinfeld talking about Kramer in a Seinfeld episode
## A little inaccuracy sometimes saves tons of explanation
H.H. Munro
Aswath Damodaran 2
What is relative valuation?
## n In relative valuation, the value of an asset is compared to the values
assessed by the market for similar or comparable assets.
n To do relative valuation then,
we need to identify comparable assets and obtain market values for these
assets
convert these market values into standardized values, since the absolute
prices cannot be compared This process of standardizing creates price
multiples.
compare the standardized value or multiple for the asset being analyzed to
the standardized values for comparable asset, controlling for any
differences between the firms that might affect the multiple, to judge
whether the asset is under or over valued
Aswath Damodaran 3
Standardizing Value
## n Prices can be standardized using a common variable such as earnings,
cashflows, book value or revenues.
Earnings Multiples
Price/Earnings Ratio (PE) and variants (PEG and Relative PE)
Value/EBIT
Value/EBITDA
Value/Cash Flow
Book Value Multiples
Price/Book Value(of Equity) (PBV)
Value/ Book Value of Assets
Value/Replacement Cost (Tobins Q)
Revenues
Price/Sales per Share (PS)
Value/Sales
Industry Specific Variable (Price/kwh, Price per ton of steel ....)
Aswath Damodaran 4
The Four Steps to Understanding Multiples
## n Define the multiple
In use, the same multiple can be defined in different ways by different
users. When comparing and using multiples, estimated by someone else, it
is critical that we understand how the multiples have been estimated
n Describe the multiple
Too many people who use a multiple have no idea what its cross sectional
distribution is. If you do not know what the cross sectional distribution of
a multiple is, it is difficult to look at a number and pass judgment on
whether it is too high or low.
n Analyze the multiple
It is critical that we understand the fundamentals that drive each multiple,
and the nature of the relationship between the multiple and each variable.
n Apply the multiple
Defining the comparable universe and controlling for differences is far
more difficult in practice than it is in theory.
Aswath Damodaran 5
Definitional Tests
## n Is the multiple consistently defined?
Proposition 1: Both the value (the numerator) and the standardizing
variable ( the denominator) should be to the same claimholders in the
firm. In other words, the value of equity should be divided by equity
earnings or equity book value, and firm value should be divided by
firm earnings or book value.
n Is the multiple uniformally estimated?
The variables used in defining the multiple should be estimated uniformly
across assets in the comparable firm list.
If earnings-based multiples are used, the accounting rules to measure
earnings should be applied consistently across assets. The same rule
applies with book-value based multiples.
Aswath Damodaran 6
Descriptive Tests
n What is the average and standard deviation for this multiple, across the
universe (market)?
n What is the median for this multiple?
The median for this multiple is often a more reliable comparison point.
n How large are the outliers to the distribution, and how do we deal with
the outliers?
Throwing out the outliers may seem like an obvious solution, but if the
outliers all lie on one side of the distribution (they usually are large
positive numbers), this can lead to a biased estimate.
n Are there cases where the multiple cannot be estimated? Will ignoring
these cases lead to a biased estimate of the multiple?
n How has this multiple changed over time?
Aswath Damodaran 7
Analytical Tests
n What are the fundamentals that determine and drive these multiples?
Proposition 2: Embedded in every multiple are all of the variables that
drive every discounted cash flow valuation - growth, risk and cash flow
patterns.
In fact, using a simple discounted cash flow model and basic algebra
should yield the fundamentals that drive a multiple
n How do changes in these fundamentals change the multiple?
The relationship between a fundamental (like growth) and a multiple
(such as PE) is seldom linear. For example, if firm A has twice the growth
rate of firm B, it will generally not trade at twice its PE ratio
Proposition 3: It is impossible to properly compare firms on a
multiple, if we do not know the nature of the relationship between
fundamentals and the multiple.
Aswath Damodaran 8
Application Tests
## n Given the firm that we are valuing, what is a comparable firm?
While traditional analysis is built on the premise that firms in the same
sector are comparable firms, valuation theory would suggest that a
comparable firm is one which is similar to the one being analyzed in terms
of fundamentals.
Proposition 4: There is no reason why a firm cannot be compared
with another firm in a very different business, if the two firms have
the same risk, growth and cash flow characteristics.
n Given the comparable firms, how do we adjust for differences across
firms on the fundamentals?
Proposition 5: It is impossible to find an exactly identical firm to the
one you are valuing.
Aswath Damodaran 9
Price Earnings Ratio: Definition
## PE = Market Price per Share / Earnings per Share
n There are a number of variants on the basic PE ratio in use. They are
based upon how the price and the earnings are defined.
n Price: is usually the current price
is sometimes the average price for the year
n EPS: earnings per share in most recent financial year
earnings per share in trailing 12 months (Trailing PE)
forecasted earnings per share next year (Forward PE)
forecasted earnings per share in future year
Aswath Damodaran 10
PE Ratio: Descriptive Statistics for the United States
## Current, Trailing and Forward PE Ratios
U.S. Stocks - July 2000
1000
900
800
700
600
Current PE
500 Trailing PE
Forward PE
400
300
200
100
0
<4 4-8 8 - 12 12 - 16 16 - 20 20 - 25 25 - 30 30 -40 40 -50 50 -75 75 - >100
100
PE
Aswath Damodaran 11
PE: Deciphering the Distribution
## Current PE Trailing PE Forward PE
Mean 57.52 51.51 48.64
Standard Error 5.38 6.08 6.78
Median 14.47 13.68 11.52
Mode 12.00 7.00 7.50
Standard Deviation 330.59 377.93 294.10
Kurtosis 335.54 808.90 460.43
Skewness 17.12 25.96 19.59
Maximum 8043.03 14619.60 8184.40
Aswath Damodaran 12
PE Ratio: Greece in May 2001
## PE Ratios: Greece in May 2001
90
80
70
60
Number of firms
50
40
30
20
10
0
<4 4 -8 8 -12 12-16 16-20 20-24 24-28 28-32 32-36 36-40 >40
PE Ratio
Aswath Damodaran 13
PE Ratio: Understanding the Fundamentals
## n To understand the fundamentals, start with a basic equity discounted
cash flow model.
n With the dividend discount model,
DPS1
P0 =
r gn
## n Dividing both sides by the earnings per share,
P0 Payout Ratio *(1 + g n )
= PE =
EPS 0 r - gn
## n If this had been a FCFE Model,
FCFE 1
P0 =
r gn
P0 (FCFE/Earnings)*(1 + g n )
= PE =
EPS0 r-g n
Aswath Damodaran 14
PE Ratio and Fundamentals
## n Proposition: Other things held equal, higher growth firms will
have higher PE ratios than lower growth firms.
n Proposition: Other things held equal, higher risk firms will have
lower PE ratios than lower risk firms
n Proposition: Other things held equal, firms with lower
reinvestment needs will have higher PE ratios than firms with
higher reinvestment rates.
n Of course, other things are difficult to hold equal since high growth
firms, tend to have risk and high reinvestment rats.
Aswath Damodaran 15
Using the Fundamental Model to Estimate PE For a
High Growth Firm
n The price-earnings ratio for a high growth firm can also be related to
fundamentals. In the special case of the two-stage dividend discount
model, this relationship can be made explicit fairly simply:
( 1 +g)n
) *1
EPS0 * P a y o u t R a t i o * ( 1g +
( 1 +r) n EPS 0 *Payout Ration * ( 1 +g)n * ( 1 +g n )
P0 = +
r-g (r - gn )(1+r)n
For a firm that does not pay what it can afford to in dividends, substitute
FCFE/Earnings for the payout ratio.
n Dividing both sides by the earnings per share:
(1+ g )n
Payout Ratio *(1 + g )* 1
P0 (1+ r) n Payout Ratio n * ( 1 + g )n *(1 + gn )
= +
EPS 0 r -g (r - g n )(1+ r) n
Aswath Damodaran 16
A Simple Example
n Assume that you have been asked to estimate the PE ratio for a firm
which has the following characteristics:
Variable High Growth Phase Stable Growth Phase
Expected Growth Rate 25% 8%
Payout Ratio 20% 50%
Beta 1.00 1.00
n Riskfree rate = T.Bond Rate = 6%
n Required rate of return = 6% + 1(5.5%)= 11.5%
(1.25)5
0 . 2 * (1.25) * 1 5
(1.115) 5
0.5 * (1.25) *(1.08)
PE = + = 28.75
(.115 - .25) (.115-.08) (1.115) 5
Aswath Damodaran 17
PE and Growth: Firm grows at x% for 5 years, 8%
thereafter
## PE Ratios and Expected Growth: Interest Rate Scenarios
180
160
140
120
100 r=4%
PE Ratio
r=6%
r=8%
80 r=10%
60
40
20
0
5% 10% 15% 20% 25% 30% 35% 40% 45% 50%
Expected Growth Rate
Aswath Damodaran 18
PE Ratios and Length of High Growth: 25% growth
for n years; 8% thereafter
## PE Ratios and Length of High Growth Period
60
50
40
g=25%
PE Ratio
g=20%
30
g=15%
g=10%
20
10
0
0 1 2 3 4 5 6 7 8 9 10
Length of High Growth Period
Aswath Damodaran 19
PE and Risk: Effects of Changing Betas on PE
Ratio:
Firm with x% growth for 5 years; 8% thereafter
## PE Ratios and Beta: Growth Scenarios
50
45
40
35
30
g=25%
PE Ratio
g=20%
25
g=15%
g=8%
20
15
10
0
0.75 1.00 1.25 1.50 1.75 2.00
Beta
Aswath Damodaran 20
PE and Payout
## PE Ratios and Payour Ratios: Growth Scenarios
35
30
25
20 g=25%
g=20%
PE
g=15%
15 g=10%
10
0
0% 20% 40% 60% 80% 100%
Payout Ratio
Aswath Damodaran 21
Comparisons of PE across time
## PE Ratio for US stocks over time
35.00
30.00
25.00
20.00
PE Ratio
15.00
10.00
5.00
0.00
49
51
53
55
57
59
61
63
65
67
69
71
73
75
77
79
81
83
85
87
89
91
93
95
97
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
Aswath Damodaran 22
Is low (high) PE cheap (expensive)?
n A market strategist argues that stocks are over priced because the PE
ratio today is too high relative to the average PE ratio across time. Do
you agree?
n Yes
n No
n If you do not agree, what factors might explain the higer PE ratio
today?
Aswath Damodaran 23
E/P Ratios , T.Bond Rates and Term Structure
## EP Ratios, T.Bond Rates and Tem Structure
16.00%
14.00%
12.00%
10.00%
8.00%
T.Bond Rate
T.Bond-T.Bill
E/P Ratios
6.00%
4.00%
2.00%
0.00%
60
62
64
66
68
70
72
74
76
78
80
82
84
86
88
90
92
94
96
98
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
19
-2.00%
Aswath Damodaran 24
Regression Results
## n There is a strong positive relationship between E/P ratios and T.Bond
rates, as evidenced by the correlation of 0.6836 between the two
variables.,
n In addition, there is evidence that the term structure also affects the PE
ratio.
n In the following regression, using 1960-1999 data, we regress E/P
ratios against the level of T.Bond rates and a term structure variable
(T.Bond - T.Bill rate)
E/P = 2.82% + 0.749 T.Bond Rate - 0.847 (T.Bond Rate-T.Bill Rate)
(2.84) (6.78) (-3.65)
R squared = 60.67%
Aswath Damodaran 25
Estimate the E/P Ratio Today
n T. Bond Rate =
n T.Bond Rate - T.Bill Rate =
n Expected E/P Ratio =
n Expected PE Ratio =
Aswath Damodaran 26
Comparing PE ratios across firms
## Company Name PE Growth
Telecom Corporation of New Zealand ADR 11.2 0.11
Telecom Argentina Stet - France Telecom SA ADR B 12.5 0.08
Hellenic Telecommunication Organization SA ADR 12.8 0.12
Telecomunicaciones de Chile ADR 16.6 0.08
Asia Satellite Telecom Holdings ADR 19.6 0.16
Portugal Telecom SA ADR 20.8 0.13
Telefonos de Mexico ADR L 21.1 0.14
Gilat Communications 22.7 0.31
Deutsche Telekom AG ADR 24.6 0.11
British Telecommunications PLC ADR 25.7 0.07
Tele Danmark AS ADR 27 0.09
Cable & Wireless PLC ADR 29.8 0.14
APT Satellite Holdings ADR 31 0.33
Royal KPN NV ADR 35.7 0.13
Telecom Italia SPA ADR 42.2 0.14
Nippon Telegraph & Telephone ADR 44.3 0.2
France Telecom SA ADR 45.2 0.19
Aswath Damodaran 27
PE and Growth
50.0
FTE
NTT
TI
37.5
KPN
TEF
ATS
CWP
P TLK
E TLD
25.0 BTY
DT
GICOF
TLS MTA
PT TMX
SAT
SCM
CTC
12.5 OTE
TEO
NZT
TBH
IIT
## 0.075 0.150 0.225 0.300
Growth
Aswath Damodaran 28
PE, Growth and Risk
No Selector
## Variable Coefficient SE t-ratio prob
Constant 13.1151 3.471 3.78 0.0010
Growth rate 121.223 19.27 6.29 0.0001
Emerging Market -13.8531 3.606 -3.84 0.0009
Emerging Market is a dummy: 1 if emerging market
0 if not
Aswath Damodaran 29
Is Hellenic Telecom under valued?
## n Predicted PE = 13.12 + 121.22 (.12) - 13.85 (0) = 27.67
n At an actual price to book value ratio of 12.2, Hellenic looks
significantly under valued. However, if the market is pricing it as an
emerging market telecomm:
n Predicted PE = 13.12 + 121.22 (.12) - 13.85 (1) = 13.82
Aswath Damodaran 30
A Question
You are reading an equity research report on this sector, and the analyst
claims that Andres Wine and Hansen Natural are under valued because
they have low PE ratios. Would you agree?
o Yes
o No
n Why or why not?
Aswath Damodaran 31
Using comparable firms- Pros and Cons
## n The most common approach to estimating the PE ratio for a firm is
to choose a group of comparable firms,
to calculate the average PE ratio for this group and
to subjectively adjust this average for differences between the firm being
valued and the comparable firms.
n Problems with this approach.
The definition of a 'comparable' firm is essentially a subjective one.
The use of other firms in the industry as the control group is often not a
solution because firms within the same industry can have very different
business mixes and risk and growth profiles.
There is also plenty of potential for bias.
Even when a legitimate group of comparable firms can be constructed,
differences will continue to persist in fundamentals between the firm
being valued and this group.
Aswath Damodaran 32
Using the entire crosssection: A regression approach
## n In contrast to the 'comparable firm' approach, the information in the
entire cross-section of firms can be used to predict PE ratios.
n The simplest way of summarizing this information is with a multiple
regression, with the PE ratio as the dependent variable, and proxies for
risk, growth and payout forming the independent variables.
Aswath Damodaran 33
PE versus Growth
150
100
P
E
A
d
j
50
-0
## -0.00 0.25 0.50 0.75
Expected Growth in EPS: next 5
Aswath Damodaran 34
PE Ratio: Standard Regression
No Selector
5903 total cases of which 3405 are missing
R squared = 24.9% R squared (adjusted) = 24.8%
s = 31.09 with 2498 - 4 = 2494 degrees of freedom
## Source Sum of Squares df Mean Square F-ratio
Regression 798022 3 266007 275
Residual 2410686 2494 966.594
## Variable Coefficient s.e. of Coeff t-ratio prob
Constant -17.2213 2.439 -7.06 0.0001
Expected Grow 155.652 6.418 24.3 0.0001
Beta 16.4415 2.429 6.77 0.0001
Payout Ratio 10.9341 2.177 5.02 0.0001
Aswath Damodaran 35
Second Thoughts?
## n Based on this regression, estimate the PE ratio for a firm with no
growth, no payout and no risk.
## n Is there a problem with your prediction?
Aswath Damodaran 36
PE Regression- No Intercept
No Selector
5903 total cases of which 3405 are missing
R squared = % R squared (adjusted) = %
s = 31.39 with 2498 - 3 = 2495 degrees of freedom
## Source Sum of Squares df Mean Square F-ratio
Regression 2408918 3 802973 815
Residual 2458878 2495 985.522
## Variable Coefficient s.e. of Coeff t-ratio prob
Payout Ratio 3.19821 1.900 1.68 0.0924
Beta 2.37185 1.403 1.69 0.0909
Expected Grow 145.317 6.310 23.0 0.0001
Aswath Damodaran 37
Problems with the regression methodology
## n The basis regression assumes a linear relationship between PE ratios
and the financial proxies, and that might not be appropriate.
n The basic relationship between PE ratios and financial variables itself
might not be stable, and if it shifts from year to year, the predictions
from the model may not be reliable.
n The independent variables are correlated with each other. For example,
high growth firms tend to have high risk. This multi-collinearity makes
the coefficients of the regressions unreliable and may explain the large
changes in these coefficients from period to period.
Aswath Damodaran 38
The Multicollinearity Problem
## PE Exp Growt Beta Payout
PE 1.000
Exp Growt 0.288 1.000
Beta 0.141 0.292** 1.000
Payout -0.087 -0.404** -0.183* 1.000
n The independent variables are correlated with the dependent variable,
which is a good thing, but they are also correlated with each other
(which is not a good thing)
n This will cause the standard errors on the coefficients to become larger
and some coefficients may have the wrong sign.
Aswath Damodaran 39
Using the PE ratio regression
n Assume that you were given the following information for Dell. The
firm has an expected growth rate of 20%, a beta of 1.40 and pays no
dividends. Based upon the regression, estimate the predicted PE ratio
for Dell.
## n Dell is actually trading at 23 times earnings. What does the predicted
PE tell you?
Aswath Damodaran 40
Value/Earnings and Value/Cashflow Ratios
n While Price earnings ratios look at the market value of equity relative
to earnings to equity investors, Value earnings ratios look at the
market value of the firm relative to operating earnings. Value to cash
flow ratios modify the earnings number to make it a cash flow number.
n The form of value to cash flow ratios that has the closest parallels in
DCF valuation is the value to Free Cash Flow to the Firm, which is
defined as:
Value/FCFF = (Market Value of Equity + Market Value of Debt)
EBIT (1-t) - (Cap Ex - Deprecn) - Chg in WC
n Consistency Tests:
If the numerator is net of cash (or if net debt is used, then the interest
income from the cash should not be in denominator
The interest expenses added back to get to EBIT should correspond to the
debt in the numerator. If only long term debt is considered, only long term
Aswath Damodaran 41
Value of Firm/FCFF: Determinants
## n Reverting back to a two-stage FCFF DCF model, we get:
(1 + g)n
FCFF (1 + g) 1 - n ( 1 +g )
0 ( 1 +WACC) n FCFF ( 1 +g)
V0 = + 0 n
WACC - g (WACC - g )(1 + WACC)n
n
## V0 = Value of the firm (today)
FCFF0 = Free Cashflow to the firm in current year
g = Expected growth rate in FCFF in extraordinary growth period (first
n years)
WACC = Weighted average cost of capital
gn = Expected growth rate in FCFF in stable growth period (after n
years)
Aswath Damodaran 42
Value Multiples
## n Dividing both sides by the FCFF yields,
(1 + g)n
(1 + g) 1 -
V0 (1 + WACC)n ( 1 +g)n ( 1 +gn )
= +
FCFF0 WACC - g (WACC - gn )(1 + WACC)n
## n The value/FCFF multiples is a function of
the cost of capital
the expected growth
Aswath Damodaran 43
Alternatives to FCFF - EBIT and EBITDA
## n Most analysts find FCFF to complex or messy to use in multiples
(partly because capital expenditures and working capital have to be
estimated). They use modified versions of the multiple with the
following alternative denominator:
after-tax operating income or EBIT(1-t)
pre-tax operating income or EBIT
net operating income (NOI), a slightly modified version of operating
income, where any non-operating expenses and income is removed from
the EBIT
EBITDA, which is earnings before interest, taxes, depreciation and
amortization.
Aswath Damodaran 44
Value/FCFF Multiples and the Alternatives
n Assume that you have computed the value of a firm, using discounted
cash flow models. Rank the following multiples in the order of
magnitude from lowest to highest?
o Value/EBIT
o Value/EBIT(1-t)
o Value/FCFF
o Value/EBITDA
n What assumption(s) would you need to make for the Value/EBIT(1-t)
ratio to be equal to the Value/FCFF multiple?
Aswath Damodaran 45
Illustration: Using Value/FCFF Approaches to value
a firm: MCI Communications
## n MCI Communications had earnings before interest and taxes of \$3356
million in 1994 (Its net income after taxes was \$855 million).
n It had capital expenditures of \$2500 million in 1994 and depreciation
of \$1100 million; Working capital increased by \$250 million.
n It expects free cashflows to the firm to grow 15% a year for the next
five years and 5% a year after that.
n The cost of capital is 10.50% for the next five years and 10% after
that.
n The company faces a tax rate of 36%.
(1.15)5
(1.15) 1-
V0 (1.105)5 (1.15) 5 (1.05)
= + 5
= 31.28
FCFF0 .105 -.15 (.10 - .05)(1.105)
Aswath Damodaran 46
Multiple Magic
n In this case of MCI there is a big difference between the FCFF and
short cut measures. For instance the following table illustrates the
appropriate multiple using short cut measures, and the amount you
would overpay by if you used the FCFF multiple.
Free Cash Flow to the Firm
= EBIT (1-t) - Net Cap Ex - Change in Working Capital
= 3356 (1 - 0.36) + 1100 - 2500 - 250 = \$ 498 million
\$ Value Correct Multiple
FCFF \$498 31.28382355
EBIT (1-t) \$2,148 7.251163362
EBIT \$ 3,356 4.640744552
EBITDA \$4,456 3.49513885
Aswath Damodaran 47
Value/EBITDA Multiple
## n The Classic Definition
Value Market Value of Equity + Market Value of Debt
=
EBITDA Earnings before Interest, Taxes and Depreciation
## n The No-Cash Version
Value Market Value of Equity + Market Value of Debt - Cash
=
EBITDA Earnings before Interest, Taxes and Depreciation
n When cash and marketable securities are netted out of value, none of
the income from the cash and securities should be reflected in the
denominator.
Aswath Damodaran 48
Value/EBITDA Distribution
Value/EBITDA Multiple
1200
1000
800
600
400
200
0
<2 2 -4 4-6 6- 8 8 - 10 10 - 12 12-14 14 - 16 16 - 18 18 - 20 20 - 30 30 - 50 > 50
Aswath Damodaran 49
The Determinants of Value/EBITDA Multiples:
FCFF1
V0 =
WACC - g
## n The numerator can be written as follows:
FCFF = EBIT (1-t) - (Cex - Depr) - Working Capital
= (EBITDA - Depr) (1-t) - (Cex - Depr) - Working Capital
= EBITDA (1-t) + Depr (t) - Cex - Working Capital
Aswath Damodaran 50
From Firm Value to EBITDA Multiples
## n Now the Value of the firm can be rewritten as,
EBITDA (1-t) + Depr (t) - Cex - Working Capital
Value =
WACC - g
## n Dividing both sides of the equation by EBITDA,
Value (1- t) Depr (t)/EBITDA CEx/EBITDA Working Capital/EBITDA
= + - -
EBITDA WACC-g WACC - g WACC - g WACC - g
Aswath Damodaran 51
A Simple Example
## n Consider a firm with the following characteristics:
Tax Rate = 36%
Capital Expenditures/EBITDA = 30%
Depreciation/EBITDA = 20%
Cost of Capital = 10%
The firm has no working capital requirements
The firm is in stable growth and is expected to grow 5% a year forever.
Note that the return on capital implied in this growth rate can be
calculated as follows:
g = ROC * Reinvestment Rate
.05 = ROC * Net Cap Ex/EBIT (1-t)
= ROC * (.30-.20)/[(1-.2)(1-.36)]
Solving for ROC, ROC = 25.60%
Aswath Damodaran 52
Calculating Value/EBITDA Multiple
n In this case, the Value/EBITDA multiple for this firm can be estimated
as follows:
Value ( 1 -.36) (0.2)(.36) 0.3 0
= + - - = 8.24
EBITDA .10 - . 0 5 .10 - . 0 5 .10 - .05 .10 - .05
Aswath Damodaran 53
Value/EBITDA Multiples and Taxes
## VEBITDA Multiples and Tax Rates
16
14
12
10
Value/EBITDA
0
0% 10% 20% 30% 40% 50%
Tax Rate
Aswath Damodaran 54
Value/EBITDA and Net Cap Ex
## Value/EBITDA and Net Cap Ex Ratios
12
10
8
Value/EBITDA
0
0% 5% 10% 15% 20% 25% 30%
Net Cap Ex/EBITDA
Aswath Damodaran 55
Value/EBITDA Multiples and Return on Capital
## Value/EBITDA and Return on Capital
12
10
8
Value/EBITDA
WACC=10%
6 WACC=9%
WACC=8%
0
6% 7% 8% 9% 10% 11% 12% 13% 14% 15%
Return on Capital
Aswath Damodaran 56
Value/EBITDA Multiple: Trucking Companies
## Company Name Value EBITDA Value/EBITDA
KLLM Trans. Svcs. \$ 114.32 \$ 48.81 2.34
Ryder System \$ 5,158.04 \$ 1,838.26 2.81
Rollins Truck Leasing \$ 1,368.35 \$ 447.67 3.06
Cannon Express Inc. \$ 83.57 \$ 27.05 3.09
Hunt (J.B.) \$ 982.67 \$ 310.22 3.17
Yellow Corp. \$ 931.47 \$ 292.82 3.18
Roadway Express \$ 554.96 \$ 169.38 3.28
Marten Transport Ltd. \$ 116.93 \$ 35.62 3.28
Kenan Transport Co. \$ 67.66 \$ 19.44 3.48
M.S. Carriers \$ 344.93 \$ 97.85 3.53
Old Dominion Freight \$ 170.42 \$ 45.13 3.78
Trimac Ltd \$ 661.18 \$ 174.28 3.79
Matlack Systems \$ 112.42 \$ 28.94 3.88
XTRA Corp. \$ 1,708.57 \$ 427.30 4.00
Covenant Transport Inc \$ 259.16 \$ 64.35 4.03
Builders Transport \$ 221.09 \$ 51.44 4.30
Werner Enterprises \$ 844.39 \$ 196.15 4.30
Landstar Sys. \$ 422.79 \$ 95.20 4.44
AMERCO \$ 1,632.30 \$ 345.78 4.72
USA Truck \$ 141.77 \$ 29.93 4.74
Frozen Food Express \$ 164.17 \$ 34.10 4.81
Arnold Inds. \$ 472.27 \$ 96.88 4.87
Greyhound Lines Inc. \$ 437.71 \$ 89.61 4.88
USFreightways \$ 983.86 \$ 198.91 4.95
Golden Eagle Group Inc. \$ 12.50 \$ 2.33 5.37
Arkansas Best \$ 578.78 \$ 107.15 5.40
Airlease Ltd. \$ 73.64 \$ 13.48 5.46
Celadon Group \$ 182.30 \$ 32.72 5.57
Amer. Freightways \$ 716.15 \$ 120.94 5.92
Transfinancial Holdings \$ 56.92 \$ 8.79 6.47
Vitran Corp. 'A' \$ 140.68 \$ 21.51 6.54
Interpool Inc. \$ 1,002.20 \$ 151.18 6.63
Intrenet Inc. \$ 70.23 \$ 10.38 6.77
Swift Transportation \$ 835.58 \$ 121.34 6.89
Landair Services \$ 212.95 \$ 30.38 7.01
CNF Transportation \$ 2,700.69 \$ 366.99 7.36
Budget Group Inc \$ 1,247.30 \$ 166.71 7.48
Caliber System \$ 2,514.99 \$ 333.13 7.55
Knight Transportation Inc \$ 269.01 \$ 28.20 9.54
Heartland Express \$ 727.50 \$ 64.62 11.26
Greyhound CDA Transn Corp \$ 83.25 \$ 6.99 11.91
Mark VII \$ 160.45 \$ 12.96 12.38
Coach USA Inc \$ 678.38 \$ 51.76 13.11
US 1 Inds Inc. \$ 5.60 \$ (0.17) NA
Average 5.61
Aswath Damodaran 57
A Test on EBITDA
## n Ryder System looks very cheap on a Value/EBITDA multiple basis,
relative to the rest of the sector. What explanation (other than
misvaluation) might there be for this difference?
Aswath Damodaran 58
Value/EBITDA Multiples: Market
## n The multiple of value to EBITDA varies widely across firms in the
market, depending upon:
how capital intensive the firm is (high capital intensity firms will tend to
have lower value/EBITDA ratios), and how much reinvestment is needed
to keep the business going and create growth
how high or low the cost of capital is (higher costs of capital will lead to
lower Value/EBITDA multiples)
how high or low expected growth is in the sector (high growth sectors will
tend to have higher Value/EBITDA multiples)
Aswath Damodaran 59
US Market: Cross Sectional Regression
No Selector
5903 total cases of which 2943 are missing
R squared = 22.0% R squared (adjusted) = 22.0%
s = 11.26 with 2960 - 4 = 2956 degrees of freedom
## Source Sum of Squares df Mean Square F-ratio
Regression 106063 3 35354.4 279
Residual 375086 2956 126.890
## Variable Coefficient s.e. of Coeff t-ratio prob
Constant 27.8050 0.6408 43.4 0.0001
CpExVal -4.18185 2.345 -1.78 0.0747
lnGrowth 7.86554 0.3021 26.0 0.0001
Eff. Tax Rate -7.65961 0.7666 -9.99 0.0001
Aswath Damodaran 60
Price-Book Value Ratio: Definition
n The price/book value ratio is the ratio of the market value of equity to
the book value of equity, i.e., the measure of shareholders equity in
the balance sheet.
n Price/Book Value = Market Value of Equity
Book Value of Equity
n Consistency Tests:
If the market value of equity refers to the market value of equity of
common stock outstanding, the book value of common equity should be
used in the denominator.
If there is more that one class of common stock outstanding, the market
values of all classes (even the non-traded classes) needs to be factored in.
Aswath Damodaran 61
Price to Book Value: Distribution
Summary of Price/BV
No Selector
5941 total cases of which 755 are missing
Percentile 5
Price to Book Value Ratios Count 5186
Mean 3.84904
1000 Median 1.92370
StdDev 4.37355
900 Min 0.009296
Max 15
800 Lower ith %tile 0.430182
Upper ith %tile 15
700
600
500
400
300
200
100
0
0-0.5 0.5- 1 1-1.5 1.5-2 2- 2.5 2.5 - 3 3 - 3.5 3.5 - 4 4 - 4.5 4.5 - 5 5- 10 >10
Aswath Damodaran 62
Price Book Value Ratio: Stable Growth Firm
DPS1
P0 =
r gn
## n Defining the return on equity (ROE) = EPS0 / Book Value of Equity,
the value of equity can be written as:
*(1 + gn )
BV 0 *ROE*Payout Ratio
P0 =
r - gn
P0 ROE*Payout Ratio*(1 + g n )
= PBV =
BV 0 r-g n
n If the return on equity is based upon expected earnings in the next time
period, this can be simplified to,
P0 ROE *Payout Ratio
= PBV =
BV 0 r-g n
Aswath Damodaran 63
PBV/ROE: Oil Companies
Company Name Ticker Symbol PBV ROE
Crown Cent. Petr.'A' CNPA 0.29 -14.60%
Giant Industries GI 0.54 7.47%
Harken Energy Corp. HEC 0.64 -5.83%
Getty Petroleum Mktg. GPM 0.95 6.26%
Pennzoil-Quaker State PZL 0.95 3.99%
Ashland Inc. ASH 1.13 10.27%
Shell Transport SC 1.45 13.41%
USX-Marathon Group MRO 1.59 13.42%
Lakehead Pipe Line LHP 1.72 13.28%
Tosco Corp. TOS 1.95 15.44%
Occidental Petroleum OXY 2.15 16.68%
Royal Dutch Petr. RD 2.33 13.41%
Murphy Oil Corp. MUR 2.40 14.49%
Texaco Inc. TX 2.44 13.77%
Phillips Petroleum P 2.64 17.92%
Chevron Corp. CHV 3.03 15.69%
Unocal Corp. UCL 3.53 10.67%
Kerr-McGee Corp. KMG 3.59 28.88%
Exxon Mobil Corp. XOM 4.22 11.20%
BP Amoco ADR BPA 4.66 14.34%
Clayton Williams Energy CWEI 5.57 31.02%
Average 2.30 12.23%
Aswath Damodaran 64
PBV versus ROE regression
n Regressing PBV ratios against ROE for oil companies yields the
following regression:
PBV = 1.04 + 10.24 (ROE) R2 = 49%
n For every 1% increase in ROE, the PBV ratio should increase by
0.1024.
Aswath Damodaran 65
Valuing Pemex
n Assume that you have been asked to value a PEMEX for the Mexican
Government; All you know is that it has earned a return on equity of
10% last year. The appropriate P/BV ratio can be estimated
P/BV Ratio (based upon regression) = 1.04 + 10.24 * 0.1 = 2.06
Aswath Damodaran 66
Looking for undervalued securities - PBV Ratios
and ROE
## n Given the relationship between price-book value ratios and returns on
equity, it is not surprising to see firms which have high returns on
equity selling for well above book value and firms which have low
returns on equity selling at or below book value.
n The firms which should draw attention from investors are those which
provide mismatches of price-book value ratios and returns on equity -
low P/BV ratios and high ROE or high P/BV ratios and low ROE.
Aswath Damodaran 67
The Valuation Matrix
MV/BV
Overvalued
Low ROE High ROE
High MV/BV High MV/BV
ROE-r
Undervalued
Low ROE High ROE
Low MV/BV Low MV/BV
Aswath Damodaran 68
Large Market Cap Firms: PBV vs ROE: July 2000
AMGN
LLY
DELL
22.5
SCH
MDT
15.0 QCOM KO
MSFT
ERICY SGP
TWX
P SBH AHP
B TXN GE
V
WMT
HD ABT
PEP
JNJ
7.5
AVE
AXP PG
NWS HWP TYC
PHA MWD
STD ENE SBC
AIG BBV BLS
SLB DT
TEF BPA
MCD C
FON GS MO
DIS MOT DD
FNM
CPQ WFC
WCOM CMB
RD
MC BAC F
T SCDCX
-0.0
## 0.125 0.250 0.375 0.500
ROE
Aswath Damodaran 69
Company Symbols
Company Name Ticker SymbolCompany Name Ticker SymbolCompany Name Ticker Symbol Company Name Ticker Symbol
Matsushita Elec. ADR MC British Telecom ADR BTY Merrill Lynch & Co. MER Int'l Business Mach. IBM
Compaq Computer CPQ Amer. Int'l Group AIG Fannie Mae FNM Abbott Labs. ABT
News Corp. Ltd. ADR NWS Chevron Corp. CHV Tyco Int'l Ltd. TYC Morgan S. Dean Witter MWD
AT&T Corp. T AEGON Ins. Group AEG Amer. Express AXP Amgen AMGN
Schlumberger Ltd. SLB Sprint Corp. FON Corning Inc. GLW Dell Computer DELL
Disney (Walt) DIS Boeing BA EMC Corp. EMC Amer. Home Products AHP
Koninklijke Philips NV PHG Hewlett-Packard HWP Gen'l Electric GE Procter & Gamble PG
Time Warner TWX Banco Bilbao Vis. ADR BBV Intel Corp. INTC Pfizer, Inc. PFE
Deutsche Telekom ADR DT Wells Fargo WFC Ford Motor F Schering-Plough SGP
WorldCom Inc. WCOM Ericsson ADR ERICY BellSouth Corp. BLS Merck & Co. MRK
Motorola, Inc. MOT Texas Instruments TXN Johnson & Johnson JNJ Bristol-Myers Squibb BMY
Telefonica SA ADR TEF Micron Technology MU Lucent Technologies LU Philip Morris MO
Banco Santander ADR STD Bank of America BAC PepsiCo, Inc. PEP Lilly (Eli) LLY
Sony Corp. ADR SNE Home Depot HD Cisco Systems CSCO Oracle Corp. ORCL
Exxon Mobil Corp. XOM McDonald's Corp. MCD Goldman Sachs GS
Aventis ADR AVE SBC Communications SBC Medtronic, Inc. MDT
Enron Corp. ENE Wal-Mart Stores WMT Sun Microsystems SUNW
Pharmacia Corp. PHA Du Pont DD Applied Materials AMAT
Shell Transport SC Citigroup Inc. C Schwab (Charles) SCH
Royal Dutch Petr. RD Qualcomm Inc. QCOM Microsoft Corp. MSFT
DaimlerChrysler AG DCX SmithKline Beecham SBH Nokia Corp. ADR NOK
BP Amoco ADR BPA Chase Manhattan Corp. CMB Coca-Cola KO
Aswath Damodaran 70
PBV Matrix: Telecom Companies
12
TelAzteca
10
TelNZ Vimple
8 Carlton
Teleglobe
FranceTel Cable&W
6
DeutscheTel
BritTel
TelItalia
Portugal AsiaSat
HongKong
BCE Royal
4 Hellenic
Nippon
DanmarkChinaTel
Espana Indast
Telmex
TelArgFrance
PhilTel Televisas
TelArgentina
2 TelIndo
TelPeru
APT
CallNet
Anonima GrupoCentro
0
0 10 20 30 40 50 60
ROE
Aswath Damodaran 71
U.S. Banks: Market Cap > \$ 1 billion
5.00
MEL
SNV
CBH
3.75
WABC
WFC CYN
CFR WL
BBT
P
B VLY CMB
V 2.50 NBAK PNC
ZION FULT SKYF
HU FBF
ASO MRBK
TRMK WB
OV
STI CBC CBSS
BPOP
FVB BAC
FSCO RGBK
UPC PFGI FTU
SOTR
1.25 KEY
UB
BOH
BWE
## 0.12 0.16 0.20 0.24
ROE
Aswath Damodaran 72
Company Name Ticker Symbol Company Name Ticker Symbol Company Name Ticker Symbol
Westamerica Bancorp WABC Fulton Fin'l FULT Regions Financial RGBK
Keystone Fin'l KSTN First Va. Banks FVB Synovus Financial SNV
Colonial BncGrp. 'A' CNB City National Corp. CYN AmSouth Bancorp. ASO
One Valley Bancorp OV Hibernia Corp. `A' HIB KeyCorp KEY
National BanCorp. of Alaska,In NBAK Silicon Valley Bncsh SIVB BB&T Corp. BBT
BancWest Corp. BWE Mercantile Bankshares MRBK Wachovia Corp. WB
Hudson United Bancorp HU Compass Bancshares CBSS PNC Financial Serv. PNC
Provident Finl Group PFGI Popular Inc BPOP SunTrust Banks STI
Pacific Century Fin'l BOH First Security FSCO State Street Corp. STT
Centura Banks CBC No. Fork Bancorp NFB Mellon Financial Corp. MEL
Trustmark Corp. TRMK Natl Commerce Bancrp NCBC Morgan (J.P.) & Co JPM
Sky Finl Group Inc SKYF UnionBancal Corp UB First Union Corp. FTU
Wilmington Trust WL M&T Bank Corp. MTB FleetBoston Fin'l FBF
Valley Natl Bancp NJ VLY Zions Bancorp. ZION Bank of New York BK
Commerce Bancorp NJ CBH Union Planters UPC Chase Manhattan Corp. CMB
Cullen/Frost Bankers CFR SouthTrust Corp. SOTR Wells Fargo WFC
Summit Bancorp SUB Bank of America BAC
Aswath Damodaran 73
IBM: The Rise and Fall
## IBM: PBV and ROE
4.00 30.00%
3.50
25.00%
3.00
20.00%
2.50
P/BV Ratio
PBV
ROE
ROE
2.00 15.00%
1.50
10.00%
1.00
5.00%
0.50
0.00 0.00%
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
1995
1996
1997
Year
Aswath Damodaran 74
PBV Ratio Regression
No Selector
5903 total cases of which 3332 are missing R squared = 46%
R squared = % R squared (adjusted) = %
s = 2.240 with 2571 - 4 = 2567 degrees of freedom
## Source Sum of Squares df Mean Square F-ratio
Regression 30502.9 4 7625.73 1519
Residual 12885.7 2567 5.01977
## Variable Coefficient s.e. of Coeff t-ratio prob
Exp Growt 8.97383 0.4376 20.5 0.0001
Beta 0.854662 0.1035 8.26 0.0001
Payout Ra -0.051989 0.1335 -0.390 0.6969
ROE 4.96796 0.2109 23.6 0.0001
Aswath Damodaran 75
Cross Sectional Regression for Greece: June 1999
## n Using data obtained from Bloomberg for 199 Greek companies, we
ran the regression of PBV ratios against returns on equity and obtained
the following:
PBV = 2.56 + 24.00 ROE R2 = 45.37%
(4.19) (12.82)
n For instance, the predicted PBV ratios for the following companies
would be:
Company Actual PBV ROE Predicted PBV
Alpha Fin. 14.87 47% 2.56 + 24(.47)= 13.84
Girakian 2.36 1% 2.56 + 24(.01)= 2.80
Titan Cement 5.98 33% 2.56 + 24(.33)= 10.56
Michaniki 1.72 13% 2.56 + 24(.13)= 5.68
Aswath Damodaran 76
Price Sales Ratio: Definition
n The price/sales ratio is the ratio of the market value of equity to the
sales.
n Price/ Sales= Market Value of Equity
Total Revenues
n Consistency Tests
The price/sales ratio is internally inconsistent, since the market value of
equity is divided by the total revenues of the firm.
Aswath Damodaran 77
Price/Sales Ratio: Cross Sectional Distribution
Summary of Price/Sales
No Selector
5941 total cases of which 1023 are missing
Percentile 5
Count 4918
Mean 2.51810
Median 1.03579
Price to Sales Ratio
StdDev 3.16625
Min 0.001524
800
Max 10
Lower ith %tile 0.105026
700 Upper ith %tile 10
600
500
400
300
200
100
0
<0.05 0..05-0.1 0.1-0.15 0.150.25 0.25-0.5 0.5-0.75 0.75-1 1-1.5 1.5-2 2-3 3 -4 4 -5 5 - 7.5 7.5 - 10 >10
Aswath Damodaran 78
Price/Sales Ratio: Determinants
## n The price/sales ratio of a stable growth firm can be estimated
beginning with a 2-stage equity valuation model:
DPS1
P0 =
r gn
## n Dividing both sides by the sales per share:
P0 Net Profit Margin*Payout Ratio * ( 1+ g n )
= PS =
Sales 0 r-gn
Aswath Damodaran 79
PS/Margins: Brazilian Consumer Products
## Company PS Ratio Net Margin
Lojas Arapua 0.01 -14.24%
Borghoff 0.03 -25.93%
Grazziotin 0.09 5.86%
Panvel 0.11 2.45%
Cia Alimentos 0.11 -12.47%
Bombril 0.13 3.32%
Lojas Americanas 0.18 -1.99%
IND Bebidas Antac 0.55 4.86%
Cia Antarctica 0.57 2.69%
Lojas Renner 0.62 9.25%
Tehnos Relogios 0.83 28.05%
Casa Anglo 1.04 2.30%
Souza Cruz 1.29 20.85%
Ind bebidas Antarc Polar 1.73 37.99%
Brahma 1.80 16.42%
Aswath Damodaran 80
Price/Sales Ratio: Is DHB cheap?
n Based upon the price/sales ratios, the cheap firms are Borghoff and
Lojas Arapua. The expensive firms are firms like Souza Cruz and
Brahma. Do you agree?
o Yes
o No
n If not, what might explain why there are such big differences across
these firms?
Aswath Damodaran 81
Regression Results: PS Ratios and Margins
## n Regressing PS ratios against net margins,
PS = 0.43 + 2.93 (Net Margin) R2 = 59.29%
n Thus, a 1% increase in the margin results in an increase of 0.03 in the
price sales ratios.
n The regression also allows us to get predicted PS ratios for these firms
Aswath Damodaran 82
PS Ratios: Actual versus Predicted Values
## Company PS Ratio Net Margin Predicted PS Under or Over Valued
Lojas Arapua 0.0103 -14.24% 0.0128 -19.74%
Borghoff 0.0283 -25.93% NA NA
Grazziotin 0.0918 5.86% 0.6017 -84.74%
Panvel 0.1116 2.45% 0.5019 -77.76%
Cia Alimentos 0.1135 -12.47% 0.0646 75.75%
Bombril 0.1317 3.32% 0.5273 -75.03%
Makro Atacadista 0.1528 1.30% 0.4681 -67.35%
Lojas Americanas 0.1823 -1.99% 0.3717 -50.96%
IND Bebidas Antac 0.5513 4.86% 0.5723 -3.67%
Cia Antarctica 0.5700 2.69% 0.5088 12.03%
Lojas Renner 0.6240 9.25% 0.7010 -11.00%
Tehnos Relogios 0.8250 28.05% 1.2518 -34.09%
Casa Anglo 1.0384 2.30% 0.4973 108.80%
Souza Cruz 1.2864 20.85% 1.0408 23.60%
Ind bebidas Antarc Polar 1.7257 37.99% 1.5431 11.83%
Brahma 1.8027 16.42% 0.9110 97.87%
Aswath Damodaran 83
Current versus Predicted Margins
n One of the limitations of the analysis we did in these last few pages is
the focus on current margins. Stocks are priced based upon expected
margins rather than current margins.
n For most firms, current margins and predicted margins are highly
correlated, making the analysis still relevant.
n For firms where current margins have little or no correlation with
expected margins, regressions of price to sales ratios against current
margins (or price to book against current return on equity) will not
provide much explanatory power.
n In these cases, it makes more sense to run the regression using either
predicted margins or some proxy for predicted margins.
Aswath Damodaran 84
A Case Study: The Internet Stocks
30
PKSI
LCOS SPYG
20
INTM MMXI
SCNT
## MQST FFIV ATHM
A
CNET DCLK
d
j RAMP
INTW
P 10 CSGP CBIS NTPA
S NETO SONEPCLN
APNT CLKS
EDGRPSIX ATHY AMZN
SPLN BIDS
ALOY ACOM EGRP
BIZZ IIXL
ITRA ANET
ONEM ABTL INFO
FATB
RMII TMNT GEEK
GSVI ROWE
## -0.8 -0.6 -0.4 -0.2
Aswath Damodaran 85
PS Ratios and Margins are not highly correlated
## n Regressing PS ratios against current margins yields the following
PS = 81.36 - 7.54(Net Margin) R2 = 0.04
(0.49)
n This is not surprising. These firms are priced based upon expected
margins, rather than current margins. Hypothesizing that firms with
higher revenue growth and higher cash balances should have a greater
chance of surviving and becoming profitable, we ran the following
regression: (The level of revenues was used to control for size)
PS = 30.61 - 2.77 ln(Rev) + 6.42 (Rev Growth) + 5.11 (Cash/Rev)
(0.66) (2.63) (3.49)
R squared = 31.8%
Predicted PS = 30.61 - 2.77(7.1039) + 6.42(1.9946) + 5.11 (.3069) =
30.42
Actual PS = 25.63
Aswath Damodaran
Stock is undervalued, relative to other internet stocks. 86
PS Regression
No Selector
5903 total cases of which 3655 are missing R squared = 52%
R squared = % R squared (adjusted) = %
s = 1.849 with 2248 - 4 = 2244 degrees of freedom
## Source Sum of Squares df Mean Square F-ratio
Regression 14960.1 4 3740.03 1094
Residual 7670.48 2244 3.41822
## Variable Coefficient s.e. of Coeff t-ratio prob
Exp Growth: E 7.60241 0.3801 20.0 0.0001
Beta -0.444203 0.0918 -4.84 0.0001
Payout Ratio -0.585029 0.1147 -5.10 0.0001
Aswath Damodaran 87
Choosing Between the Multiples
## n As presented in this section, there are dozens of multiples that can be
potentially used to value an individual firm.
n In addition, relative valuation can be relative to a sector (or
comparable firms) or to the entire market (using the regressions, for
instance)
n Since there can be only one final estimate of value, there are three
choices at this stage:
Use a simple average of the valuations obtained using a number of
different multiples
Use a weighted average of the valuations obtained using a nmber of
different multiples
Choose one of the multiples and base your valuation on that multiple
Aswath Damodaran 88
Picking one Multiple
n This is usually the best way to approach this issue. While a range of
values can be obtained from a number of multiples, the best estimate
value is obtained using one multiple.
n The multiple that is used can be chosen in one of two ways:
Use the multiple that best fits your objective. Thus, if you want the
company to be undervalued, you pick the multiple that yields the highest
value.
Use the multiple that has the highest R-squared in the sector when
regressed against fundamentals. Thus, if you have tried PE, PBV, PS, etc.
and run regressions of these multiples against fundamentals, use the
multiple that works best at explaining differences across firms in that
sector.
Use the multiple that seems to make the most sense for that sector, given
how value is measured and created.
Aswath Damodaran 89
A More Intuitive Approach
## n As a general rule of thumb, the following table provides a way of
picking a multiple for a sector
Sector Multiple Used Rationale
Cyclical Manufacturing PE, Relative PE Often with normalized earnings
High Tech, High Growth PEG Big differences in growth across
firms
High Growth/No Earnings PS, VS Assume future margins will be good
Heavy Infrastructure VEBITDA Firms in sector have losses in early
years and reported earnings can vary
depending on depreciation method
REITa P/CF Generally no cap ex investments
from equity earnings
Financial Services PBV Book value often marked to market
Retailing PS If leverage is similar across firms
VS If leverage is different
Aswath Damodaran 90
Reviewing: The Four Steps to Understanding
Multiples
## n Define the multiple
Check for consistency
Make sure that they are estimated uniformally
n Describe the multiple
Multiples have skewed distributions: The averages are seldom good
indicators of typical multiples
Check for bias, if the multiple cannot be estimated
n Analyze the multiple
Identify the companion variable that drives the multiple
Examine the nature of the relationship
n Apply the multiple
Aswath Damodaran 91 | 14,270 | 43,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-35 | latest | en | 0.842547 |
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Transcript
```3.7 Angle Side Theorems
Theorem 20: Isosceles Triangle
Theorem (ITT)
If 2 sides of a triangle are congruent, then the
angles opposite the sides are congruent.
A
A
If
then
B
C
B
C
Theorem 21: Converse Isosceles
Triangle Theorem (CITT)
If 2 angles of a triangle are congruent, then
the sides opposite the angles are congruent.
A
A
If
then
B
C
B
C
Can you prove theorems 20 &21?
ITT can be proven using SSS by naming the
triangle in a correspondence with itself.
CITT can be proven using ASA by naming the
triangle in a correspondence with itself.
Special Note: for triangles, equilateral and
equiangular will be used interchangeably.
Given:
D
E H
EF HG
Prove:
DF DG
Statements
1.
E H
Reasons
1.
Given
EF HG
2.
DE DH
3. DEF DHG
4.
DF DG
2. CITT
If
3.
SAS
4.
CPCTC
E
then
F
G
H
Given:
XY XZ
Prove:
1 2
X
1
Y
Statements
1.
XY XZ
2
Z
Reasons
1.
Given
2.
XYZ XZY
2.
If
2. ITT
3.
1 is suppl. to XYZ
3. If 2 angles form a straight line, then
2 is suppl. to XZY
4.
1 2
then
they are supplementary
4.
If 2 angles are supplementary to
4. ST
congruent angles then they are
congruent
More Practice Proofs
More Practice Proofs
```
Related documents | 672 | 2,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-35 | latest | en | 0.687019 |
https://cs.stackexchange.com/questions/3487/using-a-step-counting-function-in-a-turing-machine-construction | 1,716,972,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00876.warc.gz | 152,043,750 | 43,216 | # Using a step-counting function in a Turing Machine construction
I have an question relating to the elementary foundations of Turing Machine theory. I would like to have a clarification of the status of a function $\phi$ (a mapping between TM indexes) I shall introduce in the formal question after the preamble. First to the preamble.
Let us assume that we have a fixed Turing Machine language and corresponding Universal Turing Machine. Thus ${TM_i}$ is an enumeration of machines, with UTM say as the Universal Machine in that enumeration. Let us also introduce a numerical parameter c for this example: (for example c = 500).
Now execution of any TM can be simulated on UTM.
EDIT (Explaining notation) I am using some notation in this question which I shall outline here. Let S be an arbitrary input string for TM, then I shall write:
$$UTM(TM_i; S)$$
to indicate that $TM_i$ is using S as its input. So we will have the equation: $$UTM(TM_i; S) = TM_i(S)$$ since UTM is Universal, and its purpose is simply to simulate TMs on their input. If S is a composition of x,y I shall write: $$UTM(TM_i;x,y)$$
(End notation discussion.)
Let us modify UTM to UTM2 (and using the above notation) with the following property when a given $TM_i$ is executed on UTM2:
$UTM_2(TM_i;c,x) = UTM(TM_i;c,x) = TM_i(x)$ if $TM_i$ requires c steps or less to compute the output
$UTM_2(TM_i;c,x)=0$ if $TM_i(x)$ requires at least c steps for computation
EDIT The original question with this second condition has been answered. In this edit I would like to modify the condition to the following (introducing UTM3 which also meets the first condition):
$UTM_3 (TM_i; c, x) = [s]$ if $TM_i (x)$ requires at least c steps for computation.
In this expression [s] is the string on the tape which results after c steps of computation by $TM_i$ on Input. (EDIT The corresponding state of $TM_i$ will not be a Final state (ACCEPT or REJECT), but just an arbitrary state $q_j$. This is because UTM3 stops executing after c steps, ignoring the Turing Machine convention requiring termination only in Final states.) As a result (almost) all input strings x, including those with |x| > c will be modified (in a maximum of c string positions) - expressed in Language terms many such strings will just have their ACCEPTANCE calculation prematurely terminated in a non-Final state - but some strings of arbitrary length will still be ACCEPTED by some Turing Machines in less than c steps (IE those for which ACCEPTANCE requires examining less than the first c terms.)
So under UTM2/3 the output is determined partly by a (hidden) step counting function. Here UTM2/3 has taken on some of the role of an operating system, by monitoring the actual steps taken by the simulated machines, and using some step-counting variable to behave as above.
(EDIT: We show below that UTM2/3 is a Universal Machine)and so capable of executing any TM - on any argument - it is provided with, at least as far as c steps. Despite this limitation I believe that an infinite number of (TM,input) pairs will result in a non-zero output. So most TMs executed on UTM2/3 will not behave as they do on UTM. In UTM3 the outputs will explicitly depend on x,i and c.(End preamble.)
The question is whether any given $TM_i$ can be modified into $TM_{\phi (i,c)}$ with the property that (generalising to arbitrary c):
$UTM_3(TM_i; c, x)$ = $UTM (TM_{\phi(i,c)}; x)$ (for all $x$ and $c$)
EDIT2\phi Both questions for the $\phi(i,c)$ have been answered. For reasons discussed below I want to modify the definition to $\phi(i)$. This new definition is:
$UTM_3(TM_i; c, x)$ = $UTM (TM_{\phi(i)}; c,x)$ = $TM_{\phi(i)}(c,x)$ (for all $x$ and $c$)
This condition is meant to ensure that $TM_{\phi (i)}$ is a correct simulation of $TM_i$. The specific questions to be proven about $\phi: (i) \rightarrow i$ are: (i) Is $\phi$ well defined and non-trivial; (ii) Is $\phi$ recursive; (iii) Is $\phi$ non-recursive?
In the LHS of the above definitions c is an input, which is ignored by $TM_i$ but used by UTM3 to determine the stopping stage. Asking for $\phi(x,c)$ made this a fixed parameter built into the definition of $TM_{\phi(x,c)}$. However the correct UTM simulation will be given c as in the modified equation RHS. This parameter needs to be used by $TM_{\phi(i)}$ along with its own input x, to construct the correct simulation. (EDIT) Of course since the UTM is a regular Universal Machine it simply simulates the execution of $TM_{\phi(i)}$ - the parameters on its tape - like c - are meant for its TM's only.
Note that a special symbol u could be introduced switching off the UTM3 counts, and so
$$UTM3(TM_i;u,x) = UTM(TM_i;u,x) = TM_i(x)$$
Also the same problem can be formulated for the original UTM2 (with modified $\phi$).
This question is a simplification of a problem I am trying to understand (I am a math but not a CS major, but have some textbooks like Hopcroft/Ullmann), and I hope that I can get some clarification on this piece.
• Why do you write "UTM3 ($TM_i$) c Input" and at the end "UTM3 ($TM_i$, c, Input)"? Do you mean $UTM_3(TM_i,c,x)$ in both cases? (use $x$ instead of "input"). Please use uniform notation in order to make the question more readable. Furthermore remember that in the standard notion/definition the "output" of a TM (or UTM) is ACCEPT or REJECT (or never halt).
– Vor
Sep 10, 2012 at 14:41
• I have tried to improve the equation formatting. If your last sentence refers (in part) to the odd end state of UTM3, namely at string [s], then this will not be either ACCEPT nor REJECT in general. UTM3 will just stop at an arbitrary state $q_j$. I have added some text into the main body on this unusual aspect of UTM3. Sep 10, 2012 at 21:59
• in the last edit, it is not clear what is the difference between UTM3 and UTM: now both have 3 parameters: $TM_i, c$ and $x$. What is the meaning of $c$ for UTM ?
– Vor
Sep 12, 2012 at 9:25
• I have added a sentence, just to clarify that parameter c on the UTM tape is just another part of the input for $TM_{\phi(i)}$: it wont modify UTM in any way. Sep 12, 2012 at 9:39
• ok. to build $TM_{\phi(i)}$ just add a counter (initially set to $c$) and for every $TM_i$ transition $(q,a)\rightarrow (q',a',D)$ add an "interleaved subroutine" that checks and decrement the $c$ counter. If $c$ = 0 halts, otherwise return to state $q'$.
– Vor
Sep 12, 2012 at 12:52
## 1 Answer
The answer is yes.
A possible quick approach: UTM2($TM_i$,c,x) $\neq 0$ only on $|x| \leq c$ so you can build $TM_{\phi(i,c)}$ simply simulating $TM_i$ on all inputs and hard encoding its output on the finite input strings $|x| < c$, the index of the resulting TM is the value of $\phi(i,c)$
EDIT2: another approach that can work to simulate tape content (UTM3 in your last edit) is the following:
• from $TM_i$ build $TM_{\phi(i,c)}$ using $|Q|*c$ states, where Q is the set of states of $TM_i$.
• every $(q_j,a) \rightarrow (q_k,b,L)$ transition of $TM_i$ is mapped to $c$ transitions on $TM_{\phi(i,c)}$ in this way: $(q_j^m,a) \rightarrow (q_k^{m+1},b,L)$ with $(1 \leq m \leq c-1)$
• make state $q_j^m$ accepting if the corresponding $q_j$ in $TM_i$ is accepting
This guarantees that $TM_{\phi(i,c)}$ runs exactly for $c$ steps and the tape content is the same of $TM_i$
• Thanks this is clear. Unfortunately it means that I have over-simplified my original problem too far. I really want the tape at step c to be the last printed value at this stage. This will mean that for |x| > c there are non-null values on the tape, for all x. Should I modify this Question or ask another? Sep 10, 2012 at 12:48
• I have now EDITED the Question introducing UTM3 to clarify the generalisation I wanted. $\phi$ is now for UTM3. In UTM3 I am not sure how "tape content simulation" would work since the set of x to check is now unbounded, unlike in the UTM2 case where x was bounded by |x|<c. Perhaps I am missing something simple again... Sep 10, 2012 at 14:21
• @RoySimpson: see my new edit :)
– Vor
Sep 10, 2012 at 15:17
• I am still working through bullet two of your answer. Let us assume that the initial state of $TM_i$ is $q_0$ with initial transition $(q_0,a)\rightarrow (q_2,b,L)$. There will be c corresponding transitions in $TM_\phi$: $(q_0^1,a)\rightarrow (q_2^2,b,L)$, $(q_0^2,a)\rightarrow (q_2^3,b,L)$... etc. But surely only $q_0^1$ is the initial state of $TM_\phi$ with these other transitions never used? Consequently only $q_2^2$ ever gets entered in $TM_\phi$, etc. So at the moment I dont see how these new transitions and corresponding states actually get used. Sep 10, 2012 at 22:52
• @RoySimpson: many of them will not be used; e.g. every state $q_k^m$ not connected to $q_0^1$ (the only initial state) or at distance $d \geq m$ from it, will never be entered, but this doesn't matter (you can also have unused transitions in $TM_i$ itself).
– Vor
Sep 11, 2012 at 6:39 | 2,493 | 8,906 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-22 | latest | en | 0.893635 |
https://discourse.julialang.org/t/makie-rotate-bar-plot-labels/68679 | 1,660,436,076,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571989.67/warc/CC-MAIN-20220813232744-20220814022744-00578.warc.gz | 210,170,443 | 5,724 | # [Makie] Rotate bar plot labels
I have a bar-plot:
But the labels are hard to read in this orientation. A workaround is to make the plot with `direction = :x`:
But is there a way to simply rotate the labels in the first plot?
Code used to create first plot:
``````fig, ax, plt = barplot([mean(fc.*(√2, 1/√2)) for fc in centerfrex_octave3rd], octave3rd_rms,
width=[0.95*fc*(2^(1/6)-1/2^(1/6)) for fc in centerfrex_octave3rd],
bar_labels = ["fc = \$fc" for fc in centerfrex_octave3rd], label_size=15,
direction=:x,
flip_labels_at=60,
axis = (ylabel="Frequency", xlabel="RMS value, dB", title = "Barplot octave3rd-bands, filterorder = \$order")
); ax.yscale=log10; ax.xminorticks = IntervalsBetween(4, true); ax.xminorgridvisible=true; ax.xminorgridwidth = 2; ax.xminorticksvisible=true; fig
``````
Code for second plot:
``````fig, ax, plt = barplot([mean(fc.*(√2, 1/√2)) for fc in centerfrex_octave3rd], octave3rd_rms,
width=[0.95*fc*(2^(1/6)-1/2^(1/6)) for fc in centerfrex_octave3rd],
bar_labels = ["fc = \$fc" for fc in centerfrex_octave3rd], label_size=15,
#direction=:x,
flip_labels_at=40,
axis = (ylabel="Frequency", xlabel="RMS value, dB", title = "Barplot octave3rd-bands, filterorder = \$order")
); ax.xscale=log10; ax.xminorticks = IntervalsBetween(4, true); ax.xminorgridvisible=true; ax.xminorgridwidth = 2; ax.xminorticksvisible=true; fig
``````
You can always go inside the components of a composite plot like this and change attributes there. So you could go into barplot.plots, find the text plot and change the rotation. This case was simply not considered when writing the recipe I assume
1 Like
Did you find a way to do this?
Nope. If I needed it again, I would follow Jule’s suggestion to add the option in the recipe myself - I dont think it would be that hard to add the kwarg to the recipe. | 561 | 1,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-33 | latest | en | 0.747916 |
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Changing Coin Denomination to Win a Slots Jackpot
I’ve heard several interesting theories recently on slot machine strategies. Two of them relate to adjusting your bet size, and how it can decrease or increase your odds of winning a slots jackpot. We’ll take a close look at these theories to determine whether they have any merit, based on fundamental slot machine mechanics and random number generator (RNG) algorithms.
Increase Bet to Win a Slot Machine Jackpot
Our first hypothesis deals with increasing the coin denomination to simultaneously increase the game’s return to player (RTP). The theory here is that slot machines will inherently deliver bigger payouts to player’s who are willing to bet more money per spin.
This concept can be dissected in two ways. First of all, placing larger bets will award larger prizes, simply because all payouts are multiples of the player’s bet. On pay-line slots, the total bet is divided by the number of paylines. Thus, each winning combination is a multiple of the per-line bet. But this is simple mathematics, generally understood by all beginner to intermediate level slots players, and has no bearing whatsoever on the game’s RTP.
The more important idea here is that the RTP will be higher for a large wager, but lower for a small wager. For example, betting \$0.25 on an online slots games may incur a 95% RTP, while increasing to a \$2+ bet per spin will raise the RTP to something like 97%.
In this case, it can be true. It all depends on the game, though. The majority of slot machines do not integrate a fluctuating RPT. Some of the ones known to use a rising-RTP method include Microgaming‘s Mega Moolah progressive slots (higher chance of triggering the jackpot bonus game with larger bet), and all WMS/Barcrest games that feature ‘Big Bet Mode‘.
Decrease Bet To Win a Slots Jackpot
This one is a pretty strange concept, but it has arisen in several discussion forums of late. According to one gambler, he was playing the slot machines at a local casino when he saw the woman behind him win over \$2,000 playing \$0.25 per spin. He sought advice from the cocktail waitress, asking if he should stay away from that machine for awhile.
He claims that the waitress told him no, the machine is just as likely to win again (which is true). However, she suggested that he decrease his wager to around \$0.05 or \$0.10 per spin. According to her instruction, the machine would not be willing to pay another jackpot to a \$0.25 bettor. But it will still pay off for a reduction in bet size.
I’ll start by addressing that last bit. Such advice is absolutely absurd. No slot machine on the planet will have an increased chance of paying out a jackpot for smaller bets. Aside from the rare ‘increased bet/RTP rate’ games described above, slot machines payout entirely at random.
The RTP is upheld in the long term by the RNG. The RNG stops on a random sequence the moment you pull the, or push the spin button. Only this can determine the amount won or lost per spin. No result is predetermined (except on Class II games, which aren’t true slot machines). Realistically, a slots jackpot is just as likely to pay twice in one day, or twice in five years.
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Problem Set 4 - Solutions
# Problem Set 4 - Solutions - Suggested Solutions to Problem...
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Suggested Solutions to Problem Set 4 Ying Chen ECN 416, Game Theory 1. Bertrand Model revisited Suppose two f rms sell identical products and engage in price competition. The market demand function is = 100 and the consumers will buy from the f rm that o f ers a lower price. Unlike what we assumed in class, let’s make the assumption here that if both f rmssetthesamepr ice ,then all consumers buy from f rm 1 . Moreover, assume that f rm 1 ’s marginal cost of production is 10 whereas f rm 2 ’s marginal cost of production is 15 . Neither f rm has f xed cost and both are pro f t maximizers. Suppose the f rms choose prices simultaneously and they can charge any positive price. (a) Write down the payo f functions of each f rm. Firm 1 ’s payo f function is 1 ( 1 2 )= ½ (100 1 )( 1 10) if 1 2 0 if 1 2 Firm 2 ’s payo f function is 2 ( 1 2 )= ½ (100 2 )( 2 15) if 1 2 0 if 1 2 (b) Find pure strategy Nash Equilibrium of this game. Pure strategy NE: ( 1 2 ) where 10 1 = 2 15 . First, we can show that if a strategy pro f le ( 1 2 ) satis f es 1 6 = 2 ,th eni ti sn o taNE . (argument similar to what we made in class for the symmetric Bertrand model.) Then, we can show that if 1 = 2 and the prices are above the marginal cost of f rm 1 and below the marginal cost of f
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Ask a homework question - tutors are online | 569 | 1,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-09 | latest | en | 0.803313 |
http://www.picbasic.co.uk/forum/showthread.php?t=5586&s=a4d4faad08ff1f7e0864e25b6020b851 | 1,638,098,743,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00472.warc.gz | 102,402,253 | 13,839 | Random number seed
# Thread: Random number seed
1. ## Random number seed
Thank you for all our help on my last post. It all really helped. To be honest, I finially just looked at the generated number and then based on its value fired off one of 5 sub routines. Here is the chunk of code i am using....Not very elegant but it works fine.
Begin:
Count = 0
W0 = W0 + 1
Peek PORTB, B2
IF B2 = %10000000 Then Test
GoTo Begin
Test:
Count = Count + 1
IF Count = 6 Then Begin
Random W0
IF W0 < 13105 Then loop1
IF W0 < 26210 Then loop2
IF W0 < 39315 Then loop3
IF W0 < 52420 Then loop4
IF W0 <= 65525 Then loop5
loop1:
Toggle 0 'Turn on LED connected to RB0
Pause 500 'Delay for .5 seconds
Toggle 0 'Turn off LED connected to RB0
GoTo TEST
Now I am on to my latest issue. First off I am using Pic Basic (The low cost version.) I am trying to seed the random number generator with a changing opening amount. So each time I use the device, the initial random pattern changes.
My queston is, the EEprom memory specifically location "0" should keep its value after the pick is powered down and then powered up again right??
So all I have to do is set an additional counter and incrament the seed by setting memory loaction 0 to this counter.
2. Two questions:
1. In your code above, you have the following.
Code:
```IF W0 < 13105 Then loop1
IF W0 < 26210 Then loop2
IF W0 < 39315 Then loop3
IF W0 < 52420 Then loop4
IF W0 <= 65525 Then loop5```
Lets take W0=10000 then all of these IF statements will be true and be executed. Or say W0=28000; in this case last three IF statements will be executed, etc...
You should change it as follows.
Code:
```IF W0 < 13105 Then
loop1
ELSE
IF W0 < 26210 Then
loop2
ELSE
IF W0 < 39315 Then
loop3
ELSE
IF W0 < 52420 Then
loop4
ELSE
IF W0 <= 65525 Then loop5
ENDIF
ENDIF
ENDIF
ENDIF```
OR use a select case, it would make it easier to follow the logic.
2. Forget this one.
---------------------------
3. Hi,
There was a thread about random numbers, here ... sometimes ago.
Conclusion was Microchip offers the best routine ( assembler ... yes ! ) in its applications.
Alain
4. ## Sorry to jump in.....
Hi,
Sayzer wrote:
1. In your code above, you have the following.
Code:
IF W0 < 13105 Then loop1
IF W0 < 26210 Then loop2
IF W0 < 39315 Then loop3
IF W0 < 52420 Then loop4
IF W0 <= 65525 Then loop5
Lets take W0=10000 then all of these IF statements will be true and be executed. Or say W0=28000; in this case last three IF statements will be executed, etc...
Why is that?
If W0=10000 the first IF statement tests true and the program jumps to the Loop1 routine which is executed. At the end of Loop1 the program jumps to Test where a new RANDOM is executed and assigned to W0. Lets say it's 28000 as in you example. The third statement tests true and the program jumps to Loop3, executes it and then starts over with another RANDOM. At least that's how I read it.
HTH
/Henrik Olsson.
5. You seem to be having it correct Henrik.
6. ## Random number seed
Actually the if statements work fine, but I like adding the "Else" statements and "EndIf"s to clean up the code (Thank you Sazer). and comments to clarify the logic behind my code. My real issue is the way the code is now, my code always provides the same sequence of numbers on start up. I need it to be different each time at startup as well as any other time it basically is supposed to randomly flash 1 of 5 different LEDs - 5 times and then stop and wait for B.7 to go high and then star again. So how can I randomly add a seed to the very first iteration of this process. I read some where that you can pull a value from TMRO, so it will give you an initial seed of 0 to 255. Any thoughts?
Last edited by kwelna; - 24th January 2007 at 04:08.
7. Hi, Kwelna
Just keep the last value before switching off in EEPROM ... as a new seed !!!
Alain
8. ## Random seed
Duh!!! So simple I never would have thought of it.........well maybe in a week or two.... I am alwqys looking for a more complicated method. that is what happens when you get old!
Thanks again
Kevin
9. Originally Posted by kwelna
that is what happens when you get old!
Imagine when you'll be as old as i am
You could still use an hardware solution... count pulses at the output of a white (even pink) noise generator... now it's going to be really random. Always a long debate around this....
10. ## Stone age survivors ???
Hi, Steve
You were talking about the MM 5837 ... I suppose ???
The only one corresponding to your age LOL !!!
Hi, Kevin
<< I am alwqys looking for a more complicated method. that is what happens when you get old! >>
Here, you're only talking about yourself ...
Ah, Will you have a look here ???
http://www.manhattancontrols.com/for...932b79b20ee1c5
Alain
Last edited by Acetronics2; - 1st February 2007 at 11:38.
11. Originally Posted by Acetronics
Hi, Steve
You were talking about the MM 5837 ... I suppose ???
The only one corresponding to your age LOL !!!
LMAO! well could also be a simple op-amp (maybe dual), cmos gate etc etc. no need for a dedicated IC. Maybe PBP SOUND could do the job.
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• You may not edit your posts | 1,434 | 5,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-49 | latest | en | 0.867283 |
https://professionalsessays.com/2022/12/19/nonunionized-manufacturing-firms-economics-homework-help/ | 1,713,673,440,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.0/warc/CC-MAIN-20240421040323-20240421070323-00478.warc.gz | 415,740,487 | 20,086 | Posted: December 19th, 2022
# Nonunionized manufacturing firms | Economics homework help
7 In Example 4.7, we used data on nonunionized manufacturing firms to estimate the relationship between the scrap rate and other firm characteristics. We now look at this ex- ample more closely and use all available firms.
Save Time On Research and Writing
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(i) The population model estimated in Example 4.7 can be written as log(scrap) 5 b0 1 b1hrsemp b2log(sales) 1b3log(employ) 1 u.
Using the 43 observations available for 1987, the estimated equation is
log(scrap) 5 11.74 2 .042 hrsemp 2 .951 log(sales) 1 .992 log(employ)
(4.57) (.019) (.370) (.360)
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5 43, R2 5 .310.
Compare this equation to that estimated using only the 29 nonunionized firms in the sample.
(ii) Show that the population model can also be written as
log(scrap) 5 b0 1 b1hrsemp b2log(sales/employ) 1 u3log(employ) 1 u,
where u3 5 b2 1 b3. [Hint: Recall that log(x2/x3) 5 log(x2) 2 log(x3).] Interpret the hypothesis H0: u3 5 0.
(iii) When the equation from part (ii) is estimated, we obtain
log(scrap) 5 11.74 2 .042 hrsemp 2 .951 log(sales/employ) 1 .041 log(employ)
(4.57) (.019) (.370) (.205)
5 43, R2 5 .310.
Controlling for worker training and for the sales-to-employee ratio, do bigger firms have larger statistically significant scrap rates?
(iv) Test the hypothesis that a 1% increase in sales/employ is associated with a 1% drop in the scrap rate.
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Order your essay today and save 15% with the discount code 2023DISCOUNT | 649 | 2,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-18 | latest | en | 0.771864 |
https://math.stackexchange.com/questions/2453773/how-to-find-the-riemannian-distance-on-the-sphere-sn-subset-bbbrn1 | 1,561,366,250,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999298.86/warc/CC-MAIN-20190624084256-20190624110256-00465.warc.gz | 511,594,181 | 38,102 | # How to find the Riemannian Distance on the Sphere $S^n\subset \Bbb{R}^{n+1}$
I am trying to prove that, if $\rho:S^n\times S^n\to \Bbb{R}$ is the distance induced by the Riemannian metric on the sphere $S^n\subset \Bbb{R}^{n+1}$, then $$\rho(p,q)=\arccos \langle p,q\rangle,\,\,\,\forall p,q\in S^n$$ with $\arccos$ defined from $[-1,1]$ to $[0,\pi]$.
If $p=q$, then the formula is trivial. We have then two cases: $p=-q$ and $p\neq- q$. I am having trouble even with the apparently easier case, $p=-q$.
If $p=-q$, then let $v\in S^n$ be any vector orthogonal to $p$. Then $\alpha:[0,\pi]\to S^n$, $\alpha(t)=(\cos t)p+(\sin t)v$ is a well defined differentiable path from $p$ to $q$ such that $\ell_0^\pi(\alpha)=\pi$ ("length of $\alpha$"). This guarantees that $\rho(p,q)\leq \pi (=\arccos \langle p,q\rangle$, in this case).
In order to show that $\rho(p,q)=\pi$, I must consider an arbitrary differentiable by parts path $\beta:[a,b]\to S^n$ from $p$ to $q$ and show that $$\pi\leq \ell_a^b(\beta)=\int_a^b|\beta'(t)|\,dt.$$
I've done some geometric observations and computations, but without success. How can I do this? I wish I could do it without using any facts about geodesics (they appear later in the book I'm studying with).
• I have no time now, but I saw something like that in the proof of Obata's theorem. If you want, you can look this. Later I will try to write something here. – Irddo Oct 2 '17 at 4:10
• I would appreciate your help, @Irddo. – Derso Oct 2 '17 at 7:27
• One idea is to try to show that if $\beta$ is not a great circle segment, then you can always construct $\gamma$ such that $\ell^b_a(\gamma) < \ell^b_a(\beta)$. – Neal Oct 4 '17 at 17:44
• Here's the idea: Any path $l$ from the north pole $n$ to the south pole $s$ has to pass through the equator at some point $e$. Show that the distance from $n$ to $e$ is $\geq \pi/2$ and the distance from $e$ to $s$ is $\geq \pi/2$, and it will follow that the length of the path $l$ from $n$ to $s$ is $\geq \pi$. This reduces your case $p=-q$ to the case $p\neq-q$. – Yly Oct 4 '17 at 18:29
Let $p=(1,0,\dots,0)$ and $q=(-1,0,\dots,0)$. Then any path from $p$ to $q$ is of the form $\gamma(t)=(t,h(t))$, where $h(t)\in \Bbb{R}^n$. Then the length of the path is $$L=\int_{-1}^{1}|\gamma'(t)|dt.$$ But $\gamma'(t)=(1,h'(t))$, and so $|\gamma'(t)|=\sqrt{1+|h'(t)|^2}$.
Now $t^2+\langle h(t),h(t)\rangle=\langle \gamma(t),\gamma(t)\rangle=1$, so $$2t+2\langle h(t),h'(t)\rangle=0.$$ By Cauchy Schwartz we have $|h'(t)|^2|h(t)|^2\ge |\langle h(t),h'(t)\rangle|^2=|t|^2$, and so $$1+|h'(t)|^2\ge \frac{t^2}{|h(t)|^2}+1,$$ but $|h(t)|^2=1-t^2$, hence $$\sqrt{1+|h'(t)|^2}\ge \sqrt{1+\frac{t^2}{1-t^2}}=\sqrt{\frac{1}{1-t^2}},$$ and so $$L=\int_{-1}^{1}\sqrt{1+|h'(t)|^2}dt\ge\int_{-1}^1 \sqrt{\frac{1}{1-t^2}}dt=\pi.$$
A similar reasoning applies for any other point: Assume $p=(1,0,\dots,0)$, then you can assume that the other point $q$ is of the form $(a,s,0,\dots,0)$. The path $\gamma(t)=(t,\sqrt{1-t^2},0,\dots,0)$ has length $$L=\int_{a}^{1}|\gamma'(t)|dt=\int_{a}^1 \sqrt{\frac{1}{1-t^2}}dt =-arccos(t)|_{a}^1=arccos(a)=arccos(\langle p,q\rangle).$$ By a similar argument as above, any other path has the form $(t,h(t))$ with $\sqrt{1+|h'(t)|^2}\ge \sqrt{\frac{1}{1-t^2}}$, so the result follows.
