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https://www.physicsforums.com/tags/ac-generator/ | 1,716,963,526,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00291.warc.gz | 810,817,620 | 25,623 | # What is Ac generator: Definition and 48 Discussions
In electricity generation, a generator is a device that converts motive power (mechanical energy) into electrical power for use in an external circuit. Sources of mechanical energy include steam turbines, gas turbines, water turbines, internal combustion engines, wind turbines and even hand cranks. The first electromagnetic generator, the Faraday disk, was invented in 1831 by British scientist Michael Faraday. Generators provide nearly all of the power for electric power grids.
The reverse conversion of electrical energy into mechanical energy is done by an electric motor, and motors and generators have many similarities. Many motors can be mechanically driven to generate electricity; frequently they make acceptable manual generators.
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1. ### Understanding Electric Shock in Floating-Neutral and Floating-Ground Scenarios
Hello, let's say the AC generator and the washing machine, with a live metal frame due to a fault, are floating-neutral and floating-ground. These are many combinations where both get and get not bonded/grounded. Can anybody tell us in which case the person gets an electric shock when touches...
3. ### AC generator, current and ground
Hi, Let's say we have a basic AC generator, where a single wire is rotated within two magnets: Due to the electromotive force, a voltage is generated so that Va - Vb = 120V. Let's now ground Vb side directly to earth, while leaving Va still open: We can say that Vb = Vc = 0V and Va...
4. ### Magnets with Halbach array in an AC generator
As a project, I am demonstrating electromagnetic induction by making an AC generator. The magnets I currently have access to are very strong neodymimium countersunk magnets, but I noticed that they utilise a Halbach array. I am aware that this means the magnet has alternating N-S poles spaced...
Imagine a magnet moving up and down so that its flux 'B' cuts the copper rod to produce an alternating emf, suppose if the movement is fast enough such that its frequency equals to the electron spin resonance frequency given by F = B x 2.8 Mhz per gauss, neglecting skin effect, more copper...
7. ### Maximum Induced Voltage of an AC Generator
I would like to ask a question about the induced voltage of an AC generator. So, according to a graph I found on Google, the maximum induced voltage is reached when the loop is parallel to the loop. (The graph shown below.) Then I was wondering, what if the loop is initially (when it is at 0...
8. ### Determining Number of Turns for low voltage AC Generator
Ok, I did a quick search, and could not find answer. I need to determine how many turns / windings of 24awg magnet wire are required to generate no less than 5VAC no greater than 20VAC single phase using 3 of the N45 Neo Magnets rated at 13800 gauss each with 275-pound pull. Y'all give me a clue...
9. ### Magnetic Field of an AC generator
Hi Please help me in understanding the kind the magnetic field used in an AC generator. Most of the figures (one shown in attached file) are shown with concave poles. Hence the field has to be radial. In such a field the angle between plane of coil and the magnetic field remains zero all the...
10. ### AC generator output voltage to grid question
Hi, In a given area there are different types of power plants , let's assume they are (coal, nuclear, wind and solar) Now as I have learned most of these plants (except solar and some others) use synchronous generators as the last step between mechanical energy conversion into electricity. But...
11. ### AC generator -- Understanding coil voltages as function of position
Homework Statement In an ac generator Voltage is maximum when the coil is parallel to the direction of magnetic field (B) Voltage is minimum (0) when the coil is perpendicular to the direction of magnetic field (B) I'm trying to understand why. Homework EquationsThe Attempt at a Solution A...
12. ### Understanding the Winding and Voltages of an AC Generator: A Hyundai Unit
Hi. Can anyone tell me how my AC generator is constructed. It is a Hyundai unit, the main windings consist of 4 wires in parallel, 2 sets of 6 coils, each one of the set is maybe 8 or 10 turns. There is a secondary pair of 2 windings each. This is centre-tapped, so I guess to provide exciter...
13. ### Which drawing represents an AC generator?
Homework Statement A magnet bar has been installed in the vertical rotation axis of a coil. Which of these drawings represents an AC generator? 2. Relevant rules Fleming's right-hand rule.The Attempt at a Solution Fleming's right-hand rule assumes that the coil is in motion, but in the...
14. ### Current Rating of an AC Generator
Hello, I'm a bit confused on the term current rating in an ac generator. Is it the one in that I can get in the formula: kW= V*I*PF*sqrt(3) ? or the one in kW= V*I*sqrt(3)? The problem is the data that I have is V, PF, kVA only. I can compute the kW using those data but I'm abit confused on...
15. ### Find the maximum voltage of an AC generator
Homework Statement The sinusoidal voltage output from the AC generator oscillates between 500 V and -500 V. The circuit for an ideal transformer is constructed with a generator, transformer and a 30 Ω resistor . The generator makes 4 loops with the transformer and the resistor makes 9 loops...
16. ### How does rotating a magnet affect the magnetic field?
I'm doing an experiment on the effect of the speed of rotation of a rare Earth magnet on the voltage generated in a solenoid coil. I was wondering, how does rotating the magnet affect the magnetic field? The experiment setup is similar to the one below:
17. ### AC Generator With Repelling Magnets
Hi Everyone, I have 4 scenarios in question . Will The EMF(Voltage) be the same in all of the scenarios? - the difference between the two photos are the polls: one repel one attract . - i understand that the current(amps) will change due to the wire length assuming same gauge 1) Photo...
18. ### AC Generator Design: How to Increase Output Safely
If you buy a generator (gas powered etc) there is only a certain amount of watts that the generator can put out. This is because the engine only has a certain power rating (and losses). However, if one were to put a higher power engine on the generator head and remove the safety breaker for...
19. ### Induced Current in an AC generator
Homework Statement This is a simple inquiry, based on what my textbook has told me. "...the amount of induced current also depends on the angle of the conductor in relation to the external magnetic field. The induced current is at a maximum when the plane of the loop is parallel to the...
20. ### AC generator calculation question
1. Calculate the emf induced in ONE side of a coil of length 0.4m and width 0.3m, when there are 50 turns in the coil and the flux density is 96.8 mT, and the coil is rotated 30 degrees from the vertical. Frequency is 50 Hz and the long sides are the ones facing the magnets.
21. ### AC generator on grid: Governor and Amp
Hi Forum, once again I seek your help I have a very good understanding of rotating magnetic field due to stator, armature reaction, and the phasor diagram. from energy point of view it is very clear that if Gov increases, while output voltage remains locked by the gird, Amp will increase...
22. ### Electric AC Generator: Understand Magnetic Fields & Brushes
Hi! I have been studying AC generators. I am confused about why there is no current when the armature is perpendicular to the magnetic field (when it is vertical). Also, what is the purpose of the brushes? Probably a really obvious answer, but it is confusing me. Any help would be...
23. ### Max EMF in a Model AC Generator Coil
Homework Statement In a model ac generator, a 500 turn rectangular coil, 8.0 cm by 20 cm, rotates at 120 rev/min in a uniform magnetic field of 0.60 T. (a) What is the maximum emf induced in the coil? (b) What is the instantaneous value of the emf in the coil at t = (∏/32)s? Assume that...
24. ### AC generator and DC motor problem
Homework Statement For an AC generator, my teacher told us that when turning it, you will slowly manage to turn it at a constant speed only because of the induced current causing a force in the other direction to the turning motion. Why is this so? For a DC motor, my teacher told us that after...
25. ### Calculating the Power Output of an AC Generator.
Hello everyone! This is my first time posting here and I hope I can make a good first impression, and I apologize for the lack of links because of this being my first post. Currently I am working on a project for my grandfather's engineering business, and part of this project is designing a...
26. ### Significance of number of magnetic poles in ac generator
I am supposed to discuss the benefits of having high number of magnetic poles in ac generators with constant frequency. From formula it is obvious that the higher number of poles the lower the speed of rotation. But is there anything else, or what is the benefit of lower speed? The only...
27. ### Experimenting with 24VDC to AC Generator Output
Hello, I am new to this site and thought I would give this a shot. All my experience is in low volatge DC circuits and don't have a whole lot of experience dealing with AC but I am trying to learn. My boss and I are working on an experimental generator with 24VDC input but AC output...
28. ### Why is there an induced current in a dc motor?
When I put a current into a dc motor, there will be a turning effect. But I thought as this happens, won't there the an induced current that flows in the opposite direction? Bt Fleming's Right hand rule, the current will flow in the other direction. So I'm quite confused about this...
29. ### Exploring the AC Generator in an LRC Circuit
Homework Statement An AC generator in an LRC circuit produces a voltage V(t) = 1.414sin(wt) = 1.414sin(1000t) The values of inductance, capacitance, and resistance are shown in the diagram. Recall the w = 2pi*f. i made a picture of the diagram...
30. ### Step up transformer + ac generator question
Homework Statement An AC generator, originally designed to provide a peak output of +/- 155V at frequency 60Hz, is required to provide a peak output of +/- 340V at 50Hz. If the generator is connected to a transformer with 200 loops in its primary coil, how many loops should there be in the...
31. ### RMS Current in an AC Generator with Doubled Rotational Speed
Homework Statement The output voltage of an a.c. generator is \emph{V}=\emph{NBA}\omegasin(\omegat). When the voltage is connected across an inductor, the rms current in the inductor is found to be 10mA. What will Be the rms current if the rotating speed of the coil inside the generator is...
32. ### AC Generator with Resistor: Exploring Kirchhoff's Loop Rule
Homework Statement A resistor of resistance R is connected to an AC generator with EMF(t)=(EMF_max)sin((w_d)t). (a) Write Kirchhoff's Loop Rule for this circuit at a particular instant of time. Is your result a differential equation? (b) Find the current as a function of time. Now...
33. ### Is a Small AC Generator Suitable for Running Electronics?
Hi all, The country where I live, electricity outages are a part of life, so to compensate for it I have a small 2.2kW 220V AC generator in my house. The problem with it is that I've heard and found out for myself that its output is unfit for running electronics. Testing with a...
34. ### Calculating Resistance and Input Voltage in an RL Circuit with AC Generator
Homework Statement A resistance R and a 1.4 H inductance are in series across a 60 Hz AC voltage. The voltage across the resistor is 30V and the voltage across the inductor is 40V. (a). What is the resistance R ? (b). What is the AC input voltage ? Homework Equations \xi=I*Z The Attempt at...
35. ### How Does an AC Generator Produce EMF?
Homework Statement Need to draw a graph of the emf generated by a simple single-phase generator which has a coil of 200 turns. The coil is 14cm long and 9cm wide. The magnetic field in the generator is 0.15T. The generator coil is turned at a rate of 3000 revolutions per minute. Can someone...
36. ### Inductance in an AC generator.
I just have some confusion about whether or not self-inductance will occur in an AC generator. The current is alternating in both direction and magnitude rapidly when it is generated and this will in turn cause a changing magnetic field around the conductor. This magnetic field will then...
37. ### Maximum EMF in a rotating coil, in an AC generator?
Homework Statement Im currently in my last year of high school, and I'm writing a report on AC generators. I've heard a number of different things from different sources. When the coil is perpendicular to the magnetic field, maximum magnetic flux is flowing through the coil. This is then at a...
38. ### Current vs. Time: Mathematical Expression for Hippolyte Pixii's Dynamo
Consider a simple alternating current generator such as Hippolyte Pixii's dynamo. The current generated by such dynamo can be described as simple oscillating cosine signal. Assuming for simplicity that the max current generated each time a pole of Pixii's magnet passed the coil is (2/pi)0.5...
39. ### Archived Electromagnetic Induction of ac generator
Homework Statement The coil of an ac generator has an area per turn of 1.2x10-2m2 and consists of 500 turns. The coil is situated in a 0.13-T magnetic field and is rotating at an angular speed of 34 rads/s. What is the emf induced in the coil at the instant when the normal to the loop makes an...
40. ### Where do I start to determine where the problem is?
I have a 3 year old ( Chinese) Amico 3300 watt generator run by a 200cc gas engine. The engine works well and the generator turns but no juice is produced. Circuit breakers (110 & 220)are on. I don't have a manual for troubleshootin and distributor is not helpful. Where do I start, to determine...
41. ### How Does a Three-Phase AC Generator Create Voltages?
Although I am comfortable with three-phase circuits in general, visualizing the three-phase AC generator is giving me some problems. I understand that windings are placed 120 degrees apart, but that's about it. Would anyone happen to have a good animation/picture of a three-phase AC generator...
42. ### How Does Capacitance Affect Current in an AC Circuit?
Homework Statement An ac generator with a frequency of F=25 Hz and an rms voltage of 15 V is connected to a C=32 µF capacitor. Assume that the generator produces a sinusoidal waveform. (a) What is the maximum current in the circuit? mA (b) What is the current in the circuit when the...
43. ### Max EMF Induced in AC Generator?
Homework Statement The armature of an AC generator is rotating at a constant speed of 30 revolutions/second in a horizontal field of flux density 1.0Wb/m^2. The diameter of the cylindrical armature is 24cm and its length is 40cm. What is the maximum emf induced in the armature having 30 turns...
44. ### Solving AC Generator Problem: Find Voltage Output
Hello guys I am having a problem understanding the wording of a question, perhaps someone can help me explain it. The alternating voltage of a generator is represented by the equation E = Eo sin wt, where E is in volts, Eo = 346V, w = 685 rad/s, and t is in seconds. Find the...
45. ### Current/Voltage in an AC Generator
I'm slightly confused about something. In an AC generator at one point in the revolution it produces a positive voltage or current, and then in the other half of the revolution it produces a negative voltage or current...what I'm confused about is which is it producing negatively - the voltage...
46. ### Converting frequency to current for an ac generator
before anyone has a good laugh at me, I am studying environemental engineering and have to do a course that has some electronics it. i have a single phase alternating current (50 Hz) generator that produces 120 volts. I need to calculate the current from this information. can anyone help...
47. ### AC Generator Help: Get Answers for School Physics Assignment
Hey, I need to make an AC Generator for a school physics assignment. It will be marked on how much voltage it generates. I will be doing this using induction with the following equipment: * Permanent magnet * Coil/Wire * Iron Core * Split Rings/Brushes ? My questions are: * What's...
48. ### How do you make a hand powered AC generator?
Does anybody know? The basics? A procedure? Anything i'll be glad with anybody's advice. Thank you. | 3,782 | 16,601 | {"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-2024-22 | latest | en | 0.91272 |
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# day14 - Turing Machines 5th Model Click to edit Master...
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Click to edit Master subtitle style 10/4/11 Turing Machines 5th Model A Linear Memory Machine
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10/4/11 Dec 2, 2007/COT5310 © UCF (Charles E. Hughes) 22 Basic Description We will use a simplified form that is a variant of Post’s and Turing’s models. Here, each machine is represented by a finite set of states of states Q, the simple alphabet {0,1}, where 0 is the blank symbol, and each state transition is defined by a 4-tuple of form q a X s where q a is the discriminant based on current state q, scanned symbol a; X can be one of {R, L, 0, 1}, signifying move right, move left, print 0, or print 1; and s is the new state. Limiting the alphabet to {0,1} is not really a limitation. We can represent a k-letter alphabet by encoding the j-th letter via j 1’s in succession. A 0 ends each letter, and two 0’s ends a word.
10/4/11 Dec 2, 2007/COT5310 © UCF (Charles E. Hughes) 33 Base Machines R -- move right over any scanned symbol L -- move left over any scanned symbol 0 -- write a 0 in current scanned square
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https://nl.mathworks.com/matlabcentral/answers/477595-parfor-error-for-global-sa-morris-method-using-simbiology | 1,726,068,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00572.warc.gz | 397,404,881 | 27,994 | # parfor error for global SA Morris method using SimBiology
2 views (last 30 days)
emjey on 26 Aug 2019
Commented: emjey on 26 Aug 2019
hi, I am trying to use parallel toolbox to speed up my global SA with Morris method. Basically it averages elementary effects for a specific parameters across the entire parameter space. An elementary effect is calculated as EE = (f(p) - f(p+delta))/delta. For more details see Wentworth et al. J. UNCERTAINTY QUANTIFICATION, Vol. 4, pp. 266–297 (https://projects.ncsu.edu/crsc/reports/ftp/pdf/crsc-tr15-01.pdf).
So far I failed with my attempts, such as for this code - here using SimBiology 'sbioselect' and 'sbiosimulate'
clear variables; close all; clc;
format shortG
r = 5; % number of trajectories
l = 10; s = 2;
delta = l/(s*(l-1));
paramsFile = strcat('PARAMETERS/Lotka_parameters.csv');
p = length(paramsDF.parameter);
% Morris ===========================================================
d = zeros(r,p);
for i = 1:r
f = zeros(1,(p+1));
% an example for the Morris C matrix capturing one trajectory in parameter space
C = [15 0.085556 15;
15 0.005 15;
15 0.005 9.4444;
9.4444 0.005 9.4444];
% for each trajectory for p parameters, there are p+1 evaluations to be made
for j = 1:(p+1)
for k = 1:size(C,2) % columns = parameters
blub = sbioselect(m1, 'Name', paramsDF.parameter(k));
blub.Value = C(j,k);
end
[t,y] = sbiosimulate(m1);
f(j) = y(end,varNo); % looking for the endpoint as the sensitivity function
end
% Calculating elementary effects
v = zeros(1,p);
for j = 1:p
Cindex = find(diff(C(:,j)) ~= 0);
v(j) = ((f(Cindex) - f(Cindex+1))/delta;
end
d(i,:) = v;
end
%% disp("Calculating EE statistics")
... based on d(i,j)
Although the code is correct from the syntaxt point of view, at run time I get the error:
Starting parallel pool (parpool) using the 'local' profile ...
connected to 4 workers.
Error using sbiosimulate (line 136)
Expected input number 1, MOBJ, to be one of these types:
SimBiology.Model
Error in test (line 19)
parfor i = 1:r
Any comments would be very appreciated. I attach lotka.sbproj and parametr file as well.
Best, M
Florian Augustin on 26 Aug 2019
Hi,
It looks like you are running into an issue that the SimBiology model is not copied to the workers natively. In MATLAB versions R2019a and earlier, you have to use parallel.pool.Constant to send the model to the workers in your parallel pool. The following tweak to your code should do this for you.
m1ParConst = parallel.pool.Constant(m1);
parfor i = 1:r
% ...
% Get the model from the parPool constant as follows:
blub = sbioselect(m1ParConst.Value, 'Name', paramsDF.parameter(k));
% ...
end
Let me know if this doesn't solve your problem.
Best
-Florian
emjey on 26 Aug 2019
hi Florian, thanks for your suggestion, it worked.
I had to replace the other m1 as well of course and now it's 3x faster.
Great news, thanks a million!
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http://mathhelpforum.com/advanced-statistics/134177-better-estimator.html | 1,524,717,982,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948064.80/warc/CC-MAIN-20180426031603-20180426051603-00574.warc.gz | 191,693,886 | 13,216 | 1. ## better estimator
For a Uniform $\displaystyle U(0, \theta)$
The Maximum Likelyhodd Estimator of $\displaystyle \theta$ = $\displaystyle X_(n)$
Also, the sufficient statistic for $\displaystyle \theta$ is $\displaystyle X_(n)$
and E[X] = $\displaystyle \frac{\theta}{2}$
Then the estimators of $\displaystyle \theta$ are found out as
$\displaystyle \hat \theta$ = $\displaystyle 2 \bar X$
and
$\displaystyle \hat \theta = \frac{n}{n+1} \theta$
Which one is the better estimator? What is the process to find it?
I am wondering if the answer involves calcuating the variance.
2. It does. Calculate the variance for both of them and use the one that is smaller. Note that we usually cannot get rid of the $\displaystyle n$ term in the denominator of the variance, but we can decrease the coefficients and such.
3. Originally Posted by Anonymous1
It does. Calculate the variance for both of them and use the one that is smaller. Note that we usually cannot get rid of the $\displaystyle n$ term in the denominator of the variance, but we can decrease the coefficients and such.
I am finding it difficult to calculate the variance from $\displaystyle \hat \theta$ . Can you show me some beginning steps of how to start this?
4. $\displaystyle Var(\hat\theta) = 4/n Var(\bar X) = 4/n [E[\bar X^2]- (E[\bar X])^2].$
Now what is $\displaystyle E[\bar X]?$
5. Originally Posted by Anonymous1
$\displaystyle Var(\hat\theta) = 4/n Var(\bar X) = 4/n [E[\bar X^2]- (E[\bar X])^2].$
Now what is $\displaystyle E[\bar X]?$
$\displaystyle E[X] =\int_0^\theta \frac{x}{\theta}$
=$\displaystyle \frac{\theta}{2}$
and ,
$\displaystyle E[X^2] =\int_0^\theta \frac{x^2}{\theta}$
= $\displaystyle \frac{\theta^2}{3}$
So,
$\displaystyle Var(\hat\theta) = 4/n [\frac{\theta^2}{3} - \frac{\theta^2}{4}]$
= $\displaystyle \frac {\theta^2}{3n}$
is this correct?
How do I go with the second one?
6. Just to be clear,
$\displaystyle \bar X = \frac{1}{n} \sum^n x_i$
$\displaystyle E[\bar X] = \frac{1}{n} \sum^n E[x_i] = \frac{1}{n} \sum^n\frac{\theta}{2} = \frac{\theta}{2}$
This is why there is an $\displaystyle n$ in your variance.
So, $\displaystyle Var(\hat \theta) = (\frac{n}{n+1})^2 \times Var(\theta).$
Now, what is $\displaystyle Var(\theta)?$
7. Originally Posted by Anonymous1
Just to be clear,
$\displaystyle \bar X = \frac{1}{n} \sum^n x_i$
$\displaystyle E[\bar X] = \frac{1}{n} \sum^n E[x_i] = \frac{1}{n} \sum^n\frac{\theta}{2} = \frac{\theta}{2}$
This is why there is an $\displaystyle n$ in your variance.
So, $\displaystyle Var(\hat \theta) = (\frac{n}{n+1})^2 \times Var(\theta).$
Now, what is $\displaystyle Var(\theta)?$
Thank you. I think it becomes clearer now.
So,
It is $\displaystyle E[\bar X]$.
Likewise,
$\displaystyle E[\bar X^2] = \frac{1}{n^2} \sum^n E[x_i^2] = \frac{1}{n^2} \sum^n\frac{\theta^2}{3} = \frac{\theta^2}{3n}$
?
As, you have asked for the second one,
$\displaystyle Var(\theta) = \frac{\theta^2}{12}$
Is it?
8. Accidental double post. Is there a way to delete these things?
9. Originally Posted by harish21
$\displaystyle E[\bar X^2] = \frac{1}{n^2} \sum^n E[x_i^2] = \frac{1}{n^2} \sum^n\frac{\theta^2}{3} = \frac{\theta^2}{3n}$
I think it is important to notice that the second moment brings in the $\displaystyle n.$ This means our sample size is always going to effect our dispersion.
Now to finish the variance of the second estimator multiply $\displaystyle Var(\theta)$ by all that $\displaystyle n$ stuffs.
One of your estimators was found by MLE and the other by the method of moments. Which one is better? Why?
10. Originally Posted by Anonymous1
I think it is important to notice that the second moment brings in the $\displaystyle n.$ This means our sample size is always going to effect our dispersion.
Now to finish the variance of the second estimator multiply $\displaystyle Var(\theta)$ by all that $\displaystyle n$ stuffs.
One of your estimators was found by MLE and the other by the method of moments. Which one is better? Why?
For the variance of the second estimator, isnt it sufficient enough to state that:
$\displaystyle Var(\hat \theta) = (\frac{n}{n+1})^2 \times \frac{{\theta}^2}{12}.$
1 Variance of a constant is zero...$\displaystyle V(\theta)=0$
you want $\displaystyle V(\hat\theta)$
2 Moreover if $\displaystyle \hat\theta$ is your largest order stat then it's less than $\displaystyle \theta$
so to make it unbiased you need to mulitiply by a number bigger than 1, not smaller.
So you probably want $\displaystyle {n+1\over n}X_{(n)}$
vs. the method of moment estimator $\displaystyle 2\bar X$
Originally Posted by harish21
For a Uniform $\displaystyle U(0, \theta)$
The Maximum Likelyhodd Estimator of $\displaystyle \theta$ = $\displaystyle X_(n)$
Also, the sufficient statistic for $\displaystyle \theta$ is $\displaystyle X_(n)$
and E[X] = $\displaystyle \frac{\theta}{2}$
Then the estimators of $\displaystyle \theta$ are found out as
$\displaystyle \hat \theta$ = $\displaystyle 2 \bar X$
and
$\displaystyle \hat \theta = \frac{n}{n+1} \theta$
Which one is the better estimator? What is the process to find it?
I am wondering if the answer involves calcuating the variance.
12. Originally Posted by matheagle
1 Variance of a constant is zero...$\displaystyle V(\theta)=0$
you want $\displaystyle V(\hat\theta)$
2 Moreover if $\displaystyle \hat\theta$ is your largest order stat then it's less than $\displaystyle \theta$
so to make unbiased you need to mulitiply by a number bigger than 1, not smaller.
So you probably want $\displaystyle {n+1\over n}X_{(n)}$
vs. the method of moment estimator $\displaystyle 2\bar X$
Matheagle,
How is $\displaystyle V(\theta) = 0??$
13. Originally Posted by harish21
Matheagle,
How is $\displaystyle V(\theta) = 0??$
you're confusing parameters and statistics.
parameters are unknown CONSTANTs
statistics are random variables
$\displaystyle \theta$ is our unknown constant/parameter that we are estimating
with our stats $\displaystyle \bar X$ and $\displaystyle X_{(n)}$
14. Originally Posted by matheagle
you're confusing parameters and statistics.
parameters are unknown CONSTANTs
statistics are random variables
$\displaystyle \theta$ is our unknown constant/parameter that we are estimating
with our stats $\displaystyle \bar X$ and $\displaystyle X_{(n)}$
So the two estimates, of which, we are trying to find a better one are :
$\displaystyle Var(\hat\theta) = 4/n Var(\bar X) = 4/n [E[\bar X^2]- (E[\bar X])^2].$
vs.
$\displaystyle Var(\hat\theta)={n+1\over n}X_{(n)}$
??
are these the ones that are to be compared?
15. I took off the variance.
BUT in order to determine if that (n+1)/n in front of the largest order stat is correct you will need the density of that order stat.
Originally Posted by harish21
So the two estimates, of which, we are trying to find a better one are :
$\displaystyle Var(\hat\theta) = 4/n Var(\bar X) = 4/n [E[\bar X^2]- (E[\bar X])^2].$
vs.
$\displaystyle \hat\theta={n+1\over n}X_{(n)}$
??
are these the ones that are to be compared?
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# Algebra Final Exam Study Guide
## Chapter 1 - Vectors
Draw vectors in R2 and R3 as well as graphically represent a sum, difference, or scalar
multiple of a vector .
Compute sums , differences, scalar multiples, length, and dot products of vectors algebraically.
Two definitions of dot product and be able to find the angle between two vectors.
Properties of the dot product (Thm 1.2) and length (Thm 1.3).
• Definition of orthogonal, both geometrically and in terms of dot product.
• Compute the projection of a vector onto another vector and be able to represent the
projection geometrically.
Equations of lines (vector and parametric) and planes (vector and general).
## Chapter 2 - Systems of Linear Equations
• Use Gauss- Jordan elimination (both by hand and with a calculator) to solve a system
of linear equations .
• Find the span of a set of vectors.
• Determine if a set of vectors is linearly independent .
## Chapter 3 - Matrices
Find the sum , difference , or product of two matrices. Find the scalar multiple, power
or transpose of a matrix.
• Definition of symmetric and skew-symmetric.
• Properties of matrix addition and scalar multiplication (Thm 3.2), matrix multiplication
(Thm 3.3), and the transpose (Thm 3.4).
• Definition of the inverse of a matrix.
• Properties of the inverse (Thm 3.9)
• Fundamental Theorem of Invertible Matrices (Thm 3.12, 3.27, 4.17)
• Find the inverse of a 2 × 2 matrix using special formula or the inverse of an n × n
matrix using the Gauss- Jordan method .
• Definition of subspace of Rn, determine if a set is a subspace of Rn.
• Definition of basis, dimension, rank and nullity. For a set of vectors, find a basis and
its dimension. Find the rank and nullity of a matrix.
• The Rank Theorem (Them 3.26).
• Definition of linear transformation (from Rn to Rm) and matrix transformation.
• Every linear transformation T : Rn -> Rm is a matrix transformation and vice versa.
(Thm 3.30, 3.31) Find the standard matrix of a linear transformation.
• Find the composition and inverse of a linear transformation using the standard matrix
of the transformation.
## Chapter 4 - Eigenvalues and Eigenvectors
• Definition of eigenvalue, eigenvector, eigenspace.
• Compute the determinant of and n × n matrix.
• Properties of the determinant (Thm 4.3,4.7,4.8,4.9,4.10).
• Find eigenvalues, eigenvectors, and eigenspaces of an n × n matrix.
• Definition of algebraic and geometric multiplicity of an eigenvalue.
• Definition of similarity, properties of similarity (Thm 4.21, 4.22)
• Definition of diagonalizable.
• Determine if a matrix is diagonalizable and if it is, find an invertible matrix P and a
diagonal matrix D so that P−1AP = D.
## Chapter 5 - Orthogonality
• Definition of orthogonal set, orthogonal basis, orthonormal set and orthonormal basis.
• Definition of orthogonal matrix, properties of orthogonal matrices (Thm 5.6,5.7,5.8).
• Definition of orthogonal complement.
• Compute the orthogonal complement of a subspace W of Rn.
## Chapter 6 - Vector Spaces
• Definition of vector space (10 axioms). Determine if a set with two given operations is
a vector
space.
• Properties of vector spaces (Thm 6.1)
• Definition of subspace. Determine if a set is a subspace of a given vector space.
• Find the span of a set of vectors.
• Determine if a set of vectors is linearly independent.
• Definition of basis and dimension of a vector space.
• Properties of basis/span/L-I of vector space (Thm 6.10).
• Definition of linear transformation on a vector space.
• Properties of linear transformation (Thm 6.14).
• Definition of composition and inverses of linear transformations.
• Definition of kernel and range of a linear transformation. Be able to find the kernel
and range of a linear transformation.
• Definition of rank and nullity of a linear transformation. Be able to find the rank and
nullity of a linear transformation.
• The Rank Theorem (Thm 6.19).
• Definition of one-to-one and onto linear transformation. Determine if a linear transformation
is one-to-one and/or onto.
• Definition of isomorphism. Determine if a transformation is an isomorphism. Determine
if two vector spaces are isomorphic.
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https://www.arxiv-vanity.com/papers/1006.2610/ | 1,628,124,071,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155268.80/warc/CC-MAIN-20210805000836-20210805030836-00426.warc.gz | 640,604,426 | 44,713 | # Functions which are PN on infinitely many extensions of Fp, p odd.
Elodie Leducq
Abstract : Jedlicka, Hernando and McGuire have proved that Gold and Kasami functions are the only power mappings which are APN on infinitely many extensions of . For an odd prime, we prove that the only power mappings such that which are PN on infinitely many extensions of of are those such that , l positive integer. As Jedlicka, Hernando and McGuire, we prove that has an absolutely irreducible factor by using Bézout’s Theorem.
## 1 Introduction
In [7] and [5], the authors are interested in integers such that the function is almost perfectly nonlinear (APN) on infinitely many extensions of . Using similar methods, we study here the case of perfectly nonlinear (PN) functions over finite fields of odd characteristic.
Let be a prime number, a positive integer, and a finite field with elements. We recall the following definition :
###### Définition 1.1
We say that the function is APN over if :
∀a,b∈Fq,a≠0,|{x∈Fq,ϕ(x+a)−ϕ(x)=b}|≤2
and if, furthermore, there exists a pair such that we have equality.
Jedlicka, Hernando and McGuire prove that in the case of characteristic 2, the only integers such that is APN on infinitely many extensions of are (Gold) and (Kasami). They use the fact that a function is APN over if and only if the rational points in of are points such that or . This can happen only if has no absolutely irreducible factor over .
In characteristic 2, if then . So, there is no PN function (see definition below). Furthermore, to prove that is APN, it is sufficient to prove that
∀a,b∈Fq,a≠0,|{x∈Fq,ϕ(x+a)−ϕ(x)=b}|≤2.
On the contrary, in odd characteristic, we do not know the relationship between the two solutions of (if there exist two). So, it seems to be difficult to adapt this method to APN functions in odd characteristic. Nevertheless, we can try on PN functions :
###### Définition 1.2
If is odd, a function is PN over if for all and all
|{x∈Fq,ϕ(x+a)−ϕ(x)=b}|=1.
Equivalently, a function is PN over if for all , the only rational points in of
ϕ(x+a)−ϕ(x)−ϕ(y+a)+ϕ(y)=0
are points such that .
From now, , . We only have to consider the case where in the definition above (see [3]).
###### Remark 1.3
If is odd then, 0 and 1 are solutions of . So is not PN over for any .
The only known PN power mappings are the following :
###### Proposition 1.4
Let a power mapping. Then is PN on for
1. ,
2. where is an integer such that is odd [1, 2],
3. where and is an odd integer such that [1].
We set . Since divides , we define .
We can assume that . Indeed, if is PN over and then is also PN over .
###### Proposition 1.5
If has an absolutely irreducible factor over then is not PN on for sufficiently large.
Proof : Assume that has an absolutely irreducible factor over , denoted by . If with , then , . Hence,
−m(y+1)m−1+mym−1=∂f∂y(x,y)=2(y−x)˜Q(x,y)+(y−x)2∂˜Q∂y(x,y).
So, we get that for all , which is impossible since . Let be the degree of . Since , is not the null polynomial. So there are at most rational points of such that .
On the other hand, if we denote by P the number of affine rational points of on , we have (see [8, p. 331]):
|P−pn|≤(s−1)(s−2)√pn+s2.
Hence, for sufficiently large, has a rational point in such that and is not PN over .
From now, we are interested in the case where . We denote by the greatest integer such that divides and we set
d:=gcd(m−1,pl−1)=gcd(m−1pl,pl−1).
By Proposition 1.4, the only known functions such that which are PN on infinitely many extensions of are those such that . We want to prove that there does not exist any other.
###### Theorem 1.6
Let be an integer such that , and
. Assume that . Then has an absolutely irreducible factor over .
###### Corollary 1.7
The only such that is PN on infinitely many extensions of are .
Proof : By Theorem 1.6 and Proposition 1.5, we only have to treat the case where . Then which is odd; so is not PN on all extensions of
###### Remark 1.8
After reading the manuscript of this paper, McGuire informed me that the result in this paper has already been proved by Hernando and himself and also by R. Coulter. But as far as I know, it has never been published.
Now, we only have to prove Theorem 1.6. The method of Jedlicka, Hernando and McGuire is, using Bézout’s Theorem, to prove that has an absolutely irreducible factor over because it has not enough singular points. In Part 2, we study singular points of and their multiplicity. In Part 3, we bound the intersection number (see [4] for definition) where is a singular point of and , are such that . In part 4, we prove Theorem 1.6. Finally, we consider briefly the case where .
## 2 Singularities of h
###### Proposition 2.1
The singular points of are described in Table 1.
The proof of this theorem follows from Lemmas 2.2 to 3.12 and their corollaries.
### 2.1 Singular points at infinity
We denote by (respectively ) the homogenized form of (respectively ). We denote by (respectively ) the dehomogenized form of (respectively ) relative to .
Let . Then,
⎧⎪ ⎪⎨⎪ ⎪⎩Fx=m(x+z)m−1−mxm−1Fy=−m(y+z)m−1+mym−1Fz=m(x+z)m−1−m(y+z)m−1.
At infinity (), and
Fz(x,y,0)=m(xm−1−ym−1).
So is a singular point of F if and only if . If then ; so and we have to study the solutions of
xm−10=1. (1)
Equation (1) is equivalent to . Since , there are solutions at (1) and is the only one such that .
Now, we want to find the multiplicity of these singularities :
˜F(x+x0,z) =(x+x0+z)m−(x+x0)m−(z+1)m+1 =m∑k=2(mk)(x+z)kxm−k0−m∑k=2(mk)xkxm−k0−m∑k=2(mk)zk.
Since , for all , . Consider the terms of degree of :
1z(mpl)(xm−pl0(x+z)pl−xm−pl0xpl−zpl)=(mpl)(xm−pl0−1)zpl−1.
This term vanishes (which means that is a singular point of multiplicity greater than ) if and only if
xm−pl0=1
that is to say if and only if
xd0=1.
Now, consider the terms of degree of :
1z(mpl+1)(xm−pl−10(x+z)pl+1−xm−pl−10xpl+1−zpl+1) =(mpl+1)(xm−pl−10xpl+xm−pl−10xzpl−1+(xm−pl−10−1)zpl).
Since , singular points of of multiplicity greater than have multiplicity .
We have just proved the following lemma :
###### Lemma 2.2
Let such that . The point is a singular point of with multiplicity
{plif ωd=1, ω≠1pl−1otherwise.
Furthermore, has singular points at infinity.
### 2.2 Affine singular points
We have :
{fx=m(x+1)m−1−mxm−1fy=−m(y+1)m−1+mym−1.
So,
(x0,y0) singular point of f ⇔⎧⎪ ⎪⎨⎪ ⎪⎩f(x0,y0)=0(x0+1)m−1=xm−10(y0+1)m−1=ym−10 ⇔⎧⎪ ⎪⎨⎪ ⎪⎩xm−10(x0+1)−xm0−ym−10(y0+1)+ym0=0(x0+1)m−1=xm−10(y0+1)m−1=ym−10 ⇔⎧⎪ ⎪⎨⎪ ⎪⎩xm−10=ym−10(x0+1)m−1=xm−10(y0+1)m−1=ym−10.
###### Lemma 2.3
Affine singular points of are points satisfying
(x0+1)m−1=xm−10=ym−10=(y0+1)m−1.
From Lemma 2.3, we get that , . Since divides ,
(2)
There are at most solutions to the second equation of (2). Let be one of these solutions, we want to know the number of such that is a singular point of .
We write with , , . Then,
(y0+1)m−1pl=ym−1pl0 ⇔b∏j=1(y0+1)mjpij−l=ym−1pl0 ⇔∗∑0≤kj≤mj(b∏j=1(mjkj))y∑bj=1kjpij−l0=0
where * indicates that this sum runs over all possible b-uples except . We multiply by and we set :
∗∑0≤kj≤mjj≠b(b−1∏j=1(mjkj))αy∑b−1j=1kjpij−l0 +mb−1∑kb=0∑0≤kj≤mjj≠b(b∏j=1(mjkj))ym−1pl−(mb−kb)pib−l+∑b−1j=1kjpij−l0=0.
The degree of this polynomial in is
m−1pl−pib−l+b−1∑j=1mjpij−l=2m−1pl−(mb+1)pib−l.
###### Lemma 2.4
The number of affine singularities of is at most :
(m−1pl−1)(2m−1pl−(mb+1)pib−l)
where with , , .
Now, we study the multiplicity of affine singularities :
f(x+x0,y+y0) =(x+x0+1)m−(x+x0)m−(y+y0+1)m+(y+y0)m =m∑k=2(mk)xk(x0+1)m−k−m∑k=2xkxm−k0 −m∑k=2(mk)yk(y0+1)m−k+m∑k=2ykym−k0.
Since , for all , . So is a singularity of multiplicity at least . Consider the terms of degree :
(mpl+1)(((x0+1)m−pl−1−xm−pl−10)xpl+1−((y0+1)m−pl−1−ym−pl−10)ypl+1).
Since is a singular point, and . So,
(x0+1)m−pl−1−xm−pl−10=0 ⇔(x0+1)pl((x0+1)m−pl−1−xm−pl−10)=0 ⇔−xm−pl−10=0.
Hence, affine singularities have multiplicity at most . Then we look at terms of degree :
(mpl)(((x0+1)m−pl−xm−pl0)xpl−((y0+1)m−pl−ym−pl0)ypl).
However,
(x0+1)m−pl−xm−pl0=0 ⇔(x0+1)pl((x0+1)m−pl−xm−pl0)=0 ⇔(x0+1)m−1(x0+1)−xm0−xm−pl0=0 ⇔xm−pl0(xpl−10−1)=0 ⇔x0∈F∗pl.
We can do the same for .
###### Lemma 2.5
There are at most :
• affine singularities of such that . They have multiplicity ( for );
• affine singularities of such that . They have multiplicity ( for );
• affine singularities of such that et , . They have multiplicity (for and );
• affine singularities of such that and or . They have multiplicity (for and ).