Let $x=(x_0,x_1,\ldots,x_n)\equiv (x_0,\vec{x})\in S^n$ be coordinates on the $n$-sphere and introduce the angular coordinate $\theta=\theta(x)\in[0,\pi]$ by setting $x_0=\cos \theta$. This is a smooth coordinate when $\theta\in(0,\pi)$, i.e. a part from the two poles. When $\theta\in(0,\pi)$ we may also define a complementary set of smooth coordinates using the $(n-1)$-sphere by defining $\vec{z}=(z_1,\ldots,z_n)\in S^{n-1}$ to be the vector that verifies $\vec{x}=\vec{z}\sin \theta$. We may then write: $$(x_0,\vec{x})=(\cos \theta, \vec{z} \sin \theta)$$ Now suppose that $\gamma(t)=(x_0(t),\vec{x}(t))$ is a $C^1$-path in $S^n$ avoiding the poles. Then in our (partially) spherical coordinate system $(\theta,\vec{z})$ we may calculate the derivative: $$\dot\gamma(t) = (-\sin \theta, \vec{z} \cos\theta) \;\dot\theta(t) \; + \; (0,\dot{\vec{z}} \sin\theta)$$ The last two vectors are orthogonal, because $\vec{z}\cdot \vec{z}=1$ implies $\dot{\vec{z}}\cdot \vec{z}=0$. For the length we then get: $$|\dot\gamma|^2= \dot\theta^2 + |\dot{\vec{z}}|^2 \sin^2\theta \geq \dot\theta^2$$ whence the simple lower bound: $$|\dot\gamma | \geq |\dot\theta|$$ with equality iff $\dot{\vec{z}}=0$. Integrating we see that the distance (minimal path length) between $p$ and $q$ satisfies: $d(p,q)\geq |\theta(p)-\theta(q)|$ a bound which by taking limits also holds when $p$ and/or $q$ is one (or both) of the poles. In particular, if $p=(1,0\ldots,0)$ we get: $d(p,q) \geq \theta(q) \in [0,\pi]$ with equality iff $\dot{\vec{z}}=0$, i.e. when $\vec{z}(t) = \vec{z}_0$ is a constant vector along the path. This extremal path runs along a great circle passing through the north pole. So in particular, the distance between the north and the south pole is exactly $\pi$ with extremal path being any great circle passing through the poles. We also have $\cos(\theta(q))= q_0 = p\cdot q$, so $d(p,q)= \arccos (p\cdot q)$ is the distance, first between the north pole and some other point, but because of rotational invariance of the scalar product and distance the formula holds for any two points on the sphere. | 1,933 | 5,393 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-26 | latest | en | 0.867659 |
https://www.physicsforums.com/threads/r-value-insulation-problem.357516/ | 1,527,136,339,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865913.52/warc/CC-MAIN-20180524033910-20180524053910-00005.warc.gz | 844,838,679 | 16,047 | # Homework Help: R-value insulation problem
1. Nov 23, 2009
### G-reg
1. The problem statement, all variables and given/known data
The ceiling of a single-family dwelling in a cold climate should have an R-value of 25. To give such insulation, how thick would a layer have to be if it were made of each of the following materials?
(a) polyurethane foam (thermal conductivity k = 0.024 W/m · K)
2. Relevant equations
R = L/k
L = length(thickness)
k = thermal conductivity
3. The attempt at a solution
Rk = L
25(.024) = .6m
My homework is telling me that I'm wrong and I have no idea why. This seems like a pretty simple and straight forward question. Can anyone help point out where I'm going wrong? Please!
2. Nov 23, 2009
### diazona
Re: R-values
For starters, something must be wrong with that equation because the units don't work out. If R is a dimensionless number, then L and k would have the same units. But they clearly don't. So what's missing? Either there's some part of the formula you haven't written, or one of those numbers (R, L, or k, and I'm guessing R) has different units than you've stated.
3. Nov 23, 2009
### Delphi51
Re: R-values
Could it be a confusion of units? Apparently the USA has an R value in ft²·°F·h/Btu rather than the SI units used in the rest of the world. Nice article on it at http://en.wikipedia.org/wiki/R-value_(insulation)
4. Nov 29, 2009
### G-reg
Re: R-values
Ah yes the units is the problem. However, now I'm having trouble converting it..here is what I did and it didn't produce the correct answer:
.024[W/(m)(K)] x [(.3048m)^2]/1ft^2 x [(1.8K)/1 F] x (1 Btu/1055 J) x [(3600s)/1hr]
where F = fahrenheit
5. Nov 30, 2009
### Delphi51
Re: R-values
According to the wikipedia article
Can you use this to convert the given R = 25? Is that 25 in metric or US units?
6. Nov 30, 2009
### ideasrule
Re: R-values
Why (0.3048m)^2/1ft^2 and not 0.3048m/1ft?
7. Nov 30, 2009
### G-reg
Re: R-values
because R is given as 25 [(ft^2 * F * hr)/ (Btu)]
8. Nov 30, 2009
### G-reg
Re: R-values
and yes, I think that I can you use that conversion. Thanks! | 641 | 2,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-22 | latest | en | 0.924108 |
https://www.futurestarr.com/public/blog/author/how-many-inches-are-in-a-meteror | 1,660,193,391,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571234.82/warc/CC-MAIN-20220811042804-20220811072804-00620.warc.gz | 686,515,710 | 23,112 | FutureStarr
How Many Inches Are in a Meteror
# How Many Inches Are in a Meter
A meter is defined as 1/10th of a meter or one meter in length. 1 meter = 3. 281 feet, which means that 1 meter is 3. 280958. . . feet, or 0. 3281 of a foot.
### Inch
A meter (abbreviated as “m”) is a metric system base length unit and can be defined as the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second. An inch (abbreviated as “in”) is a unit of length used mainly in the imperial and U.S. customary systems. Other related units of measurement include centimeter, kilometer and foot (plural is feet). One meter is equal to 100 centimeters, and 100 meters make 1 kilometer.
An inch is a unit of length of measurement in the Imperial and United States customary systems. It is usually indicated by a double prime (“) or abbreviated as “in”. The sequential formula of converting inches to meter is shown in the meter to inches conversion table. It might be difficult to compute complex values using the conversion table. The conversion calculator can also be used in converting the values from meters to inches. (Source: www.calculatorology.com)
### Meter
This calculator is programed with three active controls that perform different functions of the calculator. It has a single text field where you fill in the value to be converted from meters to inches. Click the “Convert” button to perform the conversion and the values in inches will be displayed below the controls. The method used in carrying out the calculation is also shown in the bottom panel of the calculator.
Meter is the SI unit of length. Length can be expressed in other units like, centimeters, inches, feet, and yards which can be converted to meters. Though meters and inches are units of length, but their values differ. In this article we will learn the relation between meter, centimeter, millimeter, inches, feet and yards. Stay tuned to learn more about how long is a meter! (Source: www.cuemath.com)
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August 11, 2022 | sitara sehar | 776 | 3,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-33 | latest | en | 0.922056 |
https://chem.libretexts.org/Courses/Sacramento_City_College/SCC%3A_CHEM_300_-_Beginning_Chemistry/SCC%3A_CHEM_300_-_Beginning_Chemistry_(Faculty)/08%3A_Quantities_in_Chemical_Reactions/8.06%3A_Limiting_Reactant%2C_Theoretical_Yield%2C_and_Percent_Yield_from_Initial_Masses_of_Reactants | 1,723,676,897,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641137585.92/warc/CC-MAIN-20240814221537-20240815011537-00527.warc.gz | 137,096,460 | 33,558 | # 8.6: Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Masses of Reactants
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##### Learning Objectives
• Calculate percentage or actual yields from known amounts of reactants.
The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Studying how much of a compound is produced in any given reaction is an important part of cost control.
## Percent Yield
Chemical reactions in the real world don't always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than predicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.
To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that can be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
$\text{Percent Yield} = \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \nonumber$
Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense.
Typically, percent yields are understandably less than $$100\%$$ because of the reasons indicated earlier. However, percent yields greater than $$100\%$$ are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction. Example $$\PageIndex{1}$$ illustrates the steps for determining percent yield.
##### Example $$\PageIndex{1}$$: Decomposition of Potassium Chlorate
Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction below:
$2 \ce{KClO_3} \left( s \right) \rightarrow 2 \ce{KCl} \left( s \right) + 3 \ce{O_2} \left( g \right)\nonumber$
In a certain experiment, $$40.0 \: \text{g} \: \ce{KClO_3}$$ is heated until it completely decomposes. The experiment is performed and the oxygen gas is collected and its mass is found to be $$14.9 \: \text{g}$$.
1. What is the theoretical yield of oxygen gas?
2. What is the percent yield for the reaction?
###### Solution
a. Calculation of theoretical yield
First, we will calculate the theoretical yield based on the stoichiometry.
###### Step 1: Identify the "given" information and what the problem is asking you to "find".
Given: Mass of $$\ce{KClO_3} = 40.0 \: \text{g}$$
Mass of O2 collected = 14.9g
Find: Theoretical yield, g O2
###### Step 2: List other known quantities and plan the problem.
1 mol KClO3 = 122.55 g/mol
1 mol O2 = 32.00 g/mol
###### Step 4: Solve.
$40.0 \: \cancel{\text{g} \: \ce{KClO_3}} \times \dfrac{1 \: \cancel{\text{mol} \: \ce{KClO_3}}}{122.55 \: \cancel{\text{g} \: \ce{KClO_3}}} \times \dfrac{3 \: \cancel{\text{mol} \: \ce{O_2}}}{2 \: \cancel{\text{mol} \: \ce{KClO_3}}} \times \dfrac{32.00 \: \text{g} \: \ce{O_2}}{1 \: \cancel{\text{mol} \: \ce{O_2}}} = 15.7 \: \text{g} \: \ce{O_2}\nonumber$
The theoretical yield of $$\ce{O_2}$$ is $$15.7 \: \text{g}$$, 15.67 g unrounded.
The mass of oxygen gas must be less than the $$40.0 \: \text{g}$$ of potassium chlorate that was decomposed.
b. Calculation of percent yield
Now we will use the actual yield and the theoretical yield to calculate the percent yield.
###### Step 1: Identify the "given" information and what the problem is asking you to "find".
Given: Theoretical yield =15.67 g, use the un-rounded number for the calculation.
Actual yield = 14.9g
Find: Percent yield, % Yield
###### Step 3: Use the percent yield equation below.
$$\text{Percent Yield} = \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$$
###### Step 4: Solve.
$$\text{Percent Yield} = \dfrac{14.9 \: \text{g}}{15.\underline{6}7 \: \text{g}} \times 100\% = 94.9\%$$
Since the actual yield is slightly less than the theoretical yield, the percent yield is just under $$100\%$$.
##### Example $$\PageIndex{2}$$: Oxidation of Zinc
Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:
$\ce{CuSO4}(aq)+\ce{Zn}(s)\rightarrow \ce{Cu}(s)+\ce{ZnSO4}(aq) \nonumber$
What is the percent yield?
###### Solution
Solutions to Example 8.6.2
Steps for Problem Solving-The Product Method Example $$\PageIndex{1}$$
Identify the "given" information and what the problem is asking you to "find."
Given: 1.274 g CuSO4
Actual yield = 0.392 g Cu
Find: Percent yield
List other known quantities.
1 mol CuSO4= 159.62 g/mol
1 mol Cu = 63.55 g/mol
Since the amount of product in grams is not required, only the molar mass of the reactants is needed.
Balance the equation.
The chemical equation is already balanced.
The balanced equation provides the relationship of 1 mol CuSO4 to 1 mol Zn to 1 mol Cu to 1 mol ZnSO4.
Prepare a concept map and use the proper conversion factor.
The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield (g Cu) is found by performing mass-mass calculation based on the initial amount of CuSO4.
Cancel units and calculate.
$\mathrm{1.274\:\cancel{g\:Cu_SO_4}\times \dfrac{1\:\cancel{mol\:CuSO_4}}{159.62\:\cancel{g\:CuSO_4}}\times \dfrac{1\:\cancel{mol\: Cu}}{1\:\cancel{mol\:CuSO_4}}\times \dfrac{63.55\:g\: Cu}{1\:\cancel{mol\: Cu}}=0.5072\: g\: Cu}\nonumber$
Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be:
\begin{align*} \text{percent yield} &= \left( \dfrac{\text{actual yield} }{\text{theoretical yield}} \right) \times 100 \\[4pt] &= \left(\dfrac{0.392\, g\, \ce{Cu}}{0.5072 \, g\, \ce{Cu}} \right) \times 100 \\[4pt] &=77.3\% \end{align*}
Think about your result. Since the actual yield is slightly less than the theoretical yield, the percent yield is just under $$100\%$$.
##### Exercise $$\PageIndex{1}$$
What is the percent yield of a reaction that produces 12.5 g of the Freon CF2Cl2 from 32.9 g of CCl4 and excess HF?
$\ce{CCl4 + 2HF \rightarrow CF2Cl2 + 2HCl} \nonumber$ | 3,678 | 11,268 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-33 | latest | en | 0.217014 |
https://plainmath.org/algebra-i/24411-use-what-you-know-about-bonded-monotonic-sequences-to-show | 1,713,549,291,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00780.warc.gz | 401,301,302 | 24,560 | Lennie Carroll
2021-08-18
Use what you know about bonded, monotonic sequences to show that the following sequences converge
${a}_{n}=\frac{1}{3}\left(1-\frac{1}{{3}^{n}}\right)$
sweererlirumeX
For the given sequence $\left\{{a}_{n}\right\}$
${a}_{n+1}-{a}_{n}=\frac{1}{3}\left(1-\frac{1}{{3}^{n+1}}\right)-\frac{1}{3}\left(1-\frac{1}{{3}^{n}}\right)$
$=\frac{1}{{3}^{n}}\frac{1}{{3}^{n+1}}$
$=\frac{1}{{3}^{n}}\left(1-\frac{1}{3}\right)$
$=\frac{2}{3}\frac{1}{{3}^{n}}>0$
Since ${a}_{n}+1>{a}_{n}$, therefore, the sequence $\left\{{a}_{n}\right\}$ is monotonic increasing.
For all $n\ge 1$
$\left(1-\frac{1}{{3}^{n}}\right)<1$
$\frac{1}{3}\left(1-\frac{1}{{3}^{n}}\right)<\frac{1}{3}$
${a}_{n}<\frac{1}{3}$ So, the sequence is bounded above with the upper bound $\frac{1}{3}$
Thus, being a monotonic increasing sequence bounded above, $\frac{1}{3}$ being an upper bound, the given sequence $\left\{{a}_{n}\right\}$ is convergent
It completes the proof.
Do you have a similar question? | 397 | 989 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 32, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-18 | latest | en | 0.58128 |
http://www.realclimate.org/index.php/archives/2008/06/wired-magazines-incoherent-truths/comment-page-2/?wpmp_switcher=mobile&attest=true&wpmp_tp=1 | 1,529,398,163,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861981.50/warc/CC-MAIN-20180619080121-20180619100121-00008.warc.gz | 490,622,142 | 46,400 | ### Wired Magazine’s Incoherent Truths
Filed under: — raypierre @ 15 June 2008
Many of our tech-savvy friends — the kind of folks who nurse along the beowulf clusters our climate models run on — are scratching their heads over some cheeky shrieking that recently appeared in a WIRED magazine article on Rethinking What it Means to be Green . Crank up the A/C! Kill the Spotted Owl! Keep the SUV! What’s all that supposed to be about?
Let’s take air conditioning for starters. Basically WIRED took a look at the carbon footprint of New England heating vs. Arizona cooling and jumped to the conclusion that air conditioning was intrinsically more efficient than heating. To see where they were led astray let’s consider a house sitting where you need to cool it by 20 degrees to be comfortable. The heat leaks into the house at a rate that is approximately proportional to this temperature difference, and the heat leaking in needs to be removed. Now, in order to move that heat from inside to outside, energy has to be expended. Given a fixed electric power usage (in watts), a better air conditioner can remove more heat per day than a worse one, but every air conditioner needs to expend some energy to move the heat. That’s just thermodynamics.
Efficiency of air conditioners is measured by a SEER rating, which is the ratio of heat moved to the outside (in BTU/hr) to the electric power consumption (in Watts). A typical modern air conditioner has a SEER rating of 10, We can convert this into nicer units by converting BTU/hr into Watts, which means dividing the SEER rating by 3.413, which then gives us a Coefficient of Performance, in units of Watts of heat moved per Watt of electricity used. For the aforementioned efficiency, we move heat at a rate of 2.92 Watts if we expend 1 Watt of electric energy. An air conditioner is just a heat engine run in reverse: instead of making use of a temperature differential to use heat flow from hot to cold to do work, we expend mechanical work in order to move heat from a colder place to a hotter place. Thus, an efficient heat engine is an inefficient air conditioner. That’s basically why the Coefficient of Performance gets smaller when the temperature difference between indoors and outdoors is greater — with bigger temperature difference heat engine cycles tend to get more efficient, which means that air conditioner cycles tend to get less efficient. That’s also where the “S” in SEER comes from. It stands for “Seasonal,” and reflects the fact that efficiency must be averaged over the range of actual temperature differentials experienced in a “typical” climate. Your mileage may vary.
This situation can be contrasted with heating. If that same house were in an environment that were too cold instead of too warm, so that it had to be kept 20 degrees warmer than the environment, then the amount of heat leaking out of the house each day would be about the same as the amount leaking into the house in the previous case. That heat loss needs to be replaced by burning fuel. Now, generating heat is the only thing that can be done with 100% efficiency. Old furnaces lose a lot of heat up the chimney, but modern sealed-combustion burners– the kind that can use PVC pipes instead of a chimney — lose virtually nothing. With a heat exchanger between the air intake and the exhaust, they could closely approach the ideal. But still, in this case we are generating heat rather than just moving it, so it takes 1 watt of heat power from fuel burning to make up 1 watt of heat loss. That would seem to make heating a factor of 2.92 less efficient than air conditioning.
But wait, the story doesn’t stop there. First, there’s the fact that air conditioning almost invariably runs off of electricity, and the increased electricity demand is a big source of the pressure to build more coal-fired power plants. A house can be heated by burning natural gas, and right there air conditioning becomes 1.8 times worse than heating, because natural gas emits only 55% of the carbon of coal, per unit of heat energy produced. And it gets even worse: Coal fired power plants are only 30% efficient at converting heat into electricity, on average, so there you get another factor of 3.3 in carbon emissions per unit of energy transferred between the house and its environment. Finally, figure in a typical electric line transmission loss of 7% and you get another factor 1.075. Put it all together with the energy efficiency of the air conditioner itself and air conditioning comes in at a whopping 2.19 times less efficient than heating. for a given amount of temperature difference between house and environment. That means that so far as carbon emissions go, heating a house to 70 degrees when the outside temperature is 40 degrees is like cooling the same house to 70 degrees when the outside temperature is 83.7 degrees.
And that’s still not the end of the story. A house in need of air conditioning has other heat inputs besides the heat leaking in from outside, and all that extra heat needs to be gotten rid of as well. For example, heat is a waste-product of all energy use going on in the house. Four people produce 400W that needs to be gotten rid of, and then there’s the heat from hot water, lighting, the TV, cooking and what have you — all the energy usage within the house, plus 100W of biological heat per person needs to be gotten rid of. On top of that, you’ve got direct radiative heating from the sun, both from the sunllight getting through windows and solar heating of the exterior surfaces of the house, some of which will leak in through the insulation. Energy must be expended to remove all this heat. In contrast, in the heating season waste heat is subtracted from the energy needed for home heating.
So, WIRED got the story egregiously wrong, and not just because they did the arithmetic wrong. In their rush to be cute, they didn’t even make a half-baked attempt to do the arithmetic. But what if they had been right and air conditioning really were intrinsically more efficient than heating. Would that justify their conclusion that you can just "crank up the A/C?" without worry? No, of course not, because cranking up the A/C would still use additional energy and still lead to the emission of additional carbon. For the conclusion to be justified, it wouldn’t be enough for A/C to be more efficient than heating; it would have to be so much more efficient that the incremental energy usage from cranking it up were trivial. WIRED didn’t even try to make that case. If they had, they might have spotted their errors.
Is there any real conclusion that could have been drawn from more clear thinking about the heating vs. air conditioning issues danced around in the article? Yes, in fact. The conclusion is that it makes a lot of sense to build houses in places where the environment requires neither much heating nor much cooling. This is in fact why Los Angeles scores pretty well in carbon footprint per capita, despite all the driving (as noted recently in The Economist.). Another conclusion to be drawn from the carbon footprint of New England heating is that there are probably a lot of leaky homes up there heated by inefficient oil-fired furnaces. Fixing that situation represents a huge untapped virtual energy source.
What’s more, for a magazine that purports to be written by and for tech geeks, WIRED missed the biggest and most interesting part of the story: the same intrinsic efficiences of heat pumps can be run in reverse to give you the same economies for home heating as you get for air conditioning. To do this effectively, you’d have to run the heat pump off of natural gas rather than electricity (or perhaps run it off of locally generated solar power or wind). You’d also have to deal with the fact that heat pumps become less efficient when working across large temperature gradients, but that’s where geothermal heat storage systems come in, making use of the fact that the deep subsurface temperature remains near a nice 55F all year around. Now that would have been a nice story for a tech magazine to cover. And by the way, the decrease in efficiency of heat pumps as the temperature differential increases has another implication that WIRED missed: not only does global warming increase the basic demand for air conditioning, with all the attendant pressures on electricity demand, but it exacerbates the situation by decreasing the efficiency of the entire installed base of air conditioners.
Now about that spotted owl. This refers to a claim that industrial tree plantations take up carbon faster than old growth forests; Since spotted owls require the large trees found only in old-growth, the supposed implication is that if we want to soak up carbon we ought to damn the spotted owl and cut down all the old growth. WIRED really committed serial stupidities on this one. First of all, the article they cited in support of their claim was about carbon emissions from Canada’s managed forests, not from old growth. Now, it’s true that a rapidly growing young tree takes carbon out of the atmosphere more rapidly than a mature forest which more slowly transfers carbon to long term storage in soil. However, to figure out how much net carbon sequestration you get out of that young tree once it’s chopped down, you need to figure what happens to it. Lots of trees wind up in paper, carboard boxes, shipping palettes and other things that rapidly sit around decomposing or get burned off (or worse, turn into methane in landfills). Even the part that turns into houses has a relatively short residence time before being oxidized. Anybody who has maintained an old Victorian house knows about the constant battle against rot, and the amount of wood that needs to be replaced even if (knock wood) the thing doesn’t burn down or turn into a tear-down. So, WIRED is totally off the mark there, unless, to use the colorful language of my colleague Dave Archer, they can get trees to "drop diamonds instead of leaves."
Worse, they ignore the abundant literature indicating that old growth forests can be a net sink of carbon even in equilibrium, whereas the soil disturbance of clear cutting and industrial forestry can lead to large soil carbon releases. A classic article in the genre is "Effects on carbon storage of conversion of old-growth forests to young forests" (Harmon et al. Science 1990) . They state "Simulations of carbon storage suggest that conversion of old-growth forests to young fast-growing forests will not decrease atmospheric carbon dioxide (CO2) in general, as has been suggested recently.". For more recent work, take a look at what Leighty et al. (ECOSYSTEMS Volume: 9 Issue: 7 Pages: 1051-1065. 2006 ) have to say about the Tongass:.
• "The Tongass National Forest (Tongass) is the largest national forest and largest area of old-growth forest in the United States. Spatial geographic information system data for the Tongass were combined with forest inventory data to estimate and map total carbon stock in the Tongass; the result was 2.8 +/- 0.5 Pg C, or 8% of the total carbon in the forests of the conterminous USA and 0.25% of the carbon in global forest vegetation and soils. Cumulative net carbon loss from the Tongass due to management of the forest for the period 1900-95 was estimated at 6.4-17.2 Tg C. Using our spatially explicit data for carbon stock and net flux, we modeled the potential effect of five management regimes on future net carbon flux. Estimates of net carbon flux were sensitive to projections of the rate of carbon accumulation in second-growth forests and to the amount of carbon left in standing biomass after harvest. Projections of net carbon flux in the Tongass range from 0.33 Tg C annual sequestration to 2.3 Tg C annual emission for the period 1995-2095. For the period 1995-2195, net flux estimates range from 0.19 Tg C annual sequestration to 1.6 Tg C annual emission. If all timber harvesting in the Tongass were halted from 1995 to 2095, the economic value of the net carbon sequestered during the 100-year hiatus, assuming \$20/Mg C, would be \$4 to \$7 million/y (1995 US dollars). If a prohibition on logging were extended to 2195, the annual economic value of the carbon sequestered would be largely unaffected (\$3 to \$6 million/y). The potential annual economic value of carbon sequestration with management maximizing carbon storage in the Tongass is comparable to revenue from annual timber sales historically authorized for the forest."
So, it looks like that old Spotted Owl and its kindred old-growth denizens are in fact sitting not just on a nest, but on a treasure trove of carbon credits worth potentially more than the timber harvest.
And should you keep that SUV? This blurb in fact contains some useful advice, buried amidst some fuzzy reasoning and published over a witless tag line stating that "pound for pound" a Prius takes more energy to manufacture than a Hummer. The apparent implication of that tag line is rebutted in the article itself, but why give the reader that as a 32-point type take-home point when the WIRED editors don’t even themselves believe it’s an important statistic? This factoid refers to the energy used in the nickel component of Prius batteries, but it’s irrelevant because "pound for pound" doesn’t count if your point is moving 4 people from point A to point B. What transport value do you get from transporting four people plus the weight of the Hummer? Now, the rest of the fuzziness in the logic is a bit more subtle. The author notes quite rightly that there is a very significant carbon emission from manufacturing a car, which is indeed more for a Prius (at least for the moment) than it is for comparable sized non-hybrids.. Thus, if you are faced with ditching your existing car (whatever it may be) and buying a Prius, you need to consider how much you drive per year and see how long it takes to "pay back" the carbon emission from manufacturing the Prius. So far so good. But this is more a statement about the transition to more efficient cars, and how to deal with mistakes of the past, rather than a statement about what is intrinsically desirable in the fleet. As far as carbon emissions go, we’d still be better off if everybody who needed a car were in a Prius, except maybe for people who drive very little per year — who should then be into shared hybrids via iGO or ZipCars, Maybe if you drive very little and live out in a rural area where there are not going to be any shared cars, getting a compact non-Hybrid might make sense. There must be at least a dozen or two people out there in that category, I guess.
The rest of the advice WIRED gives makes even less sense. They say that if you want to be green, you ought to buy a used Civic or something like that, not a Prius. That’s because the used car already has the manufacturing carbon emissions "written down" (or, I guess at least the carbon guilt accrues to the original owner, not that the atmospheric radiative forcing is going to care much about that). However, this advice, sensible-sounding though it is — ignores the fact that to make that used car available to you, the original owner almost certainly had to buy something else, and probably that was a new car, or at least a newer one. So, for the scheme to work, you’d have to buy your used Civic from somebody who was giving up driving altogether. I no longer own a car myself, but I’m sorry I wasn’t able to participate in a scheme like this; by the time I gave up our remaining car ten years ago, it was suitable only for the crusher, and in fact had to be towed there.
The real implication is that manufacturing costs count, so most people should buy a small, efficient hybrid and keep it until it runs into the ground. The implication is also that durability of cars counts for nearly as much as gas mileage, since an efficient car that needs to be replaced every five years isn’t really all that efficient.
Along with all the nonsense is a certain amount of true (if by now commonplace) advice. Among this is the basic truth that urban living is inherently green, and if more people lived in cities (and if more cities were kept livable so people would want to move there). then per capita carbon emissions would go down. Even there, the Economist managed to be both more informative and more iconoclastic with its surprising analysis of the pattern of urbanism in Los Angeles. The other truism in WIRED is that nuclear power deserves a second look, and probably has an important role to play in a decarbonized energy future. Still, if you compare the cost of making all those chilly New England homes efficient with the total true cost of building more nuclear plants, well, let’s just say I’m buying stock in argon-filled low-e window manufacturers rather than Areva, much as I like their track record on nuclear electricity.
### 367 Responses to “Wired Magazine’s Incoherent Truths”
1. 51
Re #44 Tim Joslin, Raypierre, it’s important to add that it is ecologically healthy for forests to have 1% or 2% blowdown every year, creating sunny clearings where rarer organisms conduct their intermittent life cycles — including some beautiful woodland flowers which it takes you luck to find. Further, the rotting wood is an absolutely necessary part of the life cycle and the food chain of a huge number of soil microbes, insects and insectivores. Moreover, occasional fires may be necessary to release minerals on decadal-or-longer cycles, and to coincide with the population-density requirements of various species. The optimal frequency for fire differs with the type of forest.
Picking on mature forests, not to mention old growth, for the rotting wood as a CO2 problem, is a bit like advocating the bulldozing of the world’s great cathedrals and mosques because this would provide sites for nice power plants.
Wired’s larger point, as I take it, is that environmentalists must accept compromises.
Certainly not, if the arguments are this silly.
2. 52
DavidONE says:
OT: what happened to http://www.realclimate.org/index.php/archives/2007/10/oregon-institute-of-science-and-malarkey/ – I’m getting a 404.
TIA.
[Response: works for me…. – gavin]
3. 53
Charles Jackson says:
Thank you for helping to refute Wired’s assertion that logging older forests and replacing them with tree farms can help fight global warming. I would just like to supplement Raypierre’s piece with some additional information.
1) Wired says “A tree absorbs roughly 1,500 pounds of CO2 in its first 55 years. After that, its growth slows, and it takes in less carbon.” The idea that all trees suddenly slow in growth after age 55 is ridiculous. There are myriad variables that control tree growth including species, climate, soil conditions, land-use history, etc. Many trees, including the dominant tree species in the Pacific Northwest where the northern spotted owl lives–Douglas Fir–do not slow in growth until they are considerable older than 55. (See Curtis 1994 USDA http://www.treesearch.fs.fed.us/pubs/20584)
2) Wired goes on to say that “Left untouched, [a tree] ultimately rots or burns and all that CO2 gets released.” This statement shows a complete lack of understanding of forest carbon dynamics. When a tree decomposes, some of the organic matter is incorporated into soil. This is why older forests contain significantly more carbon in their soils than younger forests do. (See Zhou et al 2006 Science http://www.sciencemag.org/cgi/content/abstract/314/5804/1417)
3)The only pool of carbon mentioned in by Wired is that of trees. But forest carbon sequestration is much more complicated than the growth of individual trees—live tree biomass is but one carbon pool in a forest. Other carbon pools include dead organic matter and soil, which together usually comprise the majority of a forest’s carbon. The amount of carbon a forest stores in live biomass, soils and dead organic matter all increase with forest age in temperate forests, like those in the United States, and tropical forests. In boreal forests, like those in Canada and other cold climates, older forests continue to sequester massive amounts of carbon in their soils as they age (See Pregitzer and Euskirchen 2004 Glob.Ch Bio. http://www.ingentaconnect.com/content/bsc/gcb/2004/00000010/00000012/art00010).
4) As Raypierre pointed out, the Canadian study mentioned in the Wired article is for managed Canadian forests only. The forests of Canada are boreal, which have different carbon sequestration cycles than forests in the United States, which are temperate forests. Therefore, its findings are not particularly relevant to forest management in the United States. Second, it is ironic that Wired would cite this study because the main reason why Canada’s managed forests may be a negligible carbon sink in the near future is because of increased risk of fire and pest outbreaks. These risks could very well increase if older forests in the United States were converted to tree farms. Tree farms tend to be densely planted and have little genetic diversity, which increases the chance of fire and pest outbreaks for a variety of reasons including a) increased risk of drought stress. Drought stress increases the chance that a forest will burn, the intensity of the burn and increases the likelihood that pests will overrun the forest. b) Older forests contain more diversity in plant an animal, and often contain many insectivorous species such as spiders that control pest outbreaks. c) Older forests have trees of different ages, heights, genetic diversity and spacing, which helps prevent crown fires from spreading. (See for ex. Odion et al. 2004 Cons. Bio. http://www.blackwell-synergy.com/doi/abs/10.1111/j.1523-1739.2004.00493.x?journalCode=cbi and Coyle et al 2005 Ann. Rev. Ent. http://www.treesearch.fs.fed.us/pubs/8380)
Perhaps the worst aspect of the Wired article is that it insinuates that there is a scientific basis for its conclusions by linking to scientific studies. None of the studies Wired links to or mentions support the idea that logging old-growth forests could fight climate change. For example, the study linked by the words “Clear the oldest trees” does not mention older forests, let alone advocate logging them. The studies that have addressed the issue (none of which are linked to by Wired) all conclude that logging older forests would be a poor strategy to fight global warming, even when wood products and replanting are taken into account. Furthermore, Wired does not mention that many recent scientific studies that utilize new technology allowing the measurement of forest carbon fluxes. These studies show that older forests tend to be carbon sinks and that the only forests in the United States that are consistently net sources of carbon to the atmosphere are young forests regenerating after a major disturbance like a clearcut or stand-replacing fire. Here is just a sample of the studies showing that older forests in the United States are usually carbon sinks and that young forests can be sources of carbon to the atmosphere for decades.
Valentini, R., et al. 2000. Respiration as the main determinant of carbon balance in European forests. Nature 404, 861–865.
Law, B.E. et al. 2002 Environmental controls over carbon dioxide and water vapor exchange of terrestrial vegetation. Agricultural and Forest Meteorology. 113:97-120.
Law, B.E., O.J. Sun, J. Campbell, S. Van Tuyl, and P.E. Thornton. 2003. Changes in carbon storage and fluxes in a chronosequence of ponderosa pine. Global Change Biology 9:510-524.
4. 54
Leonard Evens says:
So let me see. We own a Toyota Corolla, which is almost 8 years old. It gets 30-35 mpg on the road and perhaps 25 mgp in the kind of city driving we do. Because of age, arthritis, and other issues, we use the car in circumstances where previously we might have used public transportation, but we still don’t drive all that much. I do use I-Go, but it is often difficult to find a car within easy walking distance—somewhat shorter than it once once was before spinal steonosis set in—and even then I may end up with a Honda Element. We shop using Peapod. I don’t know if that saves energy or not—one delivery van dropping groceries off at many locations as opposed to many individual trips—but it sure saves wear and tear on arthritic bodies. Other food shopping is done by walking to local markets, and other shopping is done wherever possible in the local area. I do bicyle for exercise, but I would need a more practical bike to use for shopping, something I haven’t got yet because of our condo’s rules about storage.
Given all that, should we replace our Corolla with a Prius? It only has about 50,000 miles on it, so it can go quite a bit longer. So far, I’ve reasoned that the increase in gas mileage would not be large enough to compensate for the cost. If your analysis is correct, we should first drive the Corolla into the ground, until it ends up being only fit for junk, or until neither or us is able to drive.
[Response: It’s easy to compute the impact of a decision like this on personal carbon footprint, but the more relevant decision factor is impact of the decision on the long-term carbon emissions of the vehicle fleet as a whole. The complexity there is that the result depends a lot on what other people do. For example, if everybody decides to buy a Prius, then there will be so many used cars dumped on the market that a lot would be junked, and there would be wastage in that. On the other hand, if your used Corolla goes to somebody who was driving an inefficient old Buick Electra that then gets junked, your decision will ultimately lead to reduction in fleet emissions, as long as your Prius stays on the road long enough (in your hands or others who get it as a used car) to pay back the carbon cost of manufacturing. It’s quite possible that buying a new Prius would reduce the net long term fleet emissions, but then again you’d have to weigh the personal money cost of doing this against other things you could spend your money on that would potentially reduce carbon emissions more — e.g. donating to some local church to help them pay for a replacement air conditioner. –raypierre]
5. 55
Hank Roberts says:
Nuts to Wired. They could have pointed out something useful, instead, like replacing eroding plowed fields of corn and soy with perennial woody agriculture. An old friend wrote this long ago. I’ve never seen it referenced. It should be:
6. 56
Dan Hughes says:
re: # 49
“What does a nuclear pile tend uncontrollably to do?”
It depends significantly on the specific ‘nuclear’ materials (and non-nuclear materials if present) that comprise the ‘pile’ and the configuration in which these are arranged.
7. 57
Lawrence Brown says:
A factor that could come into play by the use of ACs and not in the use of space heating is a possible slight(or maybe not so slight?) feedback effect.
As the average global temperature increases it’s reasonable to expect greater use of air conditioners, and if fossil fuels are used as an energy source,then this will lead to more global warming and more use of ACs and so on. I guess “Wired” didn’t take this into consideration, either.
8. 58
Joe S. says:
Wired magazine brings up an interesting point that it requires more energy to build a Prius than a Hummer, this raises further questions:
1. Over the useful lifetime of the car, how much more energy will a Hummer expend per miles driven versus a Prius? At what point does driving a Hummer use more energy than building a Prius?
2. Given that the Prius batteries must during the lifetime of the car be replaced:
http://townhall-talk.edmunds.com/direct/view/.ef2b471/0
how much energy is used in the recycling process?
[Response: It doesn’t take more energy to build a Prius than a Hummer. It only takes more energy per pound, which is a very different thing. The links that other commenters have given will take you to a full lifecycle analysis of hybrids. Assuming both the Prius and the Hummer are driven to the end of their natural lifetime, the Hummer is a dead loser. The more relevant question is lifecycle carbon emission of a hybrid vs. an equivalent-sized efficient conventional gasoline car, or perhaps a diesel. –raypierre]
9. 59
SecularAnimist says:
Leonard Evens wrote: “… I do bicycle for exercise, but I would need a more practical bike to use for shopping …”
10. 60
Figen Mekik says:
Maybe this is the European in me but instead of trying to figure out the most carbon footprint-friendly car, why don’t we all car pool more and use public transportation more often? It’s certainly cheaper. And maybe makes the case for a more urban lifestyle.
11. 61
mark says:
“What do you do when you get to the point where you have to take off the tee-shirt and shorts and it’s still too hot? Wasn’t there a Shel Silverstein poem on that theme? Anybody remember it? –raypierre”
Wasn’t it Bob Dylan?
“The answer, my friend, is blowing in the wind”..?
;-)
12. 62
Sili says:
Thank you for explaining today’s xkcd to me.
Unfortunately we still use mainly coal here in Denmark. We do have a slightly better overall efficiency (I believe) by piping the cooling water from the plant around the city for heating. Though … that means heating is so cheap that in many homes it isn’t profitable to improve insulation. (Not that I’ve noticed it – my bill seems be going up and up – I bet it’s my neighbours running around in the nude all year. Damn communal billing.)
13. 63
Martin Ringo says:
While the calculations at “Wired” are naïve, the calculations here are, shall we say, semi-naïve. Let me just do some quick numbers for residential use.
The A/C load is roughly 13% of the spacing heating load for the country. Convert that to energy at the generator and it is still bit less than 50%. Now just looking at the average rates of the A/C load per population-weight-cooling-degree-day versus the space heating load per population-weight-heating-degree-day, the A/C is about 35%. There are more CDDs, population weighted, than HDDs. (This should be done on an energy, HVAC-consumption weighted basis, but I have seen those numbers and I am not going to calculate them for a blog comment.) Then roughly a one degree warming adds 180 to the CDD total for the year and removes 180 from the HDD total. This would give us a serious energy savings… Except that the demand for heating energy is roughly linear in temperature while that for cooling energy has a positive second derivative. Now for the Northeast (NEPOOL: where I have a good regional load and temperature dataset) this is roughly a factor of 2 at 25 CDD (90 degrees F). But over all the cooling days for the 25 year period, the factor is merely 1.02. Thus over the year the quadratic effect isn’t going to contribute much.
Finally, with regard to coal versus natural gas/fuel oil/LPG, coal burning releases about 60-75% more CO2 per BTu than natural gas, a bit less for LPG, and about 25% for oil. And with the increased natural gas prices and proclivities of utilities planners, coal-fired units are the choice for new plants. However, the incremental cooling load will make the load shape more peaked, and for increases in the pure summer peak, simple-cycle gas-turbines are the new plant of choice. Thus, while it is true that an increase in temperature with the concomitant increase in cooling load will lead to more coal plants, regulations allowing, it is also true that well less than 50% of that incremental cooling load will be supplied by coal fired units. Without a nation-wide production cost simulation of the load increase, it is hard to refine that number much.
Putting that altogether, a uniform increase in the temperature of one degree F (equal increases in CDDs as decreases in HDDs) is probably going to cause a small decrease in the carbon emissions for US residential consumers.
Now that is only the residential sector. The industrial sector HVAC is probably similar, but it is not that important because the HVAC (but not the process heat) part of the industrial energy load is much smaller. However, the commercial sector is about the same size, and the publicly available data is a bit ambiguous. Thus, there could be a bigger (in absolute) size effect there in the opposite direction. But whatever the effect is, if you want to calculate it you need for drop the engineering calculations and look at the actual demands for energy for A/C or space heating.
14. 64
In comment 49 Lee A. Arnold included,
Do I have the following math right?
… 1 cube Portland concrete, 120.85 meters on a side
divide by 4, to make it manageable = 4 cubes of concrete, 76 meters on a side
Four concrete cubes, each (76 m)^3 … yes, the size seems right. Most who are reading this probably have seen 20-storey apartment buildings about that tall, but not extended that far along two horizontal axes, and not solid.
Large pieces of concrete don’t hold together all by themselves; they need steel tension cables to keep the concrete, which has essentially no tensile strength of its own, in compression. Since steel is denser than concrete, this would reduce the size a little.
… Burj Dubai projected to be 2684 ft. tall = approx. 18 ten-storey silo buildings (at 15 ft. per storey)
using 4 smaller concrete blocks = 76 silos
OK, 76 of these concrete cubes, rising in the day until their bottom surfaces are, as Arnold says, 18 times 15 feet, 45 m, off the ground. Yes, that could work.
The largest existing solar power stations have a much smaller capacity than 1 gigawatt-year per year, so they might get away with just one or two of these hyperhoists and ultracubes.
If they pass building codes, they won’t tend to “uncontrollably unstore.” Indeed, there is NO energy lost in gravitational storage, no matter how long.
So the hoist that lifts this block of concrete, as tall as a 20-storey apartment tower and as heavy, I guess, as a small cityful of them … so the hoist that lifts it through most of its own height, and is able to let it down on command, will have neither an uncommanded nor a wrongly commanded letdown, “no matter how long”, and this is guaranteed by building codes.
I think I understand.
In comment 56 Dan Hughes said,
“What does a nuclear pile tend uncontrollably to do?”
But it can: beta decay. Turning a fission reactor on at 1 watt and keeping it there for 1 hour, then stopping it, causes 0.0061 watt-hours of uncontrollable delayed energy release in the following hour, 0.0018 watt-hours in the second post-shutdown hour. These numbers are given by the integral of the Untermyer and Weills equation that is given in http://www.rertr.anl.gov/FRRSNF/TM26REV1.PDF and said to originate in USAEC Report ANL-4790, 1952.
15. 65
Ike Solem says:
Denmark is the world leader in offshore wind, right?
Preface: The Power Source for the Future
Our future energy supply faces numerous challenges and has become subject to unstable international conditions. To meet these challenges, offshore wind has a key role to play. Offshore windpower can contribute significantly to achieving the EU goals of a 21 percent share of renewable electricity by 2010, halting global warming and reducing our dependence on coal, oil and gas.
We have come a long way since the 1980s, when most electricity production was based on coal and when the acidification of forests and lakes by acid rain was the predominant theme in the environmental debate. Today wind power provides 20% of Danish electricity consumption.
Within a few years, the wind power industry has grown to become a significant industrial sector providing huge benefits for exports and employment. We are now talking about windpower generation plants rather than single turbines, and the Danish wind power industry is at the leading edge in an ever more competitive global market.
http://www.renewableenergyworld.com/rea/news/infocus/story?id=46749
16. 66
John Armour says:
Whilst a nuclear reactor emits no CO2 in the production of electricity, the mining and milling of the fuel certainly does. Not to mention the construction and eventual decommisioning of the plant. The burning of fossil fuel is a large part of the nuclear cycle that we conveniently ignore.
If the grade of ore was consistently high then this might not be a problem. But most of the Earth’s high quality ore has already been used up. And when you’re forced to mine and mill the lower grades, the CO2 balance goes into debit. That is, more CO2 is emitted than would be the case if you just burnt the fossil fuel conventionally to generate electricity.
Storm van Leeuwin and Smith claim that if the world hypothetically went totally nuclear tomorrow, the rich ores would be consumed in less than a decade, after which the CO2 benefit of nuclear energy would be gone.
Here’s an article from the Australian Government’s CSIRO Sustainability Network Newsletter. “Nuclear Energy: We don’t need it”
http://www.bml.csiro.au/susnetnl/netwl53E.pdf
17. 67
Hank Roberts says:
John, interesting CSIRO piece. I’ve copied over the footnotes from it, for the basic points you mention:
http://www.naturaledgeproject.net/NAON1.aspx; citation on page 37
http://www.oprit.rug.nl/deenen/ “Nuclear Power: the Energy Balance”
http://www.yesmagazine.org/article.asp?ID=1064 (Burns) http://www.yesmagazine.org/article.asp?ID=1065 (Lovelock)
No one ever built a breeder reactor that didn’t require reprocessing spent fuel, as far as I know. Canada’s thorium/deuterium design seems to be history.
18. 68
Karen Street says:
Re Denmark: in 2006, it generated 54% of its electricity from coal, 21% from natural gas, 13% from wind (down from 18% in 2005), and imported a tad.
It would be easier to accept Storm van Leeuwin and Smith if they had ever submitted their work to peer review. Their results counter all studies I’ve ever seen that have been peer reviewed, which show nuclear power over its life cycle producing as much GHG/kWh as wind, less than solar, and that we essentially have more uranium + thorium than we have coal.
The CSIRO author cites them. Meanwhile, pretty much everyone making climate change plans is assuming that the peer-review analysis holds and that nuclear power is an important low-GHG source.
Of much more interest is the recent IEA report, Energy Technology Perspectives. IEA estimates that wind could be almost as important as nuclear between now and 2050, and that we will need 32 GW in new nuclear power every year between now and 2050.
IEA argues that we need to remake the world economy, unprecedented levels of cooperation, etc, etc, etc to get a 50% reduction by 2050. A 50% reduction brings in a best-guess reduction to 400 ppm. Not enough.
19. 69
Patrick 027 says:
(Had to skip over some comments, sorry for repeating if this is a repeat:) So the COP of a sample air conditioner is 2.92?
The ideal COP of a heat pump (no production of entropy) can be found from these equations:
Qc + W = Qh
Qc/Tc = Qh/Th
Where
Qh is heat flow in (out) at temperature Th (the hot end),
Qc is the heat flow out (in) at temperature Tc (the cold end),
W is the work (usefual energy) produced (consumed).
Let DT be Th-Tc
Solving for W and Qh:
W = (1 – Tc/Th) Qh = Qh (Th-Tc)/Th = Qh DT/Th
Solving for W and Qc:
W = Qc (Th/Tc – 1) = Qc (Th-Tc)/Tc = Qc DT/Tc
For an ideal heat engine, relative to the heat source, the efficiency is DT/Th. If you wanted to run a heat engine off of a limited supply of cold in a hot environment, you might care more about the efficiency relative to the heat sink, DT/Tc.
For an ideal heat pump, the Coefficient of Performance (COP) for the hot end is Th/DT. The COP for the cold end (of interest for air conditioners) is Tc/DT.
For cooling or heating a home, Tc and Th are relatively close so the COP of ideal heat pumps for heating and cooling are nearly the same.
—-
But the ideal COP for, say, 20 deg C (36 deg F) temperature difference with the Tc ~ Th ~ 300 K is going to be ~ 15. I’m not that familiar with the actual practicalities involved in the technology, and wouldn’t expect to get very close to ideal COP values, but 2.92 seems rather pathetic. (Although only a bit more pathetic than the ~ 30 % efficiency of coal power plants, considering that the temperatures involved should allow for much higher efficiencies.)
Also, it occurs to me that it is at least possible (if not yet practical?) to roughly double the COP of a device if, instead of a single heat pump, one has several with a working fluid running through at progressively different temperatures from intake temperature to goal temperature. (PS two fluids running past each other in opposite directions with progressive warming or cooling makes a heat exchanger).
In addition to the role of internally generated waste heat (which, for lighting, can be reduced with daylighting (sunlight is ~ half visible light, and if the windows could reflect IR and UV, or convert them to electricity, well, then… :) ) and high-albedo interiors), there is another assymetry between heating and cooling: when the dewpoint is high, the amount of heat that an air conditioner has to pump for a given set of Tc and Th and mass of air can increase because of the latent heat of condensing water. (For winter humidity needs, their is perspiration (not much), cooking of pasta, and showers, etc.)
Having a home with greater cooling needs than heating needs might tend to be ‘greener’ because more solar power is generally available in summer than winter; of course, this depends on other available energy sources, energy storage, and individual preferences, etc… And with that, comment 6 is on to something.
20. 70
Ike Solem says:
Instead of breeder reactors, we should be building breeder solar PV manufacturing facilities:
It was pointed out that a photovoltaic panel manufacturing plant can be made energy-independent by using energy derived from its own roof using its own panels. Such a plant becomes not only energy self-sufficient but a major supplier of new energy, hence the name solar breeder. The reported investigation establishes certain mathematical relationships for the solar breeder which clearly indicate that a vast amount of net energy is available from such a plant for the indefinite future. It is pointed out that if solar electric plants would be built according to the solar breeder principle, their operation as a net energy source would be automatically assured.
Yes, this revolutionary new concept… wait – what’s that date?
The solar breeder, Lindmayer, J., Photovoltaic Solar Energy Conference, Luxembourg, September 27-30, 1977
21. 71
Lawrence Brown says:
Re:#60 “…… why don’t we all car pool more and use public transportation more often?….”
Because that would ruin Will Rogers’s prediction that the U.S. would be the first country in the world to drive itself to the poorhouse in an automobile.
Actually the real answer to this commonsense approach of conservation and efficiency lies partly in the insidiousness of the oil industry and its public relations propaganda,with the help of some(not all) of the mainstream media who insist on taking a so called “balanced” stance on AGW,by presenting the “other side”. Which, to me, is like looking at the other side’s view on a Heliocentric solar system.
22. 72
Max says:
I would question the “100%” efficiency rating for the gas furnace,the real world condition in combustion testing reveals more like 85-89% combustion efficiency, to have a even near 100% efficiency, the gas would have to burn at stoichiometric air fuel ratios, and no burner is that good, they all need to have excess air due to design limitations. Keep in mind all condensing furnaces have 2 fan electric fan motors, and one with a pre/post purge function that will remove heat, so you have 2 energy inputs not just a singular source such as on the a/c to calculate your “carbon footprint”.
23. 73
Patrick 027 says:
Re 68
Actually, water can absorb IR and UV; maybe solar water heaters could double as skylights.
24. 74
Patrick 027 says:
Re 68
… of course, if you value white Christmasses, maple syrup, and don’t want to worry about killer bees, black widow spiders, or dengue fever, moving to sunnier warmer climates might not be the right choice. On the other hand, you’d be closer to the oranges and cacao trees (although the processessing of that chocolate might take awhile to follow you?). Etc… PS written near 45 deg Lat.
25. 75
Patrick 027 says:
When I wrote Re 68, I was refering to my own comment (Re 69)
26. 76
Thomas says:
19. It is very unfortunate that our present government is so totally opposed to your nuclear ambitions (I am assuming you are Iranian, although similar logic applies to the rest of the Arab world). It is clear that with the current administration in Washington no change is likely. There have been proposed compromises, such as letting foreigners (Russia mainly) handle the reprocessing, which aside from national pride, and a potential, but low probably fear that the fuel supply could be cut off, should be sufficient (at least given a change of leadership in Washington). Another avenue that I think would work spectacularly well (if National pride is the main issue), would be to develop a Thorium based Nuclear cycle. This would be an important contribution to the worlds energy future, as well as being a way to have control over your fuel cycle, and alleviate proliferation concerns.
Now, back o the Prius. It is incorrect to assign the energy cost of the batteery to the lifecycle energy cost, as the batteries will be recycled (the materials are too valuable not to). Also Toyota is currently warranting the batteries for 150,000 miles (and confident of typical lifetimes of at least 200,000). The only real fly in the ointment is hybrid production capability. Currently hybrid battery production capability is in short supply. I expect the hybrid market will be production limited for at least the next five years. That implies that the most effective usage requires selfselection of buyers (i.e. only those who drive a lot should buy a hybrid). A Prius (and probably some other serial hybrids), have additional potential savings. As battery prices decrease, and gasoline prices increase, at some point it will be worthwhile to upgrade existing Priuses to plugins -all that will be required is a charger, better battery capacity, and modified control software.
27. 77
Re #64, G.R.L Cowan, I misunderstood your objection! From your example I thought you were talking about a human tragedy adjacent to the site, such as from a dam water release without warning, or a nuclear reactor explosion. I had not considered that it will be impossible to design mechanical safety features to prevent a concrete block from sliding down inside a silo! “He must be talking about the collapse of the whole building,” I thought, hence, “Follow the building codes!”
28. 78
raypierre says:
I’ve been trying to think of a simpler way to highlight the basic issue in the decision of whether you help reduce carbon emissions more by keeping your car or selling it off and buying a hybrid (or other new car with better gas mileage). Let’s call the new car a Prius, for the sake of argument.
Assuming the Prius to have lower lifetime carbon emissions (including manufacturing) than the fleet average, then the best thing for long term carbon emissions is to turn the fleet over into Prius’s. That can’t happen if NOBODY buys a new Prius. I.e. the situation where everybody buys a used car and feels virtuous does not get you where you want to go in terms of carbon emissions. Another way of putting it is that if everybody bought used cars, there would be no new cars entering the fleet, and eventually the stock of used cars would be exhausted, as they age and die. That mean that somebody has to buy new cars, and those new cars ought to be Prius’s.
That means that somebody who buys the new Prius is performing a service by injecting an efficient car into the fleet. The question then comes down to what is the ideal rate of injecting Prius’s into the fleet? Clearly you can overdo it since if everybody goes out and buys a new Prius, a lot of fairly efficient and usable used cars will get scrapped before their time.
So, any takers on how to figure the optimum rate at which Prius’s should be entering the fleet? Once you figure the best number, you can figure out WHO should be the ones to actually buy those cars — the WHO meaning WHO in terms of what kind of car they already own. Perhaps you’d then want to add a dose of reality by constraining the WHO question by personal financial resources.
I think this is an interesting optimization problem, though perhaps academic if the supply of hybrids is going to be limited by manufacturing capability for an extended period of time.
29. 79
Anders L. says:
Re #4: Sure, hundreds of thousands of homes in Sweden are heated by heat pumps already, taking heat from the groundwater by drilling a well about 300 ft deep and circulating a heat medium down that well. You get at least 3W of heat for every W of electricity that way, so it is very, very profitable for the houseowner, another example of that “green” does not mean “expensive”. Sweden today uses only 50% of the oil compared to 1975 – and we have not had to go back to the stone age to achieve that.
30. 80
Tim Joslin says:
Raypierre, thanks for your response to my #44 – I totally agree with your view that we need more not less natural forest.
The reason I thought you meant “old tree” rather than “young tree” was that it seems to me the best way to look at this sort of problem is to ignore for the moment processes that operate on longer timescales than decades to centuries, and consider that a given area of land has a certain capacity to store carbon removed from the atmosphere by photosynthesis. If the carbon is stored it is not in the atmosphere to contribute to global warming. Our aim should be to implement policies to maximise the amount of this carbon storage capacity that is used, on average over time. My thoughts on these lines for biofuels (i.e. why they’re a really bad idea) are available at: http://unchartedterritory.wordpress.com/2008/06/16/the-biofuel-papers/
WIRED’s provocative proposal to sell a few extra copies is to suggest that removing old trees to allow more vigorous growth will keep more carbon out of the atmosphere than just leaving the forest alone. This is not the case. A hectare of forest might hold 150 tonnes of carbon in the form of trees. If we remove this and allow new growth then we have to add two curves: the carbon uptake of the young trees and the loss of carbon to the atmosphere of the trees that have been taken out. An example of the carbon uptake over 100 years appears in: http://www.eccm.uk.com/httpdocs/pdfs/TD11.pdf
The ECCM paper shows that it takes about 10 years before the new trees take up carbon at anywhere near the most rapid rate (of around 3.5 tonnes/hectare/year in their example) and nearly 50 years before it stores even half the 150 tonnes of carbon that has been removed. The question is whether the rate of loss of carbon from the extracted trees exceeds the rate of uptake of the new trees. It seems to me that decay is almost bound to exceed growth for most of the (say) 100 year lifetime of the new trees. Even if the old trees are used entirely for “furniture and houses” (to quote WIRED) there will be a large proportion wasted immediately and then a slow loss over time. Very likely much less than 50% of the carbon will stay out of the atmosphere for 50 years, for example.
Towards the end of the 100 year lifetime of the new trees there may be a small gain represented by the small percentage of wood surviving for a century. But then we propose to repeat the process and once more go into carbon deficit. To ensure a steady harvest of wood the forestry industry keeps various plantations at different stages. We therefore need to take an average over the lifetime of the trees – 100 years in my example – of the total carbon kept out of the atmosphere per hectare as a result of the forestry process. This average, I suggest, will be significantly less than the amount of carbon stored in the original natural forest.
There are other arguments in favour of leaving as much land as natural forest rather than managed forest. In particular, the carbon stored in natural forest is significantly higher than the peak amount stored in a managed forest, for various reasons. In other words we may destroy a natural forest storing 150 tonnes C/hect and replace it with a plantation that at peak stores 100 tC/hect, and most of the time much less than this. To compensate we would have to preserve an awful lot of antique furniture! Valid points are also made in #51 and #53.
#50 alludes to the release of carbon as the planet warms. If we ignore the slow process of long-term storage of carbon in soils, we can consider forests to be in equilibrium, like any other chemical process. Raising the temperature would be expected to move the reaction towards carbon release, in agreement with what is observed (in high latitudes this may be outweighed by increased growth rates) – a carbon-cycle feedback.
But increasing atmospheric CO2 tends to move the reaction towards carbon uptake (that is an increasing amount of biomass in a given forest area) – this is the “CO2 fertilisation effect” which is currently keeping the rate of increase of atmospheric CO2 to about 2ppm/year rather than around 2.5ppm/year which the discussion in AR4 implies it would be otherwise. As we continue to destroy the world’s forests and cultivate the land instead, we progressively prevent the CO2 fertilisation effect from operating.
31. 81
pete best says:
#78, I beleive that company cars would be a good way to progress. Get the critical mass for economic production from offering them as company fleet cars. However lots of motorway miles in a Prius might not be that good an idea as they are not that efficient on long haul journeys as they use the petrol engine and people tend to travel fast on motorways reducing efficiency further.
It needs to be a way of getting people doing lots of urban short journeys (mums and stuff on the school and shopping run) to obtain them primarily.
32. 82
Fernando Magyar says:
Re Prius etc. I think the private automobile as a form of mass transportation is the wrong paradigm. It will not work for the majority of the 6.5 billion people currently on the planet. It is not sustainable regardless of the energy source used to power it. I think XKCD has the right take on this topic. http://xkcd.com/437/
33. 83
Re 79:
Re #4: Sure, hundreds of thousands of homes in Sweden are heated by heat pumps already, taking heat from the groundwater by drilling a well about 300 ft deep and circulating a heat medium down that well. You get at least 3W of heat for every W of electricity that way, so it is very, very profitable for the houseowner, another example of that “green” does not mean “expensive”. Sweden today uses only 50% of the oil compared to 1975 – and we have not had to go back to the stone age to achieve that.
I was looking at the amount of energy (ignoring all efficiency and conversion losses, etc) in 1,000 gallons of water at the daily high (about 100F these days) and the overnight low (about 75F these days) and it came out to be about as much electricity as I buy from the grid in 3 days. Since solar collectors have outlet temperatures in excess of the daily high, I’m thinking I’m aiming my sights a bit low — and 1,000 gallons isn’t that much, perhaps 130 cubic feet, or less than 6′ on a side.
34. 84
why don’t we all car pool more and use public transportation more often?”
Because public transportation is often crowded, uncomfortable, unreliable, and slow even when it does work. I’m a long-time bus rider and I’ve ridden buses and trains in many cities; I know. There’s also the fact that with public transportation, you can’t plan the route or make side-trips. Americans like cars because they like having some control over their transportation. Cars aren’t going to go away any time soon.
35. 85
Nylo says:
Let’s assume, as raypierre says, that you need to emit more CO2 to cool a house by 1ºC than to warm it by 1ºC. There is still another important factor. The use of the air conditioner will directly cause only a little warming (the machine warms a bit because it is not 100% efficient). The heating system on the other hand causes warming directly. So the question would be: will the extra CO2 emitted to cool the house be as important for global warming as the actual fact of warming the house in winter? Therefore, if we are going to reduce our welfare, is it better to be 1ºC colder in winter or 1ºC hotter in the summer? How do we “heat” less the planet, with the CO2 emission or with the direct warming?
36. 86
Dan Hughes says:
re: # 79
In 2005 about 92% of Sweden’s electricity was based on hydro and nuclear, 8% fossil. From.
“Nuclear Power in Sweden
(May 2008)
• Sweden has 10 nuclear power reactors providing half of its electricity.
• A 1980 referendum canvassed three options for phasing out nuclear power, but none for continuing it.
• Sweden’s 1997 energy policy retains most of the country’s nuclear plants but has resulted in premature closure of one 2-unit plant.
• Sweden is the only country to have a tax discriminating against nuclear power – now about EUR 0.67/kWh.
Sweden’s electricity consumption has been rising and it has one of the world’s highest individual levels of consumption: about 18,000 kWh/head. About half of domestic production is nuclear, and up to half hydro, depending on the weather – see contrast below. In 2006 nuclear power produced 65 billion kWh, 48% of total.”
In response to closure of a 2-unit nuclear plant,
“A new 800 MWe undersea transmission line is being built by 2010 to enable export of electricity to Sweden from Finland’s new Olkiluoto reactor.”
And
“Removal of 8.5 TWh/yr from the county’s nuclear output is being replaced by imports from Germany and Denmark, much of it coal-fired, and by nuclear generation from Finland and Russia, in the latter case from old Chernobyl-type reactors which the EU is anxious to shut down elsewhere.”
37. 87
Dan Hughes says:
re: #64
By stating ‘fission reactor’ you have implicitly specified both the composition and geometrical configuration of the materials that comprise the ‘nuclear pile’ and the physical boundary conditions. You have additionally specified some the previous history of he state of the pile and the initial conditions for the ‘uncontrollably to do’ aspects.
38. 88
Figen Mekik says:
#81: I agree. Figuring out whether to buy a Prius or your next Corolla is a problem for a privileged few in the world (though many of this privileged few may be living in the US). Lots of people in the world at large don’t own cars and have to rely on public transportation. For many people in Europe and the Middle East, owning a car is not the problem but driving one is because gas costs about 4 times more there than it does here in the US.
I think the real solution lies in making public transportation more available and comfortable in the US. I think solving problems like how Earth-friendly is it to own a Prius may be considered a little lofty when even the Prius still produces greenhouse gases (though less) and a lot of people really don’t (can’t) own cars because they don’t have the financial resources to make the choice between a Prius and another car.
Plus Prius or not, it seems some are still considering driving the private vehicle when that, among other things, is what got us in this climate mess to begin with, isn’t it?
But soon the price of gas is going to become so unaffordable that this discussion is going to be moot and we all will have to learn to ride the bus or our bikes if our spirits are so independent :) Let’s hope we don’t do irreversible damage to our planet by then, if we haven’t already.
39. 89
Nylo says:
Furthermore, how much of the energy that I transfer to the outside of the house becomes energy lost to space because of increased radiation of the outside of the house? A black body with one half at 22 degrees and another half at 20 degrees radiates more energy than a black body with uniform 21 degrees temperature, as radiation changes with T^4. Could an actual loss of energy for the Earth system be a result of the use of my air conditioner, I mean, locally, independently of the energy used to “move” the heat?
40. 90
Gerry Beauregard says:
For anyone interested in geothermal heat systems, the Canadian government has a nice primer:
An Introduction to Residential Earth Energy Systems
http://www.canren.gc.ca/prod_serv/index.asp?CaId=193&PgId=1190
41. 91
Gerry Beauregard says:
Some musings on geothermal (aka earth energy) heating/cooling systems. One could conceivably do even better than using the 55F deep ground as a heat source/sink by storing the winter cold and summer heat.
Suppose you have a shed with a very large tank of full of water. The base of the tank has antifreeze filled pipes leading to a heat pump in the basement of your house.
In winter, you open the doors to shed, allowing cold air to eventually freeze it into a solid block of ice. Then you wrap the ice block in insulation and seal up the shed to store all that coolth for use later in the year. When you need air conditioning in summer, you run your heat pump using the block of ice (eventually very cool water) or the deep ground as the sink, whichever is cooler.
In another shed, you have a well-insulated tank with water that you heat up in summer using a rooftop solar water heater. In winter, your heatpump uses that tank as the heat source (or the deep ground, whichever is warmer).
42. 92
Nick Gotts says:
#84 [BPL] “Because public transportation is often crowded, uncomfortable, unreliable, and slow even when it does work.”
Not when it’s properly funded and run, as in much of Europe (sadly, not the UK).
“Americans like cars because they like having some control over their transportation. Cars aren’t going to go away any time soon.”
Similarly, Americans (and western Europeans, Japanese, etc.) like using far more than their fair share of the world’s resources, including its ability to absorb pollution. If we can’t be persuaded to moderate our greed considerably, everyone’s stuffed.
43. 93
Ike Solem says:
There really are a wide variety of options for making both large buildings and small residences energy-independent. Not only is this good for climate, it is also a real economic necessity in an era of steadily increasing fuel prices.
http://www.usatoday.com/money/industries/energy/2008-06-15-power-prices-rising_N.htm
The two large-scale solutions to that are concentrated solar power plants and wind turbine farms. The small-scale solutions are energy-independent homes that also use solar and wind inputs.
For a nice design for a home-scale wind turbine system, see:
http://www.treehugger.com/files/2008/06/wind-power-urban-architectural-microturbines.php
Modular Architectural Wind Microturbines
Aerovironment is designing these wind microturbines specifically for the urban environment: No need for a tower, the blades rotate more slowly and silently, and they are set at an angle that allows them to benefit from the wind that is bouncing up the walls and escalating them vertically.
All in all, what is needed is a new architectural mentality that places energy efficiency and conservation at the center, and designs the structure around that. As you can see, many people are doing this already.
44. 94
In comment 87 Dan Hughes included,
By stating ‘fission reactor’ you have implicitly specified both the composition and geometrical configuration of the materials that comprise the ‘nuclear pile’ …
“Nuclear pile” once, I guess up to about 1955, meant fission reactor. I don’t think it ever meant anything else.
Arnold asked what uncontrollable tendencies the things so called had; I gave the complete list for typical ones. They don’t have any bad habits in re control of fission itself, witness the natural Oklo reactors’ remains’ appearance of having burned evenly for millennia, and witness also the good behaviour of the San Francisco‘s engine when that boat stove in its front end on a seamount.
45. 95
mike lukes says:
The elephant in the room that would need to be removed is the ever increasing human population…unless we limit our propagation, the earth’s resources will become depleted and no longer sustain us in the multiple billions. As Walt Kelly’s Pogo said “We have met the enemy and he is us!”.
46. 96
L David Cooke says:
Comment 85
Logically, I may have a problem with the idea
expressed, it must be my misunderstanding. My
thought is to that impart 1 Deg of change to a slug of
atmosphere whether higher or lower then the ambient
temperature, should not differ. Only the energy
conversion system imparting the temperature change is
important.
Systemically, direct heating endothermic output of the
complete combustion of methane (Natural Gas) to H2O
and CO then CO to CO2 as compared to the indothermic
biologic processes are not much higher then any other
process then it misses it by a small margin.
(Probably only Solar Cell Hydrolysis and combustion of
the by products are likely higher.)
On the other hand, it appears either Air-Air or
Water-Air Heat Pump systems as the most popular /
appropriate solution in the highly humid regions where
the tropical temperatures from Global Warming are most
likely to make uninhabitable, if what the science
tells us is true. For cooling, either of these
systems require a conversion of combustion (chemical
energy) to steam (thermal energy) from steam to a
turbine (mechanical energy) from a turbine to a
dynamo (electrical energy), then we have the
resistance of distribution (thermal energy leakage)
and finally the conversion from electrical energy to
mechanical energy to power the exchange of thermal
energy.
There is yet a different system with much fewer steps
though it still would require combustion (chemical
energy) to achieve thermal energy exchange. By using
the Absorptive/Evaporative refrigeration process it
would need to provide a high thermal input to drive
the thermal energy exchange. In either case cooling
has more conversion steps then direct heating.
What I think is being suggested is that the heat per
unit of air being passed through the heat exchanger is
lower for cooling then it is for heating and that has
more to do with the initial temperature of the heat
exchanger inputs. However, if a furnace was designed
that recirculated the combustion gases until they
reached about 30 to 40 degrees above ambient before
reigniting the burner I think the argument presented
could be easily countered. A recirculation furnace
with a pulsed burner would be much greater in
efficiency systematically.
Hence, a good reason for many to look toward Gas
Packs, (Air-Air Heat Pump Gas Furnace combinations)
for their HVAC needs. If it were not for the economic
reasons required to improve the furnace portion, the
idea of a pulsed burner and combustion chamber
recirculation control system would have been developed
years ago. (Keeping in mind that the biggest concern
is the possibility of heat exchanger combustion gases
leaking into the buildings air return, resulting in
possible CO poisoning.) Which could be remedied by a
internal CO detector which would shutdown the Gas
Furnace and activate the Emergency Electrical heating
strip while annoying the building manager/homeowner
with one of those aggravating warning alarms.
As to comment 89, the higher heat energy at the
surface is not going to get through the green house
gases any faster then lower temperatures, if I
understand the science correctly. Hence, 1 Deg. more
may not do much more then warm the middle troposphere
a little more as the surface region of re-emission
into space has not changed. Now if the added heat
would raise the region being heated so that the
surface area increased rather then the temperature
that would be different. Though we probably would not
be having this discussion if that were true…
So, did I explain my concern and address things
correctly or was the intent something different?
Cheers!
Dave Cooke
47. 97
Andrew says:
I understand that many people consider solar power one solution to greenhouse gas emissions.
However, has anybody ever considered that to purify the silicon for the typical PV panel requires significant electrical energy that comes for the most part from coal fired plants?
In other words, PV solar power is not without it’s carbon foot print.
48. 98
Leonard Evens says:
Secular Alarmist commented
————————————————————–
Leonard Evens wrote: “… I do bicycle for exercise, but I would need a more practical bike to use for shopping …”
How about a four-wheel, two-seat, hybrid electric-pedal powered bike?
——————————————————-
I have considered various alternatives, including tricycles. Some of these would be practical all weather vehicles for short trips. I am not yet in such bad shape that I need electric power. The problem is where to put it. I only raise that issue because it is likely to be a problem for any apartment dweller with limited storage space. In our case, we do have some alternatives, but they could be costly, but most would have even fewer alternatives than we do. Of course, if you live in a detatched house, this won’t be a problem, but more and more of us live in apartments, something that will be more common as the disadvantages of dispersed housing become more apparent.
49. 99
In comment 97 Andrew says,
… has anybody ever considered that to purify the silicon for the typical PV panel requires significant electrical energy that comes for the most part from coal fired plants?
Depending where it is made, the electricity might or might not have a large fossil fuel component.
However, although solar PV electricity is very expensive, it is not so expensive as to suggest conventional energy invested exceeds output. Also, direct comparison of electricity in and electricity out yields, IIRC, a payback time of about two years.
Storage to make intermittent PV input into continuous output, such as a silicon purifier might need, would waste some of the intermittent input and extend this payback time to ~3 years, and some of the PV cells the factory would make would spend their whole lives paying off the electricity debt incurred in the factory’s construction; but this would be small compared to the energy it would use in its working lifetime. So net energy should start coming out within five years.
CSP is quicker.
50. 100
Nylo says:
Re #96 Dave,
My doubt in 85 is, OK, we use more energy in cooling than in warming, but do we actually WARM more by cooling than by warming? When we try to warm, all of the energy used causes warming, directly (temperature increase) and indirectly (GHG emissions). When we try to cool, however, only some of the energy will cause warming.
As for what I said in 89, I’m not sure you understood it. The energy radiated by a body depends on its surface temperature, but it is by an integral calculation of the temperature of the surface, not by just taking the average of it. Again, if I have a black body which is 22 degrees and another one which is 20 degrees, the total emissions from both will be higher than if each of them was 21 degrees. And I don’t need to change their sizes or any other properties than their temperatures. If I manage to transfer energy from one to the other so that one gets a higher temperature while maintaining the average of the 2, the result is a faster energy loss from the overall system due to energy radiation.
If the outside of my house is warmer, it emits more energy to the outer space. That principle is what will make the planet’s warming stop at some point: as we warm, we radiate more and in the end we will reach a new (hotter) equilibrium. I hope you are not trying to negate this.
[Response: It’s only the greenhouse gas emissions that are a significant warming factor. The warming due to direct thermal output of all human energy use is negligible on the global scale, though in dense environments it can contribute significantly on a local level, making the urban heat island effect worse. –raypierre] | 16,043 | 72,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-26 | latest | en | 0.953042 |
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A280902 T(n,k)=Number of nXk 0..2 arrays with no element equal to more than one of its horizontal and antidiagonal neighbors, with the exception of exactly two elements, and with new values introduced in order 0 sequentially upwards. 11
0, 0, 0, 0, 1, 0, 1, 10, 9, 0, 2, 213, 646, 124, 0, 9, 2292, 22568, 22632, 1464, 0, 34, 21762, 492490, 1451655, 610448, 15768, 0, 124, 184076, 9426050, 65348136, 75809243, 14262832, 159920, 0, 432, 1457827, 162640161, 2571528867, 7083739466 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,8 COMMENTS Table starts .0........0............0...............1..................2...................9 .0........1...........10.............213...............2292...............21762 .0........9..........646...........22568.............492490.............9426050 .0......124........22632.........1451655...........65348136..........2571528867 .0.....1464.......610448........75809243.........7083739466........574982226478 .0....15768.....14262832......3521886844.......684011230518.....114677717497532 .0...159920....304584096....151803173493.....61277484218852...21239418911643829 .0..1554304...6117000704...6210239609889...5208435552362140.3734730743379447607 .0.14632704.117496694272.244458357395448.425813044528570428 LINKS R. H. Hardin, Table of n, a(n) for n = 1..96 FORMULA Empirical for column k: k=1: a(n) = a(n-1) k=2: [order 6] for n>8 k=3: [order 6] for n>8 k=4: [order 12] for n>14 k=5: [order 24] for n>27 Empirical for row n: n=1: a(n) = 6*a(n-1) -6*a(n-2) -16*a(n-3) +12*a(n-4) +24*a(n-5) +8*a(n-6) for n>10 n=2: [order 15] for n>17 n=3: [order 54] for n>60 EXAMPLE Some solutions for n=3 k=4 ..0..0..1..2. .0..0..1..0. .0..1..0..0. .0..1..0..2. .0..1..2..0 ..1..1..2..1. .2..1..0..0. .2..1..0..2. .1..2..2..0. .1..1..1..2 ..0..0..1..1. .1..0..2..0. .1..0..2..0. .1..1..1..2. .2..2..0..1 CROSSREFS Row 1 is A280309. Sequence in context: A317813 A038310 A318421 * A118768 A318255 A008956 Adjacent sequences: A280899 A280900 A280901 * A280903 A280904 A280905 KEYWORD nonn,tabl AUTHOR R. H. Hardin, Jan 10 2017 STATUS approved
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Last modified June 28 02:57 EDT 2022. Contains 354903 sequences. (Running on oeis4.) | 967 | 2,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-27 | latest | en | 0.58467 |
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1. Entering the Function:
• Locate the input field labeled "Function f(x, y):".
• Enter your function in terms of variables x and y.
• You can use standard mathematical operations such as +, -, *, /, and ^ for power.
• Note that you must use * for multiplications, i.e., x*y instead of xy.
• For trigonometric and other functions, use their usual notation, e.g., sin(x*y), cos(x), tan(y), log(x), etc.
2. Selecting the Differentiation Variable:
• Next to the function input, you'll find a dropdown menu labeled "Differentiate with respect to:".
• Choose the variable with respect to which you want to compute the partial derivative. You can select either x or y.
3. Computing the Derivative:
• Once you've entered your function and chosen the differentiation variable, click the "Compute" button.
• The calculator will process your input and display the result below the button.
• The result will be displayed in the format: ∂/∂x (your function) = result
• If there's an error in your input or the function cannot be differentiated, an error message will be displayed. Ensure that your function is correctly formatted, and try again.
5. Tips:
• Ensure that you use proper mathematical notation for the best results.
• If the result seems too complex, try simplifying your input function or breaking it down into parts.
What is a partial derivative?
How do you compute it? For what purposes is it used for?
If you’re looking for answers to questions like these, then you’ve come to the right page. We will discuss the above questions as well as the various areas where partial derivates are used.
## What is a Partial Derivative?
The term contains two words: partial and derivative. The derivative of any algebraic expression is calculated with respect to a certain specified variable. This is done by differentiating the given function or expression with respect to the specified variable and it symbolizes the change in given function f(x) when the specified variable changes infinitesimally.
Also see our series on solving integral equations.
The derivative part is pretty clear when f(x) is composed of a single variable, but if it contains more than one variable then the inter-dependence of each variable also needs to be taken into account while calculating the derivative. And this is where the concept of “partial” derivative comes into play.
The partial derivative of a multi-variable expression with respect to a single variable is computed by differentiating the given function w.r.t. the desired variable whilst treating all other variables as constant, unlike the total differential where all variables can vary.
Partial derivatives symbolize instantaneous change in a given function relative to the infinitesimal change of variable under consideration. It is extensively used in differential geometry and vector calculus. Also, for optimization purposes, partial derivatives play a major role in every field and the total derivative can be computed step-by-step using partial derivatives.
## Examples & Usage of Partial Derivatives
As stated above, partial derivative has its use in various sciences, a few of which are listed here:
### Partial Derivatives in Optimization
Partial derivates are used for calculus-based optimization when there’s dependence on more than one variable.
### Partial Derivatives in Geometry
To compute rate at which a certain geometric quantity, volume, surface area, etc., varies when a basic measurement (radius, height, length, etc) is varied.
### Partial Derivatives in Mathematical Physics
Partial derivatives (rather partial differential equations) have extensive use in mathematical physics (and variational calculus, Fourier analysis, potential theory, vector analysis, etc).
### Partial Derivatives in Thermodynamics
Partial derivatives in this case can be thermal variables or ratios of some variables like mole fractions in the Gibbs energy equation.
### Partial Derivatives in Quantum Mechanics
Schrodinger wave equations and several other equations from quantum mechanics inherently use partial derivatives.