## 3 Intersection number bounds
We write ; we want to bound the intersection number for a singularity of . In the following, we use freely this lemma proved in [6] :
###### Lemma 3.1
Let be an affine curve over and be a point of of multiplicity . Then where is an homogeneous polynomial of degree ; the factors of (on an algebraic closure) are called the tangent lines of at . We write ; if and are relatively prime then . Furthermore, if has only one tangent line at , .
### 3.1 Singularities at infinity
Let a singular point of at infinity (). We write where is the multiplicity of and is the homogeneous polynomial composed of the terms of degree of . Then,
˜f(x+ω,z) =˜h(x+ω,z)(x+ω−1) =(R+˜Hmt+1+˜Hmt)(x+ω−1) where R is a polynomial of degree greater than% mt+1 =xR+(ω−1)R+x˜Hmt+(ω−1)˜Hmt+1+(ω−1)˜Hmt.
So,
• if , we have and ;
• if , .
###### Lemma 3.2
If , , is a singular point at infinity of with multiplicity then
• if , and ;
• if , .
###### Corollary 3.3
If then
It(u,v)≤(pl−12)2.
Proof : If then its multiplicity is . By Lemma 3.2,
˜Hmt=a(xpl−1+zpl−1),
and all its factors are different. So . We get the result since .
###### Corollary 3.4
If such that , then
It(u,v)≤p2l−14.
Proof : The multiplicity of is . By Lemma 3.2,
(ω−1)˜Hpl=˜Fpl=xplωm−pl−1+xzpl−1ωm−pl−1+(ωm−pl−1−1)zpl.
So all factors of are simple and . We get the result since .
###### Corollary 3.5
If with , , then
It(u,v)=0.
Proof : The multiplicity of is . By Lemma 3.2,
(ω−1)˜Hpl−1=˜Fpl−1=αzpl−1
and
˜Fpl=x˜Hpl−1+(ω−1)˜Hpl=xplωm−pl−1+xzpl−1+zpl(ωm−pl−1−1).
So, and by Lemma 3.1, .
### 3.2 Affine singularities
We write where is the multiplicity of and is the homogeneous polynomial composed of the terms of degree of .
Let be an affine singular point of with multiplicity such that . Then
f(x+x0,y+y0) =h(x+x0,y+y0)(x+x0−y−y0) =(R+Hmt+1+Hmt)(x−y) where R is a polynomial of degree greater than mt+1 =(x−y)R+(x−y)Hmt+1+(x−y)Hmt =Fmt+1+Fmt+2+…
So, and . Furthermore, for some , | 3,655 | 10,523 | {"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-2021-31 | latest | en | 0.903495 |
https://gradeup.co/mba-entrance-exams/venn-diagrams | 1,624,562,052,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00420.warc.gz | 259,026,579 | 30,573 | # Venn Diagrams for CAT, MAT, SNAP, XAT Exams
By : Raghavendra Singh
Updated : Dec 29, 2020, 16:34
Venn Diagram Questions are asked in the logical reasoning section and carry a good weightage in the CAT exam. Venn diagrams test your logical reasoning skills and IQ level. Several questions in this category use daily life examples around you. As many students come from non-maths backgrounds, the questions on venn diagrams are structured at a basic level.
Looking at the Venn diagrams definition, it is a set of diagrams that show all logical relations between the given quantity of sets. It was first invented by John Venn in 1880 and published in the Philosophical Magazine. The questions related to Venn diagrams involve overlapping circles in which each circle represents a set.
## Important Venn Diagram Topics
The important Venn diagram topics for CAT exam are:
Topics Explanation Compare and Contrast A diagram is given, and questions asked are based on it. Diagrams in Options In this type of Venn diagram question, the diagrams are given in the options, and a question regarding that is asked.
### Tips to Solve Venn Diagrams Questions
Here are some tips and tricks that you can employ while solving Venn diagram questions in CAT exam:
• When you solve Venn diagrams questions for CAT, you must draw the diagram on the answer sheet or rough paper first to understand the question. You can opt for various Venn diagrams practice questions and sample questions from previous year CAT exams to check what type of questions are assigned for Venn diagrams.
• Venn Diagrams are used to solve complex mathematical problems. If you are thorough with Venn diagrams examples and Venn diagrams test questions, you can easily ace other fields of mathematics.
• There are many types of questions in the Venn diagrams category. When the sets given are three separate items, draw three circles for them.
• In many Venn diagrams topics and Venn diagrams notes, you will see questions where all the three items are interrelated to each other. Hence, the diagram formed will be one inside the other.
• In the third kind of questions found in many Venn diagrams study guides, two items are not related to each other but are related to the third item in the set. In this case, the two small circles are drawn inside one big circle.
• In many Venn diagrams questions and solutions, you will see that the two items are related to each other and somehow related to the third item too. Hence, the diagram will be overlapping circles placed in a bigger circle.
### Importance of Venn diagrams for CAT Exam
Let’s see why Venn diagrams are important for CAT:
• Venn diagram is an essential topic of any competitive exam because it tests a candidate's logical and analytical thinking.
• It consists of a majority part of the quantitative aptitude and logical reasoning in the CAT exam. If you are a pro in Venn diagrams, then you can score very well in the logical reasoning section.
• Venn diagrams being a part of the logical reasoning section in a CAT exam are useful in checking the aptitude of the candidates.
• They also help in testing the critical thinking of the students and their ability to solve the given problem without much difficulty.
## Most Recommended Books for Venn Diagrams for CAT
There are many books available in the market with interesting Venn diagrams tricks, Venn diagrams quiz, Venn diagrams problems with solutions included. Some of the popular ones are:
Book Author GMAT Critical Reasoning Bible Powerscore Class 11th Mathematics NCERT Lucent Reasoning Arihant Publication
## Why Prepare Venn Diagrams from Gradeup?
Gradeup is a perfect platform to practice different Venn diagrams questions and answers for your CAT exam. There are interactive and engaging quizzes through which you can test your knowledge in Venn diagrams and excel. You can also download Venn diagrams questions and answers in PDF to practice. Venn Diagrams questions and answers are scoring, and many students have got good grades in their quantitative aptitude and logical reasoning sections because of Venn Diagrams.
The response of students from across the country for Gradeup has been positive and amazing. They say that different Venn diagram questions improved their logical reasoning abilities and solving complex mathematical problems. All the quizzes are structured for students to prepare them for the most challenging CAT questions.
### FAQs
1. What are the different types of Venn diagrams?
Usually, there are 5 to 7 types of Venn diagrams based on different sets. The sets range from 2 to 4.
2. What is the middle section of the Venn diagrams called?
In the schematic Venn diagram, there are three sets. When these three circles intersect with each other, they form a triangle, which is called the Reuleaux triangle.
3. What is a Venn diagram with three circles called?
The Venn diagram with three circles represents three sets of the question. Hence, the diagram is called a three-set Venn diagram. All three circles are dividing space into two halves.
4. What are the books one can refer to for Venn diagrams?
Venn Diagrams are a part of logical reasoning and mathematical ability. They are asked in various competitive exams. You can study it from different GMAT prep and CAT prep logical reasoning books for this topic.
5. What is the maximum number of sets that can be drawn in Venn diagrams?
Infinite sets are possible to draw in the form of Venn diagrams. Hence, it would help if you practiced questions related to it from the previous years papers .
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GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 support@gradeup.co | 1,199 | 5,749 | {"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} | 3.640625 | 4 | CC-MAIN-2021-25 | latest | en | 0.92444 |
https://www.mrexcel.com/board/threads/vlookup-of-a-rolling-sum.171057/ | 1,696,303,489,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511053.67/warc/CC-MAIN-20231003024646-20231003054646-00284.warc.gz | 953,064,449 | 17,700 | # Vlookup of a rolling sum
#### Mackpazz
##### New Member
I am trying to do a vlookup of a sum within a range. I have a spreadsheet with columns of numbers (dollar values) with a column of dates at the end of the range. I am trying to lookup the cooresponding date when all of the dollars are spent in a given row.
Example:
Column A - Column B - Column C
Row 1 100 - 0 - 6/30/05
Row 2 0 - 100 - 7/31/05
Row 3 150 - 0 - 8/31/05
Row 4 500 - 0 - 9/30/05
Row 5 0 - 250 - 10/31/05
Row 6 0 - 0 - 11/30/05
Totals 750 350
I want to lookup the total of column A (i.e. 750) and return the date in column C in the cooresponding month where the sum of column A matches the total, basically what month are we done spending money (i.e.9/30/05)
### Excel Facts
Fastest way to copy a worksheet?
Hold down the Ctrl key while dragging tab for Sheet1 to the right. Excel will make a copy of the worksheet.
Try this
=SUMPRODUCT(MAX((A1:A6>0)*C1:C6))
format as date
Thanks much...works great. You have several of us Excel geeks in awe!
Can you futher explain what that formula is doing? Again, it works great but we can't follow the logic.
There appears to only be one array in the SUMPRODUCT? Doesn't the MAX require two comparable results?
The statement (A1:A6>0) is a mystery? Doesn't it need an IF?
The various pieces of the formula do not work independently, but do collectively.
Again, the formula works perfectly, we are just trying to understand why it works. Thanks.
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Go back | 728 | 2,532 | {"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-2023-40 | latest | en | 0.874507 |
https://math.stackexchange.com/questions/120925/advantage-of-accepting-the-axiom-of-choice?noredirect=1 | 1,563,271,238,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524522.18/warc/CC-MAIN-20190716095720-20190716121720-00240.warc.gz | 481,081,727 | 42,283 | # Advantage of accepting the axiom of choice
What is the advantage of accepting the axiom of choice over other axioms (for e.g. axiom of determinacy)?
It seems that there is no clear reason to prefer over other axioms..
Thanks for help.
• If you're fine with these consequences of not negating the axiom of choice, then no reason. – lhf Mar 16 '12 at 13:56
• @lhf: That looks like one consequence: you can't have the GCH. The rest are just choices that are open to you. – Charles Mar 16 '12 at 14:00
Well, there is quite the long list:
1. Every set can be well ordered, so every two cardinalities are comparable.
2. Continuity by $\epsilon$-$\delta$ is equivalent to continuity via sequences.
3. There are free ultrafilters.
4. Countable unions of countable sets are countable; similarly cardinal arithmetic become definable for infinite summations and products.
5. Hahn-Banach theorem.
6. Product of compact spaces is compact.
7. Second countable implies Lindelof.
8. Every vector space has a basis.
9. Every unital ring admits a maximal ideal.
10. Every set can be given a group structure.
The list goes on and on and on. In short, however, the main reason is that mathematics has grown to accept the axiom of choice and its consequences.
It really depends on what you want to do in mathematics, if you don't do set theory you're likely to benefit more from the axiom of choice than other axioms. If you are a set theorist (or planning on becoming one) then there will be places where the axiom of choice works naturally for your advantage and others where you will know to drop it and adopt other axioms instead.
Edit: Some of the implications above do not require the entire power of the axiom of choice. Indeed some of them follow by "mere" countable choice (choice for countable families) and other follow from weaker assertions like the Ultrafilter Lemma.
Furthermore, in some cases it is also possible to do things "by hand" and prove existence of ultrafilters, ideals, basis, etc. Most of classical analysis can get away with relatively weak principles like Dependent Choice, however it can require a bit more if you want to talk about $\ell^\infty$ kind of spaces (for example, without the axiom of choice it is possible for $\ell^1$ to be reflexive).
It is also important to notice that the universe can behave as though you have the full axiom of choice for sets which are so big that you will never go beyond their size - but after that thing breaks down badly, so while the axiom of choice is negated in a very strong way the universe that most mathematician like to think about has the axiom of choice in full.
Edit II: Having a few minutes to spare, I thought I should give my two cents why people prefer it (except for the fact they got used to it by now). One of the greatest set theorists of our time once told me during a break from a class he gave that a good axiomatic system is one that you use without noting.
For example, the axiom of extensionality makes a lot of sense because we want two sets to be equal exactly when they have the same elements. So does the axiom of power set.
The axiom of choice is very natural in the sense that it allows us to extend "finite definitions" to infinite ones (a good example is that every vector space has a basis, we can write one down for finite dimensional spaces, but we cannot "write it" for every infinite dimensional space). In this sense we are very accustomed to its gentle grace and willing to accept the few peculiarities like the Banach-Tarski paradox, or a well ordering of the real numbers.
This is why some of the mathematicians who were opposed to the axiom of choice (Lebesgue, Borel) actually used weak versions of it. You just don't notice that it's there.
• I guess one should distinguish between uses of countable choice, uses of dependent choice, and uses of full choice, as the first (and second) are quite intuitive and most of classical analysis lives there, while full choice tends to be applied a bit farther away. For example, one can use Hahn-Banach to prove facts about convex subsets of ${\mathbb R}^n$, but the version of Hahn-Banach one needs here can be proved essentially by hand, without needing a strong appeal to choice. Also in many applications of Zorn's lemma, only countable choice is used when the sets involved are countable. – Andrés E. Caicedo Mar 16 '12 at 14:32
• Andres, you are correct. I'll add some words on that. – Asaf Karagila Mar 16 '12 at 14:33
• I have added this question to my "favorites", for this answer. – The Chaz 2.0 Mar 16 '12 at 14:53
• I wonder whether the absence of Banach-Tarski from your list is a coincidence. Maybe we have a preference for cozy statements? – Marc van Leeuwen Mar 16 '12 at 15:37
• @Marc: We definitely prefer cozy statements. I could add that "You cannot partition a set into strictly more parts than elements!" too! :-) – Asaf Karagila Mar 16 '12 at 15:41
One high level (non-mathematical) reason for accepting the Axiom of Choice over, say, the Axiom of Determinacy is that the Axiom of Choice is somehow more "basic." At some level it is more natural to think that all Cartesian products of nonempty sets are nonempty. It doesn't quite seem as natural to think that the game $G_A$ is determined for all $A \subseteq \omega^\omega$.
At least, I think that most mathematicians have come to grips with the idea that there are badly behaved subsets of $\mathbb{R}$: there are sets which are not Lebesgue measurable; there are sets which do not have the Baire property; etc. Why shouldn't there be sets of reals whose associated game is undetermined?
Of course, 100 years ago (or thereabouts) some mathematicians were up in arms over the possibility of well-ordering $\mathbb{R}$, and Zermelo's theorem was considered by some as a great argument against Choice. That is to say, attitudes may change.
• It does seem natural, though, that De Morgan's law can be extended to infinitely many nested quantifiers (which is equivalent to AD). – user76284 Dec 18 '18 at 18:17
A practical reason is that certain mathematical theorems are known to be equivalent to the axiom of choice. In particular, the fact that every vector space has a basis and the theorem of Tikhonov that arbitrary products of compact spaces are compact require the axiom of choice.
More importantly, th axiom of choice is equivalent to th arbitrary product of nonempty sets being nonempty. This seems to be a consequence of our intuitive notion of set. So even though w seldom use the full axiom of choice, it seems to be part of the way of thining about sets.
• It's worth mentioning that the statement that the product of compact Hausdorff spaces is compact is strictly weaker: in fact it's equivalent to the ultrafilter lemma. And I think in practice the applications of Tychonoff's theorem that I've seen have been to compact Hausdorff spaces. – Qiaochu Yuan Mar 16 '12 at 16:32
• That is true. But I cannot imagine a person that dislikes AC because of its consequences, but thinks the ultrafilter lemma is acceptable. – Michael Greinecker Mar 16 '12 at 16:47
• It is also worth mentioning that Ultrafilter Lemma implies Bananach-Tarski, so if the consequence troubling someone is that one the Ultrafilter Lemma is too much choice indeed! – Asaf Karagila Mar 16 '12 at 21:44
It allows the use of nonprincipal ultrafilters, which are needed for nonstandard analysis. (This doesn't actually need the full strength of AC, but you can't do it in ZF.)
Another reason which might not be quite so accessible to you yet is that (a statement equivalent to) the axiom of choice is used to prove that every $R$-module admits an embedding into an injective $R$-module, which allows us to have a theory of sheaf cohomology in algebraic geometry. Sheaf cohomology is something which lots of people care about- not having choice around would present major problems for almost all of them. | 1,858 | 7,919 | {"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} | 3.109375 | 3 | CC-MAIN-2019-30 | longest | en | 0.931151 |
https://www.smartstudy.com/gmat/article/2765287.html | 1,606,837,170,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674594.59/warc/CC-MAIN-20201201135627-20201201165627-00614.warc.gz | 829,394,643 | 15,372 | 相信备考GMAT写作考试的同学，大家都听说过写作的七宗罪，但是对于具体的内容大家可能还不是很了解。之所以将这些内容总结起来，是因为很多考生写作低分的原因就出在这些方面，如果我们一开始备考的时候就了解这些内容，有效的避免，那么我们的写作提分会更容易。下面小编为大家整理了详细的内容，供大家参考!
一、什么是“七宗罪”
gmat写作七宗罪是指gmat驳论文Argument题目中存在的逻辑漏洞或错误，考生根据考试中常见的逻辑漏洞和考察方向把其总结为“七宗罪”，既形象又方便参加gmat考试备考的考生进行背诵。考生在备考时可以按照七宗罪中的逻辑错误类型进行有目的的查找，一方面节省了考试时间，另一方面确保找到的逻辑错误的准确，对于提升gmat写作效率也非常有帮助。
通常考生所说的gmat作文七宗罪指的就是：第一宗罪：无因果联系，第二宗罪：样本不足，第三宗罪：错误类比(横向)第四宗罪：时地全等(没有用发展的观点看待问题)，第五宗罪：二者择一(非此即彼)，第六宗罪：可疑调查，第七宗罪：结论无据(无根据假设)。可能个别地方的叫法不完全一样，考生只要知道其表达的真正意义即可。
二、七宗罪包含哪些内容
第一宗罪：无因果联系，是指作者给出的解释和得出的结论没有因果上的联系，或者是二者毫不相关。
The author commits a fallacy of causal oversimplification. The line of the reasoning is that because A occurred before B, the former event is responsible for the latter. But this is fallacious reasoning unless other possible causal explanations have been considered and ruled out. For example, perhaps C is the cause of these events or perhaps B is caused by D.
第二宗罪 样本不足，是指给出的论据或例子不充分，不足以说明某个问题，不能支撑作者得出这样的结论。
The evidence the author provides is insufficient to support the conclusion drawn from it. One example is logically unsounded to establish a general conclusion, unless it can be shown that A1 is representative of all A. It is possible that.... In fact, in face of such limited evidence, the conclusion that B is completely unwarranted.
第三宗罪: 错误类比，即把毫不相关的两个事物拿来作比较，并由此得出一些结论。
The argument rests on the assumption that A is analogous to B in all respects. This assumption is weak, since although there are points of comparison between A and B, there is much dissimilarity as well. For example, A..., however, B.... Thus, it is likely much more difficult for B to do....
第四宗罪 时地全等，主要是指没有用发展的观点看待问题，拿过去的事例和现在的作对比，并由此得出结论。
The author commits the fallacy of “all things are equal”. The fact that happened two years ago is not a sound evidence to draw a conclusion that.... The author assumes without justification that the background conditions have remained the same at different times or at different locations. However, it is not clear in this argument whether the current conditions at AA are the same as they used to be two years ago. Thus it is impossible to conclude that....
第五宗罪 二者择一，即限制了结论的范围，按照非此即彼的观点得出结论。
The author assumes that AA and BB are mutually exclusive alternatives and there is no room for a middle ground. However, the author provides no reason for imposing an either-or choice. Common sense tells us that adjusting both AA and BB might produce better results.
第六宗罪 可疑调查，指作者提供的证据是片面的或者是有限的，不能保证结论的顺利推出。
The poll cited by the author is too vague to be informative. The claim does not indicate who conducted the poll, who responded, or when, where and how the poll was conducted. Until these questions are answered, the results of the survey are worthless as evidence for the conclusion.
第七宗罪 结论无据，指得出的结论是毫无根据的，或者是前面的假设是不能成立的。
The author falsely depends on gratuitous assumption that.... However, no evidence is stated in the argument to support this assumption. In fact, this is not necessarily the case. For example, it is more likely that.... Therefore, this argument is unwarranted without ruling out such possibility。
以上就是关于“GMAT作文七宗罪内容分析”的内容，希望通过上述内容的学习，大家能够更好的来备考GMAT写作考试，预祝各位考生能够拿到高分作文分数。
备考推荐： | 1,125 | 3,191 | {"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} | 3.4375 | 3 | CC-MAIN-2020-50 | latest | en | 0.641184 |
https://www.coursehero.com/file/6455014/ws19/ | 1,496,140,387,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463614620.98/warc/CC-MAIN-20170530085905-20170530105905-00222.warc.gz | 1,079,563,235 | 23,070 | # ws19 - Math 464 Worksheet 19 A population has logistic...