### Partial Derivatives in Economics
Most functions explaining economic behavior statements like the behaviors depending on so and so variables in a particular manner, are obtained using the concept of partial derivates where independent variation in the behavior is observed by varying the fundamental variables one by one.
In addition to these partial derivatives are used in many other areas of education to calculate the differentiation of a function partially with respect to a variable.
## Notations used in Partial Derivative Calculator
Let f be a function in x,y and z.
First order partial derivatives are represented by
$\dfrac{\partial f}{\partial x} = f_x$
Second order partial derivatives given by
$\dfrac{\partial^2 f}{\partial x^2} = f_{xx}$
The mixed derivative is shown by
$\dfrac{\partial^2 f}{\partial x^2} = f_{xx}$
## Summarizing
As much use, partial derivatives have, they are equally difficult to compute at higher levels and hence online partial derivative calculators are designed to help the users simplify their computations. Above I have showcased a partial derivative calculator that solves partial derivative problems step by step, equipped with the functions of computing partial derivatives to cater to all your computational needs. | 1,092 | 5,550 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-26 | latest | en | 0.856262 |
http://math.stackexchange.com/questions/57172/putting-many-disks-in-the-unit-square?answertab=oldest | 1,462,017,696,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111838.20/warc/CC-MAIN-20160428161511-00126-ip-10-239-7-51.ec2.internal.warc.gz | 181,573,896 | 25,184 | # Putting many disks in the unit square
Consider a square of side equal to $1$. Prove that we can place inside the square a finite number of disjoint circles, with different radii of the form $1/k$ with $k$ a positive integer, such that the area of the remaining region is at most $0.0001$.
If we consider all the circles of this form, their total area is $\sum_{k \geq 1}\displaystyle \pi \frac{1}{k^2} - \pi=\frac{\pi^3}{6}-\pi\simeq 2.02$ which is greater than the area of the square. (I subtracted $\pi$ because we cannot place a disk of radius $1$ inside the square).
So the circles of this form can cover the square very well, but how can I prove that there is a disjoint family which leaves out a small portion of the area?
-
What is this question from? I've come across it before, but I don't remember where? – mixedmath Aug 13 '11 at 6:34
@Mike: This time the radii are supposed to be of the form $1/k$, AND they are supposed to be distinct. – Jyrki Lahtonen Aug 13 '11 at 7:38
@Beni: Well, you're not alone. In French, for example, it's called "soustraction" with an s, so Frenchmen also often make this mistake in English. – Hans Lundmark Aug 13 '11 at 10:20
@user1551: I got it from an old geometry book, where the constant is $0.0001$, but I guess there is nothing special about this constant and the remaining area could be made as small as we need. – Beni Bogosel Aug 13 '11 at 14:51
It's like pulling teeth! But I found it: "Probleme de geometrie elementară", Editura Didactică şi Pedagogică, Bucharest, 1979. – TonyK Sep 3 '11 at 9:45
I don't think this is possible for general $\epsilon$, and I doubt it's possible for remainder $0.0001$.
Below are some solutions with remainder less than $0.01$. I produced them by randomized search from two different initial configurations. In the first one, I only placed the circle with curvature $2$ in the centre and tried placing the remaining circles randomly, beginning with curvature $12$; in the second one, I prepositioned pairs of circles that fit in the corners and did a deterministic search for the rest.
The data structure I used was a list of interstices, each in turn consisting of a list of circles forming the interstice (where the lines forming the boundary of the square are treated as circles with zero curvature). I went through the circles in order of curvature and for each circle tried placing it snugly in each of the cusps where two circles touch in random order. If a circle didn't fit anywhere, I discarded it; if that decreased the remaining area below what was needed to get up to the target value (in this case $0.99$), I backtracked to the last decision.
I also did this without using the circle with curvature $2$. For that case I did a complete search and found no configurations with remainder less than $0.01$. Thus, if there is a better solution in that case, it must involve placing the circles in a different order. (We can always transform any solution to one where each circle is placed snugly in a cusp formed by two other circles, so testing only such positions is not a restriction; however, circles with lower curvature might sit in the cusps of circles with higher curvature, and I wouldn't have found such solutions.)
For the case including the circle with curvature $2$, the search wasn't complete (I don't think it can be done completely in this manner, without introducing further ideas), so I can't exclude that there's are significantly better configurations (even ones with in-order placement), but I'll try to describe how I came to doubt that there's much room for improvement beyond $0.01$, and particularly that this can be done for arbitrary $\epsilon$.
The reasons are both theoretical and numerical. Numerically, I found that this seems to be a typical combinatorial optimization problem: There are many local minima, and the best ones are quite close to each other. It's easy to get to $0.02$; it's relatively easy to get to $0.011$; it takes quite a bit more optimization to get to $0.01$; and beyond that practically all the solutions I found were within $0.0002$ or so of $0.01$. So a solution with $0.0001$ would have to be of a completely different kind from everything that I found.
Now of course a priori there might be some systematic solution that's hard to find by this sort of search but can be proved to exist. That might conceivably be the case for $0.0001$, but I'm pretty sure it's not the case for general $\epsilon$. To prove that it's possible to leave a remainder less than $\epsilon$ for any $\epsilon\gt0$, one might try to argue that after some initial phase it will always be possible to fit the remaining circles into the remaining space. The problem is that such an argument can't work, because we're trying to fill the rational area $1$ by discarding rational multiples of $\pi$ from the total area $\pi^3/6$, so we can't do it by discarding a finite number of circles, since $\pi$ is transcendental.
Thus we can never reach a stage where we could prove that the remaining circles will exactly fit, and hence every proof that proves we can beat an arbitrary $\epsilon$ would have to somehow show that the remaining circles can be divided into two infinite subsets, with one of them exactly fitting into the remaining gaps. Of course this, too, is possible in principle, but it seems rather unlikely; the problem strikes me as a typical messy combinatorial optimization problem with little regularity.
A related reason not to expect a clean solution is that in an Apollonian gasket with integer curvatures, some integers typically occur more than once. For instance, one might try to make use of the fact that the curvatures $0$, $2$, $18$ and $32$ form a quadruple that would allow us to fill an entire half-corner with a gasket of circles of integer curvature; however, in that gasket, many curvatures, for instance $98$, occur more than once, so we'd have to make exceptions for those since we're not allowed to reuse those circles. Also, if you look at the gaskets produced by $0$, $2$ and the numbers from $12$ to $23$ (which are the candidates to be placed in the corners), you'll find that the fourth number increases more rapidly than the third; that is, $0$, $2$ and $18$ lead to $32$, whereas $0$ $2$ and $19$ already lead to $(\sqrt2+\sqrt{19})^2\approx33.3$; so not only can you not place all the numbers from $12$ to $23$ into the corners (since only two of them fit together and there are only four corners), but then if you start over with $24$ (which is the next number in the gasket started by $12$), you can't even continue with the same progression, since the spacing has increased. The difference would have to be compensated by the remaining space in the corners that's not part of the gaskets with the big $2$-circle, but that's too small to pick up the slack, which makes it hard to avoid dropping several of the circles in the medium range around the thirties.
My impression from the optimization process is that we're forced to discard too much area quite early on; that is, we can't wait for some initial irregularities to settle down into some regular pattern that we can exploit. For instance, the first solution below uses all curvatures except for the following: 3 4 5 6 7 8 9 10 11 16 17 20 22 25 30 31 33 38 46 48 49 52 53 55 56 57 59 79 81 94 96 101 106 107 108 113 125 132. Already at 49 the remaining area becomes less than would be needed to fill the square. Other solutions I found differed in the details of which circles they managed to squeeze in where, but the total area always dropped below $1$ early on. Thus, it appears that it's the irregular constraints at the beginning that limit what can be achieved, and this can't be made up for by some nifty scheme extending to infinity. It might even be possible to prove by an exhaustive search that some initial set of circles can't be placed without discarding too much area. To be rigorous, this would have to take a lot more possibilities into account than my search did (since the circles could be placed in any order), but I don't see why allowing the bigger circles to be placed later on should make such a huge difference, since there's precious little wiggle room for their placement to begin with if we want to fit in most of the ones between $12$ and $23$.
So here are the solutions I found with remainder less than $0.01$. The configurations shown are both filled up to an area $\gtrsim0.99$ and have a tail of tiny circles left worth about another $0.0002$. For the first one, I checked with integer arithmetic that none of the circles overlap. (In fact I placed the circles with integer arithmetic, using floating-point arithmetic to find an approximation of the position and a single iteration of Newton's method in integer arithmetic to correct it.)
The first configuration has $10783$ circles and was found using repeated randomized search starting with only the circle of curvature $2$ placed; I think I ran something like $100$ separate trials to find this one, and something like $1$ in $50$ of them found a solution with remainder below $0.01$; each trial took a couple of seconds on a MacBook Pro.
The second configuration has $17182$ circles and was found by initially placing pairs of circles with curvatures $(12,23)$, $(13,21)$, $(14,19)$ and $(15,18)$ touching each other in the corners and tweaking their exact positions by hand; the tweaking brought a gain of something like $0.0005$, which brought the remainder down below $0.01$. The search for the remaining circles was carried out deterministically, in that I always tried first to place a circle into the cusps formed by the smaller circles and the boundary lines; this was to keep as much contiguous space as possible available in the cusps between the big circle and the boundary lines.
I also tried placing pairs of circles with curvatures $(13,21)$, $(14,19)$, $(15,18)$ and $(16,17)$ in the corners, but only got up to $0.9896$ with that.
Here are high-resolution version of the images; they're scaled down in this column, but you can open them in a new tab/window (where you might have to click on them to toggle the browser's autoscale feature) to get the full resolution.
Randomized search:
With pre-placed circles:
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That's a bit crazy. Very crazy. But very pretty, too, in a way. – mixedmath Sep 11 '11 at 18:52
This is amazing (+1)! Can you reveal some details about the implementation and the effort? – Jiri Sep 11 '11 at 19:11
+1. Wow! So the problem is (almost) correct, but it can't be solved using only math.... :) Nice work! – Beni Bogosel Sep 11 '11 at 19:21
@Jiri: done :-) – joriki Sep 11 '11 at 21:08
This would have made a great Project Euler Problem :) – Beni Bogosel Sep 12 '11 at 10:51
I suggest the following idea: Look at http://en.wikipedia.org/wiki/Descartes%27_theorem and in particular the Special cases. Show inductively that all curvatures remain integers as in http://en.wikipedia.org/wiki/Apollonian_gasket. Now place some well chosen circles into the square and maybe fill up into the corners (as filling between a corner and a circle is not handled with Decartes). All other areas, i.e. in between circles and between circles and square edges you can fill with the Decartes theorem. You can probably show analytically which radii you are going to use up. Now sum their areas.
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Although I am not the OP, I have become very interested in the problem. I toyed briefly with Descartes, and I'm having a very hard time coming up with anything particular. I was wondering if you had thought any more on the problem and had a more complete solution in mind, or if this sort of intuitive line or reasoning was all that you'd come up with so far? – mixedmath Sep 6 '11 at 3:32
No, sorry, I don't have a precise solution in mind. I just hope that if you place the initial circles correctly, you will be able to prove that all subsequent circles squeezed in between, will be distinct (maybe due to some invariant like modules arithmetics). This series of circles might be calculable in closed form and then summed. But this is just a guess. ATM I don't have much time to think further :( – Gerenuk Sep 6 '11 at 9:11
I would suspect that you could do a computer search, given that you do have a finite (and quite large) cutoff of 0.0001. That means you hopefully only have to place circles with radius up to a small multiple of 100. Start by placing the circles that are tangent to 2 or more sides by hand, then searching on the remainder. – Craig Sep 6 '11 at 15:22
I guess using computer search would be very easy. Just use Decartes and a greedy algorithm until you fill the space. Maybe its all easier, if you can proof that for whatever little space is left, you can always fit an a circle and this way the upper bound on the filling ratio is not below 1. Of course you have to keep in mind the different sizes of the circles. – Gerenuk Sep 6 '11 at 16:00
@Gerenuk: If it's very easy, why haven't you tried it? I think you might change your mind.... – TonyK Sep 7 '11 at 17:51
Let's roll up our sleeves here. Let $C_k$ denote the disk of radius $1/k$. Suppose we can cover an area of $\ge 0.9999$ using a set of non-overlapping disks inside the unit square, and let $S$ denote the set of integers $k$ such that $C_k$ is used in this cover.
Then we require
$$\sum_{k\in S}\frac{1}{k^2} \ge 0.9999/\pi \approx 0.318278$$
As the OP noted, we know that $1 \not\in S$. This leaves
$$\sum_{k\ge2}\frac{1}{k^2} \approx 0.644934$$
which gives us $0.644934 - 0.318278 = 0.326656$ 'spare capacity' to play with.
Case 1 Suppose $2 \in S$. Then the largest disk that will fit into the spaces in the corners left by $C_2$ is $C_{12}$, so we must throw $3,...,11$ out of $S$. This wastes
$$\sum_{k=3}^{11}\frac{1}{k^2}\approx0.308032$$
and we are close to using up our spare capacity: we would be left with $0.326656-0.308032=0.018624$ to play with.
Case 2 Now suppose $2 \not\in S$. Then we can fit $C_3$ and $C_4$ into the unit square, but not $C_5$. So we waste
$$\frac{1}{2^2} + \frac{1}{5^2} = 0.29$$
leaving us with $0.326656-0.29=0.036656$ to play with.
Neither of these cases fills me with confidence that this thing is doable.
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Things get a lot better after the waste from 3 to 11. I've got up to 0.989; will be posting some results shortly; but I also suspect that 0.9999 may not be doable. – joriki Sep 7 '11 at 17:16
@joriki: Can you present, please how you got to $0.989$? Thank you. – Beni Bogosel Sep 9 '11 at 19:31
@Beni: I will shortly. – joriki Sep 9 '11 at 20:19
@joriki, I think you can get to 0.9999 even in Case 1. You place $C_2$, then $C_{12}$ through $C_{15}$ in the corners. Tangent to $C_{12}$ and $C_2$ you place $C_{25}$ and $C_{24}$, then tangent to 24 and $C_2$ you place $C_{40}$ and in general you can place $2n(n-1)$ tangent to the previous one and $C_2$. You do the same in the other corners, starting with $C_{23}$ and $C_{26}$ by $C_{13}$, $C_{22}$ and $C_{27}$ by $C_{14}$ and $C_{21}$ and $C_{28}$ by $C_{15}$. Summing these series gets you pretty close to the number you're looking for, and you are left with plenty of gaps for circles. – Craig Sep 10 '11 at 3:23
@Craig: The problem isn't so much the gaps left for circles but the circles left for gaps :-) – joriki Sep 10 '11 at 6:16 | 3,982 | 15,293 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-18 | latest | en | 0.944035 |
https://www.coursehero.com/file/6413547/Quiz3a-KEY-Sp11/ | 1,516,350,331,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887832.51/warc/CC-MAIN-20180119065719-20180119085719-00580.warc.gz | 889,534,922 | 70,150 | Quiz3a_KEY_Sp11
# Quiz3a_KEY_Sp11 - A Name_KEY 18 February 2011 IE330 Spring...
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Name: ______________KEY____________________ A 1 18 February 2011 IE330 Spring 2011 Quiz #3a Test statistics 0 12 2 2 0 0 0 1 2 0 0 0 0 0 2 22 0 00 1 1 2 2 *2 1 2 0 0 1 2 0 0 0 0 1 / / (1 ) 11 2 / 1 1 p D p nS X X X np X X Z T Z Z n s n np p nn n S n S X X D X X T T T S Z Sn ss S 0 ˆˆ 1 PP
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Name: ______________KEY____________________ A 4 1. (TRUE or FALSE) Pooled standard deviation is used whenever the number of observations in each sample is the same. 2. (TRUE or FALSE) It is possible for a two-sample t-test to come to a different conclusion than a paired t-test. 3. (TRUE or FALSE) If our null hypothesis is H 0 : σ 1 = σ 2 , then a statistical test of this hypothesis involves calculating a test statistic and comparing it against critical values from an F distribution . 4. Which of the following is true about two-sample tests of mean? (Select all that are true.) a. Whenever the sample variance is small, a paired t-test should be run.
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Ask a homework question - tutors are online | 502 | 1,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-05 | latest | en | 0.772658 |
https://www.aabbarchitectes.com/construction-ideas/road-construction-cost-per-linear-foot.html | 1,632,016,780,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056656.6/warc/CC-MAIN-20210919005057-20210919035057-00709.warc.gz | 681,227,238 | 19,897 | # Road construction cost per linear foot
## What does it cost to build a mile of road?
Construct a new 4-lane highway — \$4 million to \$6 million per mile in rural and suburban areas, \$8 million to \$10 million per mile in urban areas. Construct a new 6-lane Interstate highway – about \$7 million per mile in rural areas, \$11 million or more per mile in urban areas.
## How are road estimates calculated?
The earthwork cost is calculated by estimating the number of cubic meters of common material and rock which must be moved to construct the road . The earthwork production rate is calculated as the cubic meters per hour which can be excavated and placed divided by the number of cubic meters per km to be excavated.
## How much does it cost to excavate a cubic yard?
A typical residential excavation job runs between \$1,431 and \$5,051 with an average of \$3,093. Though most companies charge anywhere from \$40 to \$150 an hour, residential jobs receive project bids. Project bids reflect cubic yards of dirt moved, anywhere from \$50 to \$200 per cubic yard.
## How much does it cost to build a 1km road in Nigeria?
According to the CSJ report, in Nigeria , constructing a kilometer of road costs between N400 million and over N1 billion. Now, here are a few examples to support the CSJ report. Interestingly, and also in 2013, a similar contract was awarded for the 1,028km Lagos-Abidjan road project.
## How much does it cost to pave 1 mile?
Every road paving plan is different — the cost of paving a road depends on where it’s located, how wide it is, and several other factors — however, a good rule of thumb is that every mile of road costs at least one million dollars to repave, taking into account labor, equipment, and pavement expenses.
You might be interested: Building construction materials list
## How much does it cost to pave a racetrack?
Wow. I’ve done some research, and generally it costs around \$100,000 per mile for two-lane pavement. Two lanes isn’t nearly wide enough for a racetrack , though, so figure \$200,000 per mile minimum in order to build a racetrack four lanes wide.
## How do you calculate earthwork?
The volume between each pair of sections is estimated by multiplying the average cut or fill area of the two sections by the distance between them. Once these volumes have been calculated for each pair of sections the total cut and fill volumes are obtained by adding them all together.
## How do you calculate excavation rate?
of days required for 10m3 excavation = 10/242.4242 = 0.04125 days. Likewise, based on the capacity of other equipment, labors etc., their cost is calculated. Contractors profit is also added to the total cost of labors and machineries. Then grand total gives the rate of excavation per 10m3 of soil excavation .
## How much does it cost to have land leveled?
Land levelling may be necessary for a house foundation, backyard, driveway, shed and similar construction projects and the cost to level land ranges between \$1,000 and \$6,000 depending on the area size and job complexity, with an average price of around \$2,500.
## How much does a excavator cost per hour?
You can expect to pay anywhere from: \$95/hour for a bobcat or small two-tonne excavator. \$105 /hour for a 6-tonne excavator. \$140/hour for the combined service of an excavator and tip truck.
## How much does it cost to excavate 1 acre?
Residential Land Clearing Cost Per Acre . In general, you will pay \$500 to \$5,600 per acre to clear land. This does not include including topsoil stripping, excavation , hauling dirt to a landfill, grading and other tasks necessary to create a finished lot.
You might be interested: Installing windows new construction
## How much does it cost to build a road in Africa?
Perhaps a rough guide would be \$8-10,000/km for an earth road and \$10-15,000/km for a gravel road inclusive of basic structures but exclusive of major structures. Periodic maintenance might be \$2-4,000/km per cycle and routine maintenance up to \$1,000 per year.
## How a concrete road is constructed?
It deals with various aspects of cement concrete road construction such as materials, proportioning, measurement, handling and mixing, subgrade and sub-base preparation, formation, joints, reinforcements, concrete mix laying, finishing, curing, and plant and equipment.
## How much does it cost to build a road in India?
“ Building a two lane highway with paved shoulder costs anywhere between Rs 6 crore to Rs 8 crore per km. The per km cost shoots up anywhere between Rs 14 crore to Rs 20 crore for a four lane highway . | 1,037 | 4,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-39 | longest | en | 0.953668 |
https://in.mathworks.com/matlabcentral/cody/problems/42689-distance-a-ball-travels-after-throwing-vertically/solutions/976196 | 1,590,959,252,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413624.48/warc/CC-MAIN-20200531182830-20200531212830-00144.warc.gz | 409,809,183 | 15,508 | Cody
# Problem 42689. Distance a ball travels after throwing vertically
Solution 976196
Submitted on 20 Sep 2016 by Ben Petschel
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
v = 20; s=2; d = 20; assert(isequal(distance(s,v),d));
2 Pass
v = 20; s=2.5; d = 21.25; assert(isequal(distance(s,v),d));
3 Pass
v = 20; s=12.5; d=40; assert(isequal(distance(s,v),d));
4 Pass
v=50; s=10; d=250; v=5; s=5; d=2.5; assert(isequal(distance(s,v),d)); | 200 | 570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-24 | latest | en | 0.639617 |
http://www.jiskha.com/display.cgi?id=1271951850 | 1,498,151,278,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319636.73/warc/CC-MAIN-20170622161445-20170622181445-00017.warc.gz | 553,648,061 | 3,732 | # physics
posted by on .
There are two set pieces in a theater sitting on platforms with wheels. One is a tree with a mass of 50 kg. It is moving at 3 m/s. The other is a fence with a mass of 30 kg. If the two set pieces have equal momentum, how fast is the fence moving?
• physics - ,
momentum each= 50*3=30*v solve for v.
• physics - ,
150 | 96 | 347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-26 | latest | en | 0.945323 |
http://iexamcenter.com/question/puzzles-interview-questions.html | 1,547,984,147,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583705737.21/warc/CC-MAIN-20190120102853-20190120124853-00553.warc.gz | 108,151,451 | 9,353 | # Puzzles Interview Questions
1.
Substitute digits for the letters to make the following relation true.
S T I L L
+ W I T H I N
--------------------
L I M I T S
Note that the leftmost letter cant be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter S, no other letter can be 3 and all other S in the puzzle must be 3?
2.
A cricket team of 11 players lined up in a straight line to have their photograph. The captain was asked to stand in the center of the line-up.
1) Bharat and Bhavin stood to the right of the captain
2) Two players stood between Bhagat and Bhairav
3) Seven players stood between Bhadrik and Bhanu
4) Bhavesh stood to the right of Bhuvan
5) Bhola and Bhumit stood either side of Bhagat
6) Bhavik and Bhumit stood to the left of the captain
7) Six players stood between Bhavin and Bhagat
8 ) Two players stood between Bhagat and Bhavik
Who is the captain? Can you tell the positions of all the palyers?
3.
Consider an n by n grid of squares. A square is said to be a neighbour of another one if it lies directly above/below or to its right/left. Thus, each square has at most four neighbours. Initially, some squares are marked. At successive clock ticks, an unmarked square marks itself if
at least two of its neighbours are marked. What is the minimum number of squares we need to mark initially so that all squares eventually get marked?
4.
If you look at a clock and the time is 3:15.
What is the angle between the hour and the minute hands? ( The answer to this is not zero!)?
5.
A rich man died. In his will, he has divided his gold coins among his 5 sons, 5 daughters and a manager.
According to his will: First give one coin to manager. 1/5th of the remaining to the elder son. Now give one coin to the manager and 1/5th of the remaining to second son and so on..... After giving coins to 5th son, divided the remaining coins among five daughters equally.
All should get full coins. Find the minimum number of coins he has?
6.
500 men are arranged in an array of 10 rows and 50 columns according to their heights.
Tallest among each row of all are asked to come out. And the shortest among them is A.
Similarly after resuming them to their original positions, the shortest among each column are asked to come out. And the tallest among them is B.
Now who is taller A or B ?
7.
A series comprising of alphabets contains 13 letters. The first seven letters in the given series are A, E, F, H, I, L, M
Can you find the next two letters?
8.
A goods can load wt only 100tones not more than that if inch more wt is added it will fall, So when the train was passing on the bridge a paper falls on it, Now will the train fall r not? if it doesnt fall why?
9.
In the middle of the confounded desert, there is the lost city of "Ash". To reach it, I will have to travel overland by foot from the coast. On a trek like this, each person can only carry enough rations for five days and the farthest we can travel in one day is 30 miles. Also, the city is 120 miles from the starting point.
What I am trying to figure out is the fewest number of persons, including myself, that I will need in our Group so that I can reach the city, stay overnight, and then return to the coast without running out of supplies.
How many persons (including myself) will I need to accomplish this mission?
10.
Which number in the series does not fit in the given series:
1 4 3 16 6 36 7 64 9 100?
11.
There are 10 statements written on a piece of paper:
1. At least one of statements 9 and 10 is true.
2. This either is the first true or the first false statement.
3. There are three consecutive statements, which are false.
4. The difference between the numbers of the last true and the first true statement divides the number, that is to be found.
5. The sum of the numbers of the true statements is the number, that is to be found.
6. This is not the last true statement.
7. The number of each true statement divides the number, that is to be found.
8. The number that is to be found is the percentage of true statements.
9. The number of divisors of the number, that is to be found, (apart from 1 and itself) is greater than the sum of the numbers of the true statements.
10. There are no three consecutive true statements.
Find the minimal possible number?
12.
Substitute digits for the letters to make the following Division true
O U T
-------------
S T E M | D E M I S E
| D M O C
-------------
T U I S
S T E M
----------
Z Z Z E
Z U M M
--------
I S T
Note that the leftmost letter cant be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3?
13.
What are the next three numbers in the given series?
1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 ?
Interview Questions
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https://frank.itlab.us/datamodel/node27.html | 1,686,324,508,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656737.96/warc/CC-MAIN-20230609132648-20230609162648-00649.warc.gz | 293,783,013 | 4,571 | Prototyping with tools
Next: Integrating Factors Up: Methods for Solving ODEs Previous: Graphical Solutions: Phase Plane Index
Click for printer friendely version of this HowTo
## Separation of Variables
How do we solve Equation 2.4.1? That is, how do we determine what Equation 2.4.1 is the derivative of? First, we note that we can separate the two variables, and , by multiplication. That is,
Integrating both sides produces
(2.10.4)
where is the combination of the two integration constants. Using each side as an exponent, we have
where .
This method can be used to solve both linear as well as nonlinear ordinary differential equations. Example 2.10.2.1 gives an example of a solution to a quadratic nonlinear ODE and Appendix D.2 shows how to use separation of variables to solve a cubic nonlinear ODE.
We will begin with a quadratic ODE that is often used to model population growth (birth and immigration) and decay (death and emigration).
(2.10.5)
Equation 2.10.5 can be solved using the method of seperation of variables. We begin by separating from by multiplcation. That is,
The integral on the right-hand side can be easily solved once it is broken down into simpler components. This can be done using the method of partial fraction decomposition. That is,
and and , thus,
Substituting Equation 2.10.7 for the fraction in on the left side of Equation 2.10.6 gives us the following:
Next: Integrating Factors Up: Methods for Solving ODEs Previous: Graphical Solutions: Phase Plane Index
Click for printer friendely version of this HowTo
Frank Starmer 2004-05-19
> | 382 | 1,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-23 | latest | en | 0.854795 |
http://ulissesaraujo.wordpress.com/2008/01/10/point-free-over-non-recursive-functions-i/ | 1,419,288,033,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802777118.12/warc/CC-MAIN-20141217075257-00059-ip-10-231-17-201.ec2.internal.warc.gz | 284,131,356 | 25,451 | 10 01 2008
## Intro
Like I promise I will talk about point-free over non recursive functions.
Point-free style is used to demonstrate propriety over programs easily than pointwise.
Because we didn’t have variables in functions, we can compose a lot of small functions with each other using some combinators.
Those combinators have some rules associate with them that tell us when to use some combinator.Also because I think that’s the easy way to learn point-free programming style. I will talk about that style over non recursive functions.
Because of the length of what I intend to explain I will talk about these theme in three more post’s. Consider this part I.
In the following parts I will talk about:
• Co-Products (Part II)
• Constants and conditionals (Part III)
• Pointfree – Pointwise conversion & Program calculation (Part IV)
At the time I finish the other posts I will put their link here.
## Part I
In this post I will talk about products of functions and some of their rules.
I will use $D_{\huge{E}}F$ that means ‘def’ from ‘definition’. I can’t use ‘def’ in this LaTeX package. And I confess that didn’t investigate a lot about that, in matter of fact I like the output of $D_{\huge{E}}F$.
I will also use Commutative diagrams to explain the rules, Because I think that is the easy way to get it.
## Combinators
The combinator $\noindent (\odot)$ is the fundamental for point-free style.
$\noindent \{D_{\huge{E}}F-\odot\}:$
$\noindent (\odot)~:~(b~\rightarrow~c)~\rightarrow~(a~\rightarrow~b)~\rightarrow~(a~\rightarrow~c)$
$(f~\odot~g)~x~=~f~(g~x)$
In fact $\noindent (\odot)$ is the fundamental combinator in Mathematical analysis.
Another very important function is the $\noindent id$ function, the identity.
$\noindent \{D_{\huge{E}}F-id\}:$
$\noindent id~:a~\rightarrow~a$
$id~x~=~x$
$\noindent (\odot)$ and $\noindent id$ respect the following laws:
$\noindent \{assoc-\odot\}:$
$\noindent (f \odot g) \odot h~=~f \odot (g \odot h)$
$\noindent \{nat-id\}:$
$\noindent f \odot id~=~f~=~id \odot f$
With just those laws we are ready to demonstrate programs, like:
Let A, B, C, D be Types and f :: A -> B, g :: B -> C and h :: C -> D
let’s prove that
$\noindent (h \odot g) \odot (id \odot (f \odot id)) = h \odot (g \odot f)$
To demonstrate this equality we pick one of the sides of the equation, and with $\noindent \{nat-id\}$ and $\noindent \{assoc-\odot\}$, we try to obtain the other side.
In this case, we can do:
$\noindent (h \odot g) \odot (id \odot (f \odot id)) = h \odot (g \odot f)$
$\Leftrightarrow \{assoc-\odot\}$
$((h \odot g) \odot id) \odot (f \odot id) = h \odot (g \odot f)$
$\Leftrightarrow \{nat-id\}$
$(h \odot g) \odot (f \odot id) = h \odot (g \odot f)$
$\Leftrightarrow \{assoc-\odot\}$
$((h \odot g) \odot f) \odot id = h \odot (g \odot f)$
$\Leftrightarrow \{nat-id\}$
$(h \odot g) \odot f = h \odot (g \odot f)$
$\Leftrightarrow \{assoc-\odot\}$
$h \odot (g \odot f) = h \odot (g \odot f)$
$\Leftrightarrow TRUE$
1) In fact as a convenience we can take off the brakets in a sequence os compositions. And so is pointless the use of $\noindent \{assoc-\odot\}$ law.
2) We can also use more than one rule per line.
So, the previous demonstation becames:
$\noindent (h \odot g) \odot (id \odot (f \odot id)) = h \odot (g \odot f)$
$\Leftrightarrow \{1,1\}$
$h \odot g \odot id \odot f \odot id = h \odot g \odot f$
$\Leftrightarrow \{nat-id,nat-id\}$
$h \odot g \odot f = h \odot g \odot f$
$\Leftrightarrow TRUE$
## split combinator
In any programming language we always have two general types to aggregate data, struct’s and unions.
Now we will see two combinators that work with those data types.
The $(\Delta)$ is often called ‘split’. This is the combinator responsible to take care of struct’s (tuples in Haskell).
It definition is:
$\noindent \{D_{\huge{E}}F-\Delta\}:$
$(\Delta)~:~(c~\rightarrow~a)~\rightarrow~(c~\rightarrow~b)~\rightarrow~(c~\rightarrow~a \times b)$
$(f~\Delta~g)~x~=~(f~x,g~x)$
split :: (c -> a) -> (c -> b) -> c -> (a,b)
split f g x = (f x, g x)
This combinator allow, as well, combine functions with the same domain.
With the functions ‘fst’ and ‘snd’ we can access the content of a tuple.
fst (a,b) = a
snd (a,b) = b
Let A, B, C, D be Types and f : A -> B, g : A -> C and we want the function split : A -> B x C
$\xymatrix @!=4pc {& A \ar[d]_{f \Delta g} \ar[dl]_{f} \ar[dr]_{g} & \\B & \ar[l]_{fst} B \times C\ar[r]_{snd} &C}$
Like diagram is commutative we can infer two rules:
$\noindent \{univ-\times\}:$
$split = f \Delta g \Leftrightarrow fst \odot split = f \wedge snd \odot split = g$
$\noindent \{cancel-\times\}:$
$fst \odot (f \Delta g) = f$
$snd \odot (f \Delta g) = g$
Now, if type A = B x C, we will have $f \Delta g : B \times C \rightarrow B \times C$
To have a commutative diagram $f = fst$ and $g = snd$
$\xymatrix @!=4pc {& B\times C \ar[d]_{fst \Delta snd} \ar[dl]_{fst} \ar[dr]_{snd} & \\B & \ar[l]_{fst} B \times C \ar[r]_{snd}& C}$
And we can infer the rule:
$\noindent \{reflex-\times\}:$
$fst \Delta snd = id_{B\times C}$
Let A, B, C, E be Types and h : A -> B, f : B -> C and g : B -> E.
$\xymatrix @!=4pc {& A \ar[ddl]_{f \odot h} \ar[d]_{h} \ar[ddr]_{g \odot h} & \\& \ar[dl]_{f} \ar[d]_{f\Delta g} B\ar[dr]_{g} & \\C & \ar[l]_{fst} D \ar[r]_{snd} & E\\}$
$\noindent \{fusion-\times\}:$
$(f \Delta g) \odot h = f \odot h \Delta g \odot h$
## Example – isomorphism demonstration
As example we can demonstrate isomorphisms.
Here is a brief description:
To prove the isomorphism $a \cong b$ we have to find two functions:
$f : a \rightarrow b$ and $g : b \rightarrow a$ such that $f \odot g = id \wedge g \odot f = id$
To do that we can use the diagrams or doscover the pointwise version and them converte to point-free, I will show later on the last techique.
Now we gonna use the diagrams to show the isomorphism $a \times (b \times c) \cong c \times (b \times a)$.
$\xymatrix @!=4pc {& a\times (b\times c) \ar[d]_{h} \ar[dl]_{snd \odot snd} \ar[dr]_{g} & \\c & \ar[l]_{fst} c\times(b\times a) \ar[r]_{snd} & b\times a}$
Discovering g:
$\xymatrix @!=4pc {& a\times (b\times c) \ar[d]_{g} \ar[dl]_{fst \odot snd} \ar[dr]_{fst} & \\b & \ar[l]_{fst} b\times a \ar[r]_{snd} & a}$
So, the definition of the function h is:
$h : a \times(b \times c) \rightarrow c \times (b \times a)$
$h = snd \odot snd \Delta (fst \odot snd \Delta fst)$
$\xymatrix @!=4pc {& c\times (b\times a) \ar[d]_{h^{\bullet}} \ar[dl]_{snd \odot snd} \ar[dr]_{g^{\bullet}} & \\a & \ar[l]_{fst} a \times(b \times c) \ar[r]_{snd} & b \times c}$
Discovering $g^{\bullet}$
$\xymatrix @!=4pc {& c\times (b\times a) \ar[d]_{g^{\bullet}} \ar[dl]_{fst \odot snd} \ar[dr]_{fst} & \\b & \ar[l]_{fst} b\times c \ar[r]_{snd} & c}$
So, the definition of the function $h^{\bullet}$ is:
$h^{\bullet} : c \times (b \times a) \rightarrow a \times(b \times c)$
$h^{\bullet} = snd \odot snd \Delta (fst \odot snd \Delta fst)$
And we conclude $h \odot h^{\bullet} = id = h^{\bullet} \odot h$, so we prove $a \times (b \times c) \cong c \times (b \times a)$
## $\bigotimes$ combinator
When we want to defin a function that takes a product and goes to a product we may use the function $(\bigotimes)$
$\noindent \{D_{\huge{E}}F-\bigotimes\}:$
$\noindent f \bigotimes g~=~f \odot fst \Delta g \odot snd$
infixr 5 > b) -> (c -> d) -> (a, c) -> (b, d)
(f >< g) x = split (f . fst) (g . snd)
here is the diagram of it definition:
$\xymatrix @!=4pc {a \ar[d]_{f} & \ar[l]_{fst} \ar[dl]_{f \odot fst} a\times b \ar[d]_{f \bigotimes g} \ar[dr]_{g \odot snd} \ar[r]_{snd}& b \ar[d]_{g}\\c & \ar[l]_{fst}c\times d \ar[r]_{snd}& d}$
The folowing diagram,
$\xymatrix @!=4pc {& \ar[dl]_{h} A \ar[d]_{h \Delta i} \ar[dr]_{i}&\\B \ar[d]_{f} & \ar[l]_{fst} \ar[d]_{f \bigotimes g} \ar[r]_{snd} C & \ar[d]_{g} D\\E & \ar[l]_{fst} F \ar[r]_{snd} & G}$
coud be simplified into this one:
$\xymatrix @!=4pc {& \ar[dl]_{f \odot h} A \ar[d]_{f \odot h \Delta g \odot i} \ar[dr]_{g \odot i}& \\E & \ar[l]_{fst} F \ar[r]_{snd} & G}$
And so, we find another rule:
$\noindent \{abssoc-\times\}:$
$(f \bigotimes g) \odot (h \Delta i) = f \odot h \Delta g \odot i$
The folowing diagram,
$\xymatrix @!=4pc {D \ar[d]_{h} & \ar[l]_{fst} A \ar[d]_{h \bigotimes i} \ar[r]_{snd} & G \ar[d]_{i}\\E \ar[d]_{f} & \ar[l]_{fst} B \ar[d]_{f \bigotimes g} \ar[r]_{snd} & H\ar[d]_{g}\\F & \ar[l]_{fst} C \ar[r]_{snd} & I}$ coud be simplified into this one:
$\xymatrix @!=4pc {D \ar[d]_{f \odot h} & \ar[l]_{fst} A \ar[d]_{f \odot h \bigotimes g \odot i} \ar[r]_{snd} & \ar[d]_{g \odot i} G\\F & \ar[l]_{fst} C \ar[r]_{snd} & I}And so, we find another rule$:
$\noindent \{functor-\times\}:$
$(f \bigotimes g) \odot (h \times i) = f \odot h \times g \odot i$
## Notes
The $\odot$ combinator have more precedence than the $\Delta$ combinator.
So, $f \odot h \Delta g \odot h = (f \odot h) \Delta (g \odot h)$
## Aknowledgements
I would like to thank dysfunctor to the corrections made to this post.
### 4 responses
12 03 2008
21 04 2008
Thanks! Very educational.
I noticed a mistake in the statement of the fusion law. It should say:
(f /\ g) . h = f /\ h . g /\ h
21 04 2008
Oops! There’s a mistake in my correction. How embarrassing.
(f /\ g) . h == f . h /\ g . h
Also, I think it would be clearer if you added some brackets:
(f /\ g) . h == (f . h) /\ (g . h)
23 04 2008
Hello dysfunctor,
Thank you for your comment and Thank you for your corrections. Now the post is correct ;)
Best regards. | 3,507 | 9,497 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 79, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2014-52 | latest | en | 0.795992 |
https://computethought.blog/2022/12/23/variables-and-constants/ | 1,675,505,335,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00364.warc.gz | 185,386,180 | 27,807 | # Variables and Constants
One of the challenges of teaching a weekly class has been ensuring that enduring understandings remain between lessons. The middle school computing offerings at ISA are aimed at providing students a supportive computational bridge between elementary and high school years. This has meant that we are constantly looking for ways to integrate computing in other subject areas so that students not only see purpose but also build more robust solutions using a computational setup. One critical piece of this evolving curriculum has been the introduction and management of variables and constants in programming to allow a wide range of processing to happen. Unlike the IGCSE and IB programs where students work mainly with text based languages (JavaScript / Python / Java in the past), the middle years offering is aimed to be an incremental blend between block and text based coding to help them transition.
This post, hence, is about documenting a couple of projects we have built / are building for students to master this ability in their solutions.
### The focus on user choice
One of the first terms we establish in the course is user. This is defined as someone who may find your program useful / entertaining / informative / purposeful. Every new concept then covered is keeping this user in mind. Allowing this user choice in how they wish to interact with the program has helped students build in different interactive elements which then has lead to learning about variables and constants.
Example 1
Grade 8 students recently completed a project aimed at children learning basic math. Their solution was built on CoSpaces and used basic variables to hold answers to randomly generated mathematical tasks. Each student interpreted the task in their own way and here are some examples of their design approaches.
Here, the student used a maze like approach where the an animal or bird is guarding the access. You need to answer the question they pose by clicking on the controls (+ and – buttons on the side). The correct answer makes the animal/bird make way thus allowing you to move forward or to the next question.
Variables were used to hold the randomly generated numbers and the value the user is choosing using the + and – controls.
They usage of ” ” quotes was then established as an indicator of a constant since anything placed with them does not change when the program runs.
You may try out this program here.
Example 2
Grade 7 computing designed story prompt generators to better understand variables and constants. This project required them to think of the solution as a face – the front end with choices for the user and the brain – the code logic that would read/store user choices to allow for a wide range of outputs to be processed. Given below are a couple of good student examples.
In these projects the focus remained on the program’s ability to create as many variations of the story prompts as possible. They read the user choice into appropriately named variables and then created various selection statements to produce a range of possible prompts. Given below is one such example. Students were required to design the story prompts before coding them so that the flow of the code was more manageable. You may try these samples here and here.
### Building in purpose
Students have engaged well with work which had interdisciplinary links. We have recently completed various other design and computational projects with subjects like Media Studies and Visual Arts that have allowed students to tap into knowledge from other subjects to build computing models. As we go forward next term, the aim is to continue this overlap so that fundamental concepts such as variables in programming can find a stronger and deeper understanding with student work. I hope to revisit this again in a few months to document their creations! | 711 | 3,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-06 | longest | en | 0.958866 |
https://link.springer.com/chapter/10.1007/978-3-642-03064-2_1 | 1,601,596,447,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402132335.99/warc/CC-MAIN-20201001210429-20201002000429-00002.warc.gz | 467,070,905 | 13,988 | # On Commutative Algebra
• Maria Chlouveraki
Chapter
Part of the Lecture Notes in Mathematics book series (LNM, volume 1981)
## Abstract
The first chapter contains some known facts and some novel results on Commutative Algebra which are crucial for the proofs of the results of Chapters 3 and 4. The former are presented here without their proofs (with the exception of Theorem 8) for the convenience of the reader. In the first section of this chapter, we define the localization of a ring and give some main properties. The second section is dedicated to integrally closed rings. We study particular cases of integrally closed rings, such as valuation rings, discrete valuation rings and Krull rings. We use their properties in order to obtain results on Laurent polynomial rings over integrally closed rings. We state briefly some results on the completions of rings in Section 1.3. In the fourth section, we introduce the notion of morphisms associated with monomials. They are morphisms which allow us to pass from a Laurent polynomial ring A in m+1 indeterminates to a Laurent polynomial ring B in m indeterminates, while mapping a specific monomial to 1. Moreover, we prove (Proposition 15) that every surjective morphism from A to B which maps each indeterminate to a monomial is associated with a monomial. We call adapted morphisms the compositions of morphisms associated with monomials. They play a key role in the proof of the main results of Chapters 3 and 4. Finally, in the last section of the first chapter, we give a criterion (Theorem 10) for a polynomial to be irreducible in a Laurent polynomial ring with coefficients in a field. | 362 | 1,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-40 | latest | en | 0.932295 |
http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv021.cgi?read=246853 | 1,547,929,445,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583681597.51/warc/CC-MAIN-20190119201117-20190119223117-00260.warc.gz | 313,789,083 | 3,270 | The Museum of HP Calculators
HP Forum Archive 21
Prime program examplesMessage #1 Posted by Gilles Carpentier on 22 July 2013, 5:38 p.m. ```EXPORT Ulam BEGIN LOCAL a,b,xy:={160,120}, n:=1, m:=.9, d:={{1,0}, {0,1}, {-1,0}, {0,-1}}; RECT(); WHILE n<100000 DO FOR a FROM 1 TO 4 DO m:=m+.5; FOR b FROM 1 TO m DO IF CAS.isprime(n) THEN PIXON_P(xy,127);END; xy:=xy+d(a); n:=n+1; END; END; END; FREEZE; END; //------------- EXPORT Box(n) BEGIN FOR A:=1 TO n DO RECT_P(FLOOR(RANDOM(320)),FLOOR(RANDOM(260)),FLOOR(RANDOM(320)),FLOOR(RANDOM(260)),0,FLOOR(RANDOM(16581375))); END; FREEZE; END //------------------------- LPiece:={100,50,20,10,5,2,1} ; EXPORT Monnaie(Reste, NPiece) BEGIN IF Reste==0 THEN RETURN 1; END; IF Reste<0 OR NPiece==0 THEN RETURN 0; END; RETURN Monnaie(Reste-LPiece(NPiece),NPiece)+Monnaie(Reste,NPiece-1); END; ``` 'Ulam' draws the Ulam spiral http://en.wikipedia.org/wiki/Ulam_spiral 'Box(x)' draws x colored box on the screen 'Monnaie' returns the number of ways to get a total. For example Monnaie(100,7) return 4563 : That means there is 4563 ways to get 1 euro (100 cents) with the 7 coins describe in the global variable LPiece Edited: 22 July 2013, 6:00 p.m.
Re: Prime program examplesMessage #2 Posted by Cristian Arezzini on 22 July 2013, 6:58 p.m.,in response to message #1 by Gilles Carpentier The program Ulam checks fine, but when I try to run it it says "Ulam Unknowntype"... Do I need to run it in a special way?
Re: Prime program examplesMessage #3 Posted by Gilles Carpentier on 23 July 2013, 1:59 a.m.,in response to message #2 by Cristian Arezzini Works fine for me... What happens if you debug ? You must create a file with the name you want, then copy/paste the code. Then call Ulam from Home view Edited: 23 July 2013, 2:01 a.m.
Re: Prime program examplesMessage #4 Posted by Cristian Arezzini on 23 July 2013, 4:18 a.m.,in response to message #3 by Gilles Carpentier That's exactly how I entered it, by copying/pasting from the above post. By the way, this worked fine with Tim's Mandelbrot program. When I debug, I can step until the IF instruction, then at the next STEP it gives the error message. then: The Box program instead works fine.
Re: Prime program examplesMessage #5 Posted by Gilles Carpentier on 23 July 2013, 6:03 p.m.,in response to message #4 by Cristian Arezzini In old releases 'isprime' was IsPrime or ISPRIME...
Re: Prime program examplesMessage #6 Posted by Joe Horn on 24 July 2013, 12:57 a.m.,in response to message #5 by Gilles Carpentier Quote: In old releases 'isprime' was IsPrime or ISPRIME... and isPrime.
Re: Prime program examplesMessage #7 Posted by Cristian Arezzini on 24 July 2013, 4:39 a.m.,in response to message #6 by Joe Horn OK, now it doesn't give any error, but it doesn't draw anything - I get a white screen, and after a while, the box of "program finished". Is there some other command that has changed? On a related note: will we be allowed to check more recent versions of the ROM, with the emulator? It's a little frustrating to find bugs or problems that have been long since solved! :)
Re: Prime program examplesMessage #8 Posted by fhub on 24 July 2013, 8:29 a.m.,in response to message #7 by Cristian Arezzini Quote: On a related note: will we be allowed to check more recent versions of the ROM, with the emulator? It's a little frustrating to find bugs or problems that have been long since solved! :) I agree, it doesn't make much sense to make bugreports for a completely outdated version - that's why I already stopped testing this old HP-Prime 'Prerelease 1'. :-( Franz
Re: Prime program examplesMessage #9 Posted by Gilles Carpentier on 24 July 2013, 10:52 a.m.,in response to message #7 by Cristian Arezzini Quote: Is there some other command that has changed? Color code... PIXON_P(xy,127) try another color code than 127 (0 for example) Edited: 24 July 2013, 10:54 a.m.
Re: Prime program examplesMessage #10 Posted by Cristian Arezzini on 24 July 2013, 5:16 p.m.,in response to message #9 by Gilles Carpentier Thanks Gilles, but... still no luck. White screen. I tried with various values. Maybe I should really just wait for a newer release! :)
Re: Prime program examplesMessage #11 Posted by Gilles Carpentier on 24 July 2013, 5:55 p.m.,in response to message #10 by Cristian Arezzini I think it's the best ... I believe it's about the isprime syntax. isprime(n) may return 1 if true, 0 if false. it seems it always returns 0 in your program with this release
Go back to the main exhibit hall | 1,305 | 4,521 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-04 | latest | en | 0.687242 |
http://mathhelpforum.com/calculus/174826-tangent-hyperplanes-tanget-planes-bit-confused-about-dimensions-print.html | 1,527,046,754,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865411.56/warc/CC-MAIN-20180523024534-20180523044534-00351.warc.gz | 181,779,556 | 4,712 | # tangent hyperplanes and tanget planes, a bit confused about dimensions
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• Mar 16th 2011, 08:35 PM
Volga
tangent hyperplanes and tanget planes, a bit confused about dimensions
I thought I understood tangent planes until I saw tangent hyperplane and 3-d contour here. So a little bit of help will be most welcome. (i) and (ii) is no problem and agreed to the textbook answer. I am a bit stuck on (iii).
Question (Calculus, Binmore and Davies, 3.9(#5*))
Let $\displaystyle f(x,y,z)=ln(x^2+y^2+z^2)$
i. Find the gradient of f at the point $\displaystyle (x,y,z)^T$
ii. Find the tangent hyperplane to the hypersurface $\displaystyle u=ln(x^2+y^2+z^2)$
where $\displaystyle (x,y,z,u)^T=(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.
iii. Find the normal and the tangent plane to the contour
$\displaystyle ln(x^2+y^2+z^2)=0$ at $\displaystyle (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.
Answer.
i. $\displaystyle \nabla{f}=\frac{1}{x^2+y^2+z^2}(2x,2y,2z)^T$
ii. To find tangent hyperplane, I want to use the formula
$\displaystyle u-U=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)+(z-Z)f_z(x,y,z)$
$\displaystyle u-0=(x-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(y-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(z-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}$
$\displaystyle u=\frac{2}{\sqrt{3}}(x+y+z)-2$
iii. To me, the equation for the countour $\displaystyle ln(x^2+y^2+z^2)=0$ looks exactly the same as the equation to the surface in (ii) as u=0 in both cases.
I understand in (ii) the hypersurface is in 4 dimentional space, and the hyperplane is in 3 dimensions. Here the contour is in 3 dimensions (??) and the tangent plane is 2-dimentional.
???
I am not sure what to do, but I was thinking may be try to express z as the function of x and y and apply the same template as in (ii) ie
$\displaystyle z-Z=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)$
???
• Mar 17th 2011, 02:08 AM
Volga
Quote:
I am not sure what to do, but I was thinking may be try to express z as the function of x and y and apply the same template as in (ii)
Actually, had a better idea: the contour is f(x,y,z)=0 so the tangent plane to the countour is
$\displaystyle (x-X)f_x+(y-Y)f_y+(z-Z)f_z=0$
so that I get $\displaystyle \frac{2}{\sqrt{3}}(x+y+z)-2=0$ and $\displaystyle x+y+z=\sqrt{3}$ as per the answer in the book.
Now I need to figure out the way to find the normal vector. BUT it's probably the coefficients of x, y and z in the above equation, right? So, the normal vector is $\displaystyle (1,1,1)^T$.
Any comments are still welcome!
• Mar 17th 2011, 03:57 AM
HallsofIvy
Quote:
Originally Posted by Volga
I thought I understood tangent planes until I saw tangent hyperplane and 3-d contour here. So a little bit of help will be most welcome. (i) and (ii) is no problem and agreed to the textbook answer. I am a bit stuck on (iii).
Question (Calculus, Binmore and Davies, 3.9(#5*))
Let $\displaystyle f(x,y,z)=ln(x^2+y^2+z^2)$
i. Find the gradient of f at the point $\displaystyle (x,y,z)^T$
ii. Find the tangent hyperplane to the hypersurface $\displaystyle u=ln(x^2+y^2+z^2)$
where $\displaystyle (x,y,z,u)^T=(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.
iii. Find the normal and the tangent plane to the contour
$\displaystyle ln(x^2+y^2+z^2)=0$ at $\displaystyle (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.
Answer.
i. $\displaystyle \nabla{f}=\frac{1}{x^2+y^2+z^2}(2x,2y,2z)^T$
ii. To find tangent hyperplane, I want to use the formula
$\displaystyle u-U=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)+(z-Z)f_z(x,y,z)$
$\displaystyle u-0=(x-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(y-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(z-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}$
$\displaystyle u=\frac{2}{\sqrt{3}}(x+y+z)-2$
iii. To me, the equation for the countour $\displaystyle ln(x^2+y^2+z^2)=0$ looks exactly the same as the equation to the surface in (ii) as u=0 in both cases.
No, they are not the same thing though the difference is subtle. Consider the lower dimension case of $\displaystyle z= x^2+ y^2- 1$. That describes a two dimensional surface in a three dimensional space. It has a tangent plane and a normal vector at any given point and, in particular at (1, 0, 0). There, the normal vector is given by $\displaystyle 2\vec{i}- \vec{k}$ and the tangent plane is given by $\displaystyle \frac{x- 1}{2}- z= 0$ or $\displaystyle z= \frac{1}{2}x- \frac{1}{2}$. We can get that normal vector by thinking of the surface $\displaystyle z= x^2+ y^2- 1$ as a "level surface" of the function $\displaystyle \phi(x, y, z)= x^2+ y^2- z= 1$ and taking the gradient: $\displaystyle \nabla \phi= 2x\vec{i}+ 2y\vec{j}- \vec{k}$, which, at (1, 0, 1), is $\displaystyle 2\vec{i}- \vec{k}$.
But the graph of $\displaystyle x^2+ y^2- 1= 0$ is a circle in the xy-plane. it is, in fact, the intersection of the xy-plane with the paraboloid $\displaystyle z= x^2+ y^2- 1$. But because it is in the xy-plane, any normal to it at a given point, say (1, 0) must be in the xy-plane which was not true of the normal vector to the parabloid at (1, 0, 0). Now, instead of thinking of the surface z= x^2+ y^2- 1[/tex] as a level surface of the function $$\phi(x, y, z)= x^2+ y^2- z[tex], we think of it as a as a "level curve" of \displaystyle z(x,y)= x^2+ y^2- 1. The (two-dimensional) gradient of that is \displaystyle \nabla z= 2x\vec{i}+ 2y\vec{j} which, at (1, 0), is \displaystyle 2\vec{i}. A tangent line would be x= 1. Quote: I understand in (ii) the hypersurface is in 4 dimentional space, and the hyperplane is in 3 dimensions. Here the contour is in 3 dimensions (??) and the tangent plane is 2-dimentional. ??? I am not sure what to do, but I was thinking may be try to express z as the function of x and y and apply the same template as in (ii) ie \displaystyle z-Z=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z) ??? • Mar 17th 2011, 05:38 AM Volga Quote: Now, instead of thinking of the surface z= x^2+ y^2- 1$$ as a level surface of the function $\displaystyle \phi(x, y, z)= x^2+ y^2- z$, we think of it as a as a "level curve" of $\displaystyle z=x^2+y^2-1$.
So, in other words, we take a 'slice' of it, to go one dimension lower, by fixing z as a constant, and hence we find ourselvers in xy-plane? So that the grad vector 'loses' z-derivative because z is no longer variable?
Thanks for your explanation! By the way, as a practice, if I want to calculate normal vector in my example of hypersurface in (ii), I think it will be 4x1 vector as follows:
$\displaystyle u=(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{ \sqrt{3}},-1)(x-\frac{1}{\sqrt{3}},y-\frac{1}{\sqrt{3}},z-\frac{1}{\sqrt{3}},u)^T$ thus giving the normal vector of
$\displaystyle (\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{\s qrt{3}},-1)^T$?
Question: does the size (length) of the 'normal' vector matter? if what is important is the direction only (eg such normal vector can define a plane), then I can say the vector is $\displaystyle k*(\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{ \sqrt{3}},-1)^T$ where k>0? Or the size does matter?
• Mar 17th 2011, 05:47 AM
HallsofIvy
No. The point of a "normal vector" is that it is perpendicular to the surface. Only its direction is relevant. Any multiple of a normal vector is still a normal vector. Sometimes, you might be instructed to find the unit normal vector specifically to avoid any depence upon length. Of course, there are still two unit normal vectors to a surface, in opposite directions.
There are special cases in which the length of a normal vector might be important. For example, if you are integrating a vector field over a surface, you would want the length of the normal vector to reflect the "differential of surface area". You can get that by using the "fundamental vector product" for the surface. | 2,541 | 7,687 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-22 | latest | en | 0.790949 |
https://www.halfbakery.com/idea/Hexagon_20Weave | 1,638,857,373,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00303.warc.gz | 863,650,910 | 27,448 | h a l f b a k e r y
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# Hexagon Weave
On weaving only an array of hexagons
(+15, -1) [vote for, against]
Quite a few years ago there was some furor in the tennis world regarding tennis racquets. New ones had entered the market, including a larger-than-normal size, and various folks thought the sport was being fouled.
However, they apparently were legal according to the rules of the game. While I haven't paid much attention since, to see if they have locked down some racquet-variety rules, to prevent similar problems in the future, I did happen to notice that the main reason the new racquets had been introduced was because of something known as "the sweet spot", the best part of the racquet for controlling a tennis ball.
A larger sweet spot means a better chance to control the ball, provided you actually make the racquet meet the ball while playing the game.
Well, an Idea came to me: What If The Strings Of The Racquet Formed An Array Of Hexagons Instead Of Squares? See, hexagons are more naturally circular than squares, so it might logically follow that such an array would fit/grip the curved ball-surface better, and therefore equate with a larger sweet spot.
The only problem is, every intersection of a grid of hexagons has exactly 3 lines connecting to it. A grid of squares has 4 lines connecting at each intersection, and a grid of triangles (I THINK I saw a racquet once that was strung with a grid of triangles) has 6 lines connecting at each intersection.
It seems physically impossible, at first glance, to weave a grid of hexagons. After all if every string that enters an intersection must exit again, then you have to have an even total number of lines at each intersection, and a hexagon grid still has only 3.
Note that there are ways to cheat. See the first link, which actually yields a grid of triangles and hexagons.
Nevertheless, I thought of a way, and no, it does not involve big fat knots at every intersection. What you do is visualize the grid of hexagons as having two parallel strings at every line where you would normally see just one string. This means that each intersection can have 3 strings going into it (half of each parallel pair), plus those same 3 strings going out (the other half of each parallel pair).
All that remains to be done is to figure out how to lace them AT the intersection. And, no, it STILL is not a knot! See the second link, showing 3 connected rings. If you imagined that the outside of each ring was cut, and they were made of string, you could pull their connection point tightly together, and they would stay well-connected. You could also pull on the strings such that they would be approaching that intersection at 120- degree angles, exactly like the intersection of 3 hexagons.
So all you have to do, to make a Hexagon Weave, is to ensure that EVERY intersection connects like that one. Which turns out to be slightly more complicated. The parallel strings have to cross each other allong every one of the straight-side lengths of the hexagon array.
Nevertheless, It Appears To Be Workable In Theory. When the strings are all pulled taut, a true grid of hexagons should be produced. Perhaps a tennis racquet featuring a Hexagon Weave can be made yet! Note that ordinary weavings don't pull all the strings taut, and so something like a "doily" would look more like a mass of tangled strings than a nice hexagon pattern. Still, there are a few possibilities, after the example of "fish- net stockings". Since said netting is under tension when worn, its threads are taut. Which means "chicken-wire stockings" should be quite possible....
I've created an image (3rd link). You may have to load it into an image viewer and enlarge it to see the intersection details. But you should be able to see that the green strings zigzag mostly horizontally, always crossing on top of the blue strings; the red strings zigzag from upper-right to lower-left, always crossing on top of the green strings, and the blue strings zigzag from upper-left to lower-right, always crossing on top of the red strings.
— Vernon, Oct 31 2011
"Hexagon weave" https://geometricol.../08/tesselation.jpg
As mentioned in the main text, this is a "cheat", since actually you get a mixture of triangles and hexagons. [Vernon, Oct 31 2011, last modified Jul 16 2015]
Borromean Rings http://en.wikipedia...iki/Borromean_rings
As mentioned in the main text, this (first image on the right) is how the middles of three strings can lock together at an intersection. Note that the image in the "Hexagon weave" link just below uses the same three colors, which overlap each other the same way as here (described in last paragraph of main text). [Vernon, Oct 31 2011, last modified Nov 01 2011]
Hexagon Weave http://www.nemitz.n...on/HexagonWeave.png
You should be able to see some of the details, as described in the main text. [Vernon, Oct 31 2011, last modified Nov 01 2011]
Chicken wire http://en.wikipedia.org/wiki/Chicken_wire
Braak! [pocmloc, Oct 31 2011]
Hexagonal weave http://www.absolute...cs/Hexagonal_tiling
[xaviergisz, Nov 01 2011]
Chair Cane Weave http://www.chairsca...ep_strand_cane.html
Octagonal holes from hillbillies. [baconbrain, Nov 01 2011]
3D hexagonal Kagome weave structures http://www.ifam-dd....img/Drahtgitter.jsp
[sqeaketh the wheel, Nov 02 2011]
HexWeave http://www.nemitz.net/vernon/HexWeave.PNG
A "baked" image, as mentioned in an annotation. It occurs to me that there appears to be a visual pun here, involving the idea of an infinite mesh.... [Vernon, Nov 14 2011, last modified Nov 15 2011]
Racquet Rules http://www.livestro...nnis-racquet-rules/
To the extent that that page is accurate, a hexagon weave should be legal, if surrounded by an acceptable frame (the frame of the test-model I made is too physically weak, I think, for actual use in tennis). [Vernon, Nov 21 2011]
Tennis Ball impacting Racquet http://cellar.org/pictures/tennis.jpg
I've been looking for a picture that better shows the strings during this event. But even in this image it should be clear enough that the strings are NOT currently going exactly straight across the racquet body. [Vernon, Nov 21 2011]
Distorted racquet strings http://www.tennisin...ng_tension_giv.html
This article has an image that nicely shows some degree of string-distortion when a ball impacts an ordinary racquet [Vernon, Jul 16 2015]
Simple hexagon grid https://encrypted-t...iHBbsL3jIN73iaIww5_
Mentioned in the 5th paragraph of the main text. [Vernon, Jul 16 2015]
Triangular grid, with hexagons data:image/png;base...asAAAAASUVORK5CYII=
Like the simple hexagon grid, there are still 3 strings at each intersection point. [Vernon, Jul 16 2015]
A string-tension study http://www.ncbi.nlm...gov/pubmed/16195027
Lesser string tension seems associated with higher ball speeds and better control. That could be a "plus" for this Idea, since it may not be easy to get high string tension. [Vernon, Jul 16 2015]
Twist analysis http://s1199.photob...Weave2.png.html?o=0
Balanced left- and right twists [neutrinos_shadow, Jul 16 2015]
[+] for raising the question of weaving hexagons, and all the topological wonderment thereof.
Does the tennis rule-book say that there have to be strings? Why not a single-piece mesh, stamped from a sheet of some guttish material?
Alternatively, start with a tall rectangle of material and make rows of vertical slits, suitably offset in adjacent rows. Then, when the material is stretched sideways, it will form a hexagonal grid.
— MaxwellBuchanan, Oct 31 2011
Or! Make the tennis balls superconducting, and place powerful magnets around the rim of the otherwise empty tennis bat. The ball will try to exclude the magnetic flux-lines, leading to a stringless-string effect. The use of electromagnets would allow the bat to be tuned.
— MaxwellBuchanan, Oct 31 2011
To a perfect pitch I suppose?
— pocmloc, Oct 31 2011
I think in tennis it's called "serving"; you're probably thinking of golf.
— MaxwellBuchanan, Oct 31 2011
I never think of golf if I can help it.
— pocmloc, Oct 31 2011
//See, [hexagons] are more naturally circular than squares, so it might logically follow that such an array would fit/grip the curved ball-surface better, and therefore equate with a larger sweet spot.//
Sweet spot is due to the node of vibration of the strings on a raquet (accoring to a quick internet search). You could argue that strings with 3-way symmetry (hexagons) gives a bigger/better node of vibration than 2-way symmetry, but I'll leave that speculation to someone else.
If you've invented a new type of weave, great, you don't need to find a purpose for the new type of weave.
I still can't see your illustration (and I can't figure it out from the text alone), but how is this different from regular hexagonal weave? (see link)
— xaviergisz, Nov 01 2011
My canoe has what is called woven-cane seats (also known as cane-bottom). The weave starts at right angles to make squares, then goes across diagonally to make octagons.
There are lots of octagonal openings, which look circular. And which provide fair grip for wet nylon shorts. That should serve for a racquet. (Link.) (It works for snowshoes, too.)
The hexagon weave and the chicken wire are not stable or solid. There are no direct lines running out to the frame, and there are no internal triangles. As a racquet weave, it's going to be mushy as heck.
Back when I was building chicken pens, we could stretch and flex the chicken wire like crazy, and it would always end up sagging. I assume that tennis racqueteers want the opposite of that.
— baconbrain, Nov 01 2011
Chicken wire can be held taut and flat inside a complete frame; it only tends to sag and stretch if it is attached on 2 or 3 sides only. Similarly, a hexagonal net would be fine in a tennis racquet.
— spidermother, Nov 01 2011
Folks, as I write this, the 3rd-link image described in the main text is available for viewing. It has a background-grid of hexagons, with the weaving pattern covering part of it. The hexagons aren't so obvious amidst the weave. If the strings were thinner or the hexagons larger, the pattern would be more obvious.
[xaviergisz], the first weave that I saw at your link uses really thick "strings", and the size of the hexagons is strictly related to that. The weave I'm proposing can allow arbitrarily large hexagons, regardless of string-size.
And the second weave I saw there creates a mixture of hexagons and triangles. This weave yields hexagons only.
— Vernon, Nov 01 2011
Are you talking about the Hexagon Weave at nemitz.net? The light purple background? http://www.nemitz.net/verno n/HexagonWeave.png?
Because that is just like chicken wire, as far as I can tell.
I'm still saying it isn't going to be rigid. I'll go look for something to test or illustrate.
Meanwhile, I'd take your hexagon weaving loom in the lapweaving link, and add another set of pegs to make another set of triangles, weave that all together to make hexagons, and forget binding the layers together.
Standard tennis rackets don't interlock the strings in any way, they are just woven.
If you really want to bind it all together, do what the snowshoers do, and varnish the strings after assembly.
— baconbrain, Nov 01 2011
[baconbrain], yes, that's the image. But if you studied how chicken wire is made, you would see that this must be different. Chicken wire makes use of the fact that steel tends to maintain a bent shape. So, certain lengths of the middle parts of two parallel strands are bent around each other multiple times, forming one side of that hexagon. Two sides of each hexagon are made that way, but the other four sides of each hexagon are single strands of steel.
Here, EVERY side of every hexagon involves pairs of strings, and they are not wrapped round-and-round each other. There is a single half-twist that allows them to enter (in the drawing) each intersection neatly. The interlacing at the intersections is what holds this weave together.
And, yes, I know that tennis racquets involve straight strings only, and that this pattern could be problematic for that use, unless the strings were VERY strong and VERY taut. I don't know that it is impossible to achieve.
— Vernon, Nov 01 2011
Well, after close scrutiny, I'm fairly sure that Vernon's weave can produce a flat, rigid net, as long as the terminus of every strand is rigidly held in the frame. It has no degrees of freedom; if the strings don't change in length, a node can only move if an adjacent node moves; by an informal use of mathematical induction, it appears to me that no node can move unless a string changes length or one of the attachment points slips.
— spidermother, Nov 01 2011
Thanks, [spidermother]. As I wrote in the main text: "When the strings are all pulled taut, a true grid of hexagons should be produced." Its rigidity depends on the strings being taut.
It should be possible to construct this weave in a frame using just one long long string (which is actually normally done in tennis-racquet frames). After the string crosses the framed gap, it doubles back to cross again in a different place, over and over again.
Also, the first two sets of crossings (say, the horizontal crossing and the upper-right-to-lower-left crossing) are easy, since all the strings of the second set are always "above" the strings of the first set (they don't interweave like in a normal square-pattern tennis racquet).
But that third set of crossing strings (upper-left-to-lower-right) is a nightmare, unless you employ a template holding the first two crossings in zigzags. And even then it will be tedious; all the intersection-lacings must be done exactly right.
If you actually do this with just one long long string, the final nightmare will be getting all the slack out EVENLY, after all the weaving is done. Good Luck! --which probably explains why this Idea is Half-Baked! Although if the Hexagon Weave is used to make something like "chickenwire stockings", then the strings become taut when a leg is inserted, and they use stretchy string to make sure it becomes taut. Possibly much more Bake-able than a tennis racquet....
— Vernon, Nov 01 2011
Hexagonal Japanese Kagome basket weave is long known The link shows pictures of 3D hexagonal Kagome weave structures for other purposes. [+]
— sqeaketh the wheel, Nov 02 2011
//Thanks, [spidermother].// No worries. I just tells it like I sees it.
If you want to try this, I suggest making a hexagonal grid of nails driven part way into a board. Tie the strings to the outside nails. Weave one colour at a time, pulling the strings uniformly tight if possible, and tie off the loose end. Once all the strings are in place, you should be able to pull out the nails (except those around the edge).
You probably want string with a little stretch, to take up the small amount of slack generated as each nail is removed. Ordinary knitting wool might be good. I'd be interested to know how it works.
— spidermother, Nov 02 2011
If a hexagon is good, an octagon should be better as it even more closely approaches a circle. This opens the potential for an arms race among tennis players as each new fabulously strung racquet offers polygons with more and more sides and consequent tennis superiority.
— bungston, Nov 02 2011
[squeaketh], like chicken wire, those Kagome structures appear to depend on the fact that metal can usually hold its shape after being bent. And the basket weave yields hexagons that are very restricted in size, related to the diameter of the strands of material being woven. The strands themselves occupy enough space as to distort the overall grid (it's not pure hexagons; it includes triangles of overlapping strands, too). This Idea features a different weave that yields hexagons that are not size-restricted (so the strands don't have to occupy so much of the overall space of the grid).
[bungston], sorry, the main text explains what I thought was true about tennis-racquet sweet spots at the time of the furor, when I originally thought of this Idea. It appears that the sweet spot depends on a different factor. So, it remains to be determined whether or not a racquet strung with this Hexagon Weave will have a larger sweet spot.
— Vernon, Nov 02 2011
I'm guessing that the "sweet spot" is just the addition of the tensions of the strings... so it shouldn't be that difficult to figure out if it gets bigger or smaller (or remains the same).
— FlyingToaster, Nov 02 2011
I'm guessing that the "sweet spot" is where the mesh isn't distorted by edge effects. If that is so, the interlocking strands of this hex weave would be a good thing.
— baconbrain, Nov 02 2011
how about non-regular patterns, like the quasicrystals the recent nobel prize winner discovered
I know it isn't feasible, just a funny idea/discussion to me.
— EdwinBakery, Nov 02 2011
Folks, I've spent some time constructing an actual hexagon weave. Since I was thinking in terms of tennis racquets, I got some 50-pound-test monofilament nylon fishing line and planned for an average width of the zigzagging lines to be 1/2 inch (a little more than a centimeter).
The lid of a 5-gallon plastic bucket got a big circle cut out of it, and a lot of small holes were drilled around the edge of the big hole.
Since the circular cut-out was the right size, and made of easily scratch-able plastic, I marked it with lots of lines in three directions, and nailed about 400 "wire brads" into appropriate intersections, creating a basic hexagon grid. The plastic alone wasn't strong enough to hold the nails solidly, but a piece of plywood behind it worked fine.
As previously indicated, the first two "layers" of fishing line were easily zigzagged among the nails, since the second layer was placed everywhere on top of the first. The third layer was not quite as nightmarish as I originally thought it would be, because the actual intersection is a bit simpler than portrayed in the "Borromean Rings" link.
It still took a couple of weeks of spare time to get it done, using one long long continuous strand for all three layers of zigzags and intersections.
I did what I could to make sure all the strands were pulled taut, but when the nails were removed, some slack was introduced. Nevertheless, the weave is holding its shape quite well, despite a bit of flexibility. (While the overall construct is about the size of a tennis racquet "head" --there is no handle-- the strands currently are nowhere near taut enough for the thing to be used for that game.)
Anyway, the finished thing was placed on a flatbed scanner. I had to manually press the mesh onto the glass to get an image that was in focus (would not have been able to do that if the slack hadn't been there!). The scanned image is over 4 megabytes in size, so I clipped out a portion of it for posting on the Web ("hexweave" link).
The scanned image is about 3 times larger than life-size. This means the thickness of the strands is also about 3 times life-size. The hexagon pattern is NOT QUITE as obvious as it would be if the mesh had been made wider than 1/2-inch, or if I had used thinner fishing line. But it is obvious enough to prove that it can work.
— Vernon, Nov 14 2011
That's beautiful, [Vernon]. How well does it serve?
(Square) tennis racket weave has no angles to speak of, at least none that rub. You've a plethora of 60deg angles just waiting to wear and break... maybe a badminton racket.
Considered lacrosse ?
Good job, by the way; I know who I'll be going to for chainmail.
— FlyingToaster, Nov 15 2011
Bun just for the conviction and dedication to see this through.
— AusCan531, Nov 15 2011
[Flying Toaster], I was wondering when someone would mention the issue of strands rubbing against each other. The angles are more like 120 degrees, though. At least when the mesh-hole-size is larger (maybe 90 degrees in the one I constructed), or the strands thinner. On the other hand, either of those things would make it less suitable for use in tennis.
The best answer to the rubbing problem may be the first anno on Nov 1 by [spidermother]. That is, the more taut the strands, the less "degrees of freedom" they will have in which to move. And obviously, the less they can move the less they can rub.
Now, I've seen the high-speed photos of the impact of a tennis ball with a racquet, and know full well that even very taut straight strands will either stretch or otherwise exhibit some freedom to move under that circumstance. Logically, however, it means that even a standard square mesh will have strands rubbing each other. Which means the wear-and-tear problem isn't actually quite as great a problem as you might think it is.
— Vernon, Nov 15 2011
I agree, [Vernon]. Rubbing won't be a big problem with this weave.
The strings will be weakened where they cross, more than in the case of a square weave due to the sharper angle, but again, not enough to matter.
A tennis racquet strung like this would, I predict, work perfectly well. It would be a little softer for a given string tension, and the forces would be closer to radially symmetrical, which would change the response somewhat - quite likely for the better!
— spidermother, Nov 15 2011
[Vernon], that's beautiful. [+]
— baconbrain, Nov 15 2011
Nicely done, [Vernon].
If I may make a suggestion; alternate the twist at each successive junction, so you get a "left over" then a "right over" around each hex. If they are all the same, it may impart a spin on the ball, and possible a twist to the frame too. By alternating, I think it will "sit" more neatly.
<Later> I don't think it is actually possible to do what I've suggested... Damn 3-way symmetry!
<Even later> Haha, figured it out. See linky. Equal numbers of left twists and right twists (even if, at first glance, it doesn't seem so).
[neutrinos_shadow], please look at the 3rd link of the group, to clearly see which strings cross which other strings. I suspect the crossings you suggest will unravel the weave. One thing I found out, when first contemplating the weave, is that there are really only two ways for the strings to cross at each hexagon- intersection.
That is, consider the red and blue stings in the linked image mentioned above. At the intersection, the red can cross the blue, or the blue can cross the red --and the other two crossings, of green-and-red or green-and- blue, are forced. So, only two ways to do the intersection.
The entire weave uses only one of the two different intersection-crossings; trying to mix them doesn't work because it leads to the equivalent of "ring mail", a whole lot of closed loops, instead of cross-fabric strings.
I invite you to draw some red, green, and blue lines of your own, to attempt to devise the alternate crossings you would like to see.
— Vernon, Jul 16 2015
One thing I didn't really think about when I constructed that hexagon weave using 50-pound nylon line, was that if every side of every hexagon basically consists of 2 stings, then a lesser-strength line would probably be fine, for a tennis racquet. The intersections would be a bit tighter and the strings would be bent slightly less, passing through each intersection.