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Unformatted text preview: Math 464, Worksheet 19 A population has logistic growth if its difference equation model is ∆R = kR 1 − R N here k is the growth rate and N is the stable equilibrium population. (N is also called the carrying capacity.) In this worksheet you will build a model of two populations, rabbits R and foxes F , such that • If F = 0, then then R has logistic growth with k = 0.02 and N = 100. • If F = 0, then 0.002RF rabbits are eaten by foxes in each time step. • Foxes have a constant death rate d = 0.01 per fox per time step. • Foxes have a birth rate of 0.0004R per fox per time step. 1. Write difference equations for R and F . 2. Assume F = 0. Then sketch equilbiria and solutions for possible populations of R. Do this on the horizontal axis of an R-F coordinate system. 3. Assume R = 0 and do a similar sketch on the F -axis. 4. Find all other equilibria for this model. 5. For each equilibrium, linearize, compute eigenvalues, and if eigenvalues are real, find eigenvectors. 6. Try to sketch the solution curve that starts with initial populations R(0) = 2 and F (0) = 10. 7. Try again with R(0) = 150 and F (0) = 50. 1 ...
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## This note was uploaded on 10/12/2011 for the course MATH 464 taught by Professor Dougbullock during the Fall '08 term at Boise State.
Ask a homework question - tutors are online | 436 | 1,552 | {"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} | 3.265625 | 3 | CC-MAIN-2017-22 | longest | en | 0.900442 |
http://mathhelpforum.com/calculus/145701-integral-involving-logarithm-square-root.html | 1,503,320,508,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108268.39/warc/CC-MAIN-20170821114342-20170821134342-00524.warc.gz | 279,673,894 | 12,999 | # Thread: Integral involving logarithm and square root
1. ## Integral involving logarithm and square root
Hey I'm trying to solve the following integrals:
$
\int x^{i}\log(x+k+m\sqrt{x^2+n})dx,
$
for $i=0,1,2$ and suitalbe constants $k, m,$ and $n$.
I tried to find an anti-derivative via integration by parts - with no success.
Does anyone else have an idea?
Thanks a lot in any case,
DeeAna
2. I think your only option is approximation.
3. Thanks Anonymous1,
I thought so, too. Mathematica, however, was able to give me an antiderivative, which I don't completely trust. It is pretty long and also plugging in the integration command several times gave me different solutions. Then I started testing the Mathematica "anti-derivative" by differentiating and then trying to simplify, which didn't work. Then I just checked a couple of values, comparing the derivative of the "anti-derivative" with the integrand above. They were more or less the same. Lastly, I computed numerically a definite integral of the above and compared it to what I obtain when plugging in the boundaries in the Mathematica anti-derivative and subtracting. This, however, yields only in about 80 percent of the cases the same results, which is rather confusing...
But if Mathematica can give me a solution, then there should be a way to do this. Don't you think?
4. Originally Posted by DeeAna
Thanks Anonymous1,
I thought so, too. Mathematica, however, was able to give me an antiderivative, which I don't completely trust. It is pretty long and also plugging in the integration command several times gave me different solutions. Then I started testing the Mathematica "anti-derivative" by differentiating and then trying to simplify, which didn't work. Then I just checked a couple of values, comparing the derivative of the "anti-derivative" with the integrand above. They were more or less the same. Lastly, I computed numerically a definite integral of the above and compared it to what I obtain when plugging in the boundaries in the Mathematica anti-derivative and subtracting. This, however, yields only in about 80 percent of the cases the same results, which is rather confusing...
But if Mathematica can give me a solution, then there should be a way to do this. Don't you think?
You may be running into branch-cut problems with the log and square root functions in the antiderivative which might cause different answers depending on what the values of k, m, n, i are. Mathematica always uses the principal-value for a multi-valued function and that may not be the correct "analytically-extended" choice for the value of the antiderivative at the limit points. What values of k, m, n, i and the upper and lower limits give a value different than the numerical result?
5. Hey shawsend,
thanks for your explanation. I wasn't quite aware of what Mathematica does. For the definite integral, I obtained for the following choices of i,xmax, xmin, k,m and n different solutions when integrating numerically compared to the result obtained with the help of the Mathematica anti-derivative:
1) i=2:
xmax = - 0.5070..
xmin = 0.9071...
k = 0.3050...
m = 1.4142...
n = 0.5930...
2) i=2:
xmax = - 0.9071...
xmin = 0.5071...
k = - 0.3050
m = 1.4142...
n = 0.5930
3) i=1:
xmax = - 1.5071...
xmin = -0.0928...
k = -0.6949...
m = 1.4142...
n = 0.9829...
- whereas for example for all i
xmax = 0.0428..
xmin = 1.4571...
k = 0.6035...
m = 1.4142...
n = 0.8642...
or
xmax = - 0.9571...
xmin = 0.4571..
k = - 0.3964
m = 1.4142...
n = 0.6571...
yielded the correct result - or a result very close to the one obtained by numerical integration. I hope I didn't mix up the numbers and also I noticed that my xmax always seems to be smaller than my xmin.
Speaking of approximating the integral: What method do you think would be appropriate? I tried Gauss-Chebyshev quadrature. The convergence rate was pretty slow. In order to obtain an exactness of ~10e-6, I needed at least nine iteration steps. When calculting many integrals the speed is not even close to Matlab's quad-command. Is there an obvious explanation for that? What other method could be useful in this context?
Many thanks,
DeeAna
6. Let's just look at the first one for i=2 and arbitrary k,m, and n:
myanti = Integrate[
x^2*Log[x + k + m*Sqrt[x^2 + n]], x];
Now plot the real part of that between the limits:
Plot[Re[myanti /. {k -> 0.305,
m -> 1.4142, n -> 0.593}],
{x, -0.507, 0.9071}, PlotRange -> All]
Note in the plot below how there is a jump discontinuity in the plot. That is because Mathematica is using the principal values for any multi-valued function in the antiderivative and the limits in this particular case passes through a principal branch-cut. In order to evaluate an integral of a multi-valued function using its antiderivative, we must use it's analytic-extension. For example, if we were to analytically extend the top curve along it's underlying Riemann surface to the point x=0.9072 , then that would represent the analytic extension and we could then take the difference of it's values at the end points to compute the integral. However, that's tough to implement in the general sense without quite a bit of computational effort. This is a Mathematica limitation and the reason why some of your integrals agree and some do not is probably because in some cases, your limits do not pass through a principal branch-cut.
And in regards to your questions about numerical integration techniques, I'm not an expert in that. However, Mathematica has an extensive help section on "Advanced Integration Techniques" that would probably help you. You can get to it from the Integrate help page.
7. Thanks shawsend, you really helped me! I will need to do some background reading on "Riemann surfaces" and "analytical extension" along them - but at least I know now where the problems are;-) | 1,482 | 5,871 | {"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": 4, "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-2017-34 | longest | en | 0.97063 |
https://www.transtutors.com/questions/16-18-d-6-240-how-much-money-do-you-need-today-to-ensure-that-you-will-have-12-000-i-5417993.htm | 1,619,167,987,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00271.warc.gz | 1,121,678,418 | 12,188 | # 16-18 d. $6,240. How much money do you need today to ensure that you will have$12,000 in 3 years, a 1 answer below »
16-18
d. $6,240. How much money do you need today to ensure that you will have$12,000 in 3 years, assuming u can earn 4% on your savings? a. $10,668. b.$11,224. c. $11,668. d.$12,668. How much money do you need today to ensure that you will have $16,000 in 4 years, assum earn 6% on your savings? a.$10,674. b. $11,274. c.$11,674. d. $12,674. How much money will you accumulate in your retirement account if you save$5,00 and earn 6% on your investments? $101,668.$111,224. \$116,380.
Dil A
Solution:
16:
Option A is correct.
FV = 12000
N = 3
Looking for Something Else? Ask a Similar Question | 242 | 720 | {"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} | 3.78125 | 4 | CC-MAIN-2021-17 | latest | en | 0.923795 |
http://www.jiskha.com/display.cgi?id=1366212661 | 1,498,555,661,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321306.65/warc/CC-MAIN-20170627083142-20170627103142-00584.warc.gz | 567,632,959 | 3,899 | # chemistry
posted by .
A 1.147 gram sample containing an unknown amount of arsenic trichloride and the rest inerts was dissolved into a NaHCO3 and HCl aqueous solution. To this solution was added 1.690 grams of KI and 50.00 mL of a 0.00821 M KIO3 solution. The excess I3– was titrated with 50.00 mL of a 0.02000 M Na2S2O3 solution. What was the mass percent of arsenic trichloride in the original sample?
• chemistry -
Note: I don't use I^3-; it's the I2 that does the job.
As^3+ + I2 ==> 2I^- + As^5+
Then mols I2 initially = 0.050L x 0.0821 = 0.004105
mols I2 titrated with S2O3^2- = 0.050 x 0.02 = 0.001 which means 0.0005 mols I2 were used.
I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
mols I2 used for the As/I2 rxn = 0.004105-0.0005 = 0.003605
Then g As = mols As x atomic mass As
%As = (g As/g sample)*100 = ? | 313 | 810 | {"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} | 2.625 | 3 | CC-MAIN-2017-26 | latest | en | 0.911124 |
https://www.lmfdb.org/L/2/800/20.7/c1/0/15 | 1,726,559,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00204.warc.gz | 805,765,656 | 7,393 | # Properties
Label 2-800-20.7-c1-0-15 Degree $2$ Conductor $800$ Sign $-0.525 + 0.850i$ Analytic cond. $6.38803$ Root an. cond. $2.52745$ Motivic weight $1$ Arithmetic yes Rational no Primitive yes Self-dual no Analytic rank $0$
# Related objects
## Dirichlet series
L(s) = 1 + (−1 − i)3-s + (3 − 3i)7-s − i·9-s − 2i·11-s + (−3 + 3i)13-s + (−1 − i)17-s + 4·19-s − 6·21-s + (1 + i)23-s + (−4 + 4i)27-s − 10i·31-s + (−2 + 2i)33-s + (1 + i)37-s + 6·39-s − 10·41-s + ⋯
L(s) = 1 + (−0.577 − 0.577i)3-s + (1.13 − 1.13i)7-s − 0.333i·9-s − 0.603i·11-s + (−0.832 + 0.832i)13-s + (−0.242 − 0.242i)17-s + 0.917·19-s − 1.30·21-s + (0.208 + 0.208i)23-s + (−0.769 + 0.769i)27-s − 1.79i·31-s + (−0.348 + 0.348i)33-s + (0.164 + 0.164i)37-s + 0.960·39-s − 1.56·41-s + ⋯
## Functional equation
\begin{aligned}\Lambda(s)=\mathstrut & 800 ^{s/2} \, \Gamma_{\C}(s) \, L(s)\cr =\mathstrut & (-0.525 + 0.850i)\, \overline{\Lambda}(2-s) \end{aligned}
\begin{aligned}\Lambda(s)=\mathstrut & 800 ^{s/2} \, \Gamma_{\C}(s+1/2) \, L(s)\cr =\mathstrut & (-0.525 + 0.850i)\, \overline{\Lambda}(1-s) \end{aligned}
## Invariants
Degree: $$2$$ Conductor: $$800$$ = $$2^{5} \cdot 5^{2}$$ Sign: $-0.525 + 0.850i$ Analytic conductor: $$6.38803$$ Root analytic conductor: $$2.52745$$ Motivic weight: $$1$$ Rational: no Arithmetic: yes Character: $\chi_{800} (607, \cdot )$ Primitive: yes Self-dual: no Analytic rank: $$0$$ Selberg data: $$(2,\ 800,\ (\ :1/2),\ -0.525 + 0.850i)$$
## Particular Values
$$L(1)$$ $$\approx$$ $$0.573689 - 1.02897i$$ $$L(\frac12)$$ $$\approx$$ $$0.573689 - 1.02897i$$ $$L(\frac{3}{2})$$ not available $$L(1)$$ not available
## Euler product
$$L(s) = \displaystyle \prod_{p} F_p(p^{-s})^{-1}$$
$p$$F_p(T)$
bad2 $$1$$
5 $$1$$
good3 $$1 + (1 + i)T + 3iT^{2}$$
7 $$1 + (-3 + 3i)T - 7iT^{2}$$
11 $$1 + 2iT - 11T^{2}$$
13 $$1 + (3 - 3i)T - 13iT^{2}$$
17 $$1 + (1 + i)T + 17iT^{2}$$
19 $$1 - 4T + 19T^{2}$$
23 $$1 + (-1 - i)T + 23iT^{2}$$
29 $$1 - 29T^{2}$$
31 $$1 + 10iT - 31T^{2}$$
37 $$1 + (-1 - i)T + 37iT^{2}$$
41 $$1 + 10T + 41T^{2}$$
43 $$1 + (5 + 5i)T + 43iT^{2}$$
47 $$1 + (-3 + 3i)T - 47iT^{2}$$
53 $$1 + (-5 + 5i)T - 53iT^{2}$$
59 $$1 + 12T + 59T^{2}$$
61 $$1 - 2T + 61T^{2}$$
67 $$1 + (-1 + i)T - 67iT^{2}$$
71 $$1 + 2iT - 71T^{2}$$
73 $$1 + (1 - i)T - 73iT^{2}$$
79 $$1 - 8T + 79T^{2}$$
83 $$1 + (5 + 5i)T + 83iT^{2}$$
89 $$1 - 16iT - 89T^{2}$$
97 $$1 + (-3 - 3i)T + 97iT^{2}$$
show less
$$L(s) = \displaystyle\prod_p \ \prod_{j=1}^{2} (1 - \alpha_{j,p}\, p^{-s})^{-1}$$
## Imaginary part of the first few zeros on the critical line
−10.04601612046077669039541069461, −9.177523187138729099815204687783, −8.043369381561794008056577988471, −7.29577776853698128231862495224, −6.68522468942190639094060403261, −5.51986834619945442717582828569, −4.63355241801052051179115717180, −3.58560007532189503648850603494, −1.86629275310895654054021106238, −0.64506924347125637965437535748, 1.78704947084357478927538519404, 2.98583320373619828613211231407, 4.70724086947760161543118487461, 5.04279350927061140336413948223, 5.84321312207294119439341714655, 7.20776009761738679295576243310, 8.055300510133531601149933734628, 8.845966807037717208827539248878, 9.881633541073524618857510399796, 10.53801898130760231602550792017 | 1,572 | 3,232 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-38 | latest | en | 0.386609 |
https://dasarpai.com/dsblog/maths-for-ds | 1,695,349,076,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506320.28/warc/CC-MAIN-20230922002008-20230922032008-00760.warc.gz | 234,389,663 | 7,468 | # Mathematics for Data Scientist
To excel in the field of data science, especially as a data scientist, I would recommend you have good command over the topics mentioned below. These are the topics from mathematics and statistics. There are many YouTube channels that you can use for this purpose. Because this is 10+2 level mathematics, and it is just a matter of revision. So I am not offering any course unless there is a specific need for some group, organization.
## Linear Algebra
1. Introduction to Linear Algebra
2. Eigenvalues And Eigenvectors
3. Calculating Eigenvalues and Eigenvectors
4. Eigen decomposition of a Matrix
5. Eigenvectors: What Are They? Intuition behind.
## Vectors, Matrices & Linear Transformations
** Vector & Vector Spaces **
1. Vectors: The Basics
2. Basis Vector
3. Norm of a vector
4. Identity matrix or operator
5. Determinant of a matrix
6. Column and Null Space
7. Rank of a matrix
8. Transpose of a matrix
9. Inverse of a matrix
10. Least Squares Approximation
11. Linear Transformations
12. Matrices: The Basics
13. Matrix Operations
14. Matrix operations and manipulations
15. Dot product of two vectors
16. Linear independence of vectors
## Multivariable Calculus
1. Critical Points, Maxima and Minima
2. Differentiation
3. Functions and Derivatives
4. Functions: Primer
5. Multivariable Functions
6. Partial Derivatives
7. Taylor Series and Linearization
8. The Hessian
9. The Jacobian
10. Vector-Valued Functions
## Probability
1. Introduction to probability – probability, events, additive & multiplicative rule
2. Basics of probability – random variables, probability distribution, expected value
3. Joint and Conditional Probability
4. Probability Rules
5. Bayes’ Theorem
## Statististics
1. Descriptive statistics
2. Inferential Statistics
3. Prescriptive statistics
4. What is sampling, different sampling techniques?
5. Random Variable, Predictor, Predicted variables
6. Data Distribution (continuous, discrete, Normal/Bernoulli, standard, binomial, Poisson, etc.)
7. CDF (Cumulative Distribution Function), PDF (Probability Distribution Function)
8. Statistical Measures (mean, mode, median, max, min)
9. Measure of dispersion (range, standard deviation, variance, covariance, correlation, error deviation)
10. Central Limit Theorem (CLT)
11. What is Regression? How it works? OLS (Ordinary Least Square), Multi-linear regression.
12. Standard Error
13. Dimensionality Reduction (PCA)
14. Parameter Properties (Bias, Consistency, Efficiency)
15. Statistical tests t-test, z-test, ANOVA test, Chi-Square test
16. Conditional Probability (Bayesian Theorem)
17. Type I/Type II errors
18. Hypothesis testing
19. Confidence Interval & Significance Level (alpha)
20. p-value and its interpretation
Tags:
Updated: | 662 | 2,770 | {"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} | 3.890625 | 4 | CC-MAIN-2023-40 | longest | en | 0.758602 |
https://www.reference.com/technology/uses-logic-gates-f0e25b68ce080ddf | 1,571,816,222,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829507.97/warc/CC-MAIN-20191023071040-20191023094540-00128.warc.gz | 1,010,202,653 | 18,433 | # What Are the Uses of Logic Gates?
Logic gates are digital components that typically work two levels of voltage and determine how a component conducts electricity. Logic gates use Boolean equations and switch tables.
Logic gates are used to build both digital and complex integrated circuits. There are seven types of gates:
• NOT or inverter gates: single bit input and single bit output
• AND gates: have at least two bits of input and work by a multiplication function
• OR gates: have at least two bits input and work by an addition function
• NAND gates: an AND gate with inverter attached
• NOR gates: an OR gate with inverter attached
• XOR gates: represents an exclusive OR gate
• XNOR gates: represents an exclusive NOR gate with an inverted output
Similar Articles | 161 | 778 | {"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-2019-43 | latest | en | 0.910101 |
https://socratic.org/questions/how-do-you-solve-y-2x-4-and-7x-5y-14-using-substitution | 1,579,319,204,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250591763.20/warc/CC-MAIN-20200118023429-20200118051429-00207.warc.gz | 683,499,108 | 6,146 | # How do you solve y = 2x – 4 and 7x – 5y = 14 using substitution?
Jul 31, 2016
x=2; y=0
#### Explanation:
Since $y = 2 x - 4$, let's substitute the expression $2 x - 4$ in y in the second equation:
$7 x - 5 \left(2 x - 4\right) = 14$
$7 x - 10 x + 20 = 14$
$- 3 x = 14 - 20$
$- 3 x = - 6$
$x = 2$
Now let's substitute the value 2 in x in the first equation:
$y = 2 \left(2\right) - 4 = 0$ | 173 | 400 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "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} | 4.53125 | 5 | CC-MAIN-2020-05 | longest | en | 0.71293 |
https://acqnotes.com/acqnote/careerfields/link-budget | 1,680,309,341,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00200.warc.gz | 105,082,891 | 32,606 | Space Acquisitions
A Link Budget shows all of the gains and losses from a transmitter, through the medium (free space, cable, waveguide, fiber, etc.) to the receiver in a telecommunication system. It’s used to predict the performance of a transmitter and receiver communication link to show in advance if its performance is acceptable, or if one option is better than another. It accounts for the attenuation of the transmitted signal due to propagation, as well as the antenna gains, feedline, miscellaneous losses and added margin. [1]
A simple link budget equation looks like this:
Received Power (dBm) = Transmitted Power (dBm) + Gains (dB) − Losses (dB)
The Information that is needed to perform a Link Budget include:
• The saturated EIRP and saturated flux density of the transponder.
• Satellite transponder bandwidth.
• Satellite transponder output backoff or attenuation.
• Satellite transponder input backoff or attenuation.
You will also need the following information that you and your customer can supply:
• Planned data or information rate.