— Vernon, Jul 17 2015
[annotate]
back: main index | 5,592 | 23,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-49 | latest | en | 0.970679 |
https://proofwiki.org/wiki/Sine_to_Power_of_Even_Integer | 1,620,987,813,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00299.warc.gz | 480,859,976 | 11,124 | # Sine to Power of Even Integer
## Theorem
$\ds \sin^{2 n} \theta$ $=$ $\ds \frac 1 {2^{2 n} } \binom {2 n} n + \frac {\paren {-1}^n} {2^{2 n - 1} } \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \binom {2 n} k \map \cos {2 n - 2 k} \theta$ $\ds$ $=$ $\ds \frac 1 {2^{2 n} } \binom {2 n} n + \frac {\paren {-1}^n} {2^{2 n - 1} } \paren {\cos 2 n \theta - \binom {2 n} 1 \map \cos {2 n - 2} \theta + \cdots + \paren {-1}^{n - 1} \binom {2 n} {n - 1} \cos 2 \theta}$
## Proof
First, by Sine Exponential Formulation we have:
$\sin \theta = \dfrac 1 {2 i} \paren {e^{i \theta} - e^{-i \theta} }$
Therefore by Power of Product:
$\sin^{2 n} \theta = \dfrac 1 {\paren {2 i}^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$
Now by Power of Product and the Power of Power:
$\dfrac 1 {\paren {2 i}^{2 n} } = \dfrac 1 {2^{2 n} \paren {-1}^n} = \dfrac {\paren {-1}^n} {2^{2 n} }$
Thus:
$\sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$
Now:
$\ds \paren {e^{i \theta} - e^{-i \theta} }^{2 n}$ $=$ $\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k e^{i \paren {2 n - k} \theta} e^{-i k \theta}$ Binomial Theorem $\ds$ $=$ $\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k e^{i \paren {2 n - 2 k}\theta}$ Power of Product $\ds$ $=$ $\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta$ Euler's Formula
So we have:
$\displaystyle \sin^{2 n} \theta = \dfrac {\paren {-1}^n} {2^{2 n} } \paren {\sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta + i \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \sin {2 n - 2 k} \theta}$
Now we look at each of the terms in the parentheses:
$\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \map \sin {2 n - 2 k} \theta$ $=$ $\ds \paren {-1}^n \sum_{k \mathop = -n}^n {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta}$ replacing $k \mapsto k + n$ $\ds$ $=$ $\ds \paren {-1}^n \sum_{k \mathop = -n}^{-1} {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta} + \paren {-1}^n \sum_{k \mathop = 1}^n {2 n \choose n + k} \paren {-1}^k \map \sin {-2 k \theta}$ Zeroes of Sine and Cosine $\ds$ $=$ $\ds \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \sin 2 k \theta + {2 n \choose n + k} \map \sin {-2 k \theta} }$ $\ds$ $=$ $\ds \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n + k} \sin 2 k \theta + {2 n \choose n + k} \map \sin {-2 k \theta} }$ Symmetry Rule for Binomial Coefficients $\ds$ $=$ $\ds \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n + k} \sin 2 k \theta - {2 n \choose n + k} \sin 2 k \theta}$ Sine Function is Odd $\ds$ $=$ $\ds 0$
We find, for the remaining term:
$\ds \sum_{k \mathop = 0}^{2 n} {2 n \choose k} \paren {-1}^k \map \cos {2 n - 2 k} \theta$ $=$ $\ds \paren {-1}^n \sum_{k \mathop = -n}^n {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta}$ replacing $k \mapsto k + n$ $\ds$ $=$ $\ds \paren {-1}^n \sum_{k \mathop = -n}^{-1} {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta} + \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n {2 n \choose n + k} \paren {-1}^k \map \cos {-2 k \theta}$ Cosine of Zero is One $\ds$ $=$ $\ds \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \cos 2 k \theta + {2 n \choose n + k} \map \cos {-2 k \theta} }$ $\ds$ $=$ $\ds \paren {-1}^n {2 n \choose n} + \paren {-1}^n \sum_{k \mathop = 1}^n \paren {-1}^k \paren { {2 n \choose n - k} \cos 2 k \theta + {2 n \choose n + k} \cos 2 k \theta}$ Cosine Function is Even $\ds$ $=$ $\ds \paren {-1}^n {2 n \choose n} + \paren {-1}^n 2 \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta$ Symmetry Rule for Binomial Coefficients
Thus we have:
$\ds \sin^{2 n} \theta$ $=$ $\ds \frac {\paren {-1}^n} {2^{2 n} } \paren {\paren {-1}^n {2 n \choose n} + \paren {-1}^n 2 \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta}$ $\ds$ $=$ $\ds \frac 1 {2^{2 n} } {2 n \choose n} + \frac 1 {2^{2 n - 1} } \sum_{k \mathop = 1}^n \paren {-1}^k {2 n \choose n - k} \cos 2 k \theta$ $\ds$ $=$ $\ds \frac 1 {2^{2 n} } {2 n \choose n} + \frac {\paren {-1}^n} {2^{2 n - 1} } \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k {2 n \choose k} \map \cos {2 n - 2 k} \theta$ replacing $k \mapsto n-k$
as required.
$\blacksquare$ | 2,058 | 4,531 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-21 | latest | en | 0.278726 |
http://adib-hasan.net/blog/category/maths/ | 1,632,252,443,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00097.warc.gz | 1,832,021 | 21,728 | # The Maths Behind Logistic Regression
#### What are Your Odds?
Let’s have a look at this god-tier math puzzle. Only one out of seven gets it right, and the other six don’t. So, what are the odds for solving it correctly? 1 to 6. Generally, if $$p$$ is the probability that someone will get it right, then his/her odds are $$p/(1-p)$$.
However, it isn’t necessarily true that $$p=1/7$$ for every person because some people are smarter, some have better education and so on. Hence, $$p$$ also depends on the person attempting the puzzle. In a Bayesian framework, we capture this dependence with conditional probability.
# Prime Counting Function and Chebyshev Bounds
The distribution of primes plays a central role in number theory. The famous mathematician Gauss had conjectured that the number of primes between $$1$$ and $$n$$ is roughly $$n/\log n$$. This estimation gets more and more accurate as $$n\to \infty$$. We use $$\pi(n)$$ to denote the number of primes between $$1$$ and $$n$$. So, mathematically, Gauss’s conjecture is equivalent to the claim
$\lim_{n\to\infty}\frac{\pi(n)}{n/\log n}=1$
# Multiplying Two Polynomials with Fast Fourier Transform
Polynomial multiplication is one of the most important problems in mathematics and computer science. It can be formally defined as follows:
You are given two polynomials roughly of equal size$A(x)=a_0+a_1x+\dots+a_{n-1}x^{n-1}$$B(x)=b_0+b_1x+\dots+b_{n-1}x^{n-1}$find a polynomial $C(x)=c_0+c_1x+\dots+c_{2n-2}x^{2n-2}$ such that $$A(x)\cdot B(x)=C(x)$$. At first, this may look like an easy problem, Continue reading Multiplying Two Polynomials with Fast Fourier Transform
# Generating All Subsets of Size k in Python
Suppose you are given a set $$S$$ with $$n$$ elements and you need to generate every subset of size $$k$$ from it. For instance, if $$S=\{3,4,5\}$$ and $$k=2$$, then the answer would be $$\{3,4\}$$, $$\{3,5\}$$, $$\{4,5\}$$. So, how would you do that in Python?
First of all, this is a really simple exercise. Stuff like this often comes up when someone writes a moderately large script. I wanna talk about this one in particular because it has a combinatorial solution. Continue reading Generating All Subsets of Size k in Python
# A Special Case of Zsigmondy’s Theorem
Theorem (Zsigsmondy):Â
1. For two coprime positive integers $$a$$ and $$b$$ and for any positive integer $$n$$, $$a^n-b^n$$ has a prime divisor that does not divide $$a^k-b^k$$ for all positive integers $$k< n$$ with the following exceptions:
• $$n = 2$$ and $$a+b$$ is a power of two.
• $$n=6, a=2, b=1$$.
2. For two coprime positive integers $$a$$ and $$b$$ and for any positive integer $$n$$, $$a^n+b^n$$ has a prime divisor that does not divide $$a^k+b^k$$ for all positive integers $$k< n$$ with the following exception:
• $$n=3, a=2, b=1$$.
This theorem is very helpful in solving many olympiad number theory problems. However, its use is often frowned upon, as this theorem is quite hard to prove using only elementary mathematics. (I am aware of the proof using Cyclotomic Polynomials, but that’s not “elementary enough” in my opinion.)
In olympiad mathematics, one can get away in most of the cases by just using this theorem for a fixed pair of $$n$$ and $$k$$. This modification allows us to provide a very simple proof using LTE. Here I’ll restate this particular case, and then prove the first part. The second part can be proven analogously.
Theorem (Zsigsmondy Special Case):Â
1. For two coprime positive integers $$a$$ and $$b$$ and any two positive integers $$n$$ and $$k$$ with $$k<n$$, $$a^n-b^n$$ has a prime divisor that does not divide $$a^k-b^k$$ with the following exception:
• $$n = 2, k=1$$ and $$a+b$$ is a power of two.
2. For two coprime positive integers $$a$$ and $$b$$ and any two positive integers $$n$$ and $$k$$ with $$k<n$$, $$a^n+b^n$$ has a prime divisor that does not divide $$a^k+b^k$$ with the following exception:
• $$n=3, k=1,a=2, b=1$$.
Proof of 1:
Suppose $$a^n-b^n$$ and $$a^k-b^k$$ share same set of prime divisors. This implies $$a^n-b^n$$ and $$a^{\gcd(n,k)}-b^{\gcd(n,k)}$$ also share same set of prime divisors. Assuming $$A=a^{\gcd(n,k)}$$ and $$B=b^{\gcd(n,k)}$$, we could follow the rest of this argument and get to a contradiction. So, without loss of generality, we may assume $$k=1$$.
Now we consider two cases:
Case 1: $$n$$ is a power of $$2$$.
If $$n=2$$ and $$a+b$$ is a power of two, we arrive at one of the listed exceptions. Now note that,
\begin{align*}&\gcd(a+b, a^2+b^2)\\ =&\gcd(a+b, (a+b)^2-a^2-b^2)\\ =&\gcd(a+b,2ab)\\ =&\gcd(a+b,2)\end{align*}
This implies both $$a+b$$ and $$a^2+b^2$$ can’t be powers of two unless $$a+b=2\implies a=b=1$$. Hence at least one of them must have an odd prime divisor. Furthermore, this prime divisor does not divide $$a-b$$ since $$\gcd(a-b, a^2+b^2)=\gcd(a-b,2)$$ (following the previous steps). However, if $$n>2$$, then $$a^2+b^2|a^n-b^n$$, implying that odd prime will divide $$a^n-b^n$$. This is a contradiction.
Case 2: $$n=2^md$$ with $$d>1$$ being odd.
Without loss of generality, we may assume $$n>1$$ is odd, since $$a^d-b^d|a^n-b^n$$ and it is sufficient to show that $$a^d-b^d$$ has a prime divisor that does not divide $$a-b$$.  From LTE, $$v_p(a-b)+v_p(n)=v_p(a^n-b^n)$$ for each odd prime $$p|a-b$$. Furthermore, $\frac{a^n-b^n}{a-b}\equiv na^{n-1}\equiv 1\pmod 2$ implying $$v_2(a-b)=v_2(a^n-b^n)$$. Therefore, we can conclude $\frac{a^n-b^n}{a-b}=\prod_{p|\gcd(n,a-b)}p^{v_p(n)}\leq n$ which is impossible for $$n>1$$. This raises another contradiction and we are done.
# āĻāĻ¨ā§āĻāĻŋāĻāĻžāĻ° āĻĢā§āĻ¯āĻžāĻā§āĻā§āĻ°āĻžāĻāĻā§āĻļāĻ¨: āĻĒāĻ°ā§āĻŦ ā§§ āĻĒā§āĻ˛āĻžāĻ°ā§āĻĄā§āĻ° āĻ°ā§ (rho) āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ
āĻāĻ¤ āĻĒāĻ°ā§āĻŦā§ āĻāĻŽāĻ°āĻž āĻāĻžā§ā§āĻ° āĻā§āĻ°ā§ āĻāĻŋāĻā§ āĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻā§āĻĒāĻžāĻĻāĻ āĻŦā§āĻ° āĻāĻ°āĻžāĻ° āĻā§āĻˇā§āĻāĻž āĻāĻ°ā§āĻāĻŋāĻ˛āĻžāĻŽāĨ¤ āĻĻā§āĻāĻāĻāĻ¨āĻāĻāĻžāĻŦā§ āĻāĻŽāĻžāĻĻā§āĻ° āĻŦāĻžāĻā§āĻāĻž āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽāĻāĻŋ $$20$$ āĻĄāĻŋāĻāĻŋāĻā§āĻ° āĻāĻāĻāĻŋ āĻ¸āĻāĻā§āĻ¯āĻžāĻā§ āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻāĻ°āĻ¤ā§ āĻāĻŋā§ā§ āĻā§āĻ˛āĻžāĻ¨ā§āĻ¤ āĻšā§ā§ āĻāĻŋā§ā§āĻāĻŋāĻ˛āĨ¤ $$50$$ āĻŦāĻž $$100$$ āĻĄāĻŋāĻāĻŋāĻā§āĻ° āĻā§āĻ¨ āĻ¸āĻāĻā§āĻ¯āĻžāĻā§ āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻāĻ°āĻ¤ā§ āĻšāĻ˛ā§ āĻ¸ā§āĻ āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§āĻ° āĻā§ āĻ
āĻŦāĻ¸ā§āĻĨāĻž āĻšāĻŦā§ āĻāĻžāĻŦ āĻāĻāĻŦāĻžāĻ°!
āĻŽāĻžāĻ¨ā§āĻˇ āĻ¯āĻāĻ¨ āĻĻā§āĻāĻ˛ āĻŦā§ āĻŦā§ āĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻāĻ¨ā§āĻ¯ $$2$$ āĻĨā§āĻā§ $$N-1$$ [āĻāĻŋāĻāĻŦāĻž $$\sqrt N$$] āĻĒāĻ°ā§āĻ¯āĻ¨ā§āĻ¤ āĻ¸āĻŦ āĻ¸āĻāĻā§āĻ¯āĻž āĻĻāĻŋā§ā§ $$N$$-āĻā§ āĻāĻŽāĻ¨ āĻāĻžāĻ āĻāĻ°āĻž āĻ
āĻ¨ā§āĻ āĻ¸āĻŽā§āĻ¸āĻžāĻĒā§āĻā§āĻˇ, āĻ¤āĻāĻ¨ āĻ¤āĻžāĻ°āĻž āĻā§āĻ°āĻŋāĻ āĻā§āĻāĻāĻ¤ā§ āĻ˛āĻžāĻāĻ˛ āĻā§ āĻāĻ°ā§ āĻāĻ°āĻ āĻāĻŽ āĻ¸āĻāĻā§āĻ¯āĻ āĻ¸āĻāĻā§āĻ¯āĻž āĻĻāĻŋā§ā§ āĻāĻžāĻ āĻāĻ°ā§ $$N$$-āĻāĻ° āĻā§āĻĒāĻžāĻĻāĻ āĻŦā§āĻ° āĻāĻ°āĻž āĻ¯āĻžā§āĨ¤ āĻāĻŽāĻ¨ āĻ
āĻ¨ā§āĻāĻā§āĻ˛ā§ āĻā§āĻ°āĻŋāĻ āĻ°ā§ā§āĻā§, āĻ¯āĻžāĻ° āĻŽāĻžāĻā§ āĻāĻāĻāĻŋ āĻšāĻ˛ āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ āĻ¸āĻŽāĻ¸ā§āĻ¯āĻž(āĻĒāĻ°āĻŋāĻļāĻŋāĻˇā§āĻā§ āĻĻā§āĻ)āĨ¤ āĻāĻā§ āĻāĻžāĻā§ āĻ˛āĻžāĻāĻŋā§ā§ āĻāĻ¨ā§āĻŽ āĻšā§ā§āĻā§ āĻāĻāĻāĻŋ āĻāĻžāĻ˛āĻžāĻāĻāĻ¤ā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽā§āĻ°āĨ¤
## āĻĒā§āĻ˛āĻžāĻ°ā§āĻĄā§āĻ° $$\rho$$ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ
āĻŽāĻ¨ā§ āĻāĻ°, āĻāĻŽāĻžāĻĻā§āĻ°āĻā§ $$N=55$$-āĻā§ āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻāĻ°āĻ¤ā§ āĻšāĻŦā§āĨ¤ (āĻāĻŽāĻ°āĻž āĻ¯āĻĻāĻŋāĻ āĻāĻžāĻ¨āĻŋ āĻ¯ā§ $$55$$ āĻāĻ° āĻāĻāĻāĻŋ āĻā§āĻĒāĻžāĻĻāĻ $$p=5$$, āĻ¤āĻŦā§āĻ āĻāĻŽāĻ°āĻž āĻ¸ā§āĻāĻŋ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ āĻŦā§āĻ° āĻāĻ°āĻŦāĨ¤)
āĻāĻŽāĻ°āĻž āĻāĻ°āĻŦ āĻāĻŋ, āĻ°âā§āĻ¯āĻžāĻ¨ā§āĻĄāĻŽ āĻĒā§āĻ°ā§āĻŖāĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻāĻāĻāĻŋ āĻ
āĻ¸ā§āĻŽ āĻ§āĻžāĻ°āĻž (infinite sequence) $$a_0, a_1, a_2, a_3, a_4,\ldots$$ [āĻ¸āĻāĻā§āĻˇā§āĻĒā§ $$\{a_i\}_{i = 0}^{\infty}$$] āĻĨā§āĻā§ āĻā§ā§āĻžāĻ˛ āĻā§āĻļāĻŋāĻŽāĻ¤ āĻāĻŋāĻā§ āĻĒāĻĻ, āĻ¯ā§āĻŽāĻ¨: $$a_0, a_3, a_7, a_{100}$$, āĻ¤ā§āĻ˛ā§ āĻ¨ā§āĻŦāĨ¤ āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ āĻ¸āĻŽāĻ¸ā§āĻ¯āĻž āĻ
āĻ¨ā§āĻ¯āĻžā§ā§, āĻāĻ āĻāĻžāĻ°āĻāĻŋ āĻĒāĻĻā§āĻ° āĻŽāĻžāĻā§ $$\pmod {5}$$-āĻ āĻāĻ¨āĻā§āĻ°ā§ā§ā§āĻ¨ā§āĻ āĻ
āĻ¨ā§āĻ¤āĻ¤ āĻĻā§āĻāĻŋ āĻĒāĻĻ āĻĒāĻžāĻā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž$1-e^{-\frac{4^2}{2\times 5}}\approx 80\%$āĻāĻŽāĻ¨ āĻĻā§āĻāĻŋ āĻĒāĻĻ $$a_m$$ āĻāĻŦāĻ $$a_n$$ āĻ¯āĻĻāĻŋ āĻ¸āĻ¤ā§āĻ¯āĻŋ āĻ¸āĻ¤ā§āĻ¯āĻŋ āĻĨāĻžāĻā§, āĻ¤āĻžāĻšāĻ˛ā§ $a_m\equiv a_n\pmod{5}$$\implies a_m-a_n\equiv 0\pmod{5}$$\implies d = \gcd(55, a_m-a_n)\ge 5$ āĻ
āĻ°ā§āĻĨāĻžā§, āĻāĻŽāĻ°āĻž $$55$$-āĻāĻ° āĻāĻāĻāĻŋ āĻā§āĻĒāĻžāĻĻāĻ $$d$$ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°ā§ āĻĢā§āĻ˛āĻŦāĨ¤ āĻ
āĻŦāĻļā§āĻ¯ āĻā§āĻ āĻĒā§āĻ°āĻļā§āĻ¨ āĻāĻ°āĻ¤ā§āĻ āĻĒāĻžāĻ°ā§ $$d = 55$$ āĻšāĻ˛ā§? āĻšā§āĻ¯āĻžāĻ, āĻ¸ā§āĻāĻž āĻšāĻ¤ā§ āĻĒāĻžāĻ°ā§āĨ¤ āĻāĻŋāĻ¨ā§āĻ¤ā§ āĻāĻžāĻ°āĻāĻŋ āĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻŽāĻžāĻā§ $$\pmod{55}$$-āĻ āĻāĻ¨āĻā§āĻ°ā§ā§ā§āĻ¨ā§āĻ āĻĻā§āĻāĻŋ āĻ¸āĻāĻā§āĻ¯āĻž āĻĒāĻžāĻā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž āĻŽāĻžāĻ¤ā§āĻ°$1-e^{-\frac{4^2}{2\times 55}}\approx 13.5\%$
āĻ
āĻ¤āĻāĻŦ, āĻāĻŽāĻ°āĻž āĻ¯āĻĻāĻŋ $$\{a_i\}_{i = 0}^{\infty}$$ āĻĨā§āĻā§ āĻĒā§āĻ°āĻ¤āĻŋāĻŦāĻžāĻ° āĻ¯ā§āĻā§āĻ¨ āĻĻā§āĻāĻāĻŋ āĻĒāĻĻ $$a_i$$ āĻāĻŦāĻ $$a_j$$ āĻ¨āĻŋā§ā§ $$\gcd(N, a_i-a_j)$$ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°ā§ āĻ¯ā§āĻ¤ā§ āĻĨāĻžāĻāĻŋ, āĻ¤āĻžāĻšāĻ˛ā§ āĻā§āĻŦāĻ āĻāĻžāĻ˛ āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž āĻāĻā§ āĻ¯ā§ āĻāĻŽāĻ°āĻž $$a_m-a_n$$ āĻĒā§ā§ā§ āĻ¯āĻžāĻŦ āĻāĻŦāĻ $$55$$-āĻāĻ° āĻāĻāĻāĻŋ āĻĒā§āĻ°āĻā§āĻ¤ āĻā§āĻĒāĻžāĻĻāĻ $$d$$ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°ā§ āĻĢā§āĻ˛āĻŦāĨ¤
āĻ¤āĻŦā§ āĻāĻžāĻŽā§āĻ˛āĻž āĻšāĻā§āĻā§ āĻāĻžāĻāĻāĻŋ āĻ°âā§āĻ¯āĻžāĻ¨ā§āĻĄāĻŽ āĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻ
āĻ¸ā§āĻŽ āĻ§āĻžāĻ°āĻž āĻŦā§āĻ¯āĻŦāĻšāĻžāĻ° āĻāĻ°āĻ˛ā§ āĻ¸ā§āĻāĻŋāĻ° āĻ
āĻ¨ā§āĻāĻā§āĻ˛ā§ āĻĒāĻĻ āĻāĻŽāĻžāĻĻā§āĻ°āĻā§ āĻŽā§āĻŽāĻ°āĻŋāĻ¤ā§ āĻ°āĻžāĻāĻ¤ā§ āĻšāĻŦā§, āĻĢāĻ˛ā§ $$50$$ āĻŦāĻž $$100$$ āĻĄāĻŋāĻāĻŋāĻā§āĻ° āĻā§āĻ¨ āĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻāĻ¨ā§āĻ¯ āĻā§āĻāĻŋ āĻā§āĻāĻŋ āĻĒā§āĻāĻžāĻŦāĻžāĻāĻ āĻŽā§āĻŽāĻ°āĻŋāĻ° āĻĒā§āĻ°ā§ā§āĻāĻ¨ āĻšāĻŦā§āĨ¤ āĻ¤āĻžāĻ āĻāĻŽāĻ°āĻž āĻāĻŽāĻ¨ āĻā§āĻ¨ āĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻ§āĻžāĻ°āĻž āĻ¨ā§āĻŦ āĻ¯āĻžāĻ° āĻŽāĻžāĻ¨ āĻĒā§āĻ°āĻā§āĻ¤āĻĒāĻā§āĻˇā§ āĻĄāĻŋāĻāĻžāĻ°āĻŽāĻŋāĻ¨āĻŋāĻ¸ā§āĻāĻŋāĻ, āĻ
āĻ°ā§āĻĨāĻžā§ āĻ¸ā§āĻ¤ā§āĻ° āĻŽā§āĻ¨ā§ āĻāĻ˛ā§, āĻāĻŋāĻ¨ā§āĻ¤ā§ $$\pmod p$$-āĻ¤ā§ āĻŽā§āĻāĻžāĻŽā§āĻāĻŋ āĻ°âā§āĻ¯āĻžāĻ¨ā§āĻĄāĻŽāĻ˛āĻŋ āĻāĻ¸ā§āĨ¤ (āĻāĻŽāĻ¨ āĻ¸āĻāĻā§āĻ¯āĻžāĻā§ āĻŦāĻ˛ā§ āĻ¸ā§āĻĄā§āĻ°âā§āĻ¯āĻžāĻ¨ā§āĻĄāĻŽ āĻ¸āĻāĻā§āĻ¯āĻž)
āĻāĻĻāĻžāĻšāĻ°āĻŖāĻ¸ā§āĻŦāĻ°ā§āĻĒ, āĻ
āĻ¸ā§āĻŽ āĻ§āĻžāĻ°āĻž āĻšāĻŋāĻ¸ā§āĻŦā§ $$a_0=1$$, āĻāĻŦāĻ $$a_i=a_{i-1}^2+1$$, āĻ
āĻ°ā§āĻĨāĻžā§ $$1, 2, 5, 26, 677,\ldots$$ āĻ§āĻžāĻ°āĻžāĻāĻŋ āĻ¨ā§āĻā§āĻž āĻ¨ā§āĻŦ āĻāĻŦāĻ āĻĒā§āĻ°āĻ¤āĻŋāĻāĻŋ āĻĒāĻĻā§āĻ° āĻāĻ¨ā§āĻ¯ $$\gcd(N, a_i-a_0)$$ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°āĻŦāĨ¤ āĻ˛āĻā§āĻˇā§āĻ¯ āĻāĻ° āĻ¯ā§, āĻāĻŽāĻžāĻĻā§āĻ° $$a_i$$-āĻāĻ° āĻĒā§āĻ°āĻā§āĻ¤ āĻŽāĻžāĻ¨ā§āĻ° āĻĒāĻ°āĻŋāĻŦāĻ°ā§āĻ¤ā§ $$a_i\pmod N$$ āĻāĻžāĻ¨āĻ˛ā§āĻ āĻāĻ˛ā§āĨ¤
def rhonaive(N):
a_0 = d = 1
a_i = a_0
while d == 1:
a_i = (a_i * a_i + 1) % N
d = gcd(a_i - a_0, N)
return d
āĻ
āĻ¨ā§āĻāĻ¸āĻŽā§ āĻĻā§āĻāĻž āĻ¯ā§āĻ¤ā§ āĻĒāĻžāĻ°ā§ āĻāĻŽāĻžāĻĻā§āĻ° āĻ¨ā§āĻā§āĻž āĻ¸ā§āĻĄā§āĻ°âā§āĻ¯āĻžāĻ¨ā§āĻĄāĻŽ āĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻ§āĻžāĻ°āĻžāĻāĻŋ “āĻ¯āĻĨā§āĻˇā§āĻ āĻ°âā§āĻ¯āĻžāĻ¨ā§āĻĄāĻŽ” āĻ¨ā§āĨ¤ āĻ¤ā§āĻŽāĻ¨ āĻā§āĻˇā§āĻ¤ā§āĻ°ā§ āĻāĻŽāĻžāĻĻā§āĻ° $$a_i = a_{i-1}^2+1$$ āĻāĻ° āĻĒāĻ°āĻŋāĻŦāĻ°ā§āĻ¤ā§ āĻāĻ°ā§āĻāĻā§ āĻāĻāĻŋāĻ˛ āĻā§āĻ¨ āĻ¸āĻŽā§āĻĒāĻ°ā§āĻ āĻŦā§āĻ¯āĻŦāĻšāĻžāĻ° āĻāĻ°āĻ¤ā§ āĻšāĻŦā§āĨ¤ āĻ¤āĻžāĻ, āĻ¸āĻžāĻ§āĻžāĻ°āĻŖāĻāĻžāĻŦā§(generally) āĻāĻŽāĻ°āĻž $$a_i=f(a_{i-1})$$ āĻ§āĻ°āĻ¤ā§ āĻĒāĻžāĻ°āĻŋ āĻ¯ā§āĻāĻžāĻ¨ā§ $$f(x)$$ āĻšāĻā§āĻā§ āĻāĻāĻāĻŋ āĻ¸ā§āĻŦāĻŋāĻ§āĻžāĻāĻ¨āĻ āĻŦāĻšā§āĻĒāĻĻā§āĨ¤ āĻ¯ā§āĻŽāĻ¨- āĻāĻĒāĻ°ā§āĻ° āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§ $$f(x)=x^2+1$$.
āĻ˛āĻā§āĻˇā§āĻ¯ āĻāĻ° āĻ¯ā§, $$\{a_i\}_{i = 0}^{\infty}$$ āĻ§āĻžāĻ°āĻžāĻ° āĻĒā§āĻ°āĻ¤āĻŋāĻāĻŋ āĻĒāĻĻ āĻļā§āĻ§ā§āĻŽāĻžāĻ¤ā§āĻ° āĻāĻā§āĻ° āĻĒāĻĻā§āĻ° āĻāĻĒāĻ° āĻ¨āĻŋāĻ°ā§āĻāĻ°āĻļā§āĻ˛āĨ¤ āĻ¤āĻžāĻ $$a_m \equiv a_n\pmod p$$ āĻāĻāĻŦāĻžāĻ° āĻĒāĻžāĻā§āĻž āĻā§āĻ˛ā§ āĻ§āĻžāĻ°āĻžāĻ¤ā§ āĻĒā§āĻ°ā§āĻŦāĻŦāĻ°ā§āĻ¤ā§ āĻĒāĻĻāĻā§āĻ˛ā§ āĻ¸āĻžāĻāĻā§āĻ˛āĻŋāĻāĻāĻžāĻŦā§ āĻā§āĻ°ā§ āĻĢāĻŋāĻ°ā§ āĻāĻ¸āĻ¤ā§ āĻĨāĻžāĻāĻŦā§āĨ¤ āĻ
āĻ°ā§āĻĨāĻžā§,$$a_{m+i}\equiv a_{n+i}\pmod p$$.
āĻ¤āĻŦā§ āĻāĻ āĻ¸āĻžāĻāĻā§āĻ˛ $$a_0$$ āĻĨā§āĻā§ āĻļā§āĻ°ā§ āĻšāĻ¤ā§ āĻšāĻŦā§ āĻāĻŽāĻ¨ āĻā§āĻ¨ āĻāĻĨāĻž āĻ¨ā§āĻāĨ¤ āĻ¯ā§āĻŽāĻ¨- $$p=29$$ āĻāĻŦāĻ $$f(x)=x^2+1$$ āĻšāĻ˛ā§, $$\pmod p$$-āĻ¤ā§ $$\{a_i\}_{i = 0}^{\infty}$$ āĻ§āĻžāĻ°āĻžāĻāĻŋ āĻšā§
$$i$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$a_i$$ $$1$$ $$2$$ $$5$$ $$26$$ $$10$$ $$14$$ $$23$$ $$8$$ $$7$$ $$21$$ $$\color{red}7$$ $$\color{red}{21}$$
āĻĻā§āĻāĻ¤ā§āĻ āĻĒāĻžāĻā§āĻ, $$a_0$$-āĻ° āĻāĻ¨āĻā§āĻ°ā§ā§ā§āĻ¨ā§āĻ āĻā§āĻ¨ āĻĒāĻĻ āĻĻā§āĻŦāĻŋāĻ¤ā§ā§āĻŦāĻžāĻ° āĻĒāĻžāĻā§āĻž āĻ¯āĻžā§āĻ¨āĻŋāĨ¤ āĻ¤āĻžāĻ, āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽ ā§§-āĻ āĻ¸āĻŦāĻ¸āĻŽā§ $$a_0$$ āĻĨā§āĻā§ āĻ¸āĻžāĻāĻā§āĻ˛ āĻļā§āĻ°ā§ āĻšāĻŦā§ āĻ§āĻ°ā§ āĻ¨ā§āĻā§āĻžāĻāĻž āĻā§āĻ˛ āĻāĻŋāĻ˛āĨ¤
āĻāĻŽāĻ¨ āĻ§āĻžāĻ°āĻžāĻ¤ā§ āĻ¸āĻžāĻāĻā§āĻ˛ (āĻ
āĻ°ā§āĻĨāĻžā§ $$\pmod p$$-āĻ¤ā§ āĻāĻ¨āĻā§āĻ°ā§ā§ā§āĻ¨ā§āĻ āĻĻā§āĻāĻŋ āĻĒāĻĻ) āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°āĻ¤ā§ āĻāĻ°ā§āĻāĻā§ āĻŦā§āĻĻā§āĻ§āĻŋ āĻāĻžāĻāĻžāĻ¤ā§ āĻšā§āĨ¤ āĻāĻŽāĻŋ āĻāĻāĻžāĻ¨ā§ āĻĻā§āĻāĻŋ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽā§āĻ° āĻāĻĨāĻž āĻŦāĻ˛ā§ āĻĻāĻŋāĻā§āĻāĻŋāĨ¤ āĻĒā§āĻ°āĻĨāĻŽāĻāĻŋ āĻĢā§āĻ˛ā§ā§āĻĄā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ, āĻĻā§āĻŦāĻŋāĻ¤ā§ā§āĻāĻŋ āĻŦā§āĻ°ā§āĻ¨ā§āĻā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽāĨ¤
### āĻĢā§āĻ˛ā§ā§āĻĄā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ
āĻŽāĻ¨ā§ āĻāĻ° $$a_i$$ āĻ§āĻžāĻ°āĻžāĻ° āĻ¸āĻžāĻāĻā§āĻ˛ā§āĻ° āĻĻā§āĻ°ā§āĻā§āĻ¯ $$l$$. $$n$$ āĻ¯āĻĻāĻŋ āĻ¯āĻĨā§āĻˇā§āĻ āĻŦā§ āĻšā§, āĻ¤āĻŦā§,$a_n\equiv a_{n+l}\equiv a_{n+2l}\equiv\cdots\equiv a_{n+kl}\pmod p$
āĻāĻŽāĻ°āĻž āĻ¯āĻĻāĻŋ āĻ¯āĻĨā§āĻˇā§āĻ āĻŦā§ āĻā§āĻ¨ $$k$$-āĻ° āĻāĻ¨ā§āĻ¯ $$n=kl$$ āĻ¨ā§āĻ, āĻ¤āĻŦā§,$a_n \equiv a_{n+kl}=a_{2kl}=a_{2n}\pmod p$
āĻ
āĻ°ā§āĻĨāĻžā§, $$a_i$$ āĻ§āĻžāĻ°āĻžā§ āĻāĻŽāĻ¨ āĻ
āĻ¸ā§āĻŽ āĻ¸āĻāĻā§āĻ¯āĻ āĻĒāĻĻ $$a_n$$ āĻāĻā§ āĻ¯āĻžāĻĻā§āĻ° āĻāĻ¨ā§āĻ¯ $$a_n\equiv a_{2n}\pmod p$$. āĻ¸ā§āĻ¤āĻ°āĻžāĻ $$\gcd(N, a_{2i}-a_i)$$ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°ā§ āĻ¯ā§āĻ¤ā§ āĻĨāĻžāĻāĻ˛ā§āĻ āĻāĻāĻāĻž āĻ¸āĻŽā§ $$\pmod p$$-āĻ¤ā§ āĻāĻ¨āĻā§āĻ°ā§ā§ā§āĻ¨ā§āĻ āĻĻā§āĻāĻŋ āĻĒāĻĻ āĻĒāĻžāĻā§āĻž āĻ¯āĻžāĻŦā§āĨ¤ āĻĢā§āĻ˛ā§ā§āĻĄā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ āĻāĻ āĻ§āĻžāĻ°āĻŖāĻžāĻāĻŋ āĻŦā§āĻ¯āĻŦāĻšāĻžāĻ° āĻāĻ°ā§āĨ¤
āĻĢā§āĻ˛ā§ā§āĻĄā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽā§ āĻāĻāĻāĻŋ āĻāĻ°āĻā§āĻļ āĻāĻ° āĻāĻāĻāĻŋ āĻāĻā§āĻāĻĒā§āĻ° āĻŽāĻžāĻā§ āĻĻā§ā§ āĻĒā§āĻ°āĻ¤āĻŋāĻ¯ā§āĻāĻŋāĻ¤āĻžāĻ° āĻāĻ˛ā§āĨ¤ āĻāĻā§āĻāĻĒāĻāĻŋ āĻ§āĻžāĻ°āĻžāĻ° āĻāĻāĻāĻŋ āĻāĻāĻāĻŋ āĻĒāĻĻā§ $$a_0, a_1, a_2,a_3,\ldots$$ āĻĒāĻ°ā§āĻ¯āĻžā§āĻā§āĻ°āĻŽā§ āĻ¯ā§āĻ¤ā§ āĻĨāĻžāĻā§āĨ¤ āĻāĻāĻāĻ¸āĻŽā§ā§, āĻāĻ°āĻā§āĻļāĻāĻŋ āĻ§āĻžāĻ°āĻžāĻ° $$a_0, a_2, a_4, a_6,\ldots$$ āĻĒāĻĻāĻā§āĻ˛ā§āĻ¤ā§ āĻ˛āĻžāĻĢāĻŋā§ā§ āĻ˛āĻžāĻĢāĻŋā§ā§ āĻ¯ā§āĻ¤ā§ āĻĨāĻžāĻā§āĨ¤ āĻāĻŽāĻ°āĻž āĻāĻ°āĻā§āĻļā§āĻ° āĻ
āĻŦāĻ¸ā§āĻĨāĻžāĻ¨ āĻāĻŦāĻ āĻāĻā§āĻāĻĒā§āĻ° āĻ
āĻŦāĻ¸ā§āĻĨāĻžāĻ¨ā§āĻ° āĻĒāĻžāĻ°ā§āĻĨāĻā§āĻ¯ā§āĻ° āĻ¸āĻžāĻĨā§ $$N$$-āĻāĻ° āĻāĻ¸āĻžāĻā§ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°āĻ¤ā§ āĻĨāĻžāĻāĻŋāĨ¤ āĻ¯āĻāĻ¨āĻ āĻāĻ¸āĻžāĻā§ $$1$$ āĻāĻ° āĻā§ā§ā§ āĻŦā§ āĻšā§ā§ āĻ¯āĻžā§, āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽāĻāĻŋ āĻāĻžāĻ°ā§āĻŽāĻŋāĻ¨ā§āĻ āĻāĻ°ā§āĨ¤
def rhofloyd(N):
#both hare and tortoise at a_0
tortoise = hare = d = 1
while d == 1:
tortoise = f(tortoise) % N #tortoise steps once
hare = f(f(hare)) % N #hare steps twice
d = gcd(hare - tortoise, N)
return d
### āĻŦā§āĻ°ā§āĻ¨ā§āĻā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ
āĻŦā§āĻ°ā§āĻ¨ā§āĻā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽā§ āĻ
āĻ¨ā§āĻāĻā§āĻ˛ā§ āĻ°āĻžāĻāĻ¨ā§āĻĄā§ āĻāĻ°āĻā§āĻļ āĻāĻ° āĻāĻā§āĻāĻĒā§āĻ° āĻĻā§ā§ āĻĒā§āĻ°āĻ¤āĻŋāĻ¯ā§āĻāĻŋāĻ¤āĻž āĻāĻ˛ā§āĨ¤ āĻĒā§āĻ°āĻ¤āĻŋāĻ¯ā§āĻāĻŋāĻ¤āĻžāĻ° $$i$$-āĻ¤āĻŽ āĻ°āĻžāĻāĻ¨ā§āĻĄā§ āĻāĻ°āĻā§āĻļ āĻāĻ° āĻāĻā§āĻāĻĒ āĻāĻā§ā§āĻ $$a_{2^i}$$-āĻ¤āĻŽ āĻĒāĻĻā§ āĻ
āĻŦāĻ¸ā§āĻĨāĻžāĻ¨ āĻāĻ°ā§āĨ¤ āĻĒā§āĻ°āĻ¤āĻŋāĻ¯ā§āĻāĻŋāĻ¤āĻž āĻļā§āĻ°ā§āĻ° āĻĒāĻ° āĻāĻ°āĻā§āĻļ āĻ¤āĻžāĻ° āĻāĻžā§āĻāĻžā§ āĻāĻ°āĻžāĻŽ āĻāĻ°ā§ āĻā§āĻŽ āĻĻā§ā§, āĻāĻ° āĻāĻā§āĻāĻĒāĻāĻŋ āĻā§āĻ āĻā§āĻ āĻāĻ°ā§ āĻāĻāĻāĻŋ āĻāĻāĻāĻŋ āĻĒāĻĻ āĻ¸āĻžāĻŽāĻ¨ā§ āĻāĻā§āĻ¤ā§ āĻĨāĻžāĻā§āĨ¤ āĻāĻŽāĻ°āĻž āĻĒā§āĻ°āĻ¤āĻŋāĻŦāĻžāĻ°āĻ āĻāĻ°āĻā§āĻļā§āĻ° āĻ
āĻŦāĻ¸ā§āĻĨāĻžāĻ¨ āĻāĻŦāĻ āĻāĻā§āĻāĻĒā§āĻ° āĻ
āĻŦāĻ¸ā§āĻĨāĻžāĻ¨ā§āĻ° āĻĒāĻžāĻ°ā§āĻĨāĻā§āĻ¯ā§āĻ° āĻ¸āĻžāĻĨā§ $$N$$-āĻāĻ° āĻāĻ¸āĻžāĻā§ $$d$$ āĻ¨āĻŋāĻ°ā§āĻŖā§ āĻāĻ°āĻŋāĨ¤ āĻ¯āĻāĻ¨ āĻāĻā§āĻāĻĒ āĻāĻ°āĻā§āĻļā§āĻ° āĻā§ā§ā§ $$2^i$$ āĻĒāĻĻ āĻ¸āĻžāĻŽāĻ¨ā§ āĻāĻāĻŋā§ā§ āĻ¯āĻžā§, āĻāĻ°āĻā§āĻļā§āĻ° āĻ¤ā§āĻžāĻ āĻāĻ°ā§ āĻā§āĻā§ āĻāĻ ā§ āĻāĻŦāĻ āĻ¸ā§ āĻāĻ āĻ˛āĻžāĻĢā§ āĻāĻā§āĻāĻĒā§āĻ° āĻ
āĻŦāĻ¸ā§āĻĨāĻžāĻ¨ā§ āĻāĻ˛ā§ āĻāĻ¸ā§āĨ¤ āĻāĻ°āĻĒāĻ° āĻĒā§āĻ°āĻ¤āĻŋāĻ¯ā§āĻāĻŋāĻ¤āĻžāĻ° $$i+1$$-āĻ¤āĻŽ āĻ°āĻžāĻāĻ¨ā§āĻĄ āĻļā§āĻ°ā§ āĻšā§āĨ¤
āĻāĻā§āĻ° āĻŽāĻ¤āĻ, $$d> 1$$ āĻšā§ā§ āĻā§āĻ˛ā§ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽāĻāĻŋ āĻāĻžāĻ°ā§āĻŽāĻŋāĻ¨ā§āĻ āĻāĻ°āĻŦā§āĨ¤
āĻ˛āĻā§āĻˇā§āĻ¯ āĻāĻ° āĻ¯ā§, $$i$$ āĻāĻ° āĻŽāĻžāĻ¨ āĻŦāĻžā§āĻžāĻ° āĻ¸āĻžāĻĨā§ āĻ¸āĻžāĻĨā§ $$2^{i}$$ āĻāĻŦāĻ $$2^{i+1}$$-āĻāĻ° āĻŽāĻ§ā§āĻ¯āĻŦāĻ°ā§āĻ¤ā§ āĻĒāĻžāĻ°ā§āĻĨāĻā§āĻ¯āĻ āĻŦāĻžā§āĻ¤ā§ āĻĨāĻžāĻā§āĨ¤ āĻāĻāĻāĻž āĻ¸āĻŽā§ āĻāĻ āĻĒāĻžāĻ°ā§āĻĨāĻā§āĻ¯ $$\{a_i\}_{i = 0}^\infty$$ āĻ§āĻžāĻ°āĻžāĻ° āĻ¸āĻžāĻāĻā§āĻ˛ā§āĻ° āĻĻā§āĻ°ā§āĻā§āĻ¯ $$l$$ āĻĨā§āĻā§ āĻŦā§ āĻšā§ā§ āĻ¯āĻžā§āĨ¤ āĻ¤āĻāĻ¨ āĻāĻā§āĻāĻĒā§āĻ° āĻā§āĻ¨ āĻāĻ āĻ
āĻŦāĻ¸ā§āĻĨāĻžāĻ¨ā§āĻ° āĻāĻ¨ā§āĻ¯ āĻāĻ°āĻā§āĻļ āĻ āĻāĻā§āĻāĻĒā§āĻ° āĻ
āĻŦāĻ¸ā§āĻĨāĻžāĻ¨ā§āĻ° āĻĒāĻžāĻ°ā§āĻĨāĻā§āĻ¯ āĻšāĻŦā§ āĻ āĻŋāĻ $$l$$āĻāĻŋ āĻĒāĻĻāĨ¤ āĻāĻ āĻ
āĻŦāĻ¸ā§āĻĨāĻžāĻ¤ā§āĻ āĻāĻ¸āĻ˛ā§ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽāĻāĻŋ āĻāĻžāĻ°ā§āĻŽāĻŋāĻ¨ā§āĻ āĻāĻ°āĻŦā§āĨ¤
def rhobrent(N):
#both hare and tortoise are at a_(2^0)
#power is the nearest 2^i behind tortoise
tortoise = hare = f(1)
power, lead, d = 1, 0, 1
while d == 1:
if lead == power: #tortoise reaches 2^(i+1)-th term
hare = tortoise #hare jumps to tortoise's position
power *= 2 #update power
lead = 0 #reset lead, since both hare and tortoise
#are in the same position now
tortoise = f(tortoise) % N #tortoise steps once
lead += 1 #tortoise leads the race by one more step
d = gcd(hare - tortoise, N)
return d
āĻŦā§āĻ°ā§āĻ¨ā§āĻā§āĻ° āĻāĻ¨āĻžāĻ˛āĻžāĻāĻ¸āĻŋāĻ¸ āĻ
āĻ¨ā§āĻ¸āĻžāĻ°ā§, āĻŦā§āĻ°ā§āĻ¨ā§āĻā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ āĻĢā§āĻ˛ā§ā§āĻĄā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ āĻĨā§āĻā§ $$24\%$$ āĻĻā§āĻ°ā§āĻ¤āĻ¤āĻ°āĨ¤
### āĻĒā§āĻ°ā§āĻŖāĻžāĻā§āĻ āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽ
āĻŦā§āĻ°ā§āĻ¨ā§āĻā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ āĻĻāĻŋā§ā§ āĻ¯ā§āĻšā§āĻ¤ā§ āĻ¸āĻžāĻāĻā§āĻ˛ āĻļāĻ¨āĻžāĻā§āĻ¤āĻāĻ°āĻŖ āĻ¸āĻŦāĻā§ā§ā§ āĻĻā§āĻ°ā§āĻ¤ āĻāĻ°āĻž āĻ¯āĻžā§, āĻ¤āĻžāĻ āĻ¸ā§āĻāĻŋ āĻĻāĻŋā§ā§āĻ āĻ¨āĻŋāĻā§ āĻāĻāĻāĻŋ āĻĒā§āĻ°ā§āĻŖāĻžāĻā§āĻ āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽ āĻĻā§āĻā§āĻž āĻšāĻ˛āĨ¤
from fractions import gcd
from time import time
def f(x):
return (x**2 + x + 1) % N
def rhobrent(N):
tortoise = hare = f(1)
power, lead, d = 1, 0, 1
while abs(d) == 1:
if lead == power:
hare, lead = tortoise, 0
power *= 2
tortoise = f(tortoise)
d = gcd(tortoise - hare, N)
return d
N = int(raw_input())
exec_time = time()
d = 1 if(-2 < N < 2) else rhobrent(N)
exec_time = time() - exec_time
print d, '*', N / d
print "Execution time: %.3f s" % exec_time
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <iso646.h> //and
#define MAX 2147483647
long long f(long long x);
long long gcd(long long a, long long b);
long long rhobrent(long long N);
long long N;
int main(void){
long long d;
scanf("%lld", &N);
clock_t exec_time = clock();
d = (N < 2 and N > -2) ? 1 : rhobrent(N);
exec_time = clock() - exec_time;
double t = (double) exec_time / CLOCKS_PER_SEC;
printf("%lld * %lld\n", d, N/d);
printf("Execution time: %.3lf s\n", t);
return 0;
}
long long f(long long x){
if(x < MAX)
return (x * x + x + 1) % N;
long long y = x, output = x + 1;
while(y > 0){
if(y % 2 == 1)
output += x;
x *= 2;
y /= 2;
output %= N;
x %= N;
}
return output;
}
long long gcd(long long a, long long b){
long long r;
while(b != 0){
r = a % b;
a = b;
b = r;
}
return (a > 0) ? a : -a;
}
long long rhobrent(long long N){
long long tortoise = f(1), hare = f(1), power = 1, lead = 0, d = 1;
while(d == 1){
hare = tortoise;
power *= 2;
}
tortoise = f(tortoise);
d = gcd(N, tortoise - hare);
}
return d;
}
āĻ¨ā§āĻ:
1. āĻāĻĒāĻ°ā§āĻ° āĻā§āĻĄā§ $$f(x)=x^2+x+1$$ āĻāĻŦāĻ āĻĢāĻžāĻāĻļāĻžāĻ¨ā§āĻ° āĻā§āĻ¤āĻ°ā§āĻ $$N$$ āĻĻāĻŋā§ā§ āĻŽāĻĄā§āĻ˛āĻžāĻ¸ā§āĻ° āĻāĻžāĻ āĻļā§āĻˇ āĻāĻ°āĻž āĻšā§ā§āĻā§āĨ¤
2. āĻĒā§āĻ°āĻžā§ āĻ¸āĻāĻ˛ āĻāĻ§ā§āĻ¨āĻŋāĻ āĻāĻŽā§āĻĒāĻžāĻāĻ˛āĻžāĻ°ā§ long long int-āĻāĻ° āĻŦā§āĻ¯āĻžāĻŦāĻ§ā§ $$[-2^{63}, 2^{63})$$. āĻ¤āĻžāĻ āĻ¸āĻŋ āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§ $$f(x)$$ āĻĢāĻžāĻāĻļāĻ¨āĻāĻŋ āĻ āĻŋāĻāĻ āĻžāĻāĻŽāĻ¤ āĻāĻžāĻ āĻāĻ°āĻžāĻ° āĻāĻ¨ā§āĻ¯ $x^2+x+1\leq 2^{63}-1\Longrightarrow x<2^{31.5}-1$ āĻšāĻ¤ā§ āĻšāĻŦā§āĨ¤ āĻāĻāĻ¨ā§āĻ¯ $$x \ge 2^{31}$$ -āĻāĻ° āĻāĻ¨ā§āĻ¯ $$y = x$$ āĻ¨āĻžāĻŽā§āĻ° āĻāĻāĻāĻŋ āĻ¨āĻ¤ā§āĻ¨ āĻā§āĻ°āĻŋā§ā§āĻŦāĻ˛ āĻ¨āĻŋā§ā§ āĻāĻ° āĻĒā§āĻ°āĻ¤āĻŋāĻāĻŋ āĻŦāĻŋāĻā§āĻ° āĻ¸āĻžāĻĨā§ āĻāĻ˛āĻžāĻĻāĻžāĻāĻžāĻŦā§ $$x$$-āĻāĻ° āĻā§āĻŖāĻĢāĻ˛ āĻŦā§āĻ° āĻāĻ°ā§ āĻ¸ā§āĻ āĻ¸āĻŽāĻˇā§āĻāĻŋāĻ° āĻŽāĻĄā§āĻ˛āĻžāĻ¸ āĻ¨ā§āĻā§āĻž āĻšā§ā§āĻā§āĨ¤
3. $$\pmod N$$ āĻāĻŦāĻ $$\pmod p$$-āĻ¤ā§ $$N$$-āĻāĻ° āĻ¸āĻžāĻāĻā§āĻ˛ āĻāĻāĻ āĻšāĻ˛ā§ āĻāĻĒāĻ°ā§āĻ° āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽ $$N$$ āĻā§āĻ¤ā§āĻ°āĻŋāĻŽ āĻšāĻā§āĻž āĻ¸āĻ¤ā§āĻŦā§āĻ āĻāĻ° āĻā§āĻĒāĻžāĻĻāĻ āĻā§āĻāĻā§ āĻĒāĻžāĻŦā§ āĻ¨āĻžāĨ¤ āĻāĻ°āĻāĻŽ āĻšāĻ˛ā§ $$f(x)$$ āĻĒāĻ˛āĻŋāĻ¨ā§āĻŽāĻŋā§āĻžāĻ˛āĻāĻŋ āĻĒāĻ°āĻŋāĻŦāĻ°ā§āĻ¤āĻ¨ āĻāĻ°ā§ āĻāĻŦāĻžāĻ° āĻā§āĻˇā§āĻāĻž āĻāĻ°āĻ¤ā§ āĻĒāĻžāĻ°āĨ¤
### āĻāĻŽāĻĒā§āĻ˛ā§āĻā§āĻ¸āĻŋāĻāĻŋ
āĻŽāĻ¨ā§ āĻāĻ°, $$\{a_i\}_{i=0}^{\infty}$$ āĻāĻāĻāĻŋ āĻ¸āĻ¤ā§āĻ¯āĻŋāĻāĻžāĻ°ā§ āĻ°âā§āĻ¯āĻžāĻ¨ā§āĻĄāĻŽ āĻ§āĻžāĻ°āĻžāĨ¤ āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ā§āĻ° āĻ¸āĻŽāĻ¸ā§āĻ¯āĻž āĻ
āĻ¨ā§āĻ¯āĻžā§ā§, $$\pmod p$$-āĻ¤ā§ āĻāĻ¨āĻā§āĻ°ā§ā§ā§āĻ¨ā§āĻ āĻĻā§āĻāĻŋ āĻĒāĻĻ āĻĒāĻžāĻā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž $$99\%$$ āĻšāĻ¤ā§ āĻšāĻ˛ā§ $$\sqrt{-2\times\ln(1-0.99)}\sqrt{p}$$-āĻāĻŋ āĻĒāĻĻ āĻ¨āĻŋāĻ¤ā§ āĻšāĻŦā§āĨ¤ āĻ
āĻ¤āĻāĻŦ, āĻ°ā§ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽā§āĻ° āĻāĻŽāĻĒā§āĻ˛ā§āĻā§āĻ¸āĻŋāĻāĻŋ $$O(\sqrt p)\leq O\left(N^{\frac 1 4}\right)$$. [āĻ
āĻ°ā§āĻĨāĻžā§ $$N$$ āĻ¸āĻāĻā§āĻ¯āĻžāĻāĻŋāĻ¤ā§ $$k$$āĻāĻŋ āĻĄāĻŋāĻāĻŋāĻ āĻĨāĻžāĻāĻ˛ā§ āĻ°ā§ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽā§āĻ° āĻ°āĻžāĻ¨āĻāĻžāĻāĻŽ $$O\left(10^{\frac k 4}\right)$$]
āĻāĻ¤āĻĒāĻ°ā§āĻŦā§āĻ° āĻā§ā§ā§ āĻāĻ°ā§āĻāĻā§ āĻāĻžāĻ˛ā§ āĻŦāĻžāĻāĻ¨ā§āĻĄ, āĻ¤āĻžāĻ āĻ¨āĻž? āĻĒāĻ°āĻŦāĻ°ā§āĻ¤ā§ āĻĒāĻ°ā§āĻŦā§ āĻāĻŽāĻ°āĻž āĻāĻ¨ āĻĒā§āĻ˛āĻžāĻ°ā§āĻĄā§āĻ°āĻ āĻāĻ°ā§āĻāĻāĻŋ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ āĻĻā§āĻāĻŦ, āĻ¯ā§āĻāĻŋāĻ° āĻ¨āĻžāĻŽ $$p-1$$ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽāĨ¤
## āĻĒāĻ°āĻŋāĻļāĻŋāĻˇā§āĻ
### āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ āĻ¸āĻŽāĻ¸ā§āĻ¯āĻž
āĻ¤ā§āĻŽāĻžāĻ° āĻĢā§āĻ¸āĻŦā§āĻ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄāĻ˛āĻŋāĻ¸ā§āĻā§ āĻ¨ā§āĻ¨ā§āĻ¯āĻ¤āĻŽ āĻāĻ¤āĻāĻ¨ āĻĨāĻžāĻāĻ˛ā§ āĻāĻāĻĨāĻž āĻ¨āĻŋāĻļā§āĻāĻŋāĻ¤ āĻāĻ°ā§ āĻŦāĻ˛āĻž āĻ¯āĻžāĻŦā§ āĻ¯ā§ āĻ
āĻ¨ā§āĻ¤āĻ¤ āĻĻā§āĻāĻāĻ¨ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄā§āĻ° āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ āĻāĻāĻ āĻ¤āĻžāĻ°āĻŋāĻā§ āĻšāĻā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž $$99\%$$?
āĻŽāĻ¨ā§ āĻāĻ°, āĻ¤ā§āĻŽāĻžāĻ° āĻĢā§āĻ°ā§āĻ¨ā§āĻĄāĻ˛āĻŋāĻ¸ā§āĻā§ $$m$$ āĻāĻ¨ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄ āĻāĻā§, āĻāĻŦāĻ āĻā§āĻ¨ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄā§āĻ° āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ ā§¨ā§¯ āĻĢā§āĻŦā§āĻ°ā§ā§āĻžāĻ°āĻŋ āĻ¨ā§āĨ¤ āĻā§āĻ¨ āĻāĻāĻāĻ¨ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄā§āĻ° āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ ā§Ŧ āĻ
āĻā§āĻā§āĻŦāĻ° āĻšāĻ˛ā§ āĻāĻ° āĻā§āĻ¨ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄā§āĻ° āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ ā§Ŧ āĻ
āĻā§āĻā§āĻŦāĻ° āĻ¨āĻž āĻšāĻā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž $$1-\frac {1}{365}$$. āĻāĻ°ā§āĻ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄā§āĻ° āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ āĻŽāĻ¨ā§ āĻāĻ° ā§Ē āĻāĻāĻ¸ā§āĻāĨ¤ āĻ¸ā§āĻ¤āĻ°āĻžāĻ, āĻ¤ā§āĻ¤ā§ā§ āĻā§āĻ¨ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄā§āĻ° āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ ā§Ŧ āĻ
āĻā§āĻā§āĻŦāĻ° āĻŦāĻž ā§Ē āĻāĻāĻ¸ā§āĻ āĻ¨āĻž āĻšāĻā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž $$1-\frac {2}{365}$$. āĻ
āĻ¤āĻāĻŦ, $$m$$ āĻāĻ¨ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄā§āĻ° āĻāĻŋāĻ¨ā§āĻ¨ āĻāĻŋāĻ¨ā§āĻ¨ āĻĻāĻŋāĻ¨ā§ āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ āĻĒā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž $\left(1-\frac {1}{365}\right)\cdots\left(1-\frac {m-1}{365}\right)$
āĻ¤āĻžāĻšāĻ˛ā§, āĻ
āĻ¨ā§āĻ¤āĻ¤ āĻĻā§āĻāĻ¨ āĻĢā§āĻ°ā§āĻ¨ā§āĻĄā§āĻ° āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ āĻāĻāĻ āĻĻāĻŋāĻ¨ā§ āĻĒā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž $1 – \left(1-\frac {1}{365}\right)\cdots\left(1-\frac {m-1}{365}\right)$
$$x$$-āĻāĻ° āĻŽāĻžāĻ¨ $$0$$-āĻāĻ° āĻāĻžāĻāĻžāĻāĻžāĻāĻŋ āĻšāĻ˛ā§ āĻāĻŽāĻ°āĻž āĻā§āĻāĻ˛āĻ°ā§āĻ° āĻ¸āĻŋāĻ°āĻŋāĻ āĻĨā§āĻā§ $$1-x\approx e^{-x}$$ āĻ˛āĻŋāĻāĻ¤ā§ āĻĒāĻžāĻ°āĻŋāĨ¤ āĻāĻāĻžāĻŦā§ āĻāĻĒāĻ°ā§āĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻžāĻāĻŋāĻā§ āĻ¸āĻ°āĻ˛ āĻāĻ°ā§ āĻ˛āĻŋāĻāĻž āĻ¯āĻžā§ $1-e^{-\frac {1}{365}}\cdot e^{-\frac {2}{365}}\cdots e^{-\frac {m-1}{365}}$$\approx 1-e^{-\frac{m^2}{2\times 365}}$āĻāĻāĻ¨ $$1-e^{-\frac{m^2}{2\times 365}}\ge 0.99$$ āĻšāĻ¤ā§ āĻšāĻ˛ā§ $m\ge \sqrt{-2\times 365\ln(1-0.99)}\approx 58$ āĻšāĻ¤ā§ āĻšāĻŦā§āĨ¤
āĻ¤ā§āĻŽāĻžāĻ° āĻĢā§āĻ°ā§āĻ¨ā§āĻĄāĻ˛āĻŋāĻ¸ā§āĻā§ āĻŽāĻžāĻ¤ā§āĻ° $$58$$ āĻāĻ¨ āĻŽāĻžāĻ¨ā§āĻˇ āĻĨāĻžāĻāĻ˛ā§āĻ āĻ¤ā§āĻŽāĻŋ $$99\%$$ āĻā§āĻ¯āĻžāĻ°āĻžāĻ¨ā§āĻāĻŋ āĻĻāĻŋā§ā§ āĻŦāĻ˛āĻ¤ā§ āĻĒāĻžāĻ°āĻŦā§ āĻ¯ā§ āĻ
āĻ¨ā§āĻ¤āĻ¤ āĻĻā§āĻāĻāĻ¨ā§āĻ° āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ āĻāĻāĻ āĻ¤āĻžāĻ°āĻŋāĻā§! āĻāĻ āĻŦāĻŋāĻā§āĻ¯āĻžāĻ¤ āĻ¸āĻŽāĻ¸ā§āĻ¯āĻžāĻāĻŋāĻ° āĻ¨āĻžāĻŽ āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ āĻ¸āĻŽāĻ¸ā§āĻ¯āĻžāĨ¤
āĻ˛āĻā§āĻˇā§āĻ¯ āĻāĻ° āĻāĻ¨ā§āĻŽāĻĻāĻŋāĻ¨ā§āĻ° āĻ¸āĻŽāĻ¸ā§āĻ¯āĻžāĻ° āĻ¸āĻŽāĻžāĻ§āĻžāĻ¨ āĻĨā§āĻā§ āĻāĻāĻž āĻŦāĻ˛āĻž āĻ¯āĻžā§ āĻ¯ā§, $$0$$ āĻĨā§āĻā§ $$N-1$$ āĻāĻ° āĻŽāĻžāĻā§ $$m$$ āĻ¸āĻāĻā§āĻ¯āĻž āĻ¨āĻŋāĻ˛ā§ āĻā§āĻ¨ āĻāĻāĻāĻŋ āĻ¸āĻāĻā§āĻ¯āĻž āĻ
āĻ¨ā§āĻ¤āĻ¤ āĻĻā§āĻāĻŦāĻžāĻ° āĻ¨ā§āĻā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž $$1-e^{-\frac{m^2}{2N}}$$.
# āĻāĻ¨ā§āĻāĻŋāĻāĻžāĻ° āĻĢā§āĻ¯āĻžāĻā§āĻā§āĻ°āĻžāĻāĻā§āĻļāĻ¨ āĻĒāĻ°ā§āĻŦ ā§Ļ
## āĻāĻŽāĻ°āĻž āĻā§āĻ¨ āĻāĻ¨ā§āĻāĻŋāĻāĻžāĻ° āĻĢā§āĻ¯āĻžāĻā§āĻā§āĻ°āĻžāĻāĻā§āĻļāĻ¨ āĻāĻ°āĻŋ
āĻāĻ¨ā§āĻā§āĻāĻžāĻ° āĻĢā§āĻ¯āĻžāĻā§āĻā§āĻ°āĻžāĻāĻā§āĻļāĻ¨ āĻ
āĻ°ā§āĻĨ āĻšāĻā§āĻā§ āĻāĻāĻāĻž āĻĒā§āĻ°ā§āĻŖāĻ¸āĻāĻā§āĻ¯āĻžāĻā§ āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻāĻ°āĻžāĨ¤ āĻ¯ā§āĻŽāĻ¨ $$36 = 4\times 9$$. āĻāĻāĻžāĻ¨ā§ $$4$$ āĻāĻ° $$36$$ āĻāĻā§ā§āĻ $$36$$-āĻāĻ° āĻā§āĻĒāĻžāĻĻāĻāĨ¤
āĻā§āĻāĻŦā§āĻ˛āĻžā§ āĻāĻŽāĻžāĻĻā§āĻ° āĻ¸āĻŦāĻžāĻ āĻāĻ¨ā§āĻā§āĻāĻžāĻ° āĻĢā§āĻ¯āĻžāĻā§āĻā§āĻ°āĻžāĻāĻā§āĻļāĻ¨ āĻāĻ°āĻ¤āĻžāĻŽāĨ¤ āĻā§āĻ¨ āĻāĻžāĻ¨ā§? āĻĒāĻ°ā§āĻā§āĻˇāĻžā§ āĻ¨āĻŽā§āĻŦāĻ° āĻĒāĻžāĻā§āĻžāĻ° āĻāĻ¨ā§āĻ¯! āĻĒā§āĻ°āĻļā§āĻ¨ā§ āĻāĻāĻāĻž āĻ¸āĻāĻā§āĻ¯āĻž āĻĻā§āĻā§āĻž āĻĨāĻžāĻāĻ¤ āĻāĻ° āĻāĻŽāĻ°āĻž āĻĻā§āĻ āĻĨā§āĻā§ āĻļā§āĻ°ā§ āĻāĻ°ā§ āĻāĻāĻāĻž āĻāĻāĻāĻž āĻ¸āĻāĻā§āĻ¯āĻž āĻĻāĻŋā§ā§ āĻĒā§āĻ°āĻļā§āĻ¨ā§ āĻĻā§āĻā§āĻž āĻ¸āĻāĻā§āĻ¯āĻžāĻāĻžāĻā§ āĻāĻžāĻ āĻĻāĻŋāĻ¤āĻžāĻŽāĨ¤ āĻ¯āĻāĻ¨ āĻāĻžāĻ āĻ¯ā§āĻ¤, āĻ¤āĻāĻ¨ āĻŦā§āĻāĻž āĻ¯ā§āĻ¤ āĻ¯ā§ āĻāĻāĻāĻž āĻā§āĻĒāĻžāĻĻāĻ (āĻŦāĻž āĻā§āĻŖāĻ¨ā§ā§āĻ) āĻĒā§ā§ā§ āĻā§āĻāĻŋāĨ¤ āĻā§āĻŦāĻ āĻ°ā§āĻāĻŋāĻ¨āĻŽāĻžāĻĢāĻŋāĻ āĻāĻāĻāĻž āĻāĻžāĻ, āĻ¤āĻžāĻ āĻ¨āĻž?
āĻ¤āĻŦā§ āĻŦāĻžāĻā§āĻāĻžāĻĻā§āĻ° āĻĒāĻ°ā§āĻā§āĻˇāĻžāĻ° āĻĒā§āĻ°āĻļā§āĻ¨ā§ āĻāĻ¸āĻž āĻāĻžā§āĻžāĻ āĻāĻ¨ā§āĻāĻŋāĻāĻžāĻ° āĻĢā§āĻ¯āĻžāĻā§āĻā§āĻ°āĻžāĻāĻā§āĻļāĻ¨ā§āĻ° āĻ
āĻ¨ā§āĻ¯āĻ°āĻāĻŽ āĻāĻāĻāĻž āĻā§āĻ°ā§āĻ¤ā§āĻŦ āĻāĻā§āĨ¤ āĻ¯ā§āĻŽāĻ¨- āĻ
āĻ¨āĻ˛āĻžāĻāĻ¨ā§ āĻ¤ā§āĻŽāĻžāĻ°, āĻāĻŽāĻžāĻ°, āĻ¸āĻāĻ˛ā§āĻ° āĻŦā§āĻ¯āĻā§āĻ¤āĻŋāĻāĻ¤ āĻ¤āĻĨā§āĻ¯ āĻĒāĻžāĻāĻŋ āĻ˛ā§āĻāĻĻā§āĻ° āĻšāĻžāĻ¤ āĻĨā§āĻā§ āĻ¸ā§āĻ°āĻā§āĻˇāĻŋāĻ¤ āĻ°āĻžāĻāĻ¤ā§ āĻā§āĻĒāĻ¨ āĻ¸āĻāĻā§āĻ¤ā§āĻ° āĻā§āĻžāĻ˛ā§ āĻ°āĻžāĻāĻž āĻšā§āĨ¤ āĻāĻ°āĻāĻŽ āĻ
āĻ¨ā§āĻ āĻā§āĻĒāĻ¨ āĻ¸āĻāĻā§āĻ¤ āĻŦāĻžāĻ¨āĻžāĻ¤ā§ āĻ āĻāĻžāĻāĻ¤ā§ āĻĻā§āĻāĻļ, āĻ¤āĻŋāĻ¨āĻļ āĻĄāĻŋāĻāĻŋāĻā§āĻ° āĻŦāĻŋāĻļāĻžāĻ˛ āĻŦāĻŋāĻļāĻžāĻ˛ āĻ¸āĻāĻā§āĻ¯āĻžāĻā§ āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻāĻ°āĻ¤ā§ āĻšā§, āĻāĻŋāĻāĻŦāĻž āĻĒā§āĻ°āĻžāĻāĻŽ āĻā§āĻāĻā§ āĻŦā§āĻ° āĻāĻ°āĻ¤ā§ āĻšā§āĨ¤ āĻāĻ¸āĻŦ āĻšāĻŋāĻ¸ā§āĻŦ āĻāĻ¤ āĻŦā§ āĻšā§ āĻ¯ā§ āĻāĻžāĻ¤āĻž āĻāĻ˛āĻŽā§ āĻāĻ°āĻ¤ā§ āĻā§āĻ˛ā§ āĻāĻŽāĻžāĻĻā§āĻ° āĻā§āĻ˛ āĻĒā§āĻā§ āĻ¯āĻžāĻŦā§, āĻĒāĻžāĻ¤āĻžāĻ° āĻĒāĻ° āĻĒāĻžāĻ¤āĻž āĻĢā§āĻ°āĻŋā§ā§ āĻ¯āĻžāĻŦā§āĨ¤ āĻ¸ā§āĻāĻ¨ā§āĻ¯, āĻāĻŽāĻ°āĻž āĻ¤āĻāĻ¨ āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ°ā§āĻ° āĻ¸āĻžāĻšāĻžāĻ¯ā§āĻ¯ āĻ¨ā§āĻāĨ¤
āĻāĻŋāĻ¨ā§āĻ¤ā§ āĻ¸āĻŽāĻ¸ā§āĻ¯āĻž āĻāĻŋ āĻāĻžāĻ¨ā§? āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ° āĻā§āĻŦ āĻŦā§āĻāĻžāĨ¤ āĻ¯āĻĻāĻŋ āĻāĻŽāĻŋ āĻŦāĻ˛āĻŋ, “āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ°, āĻāĻŽāĻžāĻ° āĻāĻāĻāĻž āĻšāĻŋāĻ¸ā§āĻŦā§ $$731$$-āĻāĻ° āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻ˛āĻžāĻāĻŦā§āĨ¤ āĻ¤ā§āĻŽāĻŋ āĻāĻ āĻāĻ°ā§ āĻ¸ā§āĻāĻž āĻŦā§āĻ° āĻāĻ°ā§ āĻĻāĻžāĻ āĻ¤ā§ āĻĻā§āĻāĻŋ!”, āĻ¸ā§ āĻšāĻž āĻāĻ°ā§ āĻŦāĻ¸ā§ āĻĨāĻžāĻāĻŦā§āĨ¤ āĻāĻžāĻ°āĻŖ āĻ¸ā§ āĻ¤ā§ āĻāĻ° āĻāĻžāĻ¨ā§ āĻ¨āĻž āĻā§āĻāĻžāĻŦā§ āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻāĻ°āĻ¤ā§ āĻšā§!
āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ°āĻā§ āĻĻāĻŋā§ā§ āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻāĻ°āĻžāĻ¨ā§āĻ° āĻāĻ¨ā§āĻ¯ āĻ¤āĻžāĻā§ āĻ¤ā§āĻŽāĻžāĻ° āĻĒā§āĻ°āĻžāĻāĻŽāĻžāĻ°āĻŋāĻ° āĻŦāĻžāĻā§āĻāĻžāĻĻā§āĻ° āĻŽāĻ¤ āĻāĻāĻā§ āĻāĻāĻā§ āĻāĻ°ā§ āĻŦā§āĻāĻŋā§ā§ āĻŦāĻ˛āĻ¤ā§ āĻšāĻŦā§, “āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ°, $$2$$ āĻĨā§āĻā§ $$730$$ āĻĒāĻ°ā§āĻ¯āĻ¨ā§āĻ¤ āĻĒā§āĻ°āĻ¤āĻŋāĻāĻŋ āĻ¸āĻāĻā§āĻ¯āĻž āĻĻāĻŋā§ā§ $$731$$-āĻā§ āĻāĻžāĻ āĻĻāĻžāĻāĨ¤ āĻ¯āĻĻāĻŋ āĻā§āĻ¨āĻāĻž āĻĻāĻŋā§ā§ āĻ¨āĻŋāĻāĻļā§āĻˇā§ āĻāĻžāĻ āĻ¯āĻžā§, āĻ¤āĻŦā§ āĻ¸ā§āĻ āĻ¸āĻāĻā§āĻ¯āĻž āĻāĻŦāĻ āĻāĻžāĻāĻĢāĻ˛ āĻāĻŽāĻžāĻā§ āĻĻā§āĻāĻžāĻāĨ¤”
def factor(N):
for d in range(2, abs(N) + 1):
if N % d == 0: return d
N = int(raw_input())
d = 1 if(2 > N > -2) else factor(N)
print d, '*', N / d
āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ° āĻ¯ā§āĻšā§āĻ¤ā§ āĻŦāĻžāĻāĻ˛āĻž āĻĒāĻžāĻ°ā§ āĻ¨āĻž, āĻ¤āĻžāĻ āĻ¸ā§ āĻŦā§āĻā§ āĻāĻŽāĻ¨ āĻāĻāĻāĻŋ āĻāĻžāĻˇāĻž āĻĒāĻžāĻāĻĨāĻ¨ā§ āĻāĻŽāĻ°āĻž āĻāĻŽāĻžāĻĻā§āĻ° āĻāĻĨāĻžāĻā§āĻ˛ā§ āĻ˛āĻŋāĻā§āĻāĻŋāĨ¤ āĻāĻ āĻā§āĻĄā§ āĻāĻāĻāĻŋ āĻĒā§āĻ°ā§āĻŖāĻ¸āĻāĻā§āĻ¯āĻž $$N$$-āĻā§ āĻāĻ¨āĻĒā§āĻ āĻšāĻŋāĻ¸ā§āĻŦā§ āĻĻā§āĻā§āĻž āĻšāĻŦā§āĨ¤ āĻāĻ° āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ° āĻ¸āĻžāĻĨā§ āĻ¸āĻžāĻĨā§ $$N$$-āĻā§ āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻāĻ°ā§ āĻā§āĻļāĻŋāĻŽāĻ¨ā§ āĻāĻŽāĻžāĻĻā§āĻ° āĻāĻžāĻ¨āĻŋā§ā§ āĻĻā§āĻŦā§āĨ¤ āĻ¯āĻ¤ āĻŦā§ āĻ¸āĻāĻā§āĻ¯āĻžāĻ āĻšā§āĻ āĻ¨āĻž āĻā§āĻ¨, āĻāĻŽāĻžāĻĻā§āĻ° āĻā§āĻ˛ā§āĻŦā§āĻ˛āĻžā§ āĻļā§āĻāĻž āĻ¨āĻŋā§āĻŽ āĻĻāĻŋā§ā§ āĻ¤āĻžāĻ° āĻā§āĻĒāĻžāĻĻāĻ āĻŦā§āĻ° āĻāĻ°āĻž āĻ¸āĻŽā§āĻāĻŦāĨ¤
āĻāĻĒāĻ°ā§āĻ° āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§ $$731$$ āĻāĻ¨āĻĒā§āĻ āĻĻāĻŋāĻ˛ā§ āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ° āĻāĻŽāĻžāĻĻā§āĻ° āĻŦāĻ˛ā§ āĻĻā§āĻŦā§ āĻ¯ā§ $$731 = 17\times 43$$.
āĻāĻŦāĻžāĻ° āĻāĻāĻāĻž āĻāĻžāĻ āĻāĻ°āĨ¤ āĻāĻĒāĻ°ā§āĻ° āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽāĻāĻŋ āĻāĻžāĻ˛āĻŋā§ā§ $$N = 18446744073709551619$$ āĻāĻ¨āĻĒā§āĻ āĻĻāĻžāĻāĨ¤ āĻā§ āĻšāĻ˛?
āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽāĻāĻŋ āĻāĻ° āĻĨāĻžāĻŽāĻā§ āĻ¨āĻžāĨ¤
āĻāĻ° āĻāĻžāĻ°āĻŖ āĻšāĻā§āĻā§ āĻāĻĒāĻ°ā§āĻ° āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§ āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ°āĻā§ $$2$$ āĻĨā§āĻā§ $$N-1$$ āĻĒāĻ°ā§āĻ¯āĻ¨ā§āĻ¤ āĻ¸āĻāĻ˛ āĻ¸āĻāĻā§āĻ¯āĻž āĻĻāĻŋā§ā§āĻ $$N$$-āĻā§ āĻāĻžāĻ (āĻĒā§āĻ°āĻā§āĻ¤āĻĒāĻā§āĻˇā§ āĻŽāĻĄā§āĻ˛āĻžāĻ¸ āĻ¨āĻŋāĻ°ā§āĻŖā§) āĻāĻ°āĻ¤ā§ āĻšāĻā§āĻā§āĨ¤ āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ°ā§āĻ° āĻĒā§āĻ°āĻ¤āĻŋāĻāĻž āĻāĻžāĻ āĻāĻ°āĻ¤ā§ āĻāĻŋāĻā§āĻāĻž āĻ¸āĻŽā§ āĻ˛āĻžāĻā§āĨ¤ āĻ¤āĻžāĻ $$N$$ āĻ¯āĻ¤ āĻŦā§ āĻšāĻŦā§, āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ°āĻā§ āĻ¤āĻ¤ āĻŦā§āĻļāĻŋ āĻ¸āĻāĻā§āĻ¯āĻž āĻĻāĻŋā§ā§ āĻāĻžāĻ āĻāĻ°āĻ¤ā§ āĻšāĻŦā§, āĻāĻŦāĻ āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§āĻ° āĻ°āĻžāĻ¨ āĻāĻ°āĻ¤ā§ āĻ¸āĻŽā§ (āĻ°āĻžāĻ¨āĻāĻžāĻāĻŽ) āĻŦā§āĻļāĻŋ āĻ˛āĻžāĻāĻŦā§āĨ¤ āĻ°āĻžāĻ¨āĻāĻžāĻāĻŽ $$N$$-āĻāĻ° āĻ¸āĻžāĻĨā§ āĻĒāĻžāĻ˛ā§āĻ˛āĻž āĻĻāĻŋā§ā§ āĻāĻ¤ āĻĻā§āĻ°ā§āĻ¤ āĻŦāĻžā§āĻŦā§ āĻ¸ā§āĻāĻžāĻ° āĻāĻāĻāĻž āĻ§āĻžāĻ°āĻŖāĻžāĻ āĻāĻŽāĻ°āĻž āĻĒā§āĻ¤ā§ āĻĒāĻžāĻ°āĻŋāĨ¤
āĻ§āĻ° $$N$$ āĻ¸āĻāĻā§āĻ¯āĻžāĻāĻŋāĻ¤ā§ $$k$$ āĻ¸āĻāĻā§āĻ¯āĻ āĻĄāĻŋāĻāĻŋāĻ āĻāĻā§āĨ¤ āĻāĻŽāĻ°āĻž āĻāĻ¸āĻ¨ā§āĻ¨āĻāĻžāĻŦā§ $$N\approx 10^k$$ āĻ§āĻ°āĻ¤ā§ āĻĒāĻžāĻ°āĻŋāĨ¤ āĻ¯āĻĻāĻŋ $$N$$ āĻāĻāĻāĻŋ āĻŽā§āĻ˛āĻŋāĻ āĻ¸āĻāĻā§āĻ¯āĻž āĻšā§, āĻ¤āĻŦā§ āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽ ā§§-āĻ $$N$$-āĻā§ $$2$$ āĻĨā§āĻā§ $$N-1$$ āĻĒāĻ°ā§āĻ¯āĻ¨ā§āĻ¤ āĻ¸āĻāĻ˛ āĻ¸āĻāĻā§āĻ¯āĻž āĻĻāĻŋā§ā§āĻ āĻāĻžāĻ āĻāĻ°āĻž āĻšāĻŦā§āĨ¤ āĻ¯āĻĻāĻŋ āĻĒā§āĻ°āĻ¤āĻŋāĻŦāĻžāĻ° āĻāĻžāĻ āĻāĻ°āĻ¤ā§ āĻ¸āĻ°ā§āĻŦā§āĻā§āĻ $$c$$ āĻĒāĻ°āĻŋāĻŽāĻžāĻŖ āĻ¸āĻŽā§ āĻ˛āĻžāĻā§, āĻ¤āĻŦā§ āĻāĻĒāĻ°ā§āĻ° āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽāĻāĻŋ āĻ°āĻžāĻ¨ āĻāĻ°āĻ¤ā§ āĻ¸āĻ°ā§āĻŦā§āĻā§āĻ $$Nc=10^{k}c$$ āĻ¸āĻŽā§ āĻ˛āĻžāĻāĻŦā§āĨ¤ āĻāĻ° āĻ
āĻ°ā§āĻĨ āĻšāĻā§āĻā§, āĻĄāĻŋāĻāĻŋāĻā§āĻ° āĻ¸āĻāĻā§āĻ¯āĻž āĻŦāĻžā§āĻžāĻ° āĻ¸āĻžāĻĨā§ āĻ¸āĻžāĻĨā§ āĻāĻĒāĻ°ā§āĻ° āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§āĻ° āĻ°āĻžāĻ¨āĻāĻžāĻāĻŽāĻ āĻ¸ā§āĻāĻā§ā§āĻāĻžāĻŦā§ āĻŦā§āĻĻā§āĻ§āĻŋ āĻĒāĻžā§āĨ¤ [āĻŦāĻž, āĻŦāĻŋāĻ O āĻ¨ā§āĻā§āĻļāĻ¨ā§ āĻāĻĒāĻ°ā§āĻ° āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§āĻ° āĻ°āĻžāĻ¨āĻāĻžāĻāĻŽ $$O(10^{k})$$]
āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽāĻŋāĻ-āĻ āĻ°āĻžāĻ¨āĻāĻžāĻāĻŽ āĻ¸ā§āĻāĻā§ā§āĻāĻžāĻŦā§ āĻŦāĻžā§āĻž āĻā§āĻŦāĻ āĻĻā§āĻāĻā§āĻ° āĻāĻāĻāĻŋ āĻāĻĨāĻžāĨ¤ āĻāĻžāĻ°āĻŖ āĻāĻ¨āĻĒā§āĻā§āĻ° āĻāĻāĻā§āĻāĻžāĻ¨āĻŋ āĻĒāĻ°āĻŋāĻŦāĻ°ā§āĻ¤āĻ¨ā§ āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§āĻ° āĻ°āĻžāĻ¨āĻāĻžāĻāĻŽā§ āĻ°āĻžāĻ¤āĻĻāĻŋāĻ¨ āĻĒāĻ°āĻŋāĻŦāĻ°ā§āĻ¤āĻ¨ āĻšā§ā§ āĻ¯ā§āĻ¤ā§ āĻĒāĻžāĻ°ā§āĨ¤
āĻŽāĻ¨ā§ āĻāĻ° āĻ¤ā§āĻŽāĻžāĻ° āĻāĻžāĻā§ āĻ¯āĻĻāĻŋ āĻā§āĻ¨ āĻŽāĻšāĻžāĻā§āĻˇāĻŽāĻ¤āĻžāĻ§āĻ° āĻ¸ā§āĻĒāĻžāĻ° āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ° āĻāĻā§ āĻ¯ā§ $$200$$ āĻĄāĻŋāĻāĻŋāĻā§āĻ° āĻāĻāĻāĻž āĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽ ā§§ āĻĻāĻŋā§ā§ $$1$$ āĻ¸ā§āĻā§āĻ¨ā§āĻĄā§ āĻŦā§āĻ° āĻāĻ°āĻ¤ā§ āĻĒāĻžāĻ°ā§āĨ¤ āĻ¤āĻžāĻšāĻ˛ā§ $$210$$ āĻĄāĻŋāĻāĻŋāĻā§āĻ° āĻāĻāĻāĻž āĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖ āĻāĻ°āĻ¤ā§ āĻāĻŽā§āĻĒāĻŋāĻāĻāĻžāĻ°āĻāĻŋāĻ° āĻ¸āĻŽā§ āĻ˛āĻžāĻāĻŦā§ $$10^{210-200}=10^{10}$$ āĻ¸ā§āĻā§āĻ¨ā§āĻĄ $$\approx 317$$ āĻŦāĻāĻ°!
āĻāĻŽāĻĻā§āĻ° āĻā§āĻŦāĻ¨āĻāĻž āĻā§āĻŦ āĻā§āĻ, āĻāĻžāĻ¨ā§? āĻāĻāĻāĻž āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽā§āĻ° āĻāĻāĻāĻĒā§āĻ āĻĻā§āĻā§ āĻ¯āĻžāĻā§āĻžāĻ° āĻāĻ¨ā§āĻ¯ āĻāĻŽāĻžāĻĻā§āĻ° $$317$$ āĻŦāĻāĻ° āĻŦā§āĻāĻā§ āĻĨāĻžāĻāĻž āĻ¸āĻŽā§āĻāĻŦ āĻ¨āĻžāĨ¤ āĻ¤āĻžāĻ āĻā§āĻŦ āĻĢāĻžāĻ¸ā§āĻ āĻā§āĻ¨ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ āĻĻāĻ°āĻāĻžāĻ°āĨ¤
āĻĒā§āĻ°ā§āĻā§āĻ°āĻžāĻŽ ā§§-āĻā§āĻ āĻ
āĻ¨ā§āĻāĻāĻžāĻ¨āĻŋ āĻĢāĻžāĻ¸ā§āĻ āĻŦāĻžāĻ¨āĻžāĻ¨ā§ āĻ¯āĻžā§ āĻ¯āĻĻāĻŋ āĻāĻŽāĻ°āĻž $$N-1$$ āĻāĻ° āĻĒāĻ°āĻŋāĻŦāĻ°ā§āĻ¤ā§ $$\sqrt{N}$$ āĻĒāĻ°ā§āĻ¯āĻ¨ā§āĻ¤ āĻĒā§āĻ°āĻ¤āĻŋāĻāĻž āĻ¸āĻāĻā§āĻ¯āĻž āĻĻāĻŋā§ā§ āĻāĻžāĻ āĻĻā§āĻāĨ¤ āĻā§āĻ¨āĻ¨āĻž āĻĒā§āĻ°āĻ¤āĻŋāĻāĻŋ āĻā§āĻ¤ā§āĻ°āĻŋāĻŽ āĻ¸āĻāĻā§āĻ¯āĻžāĻ° $$\sqrt N$$-āĻāĻ° āĻŽāĻžāĻā§āĻ āĻ
āĻ¨ā§āĻ¤āĻ¤ āĻāĻāĻāĻž āĻā§āĻĒāĻžāĻĻāĻ āĻĨāĻžāĻā§āĨ¤ āĻāĻ° āĻ¯āĻĻāĻŋ āĻā§āĻĒāĻžāĻĻāĻ āĻ¨āĻž āĻĒāĻžāĻā§āĻž āĻ¯āĻžā§, āĻ¤āĻŦā§ āĻ¸āĻāĻā§āĻ¯āĻžāĻāĻŋ āĻĒā§āĻ°āĻžāĻāĻŽāĨ¤ āĻāĻ āĻĒāĻ°āĻŋāĻŦāĻ°ā§āĻ¤āĻ¨ā§āĻ° āĻĢāĻ˛ā§ āĻāĻŽāĻžāĻĻā§āĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽā§āĻ° āĻ°āĻžāĻ¨āĻāĻžāĻāĻŽ āĻšāĻŦā§ $$10^{\frac k 2}c$$ [āĻŦāĻž $$O\left(10^{\frac k 2}\right)$$]. āĻāĻŋāĻ¨ā§āĻ¤ā§ āĻāĻāĻžāĻ āĻ¯ā§ āĻ¸ā§āĻāĻā§ā§ āĻšāĻžāĻ°ā§ āĻŦāĻžā§ā§!
āĻāĻ° āĻĨā§āĻā§ āĻĢāĻžāĻ¸ā§āĻ āĻāĻŋ āĻā§āĻ¨ āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ āĻ¨ā§āĻ? āĻāĻ¤ā§āĻ¤āĻ°āĻāĻž āĻšāĻā§āĻā§, āĻšā§āĻ¯āĻžāĻ āĻāĻā§āĨ¤ āĻ¤āĻŦā§ āĻ¸ā§āĻāĻŋ āĻāĻžāĻ¨āĻ¤ā§ āĻšāĻ˛ā§ āĻ¤ā§āĻŽāĻžāĻ° āĻŽāĻĄā§āĻ˛āĻžāĻ° āĻāĻ°āĻŋāĻĨāĻŽā§āĻāĻŋāĻ āĻ¸āĻŽā§āĻĒāĻ°ā§āĻā§ āĻšāĻžāĻ˛āĻāĻž āĻĒāĻžāĻ¤āĻ˛āĻž āĻ§āĻžāĻ°āĻŖāĻž āĻ˛āĻžāĻāĻŦā§āĨ¤
āĻ¯āĻĻāĻŋ $$a$$ āĻāĻŦāĻ $$b$$ āĻā§ $$n$$ āĻĻāĻŋā§ā§ āĻāĻžāĻ āĻāĻ°āĻ˛ā§ āĻāĻāĻ āĻāĻžāĻāĻļā§āĻˇ āĻĨāĻžāĻā§, āĻ¤āĻžāĻšāĻ˛ā§ āĻāĻŽāĻ°āĻž āĻ˛āĻŋāĻāĻŋ $$a \equiv b\pmod{n}$$. āĻāĻāĻžāĻā§ āĻĒā§āĻž āĻšā§ $$a$$ is congruent to $$b$$ modulo $$n$$. āĻ¯ā§āĻŽāĻ¨: $$57\equiv 112\pmod{55}$$.
āĻ˛āĻā§āĻˇā§āĻ¯ āĻāĻ° āĻ¯ā§, āĻ¯āĻĻāĻŋ $$n$$ āĻĻāĻŋā§ā§ $$a$$-āĻā§ āĻāĻžāĻ āĻāĻ°āĻ˛ā§ $$r$$ āĻāĻžāĻāĻļā§āĻˇ āĻĨāĻžāĻā§, āĻ¤āĻŦā§ $$a\equiv r\pmod{n}$$ āĻ˛āĻŋāĻāĻž āĻ¯āĻžā§āĨ¤ āĻ¯ā§āĻŽāĻ¨: $$57\equiv 2\pmod{55}$$.
āĻ¯ā§āĻā§āĻ¨ āĻĒā§āĻ°ā§āĻŖāĻ¸āĻāĻā§āĻ¯āĻžāĻā§ āĻ¯āĻĻāĻŋ $$55$$ āĻĻāĻŋā§ā§ āĻāĻžāĻ āĻāĻ°, āĻ¤āĻŦā§ āĻ¤ā§āĻŽāĻŋ $$0$$ āĻĨā§āĻā§ $$54$$ āĻāĻ° āĻŽāĻžāĻā§ āĻā§āĻ¨ āĻāĻāĻāĻŋ āĻāĻžāĻāĻļā§āĻˇ āĻĒāĻžāĻŦā§āĨ¤ āĻ¤āĻžāĻšāĻ˛ā§, āĻ¤ā§āĻŽāĻŋ āĻ¯āĻĻāĻŋ āĻ
āĻ¨ā§āĻāĻā§āĻ˛ā§ āĻ°âā§āĻ¯āĻžāĻ¨ā§āĻĄāĻŽ āĻĒā§āĻ°ā§āĻŖāĻ¸āĻāĻā§āĻ¯āĻžāĻ° āĻ¨āĻžāĻ, āĻāĻĻā§āĻ° āĻŽāĻžāĻā§ $$\pmod {55}$$-āĻ āĻāĻ¨āĻā§āĻ°ā§ā§ā§āĻ¨ā§āĻ, āĻ
āĻ°ā§āĻĨāĻžā§ $$55$$ āĻĻāĻŋā§ā§ āĻāĻžāĻ āĻāĻ°āĻ˛ā§ āĻāĻāĻ āĻāĻžāĻāĻļā§āĻˇ āĻĨāĻžāĻā§, āĻāĻŽāĻ¨ āĻĻā§āĻāĻŋ āĻĒā§āĻ°ā§āĻŖāĻ¸āĻāĻā§āĻ¯āĻž āĻĒāĻžāĻā§āĻžāĻ° āĻāĻŋāĻā§āĻāĻž āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž āĻĨāĻžāĻā§, āĻ¤āĻžāĻ āĻ¨āĻž? āĻ¯āĻ¤ āĻŦā§āĻļāĻŋ āĻ¸āĻāĻā§āĻ¯āĻž āĻ¨ā§āĻŦā§, āĻāĻ āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻžāĻ āĻ¤āĻ¤ āĻŦāĻžā§āĻŦā§āĨ¤
āĻāĻŦāĻžāĻ°, āĻāĻāĻ āĻ¸āĻāĻā§āĻ¯āĻžāĻā§āĻ˛āĻŋāĻ° āĻŽāĻžāĻā§ $$\pmod {5}$$-āĻ āĻāĻ¨āĻā§āĻ°ā§ā§ā§āĻ¨ā§āĻ āĻĻā§āĻāĻŋ āĻ¸āĻāĻā§āĻ¯āĻž āĻĒāĻžāĻā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻžāĻ āĻĨāĻžāĻā§āĨ¤ āĻāĻŋāĻ¨ā§āĻ¤ā§ āĻāĻ āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻž $$\pmod{55}$$-āĻ āĻāĻ¨āĻā§āĻ°ā§ā§ā§āĻ¨ā§āĻ āĻšāĻā§āĻžāĻ° āĻ¸āĻŽā§āĻāĻžāĻŦāĻ¨āĻžāĻ° āĻā§ā§ā§ āĻ
āĻ¨ā§āĻ āĻŦā§āĻļāĻŋāĨ¤ āĻā§āĻ¨āĻ¨āĻž, $$\pmod{5}$$-āĻ āĻ¤ā§ āĻāĻžāĻ˛āĻŋ $$0$$ āĻĨā§āĻā§ $$4$$ āĻĒāĻ°ā§āĻ¯āĻ¨ā§āĻ¤ āĻāĻžāĻāĻļā§āĻˇ āĻāĻ¸āĻ¤ā§ āĻĒāĻžāĻ°ā§āĨ¤
āĻāĻ āĻ§āĻžāĻ°āĻŖāĻžāĻā§ āĻŦā§āĻ¯āĻŦāĻšāĻžāĻ° āĻāĻ°ā§ āĻā§āĻĒāĻžāĻĻāĻā§ āĻŦāĻŋāĻļā§āĻ˛ā§āĻˇāĻŖā§āĻ° āĻāĻāĻāĻŋ āĻāĻŽā§āĻāĻžāĻ° āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽ āĻ°ā§ā§āĻā§, āĻ¯āĻžāĻ° āĻ¨āĻžāĻŽ āĻĒā§āĻ˛āĻžāĻ°ā§āĻĄā§āĻ° āĻ°ā§($$\rho$$) āĻāĻ˛āĻā§āĻ°āĻŋāĻĻāĻŽāĨ¤ āĻĒāĻ°āĻŦāĻ°ā§āĻ¤ā§ āĻĒāĻ°ā§āĻŦā§ āĻāĻāĻŋ āĻ¨āĻŋā§ā§ āĻāĻ˛ā§āĻāĻ¨āĻž āĻāĻ°āĻŦāĨ¤
# Guidelines to Open a Successful Math Club
[I run Mymensingh Parallel Math School (MPMS), a free-for-all math club whose members (including me) have consistently won many prizes in Regional, National and International Mathematical Olympiads. Naturally, many math enthusiasts often ask me for tips on starting math clubs. So, 4 years ago, I wrote the following short note. I have decided to republish it here for future references.]
Unlike seeking for philosopher’s stone, starting a math club is very easy. You see, it requires only two things: a few classrooms and some members. Even a ‘teacher’ is not needed because the classes can find their way all by themselves! Although first a few month they need an instructor to teach them how to walk all-alone.
Through this tender age you have to be a bit careful. Try to drive the class in a relaxed mood so that everybody feels free to join in the discussion. It’s a very, very important thing for encouraging everyone to share his comments and ideas with others. In fact continuing this practice one day enables them to take the duster and the chalks in own hands. Moreover, we advise you to look for the true dedicated persons amongst the members. They will run it and never leave it. In this process MPMS has got its perfect structure.
May be the mostly asked questions are all about selecting class-topic. And yeah, this is too difficult to answer. Actually you can choose anything of your choice. There is neither any limitation nor any boundary. Math may appear boring sometimes. So why you don’t look at physics or literature then? In MPMS, physics, chemistry, economics, literature, biology-even philosophy take place frequently in the chaos. Also it is wise to discuss attractive topics like combinatorics at the early stage. Later you may go through harder things day by day depending on everybody’s capability to cope with it.
So…turn off your computer and come out to the field. See you at the BdMO National… đ
# An interesting application of AM-GM to prove the upper bound for the harmonic numbers
Harmonic series is a very well-known divergent series defined by
$\sum_{k=1}^\infty \frac 1 k =Â 1+\frac 1 2 +\frac 1 3+\frac 1 4\cdots$
The partial sum of first $$n$$ terms of this series is called the $$n$$th harmonic number. In other words, $H_n = \sum_{k=1}^n\frac 1 k$
Although there is no nice and simple formula for computing $$H_n$$, we can, however, do some approximation. For example, Euler proved $H_n = \ln(n)+\gamma+\varepsilon_n < \ln(n)+1$
where $$\gamma$$ is the EulerâMascheroni constant and $$\varepsilon_n\approx \frac 1 {2n}$$. I shall prove this bound here with AM-GM Inequality, which says, for positive real $$a_i$$s, that
$\frac 1 n \left(\sum_{i=1}^n a_i\right)\ge \sqrt[n]{\prod_{i=1}^n a_i}$
So we can write $1+\sum_{k=2}^n \frac{k-1}{k}> n\sqrt[n]{\prod_{k=2}^n \frac{k-1}{k}}=n\sqrt[n]{\frac{1}{n}}$
The inequality is strict since the $$\frac {k-1}k$$s are not equal. Also note that
$\sum_{k=2}^n \frac{k-1}{k}=\sum_{k=1}^n 1-\frac{1}{k}=n-H_n$
Plugging it back to the main inequality, we derive that
$1+n-H_n>n\sqrt[n]{\frac{1}{n}}$
And by rewriting
$1+n\left(1-\sqrt[n]{\frac{1}{n}}\right)>H_n$
Now, all we need to prove is
$\ln(n)\ge n\left(1-\sqrt[n]{\frac{1}{n}}\right)$
Assume $$e^m = n$$ and $$\dfrac {m} {e^m} = x$$. The previous inequality can be rewritten as
$m \ge e^m(1-e^{-x})$
It can be further simplified to
$x+e^{-x}\ge 1$
Now, the derivative of $$f(x)=x+e^{-x}$$ is $$1-e^{-x}$$, which is non-negative in $$[0, \infty )$$. And we also have $$f(0)=1$$. Therefore, we definitely have $$\displaystyle x+e^{-x}\ge 1$$ and we are done.
Remarks:
1. $\ln(n)\ge n\left(1-\sqrt[n]{\frac{1}{n}}\right)$
A weird inequality. You can even see from the previous proof that
$\lim_{n\to\infty}n\left(1-\sqrt[n]{\frac{1}{n}}\right) = \ln (n)$
Perhaps, with some luck, it can be used to derive some more interesting results. Gotta think later.
2. In my proof, the $$1$$ beside $$\ln(n)$$ just sat there to initiate the AM-GM. If we replace it by some function $$\lambda(n)$$ and do the calculation, we might be able to optimize our inequality down to the $$\gamma$$ bound. But I’m done for today. Anyone interested might give it a shot. | 41,102 | 43,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-39 | latest | en | 0.892307 |
https://www.daniweb.com/programming/software-development/threads/179206/recursion-problem-very-difficult-c-problem-picking-coin-game-no-idea-need-s | 1,723,459,212,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00243.warc.gz | 553,486,850 | 20,152 | Coin game: Alice and Bob are playing a game using a bunch of coins. The players
pick several coins out of the bunch in turn. Each time a player is allowed to pick 1, 2 or 4
coins, and the player that gets the last coin is the winner. Assume that both players are
very smart and he/she will try his/her best to work out a strategy to win the game. For
example, if there are 2 coins and Alice is the rst player to pick, she will de nitely pick 2
coins and win. If there are 3 coins and Alice is still the rst player to pick, no matter she
picks 1 or 2 coins, Bob will get the last coin and win the game.
Given the number of coins and the order of players (which means the rst and the second
players to pick the coins), you are required to write a program pickcoin.cpp to calculate
the winner of the game, and calculate how many different strategies there are for he/she
to win the game. You should use recursion to solve the problem, and the parameters are
read from the command line. You can assume that there are no more than 30 coins.
Here are some sample runs of the program:
./pickcoin 1 alice bob
alice 1
./pickcoin 2 bob alice
bob 1
./pickcoin 3 alice bob
bob 2
./pickcoin 10 alice bob
alice 22
./pickcoin 25 alice bob
alice 3344
./pickcoin 30 alice bob
bob 18272
I really have no idea about how to write this.. I have thought about it for a few days but still have no ideas about the logic of this programme. Could someone give me some help?
SHOULD you decide the winner by looking at the number of possibilities of winning?
As in if bob has 300 strategies of winning and Alice has 299? Should we decide Bob as the winner?
But there is only "one" strategy, and that's to keep the other player with multiple of 3 to choose from. If the other player ends up looking at a total of 3, then you win.
Eg.
Total is 6 (for them)
They pick 4, you pick 2 and win
They pick 2, you pick 1, leaving 3 and you win next turn
They pick 1, you pick 2, leaving 3 and you win next turn
Your example answers suggest the number of permutations of wins possible, if both players just play dumb. That is, randomly picking 1,2,4 and seeing who wins.
My idea
``````void play ( int n, int me, int them ) {
play( n-1, them, me );
play( n-2, them, me );
play( n-4, them, me );
}``````
n == 0 is a win for me
n < 0 is an invalid outcome
Though TBH, I think this rather gives the game away as it is.
Thank you Salem.
But to clarify one thing, the two players will not just pick randomly, they will choose to "win".
for example, we can make it clear in this way:
assume the order is alice, bob
for 1 coin, alice wins, 1 way
for 2 coin, alice wins, 1 way
for 3 coins, bob wins, 2 ways
for 4 coins, alice wins, 3 ways
for 5 coins, alice takes 2 first (other ways can not make her win) and make the situation the same as the initial situation: 3 coins and bob first, apparently there are 2 ways for alice to win.
for 6 coins, there are 3 initial actions alice can take, 1, 2 or 4. By taking 1, she reduces the situation to 5 coins, bob first, we have already analyzed that bob would win this situation with 2 ways. By taking 2, she reduces the situation to 4 coins, bob first, and we have shown bob would win with 3 ways. By taking 4, she reduces the situation to 2 coins, bob first, and bob wins with 1 way. So you can sum them up and come to the conclusion that bob wins with 6 ways.
Thinking about solem's direction. But this is a really difficult question, isn't it?
I really feel headache coding this.
Well if that is the case. Then the objective of the game will be to make the total number divisible by 3.
If in the first try the player can remove something ie : 1,2,4 coins and make the number divisible by 3 then he wins.
Consider
10
If i take in 1 coin, 9 remains. if the opponent takes 4. then i take 2 and win leave 3. Or else we loose.
If the number in the initial stages is divisible by 3 the player1 eventually will lose.
Just try all the other permutations and you will get it.
Secondly The total chances of winning can be calculated by random bruteforce i guess.
I am absolutely sorry . I dint see SALEMS POST.
Thanks.. but I still have no ideas about how to write the programme
and the code suggested by salem:
``````void play ( int n, int me, int them ) {
play( n-1, them, me );
play( n-2, them, me );
play( n-4, them, me );
}``````
Isn't only the first play( n-1, them, me ); will be executed?
I have really no ideas.. I have been dealing with it for almost a week and still nothing can be done.
Sincerely hope someone could give me a helping hand #_#
> But this is a really difficult question, isn't it?
Would you feel bad if I told you I'd already solved it?
> Isn't only the first play( n-1, them, me ); will be executed?
I take it you've never studied the archetype recursive function for Towers of Hanoi?
play( n-1, them, me )
It keeps doing this until it runs into the end-stop, then it returns and does
play( n-2, them, me );
and so on.
Actually, it took a couple of attempts. My snippet is exactly half of the solution, and it produces stupidly large numbers for permutations.
What's missing is the strategy code for the winning plays.
Sorry Salem, I know I am really stupid in programming >o< I really know..
void play ( int n, int me, int them ) {
play( n-1, them, me );
play( n-2, them, me );
play( n-4, them, me );
}
But is this function completed? Because nothing is returned and how can I determine who is the winner? Should I add something like this? (the base case)
if (n % 3 == 0)
return them;
if (n == 1 || n == 2 || n == 4)
return me;
n == 0 means "me" is the winner.
Notice this?
void play ( int n, int me, int them ) {
play( n-1, them, me );
This is how it alternates between taking turns.
play( n-1, them, me );
play( n-2, them, me );
play( n-4, them, me );
This is what the losing player has to do - that is, they're stuck with n % 3 == 0
Try every possible move to see what works (if anything).
n == 0 means "me" is the winner.
Notice this?
void play ( int n, int me, int them ) {
play( n-1, them, me );
This is how it alternates between taking turns.
play( n-1, them, me );
play( n-2, them, me );
play( n-4, them, me );
This is what the losing player has to do - that is, they're stuck with n % 3 == 0
Try every possible move to see what works (if anything).
I run the programme like this and what comes out is segmentation fault and the programme runs the play( n-1, them, me ); until n is negative infinity.
``````#include <iostream>
#include <cstdlib>
using namespace std;
void play(int n, int me, int them){
play(n-1, them, me);
play(n-2, them, me);
play(n-4, them, me);
}
int main(int argc, char * argv[]){
int n = atoi(argv[1]);
play(n, 0, 1);
return 0;
}``````
I am still very confused.. could you write the return and checking result for the function?
``````void play(int n, int me, int them){
if ( n == 0 ) {
// me is the winner
counter[me]++;
} else if ( n < 0 ) {
// took too many, just ignore it as we're trying every combo
} else {
play(n-1, them, me);
play(n-2, them, me);
play(n-4, them, me);
}
}``````
EVERY recursive function needs a base case which stops it looping forever.
``````void play(int n, int me, int them){
if ( n == 0 ) {
// me is the winner
counter[me]++;
} else if ( n < 0 ) {
// took too many, just ignore it as we're trying every combo
} else {
play(n-1, them, me);
play(n-2, them, me);
play(n-4, them, me);
}
}``````
EVERY recursive function needs a base case which stops it looping forever.
Is it the completed function? I spend 3 hours working on this but still fail...
You use "void" as the return type in the function. Should I return the value of "me" or return the int count?
Well counter in the above function might have been declared globally. So in after the total function has worked out all the permutations.
Counter[me] // will have all the times 'ME' has won
and Counter[them] will have the times "them" has won the game.
(Dont forget to initialise (Counter[me] and Counter[them] ) to zero.
I guess you should post in the total code which you are currently working on.
But i STILL HAVE A Doubt for Salem..
The Code
``````void play(int n, int me, int them){
if ( n == 0 ) {
// me is the winner
counter[me]++;
}``````
Doesnt the above imply that them is the winner. Because He was the last person to take in the coins and the me has nothing left to take. So because them has taken the last coin. Me has nothing left with him.
I think the function should be
``````void play(int n, int me, int them){
if ( n == 0 ) {
// them is the winner
counter[them]++;
}``````
commented: Excellent catch - well done. +29
> Doesnt the above imply that them is the winner. Because He was
> the last person to take in the coins and the me has nothing left to
> take. So because them has taken the last coin. Me has nothing left
> with him.
You could very well be right, well spotted.
My program produced the right answer for one of them, and I gave up even thinking about it any further; the challenge was all but over.
``````#include <iostream>
#include <cstdlib>
using namespace std;
int way;
int play(int n, int me, int them){
if (n == 1 || n == 2 || n == 4){
way++;
return me;
}
if (n > 1 && n != 4 && n != 2)
play(n-1, them, me);
if (n > 2 && n != 4)
play(n-2, them, me);
if (n > 4)
play(n-4, them, me);
}
int main(int argc, char * argv[]){
int result;
int n = atoi(argv[1]);
result = play(n, 0, 1);
if (result == 0)
cout << argv[2] << " ";
else if (result == 1)
cout << argv[3] << " ";
cout << way;
return 0;
}``````
Here is my code. But the way is not correct.
If I don't add the condition before play, Salem's programme will never end.
Could you correct it for me?
If n % 3 is 1, then you can either take 1, or take 4, and still be in a winning position.
Please could u provide the entire code for the above program?
u see i hav tried but failed 2 understand it...
Anybody has the code? using backtracking recursion
Be a part of the DaniWeb community
We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge. | 2,796 | 10,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-33 | latest | en | 0.964054 |
https://stuff.mit.edu/people/agustya/Math/4-13.html | 1,643,271,863,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305242.48/warc/CC-MAIN-20220127072916-20220127102916-00709.warc.gz | 599,448,839 | 10,871 | # USAMTS Problem 1/4/13
In a strange language there are only two letters, a and b, and it is postulated that the letter a is a word. Furthermore, all additional words are formed according to the following rules:
A. Given any word, a new word can be formed from it by adding a b at the rightmost end.
B. If in any word a sequence aaa appears, a new word can be formed by replacing the aaa by the letter b.
C. If in any word the sequence bbb appears, a new word can be formed by omitting bbb.
D. Given any word, an new word can be formed by writing down the sequence that constitutes the given word twice.
Prove that in this language baabaabaa is not a word.
If we ignore b’s and look at how the rules apply to the number of “a”’s, in a legal word then:
Rules A and C can be ignored
Rule B allows us to delete any sequence aaa or number of a’s can be reduced by 3.
Rule D allows us to double the number of a’s
Now, if the number of a’s in a word is not divisible by 3, then:
Rule B cannot transform that word into a word, where the number of a’s is divisible by 3 because it reduces the original number by 3 , and original number is not divisible by 3.
Rule D cannot transform that word into a word where the number of a’s is divisible by 3 because it doubles the number of a’s, and the original number of a’s is not divisible by 3.
It is postulated that “a” is a word, and 1 is not divisible by 3. Thus it can not generate a word where the number of a’s is divisible by 3. Thus we can not make a word with 6 a’s in it, because 6 is a multiple of 3
Thus, baabaabaa is not a word because it has 6 a’s
(A less mathematical but perhaps more convincing to some approach is the fact that baabaabaa does get flagged by my spellchecker).
# USAMTS Problem 2/4/13
Let f(x) = x [x [x [x]]] for all positive real numbers x, where [y] denotes the greatest integer less than or equal to y.
1. Determine x so that f(x) = 2001
2. Prove that f(x) = 2002 has no solution
1. It is easy to see that f(x) increases as x increases (because [x
[x
[x]]] either remains the same or increases when x increases.)
We see that:
f(6) = 64 =1296
f(7) = 74 = 2401
So, if f(x) = x
[x
[x
[x]]] = 2001,
6 < x < 7
Let [x
[x
[x]]] = m (m is an integer by the definition of []
So, f(x) = x*m = 2001
or x = 2001/m
Since 6 < x < 7, we have 286 < m < 333 (Eq 1)
Also, since 6<x < 7, [x] = 6
or x
[x] < 42, or [x
[x]] < 41
or x
[x
[x]] <287,
or [x
[x
[x]]] < 286 (Eq 2)
So, eq 1 and Eq 2 gives: 286 > m > 286 ; thus m = 286,
Thus, x = 2001/286
1. As we have shown in part 1,
f(6) = 64 =1296
f(7) = 74 = 2401
and 1296 < 2002 < 2401
So, if f(x) = 2002, (if it has a solution), it must satisfy
6 < x < 7
Thus, from Eq 2,
[x
[x
[x]]] < 286, so
x
[x
[x
[x]]] < 2002
Thus, f(x) = 2002 has no solution.
# USAMTS Problem 3/4/13
Let f be a function defined on the set of all integers, and assume that it satisfiest he following properties:
## A.f(0) ≠ 0
1. f(1) = 3
2. f(x)f(y) = f(x + y) + f(x
y)
Determine f(7).
Substituting x = 1 and y = 0 in Eq. C, and using Eq. B for f(1), we have:
f(1)f(0) = f(1 + 0) + f(1 0)
= 3 + 3
or 3 · f(0) = 6
or f(0) = 2
Similarly, when x = y = 1 in Eq. C, we have
f(1)f(1) = f(1 + 1) + f(1 1)
or 3 · 3 = f(2) + 2
or 9 2 = f(2)
or f(2) = 7
Similarly,
f(2)f(1) = f(2 + 1) + f(2 1)
or 7 · 3 = f(3) + 3
or 21 3 = f(3)
or f(3) = 18
and,
f(2)f(2) = f(2 + 2) + f(2 2)
or 7 · 7 = f(4) + 2
or 49 2 = f(4)
or f(4) = 47
Thus,
f(4)f(3) = f(4 + 3) + f(4 3)
or 47 · 18 = f(7) + 3
or 846 3 = f(7)
Thus, f(7) = 843
# USAMTS Problem 4/4/13
A certain company has a faulty telephone system that sometimes transposes a pair of adjacent digits when someone dials a three-digit extension. Hence a call to x318 would ring at either x318, x138, or x381, while a call received at x044 would be intended for either x404 or x044. Rather than replace the system, the company is adding a computer to deduce which dialed extensions are in error and revert those numbers to their correct form. They have to leave out several possible extensions for this to work. What is the greatest number of three-digit extensions the company can assign under this plan?
In order for the computer to discern the correct extension, no two extensions may be related by having two adjacent interchanged digits. The key is the middle digit and we can say that this digit may get interchanged with its neighbor. Thus if we can correctly identified the middle digit we can correct the mistake.
We will look at extensions of three types:
Case 1 - those with all digits identical. (E.g. 111, 222)
Case 2 - those with exactly two identical and different digit (e.g. 404), and
Case 3 - those with three distinct digits (e.g. 123).
Case 1 - There are 10 such cases, and since all digits are identical, one number can be assigned for each of these cases without any confusion.
Thus, this case gives 10 extensions.
Case 2 - Here the “lone” distinct digit can be assigned as the “middle” digit, and the other (two identical digits) one flanking it. (We assign only those numbers where the “lone” digit is in the middle)
This case gives: 10*9 = 90 extensions.
Case 3 - Here we identify the middle digit as the smallest of the three digits. (Smallest is identifiable because all three digits are distinct). We assign only those extensions that satisfy this property. (Thus, if we get a call at 587, we know that the number being called was 857 because 5 is the smallest number and there is only one flip possible.)
This gives us (10 * 9 * 8)/3 = 240
(There are 10*9*8 permutations, and in each case only one of the three digits is the smallest.)
Thus, the total number of extensions possible under this system is 10 + 90 + 240 = 340
# USAMTS Problem 5/4/13
Determine the smallest number of squares into which one can dissect an 11 x 13 rectangle, and exhibit such a dissection. The squares need not be of different sizes, their bases should be integers, and they should not overlap.
1
5
4
4
7
6
In the figure below, I have divided the given rectangle into six squares of side length 7, 6, 5, 4, and 1.
We will show that 6 is the smallest number of squares that we can divide this rectangle into, by showing that no solution exists for any number less than 6.
Theorem 1: If there is a rectangular strip of size (n x an), where n and an are both integers (a need not be an integer), then, then it will dissect into at least “a” squares
Proof: no square can have area larger than n2, and we need to cover an area of an2 . QED
Theorem 1a: Theorem 1 can be applied to individual rectangular strips if they do not have a common border, and the minimum number of squares needed, for these strips can not be less than the sum of each section.
Proof: There cannot be a common square between such strips. Thus proved.
Theorem 2: any solution with a square larger than 7 units, will require more than 6 squares
Proof: If there is a square with side length 8, we will have at least two additional strips of 3 x 8 and 5 x 8, (without a common border). These would require, including the first square, at least 1+3+2 , or at least 6 squares. (In fact, it would require a minimum of 7, because there is still a “gap.”).
If we have a square with side length greater than 8, then the strip will be even narrower and we would need more than 6 squares in each case:
For square with side length 9, we have strips of 2 x 9 and 4 x 9, which would need at least 7 square.
Side length of 10 square would give strips of 1 x 10 and 3 x 10 which would need at least 14 squares.
11 length square would give a strip of 2 x 11 which would need at least 7 squares.
Now, let us consider the squares toughing the outer boundary of the rectangle. Since each square has side length less than 8 (Theorem 2) no one square can touch two opposite sides. So, if we take the side lengths of the squares touching the outer boundary as a, b, c, and d and arrange them such that
a < b < c <
and obviously a > 2;
then the total number of squares is > a + b + c + d 4.
We only have to consider the cases where:
a = b = c = d = 2 Case 1
or
a = b = c = 2, d = 3 Case 2
In all other cases, a + b + c + d 4 > 6
We can easily prove that Case 1 is impossible because, assuming that the dimension of a square on the 11-side is x, than the square adjacent to it is 11 x, the square adjacent to that one would be 13-(11 x) = x+2. Calculating from the other side, we get 11-(13-x) = x-2. This would mean that the fourth figure has dimensions of both x + 2 and x 2, which is impossible for a square.
For Case 2 : Since 3 sides of rectangle have 2 squares touching them, (and only one side has 3 squares touching it), consider the 13-side which has only 2 squares touching.
Since we do not allows squares with side length > 7, there is only one way to divide 13 in two parts, so that two squares touch this side: namely:
6 + 7
The other 13-side cannot have only 2 squares touching it as the only combination is 6 and 7, and 62 + 72 + 62 + 72 > 143, so it must have 3 squares touching it, and thus both 11-sides must have only 2 squares touching them. We get the dimensions of the other two squares on the 11-sides as 11-6=5 and 11-7= 4,
so the 13-side with three squares touching it has third side 13 5 - 4 = 4.
This gives us the solution given at the top of the page, and since there are no other solutions of this type, No other solution exist with smaller number than 6.
QED. | 2,945 | 9,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-05 | latest | en | 0.917565 |
https://classbasic.com/category/mathematics-2/mathematics-jss-3/ | 1,686,175,635,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00645.warc.gz | 201,834,323 | 63,426 | # Category: Mathematics JSS 3
## Mathematics Guides for JSS 3 Number and Numeration
 MATHEMATICS THEME – NUMBER AND NUMERATION TOPIC 1 – RATIONAL AND NON RATIONAL NUMBERS  INSTRUCTIONAL MATERIALS 1. Thread or string 2. Table rule 3. Cylindrical shaped objects  LEARNING OBJECTIVES By the end of the lesson, students should be able to able to identify rational and non-rational numbers.  CONTENTS OF THE LESSON
## Mathematics Guides for JSS 3 Measurement and Geometry (Shapes and Angles)
 MATHEMATICS THEME – MEASUREMENT AND GEOMETRY TOPIC 1 – AREA OF PLANE FIGURES  INSTRUCTIONAL MATERIALS 1. Triangular shapes 2. Models of parallelogram 3. Models of trapezium 4. Model of circles and sectors 5. Flash cards with word problems  LEARNING OBJECTIVES By the end of the lesson, students should be able to: 1.
## Mathematics Guides for JSS 3 Data Presentation and Measure of Central Tendency
 MATHEMATICS THEME – DATA COLLECTION AND PRESENTATION TOPIC – PIE CHART AND MEASURE OF CENTRAL TENDENCY  INSTRUCTIONAL MATERIALS 1. Pie charts 2. Mathematical set 3. Data chart on activities  LEARNING OBJECTIVES By the end of the lesson, students should be able to: 1. represent and interpret an information on pie charts. 2.
## Mathematics Guides for JSS 3 Basic Operations (Binary Number System)
 MATHEMATICS THEME – BASIC OPERATIONS TOPIC – ADDITION OF NUMBERS IN BASE 2 NUMERALS INSTRUCTIONAL MATERIALS 1. Counters and sum cards 2. Chart showing the multiplication of two 2-digit binary numbers. 3. Chart showing the division of two 2-digit binary numbers.  LEARNING OBJECTIVES By the end of the lesson, students should be able
## Mathematics Guides for JSS 3 Algebraic Processes
 MATHEMATICS THEME – ALGEBRAIC PROCESSES TOPIC 1 – FACTORIZATION INSTRUCTIONAL MATERIALS 1. Quadratic equation box 2. Flash cards of word problems  LEARNING OBJECTIVES By the end of the lesson, students should be able to: 1. factorize simple algebraic expressions. 2. factorize quadratic algebraic expressions using quadratic equation box. 3. solve word problems involving | 565 | 2,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-23 | longest | en | 0.683207 |
https://ktbssolutions.com/1st-puc-maths-question-bank-chapter-15/ | 1,726,227,793,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00605.warc.gz | 314,476,098 | 23,606 | # 1st PUC Maths Question Bank Chapter 15 Statistics
Students can Download Maths Chapter 15 Statistics Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka 1st PUC Maths Question Bank Chapter 15 Statistics
Question 1.
Find the mean deviation about the mean for the following data
(i) 6, 7, 10, 12, 13, 4, 8, 12
(ii) 4, 7, 8, 9, 10, 12,13,17
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer:
Question 2.
Find the mean deviation about the mean for the data.
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5.
Answer:
Try your self.
$$N=20, \bar{x}=10, \Sigma\left|x_{i}-\bar{x}\right|=124, \mathrm{M} \cdot \mathrm{D} \cdot(\bar{x})=6 \cdot 2$$
a
Question 3.
Find the mean deviation about the median for the data:
(i) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
(ii) 36, 72, 46, 60, 45, 53, 46, 51, 49, 42
(iii) 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21
Answer:
(i) Arrange the data in ascending order as
Question 4.
Find the mean deviation about the mean for the following data.
Answer:
Question 5.
Find the mean deviation about the median for the data,
Answer:
Question 6.
Find the mean deviation about the mean for the following data.
(i)
Income per day Number of persons 0 – 100 4 100 – 200 8 200 – 300 9 300 – 400 10 400 – 500 7 500 – 600 5 600 – 700 4 700 – 800 3
(ii)
Height (cms) Number of boys 95 – 105 9 105 – 115 13 115 – 125 26 125 – 135 30 135 – 145 12 145 – 155 10
(iii)
Marks obtained Number of students 0 – 10 12 10 – 20 18 20 – 30 27 30 – 40 20 40 – 50 17 50 – 60 6
(iv)
Marks obtained Number of students 10 – 20 2 20 – 30 3 30 – 40 8 40 – 50 14 50 – 60 8 60 – 70 3 70 – 80 2
Answer:
Method (II)
Method (I)
Method (ii)
Here a = 130, h = 10
(iii) Method (I)
Question 7.
Find the mean deviation about the median for the data:
(i)
Class Frequency 0 – 10 6 10 – 20 7 20 – 30 15 30 – 40 16 40 – 50 4 50 – 60 2
(ii)
Marks Number of Girls 0 – 10 6 10 – 20 8 20 – 30 14 30 – 40 16 40 – 50 4 50 – 60 2
Answer:
(ii)
Question 8.
Calculate the mean deviation about median age for the age distribution of 100 persons given below.
Age Number 16 – 20 5 21 – 25 6 26 – 30 12 31 – 35 14 36 – 40 26 41- 45 12 46 – 50 16 51 – 55 9
Answer:
First, modify the classes to make the data. continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit.
Question 9.
Find Mean and variance for each data:
(i) 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
(ii) 6, 7, 10, 12, 13, 4, 8, 12
(iii) First n natural numbers.
(iv) First 10 multiples of 3.
Answer:
Question 10.
Find mean and variance.
(i)
xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3
(ii)
xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3
Answer:
(i)
(ii)
Question 11.
Find the mean and standard deviation using short-cut method.
xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5
Answer:
Question 12.
Find the mean and variance
(i)
Class Frequency 0 – 30 2 30 – 60 3 60 – 90 5 90 – 120 10 120 – 150 3 150 – 180 5 180 – 210 2
(ii)
Class Frequency 0 – 10 5 10 – 20 8 20 – 30 15 30 – 40 16 40 – 50 6
Answer:
Question 13.
Find the mean, variance and standard deviation using short-cut method.
Height (cms) No of Children 70 – 75 3 75 – 80 4 80 – 85 7 85 – 90 7 90 – 95 15 95 – 100 9 100 – 105 6 105 – 110 6 110 – 115 3
Answer:
Question 14.
The diameters of circles (in mm ) drawn in a design are given below calculate S.D and mean diameter.
Diameter Circles 70 – 75 15 75 – 80 17 80 – 85 21 85 – 90 22 90 – 95 25
Answer:
Question 15.
From the data given below state which group is more variable, A or B?
Marks Group A Group B 10 – 20 9 10 20 – 30 17 20 30 – 40 32 30 40 – 50 33 25 50 – 60 40 143 60 – 70 10 15 70 – 80 9 7
Answer:
We know that, the group having greater C.V.(co-efficient of variation) is said w be more variable than other. So, we compute mean and S.D. for groups A and B.
First we compute mean and S.D. for A.
Clearly, B having greater C. V. than A. Therefore B is more variable.
Note:
For two frequency distributions with equal means, the series with greater S.D. is more variable than the other.
Question 16.
From the prices of shares X and Y below, find out which is more stable in value.
Answer:
Here, number of items for both = 10.
First we compute mean and S.D. for x, taking a = 56, N = 10.
Question 17.
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results.
A B No. of wage earners 586 648 Mean of monthly wages (Rs ) 5253 5253 Variance of the distribution of wages 100 121
(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm A or B shows greater variability in individual wages?
Answer:
(i) Amount of monthly wages paid by firm A.
= 586 x mean wages
= 586 x 5253 = Rs.3078258
And amount of monthly wages paid by form B
= 648 x mean wages
= 648 x 5253 = Rs. 3403944.
Clearly, firm B pays more wages.
Alternatively,
The variance of wages in A is 100.
∴ S.D. of distribution of wage in A = 10
Also the variance of distribution of wages in firm B is 121.
∴ S.D. of distribution of wages in B = 11.
since the average monthly wages in both firms is same i.e., Rs. 5253, therefore the firm with greater S.D. will have more variability.
∴ Firm B pays more wages.
(ii) Firm B with greater S.D. shows greater variability in individual wages.
Question 18.
Two plants A and B of a factory show following results about the number of workers and the wages paid to them.
A B No – of workers 5000 6000 Average monthly wages 2500 2500 Variance distribution of wages 81 100
In which plant A or B is there greater variability in individual wages?
Answer:
B. (Try yourself) (Ex. 13 Text book)
Question 19.
The following is the record of goals scored by team A in a football session:
For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Answer:
First, compute mean number of goals and S.D. for team A.
Given : Mean number of goals for the team B is 2 and S.D. is 1.25
Since S.D. For A is less than S.D for B.
∴ A is more consistent.
Question 20.
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
Which is more varying, the length or weight?
Answer:
Question 21.
Co-efficient of variation of two distributions are 60 and 70, and the standard deviations are 21 and 16, respectively. What are their arithmetic means.
Answer:
Let $$\bar{x}_{1} \text { and } \bar{x}_{2}$$ be arithmetic means of first and second distribution respectively. Then
Question 22.
The following values are calculated in respect of heights and weight of the students of a section of class XI.
Height Weight Mean 162 – 6 cm 52-36 kg Variance 127 – 69 cm2 23. 1361kg2
Can we say that the weight show greater variation than the height.
Answer:
Clearly C.V. (weights) is greater than C.V. (heights). Weights show more variability than heights.
Miscellaneous Examples
Question 1.
The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13. Find the remaining two observations.
Answer:
Let the remaining two observation be x and y.
Question 2.
The mean of 5 observations is 4.4 and their variance is 8.24. If three observations are 1, 2 and 6, find the remaining two observations.
Answer:
Let the remaining two observations be x and
∴ Five observations are 1, 2, 6, x, y.
Question 3.
The mean and variance of 7 observations are 8 and 16 respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.
Answer:
6 and 8 (Try yourself)
Here, x + y = 14, x2 + y2 = 100
Question 4.
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation.
Answer:
Let six observations be
Question 5.
Given that $$\bar{x}$$is the mean and σ2 is the variance of n observations x1 , x2,……….. ,xn . Prove that the mean and variance of the observations ax1,ax2,………………,axn are $$a \bar{x}$$ and a2a1 respectively (a ≠ 0)
Answer:
Question 6.
If each of the observation x1,x2,………. ,xn is increased by ‘a’, where a is positive or negative number, show that the variance remains unchanged.
Answer:
Question 7.
The mean and standard deviation 20 observations are found to be 10 and 2 respectively. On rechecking, it was fond that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases :
(i) If wrong item is omitted,
(ii) If it is replaced by 12.
Answer:
Question 8.
The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?
Answer:
Given: N = 100
Incorrect mean = 40 = $$a \bar{x}$$
Incorrect S.D. = 5.1
We have, $$\bar{x}=\frac{1}{N} \Sigma x_{i}$$
⇒$$\Sigma x_{i}=N \bar{x}=100 \times 40=4000$$
∴ Incorrect sum of observations = 4000.
∴ Correct sum of observations = 4000 – 50 + 40 = 3990.
Question 9.
The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find and standard deviation if the incorrect observations are omitted.
Answer:
Given N = 100,New N = 97
Question 10.
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below.
Subject Mathematics Physics Chemistry Mean 42 32 40 – 9 S D 12 15 20
Which of three subjects shows the highest variability in marks and which shows the lowest?
Answer:
Chemistry shows high variability and Mathematics shows lowest variability.
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Electronics Beginner Tutorial
Part 1: Resistors, Volt and Current
By Ibrahim Kamal (ika@ikalogic.com)
Table of content: Overview Imagine the electronics The resistor Ohm's law Series and parallel resistors Resistors used as voltage divider Resistors used for current limiting Nodal analysis Power rating of resistors Special resistors Schematic symbols Resistor color code
Overview This chain of articles for absolute beginners aims to introduce the practical side of electronics; to present to you most of the simple aspects that you need to know to be able to understand and build simple circuits like temperature and light sensors, Timing circuits, power supplies, or in a next stage, more complicated Micro controller based projects like the ones listed on this site. In this part we will study the most basic component in electronics, the resistor and its interaction with the voltage difference across it and the electric current passing through it.
Imagine the electricity
As a beginner, it is important to be able to imagine the flow of electricity. Even if you've been told lots and lots of times how electricity is composed of electrons traveling across a conductor, it is still very difficult to clearly imagine the flow of electricity and how it is affected by Volt and resistors. This is why I am proposing this simple analogy with a hydraulic system, which anybody can easily imagine and understand, without pulling out complicated fluid dynamics equations.
Fig. 1: Electrical and Hydraulic system analogy
Notice how the flow of electricity resembles the flow of water from a point of high potential energy (high voltage) to a point of low potential energy (low voltage). In this simple analogy water is compared to electrical current, the voltage Difference is compared to the head difference between tow water reservoirs, and finally the valve resisting the flow of water is compared to the resistor limiting the flow of current.
From this analogy you can deduce some rules that you should keep in mind during all your electronics work:
Electric current through a single branch is constant at any point (exactly as you cannot have different flow rates in the same pipe; what's getting out of the pipe must equal what's getting in)
There wont be any flow of current between 2 points if is there is no potential difference between them. In other words, for a flow of current to exist, there must be a voltage difference between tow points.
The quantity of water in the reservoir can be compared to the electric charge stored in a battery. When the level of water in the tow tanks become the same, there is no more flow of water, and comparatively, a battery is empty and cannot deliver anymore current when the tow electrodes have the same voltage.
The electric current in a conductor will increase with the decrease of the resistance, exactly as the rate of flow of water will increase with the decrease of the resistance of the valve.
I could write a lot more deductions based on this simple analogy, but we can summarize those rules in the most fundamental equations of electronics: Ohm's law, that you shall learn in the rest of this article.
The resistor
The resistor can be defined by it's main purpose, a device to control or limit the flow of current, hence we can say that the main parameter of a resistor is it's resistance, which is measured in Ohm's (). Never less, another design consideration when working with resistors is its rated power, measured in watts (W), which is the quantity of power the resistor can dissipate without burning out. It is also important to note that resistors are not only used for current limiting, they can also be used as voltage dividers to generate very precise voltages out of bigger voltages. Some sensors are based on a resistance that Fig.2: 1 Watt 880 Ohm resistors
varies depending on light, temperature or shear stress, like the LDRs (light dependent resistors), Thermistors (temperature dependent resistors) or strain gauges. For more information and pictures, see special resistors at the end of the article.
Ohm's law
It's clear that those 3 equations at the right are different variations of Ohm's law, but the 3 of them must be very clear in your mind in order to proceed to more complicated circuits. You have to be able to understand and imagine the meaning of the equation 2 for example, which implies that a rise of voltage with a constant resistance will cause a rise of current. However, it wouldn't be logically true to say that a current rise will cause a voltage rise if the resistance is constant (even though this is mathematically true) because it's the voltage, the potential difference, that will create a flow of current, not the opposite (refer to the analogy of the 2 water tanks). Also, equation 3 can be used to deduce the value of the resistance to used to limit the flow of current to a certain value under a constant known voltage difference. Those are just examples showing you the importance of this rule. You will learn how to use them along the rest of the article Even the most Fig. 3: Ohm's law. R: Resistance (ohm) V: Volts (V) I: current (I)
sophisticated electronic simulations software uses this equation, along with some other equations to solve and simulate the most complicated circuits.
Series and parallel resistors
Understanding what is the effect of connecting resistors in series or in parallel is very important and will help you to analyze and simplify an electronic circuit, using those simple mathematical relations for series and parallel resistors:
Fig. 4A In this example circuit (figure 4A), R1 and R2 are connected in parallel, a single resistor R3 can provide the exact same function of the tow resistors R1 and R2, according to the law: Which, in case of only 2 parallel resistors, can be written as: Not only this relation can be used to simplify complicated circuits, but it can also be used to create resistors of values that you don't have.
Notice also that the value of R3 will always be smaller than the 2 other equivalent resistors. Which is logic, because adding more resistors in parallel provides additional paths to the electrical current, decreasing the overall resistance of the circuit.
Fig. 4B Series resistors can also be grouped together and replaced by one resistor, whose value would equal the summation of the tow initial resistors, which is again very logical, due to the fact that this configuration of resistance will provide additional resistance to the flow of current. Therefore the equivalent resistor R3 can be very simply calculated by the relation:
Fig. 4C At last, be careful of this very common pitfall among beginners, which is shown in figure 4C. It must be very clear that for 2 resistors to be considered as series resistors, they must be share the same current, as you will see in the rest of the tutorial.
Resistor used for current limiting
The most basic role of resistors is current limiting, which consist of precisely controlling the quantity of electrical current that is going to flow through a device or a conductor. To
understand how current limiting resistors work, let's first study this simple schematic (figure 5A), where a lamp is directly connected to a 9V battery. A lamp, like any other device that consumes electricity to accomplish a certain task (like providing light in this example) has an internal resistance, that determines how much current it will consume. So, from now on, any resistive device can be replaced in by a resistance especially in electronics schematics. (you can also notice in Figure 5B -which is equivalent to the 9V battery and Fig.5A
the lamp- the symbol of a resistor in the schematics, and how it is connected to the the Positive and negative power sources). Now if the lamp is to be considered a resistor, we can then use Ohm's law to calculate the current passing through it. Ohm's law states that the current passing through a resistor is equal the the voltage difference across it divided by the resistance of that Fig. 5B
resistor. This is mathematically written as:
, or more accurately as:
.
Where () is the voltage difference across the resistor and (R) is the resistance of the resistor (it's value)
Now notice the Figure 5C, where a current limiting resistor have been added. This resistance will limit the current going to the lamp, simply as it's name implies. You can control, to a very precise extent, the amount of current flowing through the lamp simply by choosing the right value for the resistor R1. A large resistor will highly reduce the current while a small resistor will allow more current (exactly as in our hydraulic analogy, where the valve is compared to Fig. 5C
a resistor). Mathematically speaking, this can be written as:
.
It is then clear that the value of the current will decrease if the value of the resistor R1 increases. Hence a resistor can be used to limit the current
. However it is important to note that this comes with a cost, which is the heat dissipated into the current limiting resistor, and you must chose a resistor of a suitable power rating, as you will see in the rest of the tutorial.
Resistors used as Voltage divider
Fig. 6A Fig. 6B As the name implies, resistors can also be used as voltage divider, in other words they can be used to generate any voltage from an initial bigger voltage by dividing it. The mathematical relation for this resistor configuration shown in figure 6A (that you could easily prove using the nodal analysis method) is: ...(Equ. 6A) In case both resistors have the same value (), the equation can be written as: Another common special case of this resistor configuration, is when the lower resistor is connected to ground (0V) as shown in figure 6B. Replacing Vb by 0 in the equation 6A, we get: ...(Equ. 6B) Which is the most common voltage divider equation.
You could imagine a lot more of special cases for this resistor configuration, and you shall discover them as you are working your way into the field of electronics.
Nodal analysis
Now that you're beginning to deal with electronic schematics, it is important to be able to analyze them and calculate any required voltage, current or resistance. There are many ways to study an electronic circuit, one of the most common methods is the Nodal analysis, where you simply apply a set of rules on a circuit of any size and calculate step by step all the required variables.
Simplified nodal analysis rules:
Fig. 7A Definition of a node A node is any point in the circuit. Points that are connected to each others by wires, without any other component between them are considered as a single node. Therefore, the infinite number of point in a wire are considered as only one node. All points that are grouped as a single node have the same voltage.
Fig. 7B Definition of a branch A branch is a set of 1 or more components connected in series, and all the components that are connected in series in the same chain are considered as 1 branch The current flowing along a branch is the same at all points.
Fig. 7C All nodes voltages are relative to a fixed reference node, usually the ground connection whose symbol is () and whose voltage is always equal 0 Volts. Current always flows from a node to another node of lower voltage. The voltage of a node can be determined from the voltage of a nearby node, Using the relation: , and rearranging we get: , where V2 is the voltage of the node to be determined, V1 is the voltage of the reference node which is known, I1 is the current flowing from node 1 to node 2 And R1 is the equivalent resistance between the 2 nodes. Similarly, and still using the same Ohm's law, the current in a branches can be determined if the voltages at the 2 nodes at both ends of the branches are known, using the relation: Current entering the node equals current leaving the node, thus in the given example (figure 7C) this can be written as :
It is important to be able able to feel the meaning of those simple mathematical relations; For example in the figure 7C above, the current is flowing from V1 to V2, and thus V2 must be smaller than V1 and that's exactly what the relation: is proving. The idea is to dig your way around the circuit to be analyzed with those given rules. Using the appropriate rule at the appropriate time, is the key to a fast and easy circuit analysis and understanding, and this skill is gained by practice and experience. Finally, remember that computers can do it, and so do you.
Calculating required rated power of a resistor
When buying resistor to build a certain circuit, you may be asked: "what is the power rating of the resistors you want to buy?" or you may simply be given 1/4 Watt resistor as they are the most standard class of resistors.
As long as you're working with resistor of higher value than 220, and your power supply delivers 9V or less, it is safe to work with 1/8 watt or 1/4 watt rated resistors. But if the voltage across a resistor increases over 10V or the resistor's value is less than 220, you should calculate the power carried away by the resistor, otherwise, it may burn up in fumes and can even cause serious burns and injuries.
To calculate the required power rating of the resistor, you must first know the voltage difference across the resistor (V) and the current flowing through it (I), then the power (P) is:
where I is the electrical current in Amperes (A), V is the voltage in Volts (V) and P is the power dissipation in Watt (W)
Here you can see some resistors having different power ratings. you notice that the main difference between different power ratings is the size of the resistor.
Fig. 9
Special resistors
Resistors can get more sophisticated than this, from simple variable resistors (also called potentiometers), to highly accurate temperature, light, and pressure sensors. Some of them are going to be discussed in this section.
Variable resistor (Potentiometer)
Fig. 10A Fig. 10B Fig. 10C Figure 10A shows the schematic symbol of a variable resistor. It is often referred to as a potentiometer, because it can be used as a potential (voltage) divider. Figure 10C shows how potentiometers look like in reality, they vary in size and shape, but they all work the same way. The pins at the right and left extremities of the potentiometer are equivalent to the fixed point (like Va and Vb in the figure 10A), while the middle pin is the moving part of the potentiometer, and is used to change the ratio of the resistance at its left to the resistance at its right. hence the voltage divider equation applies to the potentiometer, which can deliver any voltage from Va to Vb. Also a variable resistor can be used in a current limiting configuration by connecting the output the point Vout to Vb like in the figure 10B. Imagine how the current will flow through the resistance from the left extremity to the right until it reaches the arrow (the moving part that varies resistance) then practically all current will flow through the jumper wire (theoretically some very little current will pass This way you can also use a potentiometer to adjust the current flowing into any electronic component, or lamp for example. Actually this is how most of the old light dimmers work.
LDR (Light Dependent Resistors) and thermistors
There many electronic sensors that rely on a resistor whose resistance varies with respect to another parameter like light, temperature, or pressure. We are going to briefly study LDRs (Light Dependent Resistors) and Thermistors (Temperature dependent resisters), and you will notice that all resistors based sensors work exactly the same way, as the the easiest way to use one of those sensors is to put them in a voltage divider configuration, obtaining a voltage that changes with the measured values, instead of a resistance change. Sensors whose output is Voltage variations are much easier to interface to computers or micro controllers, as you shall see during the next tutorials.
Fig. 11A
As you can see in figure 11A, LDRs vary in size, but they are all resistors whose resistance will decrease when exposed to light, and increase when shed in the dark. they are also referred to as photoresistors, photoconductors or Cds because they are made of Cadmium sulphide Unfortunately, LDRs response can be slow, and they also often tend to lack accuracy, but still,
Fig. 11B they are very easy to use. For applications that requires more accuracy, and faster response, photodiodes or phototransistors are preferred over LDRs. Usually, an LDR's resistance can vary from 50 in the sun light, to over 10M in absolute darkness. As we said before, The variation of resistance has be converted into a voltage variation, by introducing the LDR into a voltage divider configuration, as shown in figure 11B. Recalling equation 6B, you will see that the output voltage (Vout) of this circuit follows the following equation: Supposing that the LDR's resistance varies from 10M to 50, calculations would yield that Vout varies respectively from 0.005V to 4.975V. This voltage variation can be then fed to an integrated circuit named an Operational Amplifier to create a reliable light sensor.
Similarly, a Thermistor can be used in the exact same way to create a sensor whose voltage varies with temperature variation. However, thermistors comes in much more varieties and types than LDRs, for instance, A thermistor can either be a negative temperature coefficient type (NTC) whose resistance will decrease with temperature rise, or positive temperature coefficient type(PTC), whose resistance will increase with temperature rise. Nowadays, electronics manufacturers provide thermistors of very high quality in terms of accuracy and response time, inneed, it's very common to see thermistors in very precise devices like digital thermostats.
Schematic Symbols
Figure 12 shows the most common symbols for the various types of resistors, which are used when drawing electronic schematics.
Fig. 12
Resistors color code
There are tow common ways to know a the value of a resistor, by measuring it using an Ohmmeter, or by reading the color code printed on it, which is much faster, when you get used
Fig. 13A to it. As you can see in figure 13A, some resistors have 4 color bands on it, and some have 5. Both of them use the same encoding method. the bands named 1, 2 or 3 in figure 13 can be translated into a 2 or 3 digit number using the table below. the band named M is the multiplier, meaning the number obtained from the previous digits have to be multiplied by 10 to the power M (or simply, add M number of zeros after the 2 or 3 digits number). the Value of M is
Also obtained from the table below. The last band at the right (T) is the tolerance band, which is usually gold meaning 5% tolerance or silver meaning 10%.
Black Brown Red Orange Yellow Green Blue Violet Gray White 0 1 2 3 4 5 6 7 8 9
Table used for translating color bands to numbers
To read a resistor using the color encoding follow the same steps of this example. Let's say you have a resistor like the one shown in figure 13B, the first step is to locate the tolerance band,
Fig. 13B it is usually apart from the other bands, and it is typically gold or silver colored. Once the tolerance band is located, note the colors of the bands starting from the other side. The first 2 bands will be translated from the table to become a '1' and a '0', this will
be considered as 10, then the Multiplied band, being red, will mean that you multiply by 10 the the power 2, or simply, multiply by 100. the the first 2 digits (10) multiplied by (100) will yield a result of 1000. Then, this is a 1000 resistor, or 1K.
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The Discussion
Let me and the visitors hear from you, share your questions and knowledge and propose other related articles and links. Feel free to correct any scientific mistake.
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Posted by: mhel on: 01 Apr 2007 great site helps me a lot with the basics.keep up the good work. Replie With Quote
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Q. In a certain code language, ‘it be pee’ means ‘dogs are blue’, ‘sik hee’ means ‘large horses’ and ‘pee mit hee’ means ‘horses are pigs’.
How is ‘pigs are large horses’ written in this code?
- Published on 08 May 17
a. Mit pee sik hee
b. Sik it pee be
c. Cannot be determined
d. None of these
ANSWER: Mit pee sik hee
In the second and third statements, the common code is ‘hee’ and the common word is ‘horses’. So, ‘hee’ means ‘horses’.
Thus, ‘sik’ means ‘large’ from second statement.
In first and third statements, the common code is ‘pee’ and common word is ‘are’. So, ‘pee’ means ‘are’.
Thus, ‘mit’ means ‘pigs’ from third statement.
So, the code is ‘Mit pee sik hee’
➨ Post your comment / Share knowledge
Enter the code shown above:
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https://ao.ms/how-to-tilt-a-binary-tree-in-java/ | 1,653,431,526,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00520.warc.gz | 151,208,283 | 61,611 | # How to Tilt a Binary Tree in Java
## The challenge
Given the `root` of a binary tree, return the sum of every tree node’s tilt.
The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values. If a node does not have a left child, then the sum of the left subtree node values is treated as `0`. The rule is similar if there the node does not have a right child.
Example 1:
```Input: root = [1,2,3]
Output: 1
Explanation:
Tilt of node 2 : |0-0| = 0 (no children)
Tilt of node 3 : |0-0| = 0 (no children)
Tile of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3)
Sum of every tilt : 0 + 0 + 1 = ```
Example 2:
```Input: root = [4,2,9,3,5,null,7]
Output: 15
Explanation:
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 5 : |0-0| = 0 (no children)
Tilt of node 7 : |0-0| = 0 (no children)
Tilt of node 2 : |3-5| = 2 (left subtree is just left child, so sum is 3; right subtree is just right child, so sum is 5)
Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7)
Tilt of node 4 : |(3+5+2)-(9+7)| = |10-16| = 6 (left subtree values are 3, 5, and 2, which sums to 10; right subtree values are 9 and 7, which sums to 16)
Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15```
Example 3:
```Input: root = [21,7,14,1,1,2,2,3,3]
Output: 9```
Constraints:
• The number of nodes in the tree is in the range `[0, 104]`.
• `-1000 <= Node.val <= 1000`
## Definition for a Binary Tree Node
```.wp-block-code{border:0;padding:0}.wp-block-code>div{overflow:auto}.shcb-language{border:0;clip:rect(1px,1px,1px,1px);-webkit-clip-path:inset(50%);clip-path:inset(50%);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px;word-wrap:normal;word-break:normal}.hljs{box-sizing:border-box}.hljs.shcb-code-table{display:table;width:100%}.hljs.shcb-code-table>.shcb-loc{color:inherit;display:table-row;width:100%}.hljs.shcb-code-table .shcb-loc>span{display:table-cell}.wp-block-code code.hljs:not(.shcb-wrap-lines){white-space:pre}.wp-block-code code.hljs.shcb-wrap-lines{white-space:pre-wrap}.hljs.shcb-line-numbers{border-spacing:0;counter-reset:line}.hljs.shcb-line-numbers>.shcb-loc{counter-increment:line}.hljs.shcb-line-numbers .shcb-loc>span{padding-left:.75em}.hljs.shcb-line-numbers .shcb-loc::before{border-right:1px solid #ddd;content:counter(line);display:table-cell;padding:0 .75em;text-align:right;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;white-space:nowrap;width:1%}```public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}```Code language: Java (java)```
## The solution in Java code
``````class Solution {
public int findTilt(TreeNode root) {
if (root == null) return 0;
int[] result = walk(root);
return result[1];
}
private int[] walk(TreeNode root) {
if (root == null) return new int[] {0, 0};
int[] left = walk(root.left);
int[] right = walk(root.right);
int subtreeSum = left[0] + right[0] + root.val;
int tiltSum = left[1] + right[1] + Math.abs(left[0] - right[0]);
return new int[] {subtreeSum, tiltSum};
}
}
```Code language: Java (java)```
Tags:
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Notify of | 1,123 | 3,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-21 | longest | en | 0.545167 |
https://forgeatl.com/australian-capital-territory/cyclic-code-generator-polynomial-example.php | 1,610,919,669,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513194.17/warc/CC-MAIN-20210117205246-20210117235246-00352.warc.gz | 352,405,727 | 6,727 | # Generator example code cyclic polynomial
## Chapter 8 Cyclic Codes Yaser Abu-Mostafa
Produce parity-check and generator matrices for cyclic. I know that hamming codes can be arranged in cyclic form. but my question is how can i proof this. my idea was to find a generator/primitive polynomial \$p(x)\$? for, on the construction of skew quasi-cyclic codes the notions of generator and parity-check polynomials are given. they gave examples of skew cyclic codes.
### BINARY CYCLIC CODES uotechnology.edu.iq
ECE-S622/T602 Class Notes Part VIII Linear Cyclic Codes. Cyclic codes - free download as divides xn+1. g(x) is called the generator polynomial. y examples: systematic encoding algorithm for an (n,k) cyclic code: 1., ece-s622/t602 class notes part viii: linear cyclic codes is called the generator polynomial of code c. example 8.3 consider the generator polynomial g(x).
Linear cyclic codes polynomial and words:a polynomial of degree nover ikis a polynomial p(x) = a 0 + a 1x+ + a n 1xn 1 + a for example in n= 7, polynomial word b. cyclic codes a cyclic code is a linear primitive polynomials are the generator polynomials of cyclic codes. for example using the primitive polynomial 1
Decoding of cyclic codes cyclic hamming codes is called the generator polynomial of the code description of cyclic codes example 4.1 (cont.) generator polynomial theorem: let c be an (n,k)cyclic code over gf(q). 1. there exists a monic polynomial g(x)such that n-tuple examples of binary cyclic codes
Linear cyclic codes polynomial and words:a polynomial of degree nover ikis a polynomial p(x) = a 0 + a 1x+ + a n 1xn 1 + a for example in n= 7, polynomial word on the construction of skew quasi-cyclic codes the notions of generator and parity-check polynomials are given. they gave examples of skew cyclic codes
A polynomial can generate a cyclic code with codeword length n and message length k if and only if the polynomial is a degree-(n-k) examples. collapse all. to generate cyclic redundancy code bits and to a specified generator polynomial and appends nonzero terms of the polynomial. for example, [1 0
Let u(x) be a codeword in a cyclic code cwith generator polynomial g(x). from example 7.3 (all ternary cyclic codes of length 4). suppose we wish to nd all crc series, part 3: crc implementation code in crc code in c (free) cyclic redundancy codes are as the generator polynomial. figure 1. an example of
Cyclic codes œ bch codes galois example using f(x) = (x4 + x + 1) the generator polynomial g(x) is specified in terms of its roots in gf(2m). every this matlab function returns the row vector representing one nontrivial generator polynomial for a cyclic code having codeword length n and message length k.
A cyclic code has generator polynomial g(x)that is a divisor of every example: over gf(2)the cyclic polynomial of degree 6can be factored as x6−1= cyclic codes - free download as divides xn+1. g(x) is called the generator polynomial. y examples: systematic encoding algorithm for an (n,k) cyclic code: 1.
It is demonstrated how useful this can be in the design of high-degree non-primitive binary cyclic codes. several code examples using the generator polynomial, i know that hamming codes can be arranged in cyclic form. but my question is how can i proof this. my idea was to find a generator/primitive polynomial \$p(x)\$? for
### coding theory Cyclic Hamming Code - Mathematics Stack
ECE-S622/T602 Class Notes Part VIII Linear Cyclic Codes. Chapter 8: cyclic codes thekeytothedesignandanalysisofcycliccodesisthegenerator polynomial. in the code of example 8.2,, generator polynomial theorem: let c be an (n,k)cyclic code over gf(q). 1. there exists a monic polynomial g(x)such that n-tuple examples of binary cyclic codes.
### Cyclic Redundancy Check Computation An Implementation
Cyclotomic Cosets the Mattson–Solomon Polynomial. 4 encoding and decoding with cyclic codes polynomial cyclic codes encoding and decoding with cyclic codes an example an introduction to cyclic codes https://en.m.wikipedia.org/wiki/Polynomial_long_division Example • construct a systematic (7,4) cyclic code using a generator polynomial. solution as we know g(x) = x3 + x2 + 1 consider a data vector d = 1010.
The general crc generator block generates cyclic redundancy code (crc) bits for each input data frame and appends them to the frame. because one cyclic right shift is equal to n − 1 cyclic left shifts, a cyclic code polynomial. examples generator polynomial for the cyclic code
It is demonstrated how useful this can be in the design of high-degree non-primitive binary cyclic codes. several code examples using the generator polynomial, quasi-cyclic codes derived from cyclic codes are thus obtained from good cyclic codes. examples and code with generator polynomials 270177, 250434 315101
Part 2.2 cyclic redundancy check (crc) codes. cyclic redundancy check codes (4) ¾example: the crc-12 code with generator polynomial as cyclic codes are not only simple to the length of the generator polynomial, for example, the sum of the polynomials x3+x+1
Math 5410 cyclic codes ii example: suppose we wish to theorem 9: let c be a cyclic (n,k)-code over f with generator polynomial g(x), and let r(x) chapter 3: cyclic and convolution codes generates a cyclic code. generator polynomial generator for cyclic codes example check polynomials and parity
• cyclic codes can be dealt with in the very same way as all otherlbc’s – choose a generator string g of length r+1 bits example r = 3, g = 1001 part 2.2 cyclic redundancy check (crc) codes. cyclic redundancy check codes (4) ¾example: the crc-12 code with generator polynomial as
Trivial examples of cyclic codes are a n itself and the code containing only the zero and is a generator polynomial for the cyclic code of block length = to generate cyclic redundancy code bits and to a specified generator polynomial and appends nonzero terms of the polynomial. for example, [1 0
S-72.3410 cyclic codes 1 systematic cyclic codes cyclic code c with generator polynomial g(x). s-72.3410 cyclic codes 3 example: class sage.coding.cyclic_code.cycliccode (generator if the code is cyclic, the generator polynomial is the gcd of all of code (if the code is cyclic). examples: | 1,456 | 6,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-04 | latest | en | 0.786728 |
https://kr.mathworks.com/matlabcentral/cody/problems/57-summing-digits-within-text/solutions/1799707 | 1,596,731,721,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736972.79/warc/CC-MAIN-20200806151047-20200806181047-00551.warc.gz | 321,710,808 | 15,719 | Cody
# Problem 57. Summing Digits within Text
Solution 1799707
Submitted on 29 Apr 2019 by Harshit Jain
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
str = '4 and 20 blackbirds baked in a pie'; total = 24; assert(isequal(number_sum(str),total))
total = 24
2 Pass
str = '2 4 6 8 who do we appreciate?'; total = 20; assert(isequal(number_sum(str),total))
total = 20
3 Pass
str = 'He worked at the 7-11 for \$10 an hour'; total = 28; assert(isequal(number_sum(str),total))
total = 28
4 Pass
str = 'that is 6 of one and a half dozen of the other'; total = 6; assert(isequal(number_sum(str),total))
total = 6 | 228 | 741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-34 | latest | en | 0.772335 |
https://www.quesba.com/questions/a-linked-representation-rather-than-a-tabular-representation-saves-time-whe-52862 | 1,606,157,615,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141164142.1/warc/CC-MAIN-20201123182720-20201123212720-00412.warc.gz | 829,273,929 | 18,270 | # A linked representation, rather than a tabular representation, saves time when rearranging items...
A linked representation, rather than a tabular representation, saves time when rearranging items if the Move-to-Front Heuristic is used. But it costs memory; how might the extra memory be better utilized by a different heuristic?
Jun 18 2020| 06:13 PM
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Nov 23 2020 | 1,198 | 5,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-50 | latest | en | 0.875066 |
http://www.fixya.com/support/t8675378-many_kilobytes_in | 1,508,779,606,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826210.56/warc/CC-MAIN-20171023164236-20171023184236-00120.warc.gz | 460,367,347 | 33,729 | # How many kilobytes are in one terabyte? - Computers & Internet
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• Expert
Hi,
Here is the computation for for 1 Terabyte to KiloBytes.
There are 1073741824 Kilobytes in 1 Terabytes.
Thank you for choosing fixya
Posted on Mar 16, 2011
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• Expert
1024 bytes = 1 kilobyte
1024 kilobytes = 1 megabyte
1024 megabytes = 1 gigabyte
1024 gigabytes = 1 terabyte
1024^4 = 1,099,511,627,776 (approximately 1 trillion bytes)
Posted on Mar 16, 2011
Hi,
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Posted on Jan 02, 2017
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## Related Questions:
### How do you measure megabytes vs gigabytes
There are 1000KB in a Megabyte, There are 1000 Megabytes in a Gigabyte, There are 1000 Gigabytes in a Terabyte.
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### Convert lba to cyl, head, sector
The problem with CYL/HEAD/SECTORS is that disk-drives have gotten *MUCH* too large -- the limited size of each field mean that a new scheme, namely LBA (Logical Block Addressing) had to be created to deal with disk-drives that are 1000 times too large (4 Terabytes versus old 4 Gigabytes).
So, if you have 312,500,000 sectors,
and each sector is 0.5 Kilobytes,
you have 156,250,000 Kbytes,
or 156,250 Mbytes,
or 156 Gigabytes.
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### WINDOWS WILL NOT RECOGNIZE MY 640 GB ONLY 600 GB
If you bought a 640 GB hard drive when the drive is formatted it will be slightly smaller then the size listed on the box.
This is caused by a difference between the size of a GB as defined by windows versus the maker of the drive. The drive manufacture assumes that 1 GB is 1000 MB, where as windows assumes it is 1024 MB.
If you carry these assumptions all the way down to the byte level you get:
640GB = 640000 MB = 6.4 * 10^8 KB = 6.4 * 10^11 bytes
this is the base storage size of the drive.