• Modulation type (BPSK or QPSK)
• Forward error correction rate (1/2 or 3/4)
• Spread Factor – if any (use only for spread spectrum systems)
• Minimum digital signal strength (EB/No) for desired Bit Error Rate (BER) performance.
*The above information can generally be obtained from the satellite operator and customer.
Often link budget equations can become messy and complex, so there have evolved some standard practices to simplify the link budget equation
• The wavelength term is often considered part of the free space loss equation. This complexity reduction is acceptable for terrestrial communication systems, where only line of sight is considered.
• Considering all carrier wave propagation to be wavelength-independent. This is justified by the conservation of energy law that requires that the electric field decrease in power as the square of the distance regardless of frequency (in free space propagation conditions).
1. Has anything significant been left out?
1. This only comes with experience, yours or someone you trust.
2. Are any values significantly different to other similar budgets?
1. If so, why?
2. Is the difference explained?
3. Remember the creator must often guess at knowledge level of the audience.
1. Be willing to ask questions.
AcqTips:
• Link Budget information can generally be obtained from the satellite operator and customer. | 490 | 2,438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-14 | longest | en | 0.887441 |
http://onecrossradiopodcast.com/forum/page.php?9296f2=6-feet-in-meters | 1,643,182,211,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304928.27/warc/CC-MAIN-20220126071320-20220126101320-00677.warc.gz | 48,034,510 | 15,487 | Convert 6 Feet to Meters. In 1959 the international yard and pound agreement (between the United States and countries of the Commonwealth of Nations) defined a yard as being exactly 0.9144 metres, which in turn defined the foot as being exactly 0.3048 metres (304.8 mm). In this case we should multiply 6 Feet by 0.3048 to get the equivalent result in Meters: 6 Feet x 0.3048 = 1.8288 Meters. It is equal to 0.3048 m, and used in the imperial system of units and United States customary units. For a more accurate answer please select 'decimal' from the options above the result. The meter (symbol m) is the fundamental unit of length in the International System of Units (SI). Six Feet is equivalent to one point eight two nine Meters. 6 ft to m …
If you spot an error on this site, we would be grateful if you could report it to us by using the contact link at the top of this page and we will endeavour to correct it as soon as possible. Feet. The unit of foot derived from the human foot.
In 1799, France start using the metric system, and that is the first country using the metric. 6 Feet is equivalent to 1.8288 Meters. How long is 6 feet? Note: Fractional results are rounded to the nearest 1/64. Our full terms & conditions can be found by clicking here. As the base unit of length in the SI and other m.k.s.
A foot (symbol: ft) is a unit of length. It is equal to 0.3048 m, and used in the imperial system of units and United States customary units. It is defined as "the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second." In this case we should multiply 6 Feet by 0.3048 to get the equivalent result in Meters: The conversion factor from Feet to Meters is 0.3048. systems (based around metres, kilograms and seconds) the metre is used to help derive other units of measurement such as the newton, for force.
1.8288 Meters (m) Feet : A foot (symbol: ft) is a unit of length. How far is 6 feet in meters? The metre is a unit of length in the metric system, and is the base unit of length in the International System of Units (SI).
In 1959 the international yard and pound agreement (between the United States and countries of the Commonwealth of Nations) defined a yard as being exactly 0.9144 metres, which in turn defined the foot as being exactly 0.3048 metres (304.8 mm). To calculate 6 Feet to the corresponding value in Meters, multiply the quantity in Feet by 0.3048 (conversion factor). Feet to Meters formula You are currently converting Distance and Length units from Feet to Meters. To find out how many Feet in Meters, multiply by the conversion factor or use the Length converter above. It is defined as "the length of the path travelled by light in vacuum during a time interval of 1/299,792,458 of a second." The conversion factor from feet to meters is 0.3048, which means that 1 foot is equal to 0.3048 meters: 1 ft = 0.3048 m. To convert 6.6 feet into meters we have to multiply 6.6 by the conversion factor in order to get the length amount from feet to meters. The unit of foot derived from the human foot. This site is owned and maintained by Wight Hat Ltd. ©2003-2020. Note: You can increase or decrease the accuracy of this answer by selecting the number of significant figures required from the options above the result. Feet : 6 Feet (ft) =. Feet to Meters - Distance and Length - Conversion. It is subdivided into 12 inches. Meters : Note: For a pure decimal result please select 'decimal' from the options above the result. A foot (symbol: ft) is a unit of length. In 1799, France start using the metric system, and that is the first country using the metric. It is equal to 0.3048 m, and used in the imperial system of units and United States customary units.
Conversion formula. It is subdivided into 12 inches. Whilst every effort has been made to ensure the accuracy of the metric calculators and charts given on this site, we cannot make a guarantee or be held responsible for any errors that have been made. The meter (symbol: m) is the fundamental unit of length in the International System of Units (SI).
To calculate 6 Feet to the corresponding value in Meters, multiply the quantity in Feet by 0.3048 (conversion factor). | 1,031 | 4,223 | {"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-2022-05 | latest | en | 0.922888 |
http://www.math.union.edu/~dpvc/courses/2017-18/MTH115H-FA17/assignments/06-1.html | 1,544,501,801,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823565.27/warc/CC-MAIN-20181211040413-20181211061913-00149.warc.gz | 414,053,414 | 1,832 | Math 115H (Assignments)
# Homework on 9 October 2017:
[Not to be turned in]
1. For what value(s) of $t$ is $\left<3t-t^3,2t^2\right>$ a vertical vector?
2. For what value(s) of $t$ is $\left<1-\cos t,\sqrt3+\sin t\right>$ the longest? What is the greatest length, and what vector(s) achieve it?
[Hint: It might make it easier to note that since the length is never negative, it will be longest when its square is longest. Note also that since the length is a number that depends on $t$, it is a single-variable function, so you can use single-variable calculus to determine the greatest length. Finally, don't forget that $\tan(t)={\sin(t)\over\cos(t)}$.]
Math 115H (Fall 2017) web pages Created: 9 Oct 2017 Last modified: Oct 9, 2017 4:59:20 PM Comments to: dpvc@union.edu | 233 | 779 | {"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} | 3.4375 | 3 | CC-MAIN-2018-51 | latest | en | 0.896474 |
https://fr.mathworks.com/matlabcentral/answers/583214-how-can-i-write-a-matlab-code-on-digital-signals-processing | 1,716,243,310,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00889.warc.gz | 228,982,135 | 27,225 | # How can I write a Matlab code on Digital Signals Processing ?
6 vues (au cours des 30 derniers jours)
Jone Erikson le 23 Août 2020
Commenté : Rena Berman le 12 Oct 2020
How can I write a MATLAB function (not a script!) to generate a periodic waveform of total length L. Each period must be a pulse of amplitude A that lasts a total ofM samples followed by T−M samples that are zero so that the overall period is T. The result should be a squarewave. Could you please help with this code, with a brief explaination of the code
##### 2 commentairesAfficher AucuneMasquer Aucune
Stephen23 le 26 Août 2020
Modifié(e) : Stephen23 le 26 Août 2020
"How can I write a Matlab code on Digital Signals Processing ?"
How can I write a MATLAB function (not a script!) to generate a periodic waveform of total length L. Each period must be a pulse of amplitude A that lasts a total ofM samples followed by T−M samples that are zero so that the overall period is T. The result should be a squarewave. Could you please help with this code, with a brief explaination of the code
Rena Berman le 12 Oct 2020
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### Réponse acceptée
Thiago Henrique Gomes Lobato le 23 Août 2020
This should work for you, the code is almost self explanatory:
L = 1024;
Periods = 4;
M = 128;
A = 1;
figure,plot( squareWave(L,M,Periods,A) )
function signal = squareWave(L,M,Periods,A)
signal = zeros(L,1); %initialize signal with zeros
if mod(L,Periods) ~= 0
signal = -1; % False input data
end
T = L/Periods; % Get length
% Replace only non-zero values
for idx=1:Periods
signal( 1+(idx-1)*T:1+(idx-1)*T+M) = A;
end
end
##### 2 commentairesAfficher AucuneMasquer Aucune
Thiago Henrique Gomes Lobato le 23 Août 2020
Which error do you become? Here it works fine. Remember that you need to save it to a file and run, and not just evaluate it (F9).
Stephen23 le 26 Août 2020
The code above is not running Thigao
>> squareWave
Error: File: squareWave.m Line: 14 Column: 30
Function definitions are not permitted in this context.
Line 14: signal(1+(idx-1)*T:1+(idx-1)*T+M)= A;
Also, why did you assume these values for: L, M, A, and Periods?
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Translated by | 681 | 2,395 | {"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} | 3.078125 | 3 | CC-MAIN-2024-22 | latest | en | 0.644379 |
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Apr 25 comment Why doesn't L'Hospital's rule work for this limit? The rule (for this particular indeterminate form) says: If $f(x)\to \infty$ and $g(x)\to \infty$ and $\lim\frac{f'(x)}{g'(x)}$ exists, then $\lim\frac{f'x)}{g(x)}$ exists and equals $\lim\frac{f'(x)}{g'(x)}$. Apr 25 comment The sum of two numbers is 5/9… I read the last instruction as asking for $(0.4 x)\cdot (0.4y)$ Apr 25 answered Does there exist a complex function which is differentiable at one point and nowhere else continuous? Apr 25 comment Has the polynomial distinct roots? How can I prove it? @drxy I found my counterexample by just playing around with $\gcd(f,f')$, motivated by mathguy's solution for real roots. I'm sure that playing around will also allow finding counterexamples (regarding complex roots) for higher $p$ values one by one - but the handling becomes impractical. Apr 25 comment Prisoners and hats variation Do they guess only once? OR is the game repeted (with the same hats) if nobody dared to guess? Apr 25 comment $\sup\limits_{t>0}[\frac{g(t)}{\sup\limits_{t \beta$ or not? deleted 1 character in body Apr 25 comment The $\cos(\alpha-\beta)$ formula always need $\alpha > \beta$ or not? @user3270418 Where do I show the sum formula? I read $\cos(\alpha-\beta)$ in my post. Apr 25 answered Significance of derivative in finding square free decomposition Apr 25 answered Deterministic finite automaton parity bit question Apr 25 comment The $\cos(\alpha-\beta)$ formula always need $\alpha > \beta$ or not? Are you sure your argument leads to a "No"? - Oh, you seemingly answered the title question, which is the opposite of the body question ... Apr 25 comment The $\cos(\alpha-\beta)$ formula always need $\alpha > \beta$ or not? @user3270418 No, I mean what I wrote. For $\alpha=\beta=60^\circ$, we have $\cos(60^\circ)^2+\sin(60^\circ)^2=(1/2)^2+(\sqrt 3/2)^2=1/4+3/4=1$ Apr 25 answered The $\cos(\alpha-\beta)$ formula always need $\alpha > \beta$ or not? Apr 24 answered Show that there is a step function $g$ over $[a,b]$ Apr 24 comment Is there a way to write an infinite set that contains only irrational numbers without integer multiples? In the same spirit as the last, $\{\,q\sqrt 2\mid q\in \Bbb Q\cap [1,2)\,\}$ Apr 24 comment The maximum of the absolute value of a real-valued function You could simply use that $\sup\bigcup_{i\in I}A_i =\sup_{i\in I}sup A_i$ Apr 24 answered Probability of seeing a headlight getting switched on | 708 | 2,496 | {"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} | 3.0625 | 3 | CC-MAIN-2016-18 | latest | en | 0.876533 |
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# Fuzzy K-modes clustering how to find the cluster centers
I'm trying to understand fuzzy k-modes algorithm (look mainly at page 3) in order to implement it. I'm stuck at the calculation of cluster centers they said as shown in the pic
I need to know whether the following is true or false and please correct me
In order to get the center of a cluster we need to:
1. for each category in a variable calculate the sum of membership value for all the point, that the category belongs to, to this cluster
2. the highest obtained value to be set as the category of the center
-
Wl-i is the membership value of the point i to the cluster l. points will have this structure A(1000,001011,000000001) 1000 means the first variable of A has the first category on ... B(0110,100000,101010000). suppose we're working on the 1st cluster so and we knew that W1-A=0.7 and W1-B=0.6 So will we take for each category the membership of the whole point? and make the sum for each point and chose the highest??? – – Mariya Oct 19 '11 at 13:45 | 301 | 1,230 | {"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-2016-30 | latest | en | 0.925033 |
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# Faced with an estimated \$ 2 billion budget gap, the city's
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Faced with an estimated \$ 2 billion budget gap, the city's [#permalink]
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29 Sep 2005, 03:10
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Faced with an estimated \$ 2 billion budget gap, the city's mayor proposed a nearly 17 percent reduction in the amount allocated the previous year to maintain the city's major cultural institutions and to subsidize hundreds of local arts groups.
A. proposed a nearly 17 percent reduction in the amount allocated the previous year to maintain the city's major cultural institutions and to subsidize hundreds of local arts gourps.
B. proposed a reduction from the previous year of nearly 17 percent in the amount it was allocating to maintain the city's major cultural institutions and for subsidizing.
C. proposed to reduce, by nearly 17 percent, the amount from the previous year that was allocated for the maintenance of the city's major cultural institutions and to subsidize.
D. has proposed a reduction from the previous year of nearly 17 percent of the amount it was allocated for maintaining the city's major cultural institutions and to subsidize
E. was proposing that the amount they were allocating be reduced by nearly 17 percent from the previous year for maintaining the city's major cultural institutions and for the subsidization.
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29 Sep 2005, 03:22
Faced with an estimated \$ 2 billion budget gap, the city's mayor proposed a nearly 17 percent reduction in the amount allocated the previous year to maintain the city's major cultural institutions and to subsidize hundreds of local arts groups.
I'll stick with A.
IMO A uses correct parallellism.
A. proposed a nearly 17 percent reduction in the amount allocated the previous year to maintain the city's major cultural institutions and to subsidize hundreds of local arts gourps.
B, C and D are not is not parallell.
In E the pronoun 'they' is not correct.
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29 Sep 2005, 03:51
in that case, even C is parallel. consider this:
C. proposed to reduce, by nearly 17 percent, the amount from the previous year that was allocated for the maintenance of the city's major cultural institutions and to subsidize.
IMO, "reducing the budget amount" and "allocating the excess to XXXX" shud be prarallel.
KRishna
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Re: SC budget gap [#permalink]
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29 Sep 2005, 07:47
A. proposed a nearly 17 percent reduction in the amount allocated the previous year to maintain the city's major cultural institutions and to subsidize hundreds of local arts gourps.
>> No. Just doesn't sound right, too awkward. Although everything else is grammatically correct.
B. proposed a reduction from the previous year of nearly 17 percent in the amount it was allocating to maintain the city's major cultural institutions and for subsidizing.
>> No. Not parallel.
C. proposed to reduce, by nearly 17 percent, the amount from the previous year that was allocated for the maintenance of the city's major cultural institutions and to subsidize.
>> Yes, IMO. Not as awkward as A, tense is correct and is parallel.
D. has proposed a reduction from the previous year of nearly 17 percent of the amount it was allocated for maintaining the city's major cultural institutions and to subsidize
>> No. Same as B, not parallel.
E. was proposing that the amount they were allocating be reduced by nearly 17 percent from the previous year for maintaining the city's major cultural institutions and for the subsidization
>> No. Incorrect tense and parallelism.
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29 Sep 2005, 08:06
A IMO, C is wrong because it sounds like he is reducing the amount from previous year...
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Re: SC budget gap [#permalink]
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30 Sep 2005, 03:56
I go with C, as it seems to be parallel.
What is the OA?
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30 Sep 2005, 04:46
Hi laxieqv,
can you post the OA/OE for this?
KRishna
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Re: SC budget gap [#permalink]
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30 Sep 2005, 06:34
laxieqv wrote:
Faced with an estimated \$ 2 billion budget gap, the city's mayor proposed a nearly 17 percent reduction in the amount allocated the previous year to maintain the city's major cultural institutions and to subsidize hundreds of local arts groups.
A. proposed a nearly 17 percent reduction in the amount allocated the previous year to maintain the city's major cultural institutions and to subsidize hundreds of local arts gourps.
B. proposed a reduction from the previous year of nearly 17 percent in the amount it was allocating to maintain the city's major cultural institutions and for subsidizing.
C. proposed to reduce, by nearly 17 percent, the amount from the previous year that was allocated for the maintenance of the city's major cultural institutions and to subsidize.
D. has proposed a reduction from the previous year of nearly 17 percent of the amount it was allocated for maintaining the city's major cultural institutions and to subsidize
E. was proposing that the amount they were allocating be reduced by nearly 17 percent from the previous year for maintaining the city's major cultural institutions and for the subsidization.
Answer is definately A - this choice makes parallel the infinitives "to maintain" and "to subsidize".
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05 Oct 2005, 01:51
Sorry to all for my late OA due to my vacation here
Yeah, congrats to all as you all picked A~!!!
OE is indeed:
The construction the amount allocated ...to maintain...and to subsidize is parallel, while the phrase a nearly 17 percent reduction in the amount allocated the previous year is both clear and concise. In B, the phrease allocating to maintain...and for subsidizing is not parallel. The construction a reduction from the previous year of nearly 17 percent in the amount is awkward. imprecise, and excessviely wordy. Furthermore, there is no grammatical regerent for it in the phrase it was allocating. In C, the phrase proposed to reduce, by nearly 17 percent, the amount from the previous year that was allocated is unidiomatic and overly wordy. Choice C also violates parallelism with allocated for the maintenance of .....and to subsidize. In D, there is no grammatical referent for it in the phrase it was allocating : the mayor, not the city, is the subject of the clause. Choice D also violates parallelism with allocating for maintaining ...and to subsidize. In E, the progressive was proposing is unnecessary, and there is no grammatical regerent for they in the phrase they were allocating. Furthermore, for maintaining...and for the subsidization is not parallel.
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Re: Faced with an estimated \$ 2 billion budget gap, the city's [#permalink]
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05 May 2015, 07:02
Discussed here. Thread locked.
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Re: Faced with an estimated \$ 2 billion budget gap, the city's [#permalink] 05 May 2015, 07:02
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# Faced with an estimated \$ 2 billion budget gap, the city's
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,037 | 8,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-30 | latest | en | 0.932991 |
https://lexique.netmath.ca/en/identity-matrix/ | 1,723,262,443,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00736.warc.gz | 290,507,433 | 13,826 | # Identity Matrix
## Identity Matrix
Square matrix in which all the elements on the main diagonal are 1 and all the other elements are 0.