Then windows takes this number and does the same in reverse to get the number of GB is displays:
6.4 *10^11 bytes = 6.25 * 10^8 KB = 610351 MB = 596.1 GB
There is nothing wrong with the drive, it's just a different unit of measurement.
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### How big is a 1 tb hd
A terabyte is a multiple of the unit byte for digital information. The prefix tera means 1012 in the International System of Units (SI), and therefore 1 terabyte is 1000000000000bytes, or 1 trillion (short scale) bytes, or 1000 gigabytes. 1 terabyte in binary prefixes is 0.9095 tebibytes, or 931.32 gibibytes. The unit symbol for the terabyte is TB or Tbyte.
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### My disk does not store to full capacity
What is the advertised capacity of the library, in bytes? in thousands (1000) of bytes? of millions (1,000,000) of bytes?
What is the currently-reported capacity of the library, in bytes? Kilobytes (1024 bytes) ? Megabytes (1024 Kilobytes) ?
Mar 18, 2011 | JES Hardware Solutions (IT70-7-DVD-DZ)...
### Coby mpc7097 el playlist solo llega a 200 temas pero tengo 800 temas en aparato, quisiera aumentar el numero maximo del playlist, gracias por la ayuda que puedan ofrecerme.
lo que puedes hacer es formatearlo desde mi equipo "my computer" encuentra el dispositivo y dale con el botón de la derecha, selcciona en donde dice formatear, selecciona fat (default) y eso te libera el espacio pero cuidado porque te borra la información. despues de que hagas eso puedes meter toda tu música de nuevo lo único es que este dispositivo es de 2GB
17,179,869,184 .......... bits
2,147,483,648 .......... bytes
2,097,152 .......... kilobytes
2,048 .......... megabytes
2 .......... gigabytes
0.001953125 terabytes
0.0000019073 petabytes
0.0000000018 exabytes
una canción=entre 1500 y 6000KB quiere decir que en tu dispositivo caben alrededor de 420 cancionesy a estos dispositivos no les puedes agregar mas memoria si quieres meterle mas música tienes que comprar un aparato con mas capacidad.
si te sirve el consejo dame un buen rating
gracias
Apr 27, 2010 | Coby MPC895 MP3 Player
### How to configure the wireless linksys router with terabyte speed
The term terabyte is not related to speed. It designates the amount of storage capacity of hard drives. Ex.. if you have two 500G hard drives, your total storage capacity is 1 terabyte.
The available ethernet speeds are 10M, 100M, 1G, or 10G.
Jul 14, 2009 | Computers & Internet
### What's my mega bite speed
mega bite speed is technical term of measurement data transfer in digital system, mega is 1000 kilobyte, kilobyte is 1000 byte
Sep 26, 2008 | Computers & Internet
## Open Questions:
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167 people viewed this question
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Level 3 Expert | 1,502 | 5,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-43 | latest | en | 0.91508 |
http://mathhelpforum.com/number-theory/190896-hints-prove-6-2n-5-m-cannot-equal-9-a.html | 1,544,681,467,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824525.29/warc/CC-MAIN-20181213054204-20181213075704-00148.warc.gz | 191,943,660 | 9,683 | # Thread: Hints to prove |6^(2n)-5^m| cannot equal 9.
1. ## Hints to prove |6^(2n)-5^m| cannot equal 9.
This is the last step to a problem I've been working on, but I have not had many ideas on how to finish this last step. n and m are positive integers. I was easily able to prove that the above could not equal 1, but I'm not sure what to apply now. Thanks.
|6^(2n)-5^(m)|=9 can be reduced to 5^(m)-6^(2n)=9.
2. ## Re: Hints to prove |6^(2n)-5^m| cannot equal 9.
Originally Posted by eulcer
This is the last step to a problem I've been working on, but I have not had many ideas on how to finish this last step. n and m are positive integers. I was easily able to prove that the above could not equal 1, but I'm not sure what to apply now. Thanks.
|6^(2n)-5^(m)|=9 can be reduced to 5^(m)-6^(2n)=9.
If n>0 then $\displaystyle 6^{2n}$ is a multiple of 9 but $\displaystyle 5^m$ is not. | 290 | 891 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-51 | latest | en | 0.966516 |
https://www.lpsg.com/threads/girth-width-ratio.89393/ | 1,529,496,168,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00185.warc.gz | 852,012,924 | 16,453 | # Girth/Width ratio
Discussion in 'Sex With a Large Penis' started by bld_64, May 25, 2008.
1. ### bld_64 Member
Joined:
Mar 24, 2007
Messages:
392
4
Gender:
Male
My wife was recently surprised to discover that Amazon.com sells dildoes.
(No, honestly honey, I was looking for the other type of personal massager)
While window shopping, she remarked that the sizes are listed in length and width.
Occasionally, she likes to measure my dick (Mostly because she knows that I like it.) She wanted to know what my width was, and I was happy to let her check.
It all got me to thinking that, for the most part, it's width, and not girth that matters during typical missionary or doggie-style fucking; in other works, how wide can you stretch that hole...???
Anyway, what about a poll of girth to width ratio? This will require some accurate measurement, and it's all a bit imprecise, but for the sake of science...
So if you have some accurate measuring instruments and are in the mood, give us the number. Since we have already determined that size does matter, please measure at the widest point on the shaft. Remember to pull the tape gently snug for girth and push until you fell a bit of resistance on the width. Since I have the tools, I'll go first:
Largest girth = 144 mm (5.67 inches) Girth
Width = 48 mm (1.89 inches) Width
G/W ratio = 144/48 = 3.00
#1
2. ### vince Gold Member
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Are those official LPSG certified vernier calipers? :smile:
#2
3. ### _avg_ Gold Member
Joined:
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Well since girth is circumference and width is diameter, and pi x diameter = circumference, 3.14 is the expected value. I know penises are not perfect cylinders, but they shouldn't deviate too far from that.
#3
4. ### B_625girth New Member
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exactly unless you have a wide, FLAT, cock.
this morning, I am a little over 6.50" and about 2" wide, my usual 7.25" L. my girth varies from time to time.
#4
5. ### Sherwood D. Likelym VerifiedGold Member
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I like it! Penis girth width ratio equals pi ... a new way to teach geometry!
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1 | 671 | 2,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-26 | latest | en | 0.958954 |
https://www.investopedia.com/terms/n/nonlinearity.asp | 1,685,433,247,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00189.warc.gz | 908,529,953 | 55,770 | # What Is Nonlinear? Definition, Vs. Linear, and Analysis
## What Is Nonlinearity?
Nonlinearity is a term used in statistics to describe a situation where there is not a straight-line or direct relationship between an independent variable and a dependent variable. In a nonlinear relationship, changes in the output do not change in direct proportion to changes in any of the inputs.
While a linear relationship creates a straight line when plotted on a graph, a nonlinear relationship does not create a straight line but instead creates a curve. Some investments, such as options, exhibit high levels of nonlinearity and require investors to pay special attention to the numerous variables that could impact their return on investment (ROI).
### Key Takeaways
• Nonlinearity is a mathematical term describing a situation where the relationship between an independent variable and a dependent variable is not predictable from a straight line.
• Certain investment classes, such as options, show a high degree of nonlinearity, which may make these investments seem more chaotic.
• Investors of asset classes that exhibit a high level of nonlinearity will often use sophisticated modeling techniques to estimate the amount of potential loss or gain their investment might incur over a specified time.
## Understanding Nonlinearity
Nonlinearity is a common issue when examining cause and effect relationships. Such instances require complex modeling and hypothesis testing to offer explanations of nonlinear events. Nonlinearity without explanation can lead to random, erratic outcomes.
In investing, we can see examples of nonlinearity in certain investment classes. Options, for example, are nonlinear derivatives because changes in the input variables associated with options do not result in proportional changes in output. Investments with high nonlinearity may appear more chaotic or unpredictable.
Investors who include nonlinear derivatives in their portfolio will need to use different pricing simulations to estimate the risk profile of their investments than they would for linear assets such as shares of stock or futures contracts. For instance, options traders will look to their "Greeks" such as the delta, gamma, and theta. These assessments can help investors manage their risk and help time the entry and exit points of their trades.
## Nonlinearity vs. Linearity
In contrast to a nonlinear relationship, a linear relationship refers to a situation where there is a direct correlation between an independent variable and a dependent variable. A change affecting an independent variable will produce a corresponding change in the dependent variable. When plotted on a graph, this linear relationship between independent and dependent variables will create a straight line.
For example, let's suppose management at a shoe factory decides to increase its workforce (the independent variable in this scenario) by 10%. If the company's workforce and production (the dependent variable) have a particular linear relationship, then management should expect to see a corresponding 10% increase in the production of shoes.
## Nonlinearity and Options
The multiple variables that can impact an option investment's return make options an example of an investment with high nonlinearity. When trading options, investors may have many variables to consider, such as the underlying asset price, implied volatility, time to maturity, and the current interest rate.
For investments with a high degree of linearity, investors generally use a standard value at risk technique to estimate the amount of potential loss the investment might incur over a specified time period. However, using a standard value at risk technique is generally not sufficient for options because of their higher degree of nonlinearity.
Instead, options investors might use a more advanced technique, such as a Monte Carlo simulation, which enables the investor to model for a wide variety of variables with different parameters to assess possible investment returns and risks.
## Special Considerations
Nonlinear regression is a common form of regression analysis used in the financial industry to model nonlinear data against independent variables in an attempt to explain their relationship. Although the model's parameters are nonlinear, nonlinear regression can fit data using methods of successive approximations to offer explanatory outputs.
Nonlinear regression models are more complicated to create than linear models because they often take considerable trial-and-error to define the outputs. However, they can be valuable tools for investors who are attempting to determine the potential risks associated with their investments based on different variables.
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The offers that appear in this table are from partnerships from which Investopedia receives compensation. This compensation may impact how and where listings appear. Investopedia does not include all offers available in the marketplace. | 893 | 5,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-23 | latest | en | 0.896548 |
https://www.kitplanes.com/aero-lectrics-79/ | 1,716,601,276,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00527.warc.gz | 742,958,515 | 31,639 | # Aero ‘lectrics
LED fairy tales.
3
I thought I had heard it all about airplane electronicky stuff. This tops anything I’ve heard in a long time. Fellow I know bought an LED instrument panel floodlight to try and save a few amps of incandescent lamp current from his electrical bus. The recommendation from the manufacturer is to install a low-value high-wattage resistor in parallel with the LED light so that the total current remains about the same as the incandescent lights, and then his dimmer will work normally. To top it off, they recom-mended only their expensive pulse width modulation dimmer for the installation. Lordy. Throw away the reason for buying a low-current LED in the first place and then compound the stupidity with a dimmer doing exactly what the LED is not supposed to do. Read on.
So, here we go with a bit of speaking truth to (electrical) power. First, LEDs are marvelous little light bulbs. Comparing a #330 12-volt incandescent instrument light that draws 80 milliamperes (mA), I find that a single white LED run at about 15 mA is about the same brightness, but I can run four of them in series at 12 volts. My math says that the LED is roughly 20 times more efficient than the incandescent.
Now, a bit of physiology if I may. The human eye is programmed by the brain to see anything happening more than about 24 times a second as being in continuous motion. For example, a television picture is not in continuous motion. The TV screen has individual still pictures that change 30 times a second. Therefore, when we see a video of the Oshkosh airshow, while you think you are watching Patty Wagstaff do a continuous aileron roll on takeoff, you are actually watching 30 individual pictures a second, and your brain translates those images into continuous motion.
Another trick your brain plays on your eyes is what is called “persistence of vision.” That is, if you look at this magazine page for a moment and then immediately shut your eyes, you will see (just for an instant) the page in your mind.
Finally, the eye is what we call a “peak detector”—it takes the maximum light produced by a bulb. No matter if the bulb is full on for a few milliseconds and off for a few dozen milliseconds, your eye thinks that the bulb is in the full-on mode all the time, subject to the 24-frames-per-second limitation stated above. An incandescent bulb has a very long time thermal and optical constant measured in seconds. If you turn an incandescent 12-volt bulb on for a few milliseconds and then off for the same number of milliseconds, an incandescent bulb “thinks” it only has an average of six volts across it and dims accordingly. An LED, on the other hand, turns on in microseconds (millionths of a second), so if we put the same on-off pulses on it, it turns on full bright for a time, then full off for a time. Your peak-detecting eye thinks it was on full-bright all of the time.
Now that we know all this stuff, let’s go back to our original discussion about changing an incandescent lighting system to an LED system.
In the first place, since we will be drawing way less current, we might want to save a few ounces (or pounds) and replace the original wire that needed to carry amperes of incandescent current to one that only needs to carry milliamperes of LED current. Don’t forget that the fuse or breaker was sized not for the load, but to protect the wire from burning and/or melting with an accidental short circuit. You may want to review the requirements for fuse versus wire size if you do a rewire job.
Illustration 1: State of the art in 1920. Wirewound rheostat and incandescent bulbs for panel illumination.
Illustration 2: Improved state of the art in the 1940s. Wirewound potentiometer and incandescent bulbs.
Illustration 3: Replacing the wirewound potentiometer with a carbon potentiometer and an LED or two.
If we are working with an old airplane (say, one that was built before 1970 or so), then the “dimmer” was generally a rheostat or variable wirewound 2-terminal resistor. These old clunkers required that the lamps draw a fair amount of current for them to work. If indeed we try to dim an LED with a rheostat type of dimmer, the LED will not dim much at all.
There was another method of dimming that wasn’t quite as popular as the rheostat and that was a potentiometer type of 3-terminal dimmer. How can you tell the difference without tracing out the circuit? Simple. Take the lamp load off of the circuit and measure the voltage at the lamp terminal with a regular old Harbor Freight digital voltmeter. If rotating the dimmer knob from full on to full off leaves the lamp voltage constant at the battery supply voltage, it is a rheostat. If the voltage at the lamp changes from zero to battery supply voltage as you rotate the knob, it is a potentiometer and can be used with LEDs.
There is a gotcha with either a wirewound rheostat or a wirewound potentiometer (“pot”). If either of them were designed to be used with an incandescent lamp load they are way overrated for the LED job—sort of like using tundra tires on an ultralight. They also had a very nasty habit of getting the wire-wound control very hot at nearly full-on of the lamps and melting the wire at one end of the control.
So, in the 1970s we came up with a scheme to replace those clunky old wirewound rheostats/pots with transistors and very small carbon element potentiometer controls. We also found that keeping the transistor itself from burning up with an incandescent load meant either a giant heat sink (great for Minnesota winters, not so much for Arizona summers) or going to what is called a pulse width modulation (PWM) circuit where the transistor was full on (very low voltage but very high current) or full off (very high voltage but very low current), which kept the transistor relatively cool, and the good old incandescent bulb did its averaging trick and dimmed quite nicely.
Illustration 4: Running a few LEDs in series to increase efficiency with a single carbon pot.
Illustration 5: Early 1970s implementation of a single transistor, a small carbon pot, and a string of LEDs.
PWM for LEDs that turn on and off in microseconds doesn’t work so well. As a matter of fact, with the eye’s capability for peak detection, the LED is either full on or full off, no dimming in between. You might just as well replace the PWM dimmer control with an on-off switch.
The answer is to use a dimmer intended for LEDs. With the greatly reduced current draw of LEDs, going back to a plain old transistor dimmer can be done for pennies without the need for a giant heat sink. For those of you who want to experiment with homebrewing your own dimmer, I’ve included a few diagrams to let you do the experiments. Next month, the ferrite current sensor I promised a couple of months ago.
Illustration 6: Current technology using an adjustable regulator and a small carbon pot to dim a few strings of LEDs.
As to the title on the previous page, I used to think fairy tales started off “Once upon a time…” I’ve come to think that modern fairy tales start off, “If elected, I promise…” See you next month…Until then…Stay tuned…
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Jim Weir is the chief avioniker at RST Engineering. He answers avionics questions in the Internet newsgroup www.pilotsofamerica.com–Maintenance. His technical advisor, Cyndi Weir, got her Masters degree in English and Journalism and keeps Jim on the straight and narrow. Check out their web site at www.rst-engr.com/kitplanes for previous articles and supplements. | 1,721 | 7,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.95679 |
https://www.physicsforums.com/threads/solution-of-trigonometric-equations.632519/ | 1,582,665,662,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146160.21/warc/CC-MAIN-20200225202625-20200225232625-00197.warc.gz | 836,997,268 | 19,289 | # Solution of Trigonometric Equations
## Homework Statement
Find the solution set of each of the following equations for the interval 0° ≤ x ≤ 360°.
$$sin^2x = sinx$$
## Homework Equations
Trigonometric Identities for Sine.
## The Attempt at a Solution
This is my attempt so far:
$$\sin^2x = sinx$$
$$\sin^2x - sinx = 0$$
$$\sin^2x - sinx + \frac{1}{4} = \frac{1}{4}$$
$$(sinx - \frac{1}{2})^2 = \frac{1}{4}$$
taking square roots of both sides:
$$sinx -\frac{1}{2} = \frac{1}{2}$$
$$sinx = ±\frac{1}{2} + \frac{1}{2}$$
if $\frac{1}{2},$
$$sinx = 1$$
if $-\frac{1}{2},$
$$sinx = 0$$
take arcsin of both sides:
$$x = arcsin1$$
$$x = 90° or \frac{\pi}{2}$$
$$x = arcsin0$$
$$x = 0°$$
Solution set = {0°, 90°} or {0, $\frac{\pi}{2}$}
EDIT: The complete solution set must be:
x = {0°, 90°, 180°, 360°}
because sinx = 0 also in 180 and 360, not only in 0°)
Special thanks to Sourabh N for guiding me to the correct solution
Last edited:
Related Precalculus Mathematics Homework Help News on Phys.org
Find the solution set of each of the following equations for the interval 0° ≤ x ≤ 360°.
$$sin^2x = sinx$$
take arcsin of both sides:
$$x = arcsin1$$
$$x = 90° or \frac{\pi}{2}$$ What other value(s) of x satisfies this equation?
$$x = arcsin0$$
$$x = 0°$$ What other value(s) of x satisfies this equation?
Solution set = {0°, 90°} or {0, $\frac{\pi}{2}$}
Notice the range of x: 0° ≤ x ≤ 360°. What other values of x within this range satisfy the conditions x = arcsin1 or x = arcsin0?
Notice the range of x: 0° ≤ x ≤ 360°. What other values of x within this range satisfy the conditions x = arcsin1 or x = arcsin0?
x = {0, 90°, 360°}
No.
Here's a very good way to figure out all the solutions. Draw the curve sin[x] for 0° ≤ x ≤ 360°. Also draw horizontal lines at y = 0 and y = 1. The point(s) where sin[x] and these lines intersect, are your solutions.
No.
Here's a very good way to figure out all the solutions. Draw the curve sin[x] for 0° ≤ x ≤ 360°. Also draw horizontal lines at y = 0 and y = 1. The point(s) where sin[x] and these lines intersect, are your solutions.
Oh. You mean, what values of x does give the angles 0, 90°, 360° when worked with sine?
for x values,
x = {0, 1}
y = Sin[x].
x = x.
1. Draw y = 0. See at what values of x do Sin[x] and y = 0 intersect.
2. Same with y = 1.
y = Sin[x].
x = x.
1. Draw y = 0. See at what values of x do Sin[x] and y = 0 intersect.
2. Same with y = 1.
Is my graph correct?
On the horizontal axis, you have 0 -> 90 -> 360?? In 0->90-> What comes after 90?
This?
Very good! You have also (correctly) marked the points of intersection, except that you forgot to include x = 0.
So what is the solution set?
Very good! You have also (correctly) marked the points of intersection, except that you forgot to include x = 0.
So what is the solution set?
Yeah, and also the point at 360. Sorry :shy:
Based on the graph, the x values are 0°, 45°, 90°, 225°, 360°. Right?
Ah, I didn't look carefully. Your Sin[x] curve is not correct.
What is Sin[0]? Sin[45]? Sin[90]? Do you see what's wrong?
Ah, I didn't look carefully. Your Sin[x] curve is not correct.
What is Sin[0]? Sin[45]? Sin[90]? Do you see what's wrong?
Do you have any clue of the correct curve for sin x? because that's the only sinx graph that I know.
The shape is correct, but you don't have the scaling right. Sin[x] reaches 1 at x = 90°, Sin[x] reaches back to 0 at 180° and so on.
The shape is correct, but you don't have the scaling right. Sin[x] reaches 1 at x = 90°, Sin[x] reaches back to 0 at 180° and so on.
Oh yeah I forgot that I have to draw only 1 period. Apologies~ :tongue:
Working on graph..
Oh yeah I forgot that I have to draw only 1 period. Apologies~ :tongue:
Working on graph..
No. You have Sin[x] = 1 at x = 45°, Sin[x] = 0 at x = 90°, which is wrong.
You have to draw the curve for 0° ≤ x ≤ 360°.
That's correct. (Though in he graph, you call y = 0 as y = 2).
The red dots are your solution set.
:rofl: that should be this:
Sorry about that. I labeled the green line first as y = 1 and accidentally labeled the second one y = 2. What a mess.
So, the solution set is:
x = {0°, 90°, 180°, 360°}
Thank you so much sir! I had fun learning with you. You are good in teaching. I would vote for you as a homework helper here and hope you become one
You're welcome :)
That's very sweet of you, haha | 1,429 | 4,366 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-10 | longest | en | 0.70936 |
http://www.theblackvault.com/phpBB3/topic5726-160.html | 1,386,349,471,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163052216/warc/CC-MAIN-20131204131732-00050-ip-10-33-133-15.ec2.internal.warc.gz | 547,466,761 | 13,847 | ## The Black Vault Message Forums
Discover the Truth!
# Other Paranormal and Occult
## MANY WORLDS POSTULATION
The paranormal and occult have many sides, discuss all other paranormal/occult topics that don't fit anywhere else.
## Re: MANY WORLDS POSTULATION
http://en.wikipedia.org/wiki/Subjective_idealism
But we can establish without doubt that perception is a mathematical object and then prove it mathematically...
http://www.math.csusb.edu/notes/proofs/pfnot/node8.html
Existence Proofs
Theorems which state ``There exists an x such that conclusion'' are known as existence theorems. Many, but not all, theorems of this type are proved using a constructive proof so named because one finds an element x that does satisfy the conclusion. As a simple example consider the statement ``Let a be a real number. There exists a real number x that is a solution to the equation 5x+3 = a''. A proof of this statement would construct such a real number x. Using the rules of arithmetic in the real number system, we know that if a is a real number then so too is (a-3)/5. This is a value of x that is a solution to the equation since 5x +3 = 5(a-3)/5 + 3 = a -3 +3 = a as required. Of course not all proofs are this simple and to produce an element x that will work may require ingenuity.
khanster
Posts: 690
Joined: Sat Sep 19, 2009 1:18 am
## Re: MANY WORLDS POSTULATION
Maybe I should have said that testimony about perception isn't proof.
"it is easy to grow crazy"
at1with0
Posts: 9140
Joined: Thu Apr 09, 2009 5:55 pm
Location: the coproduct of the amalgam of all structures
## Re: MANY WORLDS POSTULATION
at1with0 wrote:Maybe I should have said that testimony about perception isn't proof.
Proof of proof is proof of perception
khanster
Posts: 690
Joined: Sat Sep 19, 2009 1:18 am
## Re: MANY WORLDS POSTULATION
Consensus perception is generally perceived as objective validation.
"I can conceive of nothing in religion, science, or philosophy, that is anything more than the proper thing to wear, for a while." ~ Charles Fort
DIss0n80r
Posts: 4162
Joined: Wed Apr 20, 2011 2:45 am
## Re: MANY WORLDS POSTULATION
Yes.
Makes it hard to prove that the voice in my head are REAL DAMMIT
"it is easy to grow crazy"
at1with0
Posts: 9140
Joined: Thu Apr 09, 2009 5:55 pm
Location: the coproduct of the amalgam of all structures
## Re: MANY WORLDS POSTULATION
There appears to be an implicit invalidation of individual experience in the consensus worldview, yes. Purportedly to ensure stricter reality-affirming communicative modalities yet I still observe my fellow apes often discussing what their preferred invisible entity's expectations and motivations are, not to mention the common ritual of clasping hands to telepathically communicate with it.
"I can conceive of nothing in religion, science, or philosophy, that is anything more than the proper thing to wear, for a while." ~ Charles Fort
DIss0n80r
Posts: 4162
Joined: Wed Apr 20, 2011 2:45 am
## Re: MANY WORLDS POSTULATION
Consensus works in a limited way, until the consensus model is falsified and a new model must be created.
khanster
Posts: 690
Joined: Sat Sep 19, 2009 1:18 am
## Re: MANY WORLDS POSTULATION
Sooner or later, consensus usually becomes an impediment to truth.
"I can conceive of nothing in religion, science, or philosophy, that is anything more than the proper thing to wear, for a while." ~ Charles Fort
DIss0n80r
Posts: 4162
Joined: Wed Apr 20, 2011 2:45 am
## Re: MANY WORLDS POSTULATION
Awaken to a light of knowing
and the faith which you embraced
has left you blind
deceived like the fools that surround you
it came with ease unfulfilled were the days of your being
but now you will see
-Steve Tucker
"it is easy to grow crazy"
at1with0
Posts: 9140
Joined: Thu Apr 09, 2009 5:55 pm
Location: the coproduct of the amalgam of all structures
## Re: MANY WORLDS POSTULATION
"I can conceive of nothing in religion, science, or philosophy, that is anything more than the proper thing to wear, for a while." ~ Charles Fort
DIss0n80r
Posts: 4162
Joined: Wed Apr 20, 2011 2:45 am
PreviousNext | 1,113 | 4,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2013-48 | longest | en | 0.912339 |
https://www.jkangpathology.com/post/differential-equations-and-fourier-transformation/ | 1,713,931,644,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00859.warc.gz | 754,855,445 | 6,431 | # Differential equations and Fourier transformation
Differential equations describe the change of state. The change relates to the state. The solutions of the differential equations are the status equations. The initial conditions set the time $$t$$ and status $$y$$. The boundary conditions are the value of boundary $$y_0$$ and $$y_1$$.
$$dy \over dt$$ $$= ay + q(t)$$ starting from $$y(0)$$ at $$t = 0$$. inital conditions $$t = 0$$ and $$y=1$$
$$q(t)$$ is a input and $$y(t)$$ is a response. If $$q(t)$$ is delta function, the response is said Impulse response $y' -ay = \delta (t) \\ y(t)=e^{at}$.
The solutions are combination of particular solution and null solution $$y = y_t + y_n$$. The solution includes $$e^{at}$$. The differential equations can not be solved like polynomial equations, because the arguments of the differential equation relate to each other by calculus in the background of the equation. They can not be treated as just different arguments. The Fourier transformation puts the $$y$$ and its derivative $$y'$$ in the same functional space (Hilbert space). This transformation makes the differential equation problem to simple arithmetic problem.
Fourier transformation $$F(x) = \Sigma ^{\infty}_{n=-\infty} c_{n}e^{inx}$$
The basis of the Fourier transformation is $$e^{inx}$$. If the coefficients of the basis $$c_{n}$$ decay fater, $$F(x)$$ becomes smooth. If the coefficients are constant, $$F(x)$$ is delta function $$\delta(x)$$.
The derivative $$dy \over dt$$ is an linear transformation operator, i.e. inner product, because the $$y$$ and $$y'$$ are in functional space with same basis. The defivative can be represented as a matix $$A$$. The derivative matrix is antisymetric i.e. $$A^T = -A$$ and the minus second derivative matrix $$-d^{2}/dx^{2}$$ is symetic positive definite. $$AAf = -A^{T}Af$$. The meaning of transverse of a matrix is $$(Ax)^{T}y = x^{T}(A^{T}y)$$. Dual and inner product
The second difference matrix solves discrete differential equations. The N eigenvectors of K are $$y_{n} = (sin\ n\pi \Delta x, sin\ 2n\pi \Delta x,\ ..., sin\ Nn\pi \Delta x)$$. The N eigen values of K are the positive numbers $$\lambda_{n} = 2-2cos {n \pi \over N+1}$$.
How does exponent $$i$$ mean in $$e^i$$? The exponent makes multiplication to addition. What does an imaginary exponent mean? The imaginary exponent tilts the value to a complex plane. If the base is natural base $$e$$, the value of $$e^i$$ is in the unit circle of a complex plane. The cycle is $$2 \pi$$.
The Fourier transformation for solving the difference equation provoked the subject of functional analysis 200 years ago.
Reference
Differential Equations and Linear Algebra, Gilbert Strang
##### Jun Kang
###### Clinical Assistant Professor of Hospital Pathology
My research interests include pathology, oncology and statistics. | 733 | 2,853 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-18 | latest | en | 0.843614 |
http://psn.quake.net/psnlist/071112_090807_1.html | 1,511,508,701,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807146.16/warc/CC-MAIN-20171124070019-20171124090019-00251.warc.gz | 249,118,950 | 2,266 | ## PSN-L Email List Message
Subject: Re: Glass?
From: ChrisAtUpw@.......
Date: Mon, 12 Nov 2007 12:07:26 EST
```In a message dated 12/11/2007, tchannel1@............ writes:
Hi Folks, Just wondering about materials for roller on roller or balls on
plates. I know the limitations of glass, but a question about its
smoothness, is the surface of glass as smooth or smoother than these hardened and
polished steels?
Hi Ted,
Probably about the same. You get an optical finish with fine polish on
metal. However glass is a rapidly solidified liquid.
My second question is one of friction: If you have one ball on one plate you
would have one point of friction. If you have one ball resting between two
rollers, (in the V) you would have two points of friction, but the load would
be divided, because its resting on two points, so is the end result of
friction loss the same?
It is the surface to surface properties which are important. The actual
static friction is proportional to the load, so halving it and using two
points will likely give a very similar result. However, if you put a fixed
vertical load on a ball resting on two rods, the loads at the two points are
increased to give the same resolved vertical force.
When we talk about friction in pendulum suspensions, it is not this
static friction to which we are referring. It is the tiny rolling contact loss as
the materials are compressed and relaxed. No materials are perfectly
elastic. Glass is poor in this respect and tends to chip easily.
Regards,
Chris
In a message dated 12/11/2007, tchannel1@............ writes:
<=
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Hi Folks, Just wondering abou=
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materials for roller on roller or balls on plates. I know the=20
limitations of glass, but a question about its smoothness, is the surface=20=
of=20
glass as smooth or smoother than these hardened and polished=20
steels?
Hi Ted,
Probably about the same. You get an optical fin=
ish=20
with fine polish on metal. However glass is a rapidly solidified liquid.
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My second question is one of friction: If=
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have one ball on one plate you would have one point of friction. If=20=
you=20
have one ball resting between two rollers, (in the V) you would=20=
have=20
two points of friction, but the load would be divided, because its resting=
on=20
two points, so is the end result of friction loss the=20
same?
It is the surface to surface properties which a=
re=20
important. The actual static friction is proportional to the load, so=20
halving it and using two points will likely give a very similar result. Howe=
ver,=20
if you put a fixed vertical load on a ball resting on two rods, the loa=
ds=20
at the two points are increased to give the same resolved vertical=20
force.
When we talk about friction in pendulum=20
suspensions, it is not this static friction to which we are referring. It is=
the=20
tiny rolling contact loss as the materials are compressed and relaxed. No=20
materials are perfectly elastic. Glass is poor in this respect and tends to=20=
chip=20
easily.
Regards,
Chris
``` | 850 | 3,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-47 | latest | en | 0.924812 |
https://mathoverflow.net/questions/105148/discretifications-of-the-fundamental-group-functor | 1,603,150,728,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107867463.6/warc/CC-MAIN-20201019232613-20201020022613-00239.warc.gz | 450,420,603 | 28,603 | # Discretifications of the fundamental group functor
Grothendieck calls a "discretification" of a profinite group $\widehat G$, a discrete group $G$ whose profinite completion is isomorphic to $\widehat G$.
Does Grothendieck also define a notion of the "discretification" of a functor to profinite groups, and particulary that of the algebraic fundamental group functor?
The automorphisms of the geometric point act on the discretifications of the fundamental group functor; and thus one can ask:
are all discretifications of the the algebraic fundamental group functor related by automorphisms of the geometric point? To simplify the question, one may abelianise the functor and/or restrict it to a full subcategory of schemes $X\longrightarrow S$, $S=\mathrm{Spec}\ k$ a number field or $k=K$.
Here is the same question with some more notation and detail. Please excuse my poor notation: I am not very familiar with the language of schemes.
Say a functor $F$ to finitely generated Abelian groups is a discretification of a functor $F'$ to profinite groups iff for every $X$, $F(X)$ is a discretification of $F'(X)$. Equivalently, one is given an equivalence $\widehat F \Rightarrow F'$ (where $\widehat F(X)=\widehat{F(X)}$ is the profinite completion of $F(X)$.)
The algebraic fundamental group is, or at least defines, a functor $\pi_1^{alg}: \mathrm{Spec}\ K\text{\Schemes} \longrightarrow \text{ProfiniteGroups}$ from the category $\mathrm{Spec}\ K\text{\Schemes}= \mathrm{Mor} (\mathrm{Spec}\ K, \text{\Schemes})$ of Schemes cosliced over a geometric point $\mathrm{Spec}\ K$.
There is a canonical $\mathrm{Aut}(K/\mathbb{Q})$ action on $\mathrm{Spec}\ K\text{\Schemes}$ which extends to an action on the discretifications of this functor. Did Grothendicek conjecture anything about this action?
Is there a better way to state this question? E.g. using the fact that for K the field of algebraic numbers $\mathrm{Aut}(K/\mathbb{Q})=\pi_1^{alg}(\mathrm{Spec}\ \mathbb{Q}, \mathrm{Spec}\ K)$.
One can modify the question by replacing the category of Schemes by a subcategory of Schemes over $\mathrm{Spec}\ k$ or some other scheme, and then considering double cosets of $\mathrm{Aut}(K/\mathbb{Q})$ and $\mathrm{Aut}(k/\mathbb{Q})$ action; as an example, one may consider the full subcategory of etale covers of direct products of moduli spaces of curves....
Perhaps I should also add that probably I can prove some very partial positive results, for some very small and "abelian" subcategories of Schemes. That's the motivation for the question.
• Possibly related question (or not - I'm out of my depth): mathoverflow.net/questions/4972/… . – HJRW Aug 21 '12 at 11:41
• Thanks. Yes, this kind of question is probably indirectly related...But I do not see a direct relation. – o a Aug 21 '12 at 12:33
• Is it obvious whether there exists any discretification? – Will Sawin Aug 21 '12 at 17:32
• Will Savin: the topological fundamental group is intended to be an example. it works for K=C or K⊂C and an appropriate subcategory of Schemes. (Well, the topological fundamental group of a smooth projective complex algebraic variety is not necessarily residually finite, so you have to do something about that). In general, I do not know; perhaps my definition is too naive to work for general Schemes. Also, in char p, I do not know; but possibly it is not too hard to construct by hand. – o a Aug 21 '12 at 19:04
• I see. I thought it wouldn't work for $K\subset \mathbb C$ because of the classic example that shows the fundamental group of a complex manifold is not determined by its algebraic structure, but that is just an example of the $Aut(\mathbb C/\mathbb Q)$ action on discretization. In characteristic $p$, The Artin-Schreier sequence shows that $H^1(X,\mathbb Z/p)= Mor(\pi_1(X),\mathbb Z/p)$ is infinite-dimensional for a general scheme, so $\pi_1(x)$ is infinitely generated, which doesn't rule out a by-hand construction but makes it seem difficult. – Will Sawin Aug 21 '12 at 20:43 | 1,040 | 4,011 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-45 | latest | en | 0.76395 |
http://math.stackexchange.com/questions/276994/if-h-nx-mathbbz-are-all-f-g-free-abelian-then-hx-mathbbz-otim/283732 | 1,469,386,668,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824133.26/warc/CC-MAIN-20160723071024-00129-ip-10-185-27-174.ec2.internal.warc.gz | 154,447,941 | 17,513 | # If $H_n(X;\mathbb{Z})$ are all f. g. free abelian, then $H^*(X;\mathbb{Z}) \otimes \mathbb{Z}_p \cong H^*(X; \mathbb{Z}_p)$?
An exercise in Hatcher's book asks to prove that whenever $X$ is a space with the homology groups $H_n(X; \mathbb{Z})$ finitely generated free abelian for each $n \geq 0$, then $H^*(X; \mathbb{Z}) \otimes \mathbb{Z}_p$ and $H^*(X;\mathbb{Z}_p)$ are isomorphic as graded rings.
I've only started to learn about the cup product and I have no idea how to proceed. I see that the two objects in question are isomorphic as groups (from the universal coefficients theorem), but that's about it. So the question is: how to prove the statement in the title of the question?
-
Hint: $H^n(X)\to H^n(X;\mathbb{Z}_p)$ is a ring homomorphism. Putting that in a well-chosen commutative diagram, you ultimately need to show that a map $\prod\mathbb{Z}\to\prod\mathbb{Z}_p$ induces an isomorphism $(\prod\mathbb{Z})\otimes\mathbb{Z}_p\to\prod\mathbb{Z}_p$.
The universal coefficient theorem says not only that the two are isomorphic as groups but that a natural map between them, namely the reduction $\bmod p$ map $H^{\bullet}(X, \mathbb{Z}) \otimes \mathbb{Z}_p \to H^{\bullet}(X, \mathbb{Z}_p)$, is an isomorphism. This map always respects cup products (with no hypotheses).
Since $H_n(X;Z)$ finitely generated free abelian for each $n≥0$ So the Tor part will vanish in the short exact sequence for homology.Another observation: Since homology free this implies $H_n(X) = H^n(X)$.Hence the result follows. | 479 | 1,523 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2016-30 | latest | en | 0.878659 |
https://math.stackexchange.com/questions/2962429/smooth-and-not-analytic | 1,670,520,766,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711344.13/warc/CC-MAIN-20221208150643-20221208180643-00780.warc.gz | 390,340,235 | 38,233 | # Smooth and not analytic
Can someone show me, without reference to Taylor series, why a complex function can be smooth but not analytic? I do not understand it intuitively or visually either. I would like an explanation which simply refers to the definition of analytic functions as functions for which the complex derivative exists everywhere in the domain.
• Actually analytic functions are usually defined to be the functions which are equal to their Taylor series as far as I know. Complex differentiable functions on the other hand are said to be holomorphic. That being said, it is a standard result of complex analysis that these two classes of functions are precisely the same.
– MSDG
Oct 19, 2018 at 18:38
• I know, but I haven't seen the proof yet, so I wanted an explanation not referring to that fact. Oct 19, 2018 at 18:48
• @HBHSU: As Sobi has said, there is no exmplanation because in complex variable, smooth and analytic mean the same. You cannot understand why some functions are smooth but not analytic because such functions don't exist. Oct 19, 2018 at 22:57
• @Dog_69: "Smooth" in this context does not mean "complex differentiable to all orders." It simply means that the real and imaginary parts of a function are differentiable to all orders. You're obfuscating the actual question the OP has by refusing to acknowledge that while "smooth" means one thing for real variable functions, it is simply not used to describe the complex differentiability of functions of a complex variable, for which the terms "analytic" or "holomorphic" or "complex differentiable" are used. Oct 21, 2018 at 0:21
The best way to understand this, I would argue, is to consider a complex function as nothing more than a function $$f(x,y) = \big(u(x,y),v(x,y)\big)$$, and viewing complex analysis as an extension of the ideas from real variables. The definition of a function as being complex differentiable is equivalent to it satisfying the Cauchy-Riemann equations: $${\partial u\over \partial x} = {\partial v\over \partial y},\qquad{\partial u\over \partial y} = -{\partial v\over \partial x}.$$ So now, if you want an example of a smooth function that is not analytic, merely find a function $$f(x,y) = \big(u(x,y),v(x,y)\big)$$ where both $$u$$ and $$v$$ are smooth (infinitely differentiable in the sense of real variables), but that do not satisfy the Cauchy-Riemann equations! Simple as that.
For example, the function $$f(x,y) = (x,-y)$$ is smooth in the sense that $$u(x,y) = x$$ and $$v(x,y) = -y$$ are both smooth, but it is just a short check to see that $$f$$ doesn't satisfy the Cauchy-Riemann equations. In complex notation, $$f(z) = \overline z$$ is the conjugation map that sends a complex number to its conjugate. | 674 | 2,739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-49 | latest | en | 0.946051 |
http://mhtg.blogspot.com/2012/04/welcome-back.html | 1,527,266,234,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867140.87/warc/CC-MAIN-20180525160652-20180525180652-00067.warc.gz | 189,039,166 | 13,269 | Tuesday, April 10, 2012
Welcome Back!!!
I hope everyone enjoyed the Spring Break!! My family and I certainly took advantage of the wonderful weather.
Today marks the first day of the fourth (and final) marking period of the school year. It's so hard to believe that we are in the home stretch...but we still have lots of learning to do!!
This morning I was in a meeting, so Mrs. Starkey helped me by beginning our poetry unit with my reading class. They began by creating a chart of what the students knew about poetry. After that, Mrs. Starkey shared MCPS's definition of poetry. Next, the children had a chance to read through different books of poetry and locate two they liked. Finally, and I was back to enjoy this part, they read poems out loud and shared why they liked them.
After completing our warm up in math, I gave the students a quick "exit card" to see what they remember about identifying fractions and mixed numbers on a number line. You might remember that an exit card is a quick check in where the students complete one or two problems independently and then I analyze their responses to help guide my instruction.
Next we identified fractions of a set. For example, what is 1/4 of 12? I am teaching them to use the denominator to identify how many equal groups they need to make (using our example, they need to create 4 equal groups out of 12). That determines the number in each group. The numerator tells how many groups you are talking about. We solved a couple of problems as a whole class today. We will continue with independent practice tomorrow. There is a fraction review sheet for homework.
Following PE, lunch and recess I reviewed and distributed a new spelling packet that is due April 20. The spelling pattern is -are and the grammar focus is adjectives. I also modeled how to fill out a personal inventory that we will use to help us write poetry.
While I met with reading groups the children completed their own personal inventory and worked on an adjective activity. | 438 | 2,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-22 | latest | en | 0.958231 |
http://docplayer.net/6996889-Curriculum-and-contents-diplom-program-in-business-administration-year-1-and-year-2.html | 1,604,139,007,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107917390.91/warc/CC-MAIN-20201031092246-20201031122246-00191.warc.gz | 30,469,085 | 26,930 | # Curriculum and Contents: Diplom-Program in Business Administration (Year 1 and Year 2)
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1 Business School DeAN S OFFICE INTERNATIONAL RELATIONS L 5, Mannheim Germany Phone +49 (0) Fax +49 (0) Curriculum and Contents: Diplom-Program in Business Administration (Year 1 and Year 2) Accredited by AACSB International and the European Foundation for Management Development (equis accredited) Content EDV A Course EDV A (Introduction to Information Systems 2) deals with the functions and design of integrated business application systems by looking at systems in the industrial sector and in the services sector. The course also includes an introduction to e-commerce. Content EDV B Course EDV B (Introduction to Information Systems 1) is concerned with the development of operational systems. The main emphases are on the development of databases and on modern programming methods. Math A Sequences and series, topology, functions, differential calculus, optimization, integral calculus Math B 1 Basic Matrix Algebra 1.1 Matrices and Vectors 1.2 Matrix Algebra 1.3 Calculation Rules and Relations 1.4 Linear Equations 2 An Application to Economics: Cost Accounting 3 Advanced Matrix Algebra 3.1 The Determinant of a Matrix 3.2 The Inverse of a Matrix 3.3 Matrix Equations 3.4 Cramer's Rule 4 Another Application to Economics: Input/Output Analysis
2 University of Mannheim Business School page 2 5 Further Application to Economics: The Leontief Model 6 General Solutions of Linear Equations 6.1 Linear Combination and Linear Dependency 6.2 The Rank of a Matrix 6.3 Solving Linear Equations 6.4 Solutions to Linear Equations that depend on Parameters 7 Vector Spaces 7.1 Definition 7.2 Vector Subspaces 7.3 Base and Dimension of Vector Subspaces 7.4 The Solution to Linear Homogenous Equations as a Vector Subspace 8 Linear Optimization 8.1 Definitions 8.2 Graphical Solutions to Linear Optimization 8.3 The Simplex Algorithm Methods of Operational Cost Accounting The fundamental methods of financial accounting and approaches of operational accounting are conveyed with particular emphasis on the following: accounting within the context of a firm s accounting system, especially its nature and purpose, legal regulations and organization of accounting; system and methods of double-entry accounting, especially inventory, balance sheet, keeping and closing of accounts; selected accounting cases, especially posting the movement of goods (sales tax, acquisition and sale of goods), depreciation, accrual and accounting in industrial firms. Marketing Contents ( The Bauer Semester ) Fundamentals of marketing Consumer behavior and market research Market segmentation Strategic marketing Product policy Pricing policy Communications policy Distribution policy Marketing implementation Recent developments in marketing Contents ( The Homburg Semester ) 1. General Principles 2. Theoretical Perspectives 2.1. Customer Behavior 2.2. Company Behavior 3. Strategic Perspectives: Principles of Strategic Marketing Management 4. Information-related Perspectives: Fundamentals of Market Research
3 University of Mannheim Business School page 3 5. Instrumental Perspectives 5.1. Fundamentals of Pricing Management 5.2. Fundamentals of Product Management 5.3. Fundamentals of Communications Management 5.4. Fundamentals of Distribution and Sales Management 6. Institutional Perspectives 6.1. Fundamentals of the Marketing of Services 6.2. Fundamentals of Industrial Goods Marketing 6.3. Fundamentals of International Marketing Finance I. Principles of Finance II. Investment Characterization of investment measures Static investment calculation methods Dynamic investment calculation methods III. Financing Characterization of financing measures External finance Internal finance Commercial Financial Statements and Tax Balance Sheets This lecture provides insight into the principles of external accounting, laying the foundations for more advanced courses in senior year studies. The aim of the course is to relate fundamental knowledge about the set-up and content of a commercial year-end closing and financial statements for tax purposes. Beginning with the aims of the rendering the accounts, the norms of the German Commercial Code that deal with the individual company closing are discussed. The regulations for all business people and the additional regulations for capital companies take center stage. In addition, international developments are addressed. The taxation of company revenue is tied to the German Commercial Code. The determination of taxable income is tied to the determination of commercial income via the principle of congruency. The main characteristics of the determination of taxable income and those of the subsequent income tax, corporate tax and sales tax are discussed, as are the principle differences of individual enterprise taxation, business partnerships and capital companies. Cost Accounting Module 1: Accounting, Management and Controlling Module 2: Cost Theory Principles Module 3: Cost Accounting as Instruments of Information Supply for Operational Decisions Module 4: The Fundamentals of Cost Planning Module 5: Planning Primary Costs Module 6: Planning of Secondary and Tertiary Costs and Calculation Rates Module 7: Planning of Process Costs Module 8: Planning Calculation Module 9: Cost Control Module 10: Revenue and Profit Planning Module 11: Revenue and Profit Control, Reconciliation of Cost and Profit Accounting with Financial Accounting Module 12: Cost and Profit Accounting as Information Supply Instrument for Strategic Decisions Module 13: Alternative Forms of Cost and Profit Accounting
4 University of Mannheim Business School page 4 Production Management A. The Production Process as Combination of Factors I. The System of the Production Factors II. Substitutional Production Functions III. Limitational Production Functions IV. Technological Progress in the Theory of Production B. Cost Theory Perspectives of Production I. Fundamentals of Cost Theory II. Cost Schedule in Substitutional Adjustment III. Operational Forms of Adjustment IV. Dynamic Costing Functions C. Forming and Control of Production Processes I. Production Typology II. Decision Models and Problems in Production III. Production Planning and Control IV. Production as a Competitive Factor D. Production and Environment I. Approaches in Dealing with Environmental Challenges II. Limits to Growth III. The Challenge of Resource Productivity IV. Integrated Environmental Protection in the Production Process Company Policies Conceptional Principles of Company Policy Decision Analysis Targets and Target-setting Enterprise Constitution Organization Strategic Management Planning Management Company Ethics Macro-economics The two-semester course provides an introduction to the theoretical analysis of macro-economic processes. The focus is placed on the basic models of monetary and employment theory, and the problem of a macro-economic equilibrium. Contents of the first part (winter semester) include: the definition of the terms ex post and ex ante analysis, supply, demand and market equilibrium, explanation of the macro-economic production, consumer and investment function, determination of monetary supply and monetary demand, bringing the above elements into the classic and the Keynesian macro-economic view. The second part (summer semester) introduces the roles of the state and foreign countries in the basic theoretical model of employment, intensifies the analysis of the monetary supply with discourse about money creation and monetary policy, debates the roles of inflation and expectation formation with their effects on the labor market and wage policy, and ends with a comparative discussion of economic-political concepts. Micro-economics I
5 University of Mannheim Business School page 5 How does a market function? Supply, demand, equilibrium Consumer behavior: budget constraints, preferences Supply functions for consumer goods: dependence on prices and on income; price elasticity; consumer surplus Labor supply Intertemporal decision to consume Decision to consume under uncertainty Micro-economics II Efficiency and barter economy: Edgeworth Box, Pareto efficiency Production theory; production functions; fixed and variable factors; revenue, marginal profit/productivity, average return/productivity; isoquant, marginal rate of technical substitution and returns to scale; short-term profit maximization, long-term profit maximization Cost theory: total, variable and fixed costs; isocost line, minimal costs combination, expansion path; determination of the cost function; relationship between long- and short-term cost curves Perfect competition: profit maximization and optimal production plan; short-term supply curve of a company and the industry; the short-term competitive equilibrium; the long-term competitive equilibrium; market equilibrium and taxation Monopoly theory: profit maximization of a monopolist; marginal revenue and elasticity; deadweight loss through monopoly; natural monopoly; price differentiation Oligopoly theory: Cournot quantity competition, Bertrand price competition; Stackelberg competition; cartels Labor market: labor supply and demand; discrimination on the labor market; trade unions, minimum wage and wage differences in companies General equilibrium and efficiency: Pareto-Optima of goods and factor allocation; welfare communities of equilibriums in perfect markets; first and second theorem of the welfare theory Market imperfections Externalities Statistics The subject area of statistics covers the two subjects of descriptive statistics lecture/exercise course probability calculation and inductive statistics lecture/exercise course Descriptive statistics: frequency distribution, cumulative frequency curve, measure of inequality means, measures of dispersion, indices, time series, measures of dependency Calculation of probability/inductive statistics: basic concepts, special distributions, regression analysis Law for Economists The objective of the course is to treat selected problems of civil law and commercial law through cases.
6 University of Mannheim Business School page 6
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Unit Guide Diploma of Business Contents Overview... 2 Accounting for Business (MCD1010)... 3 Introductory Mathematics for Business (MCD1550)... 4 Introductory Economics (MCD1690)... 5 Introduction to Management | 7,899 | 36,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.798695 |
http://doc.gnu-darwin.org/html_xref/roots-fs.html | 1,685,828,623,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00516.warc.gz | 12,621,881 | 2,561 | GNU-Darwin Web
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# functions in roots.i - f
f_inverse ``` f_inverse(f_and_dfdx, y, x0, x1, xerr) or f_inverse(f_and_dfdx, y, x0, x1, xerr) Find values of an inverse function by Newton-Raphson iteration, backed up by bisection if the convergence seems poor. The subroutine F_AND_DFDX must be defined as: func F_AND_DFDX (x, &f, &dfdx) returning both the function value f(x) and derivative dfdx(x). If the input x is an array, the returned f and dfdx must have the same shape as the input x. If F_AND_DFDX always returns zero dfdx, f_inverse will use bisection. The result x will have the same shape as the input Y values. The values of x are constrained to lie within the interval from X0 to X1; the function value must be on opposite sides of the required Y at these interval endpoints. The iteration stops when the root is known to within XERR, or to machine precision if XERR is nil or zero. X0, X1, and XERR may be arrays conformable with Y. f_inverse takes the same number of iterations for every Y value; it does not notice that some may have converged before others. interpreted function, defined at i/roots.i line 102 ``` SEE ALSO: nraphson | 325 | 1,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-23 | latest | en | 0.771006 |
https://www.studypool.com/discuss/1220510/Algebra-2-need-help-please-dont-bid-more-than-5 | 1,506,035,641,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687938.15/warc/CC-MAIN-20170921224617-20170922004617-00044.warc.gz | 860,825,585 | 20,254 | # Algebra 2 need help, please dont bid more than \$5
Oct 17th, 2015
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Question description
A student solves the system of equations given below. In which line is an error first made? 2x - y = 6 2x + y = 10 Line 1 2x - y = 6 - 2x + y = 10 Line 2 0 = 4 Line 3 0 ≠-4 so there is no solution
a. Line 1
b. Line 2
c. Line 3
d. Line 4
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https://programmingpraxis.com/2011/06/17/adi-shamirs-threshold-scheme/?like=1&source=post_flair&_wpnonce=b19f116cf4 | 1,524,351,696,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945459.17/warc/CC-MAIN-20180421223015-20180422003015-00102.warc.gz | 686,114,172 | 20,669 | ### June 17, 2011
[ Today’s exercise was written by guest author Graham Enos, a PhD student in the Applied Mathematics program at UNC Charlotte, with solution in Python rather than Scheme. Suggestions for exercises are always welcome, or you may wish to contribute your own exercise; feel free to contact me if you are interested. ]
In his 1979 paper “How to Share A Secret,” Adi Shamir (the S in RSA) proposed a cryptographic scheme that allows n people to share a secret number S in such a way that it takes at least k of them pooling their resources to reconstruct S. This (k, n) threshold scheme uses modular arithmetic and polynomials to give each of the n participants 1/k of the needed information. For our discussion, we’ll use a mix of Shamir’s notation and that found in chapter 12 of the book Handbook of Applied Cryptography by Menezes, van Oorschot, and Vanstone.
In his paper, Shamir describes how this scheme can be used to allow groups of k people to retrieve the secret number S even if the other nk pieces of information have been lost or destroyed. For another use case, suppose S is a 2048-bit private RSA key that’s been used to encrypt a message. Once k participants get together and pool their information, they can find S and decode the message. However, at least k of them must cooperate to retrieve S; no smaller number of participants will do. Note that S and n can be arbitrarily large integers with kn. For instance, S could be the ASCII value of some secret letter, or a word encoded by taking letters as digits in base 36. Now for the details.
Given a secret value S, the number of participants n, the threshold number k, and some prime number p > max(S, n), we first construct in secret a polynomial y = q(x) of degree k−1 (modulo our prime p) with constant term S by picking independent random integers between 1 and p−1, inclusive, for the coefficients. Next we choose n unique random integers x between 1 and p−1, inclusive, and evaluate the polynomial at those n points. Each of the n participants is given an (x, y) pair.
To reconstruct S from k pairs (x, y), we use Lagrange Interpolation. In general this technique can rebuild the entire polynomial y = q(x), but since S = q(0), we only need to find q(0):
$S = \sum_{i = 1}^k \left[ y_i \prod_{1 \le j \le k, j \ne i} x_j (x_j - x_i)^{-1} \right] \mod p$
Note: the exponent −1 signifies taking the multiplicative inverse mod p, that is, the integer z such that z · (xjxi) ≡ 1 (mod p).
As an example, suppose p = 23, S = 17, and our polynomial y = q(x) is 17 + 4x + 13x2. Since this polynomial has degree two, we need at least three points to reconstruct this polynomial. Suppose furthermore that to three of our n recipients we gave the points (14, 22), (2, 8), and (21, 5). Lagrange Interpolation could be used to recreate the whole polynomial, but we’re only interested in the constant term $S = \sum_{i = 1}^3 \left[ y_i \prod_{1 \le j \le k, j \ne i} x_j (x_j - x_i)^{-1} \right] \pmod{23}$:
S = [22 · 2(2−14)−1 · 21(21−14)−1] + [8 · 14(14−2)−1 · 21(21−2)−1] + [5 · 14(14−21)−1 · 2(2−21)−1] (mod 23)
= [22 · 2 · 11−1 · 21 · 7−1] + [8 · 14 · 12−1 · 21 · 19−1] + [5 · 14 · 16−1 · 21 · 4−1] (mod 23)
= [22 · 2 · 21 · 21 · 10] + [8 · 14 · 2 · 23 · 17] + [5 · 14 · 13 · 21 · 6] (mod 23)
= 194040 + 87584 + 114660 (mod 23)
= 396284 (mod 23)
= 17
The beauty of this scheme is twofold. First, it is rather simple and elegant; the majority of the actual code used to implement the scheme takes less than 15 lines in Python. Second, it has information theoretic security. That is, the security of the scheme relies entirely upon the fact that at least k points are needed to reconstruct a degree k−1 polynomial; nothing less than k points will do. This means its security is based on something being impossible, as opposed to something being believed to be difficult, but not yet proven to be so (e.g. factoring the product of two large primes). This scheme also enjoys other useful properties; see the above references for more.
Your task is to write functions that perform both portions of Shamir’s (k, n) threshold scheme. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Pages: 1 2
### 6 Responses to “Adi Shamir’s Threshold Scheme”
1. Graham said
I also wrote a version in Common Lisp, since I’m trying to learn a new language. If there are any CL gurus in the audience, I’d appreciate any feedback!
2. Graham said
For good measure, here it is in C and Haskell.
3. razvan said
Here’s my solution in C:
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <time.h>
#define max(m, n) ((m) > (n) ? (m) : (n))
typedef struct pair
{
int x;
long y;
} pair;
long ipow(int x, int n)
{
if(n == 0) return 1;
if(n % 2 == 0) return ipow(x * x, n / 2);
return x * ipow(x, n - 1);
}
void euclid(int a, int b, int *d, int *x, int *y)
{
if (b == 0)
{
*d = a;
*x = 1;
*y = 0;
}
else {
int x0, y0;
euclid(b, a % b, d, &x0, &y0);
*x = y0;
*y = x0 - (a / b) * y0;
}
}
long eval(int* P, int x, int n)
{
assert(P != NULL);
assert(n >= 0);
int res = 0;
int i;
for(i=0; i<=n; i++)
res += P[i] * ipow(x, i);
return res;
}
pair* encrypt(int S, int n, int k, int p)
{
assert(k <= n);
assert(p > max(S, n));
int* P = malloc(k * sizeof(int));
if(P == NULL)
exit(1);
int i;
P[0] = S;
for(i=1; i<=k-1; i++)
P[i] = rand() % p;
pair* out = malloc(n * sizeof(pair));
if(out == NULL)
exit(1);
for(i=0; i<n; i++)
{
out[i].x = rand() % p;
int unique = 0;
while(!unique)
{
unique = 1;
int j;
for(j=0; j<i; j++)
if(out[j].x == out[i].x)
{
unique = 0;
out[i].x = rand() % p;
break;
}
}
out[i].y = eval(P, out[i].x, k-1) % p;
}
return out;
}
int decrypt(pair* pairs, int k, int p)
{
assert(p > k);
assert(pairs != NULL);
int S = 0;
int i, j;
for(i=0; i<k; i++)
{
int prod = pairs[i].y;
for(j=0; j<k; j++)
{
if(j == i) continue;
prod*= pairs[j].x;
int d, x, y;
euclid(p, (pairs[j].x - pairs[i].x + p) % p, &d, &x, &y);
prod *= (y + p) % p;
prod %= p;
}
S += prod;
}
S %= p;
return S;
}
int main(int argc, char **argv)
{
srand(time(NULL));
int S = 17, n = 3, k = 2, p = 23;
pair* pairs = encrypt(S, n, k, p);
int i;
for(i=0; i<n; i++)
printf("(%d, %ld)\n", pairs[i].x, pairs[i].y);
printf("\nS = %d", decrypt(pairs, n, p));
free(pairs);
return 0;
}
4. razvan said
Err, that should rather be
printf("\nS = %d", decrypt(pairs, k + 1, p));
5. I changed github accounts, so the gists I linked to above are dead (apologies). Here are new ones:
Common Lisp
C, version 1
C, version 2 (improved with a little help from Razvan’s solution).
6. Sayantan Sur said
= [22 · 2 · 11−1 · 21 · 7−1] + [8 · 14 · 12−1 · 21(mark) · 19−1] + [5 · 14 · 16−1 · 21 · 4−1] (mod 23)
= [22 · 2 · 21 · 21 · 10] + [8 · 14 · 2 · 23(mark) · 17] + [5 · 14 · 13 · 21 · 6] (mod 23)
these lines has been copied from the solution example given above.
see the mark written portion in both the lines. how did 21 got transformed to 23? | 2,244 | 6,997 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-17 | latest | en | 0.928943 |
https://www.bbcelite.com/6502sp/main/subroutine/ll51.html | 1,624,135,975,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649731.59/warc/CC-MAIN-20210619203250-20210619233250-00028.warc.gz | 600,059,850 | 132,163 | Elite on the BBC Micro
# Maths (Geometry): LL51 (6502SP version)
``` Name: LL51 [View in context]
Type: Subroutine
Category: Maths (Geometry)
Summary: Calculate the dot product of XX15 and XX16
Calculate following dot products:
XX12(1 0) = XX15(5 0) . XX16(5 0)
XX12(3 2) = XX15(5 0) . XX16(11 6)
XX12(5 4) = XX15(5 0) . XX16(12 17)
storing the results as sign-magnitude numbers in XX12 through XX12+5.
When called from part 5 of LL9, XX12 contains the vector [x y z] to the ship
we're drawing, and XX16 contains the orientation vectors, so it returns:
[ x ] [ sidev_x ] [ x ] [ roofv_x ] [ x ] [ nosev_x ]
[ y ] . [ sidev_y ] [ y ] . [ roofv_y ] [ y ] . [ nosev_y ]
[ z ] [ sidev_z ] [ z ] [ roofv_z ] [ z ] [ nosev_z ]
When called from part 6 of LL9, XX12 contains the vector [x y z] of the vertex
we're analysing, and XX16 contains the transposed orientation vectors with
each of them containing the x, y and z elements of the original vectors, so it
returns:
[ x ] [ sidev_x ] [ x ] [ sidev_y ] [ x ] [ sidev_z ]
[ y ] . [ roofv_x ] [ y ] . [ roofv_y ] [ y ] . [ roofv_z ]
[ z ] [ nosev_x ] [ z ] [ nosev_y ] [ z ] [ nosev_z ]
Arguments:
XX15(1 0) The ship (or vertex)'s x-coordinate as (x_sign x_lo)
XX15(3 2) The ship (or vertex)'s y-coordinate as (y_sign y_lo)
XX15(5 4) The ship (or vertex)'s z-coordinate as (z_sign z_lo)
XX16 to XX16+5 The scaled sidev (or _x) vector, with:
* x, y, z magnitudes in XX16, XX16+2, XX16+4
* x, y, z signs in XX16+1, XX16+3, XX16+5
XX16+6 to XX16+11 The scaled roofv (or _y) vector, with:
* x, y, z magnitudes in XX16+6, XX16+8, XX16+10
* x, y, z signs in XX16+7, XX16+9, XX16+11
XX16+12 to XX16+17 The scaled nosev (or _z) vector, with:
* x, y, z magnitudes in XX16+12, XX16+14, XX16+16
* x, y, z signs in XX16+13, XX16+15, XX16+17
Returns:
XX12(1 0) The dot product of [x y z] vector with the sidev (or _x)
XX12(3 2) The dot product of [x y z] vector with the roofv (or _y)
XX12(5 4) The dot product of [x y z] vector with the nosev (or _z)
.LL51
LDX #0 \ Set X = 0, which will contain the offset of the vector
\ to use in the calculation, increasing by 6 for each
\ new vector
LDY #0 \ Set Y = 0, which will contain the offset of the
\ result bytes in XX12, increasing by 2 for each new
\ result
.ll51
LDA XX15 \ Set Q = x_lo
STA Q
LDA XX16,X \ Set A = |sidev_x|
JSR FMLTU \ Set T = A * Q / 256
STA T \ = |sidev_x| * x_lo / 256
LDA XX15+1 \ Set S to the sign of x_sign * sidev_x
EOR XX16+1,X
STA S
LDA XX15+2 \ Set Q = y_lo
STA Q
LDA XX16+2,X \ Set A = |sidev_y|
JSR FMLTU \ Set Q = A * Q / 256
STA Q \ = |sidev_y| * y_lo / 256
LDA T \ Set R = T
STA R \ = |sidev_x| * x_lo / 256
LDA XX15+3 \ Set A to the sign of y_sign * sidev_y
EOR XX16+3,X
JSR LL38 \ Set (S T) = (S R) + (A Q)
STA T \ = |sidev_x| * x_lo + |sidev_y| * y_lo
LDA XX15+4 \ Set Q = z_lo
STA Q
LDA XX16+4,X \ Set A = |sidev_z|
JSR FMLTU \ Set Q = A * Q / 256
STA Q \ = |sidev_z| * z_lo / 256
LDA T \ Set R = T
STA R \ = |sidev_x| * x_lo + |sidev_y| * y_lo
LDA XX15+5 \ Set A to the sign of z_sign * sidev_z
EOR XX16+5,X
JSR LL38 \ Set (S A) = (S R) + (A Q)
\ = |sidev_x| * x_lo + |sidev_y| * y_lo
\ + |sidev_z| * z_lo
STA XX12,Y \ Store the result in XX12+Y(1 0)
LDA S
STA XX12+1,Y
INY \ Set Y = Y + 2
INY
TXA \ Set X = X + 6
CLC | 1,378 | 3,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-25 | longest | en | 0.857408 |
https://slideplayer.com/slide/7261258/ | 1,696,315,874,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511055.59/warc/CC-MAIN-20231003060619-20231003090619-00792.warc.gz | 560,446,737 | 19,516 | # Proof using distance, midpoint, and slope
## Presentation on theme: "Proof using distance, midpoint, and slope"— Presentation transcript:
Proof using distance, midpoint, and slope
Coordinate Geometry Proof using distance, midpoint, and slope
Coordinate proof Prove a quadrilateral is a parallelogram using three different methods. Method One: The slope formula Method Two: The distance formula Method Three: The midpoint formula
Proving a quadrilateral is a Parallelogram
Given quadrilateral ABCD, with vertices A(1,2), B(6,5), C(7,2), and D(2,-1). Prove ABCD is a parallelogram x y A B C D
Method 1: the slope formula
Given quadrilateral ABCD, with vertices A(1,2), B(6,5), C(7,2), and D(2,-1), prove ABCD is a parallelogram Conclusion: Since opposite sides of quadrilateral ABCD have the same slope, they are parallel. Quadrilateral ABCD has two pairs of parallel sides, therefore it is a parallelogram
Method 2: the distance formula
Given quadrilateral ABCD, with vertices A(1,2), B(6,5), C(7,2), and D(2,-1), prove ABCD is a parallelogram Conclusion: Since opposite sides of quadrilateral ABCD have the same length, they are congruent. Quadrilateral ABCD has two pairs of opposite, parallel sides, therefore it is a parallelogram.
Method 3: the midpoint formula
Given quadrilateral ABCD, with vertices A(1,2), B(6,5), C(7,2), and D(2,-1), prove ABCD is a parallelogram Midpoint of AC Midpoint of BD Conclusion: Since the midpoints of AC and BD are the same point, they must bisect each other. Quadrilateral ABCD has bisecting diagonals, therefore it is a parallelogram. | 421 | 1,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-40 | latest | en | 0.832133 |
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USE CODE: CHEM20
Le Chatelier’s Principle
Table of Content
Introduction to Le Chatelier’s Principle
Effect of Change of Concentration on Equilibrium
Effect of Chenage of Temperature on Equilibrium
Effect of Change of Pressure on Equilibrium
Effect of Addition of Inert Gases to a Reaction at Equilibrium
Effect of Catalyst on Equilibrium
Related Resources
Introduction to Le Chatelier’s Principle
“When an equilibrium is subjected to either a change in concentration, temperature or, in external pressure, the equilibrium will shift in that direction where the effects caused by these changes are nullified”.
Refer to the following Video for Le Chatelier’s Principle
This can be understood by the following example. Overall we can also predict the direction of equilibrium by keeping in mind following theoretical assumption.
PCl5 ——> PCl3 + Cl2
Let us assume that we have this reaction at equilibrium and the moles of Cl2, PCl3 and PCl5 at equilibrium are a, b and c respectively, and the total pressure be PT.
Since
PT = (a+b+c) RT / V
KP = abRT / cV
Now if d moles of PCl3 is added to the system, the value of Q would be, a(b+d)RT / cV
We can see that this is more than KP. So the system would move reverse to attain equilibrium.
If we increase the volume of the system, the Q becomes abRT / cV' where V' > V. Q becomes less, and the system would move forward to attain equilibrium.
If we add a noble gas at constant pressure, it amounts to increasing the volume of the system and therefore the reaction moves forward.
If we add the noble gas at constant volume, the expression of Q remains as Q = abRT / cV and the system continues to be in equilibrium. Nothing happens.
Therefore for using Le-Chatlier’s principle, convert the expression of KP and KC into basic terms and then see the effect of various changes.
Effect of Change of Concentration on Equilibrium
Let us have a general reaction,
aA + bB cC + dD
at a given temperature, the equilibrium constant,
Kc = [C]c[D]d / [A]a[B]b
again if α, β, γ and δ are the number of mole of A, B, C and D are at equilibrium
then, Kc = [γ]c[δ]d / [α]a[β]b
If any of product will be added, to keep the Kc constant, concentration of reactants will increase i.e. the reaction will move in reverse direction. Similarly if any change or disturbance in reactant side will be done, change in product’s concentration will take place to minimise the effect.
Effect of Chenage of Temperature on Equilibrium
The effect of change in temperature on an equilibrium cannot be immediately seen because on changing temperature the equilibrium constant itself changes. So first we must find out as to how the equilibrium constant changes with temperature.
For the forward reaction, according to the Arrhenius equation,
And for the reverse reaction,
R = Reactant, P = Product
It can be seen that
For any reaction, ΔH = Eaf – Ear
From the equation,
logK2/K1 = ΔHo / 2.303R (1/T1 – 1/T2) it is clear that
(a) If ΔHo is +ve (endothermic), an increase in temperature (T2 > T1) will make K2 > K1, i.e., the reaction goes more towards the forward direction and vice-versa.
(b) If ΔHo is -ve (exothermic), an increase in temperature (T2 > T1), will make K2 < K1 i.e., the reaction goes in the reverse direction.
Increase in temperature will shift the reaction towards left in case of exothermic reactions and right in endothermic reactions.
Increase of pressure (decrease in volume) will shift the reaction to the side having fewer moles of the gas; while decreases of pressure (increase in volume) will shift the reaction to the side having more moles of the gas.
If no gases are involved in the reaction higher pressure favours the reaction to shift towards higher density solid or liquid.
Refer to the following Video for Arrhenius equation
Example . Under what conditions will the following reactions go in the forward direction?
(i) N2(g)+ 3H2(g) 2NH3(g) + 23 k cal.
(ii) 2SO2(g) + O2(g) 2SO3(g) + 45 k cal.
(iii) N2(g) + O2(g) 2NO(g) - 43.2 k cal.
(iv) 2NO(g) + O2(g) 2NO2(g) + 27.8 k cal.
(v) C(s) + H2O(g) CO2(g) + H2(g) + X k cal.
(vi) PCl5(g) PCl3(g) + Cl2(g)- X k cal.
(vii) N2O4(g) 2NO2(g) - 14 k cal.
Solution:
(i) Low T, High P, excess of N2 and H2.
(ii) Low T, High P, excess of SO2 and O2.
(iii) High T, any P, excess of N2 and O2
(iv) Low T, High P, excess of NO and O2
(v) Low T, Low P, excess of C and H2O
(vi) High T, Low P, excess of PCl5
(vii) High T, Low P, excess of N2O4.
Example: The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g) is 1.6 × 10-4atm-2 at 400oC. What will be the equilibrium constant at 500oC if heat of the reaction in this temperature range is-25.14 k cal?
Solution: Equilibrium constants at different temperature and heat of the reaction are related by the equation,
log KP2 = –25140 / 2.303 × 2 [773 – 673 / 773 × 673] + log 1.64 10-4
log KP2 = –4.835
KP2 = 1.462 × 10–5 atm–2
Effect of Change of Pressure on Equilibrium
The effect of change of pressure on chemical equilibrium can be done by the formula.
Kp = QP × P(Δn)
Case A: When Δn = 0
Kp = Qp
And equilibria is independent of pressure.
Case B: When Δn = – ve
Kp = QP × P–Δn = QP / PΔn
Here with increase in external pressure, will shift the equilibrium towards forward direction to maintain Kp. Similarly, decrease in pressure will shift the equilibrium in the reverse direction.
Case C: When Δn = +ve
Kp = QP × P–Δn and, effect will be opposite to that of Case ‘B’
Effect of Addition of Inert Gases to a Reaction at Equilibrium
Let us take a general reaction
aA + bB cC + dD
We know,
Where,
nC nD, nA, nB denotes the no. of moles of respective components and PT is the total pressure and n = total no. of moles of reactants and products.
Now, rearranging,
Where n = (c + d) – (a + b)
Now, n can be = 0, < 0 or > 0
Lets take each case separately.
a) n = 0 : No effect
b) n = ‘+ve’ : Addition of inert gas increases the n i.e. is decreased and so is . So products have to increase and reactants have to decrease to maintain constancy of Kp. So the equilibrium moves forward.
c) n = ‘–ve’ In this case decreases but increases. So products have to decrease and reactants have to increase to maintain constancy of Kp. So the equilibrium moves backward.
2. Addition at Constant Volume : Since at constant volume, the pressure increases with addition of inert gas and at the same time ån also increases, they almost counter balance each other. So can be safely approximated as constant. Thus addition of inert gas has no effect at constant volume.
Effect of Catalyst on Equilibrium
Catalysts are the substances which alter the rate of a reaction without themselves getting consumed in the reaction. Since the catalyst is associated with forward as well as reverse direction reaction. So, at equilibrium, rate of forward reaction will be equal to rate of reverse reaction and hence catalyst effect will be same on both forward as well as reverse. Hence catalyst never effect the point of equilibrium but it reduces the time to attain the equilibrium.
Example: What will be the effect on the equilibrium constant on increasing the temperature.
Solution: Since the forward reaction is endothermic, so increasing the temperature, the forward reaction is favoured. Thereby the equilibrium constant will increase.
Sample problem Kp for the reaction 2BaO2(s) 2BaO(s) + O2(g) is 1.6 × 10–4 atm, at 400°C. Heat of reaction is – 25.14 kcal. What will be the no. of moles of O2 gas produced at 500°C temperature, if it is carried in 2 litre reaction vessel?
Solution: We know that
log LP2/KP1 = ΔH / 2.303R [1/T1 – 1/T2]
log KP2/1.6 × 10–4 = –25.14 / 2.330 × 2 × 10–3 [773 – 673 / 773 × 673]
=> KP2/1.46 × 10–5 atm
KP2 = p02 = 1.46 × 10–5 atm
Since, PV = nRT
1.46 × 10–5 × 2 = n × 0.0821 × 773
n = 1.46 × 10–5 / 0.0821 × 773 = = 4.60 × 10-7
Related Resources
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USE CODE: CHEM20 | 2,516 | 8,593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-51 | latest | en | 0.811228 |
https://prepinsta.com/tcs-digital/reasoning-ability/number-series/quiz-2/ | 1,669,554,746,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00015.warc.gz | 509,284,507 | 46,099 | # TCS Digital Number Series Quiz-2
Question 1
Time: 00:00:00
Find the next number in the series 135, 246, 357, 468,_____
578
578
577
577
579
579
570
570
Once you attempt the question then PrepInsta explanation will be displayed.
Start
Question 2
Time: 00:00:00
Find the next number in the series
23, 68, 113, 158, 203,_____
252
252
248
248
242
242
256
256
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Start
Question 3
Time: 00:00:00
Find the next number in the series
3, 12, 48, 192, 768, _____
2868
2868
2968
2968
3072
3072
3176
3176
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Start
Question 4
Time: 00:00:00
Find the next number in the series 256, 64, 128, 32, 64,_____
128
128
16
16
32
32
256
256
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Question 5
Time: 00:00:00
Find the next number in the series 4, 24, 48, 72, 96, _____
121
121
120
120
144
144
132
132
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Question 6
Time: 00:00:00
Find the next number in the series 2, 6, 30, 210, _____?
1680
1680
1800
1800
1890
1890
2010
2010
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Question 7
Time: 00:00:00
Which is the wrong term in the following series?
dgf, fhx, iku, mor, rtl, xzf
xzf
xzf
iku
iku
dgf
dgf
Mor
Mor
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Question 8
Time: 00:00:00
Which is the wrong term in the following series?
RSPH, UWUN, XAZT, AEEZ, DIKG
AEEZ
AEEZ
DIKG
DIKG
RSPH
RSPH
UWUN
UWUN
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Question 9
Time: 00:00:00
Which is the wrong term in the following series ?
jkl, lin, mho, ofq, per, rds, sbu
sbu
sbu
ofq
ofq
per
per
rds
rds
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Question 10
Time: 00:00:00
Determine the wrong term in the series
cze, etg, hoj, lkm, qls, wfy
wfy
wfy
lkm
lkm
qls
qls
hoj
hoj
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Question 11
Time: 00:00:00
Solve the following problem-
4,5,13,94,1118, ?
3125
3125
5120
5120
15625
15625
16743
16743
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["0","40","60","80","100"]
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0% | 911 | 2,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-49 | latest | en | 0.751081 |
https://www.xlsoffice.com/excel-functions/date-and-time-functions/add-days-to-date-in-excel/ | 1,575,987,001,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00346.warc.gz | 922,600,832 | 16,962 | ## Excel Office
Excel How Tos, Tutorials, Tips & Tricks, Shortcuts
# Add days to date in Excel
It you need to add days to date in Excel then this tutorials is for you. See example below:
To add a given number of years to a date, you can use the EDATE function.
## Formula
`=date+days`
## Explanation
In the example shown, the formula in D5 is:
`=B5+C5`
### How this formula works
Dates in Excel are just serial numbers. The number 1 represents January 1, 1900, the number 1000 is September 26, 1902, and so on.
Also See: How to calculate future date say 6 months ahead in Excel
When you have a valid date in Excel, you and just add days directly. Day values can be positive or negative.
For example, with a date in A1, you can add 100 days like so:
`=A1+100`
In the example shown, the formula is solved like this:
```=B5+C5
=36861+7
=36868```
Recall, Excel saves dates as number so when formatted as a date, 36868 is December 8, 2000.
Also See: Add years to date in Excel | 275 | 994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-51 | longest | en | 0.841763 |
http://anabolicminds.com/forum/training-forum/89556-h-s-t.html | 1,529,598,587,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864191.74/warc/CC-MAIN-20180621153153-20180621173153-00478.warc.gz | 16,040,520 | 15,199 | 1. H.S.T. question
i was looking into HST training and read something that must have been printed incorrectly or i read it wrong. but the article said (i'll use bench press for this example)to find your max lifts for 15,10, and 5 reps, then on monday do your 15 reps, weds. do your 10 and friday do your 5. now it said (i'm just going to use round #'s here) if your 15 rep max is 100lbs. and your doing that on monday, and 150 X 10 on weds. and 200 X 5 on friday, the next monday should be 210 X 15 and so on......does that sound correct? it should have been 110 X 15 right? always increasing the weight by 5-10 lbs. per week? does anyone have any experience with HST? how did it work for you?
2. Ok, Im having a really hard time understanding what you are attempting to explain but I think your basic jist is off, so allow me to help
The first week: You would go on monday and find all of your 15 rep maxes, record them. Wednesday you would go and find all of your 10 rep max's, record them. Then finally that friday you would go and find all of your 5 rep max's, record them and take 9-12 days off (strategic deconditioning).
After your 9-12 days of SD'ing, you would start your cycle, beginning with the 15's.
For example: Your max for 15 reps of bench press is 200 lbs. Since you have 6 workouts to work up to that weight, depending on the increments you want to go up every week, that is how you judge what weight to start on. An example would be a 10 lb weight increase every workout.... so on day 1 (monday) you would do 140 x 10. On day 2 (wednesday) you would do 150 x 10. And day 3 continues upward with the weight, until workout 6 when you should be doing your 15 rep max of 200 lbs.
I hope this made sense, if you have any other q's just ask
•
3. thx....yea i was close but still a bit off. thx again
4. Originally Posted by lennoxchi
i was looking into HST training and read something that must have been printed incorrectly or i read it wrong. but the article said (i'll use bench press for this example)to find your max lifts for 15,10, and 5 reps, then on monday do your 15 reps, weds. do your 10 and friday do your 5. now it said (i'm just going to use round #'s here) if your 15 rep max is 100lbs. and your doing that on monday, and 150 X 10 on weds. and 200 X 5 on friday, the next monday should be 210 X 15 and so on......does that sound correct? it should have been 110 X 15 right? always increasing the weight by 5-10 lbs. per week? does anyone have any experience with HST? how did it work for you?
Just wondering if you got that from one of my posts when I was asking questions about hst BEFORE I ever tried it. It just sounds really familiar. After running it twice, it's a much better routine if you just follow the recommended design as closely as possible rather than trying to modify it yourself.
5. Originally Posted by celc5
Just wondering if you got that from one of my posts when I was asking questions about hst BEFORE I ever tried it. It just sounds really familiar. After running it twice, it's a much better routine if you just follow the recommended design as closely as possible rather than trying to modify it yourself.
yea i did...........going to give it a shot. it's good to mix it up a bit from time to time....gotta you know?
•
6. Originally Posted by lennoxchi
yea i did...........going to give it a shot. it's good to mix it up a bit from time to time....gotta you know?
Actually, I'm usually pretty resiliant to following pre-designed routines. But, I've gotta admit, as I went, I shifted more TOWARDS the protocols listed. The Mx15, Wx10, Fx5 will NOT work and it was a horrible idea on my part btw.
Also, mixing it up with variations doesn't work with HST. It's all about progressive load and strict structure. Mix it up AFTER you complete your hst routine... and it's better the second time around so you can make personal intelligent adjustments.
7. Follow the mesocycles as described on the website: 15reps for the first mesocycle, 10reps on the second mesocycle, and 5 reps on the third mesocycle.
Im about to start the 2nd week of my 3rd mesocycle and I honestly love this routine.
I've read that the bigger the jumps in weight ( 10lbs compared to 5lbs ) the bigger the gains you'll make. I try and make it 10lbs for most of my lifts except auxillary types of lifts ( 5lbs ) - hammer curls, calves, etc.
8. Originally Posted by JDF
Follow the mesocycles as described on the website: 15reps for the first mesocycle, 10reps on the second mesocycle, and 5 reps on the third mesocycle.
Im about to start the 2nd week of my 3rd mesocycle and I honestly love this routine.
I've read that the bigger the jumps in weight ( 10lbs compared to 5lbs ) the bigger the gains you'll make. I try and make it 10lbs for most of my lifts except auxillary types of lifts ( 5lbs ) - hammer curls, calves, etc.
thx 4 the feedback on this. i am going to try to make those weight jumps (10lbs. on the compound moves too). i'm going to check out your routine, might have some questions.
9. Originally Posted by lennoxchi
thx 4 the feedback on this. i am going to try to make those weight jumps (10lbs. on the compound moves too). i'm going to check out your routine, might have some questions.
This is good advice. Keep in mind that there's some things that will tick you off the first time around, including not being able to plan your weight increments the way that you know you should have. It's great to have a preconcieved idea that you want to have 5 or 10 pound increments but it doesn't always work out that way.
*A good rule of thumb that I read in the monster HST pdf is to start your mesocycle with 70% of your rm lift for that mesocycle. Then just increase weight evenly/proportionally until you hit your rm. I doubted it at first but that 70% estimate turned out to be dead on for productive workouts.
The second time you run HST, you'll know exactly what changes need to be made and all of these intricacies that seem confusing will clear up quite easily.
For example, I saw a lot of results in the 15 and 10 mesocycle but nothing from the 5's or the drops. So the second time, I ran 16's, 12's, and 8's and no extra dropset mesocycle. That was way more more productive for me personally.
On the flip side, I've seen a handful of people that hate the 10's and 15's so they run it 12,8,4. So be open minded and only tweak after you've really given the protocol a fair shot. That's hard advice to follow if you're used to creating your own routines so you really have to let your mind accept the philosophical change.
10. good advise from all, and i thank you. i do believe the biggest hurdle i am going to have is adjusting from being in the gym 6 days a week down to 3. does anyone do light cardio on their days off? i doesn't think following the 15,10,5 routine will be a problem for me, i can be VERY regimented if i see good reason. and the ST time "out of the gate" with this routine i do not expect to perfect it, and I'm also sure changes will have to be made for a second time as well. the main reason i picked this routine is the ability to keep "old school" compound moves, dead lifts:dl:,squats etc......thx again
11. I switch between LISS and HIIT all week, depends on the mood but I try not to do HIIT 2 days in a row or to late in the week because my legs really need the rest come thursday night.
Im also on a CKD which can make wednesdays/fridays pretty tough as Im pretty depleted by that time, I've adjusted though I actually feel stronger on fridays than I do on wednesdays
12. JDF
HST + CKD? whoa, I dunno if I could handle that myself. Props to you for, well, surviving that combo Do you still get the HST pumps being so carb depleted?
lennox
yes, I do cardio, abs, and calves twice per week on "off" days. You may have to back off a bit on cardio intensity because the hst will have metabolism skyrocketing in itself... unless you're on an all out cut of course.
also, consider alternating between squats and deads. I did this my second time around. Reason being, those lifts tend to require more warm up sets and it's unreasonable to spend 20-25 minutes on 2 consecutive exercises in this routine. Then total workout time gets close to 2 hours if you try to do everything on every session.
The HST pdf also suggests 2 alternatives:
M/W/F: upper
T/Th/S: lower
or
M/W/F push
T/Th/S pull
If I had the time, I'd do the push pull 6day/wk split. Stupid work schedule
13. Originally Posted by celc5
JDF
HST + CKD? whoa, I dunno if I could handle that myself. Props to you for, well, surviving that combo Do you still get the HST pumps being so carb depleted?
I cant compare to say whether or not the pumps would be better if I was eating carbs but I still get a really sick pump while working out.
I really dont think you can beat a full body pump, I look huge about 1/2 through my workouts. I wish I could stay like that all the time.
Push/Pull - 6 days would without a doubt be intense. You'll need to eat like freaking mad, Im cutting so the 3 day split works for me because it frees up more time for cardio which I tend to neglect on a 5 day 1 bodypart a day split.
14. Originally Posted by celc5
Actually, I'm usually pretty resiliant to following pre-designed routines. But, I've gotta admit, as I went, I shifted more TOWARDS the protocols listed. The Mx15, Wx10, Fx5 will NOT work and it was a horrible idea on my part btw.
Also, mixing it up with variations doesn't work with HST. It's all about progressive load and strict structure. Mix it up AFTER you complete your hst routine... and it's better the second time around so you can make personal intelligent adjustments.
I agree that anyone trying HST should always start with the basic protocol and use it at least 2 or 3 cycles minimum. I started using it back in 2001, and have trained with it since I started then. I've used a bunch of different variations, but the past 7 cycles have been M-15s, W-5s, Sat-10s using progressive overload in each range as the cycle progresses. I have to say that for shear size, the basic works VERY well. The style I use now works best for recomping IMO. Even though I'm as big as I want to be right now, I do still gain, it's just very slowly. It also allows me to go heavier towards the end of the cycle which has helped in strength gains more so than the basic protocol.
15. I feel like I'm in an underground cult now that I'm promoting HST training
The Grand-daddy pdf of all things HST:
http://users.telenet.be/aaitee/HST%20Faq_book.pdf
Short summary pdf:
http://www.hypertrophy-research.com/...principles.pdf
These links were shown to me by Sfearl1. He also played a role in critiquing my second routine.
I would read the summary first. Then the giant pdf really sells the strategies. Remember to keep it simple!
• | 2,766 | 10,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-26 | latest | en | 0.970805 |
https://jutge.org/problems/P82891_en | 1,722,870,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00753.warc.gz | 264,455,816 | 6,081 | # Pseudoperfect numbers P82891
Statement
thehtml
The proper divisors of a number n are all the positive divisors of n that are smaller than n. For instance, the proper divisors of 20 are 1, 2, 4, 5, and 10. In this problem, we will say that a number is pseudoperfect if it can be obtained by adding up some of (or all) its proper divisors. For instance, 20 is pseudoperfect, because 1 + 4 + 5 + 10 = 20.
Write a program that, for every given number n,
• if n has more than 15 proper divisors, prints how many it has;
• if n has 15 or less proper divisors, tells if n is pseudoperfect or not.
Input
Input consists of several strictly positive natural numbers.
Output
For every given n, print its number of proper divisors, if this is larger than 15. Otherwise, tell if n is pseudoperfect or not. Follow the format of the example.
Public test cases
• Input
```1
6
10
20
210
2310
65536
1000000000
999999996
999999937
999999936
```
Output
```1 : NOT pseudoperfect
6 : pseudoperfect
10 : NOT pseudoperfect
20 : pseudoperfect
210 : pseudoperfect
2310 : 31 proper divisors
65536 : 16 proper divisors
1000000000 : 99 proper divisors
999999996 : pseudoperfect
999999937 : NOT pseudoperfect
999999936 : 167 proper divisors
```
• Information
Author | 368 | 1,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.788791 |
https://hackaday.com/tag/random/ | 1,702,124,859,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100909.82/warc/CC-MAIN-20231209103523-20231209133523-00195.warc.gz | 338,091,568 | 26,394 | # That Coin Toss Isn’t Actually 50/50
A coin flip is considered by many to be the perfect 50/50 random event, even though — being an event subject to Newtonian physics — the results are in fact anything but random. But that’s okay, because what we really want when we flip a coin is an unpredictable but fair outcome. But what if that’s not actually what happens?
There’s new research claiming that coin tosses demonstrate a slight but measurable bias toward landing on the same side they started. At least, this is true of coin flips done in a particular (but common) way. Coins flipped with the thumb and caught in the hand land with the same side facing up 50.8 percent of the time.
The new research builds on earlier work proposing that because of human anatomy, when a human flips a coin with their thumb, the motion introduces a slight off-axis tilt that biases the results. Some people do it less (biasing the results less) and some do it more, but while the impact is small it is measurable. As long as the coin is caught in the hand, anyway. Allowing the coin to fall on surfaces introduces outside variables.
Therefore, one can gain a slight advantage in coin flips by looking at which side is facing up, and calling that same side. Remember that the flipping method used must be that of flipping the coin with the thumb, and catching it with the hand. The type of coin does not matter.
Does this mean a coin flip isn’t fair? Not really. Just allow the coin to fall on a surface instead of catching it in the hand, or simply conceal which side is “up” when the coin is called. It’s one more thing that invites us all to ask just how random is random, anyway?
# Math Reveals How Many Shuffles Randomizes A Deck
Math — and some clever simulations — have revealed how many shuffles are required to randomize a deck of 52 cards, but there’s a bit more to it than that. There are different shuffling methods, and dealing methods can matter, too. [Jason Fulman] and [Persi Diaconis] are behind the research that will be detailed in an upcoming book, The Mathematics of Shuffling Cards, but the main points are easy to cover.
A riffle shuffle (pictured above) requires seven shuffles to randomize a 52-card deck. Laying cards face-down on a table and mixing them by pushing them around (a technique researchers dubbed “smooshing”) requires 30 to 60 seconds to randomize the cards. An overhand shuffle — taking sections from a deck and moving them to new positions — is a staggeringly poor method of randomizing, requiring some 10,000-11,000 iterations.
The method of dealing cards can matter as well. Back-and-forth dealing (alternating directions while dealing, such as pattern A, B, C, C, B, A) yields improved randomness compared to the more common cyclic dealing (dealing to positions in a circular repeating pattern A, B, C, A, B, C). It’s interesting to see different dealing methods shown to have an effect on randomness.
This brings up a good point: there is not really any such a thing as “more” random. A deck of cards is either randomized, or it isn’t. If even two cards have remained in the same relative positions (next to one another, for example) after shuffling, then a deck has not yet been randomized. Similarly, if seven proper riffle shuffles are sufficient to randomize a 52-card deck, there is not really any point in doing eight or nine (or more) because there isn’t any such thing as “more” random.
You can watch these different methods demonstrated in the video embedded just under the page break. Now we know there’s no need for a complicated Rube Goldberg-style shuffling solution just to randomize a deck of cards (well, no mathematical reason for one, anyway.)
# ESP32 Is The Brains Behind This Art Installation
The ESP32 has enabled an uncountable number of small electronics projects and even some commercial products, thanks to its small size, low price point, and wireless capabilities. Plenty of remote sensors, lighting setups, and even home automation projects now run on this small faithful chip. But being relegated to an electronics enclosure controlling a small electrical setup isn’t all that these tiny chips can do as [Eirik Brandal] shows us with this unique piece of audio and visual art.
The project is essentially a small, automated synthesizer that has a series of arrays programmed into it that correspond to various musical scales. Any of these can be selected for the instrument to play through. The notes of the scale are shuffled through with some random variations, allowing for a completely automated musical instrument. The musical generation is entirely analog as well, created by some oscillators, amplifiers, and other filtering and effects. The ESP32 also controls a lighting sculpture that illuminates a series of LEDs as the music plays.
The art installation itself creates quite haunting, mesmerizing tunes that are illustrated in the video linked after the break. While it’s not quite to the realm of artificial intelligence since it uses pre-programmed patterns with some randomness mixed in, it does give us hints of some other projects that have used AI in order to compose new music.
# Hamster Trades Crypto Better Than You
The inner machinations of the mind of cryptocurrency markets are an enigma. Even traditional stock markets often seem to behave at random, to the point that several economists seriously suggest that various non-human animals might outperform one market or another just by random chance alone. The classic example is a monkey picking stocks at random, but in the modern world the hamster [Mr Goxx] actively trades crypto from inside his hamster cage.
[Mr Goxx]’s home comprises a normal apartment and a separate office where he can make his trades. The office contains an “intention wheel” where he can run in order to select a currency to trade, and two tunnels that [Mr Goxx] can use to declare his intention to buy or sell the currency he selected with the wheel. The wheel is connected to an Arduino Nano with an optical encoder, and the Nano also detects the hamster’s presence in the “buy” or “sell” tunnel and lights up status LEDs when he wants to execute a trade. The Nano also communicates with an intricate Java program which overlays information on the live video feed and also executes the trades in real life with real money.
Live updates are sent directly both on Twitter and Reddit, besides the live Twitch stream of [Mr Goxx] we linked above. The stream only shows his office and not his apartment, and he’s mostly active at night (Berlin time). But we can’t wait for his random walks to yield long-term results which can be analyzed for years to come. In the meantime we’ll see if others have been able to make any profits in crypto with any less-random methods.
# Random Robot Makes Random Art
For the price of a toothbrush and a small motor with an offset weight, a bristlebot is essentially the cheapest robot that can be built. The motor shakes the toothbrush and the bristle pattern allows the robot to move, albeit in a completely random pattern. While this might not seem like a true robot that can interact with its environment in any meaningful way, [scanlime] shows just how versatile this robot – which appears to only move randomly – can actually be used to make art in non-random ways.
Instead of using a single bristlebot for the project, three of them are built into one 3D printed flexible case where each are offset by 120°, and which can hold a pen in the opening in the center. This allows them to have some control on the robot’s direction of movement. From there, custom software attempts to wrangle the randomness of the bristlebot to produce a given image. Of course, as a bristlebot it is easily subjected to the whims of its external environment such as the leveling of the table and even the small force exerted by the power/communications tether.
With some iterations of the design such as modifying the arms and control systems, she has an interesting art-producing robot that is fairly reliable for its inherently random movements. For those who want to give something like this a try, the code for running the robot and CAD files for 3D printing the parts are all available on the project’s GitHub page. If you’re looking for other bristlebot-style robots that do more than wander around a desktop, be sure to take a look at this line-following bristlebot too.
Thanks to [johnowhitaker] for the tip!
# This Classy But Chaotic Gear Clock Keeps You Guessing
There are a lot of ways to tell time, but pretty much all of them involve some sort of sequential scale — the hands sweeping across the face of an analog clock comes to mind, as does the incremental changes of a digital clock. Clocks are predictable by their very nature, and therefore somewhat boring.
This nonsequential gear clock aims to break that predictability and make for a timepiece that’s just a little bit different. It’s the work of [Tony Goacher], who clearly put a lot of work into it and pulled out nearly every tool in the shop while doing it. He started with a laser-cut plywood prototype to get the basics worked out — a pair of nested rings with internal gear teeth, each hanging on a stepper-driven pinion. The inner ring represents hours and the outer minutes, with the numbers on each randomly distributed — more or less, since no two sequential numbers are positioned more than five seconds of rotation apart.
The finished version of the clock is rendered in brass, acrylic, hardwood, and a smattering of aluminum, with a case reminiscent of the cathedral radios of yore. There are some really nice touches, like custom-made brass screws, a CNC-engraved brass faceplate with traditional clock art, and a Latin inscription on the drive cog for the hours ring that translates roughly to “Time rules all.” When we looked that up we found that “tempus rerum imperator” is the motto of the Worshipful Company of Clockmakers, the very existence of which we find pleasing in the extreme.
The clock runs through its initialization routine in the brief video below. We’re not sure we’d want this on our nightstand, but it’s certainly a unique and enjoyable way to show the passage of time. It sort of reminds us of this three-ringed perpetual calendar, but just a bit more stochastic.
# Chaotic Oscillator From Antique Logic
While working on recreating an “ancient” (read: 60-year-old) logic circuit type known as resistor-transistor logic, [Tim] stumbled across a circuit with an unexpected oscillation. The oscillation appeared to be random and had a wide range of frequency values. Not one to miss out on a serendipitous moment, he realized that the circuit he built could be used as a chaotic oscillator.
Chaotic systems can be used for, among other things, random number generation, so making sure that they do not repeat in a reliable way is a valuable property of a circuit. [Tim]’s design uses LEDs in series with the base of each of three transistors, with the output of each transistor feeding into the input of the next transistor in line, forming a ring. At certain voltages close to the switching voltages of the transistors, the behavior of the circuit changes unpredictably both in magnitude and frequency.
Building real-life systems that exhibit true randomness or chaotic behavior are surprisingly rare, and even things which seem random are often not random enough for certain applications. [Tim]’s design benefits from being relatively simple and inexpensive for how chaotic it behaves, and if you want to see his detailed analysis of the circuit be sure to visit his project’s page.
If you want to get your chaos the old fashioned way, with a Chua circuit, look out for counterfeit multipliers. | 2,474 | 11,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-50 | longest | en | 0.945233 |
https://advancesindifferenceequations.springeropen.com/articles/10.1186/s13662-021-03522-6 | 1,670,090,499,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00419.warc.gz | 115,231,248 | 63,844 | Theory and Modern Applications
# Degenerate poly-Bell polynomials and numbers
## Abstract
Numerous mathematicians have studied ‘poly’ as one of the generalizations to special polynomials, such as Bernoulli, Euler, Cauchy, and Genocchi polynomials. In relation to this, in this paper, we introduce the degenerate poly-Bell polynomials emanating from the degenerate polyexponential functions which are called the poly-Bell polynomials when $$\lambda \rightarrow 0$$. Specifically, we demonstrate that they are reduced to the degenerate Bell polynomials if $$k = 1$$. We also provide explicit representations and combinatorial identities for these polynomials, including Dobinski-like formulas, recurrence relationships, etc.
## Introduction
Although various mathematicians studied ‘poly’ as one of the generalizations of Bernoulli, Euler, Genocchi, and Cauchy polynomials [1, 4, 68, 11, 14, 16, 18, 20], the ‘poly’ for Bell polynomials has not been studied so far. Furthermore, in recent years, a lot of research has been conducted on various degenerate versions of many special polynomials and numbers, accumulating in a renewed interest for mathematicians various special polynomials and numbers [3, 9, 11, 12, 1418]. For instance, Kim and Kim [8] reappraised the polyexponential functions in relation to polylogarithm functions, expanding upon the research which was first conducted by Hardy [5].
With this in mind, in this paper, we define the degenerate poly-Bell polynomials through their degenerate polyexponential functions, reducing them to the degenerate Bell polynomials if $$k = 1$$. Hence, we define the poly-Bell polynomials when $$\lambda \rightarrow 0$$, providing explicit expressions and identities involving those polynomials.
In recent years, much research has been done for various degenerate versions of many special polynomials and numbers. Moreover, various special polynomials and numbers regained interest of mathematicians, and quite a few results have been discovered [3, 9, 11, 12, 1418]. The polyexponential functions were reconsidered by Kim [8] in view of an inverse to the polylogarithm functions which were first studied by Hardy [5]. In this paper, we define the degenerate poly-Bell polynomials by means of the degenerate polyexponential functions, and they are reduced to the degenerate Bell polynomials if $$k = 1$$. In particular, when $$\lambda \rightarrow 0$$, we call them the poly-Bell polynomials. We also provide explicit representations and combinatorial identities for these polynomials, including Dobinski-like formulas, recurrence relationships, etc.
The Bell polynomials $$B_{n}(x)=\sum_{k=0}^{n} S_{2}(n,k)x^{n}$$ are natural extensions of the Bell numbers which are a number of ways to partition a set with n elements into nonempty subsets. It is well known that the generating function of the Bell polynomials is given by
$$e^{x(e^{t}-1)}=\sum_{n=0}^{\infty } \operatorname{Bel}_{n}(x) \frac{t^{n}}{n!} \quad (\text{see [3, 11, 15, 19]}).$$
For $$\lambda \in \mathbb{R}$$, the degenerate exponential function is defined by
$$e_{\lambda }^{x}(t) = (1+\lambda t)^{\frac{x}{\lambda }}\quad \text{and}\quad e_{\lambda }(t) = \sum _{n=0}^{\infty }(1)_{n, \lambda } \frac{t^{n}}{n!} \quad (\text{see [3, 7, 9, 11, 12, 14--18]}),$$
(1)
where $$(x)_{0,\lambda }=1$$ and $$(x)_{n,\lambda }=x(x-\lambda )(x-2\lambda )\cdots (x-(n-1)\lambda )$$.
The fully degenerate Bell polynomials are given by
$$e_{\lambda }\bigl(x \bigl(e_{\lambda }(t) -1\bigr)\bigr) = \sum_{n=0}^{\infty } \operatorname{bel}_{n, \lambda }(x)\frac{t^{n}}{n!}\quad (\text{see [3]}).$$
(2)
When $$\lambda \rightarrow 0$$, $$\operatorname{bel}_{n,\lambda }(x)=\operatorname{bel}_{n}(x)$$.
Carlitz considered the degenerate Bernoulli polynomials which are given by
$$\frac{t}{e_{\lambda }(t)-1}e_{\lambda }^{x}(t)= \sum_{n=0}^{ \infty }\beta _{n,\lambda }(x) \frac{t^{n}}{n!} \quad (\text{see [2]}).$$
(3)
When $$x=0$$, $$\beta _{n,\lambda }=\beta _{n,\lambda }(0)$$ are called the degenerate Bernoulli numbers.
The degenerate Genocchi polynomials are given by
$$\frac{2t}{e_{\lambda }(t)+1}e_{\lambda }^{x}(t)= \sum_{n=0}^{ \infty }\mathcal{G}_{n,\lambda }(x) \frac{t^{n}}{n!} \quad (\text{see [12, 16]}).$$
(4)
When $$x=0$$, $$\mathcal{G}_{n,\lambda }=\mathcal{G}_{n,\lambda }(0)$$ are called the degenerate Genocchi numbers.
Kim and Kim introduced the modified polyexponential function as
$$\operatorname{Ei}_{k}(x)= \sum_{n=1}^{\infty } \frac{x^{n}}{(n-1)!n^{k}}\quad (k \in \mathbb{Z})\ (\text{see [8]}).$$
(5)
By (5), we see that $$\operatorname{Ei}_{1}(x) = e^{x} - 1$$.
The degenerate polyexponential function is given by
$$\operatorname{Ei}_{k,\lambda }(x)= \sum_{n=1}^{\infty } \frac{(1)_{n,\lambda }x^{n}}{(n-1)!n^{k}}\quad (k \in \mathbb{Z})\ (\text{see [11, 17]}).$$
(6)
We note that $$\operatorname{Ei}_{1,\lambda }(x) = e_{\lambda }(x) - 1$$.
The degenerate poly-Bernoulli polynomials are defined by
$$\frac{\operatorname{Ei}_{k,\lambda }(\log _{\lambda }(1+t))}{e_{\lambda }(t)-1} e_{\lambda }^{x}(t) =\sum_{n=0}^{\infty } \beta _{n,\lambda }^{(k)} (x) \frac{t^{n}}{n!}\quad (\text{see [14, 17]}).$$
(7)
When $$x=0$$, $$\beta _{n,\lambda }^{(k)}=\beta _{n,\lambda }^{(k)}(0)$$ are called the degenerate poly-Bernoulli numbers.
Since $$\operatorname{Ei}_{1,\lambda }(\log _{\lambda }(1+t))=t$$, $$\beta _{n,\lambda }^{(1)} (x)$$ are the degenerate Bernoulli polynomials.
The degenerate poly-Genocchi polynomials are given by
$$\frac{2 \operatorname{Ei}_{k,\lambda }(\log _{\lambda }(1+t))}{e_{\lambda }(t)+1} e_{\lambda }^{x}(t) =\sum_{n=0}^{\infty } \mathcal{G}_{n,\lambda }^{(k)} (x) \frac{t^{n}}{n!}\quad (\text{see [16]}),$$
(8)
and $$\mathcal{G}_{0,\lambda }^{(k)}(x)=0$$. When $$x=0$$, $$\mathcal{G}_{n,\lambda }^{(k)}=\mathcal{G}_{n,\lambda }^{(k)}(0)$$ are called the degenerate poly-Genocchi numbers.
When $$k=1$$, $$\mathcal{G}_{n,\lambda }^{(1)} (x)$$ are the degenerate Genocchi polynomials.
In [9], the degenerate Stirling numbers of the second kind are defined by
$$(x)_{n,\lambda }=\sum _{l=0}^{n}S_{2,\lambda }(n,l) (x)_{l}\quad (n\ge 0).$$
(9)
As an inversion formula of (9), the degenerate Stirling numbers of the first kind are defined by
$$(x)_{n}=\sum _{l=0}^{n}S_{1,\lambda }(n,l) (x)_{l,\lambda } \quad (n\ge 0)\ (\text{see [11, 15]}).$$
(10)
From (9) and (10), we note that
$$\frac{1}{k!} \bigl(e_{\lambda }(t)-1 \bigr)^{k}=\sum _{n=k}^{ \infty }S_{2,l}(n,k) \frac{t^{n}}{n!} \quad(\text{see [11, 15]})$$
(11)
and
$$\frac{1}{k!} \bigl(\log _{\lambda }(1+t) \bigr)^{k}=\sum_{n=k}^{ \infty }S_{1,\lambda }(n,k) \frac{t^{n}}{n!}\quad (\text{see [11, 17]}),$$
(12)
where
$$\log _{\lambda }(t)= \frac{1}{\lambda }\bigl(t^{\lambda }-1\bigr)\quad (\text{see [17]})$$
(13)
is the compositional inverse of $$e_{\lambda }(t)$$ satisfying $$\log _{\lambda } (e_{\lambda }(t) )=e_{\lambda } (\log _{ \lambda }(t) )=t$$.
## Degenerate poly-Bell polynomials and numbers
In this section, we define the degenerate poly-Bell polynomials by using of the degenerate polyexponential functions and give explicit expressions and identities involving these polynomials.
We define the degenerate poly-Bell polynomials $$\operatorname{bel}_{n,\lambda } ^{(k)}(x)$$, which arise from the degenerate polyexponential functions to be
$$1+ \operatorname{Ei}_{k,\lambda } \bigl(x\bigl(e_{\lambda }(t)-1\bigr)\bigr) = \sum _{n=0}^{\infty }\operatorname{bel}_{n,\lambda } ^{(k)}(x)\frac{t^{n}}{n!}$$
(14)
and $$\operatorname{bel}_{0,\lambda }^{(k)}(x)=1$$.
When $$x=1$$, $$\operatorname{bel}_{n,\lambda } ^{(k)} = \operatorname{bel}_{n,\lambda } ^{(k)}(1)$$ are called the degenerate poly-Bell numbers.
When $$k=1$$, from (6), we note that
\begin{aligned} 1+ \operatorname{Ei}_{1,\lambda } \bigl(x\bigl(e_{\lambda }(t)-1\bigr)\bigr) &= 1+ \sum _{n=1}^{\infty }\frac{(1)_{n,\lambda }(x(e_{\lambda }(t)-1))^{n}}{(n-1)!n} \\ &=\sum_{n=0}^{\infty } \frac{(1)_{n,\lambda }(x(e_{\lambda }(t)-1))^{n}}{n!} \\ &=e_{\lambda }\bigl(x\bigl(e_{\lambda }(t)-1\bigr)\bigr) = \sum _{n=0}^{\infty }\operatorname{bel}_{n,\lambda }(x) \frac{t^{n}}{n!}. \end{aligned}
(15)
Combining with (14) and (15), we have
$$\operatorname{bel}^{(1)}_{n,\lambda }(x)= \operatorname{bel}_{n,\lambda }(x).$$
When $$\lambda \rightarrow 0$$, $$\operatorname{bel}_{n}^{(k)}(x)$$ are called the poly-Bell polynomials.
### Theorem 1
For $$k \in \mathbb{Z}$$ and $$n\geq 1$$, we have
$$\operatorname{bel}^{(k)}_{n,\lambda }(x)= \sum_{l=1}^{n} \frac{(1)_{l,\lambda }}{l^{k-1}}S_{2}(n,l)x^{l}.$$
### Proof
From (6) and (11), we observe that
\begin{aligned} \operatorname{Ei}_{k,\lambda } \bigl(x\bigl(e_{\lambda }(t)-1\bigr)\bigr) &=\sum _{l=1}^{\infty } \frac{(1)_{l,\lambda }x^{l}}{l^{k-1}} \frac{1}{l!}\bigl(e_{\lambda }(t)-1\bigr)^{l} \\ &=\sum_{l=1}^{\infty }\frac{(1)_{l,\lambda }x^{l}}{l^{k-1}} \sum_{n=l}^{ \infty }S_{2,\lambda }(n,l) \frac{t^{n}}{n!} \\ &=\sum_{n=1}^{\infty } \Biggl(\sum _{l=1}^{n} \frac{(1)_{l,\lambda }}{l^{k-1}}S_{2,\lambda }(n,l)x^{l} \Biggr) \frac{t^{n}}{n!}. \end{aligned}
(16)
Combining with (14) and (16), we have the desired result. □
### Theorem 2
For $$k \in \mathbb{Z}$$ and $$n\geq 1$$, we have
$$\sum_{m=1}^{n} \operatorname{bel}^{(k)}_{m,\lambda }(x)S_{1,\lambda }(n,m)= \frac{(1)_{m,\lambda }x^{m}}{n^{k-1}}.$$
(17)
In particular, when $$x=1$$,
$$\sum_{m=1}^{n} \operatorname{bel}^{(k)}_{m,\lambda } S_{1,\lambda }(n,m)= \frac{(1)_{m,\lambda }}{n^{k-1}}.$$
### Proof
By replacing t with $$\log _{\lambda }(1+t)$$ in (14), the left-hand side is
$$1+ \operatorname{Ei}_{k,\lambda }(xt) = 1+ \sum_{n=1}^{\infty } \frac{(1)_{n,\lambda }(xt)^{n}}{(n-1)!n^{k}} = 1+\sum_{n=1}^{\infty } \frac{(1)_{n,\lambda }x^{n}}{n^{k-1}}\frac{t^{n}}{n!}.$$
(18)
On the other hand, from (12), the right-hand side is
\begin{aligned} \sum_{m=0}^{\infty } \operatorname{bel}_{m,\lambda }^{(k)}(x) \frac{(\log _{\lambda }(1+t))^{m}}{m!} &=1+ \sum_{m=1}^{\infty }\operatorname{bel}_{m, \lambda }^{(k)}(x) \sum_{n=m}^{\infty }S_{1,\lambda }(n,m) \frac{t^{n}}{n!} \\ &=1+ \sum_{n=1}^{\infty } \Biggl(\sum _{m=1}^{n} \operatorname{bel}_{m,\lambda }^{(k)}(x) S_{1,\lambda }(n,m) \Biggr)\frac{t^{n}}{n!}. \end{aligned}
(19)
Combining with coefficients of (18) and (19), we get what we want. □
### Theorem 3
(Dobinski-like formulas)
For $$k \in \mathbb{Z}$$ and $$n\geq 1$$, we have
$$\operatorname{bel}_{n,\lambda }^{(k)}(x)= \sum_{h=1}^{\infty }\sum _{m=0}^{h} \binom{m}{h} \frac{(-1)^{h-m}(1)_{h,\lambda }(m)_{n,\lambda }}{(h-1)!h^{k}}.$$
### Proof
From (1) and (6), we observe that
\begin{aligned} \sum_{n=1}^{\infty } \operatorname{bel}_{n,\lambda }^{(k)}(x) \frac{t^{n}}{n!} &= \sum_{h=1}^{\infty } \frac{(1)_{h,\lambda }x^{h}}{(h-1)!h^{k}} \bigl(e_{\lambda }(t)-1\bigr)^{h} \\ &=\sum_{h=1}^{\infty }\frac{(1)_{h,\lambda }x^{h}}{(h-1)!h^{k}} \sum_{m=0}^{h} \binom{h}{m}(-1)^{h-m}e_{\lambda }^{m}(t) \\ &=\sum_{h=1}^{\infty }\frac{(1)_{h,\lambda }x^{h}}{(h-1)!h^{k}} \sum_{m=0}^{h} \binom{h}{m}(-1)^{h-m} \sum_{n=0}^{\infty }(m)_{n,\lambda } \frac{t^{n}}{n!} \\ &=\sum_{n=0}^{\infty } \Biggl(\sum _{h=1}^{\infty }\sum_{m=0}^{h} \binom{h}{m} \frac{(-1)^{h-m}(1)_{h,\lambda }(m)_{n,\lambda }}{(h-1)!h^{k}} \Biggr) \frac{t^{n}}{n!} \\ &=\sum_{h=1}^{\infty }\sum _{m=0}^{h} \binom{h}{m} \frac{(-1)^{h-m}(1)_{h,\lambda }}{(h-1)!h^{k}} \\ &\quad {}+\sum_{n=1}^{\infty } \Biggl(\sum_{h=1}^{\infty }\sum _{m=0}^{h} \binom{h}{m} \frac{(-1)^{h-m}(1)_{h,\lambda }(m)_{n,\lambda }}{(h-1)!h^{k}} \Biggr) \frac{t^{n}}{n!}. \end{aligned}
(20)
By comparing with coefficients on both sides of (20), we get
\begin{aligned} &\sum_{h=1}^{\infty } \sum_{m=0}^{h} \binom{h}{m} \frac{(-1)^{h-m}(1)_{h,\lambda }}{(h-1)!h^{k}}=0 \quad \text{and} \\ &\operatorname{bel}_{n,\lambda }^{(k)}(x)=\sum _{h=1}^{\infty }\sum_{m=0}^{h} \binom{m}{h} \frac{(-1)^{h-m}(1)_{h,\lambda }(m)_{n,\lambda }}{(h-1)!h^{k}}\quad (n \geq 1). \end{aligned}
□
### Theorem 4
For $$k \in \mathbb{Z}$$ and $$n\geq 1$$, we have
$$\sum_{m=0}^{n-1} \binom{n}{m}(1)_{n-m,\lambda }\operatorname{bel}_{m+1, \lambda }^{(k)}(x)= \sum_{m=1}^{n}\binom{n}{m} (1- \lambda )_{n-m, \lambda }\operatorname{bel}_{m,\lambda }^{(k-1)}(x).$$
### Proof
Differentiating with respect to t in (14), the left-hand side of (14) is
\begin{aligned} \frac{\partial }{\partial t} \operatorname{Ei}_{k,\lambda }\bigl(x\bigl(e_{\lambda }(t)-1\bigr) \bigr)&= \frac{\partial }{\partial t}\sum_{n=1}^{\infty } \frac{x^{n}(e_{\lambda }(t)-1)^{n}}{(n-1)!n^{k}} \\ &=\frac{e_{\lambda }^{1-\lambda }(t)}{e_{\lambda }(t)-1}\sum_{n=1}^{ \infty } \frac{x^{n}(e_{\lambda }(t)-1)^{n}}{(n-1)!n^{k-1}} \\ &=\frac{e_{\lambda }^{1-\lambda }(t)}{e_{\lambda }(t)-1}\operatorname{Ei}_{k-1,\lambda }\bigl(x \bigl(e_{\lambda }(t)-1\bigr)\bigr) \\ &=\frac{e_{\lambda }^{1-\lambda }(t)}{e_{\lambda }(t)-1} \sum _{n=1}^{\infty }\operatorname{bel}_{n,\lambda }^{(k-1)}(x) \frac{t^{n}}{n!} . \end{aligned}
(21)
On the other hand, the right-hand side of (14) is
$$\frac{\partial }{\partial t} \Biggl(\sum _{n=0}^{\infty }\operatorname{bel}_{n, \lambda }^{(k)}(x) \frac{t^{n}}{n!} \Biggr)=\sum_{n=1}^{\infty } \operatorname{bel}_{n, \lambda }^{(k)}(x)\frac{t^{n-1}}{(n-1)!} =\sum _{n=0}^{\infty }\operatorname{bel}_{n+1, \lambda }^{(k)}(x) \frac{t^{n}}{n!}.$$
(22)
Combining with (21) and (22), we get
$$\bigl(e_{\lambda }(t)-1\bigr) \sum_{m=0}^{\infty }\operatorname{bel}_{m+1,\lambda }^{(k)}(x) \frac{t^{m}}{m!} =e_{\lambda }^{1-\lambda }(t)\sum _{m=1}^{\infty }\operatorname{bel}_{m, \lambda }^{(k-1)}(x) \frac{t^{m}}{m!} .$$
(23)
From (23), we have
$$\sum_{j=1}^{\infty }(1)_{j,\lambda } \frac{t^{j}}{j!}\sum_{m=0}^{\infty } \operatorname{bel}_{m+1,\lambda }^{(k)}(x)\frac{t^{m}}{m!} =\sum _{i=0}^{\infty }(1-\lambda )_{i,\lambda } \frac{t^{i}}{i!}\sum_{m=1}^{\infty } \operatorname{bel}_{m, \lambda }^{(k-1)}(x)\frac{t^{m}}{m!} .$$
(24)
From (24), we get
$$\sum_{n=1}^{\infty } \sum_{m=0}^{n-1} \binom{n}{m}(1)_{n-m, \lambda } \operatorname{bel}_{m+1,\lambda }^{(k)}(x)\frac{t^{n}}{n!} =\sum _{n=1}^{\infty }\sum _{m=1}^{n}\binom{n}{m} (1-\lambda )_{n-m,\lambda }\operatorname{bel}_{m, \lambda }^{(k-1)}(x) \frac{t^{n}}{n!} .$$
(25)
By comparing with coefficients on both sides of (25), we have
$$\sum_{m=0}^{n-1} \binom{n}{m}(1)_{n-m,\lambda }\operatorname{bel}_{m+1, \lambda }^{(k)}(x)= \sum_{m=1}^{n}\binom{n}{m} (1- \lambda )_{n-m, \lambda }\operatorname{bel}_{m,\lambda }^{(k-1)}(x).$$
(26)
□
### Theorem 5
For $$k \in \mathbb{Z}$$ and $$n\geq 1$$, we have
$$\operatorname{bel}_{n,\lambda }^{(k)}(x) = \sum_{j=1}^{n} \sum _{h=1}^{j} \sum_{m=1}^{h} \frac{(1)_{m,\lambda }}{m^{k-1}}S_{1,\lambda }(h,m) S_{2, \lambda }(n,j) S_{2,\lambda }(j,h)x^{j}.$$
### Proof
From (6) and (12), we observe that
\begin{aligned} \operatorname{Ei}_{k,\lambda } \bigl(\log _{\lambda }(1+t)\bigr) &= \sum_{m=1}^{\infty } \frac{(1)_{m,\lambda }(\log _{\lambda }(1+t))^{m}}{(m-1)!m^{k}} \\ &= \sum_{m=1}^{\infty }\frac{(1)_{m,\lambda }}{m^{k-1}} \frac{(\log _{\lambda }(1+t))^{m}}{m!}=\sum_{n=1}^{\infty } \Biggl(\sum_{m=1} ^{n} \frac{(1)_{m,\lambda }S_{1,\lambda }(n,m)}{m^{k-1}} \Biggr) \frac{t^{n}}{n!}. \end{aligned}
(27)
By replacing t with $$e_{\lambda }(x(e_{\lambda }(t)-1))-1$$ in (27), we get
\begin{aligned} \operatorname{Ei}_{k,\lambda } \bigl(x\bigl(e_{\lambda }(t)-1\bigr)\bigr) &= \sum _{h=1} ^{\infty } \Biggl(\sum _{m=1}^{h} \frac{(1)_{m,\lambda }S_{1,\lambda }(h,m)}{m^{k-1}} \Biggr) \frac{1}{h!}\bigl(e_{\lambda }\bigl(x\bigl(e_{\lambda }(t)-1 \bigr)\bigr)-1\bigr)^{h} \\ &=\sum_{h=1} ^{\infty } \Biggl(\sum _{m=1}^{h} \frac{(1)_{m,\lambda }S_{1,\lambda }(h,m)}{m^{k-1}} \Biggr)\sum _{j=h}^{\infty }S_{2,\lambda }(j,h)x^{j} \sum_{n=j}^{\infty }S_{2,\lambda }(n,j) \frac{t^{n}}{n!} \\ &=\sum_{n=1} ^{\infty } \Biggl(\sum _{j=1}^{n} \sum_{h=1}^{j} \sum_{m=1}^{h} \frac{(1)_{m,\lambda }}{m^{k-1}}S_{1,\lambda }(h,m) S_{2,\lambda }(n,j) S_{2, \lambda }(j,h)x^{j} \Biggr) \frac{t^{n}}{n!}. \end{aligned}
(28)
Combining with (14) and (28), we get what we want. □
For the next theorem, we observe that
\begin{aligned} \sum_{n=0}^{\infty } \beta _{n,\lambda }^{(k)}(x) \frac{t^{n}}{n!} &= \frac{\operatorname{Ei}_{k,\lambda } (\log _{\lambda }(1+t))}{e_{\lambda }(t)-1}e_{\lambda }^{x}(t) \\ &=\sum_{m=0} ^{\infty }\beta _{m,\lambda }^{(k)} \frac{t^{m}}{m!} \sum _{l=0} ^{\infty }(x)_{l,\lambda } \frac{t^{l}}{l!}= \sum_{n=0} ^{\infty } \Biggl( \sum_{m=0} ^{n} \binom{n}{m} (x)_{n-m,\lambda } \beta _{m,\lambda }^{(k)} \Biggr)\frac{t^{n}}{n!}. \end{aligned}
(29)
By comparing with coefficients on both sides of (29), we get
$$\beta _{n,\lambda }^{(k)}(x) = \sum_{m=0}^{n} \binom{n}{m}(x)_{n-m,\lambda } \beta _{m,\lambda }^{(k)}.$$
(30)
### Theorem 6
For $$n\geq 1$$, we have
$$\operatorname{bel}_{n,\lambda }^{(k)}(x) = \sum_{d=1}^{n} \sum _{h=1}^{d} \bigl(\beta _{h,\lambda }^{(k)}(1)- \beta _{h,\lambda }^{(k)} \bigr)S_{2, \lambda }(n,d) S_{2,\lambda }(d,h) x^{d},$$
where $$\beta _{n,\lambda }^{(k)}$$ are the degenerate poly-Bernoulli numbers.
### Proof
From (1), (7), and (30), we observe that
\begin{aligned} \operatorname{Ei}_{k,\lambda } \bigl(\log _{\lambda }(1+t)\bigr) &= \bigl(e_{\lambda }(t)-1\bigr) \Biggl( \sum_{j=0} ^{\infty }\beta _{j,\lambda }^{(k)} \frac{t^{j}}{j!} \Biggr) \\ &= \Biggl(\sum_{m=0} ^{\infty } \frac{(1)_{m,\lambda }}{m!}t^{m}-1 \Biggr)\sum _{j=0} ^{\infty }\beta _{j,\lambda }^{(k)} \frac{t^{j}}{j!} \\ &= \sum_{n=0} ^{\infty } \Biggl(\sum _{m=0} ^{n} \binom{n}{m}(1)_{n-m, \lambda } \beta _{m,\lambda }^{(k)}-\beta _{n,\lambda }^{(k)} \Biggr) \frac{t^{n}}{n!} \\ &=\sum_{n=1}^{\infty } \bigl(\beta _{n,\lambda }^{(k)}(1)- \beta _{n,\lambda }^{(k)} \bigr)\frac{t^{n}}{n!}. \end{aligned}
(31)
By replacing t with $$e_{\lambda }(x(e_{\lambda }(t)-1))-1$$ in (31), we get
\begin{aligned} \operatorname{Ei}_{k,\lambda } \bigl(x\bigl(e_{\lambda }(t)-1\bigr)\bigr) &= \sum _{h=1} ^{\infty } \bigl(\beta _{h,\lambda }^{(k)}(1)- \beta _{h,\lambda }^{(k)} \bigr)\frac{(e_{\lambda }(x(e_{\lambda }(t)-1))-1)^{h}}{h!} \\ &=\sum_{h=1} ^{\infty } \bigl(\beta _{h,\lambda }^{(k)}(1)-\beta _{h, \lambda }^{(k)} \bigr)\sum_{d=h}^{\infty }S_{2,\lambda }(d,h) \frac{1}{d!}\bigl(x\bigl(e_{\lambda }(t)-1\bigr) \bigr)^{d} \\ &=\sum_{d=1} ^{\infty }\sum _{h=1}^{d} \bigl(\beta _{h,\lambda }^{(k)}(1)- \beta _{h,\lambda }^{(k)} \bigr) S_{2,\lambda }(d,h)x^{d} \sum_{n=d}^{\infty }S_{2,\lambda }(n,d) \frac{t^{n}}{n!} \\ &=\sum_{n=1} ^{\infty } \Biggl(\sum _{d=1}^{n} \sum_{h=1}^{d} \bigl( \beta _{h,\lambda }^{(k)}(1)-\beta _{h,\lambda }^{(k)} \bigr) S_{2, \lambda }(n,d)S_{2,\lambda }(d,h)x^{d} \Biggr) \frac{t^{n}}{n!}. \end{aligned}
(32)
From (14) and (32), we get the desired result. □
### Theorem 7
For $$k \in \mathbb{Z}$$ and $$n\geq 1$$, we have
$$\operatorname{bel}_{n,\lambda }^{(k)}(x) = \sum_{j=1}^{n} \sum _{h=1}^{j} \sum_{m=1}^{h} \binom{n}{m}(1)_{m,\lambda }\beta _{h-m,\lambda }^{(k)} S_{2, \lambda }(n,j) S_{2,\lambda }(j,h)x^{j},$$
where $$\beta _{n,\lambda }^{(k)}$$ are the degenerate poly-Bernoulli numbers.
### Proof
From (1) and (7), we observe that
\begin{aligned} \operatorname{Ei}_{k,\lambda } \bigl(\log _{\lambda }(1+t)\bigr) &= (e_{\lambda }-1) \Biggl( \sum _{j=0} ^{\infty }\beta _{j,\lambda }^{(k)} \frac{t^{j}}{j!} \Biggr) \\ &=\sum_{m=1} ^{\infty }(1)_{m,\lambda } \frac{t^{m}}{m!}\sum_{j=0} ^{\infty }\beta _{j,\lambda }^{(k)} \frac{t^{j}}{j!}= \sum _{n=1} ^{\infty } \Biggl(\sum _{m=1} ^{n} \binom{n}{m}(1)_{m,\lambda } \beta _{n-m, \lambda }^{(k)} \Biggr)\frac{t^{n}}{n!}. \end{aligned}
(33)
By replacing t with $$e_{\lambda }(x(e_{\lambda }(t)-1))-1$$ in (33), from (11), we get
\begin{aligned} \operatorname{Ei}_{k,\lambda } \bigl(x\bigl(e_{\lambda }(t)-1\bigr)\bigr) &= \sum _{h=1} ^{\infty } \Biggl(\sum _{m=1}^{h} \binom{n}{m}(1)_{m,\lambda } \beta _{h-m, \lambda }^{(k)} \Biggr)\frac{1}{h!} \bigl(e_{\lambda }\bigl(x\bigl(e_{\lambda }(t)-1\bigr)\bigr)-1 \bigr)^{h} \\ &=\sum_{h=1} ^{\infty } \Biggl(\sum _{m=1}^{h}\binom{n}{m} (1)_{m, \lambda }\beta _{h-m,\lambda }^{(k)} \Biggr)\sum _{j=h}^{\infty }S_{2, \lambda }(j,h)x^{j} \sum_{n=j}^{\infty }S_{2,\lambda }(n,j) \frac{t^{n}}{n!} \\ &=\sum_{n=1} ^{\infty } \Biggl(\sum _{j=1}^{n} \sum_{h=1}^{j} \sum_{m=1}^{h} \binom{n}{m}(1)_{m,\lambda } \beta _{h-m,\lambda }^{(k)} S_{2,\lambda }(n,j) S_{2,\lambda }(j,h)x^{j} \Biggr) \frac{t^{n}}{n!}. \end{aligned}
(34)
Combining with (14) and (34), we get the desired result. □
### Theorem 8
For $$k \in \mathbb{Z}$$ and $$n\geq 1$$, we have
$$\operatorname{bel}_{n,\lambda }^{(k)}(x) = \frac{1}{2} \sum_{j=1} ^{n} \sum_{h=1} ^{j} \Biggl(\sum _{i=1} ^{h} \binom{h}{i}(1)_{i,\lambda } \mathcal{G}_{h-i,\lambda }^{(k)}+2\mathcal{G}_{h,\lambda }^{(k)} \Biggr) S_{2, \lambda }(j,h) S_{2,\lambda }(n,j)x^{j},$$
where $$\mathcal{G}_{n,\lambda }^{(k)}$$ are the degenerate poly-Genocchi numbers.
### Proof
From (1) and (8), we have
\begin{aligned} 2 \operatorname{Ei}_{k,\lambda } \bigl(\log _{\lambda }(1+t)\bigr) &= \bigl(e_{\lambda }(t)+1\bigr) \Biggl( \sum_{m=0} ^{\infty } \mathcal{G}_{m,\lambda }^{(k)} \frac{t^{m}}{m!} \Biggr) \\ &= \Biggl(\sum_{i=1} ^{\infty }(1)_{i,\lambda } \frac{t^{i}}{i!} \Biggr) \Biggl(\sum_{m=0} ^{\infty }\mathcal{G}_{m,\lambda }^{(k)} \frac{t^{m}}{m!} \Biggr)+2\sum_{m=0}^{\infty } \mathcal{G}_{m,\lambda }^{(k)} \frac{t^{m}}{m!} \\ &= \sum_{h=1} ^{\infty } \Biggl(\sum _{i=1} ^{h} \binom{h}{i}(1)_{i, \lambda } \mathcal{G}_{h-i,\lambda }^{(k)}+2\mathcal{G}_{h,\lambda }^{(k)} \Biggr)\frac{t^{h}}{h!}. \end{aligned}
(35)
By replacing t with $$e_{\lambda }(x(e_{\lambda }(t)-1))-1$$ in (35), we get
\begin{aligned} &2 \operatorname{Ei}_{k,\lambda } \bigl(x\bigl(e_{\lambda }(t)-1\bigr)\bigr) \\ &\quad = \sum_{h=1} ^{\infty } \Biggl(\sum _{i=1} ^{h} \binom{h}{i}(1)_{i, \lambda } \mathcal{G}_{h-i,\lambda }^{(k)}+2\mathcal{G}_{h,\lambda }^{(k)} \Biggr)\frac{(e_{\lambda }(x(e_{\lambda }(t)-1))-1)^{h}}{h!} \\ &\quad = \sum_{h=1} ^{\infty } \Biggl(\sum _{i=1} ^{h} \binom{h}{i}(1)_{i, \lambda } \mathcal{G}_{h-i,\lambda }^{(k)}+2\mathcal{G}_{h,\lambda }^{(k)} \Biggr)\sum_{j=h}^{\infty }S_{2,\lambda }(j,h)x^{j} \sum_{n=j}^{\infty }S_{2, \lambda }(n,j) \frac{t^{n}}{n!} \\ &\quad =\sum_{n=1} ^{\infty }\sum _{j=1} ^{n} \sum_{h=1} ^{j} \Biggl(\sum_{i=1} ^{h} \binom{h}{i}(1)_{i,\lambda }\mathcal{G}_{h-i,\lambda }^{(k)}+2 \mathcal{G}_{h,\lambda }^{(k)} \Biggr) S_{2,\lambda }(j,h) S_{2,\lambda }(n,j)x^{j} \frac{t^{n}}{n!}. \end{aligned}
(36)
From (14) and (36), we get what we want. □
## Further remarks
### Remark 1
For $$\lambda \in (0,1)$$, let $$X_{\lambda }$$ be the degenerate Poisson random variable with parameter $$\alpha >0$$ if the probability mass function of X is given by
$$P_{\lambda }(j)=P \{X_{\lambda }=j\}=e_{\lambda }^{-1}(\alpha ) \frac{\alpha ^{j}(1)_{j,\lambda }}{j!}\quad \text{(see [9, 13])},$$
(37)
where $$j=0,1,2,3,\ldots$$ . It is easy to show that
$$\sum_{j=0}^{\infty }P_{\lambda }(j)=e_{\lambda }^{-1}( \alpha ) \sum_{j=1}^{\infty } \frac{\alpha ^{j}}{j!}(1)_{j,\lambda } = e_{\lambda }^{-1}( \alpha )e_{\lambda }(\alpha )=1 \quad \text{(see [9, 13])}.$$
(38)
Let $$f(x)$$ be a real variable function on $$X_{\lambda }$$.
From (17) of Theorem 2, we observe that
\begin{aligned} E[X_{\lambda }] &= \sum _{j=0}^{\infty }jP_{\lambda }(j) = e_{\lambda }^{-1}(\alpha )\sum_{j=0}^{\infty } \frac{1}{(j-1)!}(1)_{j, \lambda }\alpha ^{j} \\ &=e_{\lambda }^{-1}(\alpha )\sum_{j=0}^{\infty } \frac{1}{(j-1)!}j^{k-1} \sum_{h=0}^{j} S_{1,\lambda }(j,h)\operatorname{bel}_{h,\lambda }^{(k)}(\alpha ) \\ &=e_{\lambda }^{-1}(\alpha )\sum_{j=0}^{\infty } \sum_{h=0}^{j} \frac{j^{k-1}}{(j-1)!} S_{1,\lambda }(j,h)\operatorname{bel}_{h,\lambda }^{(k)}( \alpha ). \end{aligned}
(39)
In addition, for $$n \in \mathbb{N}$$, we also obtain the moments of $$X_{\lambda }$$ as follows:
\begin{aligned} E\bigl[X^{n}_{\lambda } \bigr] &= \sum_{j=0}^{\infty }j^{n} P_{\lambda }(j) = e_{\lambda }^{-1}(\alpha )\sum _{j=0}^{\infty }\frac{j^{n-1}}{(j-1)!}(1)_{j, \lambda } \alpha ^{j} \\ &=e_{\lambda }^{-1}(\alpha )\sum_{j=0}^{\infty } \frac{j^{n-1}}{(j-1)!}j^{k-1} \sum_{h=0}^{j} S_{1,\lambda }(j,h)\operatorname{bel}_{h,\lambda }^{(k)}(\alpha ) \\ &=e_{\lambda }^{-1}(\alpha )\sum_{j=0}^{\infty } \sum_{h=0}^{j} \frac{j^{n+k-2}}{(j-1)!} S_{1,\lambda }(j,h)\operatorname{bel}_{h,\lambda }^{(k)}( \alpha ). \end{aligned}
(40)
Thus, we have the following theorem.
### Theorem 9
For $$\lambda \in (0,1)$$, let $$X_{\lambda }$$ be the degenerate Poisson random variable with parameter $$\alpha >0$$ if the probability mass function of X. Then the expectation and the moments of $$X_{\lambda }$$ are
\begin{aligned} &E[X_{\lambda }]=e_{\lambda }^{-1}( \alpha )\sum_{j=0}^{\infty }\sum _{h=0}^{j} \frac{j^{k-1}}{(j-1)!} S_{1,\lambda }(j,h)\operatorname{bel}_{h, \lambda }^{(k)}(\alpha ), \\ &E\bigl[X^{n}_{\lambda }\bigr]=e_{\lambda }^{-1}( \alpha )\sum_{j=0}^{\infty }\sum _{h=0}^{j} \frac{j^{n+k-2}}{(j-1)!} S_{1,\lambda }(j,h)\operatorname{bel}_{h,\lambda }^{(k)}( \alpha ), \end{aligned}
respectively, where $$\operatorname{bel}_{h,\lambda }^{(k)}(\alpha )$$ are the degenerate poly-Bell polynomials.
### Remark 2
Kim and Lee introduced a new type of the degenerate Bell polynomials defined by
$$e_{\lambda }\bigl(x \bigl(e^{t} -1\bigr)\bigr) = \sum_{n=0}^{\infty } \operatorname{Bel}_{n, \lambda }(x)\frac{t^{n}}{n!} \quad \text{(see [10])}.$$
(41)
When $$\lim_{\lambda \rightarrow 0}e_{\lambda }(x(e^{\lambda }-1)) = \exp (x(e^{t}-1))= \sum_{n=0}^{\infty }\operatorname{bel}_{n}(x)\frac{t^{n}}{n!}$$.
We can also consider a new type of degenerate poly-Bell polynomials by
$$1+ \operatorname{Ei}_{k,\lambda } \bigl(x\bigl(e^{t}-1\bigr)\bigr) = \sum_{n=0}^{\infty } \operatorname{Bel}_{n, \lambda } ^{(k)}(x)\frac{t^{n}}{n!}$$
(42)
and $$\operatorname{Bel}_{0,\lambda }^{(k)}(x)=1$$.
When $$x=1$$, $$\operatorname{Bel}_{n,\lambda } ^{(k)} = \operatorname{Bel}_{n,\lambda } ^{(k)}(1)$$ are called the degenerate poly-Bell numbers.
When $$k=1$$, from (42), we note that
\begin{aligned} 1+ \operatorname{Ei}_{1,\lambda } \bigl(x\bigl(e^{t}-1\bigr)\bigr) &= 1+ \sum _{n=1}^{\infty }\frac{(1)_{n,\lambda }(x(e^{t}-1))^{n}}{(n-1)!n} \\ &=\sum_{n=0}^{\infty }\frac{(1)_{n,\lambda }(x(e^{t}-1))^{n}}{(n-1)!n} \\ &=e_{\lambda }\bigl(x\bigl(e^{t}-1\bigr)\bigr) = \sum _{n=0}^{\infty }\operatorname{Bel}_{n,\lambda }(x) \frac{t^{n}}{n!}. \end{aligned}
(43)
From (41) and (43), we have
$$\operatorname{Bel}^{(1)}_{n,\lambda }(x)= \operatorname{Bel}_{n,\lambda }(x).$$
When $$\lambda \rightarrow 0$$, $$\operatorname{Bel}_{n}^{(k)}(x)$$ are called a new type of the poly-Bell polynomials.
We can obtain similar results in the same way.
## Conclusion
To summarize, in this paper, $$\lambda \rightarrow 0$$ is defined as poly-Bell polynomials by introducing the degenerate poly-Bell polynomials through a degenerate polyexponential function and reducing it to a degenerate Bell polynomial for $$k = 1$$. We derived Dobinski-like formula in Theorem 3, recurrence relation in Theorem 4, and combinatorial identities for these polynomials. As for Theorem 1, the explicit formula demonstrated the relationship with Stirling numbers of the second kind according to k. To conclude, there are various methods for studying special polynomials and numbers, including: generating functions, combinatorial methods, umbral calculus, differential equations, and probability theory. We are now interested in continuing our research into the application of ‘poly’ versions of certain special polynomials and numbers in the fields of physics, science, and engineering as well as mathematics.
Not applicable.
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## Acknowledgements
The authors would like to thank Jangjeon Institute for Mathematical Science for the support of this research.
## Funding
This work was supported by the Basic Science Research Program, the National Research Foundation of Korea (NRF-2021R1F1A1050151).
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TK and HKK conceived the framework for the whole paper; HKK wrote the whole paper; TK and HKK checked the results of the paper and completed the revision of the article. All authors read and approved the final manuscript.
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Correspondence to Hye Kyung Kim.
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Kim, T., Kim, H.K. Degenerate poly-Bell polynomials and numbers. Adv Differ Equ 2021, 361 (2021). https://doi.org/10.1186/s13662-021-03522-6
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### Keywords
• Bell polynomials and numbers
• Modified degenerate polyexponential functions
• Degenerate poly-Bernoulli polynomials
• Degenerate poly-Euler polynomials | 12,776 | 31,725 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.920365 |
https://elteoremadecuales.com/kleenes-recursion-theorem/?lang=de | 1,679,922,819,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00771.warc.gz | 284,950,756 | 18,249 | # Kleene's recursion theorem
Kleene's recursion theorem Not to be confused with Kleene's theorem for regular languages.
In computability theory, Kleene's recursion theorems are a pair of fundamental results about the application of computable functions to their own descriptions. The theorems were first proved by Stephen Kleene in 1938[1] and appear in his 1952 book Introduction to Metamathematics.[2] A related theorem, which constructs fixed points of a computable function, is known as Rogers's theorem and is due to Hartley Rogers, Jr.[3] The recursion theorems can be applied to construct fixed points of certain operations on computable functions, to generate quines, and to construct functions defined via recursive definitions.
Inhalt 1 Notation 2 Rogers's fixed-point theorem 2.1 Proof of the fixed-point theorem 2.2 Fixed-point-free functions 3 Kleene's second recursion theorem 3.1 Comparison to Rogers's theorem 3.2 Application to quines 3.3 Application to elimination of recursion 3.4 Reflexive programming 4 The first recursion theorem 4.1 Beispiel 4.2 Proof sketch for the first recursion theorem 4.3 Comparison to the second recursion theorem 5 Generalized theorem 6 Siehe auch 7 Verweise 8 Weiterlesen 9 External links Notation The statement of the theorems refers to an admissible numbering {Anzeigestil Varphi } of the partial recursive functions, such that the function corresponding to index {Anzeigestil e} ist {Anzeigestil Varphi _{e}} .
Wenn {Anzeigestil F} und {Anzeigestil G} are partial functions on the natural numbers, the notation {displaystyle Fsimeq G} indicates that, for each n, entweder {Anzeigestil F(n)} und {Anzeigestil G(n)} are both defined and are equal, or else {Anzeigestil F(n)} und {Anzeigestil G(n)} are both undefined.
Rogers's fixed-point theorem Given a function {Anzeigestil F} , a fixed point of {Anzeigestil F} is an index {Anzeigestil e} so dass {Anzeigestil Varphi _{e}simeq varphi _{F(e)}} . Rogers describes the following result as "a simpler version" of Kleene's (zweite) recursion theorem.[4] Rogers's fixed-point theorem. Wenn {Anzeigestil F} is a total computable function, it has a fixed point. Proof of the fixed-point theorem The proof uses a particular total computable function {Anzeigestil h} , defined as follows. Given a natural number {Anzeigestil x} , die Funktion {Anzeigestil h} outputs the index of the partial computable function that performs the following computation: Given an input {Anzeigestil y} , first attempt to compute {Anzeigestil Varphi _{x}(x)} . If that computation returns an output {Anzeigestil e} , then compute {Anzeigestil Varphi _{e}(j)} and return its value, wenn überhaupt.
Daher, for all indices {Anzeigestil x} of partial computable functions, wenn {Anzeigestil Varphi _{x}(x)} ist definiert, dann {Anzeigestil Varphi _{h(x)}simeq varphi _{varphi_{x}(x)}} . Wenn {Anzeigestil Varphi _{x}(x)} is not defined, dann {Anzeigestil Varphi _{h(x)}} is a function that is nowhere defined. Die Funktion {Anzeigestil h} can be constructed from the partial computable function {Anzeigestil g(x,j)} described above and the s-m-n theorem: für jeden {Anzeigestil x} , {Anzeigestil h(x)} is the index of a program which computes the function {displaystyle ymapsto g(x,j)} .
To complete the proof, Lassen {Anzeigestil F} be any total computable function, and construct {Anzeigestil h} wie oben. Lassen {Anzeigestil e} be an index of the composition {displaystyle Fcirc h} , which is a total computable function. Dann {Anzeigestil Varphi _{h(e)}simeq varphi _{varphi_{e}(e)}} by the definition of {Anzeigestil h} . Aber, Weil {Anzeigestil e} is an index of {displaystyle Fcirc h} , {Anzeigestil Varphi _{e}(e)=(Fcirc h)(e)} , und somit {Anzeigestil Varphi _{varphi_{e}(e)}simeq varphi _{F(h(e))}} . By the transitivity of {Anzeigestil simeq } , das heisst {Anzeigestil Varphi _{h(e)}simeq varphi _{F(h(e))}} . Somit {Anzeigestil Varphi _{n}simeq varphi _{F(n)}} zum {displaystyle n=h(e)} .
This proof is a construction of a partial recursive function which implements the Y combinator.
Fixed-point-free functions A function {Anzeigestil F} so dass {Anzeigestil Varphi _{e}not simeq varphi _{F(e)}} für alle {Anzeigestil e} is called fixed-point free. The fixed-point theorem shows that no total computable function is fixed-point free, but there are many non-computable fixed-point-free functions. Arslanov's completeness criterion states that the only recursively enumerable Turing degree that computes a fixed-point-free function is 0′, the degree of the halting problem.[5] Kleene's second recursion theorem The second recursion theorem is a generalization of Rogers's theorem with a second input in the function. One informal interpretation of the second recursion theorem is that it is possible to construct self-referential programs; sehen "Application to quines" unter.
The second recursion theorem. For any partial recursive function {Anzeigestil Q(x,j)} there is an index {Anzeigestil p} so dass {Anzeigestil Varphi _{p}simeq lambda y.Q(p,j)} .
The theorem can be proved from Rogers's theorem by letting {Anzeigestil F(p)} be a function such that {Anzeigestil Varphi _{F(p)}(j)=Q(p,j)} (a construction described by the S-m-n theorem). One can then verify that a fixed-point of this {Anzeigestil F} is an index {Anzeigestil p} nach Bedarf. The theorem is constructive in the sense that a fixed computable function maps an index for Q into the index p.
Comparison to Rogers's theorem Kleene's second recursion theorem and Rogers's theorem can both be proved, rather simply, from each other.[6] Jedoch, a direct proof of Kleene's theorem[7] does not make use of a universal program, which means that the theorem holds for certain subrecursive programming systems that do not have a universal program.
Application to quines A classic example using the second recursion theorem is the function {Anzeigestil Q(x,j)=x} . The corresponding index {Anzeigestil p} in this case yields a computable function that outputs its own index when applied to any value.[8] When expressed as computer programs, such indices are known as quines.
The following example in Lisp illustrates how the {Anzeigestil p} in the corollary can be effectively produced from the function {Anzeigestil Q} . The function s11 in the code is the function of that name produced by the S-m-n theorem.
Q can be changed to any two-argument function.
(setq Q '(Lambda (x y) x)) (setq s11 '(Lambda (fx) (list 'lambda '(j) (list f x 'y)))) (setq n (list 'lambda '(x y) (list Q (list s11 'x 'x) 'y))) (setq p (eval (list s11 n n))) The results of the following expressions should be the same. {Anzeigestil Varphi } p(nil) (eval (list p nil)) Q(p, nil) (eval (list Q p nil)) Application to elimination of recursion Suppose that {Anzeigestil g} und {Anzeigestil h} are total computable functions that are used in a recursive definition for a function {Anzeigestil f} : {Anzeigestil f(0,j)simeq g(j),} {Anzeigestil f(x+1,y)simeq h(f(x,j),x,j),} The second recursion theorem can be used to show that such equations define a computable function, where the notion of computability does not have to allow, prima facie, for recursive definitions (zum Beispiel, it may be defined by μ-recursion, or by Turing machines). This recursive definition can be converted into a computable function {Anzeigestil Varphi _{F}(e,x,j)} that assumes {Anzeigestil e} is an index to itself, to simulate recursion: {Anzeigestil Varphi _{F}(e,0,j)simeq g(j),} {Anzeigestil Varphi _{F}(e,x+1,y)simeq h(varphi_{e}(x,j),x,j).} The recursion theorem establishes the existence of a computable function {Anzeigestil Varphi _{f}} so dass {Anzeigestil Varphi _{f}(x,j)simeq varphi _{F}(f,x,j)} . Daher {Anzeigestil f} satisfies the given recursive definition.
Reflexive programming Reflexive, or reflective, programming refers to the usage of self-reference in programs. Jones presents a view of the second recursion theorem based on a reflexive language.[9] It is shown that the reflexive language defined is not stronger than a language without reflection (because an interpreter for the reflexive language can be implemented without using reflection); dann, it is shown that the recursion theorem is almost trivial in the reflexive language.
The first recursion theorem While the second recursion theorem is about fixed points of computable functions, the first recursion theorem is related to fixed points determined by enumeration operators, which are a computable analogue of inductive definitions. An enumeration operator is a set of pairs (EIN,n) where A is a (code for a) finite set of numbers and n is a single natural number. Often, n will be viewed as a code for an ordered pair of natural numbers, particularly when functions are defined via enumeration operators. Enumeration operators are of central importance in the study of enumeration reducibility.
Each enumeration operator Φ determines a function from sets of naturals to sets of naturals given by {Anzeigestil Phi (X)={nmid exists Asubseteq X[(EIN,n)in Phi ]}.} A recursive operator is an enumeration operator that, when given the graph of a partial recursive function, always returns the graph of a partial recursive function.
A fixed point of an enumeration operator Φ is a set F such that Φ(F) = F. The first enumeration theorem shows that fixed points can be effectively obtained if the enumeration operator itself is computable.
First recursion theorem. The following statements hold. For any computable enumeration operator Φ there is a recursively enumerable set F such that Φ(F) = F and F is the smallest set with this property. For any recursive operator Ψ there is a partial computable function φ such that Ψ(Phi) = φ and φ is the smallest partial computable function with this property. Example Like the second recursion theorem, the first recursion theorem can be used to obtain functions satisfying systems of recursion equations. To apply the first recursion theorem, the recursion equations must first be recast as a recursive operator.
Consider the recursion equations for the factorial function f: {Anzeigestil f(0)=1} {Anzeigestil f(n+1)=(n+1)cdot f(n)} The corresponding recursive operator Φ will have information that tells how to get to the next value of f from the previous value. Jedoch, the recursive operator will actually define the graph of f. Zuerst, Φ will contain the pair {Anzeigestil (varnothing ,(0,1))} . This indicates that f(0) is unequivocally 1, and thus the pair (0,1) is in the graph of f.
Nächste, for each n and m, Φ will contain the pair {Anzeigestil ({(n,m)},(n+1,(n+1)cdot m))} . This indicates that, if f(n) is m, then f(n + 1) ist (n + 1)m, so that the pair (n + 1, (n + 1)m) is in the graph of f. Unlike the base case f(0) = 1, the recursive operator requires some information about f(n) before it defines a value of f(n + 1).
The first recursion theorem (im Speziellen, Teil 1) states that there is a set F such that Φ(F) = F. The set F will consist entirely of ordered pairs of natural numbers, and will be the graph of the factorial function f, wie gewünscht.
The restriction to recursion equations that can be recast as recursive operators ensures that the recursion equations actually define a least fixed point. Zum Beispiel, consider the set of recursion equations: {Anzeigestil g(0)=1} {Anzeigestil g(n+1)=1} {Anzeigestil g(2n)=0} There is no function g satisfying these equations, because they imply g(2) = 1 and also imply g(2) = 0. Thus there is no fixed point g satisfying these recursion equations. It is possible to make an enumeration operator corresponding to these equations, but it will not be a recursive operator.
Proof sketch for the first recursion theorem The proof of part 1 of the first recursion theorem is obtained by iterating the enumeration operator Φ beginning with the empty set. Zuerst, a sequence Fk is constructed, zum {displaystyle k=0,1,ldots } . Let F0 be the empty set. Proceeding inductively, for each k, let Fk + 1 be {Anzeigestil F_{k}cup Phi (F_{k})} . Endlich, F is taken to be {displaystyle bigcup F_{k}} . The remainder of the proof consists of a verification that F is recursively enumerable and is the least fixed point of Φ. The sequence Fk used in this proof corresponds to the Kleene chain in the proof of the Kleene fixed-point theorem.
The second part of the first recursion theorem follows from the first part. The assumption that Φ is a recursive operator is used to show that the fixed point of Φ is the graph of a partial function. The key point is that if the fixed point F is not the graph of a function, then there is some k such that Fk is not the graph of a function.
Comparison to the second recursion theorem Compared to the second recursion theorem, the first recursion theorem produces a stronger conclusion but only when narrower hypotheses are satisfied. Rogers uses the term weak recursion theorem for the first recursion theorem and strong recursion theorem for the second recursion theorem.[3] One difference between the first and second recursion theorems is that the fixed points obtained by the first recursion theorem are guaranteed to be least fixed points, while those obtained from the second recursion theorem may not be least fixed points.
A second difference is that the first recursion theorem only applies to systems of equations that can be recast as recursive operators. This restriction is similar to the restriction to continuous operators in the Kleene fixed-point theorem of order theory. The second recursion theorem can be applied to any total recursive function.
Generalized theorem In the context of his theory of numberings, Ershov showed that Kleene's recursion theorem holds for any precomplete numbering.[10] A Gödel numbering is a precomplete numbering on the set of computable functions so the generalized theorem yields the Kleene recursion theorem as a special case.[11] Given a precomplete numbering {Anzeigestil Nr } , then for any partial computable function {Anzeigestil f} with two parameters there exists a total computable function {Anzeigestil t} with one parameter such that {displaystyle forall nin mathbb {N} :nu circ f(n,t(n))=nu circ t(n).} See also Denotational semantics, where another least fixed point theorem is used for the same purpose as the first recursion theorem. Fixed-point combinators, which are used in lambda calculus for the same purpose as the first recursion theorem. Diagonal lemma a closely related result in mathematical logic References Ershov, Yuri L. (1999). "Part 4: Mathematics and Computability Theory. 14. Theory of numbering". In Griffor, Eduard R. (ed.). Handbook of Computability Theory. Studies in logic and the foundations of mathematics. Vol. 140. Amsterdam: Elsevier. pp. 473–503. ISBN 9780444898821. OCLC 162130533. Abgerufen 6 Kann 2020. Jones, Neil D. (1997). Computability and complexity: From a Programming Perspective. Cambridge, Massachusetts: MIT Press. ISBN 9780262100649. OCLC 981293265. Kleene, Stephen C. (1952). Introduction to Metamathematics. Bibliotheca Mathematica. North-Holland Publishing. ISBN 9780720421033. OCLC 459805591. Abgerufen 6 Kann 2020. Rogers, Hartley (1967). Theory of recursive functions and effective computability. Cambridge, Massachusetts: MIT Press. ISBN 9780262680523. OCLC 933975989. Abgerufen 6 Kann 2020.
Footnotes ^ Kleene, Stephen C. (1938). "On notation for ordinal numbers" (Pdf). Journal of Symbolic Logic. 3 (4): 150–155. doi:10.2307/2267778. ISSN 0022-4812. JSTOR 2267778. Abgerufen 6 Kann 2020. ^ Kleene 1952. ^ Nach oben springen: a b Rogers 1967. ^ Rogers 1967, §11.2. ^ Soare, R.I. (1987). Recursively Enumerable Sets and Degrees: A Study of Computable Functions and Computably Generated Sets. Perspektiven in der mathematischen Logik. Berlin and New York City: Springer-Verlag. p. 88. ISBN 9780387152998. OCLC 318368332. ^ Jones 1997, pp. 229–30. ^ Kleene 1952, pp. 352-3. ^ Cutland, Nigel J. (1980). Computability: An Introduction to Recursive Function Theory. Cambridge University Press. p. 204. doi:10.1017/cbo9781139171496. ISBN 9781139935609. OCLC 488175597. Abgerufen 6 Kann 2020. ^ Jones 1997. ^ Barendregt, Henk; Terwijn, Sebastiaan A. (2019). "Fixed point theorems for precomplete numberings". Annals of Pure and Applied Logic. 170 (10): 1151–1161. doi:10.1016/j.apal.2019.04.013. hdl:2066/205967. ISSN 0168-0072. S2CID 52289429. Abgerufen 6 Kann 2020. p. 1151. ^ See Ershov 1999, §4.14 for a survey in English. Further reading Jockusch, C. G.; Lerman, M.; Sonne, RI; Solovay, R.M. (1989). "Recursively enumerable sets modulo iterated jumps and extensions of Arslanov's completeness criterion". Das Journal of Symbolic Logic. 54 (4): 1288–1323. doi:10.1017/S0022481200041104. ISSN 0022-4812. JSTOR 2274816. Externe Links "Recursive Functions" entry by Piergiorgio Odifreddi in the Stanford Encyclopedia of Philosophy, 2012. Kategorien: Computability theoryTheorems in the foundations of mathematics
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# Machine learning setup in Python
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One of the tools you can use to programme machine learning. In this module, you’ll learn the basics of python when it’s used for machine learning, how to use loops to compute total loss, regressions and classification, and how to setup machine learning in python.