### Example
The 4 × 4 identity matrix is :
I = $$\begin{pmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}$$ | 106 | 281 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-33 | latest | en | 0.658819 |
https://www.informagiovanilerici.it/mobile/07-22-grinding-volume-calculation-in-a-ball-mill.html | 1,586,384,170,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371824409.86/warc/CC-MAIN-20200408202012-20200408232512-00469.warc.gz | 972,171,911 | 6,801 | # Grinding Volume Calculation In A Ball Mill
• ## Calculate Ball Mill Grinding Capacity
The sizing of ball mills and ball milling circuits from stoneoratory grinding tests is largely a question of applying empirical equations or factors based on
• ## How to Size a Ball Mill -Design Calculator & Formula
expressed as a percentage of the net internal mill volume How to Size a Ball Mill -Design Calculator grinding wet in open circuit 2) Ball milling
• ## grinding volume calculation in a ball mill - energietourbe
ball Grinding Capacity Calculation Of Ball Mill xiplball mill volume calculation 2016 Contact Us: calculating of volume in ball mill capacity Description
• ## TECHNICAL NOTES 8 GRINDING R P King
Figure 83 Simplified calculation of the torque required steel balls in a ball mill, Let Jt be the fraction of the mill volume that is occupied
• ## REGARDING % OF GRINDING MEDIA IN THE MILL - Page 1 of 2
11-3-2014 · re regarding % of grinding media in the mill 113 and 126 are come for calculation from grinding chart vs grinding volume in ball mill
• ## grinding volume calculation in a ball mill - energietourbe
ball Grinding Capacity Calculation Of Ball Mill xiplball mill volume calculation 2016 Contact Us: calculating of volume in ball mill capacity Description
• ## TECHNICAL NOTES 8 GRINDING R P King
Figure 83 Simplified calculation of the torque required steel balls in a ball mill, Let Jt be the fraction of the mill volume that is occupied
• ## REGARDING % OF GRINDING MEDIA IN THE MILL - Page 1 of 2
11-3-2014 · re regarding % of grinding media in the mill 113 and 126 are come for calculation from grinding chart vs grinding volume in ball mill
• ## Calculating A Ball Mill Structure - alkmaarvarennl
grinding volume calculation in a ball mill calculate ball mill grinding capacity processing the sizing of ball mills and ball milling circuits from stoneoratory
• ## volume calculation for ball mill - visualise-projecteu
ball mill filling degree calculation keithkirsten Aug 3, 2016 calculating of volume in ball mill capacity Description : grinding volume calculation in a ball ball
• ## Mill (grinding) - Wikipedia
SAG is an acronym for Semi-Autogenous Grinding SAG mills are autogenous mills but use grinding balls like a ball mill A SAG mill is usually a primary or first stage
• ## CALCULATION OF THE POWER DRAW OF DRY …
by Erdem (2002) for dry multi-component cement grinding ball mills, Calculated mill load volume with using different calculation method for Çorum cement mill
• ## Grinding in Ball Mills: Modeling and Process Control
51 BULGARIAN ACADEMY OF SCIENCES CYBERNETICS AND INFORMATION TECHNOLOGIES • Volume 12, No 2 Sofia • 2012 Grinding in Ball Mills: Modeling and Process Control
• ## Ball Mill Design/Power Calculation - LinkedIn
12-12-2016 · If P is less than 80% passing 70 microns, power consumption will be Ball Mill Power Calculation Example A wet grinding ball mill in closed circuit is to
• ## Optimization of mill performance by using - SciELO
measurement or the percentage by volume of balls in the mill the degree of grinding ball fill and pulp position for timely decision making and actions
• ## Optimization of mill performance by using - SciELO
measurement or the percentage by volume of balls in the mill the degree of grinding ball fill and pulp position for timely decision making and actions
• ## Calculation of Grinding Balls Surface Area and Volume
Quite often, there is a need for a quick and correct calculation of a grinding ball surface area and volume Necessity of such calculations may arise when choosing a
• ## CALCULATION OF BALL MILL GRINDING EFFICIENCY - Page 1 of 1
8-3-2013 · calculation of ball mill grinding efficiency dear experts please tell me how to calculate the grinding efficiency of a closed ckt & open ckt ball mill
pdf grinding media balls charge calculation in ball mill Home > Products > pdf grinding media balls charge calculation in ball mill Mobile Crushing Plant
• ## A Method to Determine the Ball Filling, in Miduk Copper
ball filling strongly affects on grinding rate, ball filling calculation due to ragged load Mill Ball (%) = 100{[(ball volume in the
• ## AMIT 135: Lesson 7 Ball Mills & Circuits – Mining Mill
The percent of mill volume occupied by grinding media, Ball Mill Size as a Replacement Grinding media wears and reduces in size at a rate dependent on the
• ## Ball charges calculators - thecementgrindingoffice
The Cement Grinding - Ball charges: This calculator gives the surface material inside the mill and proposes a modification of the ball charge in
• ## (PDF) Circulating load calculation in grinding circuits
Circulating load calculation in grinding circuits Circulating load calculation in grinding An industrial ball mill operating in closed–circuit with
• ## Calculation of energy required for grinding in a ball mill
Volume 25, Issues 1–2 Calculation of energy required for grinding in a ball mill The grinding-product size, P, in a Bond ball mill,
• ## Page 1 Ball Milling Theory - freeshellorg
involve grinding) With Lloyd's ball milling book having sold over 2000 copies, ball mills operating in just America Volume 3, Issue 5 Ball Milling Theory Page 2
• ## (PDF) Circulating load calculation in grinding circuits
Circulating load calculation in grinding circuits Circulating load calculation in grinding An industrial ball mill operating in closed–circuit with
• ## Calculation of energy required for grinding in a ball mill
Volume 25, Issues 1–2 Calculation of energy required for grinding in a ball mill The grinding-product size, P, in a Bond ball mill,
• ## ball mill capacity calculation - energietourbe
grinding capacity calculation of ball mill Crusher The optimal ball diameter in a mill Mar 16, calculation of volume in ball mill capacity jclop
• ## Ball Mill - an overview | ScienceDirect Topics
The ball mill is a tumbling mill that uses steel balls as the grinding media The length of the cylindrical shell is usually 1–15 times the shell diameter (Figure 811
• ## simple calculation in ball mill - mrsiude
Ball mill &#; A ball mill, a type of grinder, is a cylindrical device used in grinding (or mixing) materials like ores, chemicals, ceramic raw materials and paints
• ## Mill sizing method - thecementgrindingoffice
Please find below two calculators for sizing mills using the Bond and Rowland methods: Ball mill sizing: Calculator for ball mill(s) in a single stage circuit
• ## Ball Mill - RETSCH - powerful grinding and homogenization
A ball mill can grind and homogenize small sample volumes down to the nano range The Emax is an entirely new type of ball mill for high energy input
• ## Ball mill - Wikipedia
A ball mill, a type of grinder, is a cylindrical device used in grinding (or mixing) materials like ores, chemicals, ceramic raw materials and paints
• ## THE OPTIMAL BALL DIAMETER IN A MILL - Strona główna
The mill ball loading was 40% by volume, The volume of grinding samples was equal to the volume of The optimal ball diameter in a mill
• ## Design Method of Ball Mill by Sumitomo stone Co, Ltd
A ball mill is one kind of grinding machine, the mill was stopped, and small volume samples were (Eq 11) (Eq 10) Table 1 Physical properties for DEM simulation
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>> Next:Sbm Jaw Crusher China | 1,636 | 7,363 | {"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} | 2.71875 | 3 | CC-MAIN-2020-16 | latest | en | 0.688878 |
https://www.askiitians.com/forums/Integral-Calculus/consider-a-function-g-x-which-is-defined-and-diff_96296.htm | 1,680,380,436,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950247.65/warc/CC-MAIN-20230401191131-20230401221131-00068.warc.gz | 744,567,920 | 32,654 | # Consider a function g(x) which is defined and differentiable on (-8, 8) and increasing in (1,2) and decreasing elsewhere.We construct another function f (x) = g(x) - (g(x))2 + (g(x))3 . Find domain of f (x), it’s interval of monotonicity.
bharat bajaj IIT Delhi
9 years ago
f(x) = g(x) - (g(x))^2 + (g(x))^3
f'(x) = g'(x) ( 1 - 2 g(x) + 3g(x)^2)
Now, 3g(x)^2 - 2g(x) + 1 has Discriminant = 0. Hence, this means that this is either always positive or always negative.
g'(x) > 0 in the interval (1,2)
g'(x) < 0 in the interval (-8,1) U (2,8)
The domain of f(x) is same as that of g(x) which is (-8,8).
For the interval of monotonicity are :
We cannot clearly say that 3g(x)^2 - 2g(x) + 1 is positive or negative as we do not know much about g(x). Say it is positive. Hence,
f(x) is monotonically increasing in interval (1,2)
f(x) is monotically decreasing in interval (-8,1) U (2,8)
Thanks
Bharat Bajaj
IIT Delhi | 323 | 914 | {"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-2023-14 | latest | en | 0.885149 |
http://usakochan.net/download/statistics-workbook-pts-1-and-2-concepts-and-applications-for-science/ | 1,545,190,070,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376830479.82/warc/CC-MAIN-20181219025453-20181219051453-00548.warc.gz | 274,430,260 | 16,872 | ## Statistics
Concepts and Applications for Science
Author: David C. LeBlanc
Publisher: Jones & Bartlett Learning
ISBN: 9780763746995
Category: Statistics.
Page: 382
View: 622
Designed for students majoring in the life, health, and natural sciences, Statistics: Concepts and Applications for Science is a text and workbook package that introduces statistics with an important emphasis on the real-world applications of statistical reasoning and procedures. Through intensive exposure to the core concepts of statistics in the context of science, students acquire the skills and understanding they need to formulate valid research designs, implement statistical analysis, interpret data, and explain their results.
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## Statistics and Probability for Engineering Applications with Microsoft Excel
Author: William J. DeCoursey
Publisher: Amsterdam : Newnes
ISBN: 9780750676182
Category: Technology & Engineering
Page: 396
View: 1431
More than ever, American industry- especially the semiconductor industry- is using statistical methods to improve its competitive edge in the world market. It is becoming more imperative that graduate engineers have solid statistical know-how, yet engineers in industry typically are not well-prepared to use statistics and they are fuzzy about how to apply statistical tools and techniques. This valuable reference makes statistical methods easier and more accessible to engineers. Although the book can be read sequentially, like a normal textbook, it is designed to be used as a handbook, pointing the reader to the topics and sections pertinent to a particular type of statistical problem. It contains the following features: * Covers all major topics treated in a standard college engineering statistics course, but minimizes the mathematical derivations and focuses on practical applications * Uses real data sets/case studies taken from electronics, electrical engineering, and other engineering fields, such as mechanical and chemical engineering * Contains numerous software examples using the powerful statistical functions of Excel In addition, the book provides an "engineering problem solver" section that directs the reader to the relevant section of the book for the problem they are trying to solve. The accompanying CD-ROM contains the Excel data sets for the examples and case studies given in the book, along with other statistical tools and software. * Filled with practical techniques directly applicable on the job * Contains hundreds of solved problems and case studies, using real data sets * Avoids unnecessary theory
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## Introduction to Probability and Statistics Using R
Author: G. Jay Kerns
Publisher: Lulu.com
ISBN: 0557249791
Category:
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## Learning Directory
Author: N.A
Publisher: N.A
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## Differential Diagnosis for Physical Therapists- E-Book
Author: Catherine C. Goodman,John Heick,Rolando T. Lazaro
Publisher: Elsevier Health Sciences
ISBN: 0323478484
Category: Medical
Page: 784
View: 3062
Learn how to screen for red flags and when to refer clients to a medical specialist! Differential Diagnosis for Physical Therapists: Screening for Referral, 6th Edition provides a step-by-step approach to screening for systemic disease and medical conditions that can mimic neuromuscular and musculoskeletal problems. It describes both red flags and yellow flags, so you can recognize the signs and symptoms for conditions outside the scope of physical therapy practice. This edition includes new information on women’s health issues. Written by experienced PT practitioner Catherine Cavallaro Goodman, this book helps you determine whether a client’s symptoms require physical therapy or physician referral! UNIQUE! Five-step screening model is systems- and symptoms-based, and follows the standards for competency established by the American Physical Therapy Association, covering past medical history, risk factor assessment, clinical presentation, associated signs and symptoms, and review of symptoms. UNIQUE! Case studies are based on clinical experience and give real-world examples of how to integrate screening information into the diagnostic process and when to treat or refer. Evidence for the screening process is based on peer-reviewed literature, reporting on the sensitivity, specificity, and likelihood ratios of yellow (cautionary) and red (warning) flags. Key Points to Remember boxes at the end of each chapter provide quick, bulleted summaries of critical information. Quick-reference summaries include tables, boxes, follow-up questions, clinical signs and symptoms, and case examples. Screening tools and checklists are found in the book and on the Evolve website, and are downloadable and printable for use in the clinic. Quick response (QR) codes that can be scanned on a smartphone, tablet, or other mobile device provide links to valuable screening tools such as lists of questions for screening specific problems, checklists, intake forms, and assessment tests New content on women’s health expands coverage of this important topic. UPDATES reflect the most current information on screening for referral. New associate editors — John Heick and Rolando Lazaro — bring fresh insight, as respected physical therapy educators. New! Color tabs make it easier to locate chapters and topics.
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## Problem Book in the Theory of Functions: Problems in the elementary theory of functions, translated by L. Bers
Publisher: N.A
ISBN: N.A
Category: Functions
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## New Technical Books
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Publisher: N.A
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Category: Engineering
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## Understandable Statistics: Concepts and Methods
Author: Charles Henry Brase,Corrinne Pellillo Brase
Publisher: Cengage Learning
ISBN: 1337119911
Category: Mathematics
Page: 839
View: 620
UNDERSTANDABLE STATISTICS: CONCEPTS AND METHODS, Twelfth Edition, is a thorough yet accessible program designed to help you overcome any apprehensions you may have about statistics and to master the subject. The authors provide clear guidance and informal advice while showing you the links between statistics and the world. To reinforce this approach—and make the material interesting as well as easier to understand—the book integrates real-life data from a variety of sources, including journals, periodicals, newspapers, and the Internet. You'll also have opportunities to develop your critical-thinking and statistical literacy skills through special features and exercises throughout the text. The use of graphing calculators, Excel, Minitab, Minitab ExpressTM, and SPSS is covered, although not required. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.
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## Concepts of Biology
Author: Samantha Fowler,Rebecca Roush,James Wise
Publisher: N.A
ISBN: 9789888407453
Category: Science
Page: 618
View: 9755
Concepts of Biology is designed for the single-semester introduction to biology course for non-science majors, which for many students is their only college-level science course. As such, this course represents an important opportunity for students to develop the necessary knowledge, tools, and skills to make informed decisions as they continue with their lives. Rather than being mired down with facts and vocabulary, the typical non-science major student needs information presented in a way that is easy to read and understand. Even more importantly, the content should be meaningful. Students do much better when they understand why biology is relevant to their everyday lives. For these reasons, Concepts of Biology is grounded on an evolutionary basis and includes exciting features that highlight careers in the biological sciences and everyday applications of the concepts at hand.We also strive to show the interconnectedness of topics within this extremely broad discipline. In order to meet the needs of today's instructors and students, we maintain the overall organization and coverage found in most syllabi for this course. A strength of Concepts of Biology is that instructors can customize the book, adapting it to the approach that works best in their classroom. Concepts of Biology also includes an innovative art program that incorporates critical thinking and clicker questions to help students understand--and apply--key concepts.
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## Wound Care
A Collaborative Practice Manual
Author: Carrie Sussman,Barbara M. Bates-Jensen
Publisher: Lippincott Williams & Wilkins
ISBN: 9780781774444
Category: Medical
Page: 720
View: 7504
Designed for health care professionals in multiple disciplines and clinical settings, this comprehensive, evidence-based wound care text provides basic and advanced information on wound healing and therapies and emphasizes clinical decision-making. The text integrates the latest scientific findings with principles of good wound care and provides a complete set of current, evidence-based practices. This edition features a new chapter on wound pain management and a chapter showing how to use negative pressure therapy on many types of hard-to-heal wounds. Technological advances covered include ultrasound for wound debridement, laser treatments, and a single-patient-use disposable device for delivering pulsed radio frequency.
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## An Introduction to Statistical Methods and Data Analysis
Author: R. Lyman Ott,Micheal T. Longnecker
Publisher: Cengage Learning
ISBN: 1305465520
Category: Mathematics
Page: 1296
View: 6198
Ott and Longnecker's AN INTRODUCTION TO STATISTICAL METHODS AND DATA ANALYSIS, Seventh Edition, provides a broad overview of statistical methods for advanced undergraduate and graduate students from a variety of disciplines who have little or no prior course work in statistics. The authors teach students to solve problems encountered in research projects, to make decisions based on data in general settings both within and beyond the university setting, and to become critical readers of statistical analyses in research papers and news reports. The first eleven chapters present material typically covered in an introductory statistics course, as well as case studies and examples that are often encountered in undergraduate capstone courses. The remaining chapters cover regression modeling and design of experiments. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.
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## Matlab
A Practical Introduction to Programming and Problem Solving
Author: Stormy Attaway
Publisher: Butterworth-Heinemann
ISBN: 0124058930
Category: Computers
Page: 560
View: 1248
MatLab, Third Edition is the only book that gives a full introduction to programming in MATLAB combined with an explanation of the software’s powerful functions, enabling engineers to fully exploit its extensive capabilities in solving engineering problems. The book provides a systematic, step-by-step approach, building on concepts throughout the text, facilitating easier learning. Sections on common pitfalls and programming guidelines direct students towards best practice. The book is organized into 14 chapters, starting with programming concepts such as variables, assignments, input/output, and selection statements; moves onto loops; and then solves problems using both the ‘programming concept’ and the ‘power of MATLAB’ side-by-side. In-depth coverage is given to input/output, a topic that is fundamental to many engineering applications. Vectorized Code has been made into its own chapter, in order to emphasize the importance of using MATLAB efficiently. There are also expanded examples on low-level file input functions, Graphical User Interfaces, and use of MATLAB Version R2012b; modified and new end-of-chapter exercises; improved labeling of plots; and improved standards for variable names and documentation. This book will be a valuable resource for engineers learning to program and model in MATLAB, as well as for undergraduates in engineering and science taking a course that uses (or recommends) MATLAB. Presents programming concepts and MATLAB built-in functions side-by-side Systematic, step-by-step approach, building on concepts throughout the book, facilitating easier learning Sections on common pitfalls and programming guidelines direct students towards best practice
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## Introduction to Probability
Author: Charles Miller Grinstead,James Laurie Snell
Publisher: American Mathematical Soc.
ISBN: 0821894145
Category: Probabilities
Page: 510
View: 8058
This text is designed for an introductory probability course at the university level for sophomores, juniors, and seniors in mathematics, physical and social sciences, engineering, and computer science. It presents a thorough treatment of ideas and techniques necessary for a firm understanding of the subject. The text is also recommended for use in discrete probability courses. The material is organized so that the discrete and continuous probability discussions are presented in a separate, but parallel, manner. This organization does not emphasize an overly rigorous or formal view of probability and therefore offers some strong pedagogical value. Hence, the discrete discussions can sometimes serve to motivate the more abstract continuous probability discussions. Features: Key ideas are developed in a somewhat leisurely style, providing a variety of interesting applications to probability and showing some nonintuitive ideas. Over 600 exercises provide the opportunity for practicing skills and developing a sound understanding of ideas. Numerous historical comments deal with the development of discrete probability. The text includes many computer programs that illustrate the algorithms or the methods of computation for important problems. The book is a beautiful introduction to probability theory at the beginning level. The book contains a lot of examples and an easy development of theory without any sacrifice of rigor, keeping the abstraction to a minimal level. It is indeed a valuable addition to the study of probability theory. --Zentralblatt MATH
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## OpenIntro Statistics
Author: David Diez,Christopher Barr,Mine Çetinkaya-Rundel
Publisher: N.A
ISBN: 9781943450046
Category:
Page: N.A
View: 4316
The OpenIntro project was founded in 2009 to improve the quality and availability of education by producing exceptional books and teaching tools that are free to use and easy to modify. We feature real data whenever possible, and files for the entire textbook are freely available at openintro.org. Visit our website, openintro.org. We provide free videos, statistical software labs, lecture slides, course management tools, and many other helpful resources.
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## Introduction to Algorithms
Author: Thomas H. Cormen
Publisher: MIT Press
ISBN: 0262533057
Category: Computers
Page: 1292
View: 1754
A new edition of the essential text and professional reference, with substantial new material on such topics as vEB trees, multithreaded algorithms, dynamic programming, and edge-based flow.
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Author: N.A
Publisher: N.A
ISBN: N.A
Category: Education
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## Statistics in a Nutshell
Author: Sarah Boslaugh
Publisher: "O'Reilly Media, Inc."
ISBN: 1449316824
Category: Mathematics
Page: 569
View: 5079
A clear and concise introduction and reference for anyone new to the subject of statistics.
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## Diagnostic Imaging for Physical Therapists
Author: James M. Swain,Kenneth W. Bush,Juliet Wain Brosing
Publisher: Elsevier Health Sciences
ISBN: 1416029036
Category: Medical
Page: 320
View: 6262
With the ever-increasing demand on physical therapists to develop the most effective treatment interventions comes this invaluable imaging resource covering exactly what you need to know! Diagnostic Imaging for Physical Therapists gives you the knowledge to understand the basic principles of musculoskeletal imaging and how to interpret radiographic images in your physical therapy practice. This straightforward, highly illustrated text is organized by body region and covers all the fundamentals with an emphasis on standard, two-dimensional x-rays. An accompanying DVD delivers high-resolution copies of the images in the text along with interactive activities to enhance your understanding of the material. With this indispensable text, you'll recognize when diagnostic imaging is necessary, and you'll be able to interpret the results with confidence. Written specifically for PTs, this book covers the most common film images you will see in your practice and introduces you to some of the not-so-common images. UNIQUE companion DVD helps you hone your diagnostic imaging skills with high-resolution radiographic images and animations. DVD icons in the book direct you to interactive exercises including ABCs, pathologies, case studies, and quizzes that will enhance your understanding of concepts in the text. Provides you with a "systematic" basis for approaching the interpretation of standard films. The body system approach of the chapters makes it easy to find information specific to a body region. Text edited by highly respected experts in musculoskeletal rehabilitation gives you authoritative guidance on the management of musculoskeletal pathology and injury.