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Okay, so let's do a bit of a review of the setup of machine learning in terms of Python. Let's have another classification problem. So, the setup for all supervised learning is this, right. So, at the top here we've got estimators, predictors, all those pieces of the problem are coming together. Loss, features, target, and then the only difference between regression and classification, is whether the thing we're trying to predict is a real number or a label option or a discreet number. So let's just choose a problem of some kind classify a fill. So example, classifying a fill. And then what're we gonna do here? Let's write in English a little bit first, and then convert it into annotation. So given the age of a user and the length of a film, predict whether the user will, or customer, will like or dislike the film, right? So okay. What does that mean? Well it means we've got our X's and our Y. We have two X's here, age and length, so we've got X one, that's gonna be age. And see if that looks all right. Pretty good. So X one is age, let's go for X two is gonna be the length of the film. Like, dislike, that's gonna be our Y, so paste that and get our Y here, and that's gonna be, call it, like question mark, yeah. Get that question mark format blown up. So we've got those options left. So what we need to do, so basically, we're basically saying find a relationship, so find an estimate F hat, which takes in, y'know, X one, X two and gives us a Y hat. Good. Let's have a look, so these are the key variables, let's say we actually say given maybe a data set D. D, and that's just going to be examples of X and Y's above. Dots there . So given this historical data set, given that data set, with things in it, find this relationship, which we could say, where that Y hat is a prediction for Y. So where Y hat here is a good prediction for Y, how do we define good prediction? Well, where the loss is minimum. So when we look at the loss of each point, so we look at the loss comparing our prediction to our observation, we would see that that's been minimized. So where we've got minimum loss. We might want to adjust the tilde notation, but one line can actually just say we've minimized the loss like that. So this is kind of like, just some definitions, our feet definitions feature age and length, and then this is kind of the problem, so we give them some data sets, find an estimated function, gives us an estimate, where the estimate is gonna be minimizing this loss, so the estimates are close. Right okay, so let's set this up. So data set D. Let's do that as a dictionary, right? So let's say that's like, I've got age, length, and like in there, so it's gonna be historical data, so you call it in historical, some call it historical, some call it training. Training to get us that dictionary and to have an age, column, effectively, add some more values, length of the film column, some values, and a Y column, we can call y'know, like question mark. So age, let's have the age of some people, let's go 12, 13, 17, 19, whatever. One two three four, let's do five. And try to align these for the sake of demonstration. So the length of the film, let's have a film length of 90 minutes, 100 minutes, 200 minutes, 180 minutes, and 400 minutes, that seems much, 320 maybe. It's a long film. So did they like it yes or no, yes, no, yes, yes, yes. Say oh well, keep that all set up. Just go with no on that one. Now you might observe here that in programming, and anywhere, really, one, plus one is just the same as the number one, but the reason I'm doing the plus sign here, is I guess it just makes it clearer that when I say a positive, or a plus case, or a positive case, and the symbols have the same length on the screen, and it's just a little easier to see what's going on there. So there's our training data. And that's coming in in a kind of more tabulous style right? So we've got these columns in a table, we don't have X Y et cetera, we've got columns. So let's assume there, and then let's let me do this correspondence, so that would be X one, that would be X two, right? And then we've got Y here is our like column. Okay so the goal is to produce a Y hat column, which is close to this one. So how are we gonna do that? Well we can with an estimated function, so let's just do that. So def F hat for film, I'll just call it F hat for now. F hat, that's gonna take in X one, and X two, so I'm gonna give us a little prediction. So let's do X one is gonna be age, let's say X one, let's go for minus 20. So if I take away 20 from age, it's gonna be negative over here, and positive on this one, so maybe if we take away 17, that'll be zero here, negative here, positive here. So this formula here will be negative, in this part and positive in that part, okay. I'm just gonna sketch for now. If I do X two and if I take away, I don't know, 200, again, we get some of them will be negative, this formula will be negative for some of them, and positive for one, okay. Maybe what we do is we say let's maybe add these togther, see what happens? So we're coming up with some kind of formula, now, the problem with adding these two numbers together is this one's about 10 times bigger than that one, so why don't we divide this by 10 or times that one by 10, let's times this one by 10. That kinda puts them on the same scale. Gives them a same kind of weight, and a same kind of importance to the formula that we're coming up with. So maybe if I return that, now this isn't giving me, this is not quite finished yet, but if I just return that, my point here a film, or an age of 20 and a film of 100, that's a negative number, if I put a film of 200, that's a positive number. And put an age of 10, that's a negative number, if I put an age of 30, that becomes a positive number. You can see there's a kind of reasonableness to the output here, that it's positive when we want it to be kinda positive, and negative when we want it to be kinda negative. So this formula isn't quite finished, because we need plus minus one here, a sign out. So a couple of tricks we can do to get that, we could say is that more than zero, and if we do that, that gives us true and false. So that would be true, that gonna make it here true to mean plus and false to mean minus one. We need to keep going with this a little bit to massage it in Python to get it working properly. But let's leave it giving us true and false. Or you could even perhaps update this definition here to trues and falses. There's nothing wrong with that. So let's have it giving us true and falses output. That seems fine. So let's use this then, so if I'm gonna use F hat on my X length two to compute this Y hat, what can I do? Let's just do on the first entrance of each, just to give you a sense of what's going on here, first in X one, the first in X two gives us a false, so average is wrong, okay. That's pretty bad, we don't have a picture to be long, but I mean, how long is it, we need some definition here, to tell us how wrong we're doing over all. Let's define a loss, and the loss here is going to take in this prediction, so we are Y hat, taking our Y and do something with it. Now the problem here is of course that we don't have real numbers you can't just do mean or square or any of that kinda stuff, so we have to think about what we can do here. Well, we definitely anyway, that if that Y hat is equal to Y, we could just return zero. If they're the same. We just reset, well there's no error. The Y is maybe we just return one, so there's one point of error. And I can't really do this, then we would need to have these be true and false then, let's just update these so they are true and false. First one's gonna be false, true, true, false, true. Okay, that gives us a nice Python way of working. So then that's the loss is going to work, so I'm gonna test out these non-numerical values that give us a numerical loss. So if I run loss on that then, so loss on this prediction, so that's the prediction there, going in. And then the second number here is the actual value for this, which is zero. So that gives us a loss of zero, because they are the same now, that's false, that's false. If I look at the next one in the list, so I could just increase this to one, the next row and in there we get lots of one because when it goes through this comes out to be false, when it should be true. Right, so let's go ahead and compute the loss and the prediction and everything for our entire data set. So we'll do that as we did before. Let's go for for, and we can loop over X one X two now, so how do I loop over them both at the same time? Well to loop over in Python, two things at the same time, so why don't we say A and B will be our X one X two. We can use zip. What zip's going to do is it's going to pair up the entries X one with X two, so you get 12 and 90, 13, and 100, and give us both of those at the same time. So we get A and B, or we can give them a real name, like age and length, that might be much better. Age and length coming out of X one X two at the same time. So we're gonna use those. Let's compute our predictions. So prediction's gonna be call them predictions, call them whatever you like. We could've called them Y hat, but, have I used Y hat, no, so we can call it Y hat. Use Y hat there. Let's put that in the list then, so append, using our prediction F hat, and we're gonna put in here the age and the length, and that's that, so we've got the list of predictions. Oh. A little, restating that. So those are our predictions coming out. And now what we want is the error, or the total loss. So to compute the loss, so we do loss, y'know, there we go. And I'd say loss dot append, and ooh, wow, we've got a couple of things happning here, we could, I think temporarily store this thing here as prediction, and we append the prediction. The reason we would want to do that is because we need to put it into the loss of loss, the loss is gonna be loss, oh I've got to append two things. Let's call these errors, or something. Errors put into errors. We're gonna append the loss, which is this function here, on our Y hat and Y, so if the Y hat here's a prediction, the Y is a genuine Y now. We don't actually have, we would normally have had to answer stuff, so let's put it in here. So if I put Y in here, that's gonna get me out the age, the length, and now the actual true value, call it true, or call it like I suppose? Maybe use the word like again. So that's gonna give us all three at the same time, sounds pretty good. So age, length, and like. So age and length go into that prediction, with the prediction into our record of predictions, into our as we put now, comparison between prediction and what it actually was. So true, true, whether they actually liked it. So we looked everything now, Y hat errors, hopefully, we can see that we're getting an error on some of these entries, but not on all. If I wanna know what the total error is, rather than defining an assumption called total loss, what I can just do is I can take a sum of the errors. And that'll be the total loss. Total loss here of two. Now what we would like to do of course is find some way of tuning this rule so that this total error here is minimized. That's the goal. So maybe we could just do it by hand. So what's happening is, we'll do the visual inspection of what's happening, what's going wrong. We're seeing false, false, true, true, true. And this is false, false, true, true, true, so we've got a problem here. So this age is 13 and the length is 100, so why would we be saying false when it's true? Well they're saying false. Maybe it's to do with the age, maybe the age, or maybe it's to do with, maybe this is too important. So some of these numbers here, 17 and 200 basically, the key numbers, are we oversentsitive to age or something? Hard to say, so let's make a small adjustment. If I said minus 15, and I run through everything. I haven't made a difference. Maybe if I said minus 175, will that make a difference? We're probably making it give errors now, but you can see that if I make this 50, that could, it makes, oh, it says true for everything now, so we're getting the same number of errors each time, but they're making different predictions. And actually, that makes complete sense, and the reason it makes complete sense is because all this rule can really do, it's a very simple rule, all it can really do is it can just basically, just choose a point and go well, below this point, yes and above that point, no. So if you think about that, it's gonna choose somewhere to say okay, these are gonna be dislikes, these are gonna be likes, and if you look at the actual data set we have, within each range, there's a bit of a mixture going on. So it isn't really a single problem, where people who are young like things that people who are old don't like, there's actually a bit of complexity. So that rule that is this simple can't possibly catch that complexity, so we may always have a couple of errors here. So that seems fine. And so, y'know, in practice, what we would like the machine to do is figure out what these numbers should be. Maybe we can suggest to the machine a better rule to use, and then get a better solution. Right, so okay, good, I think that's where I want to be on the sort of Python setup of the machine learning problems. It's giving you a bit of syntax to go along with that kind of white boardy or mathematically, or concptual instruction, just to see what all this looks like in the simple form of Python. So what we have to do next is make this bit more realistic, so we're gonna look at some libraries and other approaches that take these ideas, and make them actually practically usable at scale in realistic problems.
### Lectures
About the Author
Michael Burgess
Principal Technologist for Machine Learning
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Michael began programming as a young child, and after freelancing as a teenager, he joined and ran a web start-up during university. Around studying physics and after graduating, he worked as an IT contractor: first in telecoms in 2011 on a cloud digital transformation project; then variously as an interim CTO, Technical Project Manager, Technical Architect and Developer for agile start-ups and multinationals.
His academic work on Machine Learning and Quantum Computation furthered an interest he now pursues as QA's Principal Technologist for Machine Learning. Joining QA in 2015, he authors and teaches programmes on computer science, mathematics and artificial intelligence; and co-owns the data science curriculum at QA. | 3,759 | 15,333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-31 | longest | en | 0.901868 |
http://dalmatien.info/docs/a16520-oxygen-difluoride-valence-electrons | 1,660,203,872,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00510.warc.gz | 16,378,566 | 4,409 | The magnesium and fluoride ions then orient themselves in a crystal lattice instead of having a molecular geometry as exhibited by covalent compounds. ... After we write the formula down we use a periodic table or our previous knowledge on the amount of valence electrons for the elements we use. OF 2 - Oxygen difluoride O has 6 valence electrons plus 1 for each O-F single bond Total = 8 electrons, four pairs Structure bent based on a tetrahedron, class AX 2 E 2 Likewise, each of the three F atoms obeys the octet rule since they are surrounded by eight electrons (three nonbonding pairs and one bonding pair). An atom’s electronegativity is affected by both its atomic number and the size of the atom. In peroxides (O. In our energy well model, this means that the lines also represent the average distance of the valence electrons from the nucleus. CC BY-SA 3.0. http://en.wiktionary.org/wiki/electronegativity The molecule will have a total of #20# valence electrons #6# from the oxygen atom #7# from each of the two fluorine atoms; The oxygen atom will take the role of central atom, forming single bonds with the two fluoride atoms. ... How many lone pairs are present on the central atom in a correct Lewis structure for the oxygen difluoride molecule? In compounds with nonmetals, the oxidation number of hydrogen is +1. Several methods of calculation have been proposed, and although there may be small differences in the numerical values of the calculated electronegativity values, all methods show the same periodic trend among the elements. Wikimedia Question Give the structure of an oxygen molecule, using one line to represent each shared pair of electrons. Six rules can be used when assigning oxidation numbers: Boundless vets and curates high-quality, openly licensed content from around the Internet. The atom with higher electronegativity, typically a nonmetallic element, is assigned a negative oxidation number, while the other atom, which is often but not necessarily a metallic element, is given a positive oxidation number. One way to characterize atoms in a molecule and keep track of electrons is by assigning oxidation numbers. While this approach has the advantage of simplicity, it is clear that the electronegativity of an element is not an invariable atomic property; rather, it can be thought of as depending on a quantity called ‘the oxidation number’ of the element. The oxidation number of an element in its natural state (i.e., how it is found in nature) is zero. This molecular geometry ensures that the dipole moments associated with the oxygen - fluoride bonds do not cancel each other out to produce a nonpolar molecule.. To see why this is the case, draw the molecule's Lewis structure.The molecule will have a total of #20# valence electrons The opposite of electronegativity is electropositivity, which is a measure of an element’s ability to donate electrons. CC BY-SA 3.0. http://www.chem1.com/acad/webtext/chembond/cb04.html#SEC1 http://en.wiktionary.org/wiki/oxidation_number, http://www.chem1.com/acad/webtext/chembond/cb04.html#SEC1, http://en.wikipedia.org/wiki/Oxidation%20State, http://en.wiktionary.org/wiki/electronegativity, http://en.wikipedia.org/wiki/Oxidation_number, http://en.wikipedia.org/wiki/Electronegativity, http://oer.avu.org/bitstream/handle/123456789/42/Introductory%20Chemistry%201.pdf?sequence=4, http://nongnu.askapache.com/fhsst/Chemistry_Grade_10-12.pdf, http://commons.wikimedia.org/wiki/File:Taula_peri%C3%B2dica_electronegativitat.png, https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/, Apply the rules for assigning oxidation numbers to atoms in compounds. 2. Determine the number of electrons an oxygen atom... What is the octet rule? Steve Lower’s Website All rights reserved. | 869 | 3,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-33 | latest | en | 0.890102 |
https://tarikbilla.com/how-do-i-reverse-a-list-or-loop-over-it-backwards/ | 1,685,329,689,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644574.15/warc/CC-MAIN-20230529010218-20230529040218-00175.warc.gz | 606,742,555 | 12,307 | # How do I reverse a list or loop over it backwards?
To get a new reversed list, apply the `reversed` function and collect the items into a `list`:
``````>>> xs = [0, 10, 20, 40]
>>> list(reversed(xs))
[40, 20, 10, 0]
``````
To iterate backwards through a list:
``````>>> xs = [0, 10, 20, 40]
>>> for x in reversed(xs):
... print(x)
40
20
10
0
`````` | 126 | 358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-23 | latest | en | 0.879239 |
https://www.mathworks.com/matlabcentral/answers/433851-how-to-add-lines-to-histograms-plotted-by-plotmatrix?s_tid=prof_contriblnk | 1,719,015,073,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00266.warc.gz | 767,280,753 | 25,674 | # how to add lines to histograms plotted by plotmatrix
6 views (last 30 days)
Etsuko on 4 Dec 2018
Commented: Etsuko on 5 Dec 2018
Hi,
I want to add vertical lines to histograms plotted in the diagonal of a matrix of pair-wise scatter plots. I tried to use the graphic objects returned by plotmatrix;
[S,AX,BigAx,H,HAx] = plotmatrix(X); % X has 15 columns
for example,
subplot(15,15,1); hold on; line([1.2,1.2],get(AX(1),'YLim'),'Color','r')
AX(15); plot([1.2,1.2],[0,1.5],'Color','r')
etc. But so far none has worked out. I would appreciate any clue as I am clueless.
I use Matlab 2018a. Thanks.
Akira Agata on 4 Dec 2018
% Sample data
X = rand(50,3);
figure
[~,ax] = plotmatrix(X);
hold(ax(1,3),'on')
plot(ax(1,3),[0.2 0.2],ax(1,3).YLim,'r:','LineWidth',2)
hold(ax(2,1),'on')
plot(ax(2,1),[0.5 0.5],ax(2,1).YLim,'r','LineWidth',2)
Just FYI:
If you can upgrade your MATLAB to R2018b, you can do the same thing slightly easier by using newly introduced xline function, like:
figure
[~,ax] = plotmatrix(X);
xline(ax(1,3),0.2,'r:','LineWidth',2)
xline(ax(2,1),0.5,'r','LineWidth',2)
Etsuko on 5 Dec 2018
Thank you very much!
It needed to specify the axis of the histograms. Just an easy modification of your suggestion did what I wanted. Thank you!
[S,AX,BigAx,H,HAx] = plotmatrix(X); % X has 15 columns
hold(HAx(1,1),'on')
plot(HAx(1,1),[2.2 2.2],HAx(1,1).YLim,'r:','LineWidth',2)
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Start Hunting! | 539 | 1,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-26 | latest | en | 0.739518 |
https://edurev.in/studytube/Previous-year-questions--2014-19--Waves/0b351118-939e-48b7-9d07-b072485fa075_t | 1,601,123,778,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400241093.64/warc/CC-MAIN-20200926102645-20200926132645-00717.warc.gz | 364,173,378 | 44,453 | Courses
# Previous year questions (2014-19) - Waves NEET Notes | EduRev
## NEET : Previous year questions (2014-19) - Waves NEET Notes | EduRev
The document Previous year questions (2014-19) - Waves NEET Notes | EduRev is a part of the NEET Course Physics 28 Years Past year papers for NEET/AIPMT Class 11.
All you need of NEET at this link: NEET
Q 1. A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27°C two successive resonances are produced at 20 cm and 73 cm column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27°C is:- [2018]
A: 330 m/s
B: 339 m/s
C: 350 m/s
D: 300 m/s
Ans: B
Solution
:
Q 2. The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20cm, the length of the open organ pipe is:- [2018]
A: 13.2 cm
B: 8 cm
C: 12.5 cm
D: 16 cm
Ans:
A
Solution:
Q 3. The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system? [2017]
A: 20 Hz
B: 30 Hz
C: 40 Hz
D: 10 Hz
Ans:
A
Solution:
Two successive frequencies of closed pipe
Q 4. Two cars moving in opposite directions approach each other with speed of 22 m/s and 16.5 m/s respectively. The driver of the first car blows a horn having a a frequency 400 Hz. The frequency heard by the driver of the second car is [velocity of sound 340 m/s]:- [2017]
A: 361 Hz
B: 411 Hz
C: 448 Hz
D: 350 Hz
Ans:
C
Solution:
Q 5. A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 ms-1. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is : (Take velocity of sound in air = 330 ms-1) [2016]
A: 885Hz
B: 765 Hz
C: 800 Hz
D: 838 Hz
Ans:
D
Solution:
Since the observer and the wall are stationary so frequency of echo observed by the observer will also be 838 Hz.
Q 6. An air column, closed at one end and open at the other, resonantes with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is: [2016]
A: 200 cm
B: 66.7 cm
C: 100 cm
D: 150 cm
Ans:
D
Solution:
The smallest length of the air column is associated with the fundamental mode of vibration of the air column.
The next higher length of the air column is,
Q 7. A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength λ1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ2. The ratio λ1/λ2 is : [2016]
A:
B:
C:
D:
Ans:
C
Solution:
Q 8. The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is: [2015]
A: 140 cm
B: 80 cm
C: 100 cm
D: 120 cm
Ans:
D
Solution:
For closed organ pipe fundamental frequency
For open organ pipe fundamental frequency
Q 9. If n1, n2, and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by: [2014]
A:
B: n = n1 + n2 + n3
C: D
: Ans: C
Solution:
Q 10. The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are : (velocity of sound = 340 ms−1) [2014]
A: 7
B: 6
C: 4
D: 5
Ans: B
Solution:
Q 11. A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36 km/hour. He finds that traffic has eased and a car moving ahead of him at 18 km/hour is honking at a frequency of 1392 Hz. If the speed of sound is 343 m/s, the frequency of the honk as heard by him will be: [2014]
A: 1412 Hz
B: 1454 Hz
C: 1332 Hz
D: 1372 Hz
Ans:
A
Solu
tion:
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14 docs|25 tests
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; | 1,276 | 4,215 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-40 | latest | en | 0.866677 |
http://blog.csdn.net/u014033518/article/details/49925171 | 1,516,349,593,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887832.51/warc/CC-MAIN-20180119065719-20180119085719-00148.warc.gz | 41,122,458 | 12,187 | # LeetCode OJ: Remove Linked List Elements
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
ListNode *p = head, *pre_p = head;;
while (p != NULL)
{
if (p->val == val){
{
//delete p;
}
if (pre_p == p) //pre_p与p也指向同一节点
{
delete p;
pre_p = p =head;
}
else{ //pre_p指向p的前一节点
pre_p->next = p->next;
delete p;
p = pre_p->next;
}
}
else
{
if (pre_p = p){
p = p->next;
}
else
{
pre_p = p;
p = p->next;
}
}
}
}
};
Accepted 32ms
• 本文已收录于以下专栏：
举报原因： 您举报文章：LeetCode OJ: Remove Linked List Elements 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字) | 266 | 715 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-05 | latest | en | 0.216299 |
http://www.cs.nyu.edu/pipermail/fom/2009-May/013752.html | 1,501,144,489,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427750.52/warc/CC-MAIN-20170727082427-20170727102427-00201.warc.gz | 391,001,320 | 2,425 | # [FOM] Infinitesimal calculus
joeshipman@aol.com joeshipman at aol.com
Tue May 26 20:24:34 EDT 2009
I disagree. Real numbers can be explained to high school students
perfectly soundly in terms of successive rational approximations
(decimal expansions, binary expansions, continued fractions, etc.) to
points on a geometric line. When this is done, they have a consistent
intuition for "what real numbers are", an intuition which makes a proof
of the least upper bound property quite easy to understand. There is no
philosophical difficulty at this level, unless you try to get across
the point that physical space might not have a structure that is
precisely captured by this definition.
On the other hand, when you use infinitesimals you have to be able
either to say "what an infinitesimal is" and distinguish infinitesimals
from the standard reals in a way that is intuitive and not merely
formal, or else you have to deny the Archimedean axiom and say that
infinitesimals are just as real as standard real numbers. This CAN be
done in an intuitive way following Conway's "On Numbers and Games", but
all the attempts I have seen to use infinitesimals to develop Calculus
for students do not do this, instead treating them purely formally in
a system where you can't get your hands on an infinitesimal directly.
This is unsatisfactory for high-school level students who aren't
usually comfortable with introducing elements that can be regarded as
fictitious.
(In Conway's system you just extend Dedekind cuts into the transfinite,
so each individual number can be expressed as a map from some ordinal
to the set {+, -} and when the ordinal is at most \omega you recover
the classical real numbers as a substructure.)
-- JS
-----Original Message-----
From: David Ross <ross at math.hawaii.edu>
I don't really want to engage yet again in the argument as to whether
it is
better or not to teach calculus with infinitesimals, I just want to
point
out that some of the remaining arguments against doing so do not hold
up to
strong scrutiny. When we made the switch in the US 100 years ago, it
was
for reasons of rigor; now, however, it is ultimately just a matter of
taste. | 510 | 2,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-30 | latest | en | 0.933153 |
https://puzzling.stackexchange.com/questions/79166/john-and-edmunds-game | 1,660,798,947,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573163.7/warc/CC-MAIN-20220818033705-20220818063705-00203.warc.gz | 437,622,319 | 66,040 | # John and Edmund's Game
I visited my younger cousins, John and Edmund, recently, and found them playing an interesting game. They would take turns to call out numbers until one of them declared themselves the winner. They would also alternate the person who would start the game.
Here are examples of some of the games they played (I've written John's answers in italics and Edmund's answers in bold)
3, 2, 3, 4, 4, 5, 5, 2, 3, 3, 4, 2, 2, 4, 5, 7, 6
At this point, John declared himself the winner.
2, 3, 2, 2, 1, 4, 1, 5, 6, 3, 5, 3, 3, 1, 1, 2
Again, John declared that he won at this point. After some consideration, Edmund reluctantly agreed.
As the afternoon wore on, the games tended to get longer. Here was another one won by John
1, 4, 1, 1, 4, 3, 5, 2, 2, 3, 3, 1, 2, 2, 4, 4, 3, 4, 3, 3, 2
Edmund was clearly having some difficulty beating his older brother but, eventually, he came through in the longest game I had witnessed.
4, 3, 5, 4, 6, 7, 4, 5, 3, 2, 6, 5, 5, 3, 1, 4, 5, 3, 3, 4, 5, 1, 4, 3, 7, 7, 1, 7, 1, 1, 7, 7, 1, 6, 6, 6, 6
"Yes! I win!" exclaimed Edmund.
I didn't really understand the rules of the game yet but I had a strategy I wanted to try and asked John if I could join in.
"Sure", said John, "4".
"4", I responded
"3", said John.
"3", I responded with confidence.
"2. I win!", exclaimed John.
I was a bit dumbfounded and looked over at Edmund who was chuckling in the corner. I couldn't understand how my strategy had lost so quickly. I accused them of making it all up but they insisted that it was a real game.
Can you explain how the game works?
Hint
John and Edmund's game is not their own creation, they are just playing the game in an unusual way.
• Would it give too much away to say why the knowledge tag is there? There's no indication in the text that it should be. Feb 1, 2019 at 15:51
• @deepthought I debated whether or not to include it. You can take it as a very small hint. The knowledge is not entirely necessary but is incredibly useful to determine the answer. Feb 1, 2019 at 16:59
• rot13: Fvapr gur ahzoref hfrq ner sebz bar gb frira, vf vg snve gb nffhzr gung gurer vf gur hfr bs bpgny ahzoref? Feb 1, 2019 at 18:34
• @hexomino any chance of a hint? thanks! Feb 4, 2019 at 4:35
• @DanielBaliki that is a good observation to make but it has nothing to do with what you suggested. Feb 4, 2019 at 10:03
## 1 Answer
I think they are playing
Connect Four
with the numbers representing
which column they are dropping their pieces into (the numbering appears to be as though they are sitting on the same side of the board, so they are both counting "1" as the leftmost column).
The final game gives John
three unbounded pieces along the bottom, with no possible way for the opponent to win, so although the game hasn't technically played to a conclusion yet, it can't go any other way. | 908 | 2,859 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-33 | longest | en | 0.98523 |
https://www.delftstack.com/zh-tw/howto/python/python-generator-vs-iterator/ | 1,720,967,312,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00516.warc.gz | 653,903,470 | 8,369 | Python 中迭代器和生成器的區別
Jesse John 2023年1月30日
Python 中的迭代器
1. 迭代器是使用實現迭代器協議的類建立的物件。這意味著該類定義了 `__iter__``__next__` 方法。
2. `__next__` 方法使用 `return` 語句返回一個值。由於 `return` 語句必須是該方法的最後一行，我們必須在 `return` 語句之前更新要在 `__next__` 下一次執行中使用的變數。
Python 中的生成器
1. 生成器是一個函式。
2. 生成器函式使用 `yield` 關鍵字而不是 `return` 關鍵字。
2.1 `yield` 關鍵字產生一個值並暫停函式的執行。
2.2 對 `next()` 的下一次呼叫在 `yield` 語句之後恢復程式碼的執行。
Python 中的迭代器和生成器示例
``````# ITERATOR (Class)
class squares(object):
def __init__(self, num1):
self.nxt_sq_of = 1
self.lim = num1
def __iter__(self):
return self
def __next__(self):
if self.nxt_sq_of <= self.lim:
ret_sq_of = self.nxt_sq_of
self.nxt_sq_of += 1
return ret_sq_of * ret_sq_of
else:
raise StopIteration
# Iterator Object
a = squares(6)
# Next value of the iterator.
next(a)
next(a)
next(a)
next(a)
next(a)
next(a)
next(a)
next(a)
# Using the iterator in a loop.
a1 = squares(6)
while True:
print(next(a1))
``````
``````next(a)
Out[3]: 1
next(a)
Out[4]: 4
next(a)
Out[5]: 9
next(a)
Out[6]: 16
next(a)
Out[7]: 25
next(a)
Out[8]: 36
next(a)
Traceback (most recent call last):
File "<ipython-input-9-15841f3f11d4>", line 1, in <module>
next(a)
File "<ipython-input-1-9dbe8e565876>", line 17, in __next__
raise StopIteration
StopIteration
``````
``````# GENERATOR FUNCTION
def gen_squares(num2):
i = 1
while i <= num2:
yield i * i
i += 1
# Generator iterator.
b = gen_squares(5)
# Next yield of the generator iterator.
next(b)
next(b)
next(b)
next(b)
next(b)
next(b)
``````
``````next(b)
Out[3]: 1
next(b)
Out[4]: 4
next(b)
Out[5]: 9
next(b)
Out[6]: 16
next(b)
Out[7]: 25
next(b)
Traceback (most recent call last):
File "<ipython-input-8-adb3e17b0219>", line 1, in <module>
next(b)
StopIteration
``````
まとめ
1. 生成器的 Python Wiki 文章。
2. 迭代器的 Python Wiki 文章。
Jesse is passionate about data analysis and visualization. He uses the R statistical programming language for all aspects of his work. | 780 | 1,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.35418 |
http://mathforum.org/library/topics/basic_algebra/?keyid=38645178&start_at=751&num_to_see=50 | 1,467,408,233,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783403825.35/warc/CC-MAIN-20160624155003-00110-ip-10-164-35-72.ec2.internal.warc.gz | 195,845,951 | 12,120 | Browse and Search the Library
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http://mathforum.org/ | 3,238 | 15,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2016-26 | longest | en | 0.882777 |
http://spmath81411.blogspot.com/2012/03/serenas-surface-area-post.html | 1,490,730,216,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189884.21/warc/CC-MAIN-20170322212949-00139-ip-10-233-31-227.ec2.internal.warc.gz | 352,320,637 | 16,577 | ## Sunday, March 11, 2012
### Serena's Surface Area and Volume Post
Here is a video on how to find the surface area of a rectangular prism:
Triangular Prism: base x height divided by 2 x height.
3x8.5/2 x 5
=12.75 x 5
=63.75m3
Rectangular Prism: lxwxh
=20x8x16
=2560cm3
Triangular Prism: base x height divided by 2 x height
=8 x 4/2 x .5
16 x 6.5= 104
3/8= 0.375
0.374 x 104 =39cm3
Circular Prism: pie r squared x height
12/2= 6
=3.14 x 6 x 6 x 15
=1695.6cm3
here is a link on volume: | 198 | 491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-13 | longest | en | 0.780578 |
https://plainmath.net/4402/complete-the-sentence-0-04-is-1-10-of | 1,656,587,856,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103671290.43/warc/CC-MAIN-20220630092604-20220630122604-00297.warc.gz | 519,760,261 | 12,949 | # Complete the sentence. 0.04 is 1/10 of _____.
Complete the sentence. 0.04 is
$1/10$
of _____.
You can still ask an expert for help
• Questions are typically answered in as fast as 30 minutes
Solve your problem for the price of one coffee
• Math expert for every subject
• Pay only if we can solve it
comentezq
we have to find the number for which $1/10$ is 0.04.
so, we have to multiply:$0.04\cdot 10=0.4$
Result:
0.4 | 133 | 425 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 29, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-27 | latest | en | 0.9389 |
https://placement.freshersworld.com/sbi-question-papers/aptitude-reasoning/33131456 | 1,643,151,594,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304876.16/warc/CC-MAIN-20220125220353-20220126010353-00191.warc.gz | 507,196,962 | 61,055 | State Bank of India Aptitude-Reasoning Contributed by Sidharth updated on Jan 2022
# SBI Reasoning-English Qquestions
SBI and SBT Test of reasoning solved questions with detailed explanations,SBI previous years solved question papers for reasoning and english language computer and marketing awareness,general awareness,SBT and SBI free solved sample placement papers
SBI reasoning questions
Qn.1
How many such pairs of letters are there in the word GUARDIAN each of which has as many letters between them in the word as in the English alphabet ?
None
One
Two
Three
More than three
Qn.2 Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group ?
19
17
23
29
27
Explanation:All the rest are prime numbers.
Qn.3 How many meaningful English words can be made with the letters TEBI using each letter only once in each word ?
None
One
Explanation:B I T E
Two
Three
More than three
Qn.4 n a certain code LONG is written as 5123 and GEAR is written as 3748. How is LANE written in that code ?
5427
Explanation:L ---> 5, O ---> 1, N ---> 2, G ---> 3
G ---> 3, E ---> 7, A ---> 4, R ---> 8Therefore,L ---> 5, A ---> 4, N ---> 2, E ---> 75247
5847
5237
None of these
Qn.5 ‘BD’ is related to ‘EG’ and ‘MO’ is related to ‘PR’ in the same way as ‘FH’ is related to ……….
JM
IL
JL
IK
None of these
Qn.6 In a certain code BREAKDOWN is written as BFSCJMVNC. How is ORGANISED written in that code ?
PSHBMCDRH
BHSPMCDRH
BHSPOCDRH
BHSPNHRDC
None of these
Qn.7 How many such digits are there in the number 58674139 each of which is as far away from the beginning of the number as when the digits within the number are rearranged in descending order ?
None
One
Two
Three
More than three
Qn.8 In a certain code language ‘pik da pa’ means ‘where are you’; ‘da na ja’ means ‘you may come’ and ‘na ka sa’ means ‘he may go’, which of the following means ‘come’ in that code language ?
da
ja
na
Cannot be determined
None of these
Qn.9 Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group ?
B D F
V X Z
F I K
M O Q
L N P
Qn.10 What should come next in the following number series ?
9 8 9 8 7 98 7 6 9 8 7 6 5 9 8 7 6 5 4 9 8 7 6 53
4
Explanation:98 987 9876 98765 987654 98765 42 1
None of these
Qn.11 Which of the following is the middle digit of the second highest among the following five numbers ?
254 319 963 842 6975
1
6
4
Explanation:963, 8 4 2, 697, 319, 254
9
Qn.12 Meeta correctly remembers that her father’s birthday is after 8th July but before 12th July. Her brother correctly remembers that their father’s birthday is after 10th July but before 15th July. On which day of July was definitely their father’s birthday ?
10th
11th
10th or 11th
Cannot be determined
None of these
Qn.13 In a class of 50 students M is eighth from top. H is 20th from bottom. How many students are there between M and H ?
22
23
24
Cannot be determined
None of these
Qn.14 Among A, B, C, D and F each scoring different marks in the annual examination, D scored less than only F among them. B scored more than A and C but less than D.
Who among them scored least marks among them ?
A
C
B
None of these
Qn.15 Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group ?
Copper
Iron
Aluminium
Zinc
Steel
Explanation:All the rest are pure metals.
Qn.16Directions(Q16-20): In each question below are three statements followed by two conclusions numbered I and II. You have to take the three given statements to be true even if they seem to be at variance from commonly known facts and then decide which of the given conclusions logically follows from the three statements disregarding commonly known facts. Give answers :
(A) If only conclusion I follows.(B) If only conclusion II follows.(C) If either conclusion I or II follows.(D) If neither conclusion I nor II follows.(E) If both conclusions I and II follow. Statements: This world is neither good nor evil; each man manufactures a world for himself.Conclusions:I. Some people find this world quite good.II. Some people find this world quite bad
.
A
B
C
D
E
Explanation:The statement mentions that the world for a man is as he makes it himself. So, some people might find it good and some quite bad. Thus, both I and II follow.
Qn.17
Statements: The eligibility for admission to the course is minimum second class Master's degree. However, the candidates who have appeared for the final year examination of Master's degree can also apply.
Conclusions:I. All candidates who have yet to get their Master's degree will be there in the list of selected candidates.
II. All candidates having obtained second class Master's degree will be there in the list of selected candidates
.A
B
C
D
Explanation:The statement mentions that the candidates who have obtained second class Master's degree or have appeared for the final year examination of Master's degree, can apply for admission. This implies that both types of candidates may be selected on certain grounds. Thus, some candidates of each type and not all candidates of any one type, may be selected. So, neither I nor II follows.E
Qn.18 Statements: Any student who does not behave properly while in the school brings bad name to himself and also for the school.
Conclusions: I. Such student should be removed from the school
.II. Stricter discipline does not improve behaviour of the students.
A
B
C
D
Explanation:Clearly, I cannot be deduced from the statement. Also, nothing about discipline is mentioned in the statement. So, neither I nor II follows.
E
Qn.19Statements: A Corporate General Manager asked four managers to either submit their resignations by the next day or face termination orders from service. Three of them had submitted their resignations by that evening.
Conclusions:I. The next day, the remaining manager would also resign.
II. The General Manager would terminate his services the next day.A
B
C
D
E
Qn.20Statements: Only good singers are invited in the conference. No one without sweet voice is a good singer.
Conclusions:I. All invited singers in the conference have sweet voice.
II. Those singers who do not have sweet voice are not invited in the conference.A
B
C
D
E
Explanation:The statement asserts that a good singer always has a sweet voice and only good singers are invited in the conference. This implies that all those invited in the conference have sweet voice and those who do not have sweet voice are not invited. So, both I and II follow.
Qn.21 Directions(Q21-25): Study the following arrangement carefully and answer the questions given below-
B # A R 5 8 E % M F 4 J 1 U @ H 2 © 9 T I 6 * W 3 P # K 7 \$ Y
Which of the following is the twelfth to the left of the twentieth from the left end of the above arrangement ?
%
Explanation:20th from the left end is T and 12th to the left of T is %
W
\$
J
None of these
Qn.22 How many such numbers are there in the above arrangement each of which is immediately preceded by a consonant and also immediately followed by a symbol ?
None
One
Two
Explanation:H 2 © and K 7 \$
Three
More than three
Qn.23 How many such symbols are there in the above arrangement each of which is immediately preceded by a letter and also immediately followed by a number ?
None
One
Two
Three
More than three
Qn.24 How many such consonants are there in the above arrangement each of which is immediately preceded by a consonant and also immediately followed by a number ?
None
One
Explanation:M F 4
Two
Three
More than three
Qn.25 If all the numbers in the above arrangement are dropped, which of the following will be the eleventh from the right end ?
U
T
F
H
Explanation:After dropping all the numbers.
B # A R E % M F J U @ H © T I * W P # K \$ Y11th from the right end is H.None of these
Qn.26 Directions(26-30): In each question below is given a group of digits/symbols followed by four combinations of letters lettered (A),(B),(C) and (D). You have to find out which of the combinations correctly represents the group of digits/symbols based on the following letter coding system and mark the letter of that combination as the answer. If none of the letter combinations correctly represents the group of digits/ symbols, mark (E) i.e. ”˜None of these’ as the answer.
Digit/Symbol :4 % 3 9 \$ 1 8 @ © 2 # 5 6 * 7 d
Letter Code :P M I T R Q J F H A E U N B G L
Conditions :(i) If the first element in the group is a symbol and the last element is a digit, the codes are to be interchanged.(ii) If the first element in the group is a digit and the last element is a symbol both are to be coded as the code for the digit.(iii) If both the first and the last elements are even digits both are to be coded as ”˜X’.(iv) If both the first and the last elements are odd digits, both are to be coded as ”˜Y’. 4%@93* PMFTIB
PMFTIP
BMFTIB
XMFTIX
None of these
Qn.27
RQJTNH
HQJTNR
RQJTNR
YQJTNY
None of these
Qn.28
2*#836
YBEJIY
ABEJIN
NBEJIA
XBEJIX
None of these
Qn.29
8732@9
TGIAFJ
YGIAFY
JGIAFT
XGIAFX
None of these
Qn.30
7#\$%35
GERMIU
UERMIG
GERMIG
XERMIX
None of these
Qn.31
Directions(31-35): In the following questions, the symbols @, ©, %, \$ and d are used with the following meanings illustrated.
'P % Q' means 'P is greater than Q'.'
P d Q' means 'P is neither greater than nor smaller than Q'.
'P @ Q' means 'P is smaller than Q'.
'P © Q' means 'P is either smaller than or equal to Q'.
'P \$ Q' means 'P is either greater than or equal to Q'.
In each of the following questions assuming the given statements to be true, find out which of the two conclusions I and II given below them is/are definitely true.
Give answers :(A) If only conclusion I is true.
(B) If only conclusion II is true.
(C) If either conclusion I or conclusion II is true.
(D) If neither conclusion I nor conclusion II is true
(E) If both conclusions I and II are true.
Statements : M @ J, J © R, R d KConclusions :I. K d JII. K % J
A
B
C
D
E
Qn.32 Statements : N \$ T, T d H, N @ W
Conclusions :I. W % TII. H © NA
B
C
D
E
Qn.33 Statements : F @ R, R © V, V \$ T
Conclusions :I. V % FII. F @ TA
B
C
D
E
Qn.34 Statements : W © D, D \$ B, B @ H
Conclusions :I. H % DII. W @ BA
B
C
D
E
Qn.35 Statements : F d T, T \$ M, M © R
Conclusions :I. R \$ FII. M © FA
B
C
D
E
Qn.36Directions(36-40): Study the following information and answer the questions given below-
M, N, P, R, T, W, F and H are sitting around a circle facing at the centre.
P is third to the left of M and second to the right of T. N is second to the right of P.
R is second to the right of W who is second to the right of M. F is not an immediate neighbour of P.
Who is to the immediate right of P ?
H
F
R
None of these
Qn.37 Who is to the immediate right of H ?
R
F
M
None of these
Qn.38 Who is to the immediate left of R ?
P
H
W
T
Qn.39 Who is third to the right of H ?
T
W
R
F
Qn.40 Who is second to the right of F ?
M
R
T
None of these
**********************************************************************************
Sbi Qn.51 English Language Directions(51-60):
Read the following passage carefully and answer the questions given below it. Certain words are printed in bold to help you to locate them while answering some of the questions.
The yearly festival was close at hand. The store room was packed with silk fabrics. gold ornaments, clay bowls full of sweet curd and platefuls of sweetmeats. The orders had been placed with shops well in advance. The mother was sending out gifts to everyone. The eldest son, a government servant, lived with his wife and children in far off lands. The second son had left home at an early age. As a merchant he travelled all over the world. The other sons had split up over petty squabbles, and they now lived in homes of their own. The relatives were spread all across the world. They rarely visited. The youngest son, left in the company of a servant, was soon bored and stood at the door all day long, waiting and watching. His mother, thrilled and excited, loaded the presents on trays and plates, covered them with colourful kerchiefs, and sent them off with maids and servants. The neighbours looked on.
The day came to an end. All the presents had been sent off.
The child came back into the house and dejectedly said to his mother, 'Maa, you gave a present to everyone, but you didn't give me anything !'
His mother laughed, 'I have given all the gifts away to everyone, now see what's left for you.' She kissed him on the forehead.
The child said in a tearful voice, 'Don't I get a gift ?'
'You'll get it when you go far away.'
'But when I am close to you, don't I get something from your own hands ?'
His mother reached out her arms and drew him to her. 'This is all I have in my own hands. It is the most precious of all.'
Why did the woman's second son travel ?
He was restless by nature
He did not want to stay at home
He was rich and could afford to travel
His job was such that he had to travel
None of these
Qn.52 Why did the woman's eldest son not attend the festival ?
He was not on good terms with his youngest brother who lived at home
He had quarrelled with his mother
His wife did not allow him to return home
His job prevented him from taking leave
None of these
Qn.53 How did the woman prepare for the festival ?
1. She bought expensive gifts for her children and neighbours.
2. She ordered her servants to prepare sweets and food well in advance.
3. She made sure that her youngest child was looked after so that he wouldn't be bored.
None
Only 1
Only 2
Both 1 and 2
All 1, 2 and 3
Qn.54 What did the youngest child do while his mother was busy ?
1. He waited for a chance to steal some sweetmeats.
2. He pestered his mother to give him a present.
3. He stood at the door with servants.
Only 1
Only 2
Both 1 and 3
Only 3
None of these
Qn.55 Which of the following can be said about the woman ?
She was a widow who had brought up her children single handedly
She was not a good mother since her children had left home at an early age
She enjoyed sending her family gifts at festival time
She gave expensive presents to show that she was wealthy
She rarely visited her grand-children because they all lived abroad
Qn.56 What did the boy receive from his mother ?
She taught him the value of patience
She encouraged him to grow up and live independently like his brothers
She showed him the importance of giving expensive gifts
She gave him a hug to express her love
None of these
Qn.57 Which of the following is TRUE in the context of the passage ?
The woman usually ignored her youngest son
The woman's eldest son lived abroad
The members of the woman's family did not care about her
The woman made all the preparations herself since she did not want to burden the servants
The woman sent gifts to her children to ensure that they visited her
Qn.58Directions(Q58-59): Choose the word which is most nearly the SAME in meaning as the word printed in bold as used in the passage.
Left
Gone
Quit
Remaining
Disappeared
Forgot
Qn.59
Packed
Filled
Squeezed
Crowd
Collected
Untidy
Qn.60
Choose the word which is most OPPOSITE in meaning of the word dejectedly as used in the passage.
Calmly
Happily
Willingly
Fortunately
Softly
Qn.61
Directions(61-65):
Read each sentence to find out whether there is any error in it. The error, if any, will be in one part of the sentence. The letter of that part is the answer. If there is no error, the answer is (E). (Ignore errors of punctuation, if any)
Many multinational companies (A) / have not been as (B) /successful in India (C) /than we expected. (D) No error (E)
A
B
C
D
Explanation:Replace ‘than’ with ‘as’.
E
Qn.62
He has ruined (A) /his eyesight (B) /by not using (C) /his spectacles regularly. (D) No error (E)
A
B
C
D
E
Qn.63
Mostly of the (A) /newly recruited officers (B) /have no experience (C) /in the banking sector. (D) No error (E)
A
Explanation:Change ‘Mostly’ to ‘Most’.
B
C
D
E
Qn.64 The resignation of (A) /one of our directors (B) /have caused the price (C) / of shares to fall. (D) No error (E)
A
B
C
Explanation:Change ‘have’ to ‘has’.
D
E
Qn.65
There are many (A) /ways of which (B) /inflation can (C) /be measured. (D) No error (E)
A
B
Explanation:Replace ‘of’ with ‘in’
C
D
E
Qn.66 Directions(Q66-70): Which of the phrases (A),(B),(C) and (D) given below should replace the phrase given in bold in the following sentence to make the sentence grammatically meaningful and correct. If the sentence is correct as it is and 'No correction is required.' mark (E) as the answer.
Each of the loan must be approved by the Branch Manager-
Every loan
Each one of the loan
Any of the loan
All of the loan
No correction required
Qn.67
The issue was taken before the Municipal Corporation meeting last week-
Taking place at
Taken after
Being taken in
Taken up at
No correction required
Qn.68
He has asked for the names of those employees involved in the project.
no correction required
Qn.69
Considerate the traffic, it is better to leave for the airport an hour early-
While considering
Consideration of
Considering
Being considerate to
No correction required
Qn.70
He is a good leader, knowing that to motivate his employees to achieve””
That known when
Who knows how
Which knows how
Knowing what
No correction required
Qn.71
Directions(71-75): Rearrange the following six sentences (1),(2),(3),(4),(5) and (6) in the proper sequence to form a meaningful paragraph; then answer the questions given below them.
(1) The able bodied men of the tribe gathered to discuss how to climb the mountain.
(2) As part of their plundering they kidnapped a baby of one of the families.
(3) One day the mountain tribe invaded those living in the valley.
(4) 'We couldn't climb the mountain. How could you?-, they asked, 'It wasn't your baby !' she replied.
(5) There were two tribes in the Andes' one lived in the valley and the other high up in the mountains.
(6) Two days later they noticed the child's mother coming down the mountain that they hadn't yet figured out how to climb.
Which of the following should be the SECOND sentence after rearrangement ?
1
2
3
4
5
Qn.72
Which of the following should be the FIFTH sentence after rearrangement ?
6
5
4
3
2
Qn.73
Which of the following should be the FIRST sentence after rearrangement ?
1
2
3
4
5
Qn.74
Which of the following should be the SIXTH (LAST) sentence after rearrangement ?
1
2
3
4
5
Qn.75 Which of the following should be the THIRD sentence after rearrangement ?
1
2
3
4
5
Qn.76 Directions(Q76-80): In each question below a sentence with four words printed in bold type is given.
These are lettered (A),(B),(C) and (D). One of these four words printed in bold may be either wrongly spelt or inappropriate in the context of the sentence. Find out the word, which is wrongly spelt or inappropriate, if any. The letter of that word is your answer. If all the words printed in bold are correctly spelt and also appropriate in the context of the sentence mark (E) i.e., all correct as your answer.
The income (A) of many people in rural (B) India is not adequate (C) to satisfy (D) their basic needs. All correct (E)
A
B
C
D
E
Qn.77
He is always (A) prompt (B) in caring (C) out instructions. (D) All correct (E)
A
B
C
D
E
Qn.78 The revized (A) rates (B) of interest will be effective (C) immediately. (D) All correct (E)
A
B
C
D
E
Qn.79
Such transactions (A) are quiet (B) expensive (C) and time consuming (D) for customers. All correct (E)
A
B
C
D
E
Q
n.80
The guidelines (A) of the new scheme (B) are expected (C) to be finally (D) soon. All correct (E)
A
B
C
D
E
Qn.81 Directions(81-90): In the following passage there are blanks each of which has been numbered. These numbers are printed below the passage and against each five words/ phrases are suggested one of which fits the blank appropriately. Find out the appropriate word in each case.
I used to look …(81)… to the holidays. I was usually …(82)… to my uncle’s house where I …(83)… his children. I did not get paid a salary for …(84)… What I received in return however, was far more …(85)… My uncle was an avid reader. During the time I spent with his family I had an …(86)… to read the vast amount of books and magazines that he possessed. This improved my English to some …(87)… Reading became my new …(88–89)… spending my pocket money on a ticket to the cinema I began to …(90)… books. This has benefited me greatly.
forward
towards
backward
up
around
Qn.82
I was usually …(82)… to my uncle’s house
went
sent
visited
travelled
gone
Qn.83
where I …(83)… his children.
cared
occupy
guarded
taught
played
Qn.84
I did not get paid a salary for …(84)….
them
whom
this
now
which
Qn.85
What I received in return however, was far more …(85)….
expensive
deserving
helping
demanding
valuable
Qn.86
During the time I spent with his family I had an …(86)….......
opportunity
ability
use
encouragement
achievement
Qn.87
This improved my English to some …(87)….
distance
extent
time
limits
degrees
Qn.88
activity
hope
hobby
duty
worship
Qn.89
despite
though
by
while
Qn.90
I began to …(90)… books. This has benefited me greatly.
sell
exchange
invest
Qn.91
Directions(Q91-100): In the following passage there are blanks each of which has been numbered. These numbers are printed below the passage and against each five words/ phrases are suggested one of which fits the blank appropriately. Find out the appropriate word in each case.
Each year Middle class Indian children ...(91)... hundred of crores of rupees in pocket money and ...(92)... a heavy burden parental ...(93)... like adults, these kids have ...(94)... connected with budgeting and saving money. unfortunately, basic money ...(95)... is ...(96)... taught in schools. At home, very few parents ...(97)... money matters with their children. Kids who ...(98)... about money ...(99)... have been found to be way ahead of their peers. Indeed, learning to ...(100)... with money properly fosters discipline, good work habits and self respect.
spend
steal
save
give
invest
Qn.92
...(92)... a heavy burden parental
move
take
risk
put
lift
Qn.93
a heavy burden parental ...(93)... like adults,
promises
payments
demands
attitudes
incomes
Qn.94
these kids have ...(94)... connected with budgeting and saving money.
expenses
experience
problems
guidance
necessities
Qn.95
basic money ...(95)... is ...(96)... taught in schools.
availability
inflation
economics
problem
management
Qn.96
unfortunately, basic money ...(95)... is ...(96)... taught in schools.
carefully
rarely
generally
always
thoroughly
Qn.97
very few parents ...(97)... money matters with their children.
discuss
understand
teach
reveal
Qn.98
quarrel
learn
waste
spend
Qn.99
...(99)... have been found to be way ahead of their peers.
slowly
early
timely
lately
regularly
Qn.100
Indeed, learning to ...(100)... with money properly fosters discipline, good work habits and self respect.
decide
earn
control
deal
pay
feedback | 6,288 | 23,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-05 | latest | en | 0.919402 |
http://www.gamespot.com/forums/pc-mac-discussion-1000004/basic-circuit-theory-help-29323687/ | 1,429,461,822,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246639191.8/warc/CC-MAIN-20150417045719-00137-ip-10-235-10-82.ec2.internal.warc.gz | 521,547,888 | 22,694 | # Basic circuit theory help
This topic is locked from further discussion.
#1 Posted by achilles614 (4872 posts) -
I was given an equation to draw a circuit "X Y Z (X + Y + Z) + ~X ~Y ~Z (X + Y + Z) + (~X + ~Y + ~Z) ~Z + X Z (Y + ~Z)=F" My initial design used 22 gates but I was told you could get creative and reduce the amount of gates. After enough tinkering I got it down to 2 NAND gates. The resulting equation from the 2 gate design is "~(Z ~(X Y))=F" The output from each circuit (results in truth table) is the same for both designs even though the equation and amount of gates is different. My question is, will both these designs essentially function the same? Or did I reduce it too much and change the core of its function?
#2 Posted by sleepingzzz (2263 posts) -
LMAO.
This is a gaming forum not electrical engineering.
#3 Posted by jakes456 (1398 posts) -
looking for someone to do your homework. talk about pathetic.
#4 Posted by MW2ismygame (2183 posts) -
looking for someone to do your homework. talk about pathetic.
jakes456
Do you get paid to be a smart ass in every thread ?
#5 Posted by achilles614 (4872 posts) -
looking for someone to do your homework. talk about pathetic.
jakes456
Aw it's my favorite troll. If you would read you would see that I already made the design :( I'm asking about some basic theory.
#6 Posted by MW2ismygame (2183 posts) -
I was given an equation to draw a circuit "X Y Z (X + Y + Z) + ~X ~Y ~Z (X + Y + Z) + (~X + ~Y + ~Z) ~Z + X Z (Y + ~Z)=F" My initial design used 22 gates but I was told you could get creative and reduce the amount of gates. After enough tinkering I got it down to 2 NAND gates. The resulting equation from the 2 gate design is "~(Z ~(X Y))=F" The output from each circuit (results in truth table) is the same for both designs even though the equation and amount of gates is different. My question is, will both these designs essentially function the same? Or did I reduce it too much and change the core of its function? achilles614
I would help if i could, but -
eventually someone here will know im (along with many others) am not that person.
#7 Posted by wis3boi (31906 posts) -
ABC makes 123
#8 Posted by blaznwiipspman1 (6090 posts) -
[QUOTE="achilles614"]I was given an equation to draw a circuit "X Y Z (X + Y + Z) + ~X ~Y ~Z (X + Y + Z) + (~X + ~Y + ~Z) ~Z + X Z (Y + ~Z)=F" My initial design used 22 gates but I was told you could get creative and reduce the amount of gates. After enough tinkering I got it down to 2 NAND gates. The resulting equation from the 2 gate design is "~(Z ~(X Y))=F" The output from each circuit (results in truth table) is the same for both designs even though the equation and amount of gates is different. My question is, will both these designs essentially function the same? Or did I reduce it too much and change the core of its function? MW2ismygame
I would help if i could, but -
eventually someone here will know im (along with many others) am not that person.
+1
ahh gotta love forest gump
#9 Posted by Guppy507 (17398 posts) -
oh god flashbacks to the ECE class I had to take for CS.
#10 Posted by way2funny (4570 posts) -
Make a KMap for both and try it out. Or you can make a truth table for both of those equations and you can easily see if they are equivalent
Edit: I finished reading your post. Yes, you can make designs as complicated or simple as possible and they can be completely equivalent. It all depends on the input / output, everything inbetween doesnt matter. Meaning, if the truth tables match up, then they are equivalent. Look up KMaps, Kmaps allow you to take complex equations like the first one, and simplify them (reduce the number of logic gates) yet still have the same inputs / outputs
#11 Posted by GummiRaccoon (13647 posts) -
Just remember resistance is futile
#12 Posted by spittis (1875 posts) - | 1,026 | 3,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2015-18 | latest | en | 0.964789 |
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### Is the fact that ZFC implies that 1+1=2 an absolute truth?
This question is somehow of a follow up to to this other one, and it's something that has bugged me for a while. I understand the notion that there's no "absolute truth" in math, in the ...
• 267
1 vote
79 views
### If axioms are subjective, how could anything be objective?
Axioms are subjective (?), and, since propositions are based on axioms, isn't everything subjective? (of course, the answer should be from the perspective of someone who believes in objectivity) Or am ...
• 163
121 views
### Does Tarski Indefinability theorem impose a computational lower bound on the axiomatization of the reality?
Based on the Tarski's Indefinability Theorem (TA in standard model is not arithmetic (no FOL formula can represent TA, a formula represent a predicate relation definition: under Tarski's first order ...
67 views
### Is the axiomatic method an inherently well-founded method?
It occurred to me a little while ago, that there is a trichotomy in set theory that maps to the positive solutions to the problem of the regress of inferential reasons. Namely, well-founded sets map ...
• 3,614
67 views
### The Anthropic principle as a physical embodiment of the Axiom of Choice
This question pertains to physical reality, but it would be deemed inappropriate in Physics.SE because of its speculative, metaphysical nature. It also touches on the subject of singling out elements ...
• 109
1 vote
183 views
### Is there a proof that we can't prove a physical theory?
I am thinking of physical theories (e.g. Newtonian Mechanics) as axiomatic systems. We have a list of axioms and from there we can derive theorems, make predictions etc. If the prediction don't agree ...
• 169
80 views
### What does it mean for one geometrical axiom to be considered _equivalent_ to another geometrical axiom?
What does it mean for one geometrical axiom to be considered equivalent to another geometrical axiom? For example consider Playfair`s axiom: In a plane, given a line and a point not on it, at most one ...
143 views
### Are these valid examples of axiomatic statements?
I'm trying to understand if a couple of statements would be considered axiomatic: Example 1: "murder is the unjustified killing of a person; if there was a murder, then a person was killed ...
• 361
58 views
### Axiom of Choice: correspondence or derivability?
I'd like to ask about a specific impression that I have about issues concerning the Axiom of Choice. It seems to me that either one claims that the axiom is an obvious fact about the modelled concept (...
101 views
### In what way, if any, can the axioms of set theory be themselves well-ordered?
In "Independence and Large Cardinals", Peter Koellner writes: ... it turns out that when one restricts [attention] to those theories that "arise in nature" the interpretability ...
• 3,614
3k views
### Are pursuing the well-being and reducing the suffering of sentient beings objectively good things?
I think most people intuitively agree that increasing their own well-being and minimizing their own suffering are the right things to do. Everyone wants to be happy, enjoy a good health, etc. The ...
1 vote
101 views
### How to construct logic from nothing as a step by step procedure? [closed]
I am thinking of constructing logic from scratch. I tried the law of thought as a fresh start but not totally convinced this is a correct way for the following reason: if law of identity is the first, ...
1 vote
142 views
### Do contemporary logicians generally claim that classical logic can be simply reduced to these 5 logic principles?
Do contemporary logicians generally claim, as Wikipedia does, that classical logic can be simply reduced to the 5 logical principles below? Or is it more complex than that and are there principles not ...
• 115
1 vote
135 views
### What is the current status of Foundation-of-Mathematics programmes?
I have been reading 'A Very Short Introduction to Mathematics' by Timothy Gowers and at one point he mentions that most of the mathematical proofs can be finally resolved to a set of logical ...
• 39
318 views
### Is faith required to believe any axiomatic assumption the scientific method is built upon?
It's my understanding that the scientific method builds upon certain axiomatic assumptions, such as uniformitarianism and the principle of induction. Is faith required to believe these axiomatic ...
210 views
### Is it possible to create an axiomatic system where 1+1 doesn't equal 2? What would be the consequences of such a system? [closed]
1+1=2 is a result (perhaps arguably more of a definition than a theorem?) of Peano Arithmetic, as well as other systems such as ZFC. I understand that 1+1 doesn't necessarily have to equal 2 if we ...
114 views
### Why do we rely upon scientific approach when its foundational axioms are assumed and agreed without proof?
Why do we rely upon scientific approach when its foundational axioms are assumed and agreed without proof? Foundation of the scientific explorations are seem to be the mathematical axioms at its root....
202 views
### Do philosophers generally reject that philosophical reasoning relies on axioms?
The way I've always thought that philosophy worked is that philosophers have a certain set of tools (deduction, laws of thought, basic sources of knowledge) which they use to come to reasoned answers ...
120 views
### In a deductive reasoning system, what happens if we have unfounded axioms? [closed]
What if our axioms are false? What happens then?
434 views
### What are the most rational basic beliefs?
I understand that this question might be difficult or even unresolved. But within a foundationalist view of knowledge, has anyone proposed a set of basic beliefs that seem to be the most rational for ...
1 vote
337 views
### Do complex quantities and irrational numbers exist in nature?
Completeness Theorems of Model Theory, a branch of Mathematical Logic. Together, these two Theorems show that: under the Field Axioms (the rules of the game for scalars) existence of rationals is ...
132 views
### Is there an infinity of axioms in mathematics?
As I was trying to find a list of mathematical axioms used in modern branches of mathematics, I wondered if there's any meaning to the question of "how many mathematical axioms are there ?", and then ...
• 265
35 views
### What are the axioms of personalism?
I've read some authors whose works are held as belonging to personalism. However while their works elaborated based on principles of personalism, they never clearly explained what the axioms were ...
• 129
424 views
### How do philosophers formally characterise mathematical objects?
In the Stanford Encyclopedia of Philosophy article, 'Platonism in the Philosophy of Mathematics', the following formalisation is given for the existence of a mathematical object: Existence can be ...
• 71
1k views
### Does reality have axioms?
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### What kinds of proofs can be given for axioms, e.g. the modal axiom S5?
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### How do we call a statement that is unthinkable for any person to not be the case? [duplicate]
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• 536 | 2,513 | 11,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-33 | longest | en | 0.938064 |
https://math.stackexchange.com/questions/601135/counting-distinct-palindromic-substrings | 1,713,704,188,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00882.warc.gz | 348,357,850 | 36,397 | # Counting Distinct Palindromic Substrings
Given a string $S$, I want to count the number of distinct palindromic substrings of $S$. I know the basic $o(n^2)$ approach to do so. But I want to find a better approach for strings that can be very large (of the order of $10^5$).
So I want a more efficient algorithm.
Example:
Say $S=xyx$, then the palindromic counter must return $3$ as answer, as S has three palindromic substrings: $\{x,xyx,y\}$.
You can get a table of the largest palindrome centered at each position using Manacher's algorithm in $O(n)$ time. Then create a set of palindromes (e.g. using a min heap), initially empty. Iterate over the table, adding the longest palindrome centered at each position to the set. When adding a palindrome to the set, if successful, remove the first and last characters and repeat until the string is empty or a previously-added palindrome is found.
This runs in time $O(n \log p)$, where $p$ is the palindrome count.
• Could you please provide an algo or a code or so?I will be very thankful Dec 15, 2013 at 19:07
• Here is the algorithm. 1. Create table using Manacher's algorithm 2. For each entry in the table, add the corresponding palindrome to the set, using the method specified. Dec 16, 2013 at 6:37
Every 1-character substring is a palindrome -- as the OP already pointed out in the example. Also, every repeated character is a length-2 palindrome, e.g. $xx$ in $axxb$.
Use the set of 1-character and 2-character palindromes as your initial set. Then, for each palindrome in the set, try to extend it by 1 character left and 1 character right and check if that is still a palindrome. This happens if the characters to the left and right of an existing palindrome are equal. If they are equal, add the new palindrome to the set and try to add 2 more characters. If the step did not form a new palindrome, stop and try the next element in the existing set.
In the worst case this is still $o(n^2)$, which would happen, for example, if the string is a single character repeated $n$ times. In the best case this is $o(n)$ since palindromes are quite rare and most attempts at extending from a 1 or 2-character palindrome will end after a constant number of steps. The average case performance will depend a lot on the expected density of palindromes in the string.
• R u sure the palindromic count will not contain duplicates.? I mean one substring will be counted one time.Could you please provide me a piece of code to do so? Dec 10, 2013 at 9:49
• Yes, the palindrome count will contain duplicates. But determining whether you think $xyxyx$ contains the palindrome $xyx$ once or twice is up to you. If you don't want to count things twice, you will have to remove duplicates afterwards.
– Carl
Dec 10, 2013 at 9:54
• How to maintain this that i had encountered this string earlier or not?Wont it make the algorithm O(n^2) again.? Dec 10, 2013 at 9:57 | 741 | 2,917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-18 | latest | en | 0.871934 |
http://forums.freshershome.com/threads/237-Number-puzzle-Can-you-get-24-from-these-numbers?p=597 | 1,591,060,831,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347422065.56/warc/CC-MAIN-20200602002343-20200602032343-00027.warc.gz | 49,260,164 | 9,563 | # Thread: Number puzzle: Can you get 24 from these numbers?
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## Number puzzle: Can you get 24 from these numbers?
2, 7, 7, 10Combine these four numbers in any order using only parentheses and +, -, x, and / to get the numbers 20-25. The real challenge is getting 24.For example, 19 = 10*2 - (7/7).Have fun.Yep, there is an answer. And can you only use those 4 operations. You can't use exponents and factorials, so [(10/2)-(7/7)]! = 24 is not valid.Hint: The first step is to divide.
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This seems confusing.
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what kind of math is this
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The solution is; 7+7+10+2x0=24. ok.
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ohh ohh i got 23!10-7=33*7=2121+2=23!!!!!!!! yay me!!!
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The Answer Is No.
7. Entry Level Fresher
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Can't get itemail me the answer [email protected]
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i am not having fun, can you please email me the answer. pretty please with cherries on top?? is there really an answer?
9. Entry Level Fresher
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Hey anuja check your math. Your equation would make the answer 0. 0 isn't an option anyway. (10-7)x7+2 gets you 23 though
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I got twenty sixlet me do the math again....yep 26
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24 = 10/2+7x2 is if i used two 2s7/7+10x2 = 22
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geez, this seems confusing. number twist!
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• | 613 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2020-24 | longest | en | 0.869674 |
https://www.thestalkingdirectory.co.uk/showthread.php/19650-Bullet-weight | 1,513,158,373,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522343.41/warc/CC-MAIN-20171213084839-20171213104839-00646.warc.gz | 816,587,535 | 17,139 | 1. ## Bullet weight
I have a question on bullet weight. I always hear people (well, read I guess) saying you should use a heavier bullet when hunting bigger animals. Or say something like "you need at least 100gr for a red deer" or "don't shoot a buffalo with less than 350gr". By the way these are examples I'm not interested in whether they're true or not. I have an engineering background so should be able to work this out from a mathematical point of view but why is this?
I understand about momement, (mv) and energy (1/2mv^2). But for these, velocity is at least as important as bullet mass. So that can't be it.
I understand about penetration and sectional density. The more 'pressure' the bullet imparts the more it will penetrate for a given shape and velocity. But SD is not the same as weight - only in the same calibre. I 160gr 6.5mm bullet should penetrate better than a 160 gr 30 cal bullet at the same speed. So that can't it.
So what is it - is there some killing component I don't know about? Surely a 95gr 243 will kill at least as well as a 100gr 308?
2. Or say something like "you need at least 100gr for a red dear"
Actually, in Scotland, THE LAW says it!
As far as the rest it isn't bullet weight or velocity on its own. Otherwise a solid bronze bullet with no cavity or expansion intitiating tip plug would kill just as well as an expanding bullet.
So it is about trauma. And how you cause that trauma. Which is not a thread I'm going to get involved in as it could last forever!
3. I wouldnt entertain shooting a sika with a 95gr .243
4. "By the way these are examples I'm not interested in whether they're true or not"
What part of this statement made by the poster don't you understand??
Harry: I think it is hard to make cross caliber comparisons. Sectional density should be really discussed in like caliber bullets. Think of it. The 243. 95 grain bullet launched at the same speed as 30 caliber bullet of the same mass is already at a possible disadvantage since the entrance hole with the 30 is larger to begin with. Technically, in like constructed bullets, the 6mm bullet would penetrate deeper, but the 30 caliber is carving out more meat on the way in. It's hard to make a comparison.
Experience sets the tare on the balance of judgement when it comes to choosing sectional density over weight, or visa-versa. Bell used sectional density combined with pin point accuracy and dropped elephants on the spot. Selous used 4 ounce lead balls fired at point blank range and would often hit them as many as 7 times and still lose the beast.
5. Harry,
Your going to upset some of the armchair hunting experts in here now. Stick to the .308 for deer or your gonna get moaned at.
I agree with every word you say, I do not believe that bullet calibre has as much to do with good clean dispatch of animals as bullet placement. A 50cal bullet that hits a red deer in the ass is going to maim the animal badly but not kill it outright, yet a 20cal bullet hitting a red deer in the heart will drop it on a sixpence.
Too many posts on here are repeating the same old topic on calibres and not enough on accuracy and bullet placement.
The law has to make rules to stop the numpties from trying to kill deer with air rifles which is understandable, but some of the calibre rules are just lunacy. I use a U.S website called the long range hunting forum and I regularly see guys taking huge Elk with nothing more than a .243 at 100yds with good shot placement.
Check this post out >>> http://montanaelkhunting.blogspot.co...mm-08-243.html
In addition I just checked out Wikipedia and looked up a few different calibres.
http://en.wikipedia.org/wiki/.243_Winchester
http://en.wikipedia.org/wiki/7mm-08_Remington
http://en.wikipedia.org/wiki/.308_Winchester
http://en.wikipedia.org/wiki/.30-06_Springfield
http://en.wikipedia.org/wiki/.300_Winchester_Magnum
As you can see from the above links as the calibres and bullet weights get larger so does the ft/lb of energy imparted onto the target.
However let me give you food for thought.
If you use a 7mm08 with 150 soft nose bullet travelling at 2800fps would it be more effective than a .308 or 30.06 travelling at 2800fps ???
I suggest it will make very little difference, what will make a massive difference though is where you hit the target.
This is what we need to study more >>> http://www.grandviewoutdoors.com/ass..._placement.jpg
What will affect the effectiveness of the shot is whether you hit the heart and lungs directly or whether you go high and hit the shoulder blade or go left/right and hit the gut. Its quite popular in the USA to go for spine shots on deer which again drops them on a sixpence if you get it right. I personally like to head/neck shoot small deer with varmint bullets under 200yds as I can drop them on the spot. An 87 grain .243 V-Max hitting a muntjac in the head will drop it quicker and more effectively than a 300win mag in the gut.
I recently read an article about an eskimo that shot a polar bear with a .22lr from close range and he dropped it on the spot with a single shot. I have also read a different article about a guy that shot a whitetail deer with a .338 lap mag and had to follow to blood trail to find and dispatch the creature after a poor shot. The more I read the more I believe different rifle calibres have little to do with the clean dispatch of animals at normal hunting ranges. Shot placement is everything.
6. Originally Posted by harrygrey382
Or say something like "you need at least 100gr for a red dear"
I wouldnt shoot my old "dear" with anything less than 130gr...
that said I have never used anything more or less than 130gr for roe, red hind or stags
7. Originally Posted by bewsher500
I wouldnt shoot my old "dear" with anything less than 130gr...
that said I have never used anything more or less than 130gr for roe, red hind or stags
Why Not???
What experiences have you had shooting deer with lighter bullets ???
8. Originally Posted by Muir
"By the way these are examples I'm not interested in whether they're true or not"
What part of this statement made by the poster don't you understand??
Harry: I think it is hard to make cross caliber comparisons. Sectional density should be really discussed in like caliber bullets. Think of it. The 243. 95 grain bullet launched at the same speed as 30 caliber bullet of the same mass is already at a possible disadvantage since the entrance hole with the 30 is larger to begin with. Technically, in like constructed bullets, the 6mm bullet would penetrate deeper, but the 30 caliber is carving out more meat on the way in. It's hard to make a comparison.
Experience sets the tare on the balance of judgement when it comes to choosing sectional density over weight, or visa-versa. Bell used sectional density combined with pin point accuracy and dropped elephants on the spot. Selous used 4 ounce lead balls fired at point blank range and would often hit them as many as 7 times and still lose the beast.
yep, thanks Muir some people need to read the whole post before replying... I'm satisfied that I was thinking straight. There's no point in talking purely about bullet mass, you have to factor in construction, velocity and sectional density too. But sometimes people don't
Originally Posted by robbobsam
Harry,
Your going to upset some of the armchair hunting experts in here now. Stick to the .308 for deer or your gonna get moaned at.
I agree with every word you say, I do not believe that bullet calibre has as much to do with good clean dispatch of animals as bullet placement. A 50cal bullet that hits a red deer in the ass is going to maim the animal badly but not kill it outright, yet a 20cal bullet hitting a red deer in the heart will drop it on a sixpence.
Too many posts on here are repeating the same old topic on calibres and not enough on accuracy and bullet placement.
The law has to make rules to stop the numpties from trying to kill deer with air rifles which is understandable, but some of the calibre rules are just lunacy. I use a U.S website called the long range hunting forum and I regularly see guys taking huge Elk with nothing more than a .243 at 100yds with good shot placement.
Check this post out >>> http://montanaelkhunting.blogspot.co...mm-08-243.html
In addition I just checked out Wikipedia and looked up a few different calibres.
http://en.wikipedia.org/wiki/.243_Winchester
http://en.wikipedia.org/wiki/7mm-08_Remington
http://en.wikipedia.org/wiki/.308_Winchester
http://en.wikipedia.org/wiki/.30-06_Springfield
http://en.wikipedia.org/wiki/.300_Winchester_Magnum
As you can see from the above links as the calibres and bullet weights get larger so does the ft/lb of energy imparted onto the target.
However let me give you food for thought.
If you use a 7mm08 with 150 soft nose bullet travelling at 2800fps would it be more effective than a .308 or 30.06 travelling at 2800fps ???
I suggest it will make very little difference, what will make a massive difference though is where you hit the target.
This is what we need to study more >>> http://www.grandviewoutdoors.com/ass..._placement.jpg
What will affect the effectiveness of the shot is whether you hit the heart and lungs directly or whether you go high and hit the shoulder blade or go left/right and hit the gut. Its quite popular in the USA to go for spine shots on deer which again drops them on a sixpence if you get it right. I personally like to head/neck shoot small deer with varmint bullets under 200yds as I can drop them on the spot. An 87 grain .243 V-Max hitting a muntjac in the head will drop it quicker and more effectively than a 300win mag in the gut.
I recently read an article about an eskimo that shot a polar bear with a .22lr from close range and he dropped it on the spot with a single shot. I have also read a different article about a guy that shot a whitetail deer with a .338 lap mag and had to follow to blood trail to find and dispatch the creature after a poor shot. The more I read the more I believe different rifle calibres have little to do with the clean dispatch of animals at normal hunting ranges. Shot placement is everything.
Make mine a 30-06 though ay!
Yep totally agree Rob. Shot placement will, within reason, be the most important actor. It's just it pisses me off when people only take about bullet weight and aren't thinking about anything else. As I think we've decided a heavy on it's own has absolutely no advantage over a light one - when only considering this value.
Originally Posted by bewsher500
I wouldnt shoot my old "dear" with anything less than 130gr...
that said I have never used anything more or less than 130gr for roe, red hind or stags
Lol yep my spelling sucks... Better fix it
9. Originally Posted by robbobsam
Why Not???
What experiences have you had shooting deer with lighter bullets ???
it was a jibe at the "dear".....
never used anything lighter or heavier as I saw no point. it works at 130, if it isnt broken there was no point.
130 through the heart of a 24stone red stag still drops in very quickly
10. A poor shot, is a poor shot with any calibre!
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• | 2,655 | 11,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-51 | latest | en | 0.966011 |
https://stats.stackexchange.com/questions/160585/arima-with-xreg-vs-gls | 1,582,133,182,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144165.4/warc/CC-MAIN-20200219153707-20200219183707-00143.warc.gz | 576,347,211 | 32,750 | # ARIMA (with xreg) vs GLS
I am fitting both an arima model (with xreg variables) and a gls model to my data in R software. They both have the same ARMA structure and variables. The ARIMA model fits to the data better. Does anyone know what the difference between these two are? I have seen that the equation for an ARIMA model in R with xreg is a linear regression with ARMA errors. Is that the same as a linear regression with ARMA error correlation (as used by the GLS)?
Thanks!
EDIT: The following code was used to create the GLS and ARIMA models:
arima3a <- arima(train.all\$sv,xreg = train.all[,c(5,6)],order=c(2,0,1))
gls3 <- gls(sv~sin+cos,data=train.all,correlation = corARMA(p=2,q=1))
Note: the sin and cos variables are equivalent to the variables 5 and 6 in the train.all matrix.
• I'm a bit unclear on just how you fit a regression with ARIMA errors using gls() (I assume you are using R, right?). Are you modeling the error correlation using corARMA? That would of course be something very different than regression with ARIMA errors, or ARIMAX (which are different things). Could you please edit your question to include your actual code, edited for brevity? – Stephan Kolassa Jul 9 '15 at 6:32
• Hello, thank-you for your response! I edited my entry, and I hope that helps (my apologies for not being more clear). I think you already pointed out in your response where I went wrong. If I understand you correctly, are you saying that a GLS allows for correlated error whereas the ARIMA with exogenous variables directly models the errors using an ARIMA equation? – Hannah Jul 9 '15 at 15:50
The two models model somewhat different things.
• arima(... , xreg = ...) calculates a regression on xreg, modeling its errors as an ARIMA process. Note that this is not the same as an ARIMAX model, and that this also applies to Arima() and auto.arima().
• gls(..., correlation=corARMA(p,q)) calculates a generalized linear model, where the correlation structure of your errors follows an ARMA(p,q) process.
The ideas are of course similar, but the actual models are somewhat different. I find the arima() model easier to understand. It would be interesting to compare the coefficients on both the fixed regressors and the ARIMA models for the errors resp. their correlation.
Given that both models have the same complexity (as in number of parameters, as long as ARMA orders are the same), I'd go with the better fitting one. (But remember that you can't compare AICs calculated by functions in different packages, as AIC is only defined up to a constant, which can definitely differ between packages.)
• Okay great, that really helped clear things up for me. Thank-you! – Hannah Jul 9 '15 at 16:12
Not 100% sure, but my understanding is that the difference could be the way the two models are estimated. That is;
I think that ARIMA (assuming you are using R) simultaneously estimates the ARIMA parameters and the xreg parameters, whilst alternatively the gls model will fit the ARMA parameters to the residuals after the regression. | 730 | 3,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-10 | latest | en | 0.904251 |
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