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## Using R for Introductory Statistics, Second Edition
Author: John Verzani
Publisher: CRC Press
ISBN: 1466590734
Category: Mathematics
Page: 518
View: 4501
The second edition of a bestselling textbook, Using R for Introductory Statistics guides students through the basics of R, helping them overcome the sometimes steep learning curve. The author does this by breaking the material down into small, task-oriented steps. The second edition maintains the features that made the first edition so popular, while updating data, examples, and changes to R in line with the current version. See What’s New in the Second Edition: Increased emphasis on more idiomatic R provides a grounding in the functionality of base R. Discussions of the use of RStudio helps new R users avoid as many pitfalls as possible. Use of knitr package makes code easier to read and therefore easier to reason about. Additional information on computer-intensive approaches motivates the traditional approach. Updated examples and data make the information current and topical. The book has an accompanying package, UsingR, available from CRAN, R’s repository of user-contributed packages. The package contains the data sets mentioned in the text (data(package="UsingR")), answers to selected problems (answers()), a few demonstrations (demo()), the errata (errata()), and sample code from the text. The topics of this text line up closely with traditional teaching progression; however, the book also highlights computer-intensive approaches to motivate the more traditional approach. The authors emphasize realistic data and examples and rely on visualization techniques to gather insight. They introduce statistics and R seamlessly, giving students the tools they need to use R and the information they need to navigate the sometimes complex world of statistical computing.
Release | 3,765 | 19,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-51 | longest | en | 0.894311 |
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# by fractions中文什么意思
• 有余数的
### 例句与用法
1. Decimal divided by fraction , when there ' s a multiple relationship between decimal and fraction , we can use reduce fractions
小数除以分数，当小数与分数的分子有倍数关系时，可用直接约分法。
2. It s a strange way of killing ! not by inches , but by fractions and hairbreadths , to beguile me with the spectre of a hope , through eighteen years
那是一种奇怪的杀人方法：不是一寸寸的，而是像头发丝那样的一丝丝地割，十八年来就用幽灵样的希望来引诱我！ ”
3. The ultraviolet absorption detector is also applied to detect the protein apexes . the components are respectively collected by fraction collection . we have gotten two absolute components
利用紫外检测对分离的物质进行检测，检测出蛋白峰，并用分部收集器对所分离的样品进行分部收集，结果共得到两个组分。 | 237 | 661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-49 | latest | en | 0.526975 |
http://www.perlmonks.org/?parent=433860;node_id=3333 | 1,527,458,226,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870470.67/warc/CC-MAIN-20180527205925-20180527225925-00082.warc.gz | 418,358,123 | 6,999 | Your skill will accomplishwhat the force of many cannot PerlMonks
### Comment on
Need Help??
I don't get it. When Purity=0 DD, doesn't claim to reevaluate correctly and it's frequently more readable. When Purity=1, it seems to work perfectly for me.
Here's something that takes your examples above, DDDumps them with Purity=1 and reevals the string and then shows the Streamer version of the original and of the evaled.
```use Data::Dumper;
use Data::Dump::Streamer;
\$Data::Dumper::Purity=1;
my (\$x,\$y);
\$x=\\$y;
\$y=\\$x;
test([\$x, \$y]);
my \$ary=[];
\$ary->[0]=\\$ary->[1];
\$ary->[1]=\\$ary->[0];
test(\$ary);
sub test
{
my \$VAR1;
my \$s = shift;
my \$d = Dumper(\$s);
my \$s2 = eval "\$d;\\$VAR1";
die \$@ if \$@;
print "original---------------\n";
print Dump(\$s);
print "DDed---------------\n";
print Dump(\$s2);
print "\n"
}
__END__
original---------------
\$ARRAY1 = [
\do { my \$v = 'V: \$ARRAY1->[1]' },
\do { my \$v = 'V: \$ARRAY1->[0]' }
];
\${\$ARRAY1->[0]} = \$ARRAY1->[1];
\${\$ARRAY1->[1]} = \$ARRAY1->[0];
DDed---------------
\$ARRAY1 = [
\do { my \$v = 'V: \$ARRAY1->[1]' },
\do { my \$v = 'V: \$ARRAY1->[0]' }
];
\${\$ARRAY1->[0]} = \$ARRAY1->[1];
\${\$ARRAY1->[1]} = \$ARRAY1->[0];
original---------------
\$ARRAY1 = [
'R: \$ARRAY1->[1]',
'R: \$ARRAY1->[0]'
];
\$ARRAY1->[0] = \\$ARRAY1->[1];
\$ARRAY1->[1] = \\$ARRAY1->[0];
DDed---------------
\$ARRAY1 = [
\do { my \$v = 'V: \$ARRAY1->[1]' },
\do { my \$v = 'V: \$ARRAY1->[0]' }
];
\${\$ARRAY1->[0]} = \$ARRAY1->[1];
\${\$ARRAY1->[1]} = \$ARRAY1->[0];
The first one comes out identical. The second one is not textually identical but if \${\$x} = \$y means the same thing as \$x = \\$y then I can't see what DD did wrong. Do they mean the same thing?
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Notices? | 979 | 3,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-22 | latest | en | 0.60406 |
https://nrich.maths.org/public/topic.php?code=134&cl=4&cldcmpid=5751 | 1,582,456,960,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00057.warc.gz | 490,453,400 | 7,984 | Resources tagged with: Symmetry
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Reflections. Coordinates. Chemistry. Visualising. Cubes & cuboids. Games. Interactivities. Nets. Complex numbers. Rotations.
There are 34 results
Broad Topics > Transformations and constructions > Symmetry
Attractive Tablecloths
Age 14 to 16 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
Cut Cube
Age 16 to 18 Challenge Level:
Find the shape and symmetries of the two pieces of this cut cube.
A Roll of Patterned Paper
Age 14 to 16 Challenge Level:
A design is repeated endlessly along a line - rather like a stream of paper coming off a roll. Make a strip that matches itself after rotation, or after reflection
Rose
Age 16 to 18 Challenge Level:
What groups of transformations map a regular pentagon to itself?
Symmetric Trace
Age 14 to 16 Challenge Level:
Points off a rolling wheel make traces. What makes those traces have symmetry?
Arclets
Age 14 to 16 Challenge Level:
Each of the following shapes is made from arcs of a circle of radius r. What is the perimeter of a shape with 3, 4, 5 and n "nodes".
Rotations Are Not Single Round Here
Age 14 to 16 Challenge Level:
I noticed this about streamers that have rotation symmetry : if there was one centre of rotation there always seems to be a second centre that also worked. Can you find a design that has only. . . .
Frieze Patterns in Cast Iron
Age 11 to 16
A gallery of beautiful photos of cast ironwork friezes in Australia with a mathematical discussion of the classification of frieze patterns.
Two Triangles in a Square
Age 14 to 16 Challenge Level:
Given that ABCD is a square, M is the mid point of AD and CP is perpendicular to MB with P on MB, prove DP = DC.
Tournament Scheduling
Age 11 to 16
Scheduling games is a little more challenging than one might desire. Here are some tournament formats that sport schedulers use.
Paint Rollers for Frieze Patterns.
Age 11 to 16
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach.
Trominoes
Age 11 to 16 Challenge Level:
Can all but one square of an 8 by 8 Chessboard be covered by Trominoes?
Sliced
Age 14 to 16 Challenge Level:
An irregular tetrahedron has two opposite sides the same length a and the line joining their midpoints is perpendicular to these two edges and is of length b. What is the volume of the tetrahedron?
A Problem of Time
Age 14 to 16 Challenge Level:
Consider a watch face which has identical hands and identical marks for the hours. It is opposite to a mirror. When is the time as read direct and in the mirror exactly the same between 6 and 7?
The Frieze Tree
Age 11 to 16
Patterns that repeat in a line are strangely interesting. How many types are there and how do you tell one type from another?
Dancing with Maths
Age 7 to 16
An article for students and teachers on symmetry and square dancing. What do the symmetries of the square have to do with a dos-e-dos or a swing? Find out more?
One Reflection Implies Another
Age 14 to 16 Challenge Level:
When a strip has vertical symmetry there always seems to be a second place where a mirror line could go. Perhaps you can find a design that has only one mirror line across it. Or, if you thought that. . . .
Prime Magic
Age 7 to 16 Challenge Level:
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find?
Classifying Solids Using Angle Deficiency
Age 11 to 16 Challenge Level:
Toni Beardon has chosen this article introducing a rich area for practical exploration and discovery in 3D geometry
Square Pizza
Age 14 to 16 Challenge Level:
Can you show that you can share a square pizza equally between two people by cutting it four times using vertical, horizontal and diagonal cuts through any point inside the square?
Encircling
Age 14 to 16 Challenge Level:
An equilateral triangle is sitting on top of a square. What is the radius of the circle that circumscribes this shape?
Maltese Cross
Age 16 to 18 Challenge Level:
Sketch the graph of $xy(x^2 - y^2) = x^2 + y^2$ consisting of four curves and a single point at the origin. Convert to polar form. Describe the symmetries of the graph.
Dicey Decisions
Age 16 to 18 Challenge Level:
Can you devise a fair scoring system when dice land edge-up or corner-up?
Holly
Age 14 to 16 Challenge Level:
The ten arcs forming the edges of the "holly leaf" are all arcs of circles of radius 1 cm. Find the length of the perimeter of the holly leaf and the area of its surface.
Eight Dominoes
Age 7 to 16 Challenge Level:
Using the 8 dominoes make a square where each of the columns and rows adds up to 8
Octa-flower
Age 16 to 18 Challenge Level:
Join some regular octahedra, face touching face and one vertex of each meeting at a point. How many octahedra can you fit around this point?
Logosquares
Age 16 to 18 Challenge Level:
Ten squares form regular rings either with adjacent or opposite vertices touching. Calculate the inner and outer radii of the rings that surround the squares.
More Dicey Decisions
Age 16 to 18 Challenge Level:
The twelve edge totals of a standard six-sided die are distributed symmetrically. Will the same symmetry emerge with a dodecahedral die?
Mean Geometrically
Age 16 to 18 Challenge Level:
A and B are two points on a circle centre O. Tangents at A and B cut at C. CO cuts the circle at D. What is the relationship between areas of ADBO, ABO and ACBO?
Pitchfork
Age 16 to 18 Challenge Level:
Plot the graph of x^y = y^x in the first quadrant and explain its properties.
Cocked Hat
Age 16 to 18 Challenge Level:
Sketch the graphs for this implicitly defined family of functions.
Folium of Descartes
Age 16 to 18 Challenge Level:
Investigate the family of graphs given by the equation x^3+y^3=3axy for different values of the constant a.
Witch of Agnesi
Age 16 to 18 Challenge Level:
Sketch the members of the family of graphs given by y = a^3/(x^2+a^2) for a=1, 2 and 3.
Flower Power
Age 11 to 16 Challenge Level:
Create a symmetrical fabric design based on a flower motif - and realise it in Logo. | 1,580 | 6,468 | {"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} | 3.125 | 3 | CC-MAIN-2020-10 | latest | en | 0.918835 |
https://brilliant.org/problems/flash-back-of-the-year-2011/ | 1,529,397,820,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861981.50/warc/CC-MAIN-20180619080121-20180619100121-00243.warc.gz | 577,871,044 | 11,088 | # Flash back of the year 2011
$If \frac{1}{\sqrt{2011+\sqrt{2011^{2}-1}}} =\sqrt{m} - \sqrt{n} ,$ where m and n are positive integers , what is the value of m + n proof is needed compulsary. according to me the poster's of solution are the solvers
× | 81 | 251 | {"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} | 2.609375 | 3 | CC-MAIN-2018-26 | latest | en | 0.848322 |
http://math260.drjimo.net/pfd/part-part-fractions/ | 1,618,562,715,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00149.warc.gz | 60,778,434 | 10,437 | # Part-Part Fractions!
We usually think of fractions as part to whole.
However, fractions can be part to part fractions. Here are a few ideas.
Stating part-to-part relationships as ratios is more common. Part-to-part relationships can be stated (written) as fractions, as well.
From the Common Core State Standards:
3 cups flour to 4 cups of sugar 3 to 4 3:4 (written as ratios)
Written as a fraction it would be 3/4.
Another way to think of it is 3/4 cup of flour per cup of sugar (only changing one word from 6.RP.2).
It was complaining about the Ministry of Education’s 28-page booklet explaining how to teach fractions. I do agree that they may have made some poor choices and didn’t explain things as well as perhaps they could have, in the part-part concept of fractions section, but I don’t feel what they write is mathematically incorrect.
The 7/2 is 7/2 non-oranges per orange. It is a force (& a poor choice), but it works. “3.5 non-oranges per orange.”
Part-to-part fractions behave like rates (and it’s no accident that the other example in 6.RP.2 is a rate!). Ironically, the units *are* the same (cups/cups). I’ve found that when dividing when the units are the same you can let them cancel *or* keep them. Keeping them may help (because you have some labels).
Finally, I’ll end with one more part-to-part fraction. The oil-gas mixture (for a chainsaw, for example) is 1 part oil to 50 parts gasoline. It might be 1 oz to 50 oz. or 1 gal to 50 gal. (1:50) As a fraction we get 1/50 ounce of oil per ounce of gas.
The 6-7, Ratios & Proportional Relationships progressions document has many good examples of part-to-part ratios. https://commoncoretools.files.wordpress.com/2012/02/ccss_progression_rp_67_2011_11_12_corrected.pdf
“For each,” “for every,” “per”
Save
Save
Save
Save | 483 | 1,828 | {"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} | 3.921875 | 4 | CC-MAIN-2021-17 | latest | en | 0.911965 |
https://agrilacorte.com/get-answers-191 | 1,674,974,863,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499700.67/warc/CC-MAIN-20230129044527-20230129074527-00291.warc.gz | 108,560,666 | 5,672 | # How to find y intercept given 2 points
Find the y-intercept with 2 points: Let (x 1, y 1) and (x 2, y 2) be the two given points. Slope of the line passing through the given points is: m = y 2-y 1 x 2-x 1. The equation of a line in slope
Solve Now
## How to Find the Y Intercept with two points
The y-intercept is 1 Step 1. Find the slope using the equation m=(y_2-y_1)/(x_2-x_1), where m is the slope, and (x_1,y_1) and (x_2,y_2) are the two points on the line. Example
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## How to Find the Equation of a Line from Two Points
Steps Calculate the slope from 2 points. For Example, Two points are (3, 5) and (6, 11) Substitute the slope (m) in the slope-intercept form of the equation. Substitute either point
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## 3 Ways to Find the Y Intercept
Finding y intercept given slope & point (old) About. Transcript. An old video where Sal finds the y-intercept of the line y=17/13x+b given that the line passes through (5,8).
Slope Intercept Form Calculator
{{@N-H2TEXT@}} | 459 | 1,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-06 | latest | en | 0.901983 |
http://www.physicsforums.com/showthread.php?p=3295241 | 1,369,150,892,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700132256/warc/CC-MAIN-20130516102852-00004-ip-10-60-113-184.ec2.internal.warc.gz | 640,076,983 | 7,563 | ## Why is the cube of a unitary operator = identity matrix?
Hi there,
If A is unitary I understand that it obeys AA+=1 because A-1=A+.
Why does A3=1?
The explanation simply says that "A just permutes the basis vectors"..
It then goes on to say that since A3=1, then eigenvalue a3=1 also, which are 1, ei.2pi.theta/3, and ei.4pi.theta/3. This would make sense to me if I knew why A3=1..
PhysOrg.com physics news on PhysOrg.com >> Iron-platinum alloys could be new-generation hard drives>> Lab sets a new record for creating heralded photons>> Breakthrough calls time on bootleg booze
What is this operator A we are talking about? Just any random Unitary operator? It doesn't seem to be true for any unitary operator...
Recognitions: Gold Member You must have a specific operator in mind because that's not true in general.
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## Why is the cube of a unitary operator = identity matrix?
Why does A3=1? The explanation simply says that "A just permutes the basis vectors"..
Because A takes each of three basis vectors into the next one, so if you apply A three times you'll get back to the original situation. For example, A3 maps ijki. | 295 | 1,171 | {"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} | 2.796875 | 3 | CC-MAIN-2013-20 | latest | en | 0.90284 |
https://byjus.com/question-answer/two-non-integer-roots-of-the-equation-x-2-3x-2-x-2-3x-6/ | 1,643,257,896,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305141.20/warc/CC-MAIN-20220127042833-20220127072833-00654.warc.gz | 206,682,271 | 21,040 | Question
# Two non - integer roots of the equation $$(x^{2}+3x)^{2}-(x^{2}+3x)-6= 0$$ are
A
12(3+11),12(311)
B
12(3+7),12(37)
C
12(3+21),12(321)
D
none of these
Solution
## The correct option is C $$\frac{1}{2}(-3+\sqrt{21}),\frac{1}{2}(-3-\sqrt{21})$$Equation is $$(x^{2}+3x)^{2}-(x^{2}+3x)-6=0$$Put $$x^{2}+3x=y$$Therefore,$$y^{2}-y-6=0$$$$(y+2)(y-3)=0$$$$y=-2,3$$When $$y=-2$$$$x^{2}+3x=-2$$$$x^{2}+3x+2=0$$$$(x+1)(x+2)=0$$$$x=-1,-2$$When, $$y=3$$$$x^{2}+3x-3=0$$$$x=\dfrac{-3\pm \sqrt{9+12}}{2}$$$$x=\dfrac{1}{2}(-3+\sqrt{21}), \dfrac{1}{2}(-3-\sqrt{21})$$Maths
Suggest Corrections
0
Similar questions
View More | 323 | 623 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-05 | latest | en | 0.406708 |
http://math.stackexchange.com/questions/31467/degree-of-differentiability-of-a-manifold-at-a-point | 1,469,752,179,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829320.91/warc/CC-MAIN-20160723071029-00019-ip-10-185-27-174.ec2.internal.warc.gz | 164,733,462 | 18,079 | # degree of differentiability of a manifold at a point
I am neither aware fully nor have studied differential geometry, but i'd like to learn it if i get to know the answer for this question. I am asking this question based on the very superficial knowledge of differential geometry i got know reading wikipedia.
Does a manifold exist whose degree of differentiability is different at different points ?
-
Yes, certainly. For example the solutions $C$ to the equation $y=|x|$ is a topological submanifold of $\mathbb R^2$. Away from the origin it's a $C^\infty$-manifold, meaning that for sufficiently small neighbourhoods $U$ of points, $U \cap C$ is $C^\infty$. But if you intersect $C$ with any neighbourhood of the origin, it's never even a $C^1$-manifold. So you can make sense of "degree of differentiability near a point". For abstract manifolds a sheafy language would be the most natural way to phrase your question.
@Gerben: the graph of any continuous function $f: \mathbb R \to \mathbb R$ is a topological submanifold of $\mathbb R^2$. In this case the function is $f(x)=|x|$. From the flavour of your argument it sounds like you're talking about the set $\{(x,y)\in \mathbb R^2 : |x|=|y| \}$ ?? This is a different kind of object. – Ryan Budney Apr 8 '11 at 17:52
@yasmar: Most definitions of differentiable structures make sense for topological manifolds -- topological manifolds are smooth $C^0$-manifolds. That means there is an atlas and the transition maps are simply homeomorphisms. But given a neighbourhood $U$ of a point $p$ in the manifold you can intersect all your charts with $U$ and then look at all the transition maps for those charts, and ask what their order of differentiability is. If it's larger than $0$, then you could say your topological manifold is $C^k$ for some $k>0$ near $p$. This is equivalent to the above submanifold def. – Ryan Budney Apr 8 '11 at 18:04 | 494 | 1,904 | {"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} | 3.328125 | 3 | CC-MAIN-2016-30 | latest | en | 0.924542 |
https://music.stackexchange.com/questions/36403/is-there-a-pattern-behind-the-numbers-of-sharp-keys-in-chromatic-scale/36408#36408 | 1,638,156,575,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358685.55/warc/CC-MAIN-20211129014336-20211129044336-00401.warc.gz | 505,159,363 | 37,468 | # Is there a pattern behind the numbers of sharp keys in chromatic scale
Here are the keys of the chromatic scale containing 12 pitches:
```| C | C# | D | D# | E | F | F# | G | G# | A | A# | B |
|---+----+---+----+---+---+----+---+----+---+----+----+
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
```
The sharp keys have numbers:
``````| Key | 1 | 3 | 6 | 8 | 10 | 13 (C#) |
|-------+-------------+---+---+---+----+---------|
| Delta | 3 (prev A#) | 2 | 3 | 2 | 2 | 3 |
``````
Why are the pitches with these numbers specifically called "sharp" or "flat"?
Is there any mathematical pattern behind it or it's just a historical coincidence?
• Aren't you in effect asking why the major scale uses numbers `0 2 4 5 7 9 11`, instead, for example, `0 2 3 5 7 8 10` (a minor scale pattern)? Sep 2 '15 at 21:12
• @hpaulj Yeah, why does it use these numbers? Very interesting. Why is the first sequence big (major) and the second - small (minor)? Sep 3 '15 at 2:16
• Don't think of 'major/minor' as 'big/small'. Don't read too much into the names; they are more historical convensions than anything else. Look at en.wikipedia.org/wiki/Mode_(music)#Modern to see how they fit in a bigger picture of `modes`. Sep 3 '15 at 5:06
Yes, there is a pattern. The initial starting point is the following two facts:
• Traditionally, the western musical scale was based on a 7-note scale, named A-G.
• Acoustically, the most basic harmony, aside from unisons and octaves, is the perfect fifth (P5), which can be very closely approximated by 7/12ths of an octave (where each 12th of an octave is called a half-step, H).
In order to maximize the occurrence of P5's, the scale is constructed so that each of the seven notes in the scale is a P5 away from another note. Since pitches "wrap around" at the octave, we use so-called "modular arithmetic" (think of adding clock times, where 11:00 + 2 hours = 1:00), denoted below by the notation "mod12". If we start from F (which happens to be 5 in your numbering scheme), and add a P5 (7 half steps) each time, this gives us the following sequence of notes:
• F = 5
• C = F+P5 = 5+7 mod12 = 0
• G = C+P5 = 0+7 mod12 = 7
• D = G+P5 = 7+7 mod12 = 2
• A = D+P5 = 2+7 mod12 = 9
• E = A+P5 = 9+7 mod12 = 4
• B = E+P5 = 4+7 mod12 = 11
This then, gives us the patterns of the non-sharp and non-flat notes. You'll notice, however, that from B back to F is a distance of 6 not 7, which corresponds to a dissonant interval called the tritone, instead of a P5. In order to address this, you have two choices: You can either replace the B with a note a P5 below F:
• B♭ = F-P5 = 5-7 mod12 = 10 = B-1
Or you can replace the F with a note a P5 above B:
• F♯ = B+P5 = 11+7 mod12 = 6 = F+1
Note that these new notes replace the original, and are either a half step below or above the note that they replace. Also note that this pattern can then be continued on indefinitely, by adding or subtracting 7 (mod12) to get the next note in the sequence.
Update: If you extrapolate and generalize the above sequence, you'll notice that any pitch can be represented by the formula:
(5 + n*7) mod 12
In this formula, the value of n tells you two important things about how this pitch is named.
• If you divide n by 7, the integer part of the division (technically, the floor) tells you how many sharps (positive) or flats (negative) the note has. For example, if n is in the range 0..6, floor(n/7) = 0 and you get the plain note names listed above. If n is in the range 7..13 (floor(n/7) = 1), you get names with single sharps. In the range 14..20 (floor(n/7) = 2), you get double sharps. In the range -7..-1 (floor(n/7) = -1), you get flats.
• The remainder of dividing n/7 gives you a number from 0-6, which gives you the letter name in the order (F, C, G, D, A, E, B).
As you point out in the comments, this sequence will eventually repeat, since it is modular arithmetic. Indeed, this is true, and it reflects a very important fact about our musical system: no note has a single unique name, but rather can be expressed using any number of different names (note names to pitches are not a one-to-one function). For example, all of the following pitch names map to the same pitch class:
• F = 5
• E♯ = (5 + 12*7) mod12 = 5
• D♯♯♯ = (5 + 24*7) mod12 = 5
• C♯♯♯♯♯ = (5 + 36*7) mod12 = 5
• G♭♭ = (5 - 12*7) mod12 = 5
• A♭♭♭♭ = (5 - 24*7) mod12 = 5
Thus, as you can see, all pitches can technically be described as a sharp or a flat. However, there will also exist a non-sharp and non-flat note name only in the case where the pitch number can be expressed with an n such that floor(n/7) == 0 (in other words, n is in the range 0..6).
• "this pattern can then be continued on indefinitely, by adding or subtracting 7 (mod12) to get the next note in the sequence." - yeah, but it will repeat itself as it's a modular arithmetics. Sep 2 '15 at 8:54
• Can we use another numbering not to start from 5? I want to get a formula for sharp and non-sharp notes. Sep 2 '15 at 8:54
• It's mildy curious that the circle of fifths has to start at `F` to get the 7 `natural` notes, even though `C` is the root of the corresponding Major scale. In a sense the circle fits the Lydian (F) mode better than the Ionian (C). Sep 3 '15 at 22:18
• @hpaulj I found the formula. You need to get f(x) = (x - 5) * 7 mod 12. If f(x) <= 6, the note is non-sharp. Otherwise it's sharp. x >= 0 and x <= 11. Sep 6 '15 at 6:45
A somewhat simpler answer, for us mere mortals. Write the note names around in a circle, as in the numbers on a clock face, in the same order that you did earlier. C can go anywhere - I put it at 12 o'clock. Start at C (no # or b), and count clockwise 7. You get to G. 1#. Go another 7, you get to D. 2#. And so on. Now back to C, this time count anti-clockwise 7. You get F. 1b. On another 7, you get Bb. 2b. Obviously (?) the Bb isn't going to be called A#, because we're now in flat territory. How you make this into an equation is up to you, the mathematician!
• Here's a picture: circle-of-fifths.net/images/circle-of-fifths.gif. There's a key at 6:00 that has two possible names, G flat or F#. But except for that one, there is a clear convention for whether to write the key signature with sharps or with flats so that you don't end up with more than 7 sharps or flats in the key signature. Sep 2 '15 at 13:18
Here's a 'formula' for finding the natural and sharp notes, expressed as Python/numpy calculations (MATLAB would do just as well). It's not a refined calculation, just an easy way to generate the numbers and group them (mixing arrays, sets and sorted lists).
``````i = np.arange(5,200,7) # numbers from 5 up, stepping by 7
natural = set((i%12)[:7]) # modulus by 12; 1st set of 7
# set([0, 2, 4, 5, 7, 9, 11])
next = set((i%12)[7:14]) # 2nd set
# set([0, 1, 3, 5, 6, 8, 10])
sharps = sorted(set(next-natural) # remove the naturals
# [1, 3, 6, 8, 10]
``````
Naturals are the 1st set of 7, sharps are the 2nd set, minus the ones we already identified as naturals (0 and 5).
If I start the count with `0`,
``````naturals: [0, 2, 4, 6, 7, 9, 11])
sharps: [1, 3, 5, 8, 10]
``````
In effect, `F G A B C D E` and `F# G# A# C# D#`.
So I can start the count anywhere, but the location of the half steps in the natural scale will shift.
https://en.wikipedia.org/wiki/Mode_(music)#Summary - describes this connection between modes and the circle.
`````` f(x) = (x - 5) * 7 mod 12.
If f(x) <= 6, the note is non-sharp.
Otherwise it's sharp.
x >= 0 and x <= 11.
In [79]: x=np.arange(0,12)
In [80]: fx=((x-5)*7)%12
In [81]: x[~(fx<7)] # the sharps
Out[81]: array([ 1, 3, 6, 8, 10])
In [82]: x[fx<7] # the naturals
Out[82]: array([ 0, 2, 4, 5, 7, 9, 11])
``````
The 7&12 are producing the circular pattern of `TTSTTTS`; the 5 is anchoring it to the Ionian (CMajor) mode.
• I'll need to read it carefully to comprehend. Sep 4 '15 at 7:47
The previous answers are good, but I think it's also musically significant that since sharps are added through the cycle of fifths means that what you're calling the "sharp keys", taken together form a pentatonic scale. Specifically C#, D#, F#, G#, and A# form the F# major pentatonic scale. | 2,666 | 8,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-49 | latest | en | 0.919094 |
https://math.stackexchange.com/questions/2891283/last-digit-of-a-triangular-number-is-the-midpoint-between-two-primes | 1,566,503,100,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317359.75/warc/CC-MAIN-20190822194105-20190822220105-00146.warc.gz | 553,783,898 | 32,449 | # Last digit of a triangular number is the midpoint between two primes
If a triangular number is even, add and subtract 1 to see if you have found two primes, as in 6 +/1 gives 5 and 7. If the triangular number is odd, add and subtract 2 to see whether or not both are primes, as in 105 +/-2 gives 103 and 107. A small sample found that two primes are found with triangular numbers 6, 15, 45, 105, 231, 465, 741, 861. Is it normal to expect so many with last digit 1 or 5? The cycle for last digits of triangular numbers has 20 terms and last digit 0 appears 4 times and last digit 8 appears 2 times. Last digit 1 appears 4 times, last digit 3 appears 2 times (never a solution), and last digit 5 appears 4 times. Will a larger sample conform to the statistically expect distribution?
• Why could a cyclic sequence with period 20 to have unexpected behaviour in the long run? Perhaps I don't understand what you're asking. – Harald Hanche-Olsen Aug 22 '18 at 18:02
• The unexpected behavior concerns the frequency of finding primes using the triangular numbers. Remember that this method asks for a prime on either side of the triangular number at a distance of either 1 or 2. The cycle gives NO guaranteed of finding a prime; rather it does show when they will NOT be found, as in multiples of 3. – J. M. Bergot Aug 22 '18 at 18:37
• Again, will a larger sample show a preponderance of last digits 1 and 5 for the midpoints between two primes as found by the above method? – J. M. Bergot Aug 22 '18 at 19:16
• Ah silly me; total failure of my reading comprehension, sorry. (Not enough caffeine, perhaps?) Well, since primes are getting farther between as numbers get bigger, I would be surprised if this is true. A few lines of code should let you check it out up to fairly large numbers, though. I'd try that before beginning to conjecture anything. – Harald Hanche-Olsen Aug 23 '18 at 8:08
• A little experimentation keeps Grand Theory healthy. Remember to test for the distance to these two primes being either one away from the midpoint or two away. If you enjoyed this question, math.stackexchange.com/questions/2890178/… – J. M. Bergot Aug 23 '18 at 18:01
I think you are right about even numbers: add $1$ or subtract $1$, since
$3 \cdot even - 1 \lor 3 \cdot even +1 = prime$ (if prefer to use positive coefficients $5$ and $7$ here instead of $1$ and $-1$ as of $0$ is regarded an even as well)
But, in case of odd progression you might consider to add $4$ to see the full pattern:
$3 \cdot odd + 2 \lor 3 \cdot odd + 4 = prime$
So for example a triangular number $15$ using the prime number form above will generate:
$13$, $17$, and also $11$ and $19$. In case of last digit $3$, for which you mentioned to have no solution, a triangular number $153$ will give you two prime numbers: $149$ and $157$, but not $151$ and $155$, since the latter one is the product of $5$ and $31$.
• Was any experiments done with even numbers ending in digit 0 or 8 to see if primes can be found for both -1 and +1? For odds ending in 1 or 5: 21,231,351,741,861,1431,3081..for 5: 5,15,45,105,465,1485,4005... – J. M. Bergot Aug 25 '18 at 17:42
• My answer is rather a suggestion, I haven't checked that for all cases. I will do it in a free time, since your question sounds interesting to me. – usiro Aug 25 '18 at 19:02
• Let there be numbers! Let there be evidence! Good luck with your search. – J. M. Bergot Aug 27 '18 at 17:52 | 953 | 3,430 | {"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} | 3.5625 | 4 | CC-MAIN-2019-35 | latest | en | 0.931415 |
https://www.statisticshowto.com/statistics-symbols/ | 1,657,124,557,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00515.warc.gz | 1,050,259,296 | 16,871 | # Statistics Symbols in Alphabetical Order
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Statistics Basics > Statistics Symbols in Alphabetical Order
## Probability and Statistics Symbols
I’ve put these statistics symbols into quasi-alphabetical order, so that they are easy to find. When I say quasi-alphabetical, I mean that I have ordered them according to what they look like. For example, the Greek letter Beta (β) looks like the letter b, so you’ll find it in the b section. Same for the letter mu(μ), which looks like the letter u.cIf a statistics symbol doesn’t look like a letter, for example the “Greater than” symbol, >, you’ll find it at the very bottom under “miscellaneous.” Most items are clickable, taking you to an article that explains the statistics symbols in more depth.
## Statistics Symbols A to Z
α: significance level (type I error).
b or b0: y intercept.
b1: slope of a line (used in regression).
β: probability of a Type II error.
1-β: statistical power.
BD or BPD: binomial distribution.
CI: confidence interval.
CLT: Central Limit Theorem.
d: difference between paired data.
df: degrees of freedom.
DPD: discrete probability distribution.
E = margin of error.
f = frequency (i.e. how often something happens).
f/n = relative frequency.
HT = hypothesis test.
Ho = null hypothesis.
H1 or Ha: alternative hypothesis.
IQR = interquartile range.
m = slope of a line.
M: median.
n: sample size or number of trials in a binomial experiment.
N: population size.
ND: normal distribution.
σ: standard deviation.
σx̅: standard error of the mean.
σp̂: standard error of the proportion.
p: p-value, or probability of success in a binomial experiment, or population proportion.
ρ: correlation coefficient for a population.
: sample proportion.
P(A): probability of event A.
P(AC) or P(not A): the probability that A doesn’t happen.
P(B|A): the probability that event B occurs, given that event A occurs.
Pk: kth percentile. For example, P90 = 90th percentile.
q: probability of failure in a binomial or geometric distribution.
Q1: first quartile.
Q3: third quartile.
r: correlation coefficient of a sample.
R2: coefficient of determination.
s: standard deviation of a sample.
s.d or SD: standard deviation.
SEM: standard error of the mean.
SEP: standard error of the proportion.
t: t-score.
μ mean.
ν: degrees of freedom.
X: a variable.
Χ2: chi-square.
x: one data value.
: mean of a sample.
z: z-score.
Miscellaneous Statistics Symbols
~ has the distribution of (source).
= equal to.
almost equal to.
> greater than.
< less than.
not equal to.
less than or equal to.
greater than or equal to.
Σ Summation.
## References
Everitt, B. S.; Skrondal, A. (2010), The Cambridge Dictionary of Statistics, Cambridge University Press.
Gonick, L. (1993). The Cartoon Guide to Statistics. HarperPerennial.
Lindstrom, D. (2010). Schaum’s Easy Outline of Statistics, Second Edition (Schaum’s Easy Outlines) 2nd Edition. McGraw-Hill Education
CITE THIS AS:
Stephanie Glen. "Statistics Symbols in Alphabetical Order" From StatisticsHowTo.com: Elementary Statistics for the rest of us! https://www.statisticshowto.com/statistics-symbols/
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Amoung all types of short circuit faults in a transmission
line which one has less number of occurance and why?
Amoung all types of short circuit faults in a transmission line which one has less number of occur..
The short circuit faults with relative occurances are..
1)L-G faults-----75-80 %
2)L-L faults----- 5-7 %
3)L-L-G faults---10-12 %
4)L-L-L-G faults- 8-10 %
L-L faults has less number of occurance because the
distance between the two lines is adequate.
Author:arbelectbit@yahoo.com
Is This Answer Correct ? 3 Yes 0 No
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• Engineering AllOther (1377) | 608 | 2,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-21 | latest | en | 0.907615 |
https://cstheory.stackexchange.com/questions/18074/simplification-of-weighted-nfa/18077 | 1,624,146,424,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487653461.74/warc/CC-MAIN-20210619233720-20210620023720-00134.warc.gz | 181,619,674 | 40,172 | # Simplification of weighted NFA
What options does one have for the simplification (meaning reduction in the number of states) of weighted NFA over the probability semiring? From my understanding one can determinize, and then minimize an automaton, but
• The minimal DFA can actually have more states than the original NFA
• Not every weighted NFA is determinizable (in fact, most of them aren't)
My aim, however, is not necessarily to minimize an automaton, but simply reduce its complexity. So, what I'm looking for is a simplification algorithm that works directly with NFA and, while it doesn't guarantee to find the optimal solution, can simplify the automata in some cases significantly, and, presumably, has time complexity polynomial in the number of states. Are there any such algorithms known in literature?
By simplification I understand either minimization or determinization. I'll try to sum up what I know about both problems, in the quite general setting of weighted automata over arbitrary semiring. The original works were done by Marcel-Paul Schützenberger (who introduced them), and you'll find a nice account of what is known about them in the book Elements of Automata Theory by Jacques Sakarovitch (also available in French): For shorter explanations, check out the lecture notes by Jacques Sakarovitch again:
For both minimization and determinization (called "sequentialization") there are nice theoretical answers. For instance, every weighted automaton over a field can be minimized in polynomial time (see for instance this lecture note).
Mayr and Clemente have shown that it is often possible to simplify NFAs. Their techniques rely on pruning the underlying labelled transition system via local approximations of trace inclusions. As far as I can tell, this technique would still apply in the weighted case.
• Thanks for the link, I wasn't aware of simplification techniques based on simulation relations. However, it's not immediately obvious to be how can one extend the concept of simulation game to the weighted case. The problem is that it's hard for Spoiler to prove that her state can't be simulated if the accepted languages induced by both her and Duplicator's states are the same and only weights are different, because in order to show that she has to aggregate weights for some word along all the accepting paths. Do you have any idea on how to fix that? – hr0nix Jun 22 '13 at 18:26
• Doesn't the locality restriction help with that? – András Salamon Jun 22 '13 at 20:28
• If by the locality restriction you mean small lookahead, then it doesn't seem to help. The problem as I see it is that even trace inclusions aren't sufficient in weighted case, not to speak about their local approximations. And the reason is that in the non-weighted case Spoiler can prove her point with a single trace, while in weighted case she needs all the traces since weights have to be aggregated. – hr0nix Jun 22 '13 at 20:49
• Hmmm, you do have a point. My intuition that this could be used may have been wrong. – András Salamon Jun 22 '13 at 21:20
Actually there is an algorithm for approximated determinization of a weighted NFA, by Aminoff Kupferman and Lampert, where the approximation factor can be determined beforehand (if I remember correctly).
See here.
• This algorithm is over the tropical semiring, not the probabilistic. Making the transition is not certain to be simple, or even possible. – Shaull Jun 19 '13 at 15:06
• Also, what about the concern that minimal DFA can have more states than the original NFA because of the exponential state space explosion? – hr0nix Jun 19 '13 at 17:30
• @Shaull is ofcourse correct, I missed that. – Shir Jun 23 '13 at 8:22
• @hr0nix, "reduce the complexity of an NFA" is not well defined. In my opinion determinization is simplification of sort. – Shir Jun 23 '13 at 8:25
• @Shir, in the question I define simplification as a reduction of the number of states. – hr0nix Jun 23 '13 at 11:53 | 919 | 3,974 | {"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-2021-25 | latest | en | 0.948981 |
http://mathhelpforum.com/calculus/40892-c4-vector-question.html | 1,481,299,263,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542712.49/warc/CC-MAIN-20161202170902-00189-ip-10-31-129-80.ec2.internal.warc.gz | 181,703,397 | 10,098 | 1. ## c4 vector question
The point A with coordinates (0,-5,11) lies on the line l1, which has equation
r=6i+19j-k+(lamda)
(i+4-2k)
The point p lies on l1 and is such that OP is perindicular to l1, where O is the origin.
Find the position vectore of point P.
Given that B had coordinates (5,15,1)
Show that the points A, P and B are collinear and find ratio AP:PB
Thanks
Bryn
2. Have you found the values of $\lambda$ that corresponds to each point a , b and p ?
Once you find them you can consider differences of values of $\lambda$ as distances between points on the line, that should help with finding the ratio.
Finding the values of $\lambda$ for a should not be a problem
for p, your given it lies on $l_{1}$ so the position of p is $\left(\begin{array}{c}6 + \lambda \\19 + 4 \lambda \\-1 - 2 \lambda\end{array}\right)$ and you're given that OP it is perpendicular to the the line $l_{1}$ so $\left(\begin{array}{c}6 + \lambda \\19 + 4 \lambda \\-1 - 2 \lambda\end{array}\right) \cdot \left(\begin{array}{c}6 \\19 \\-1\end{array}\right) = 0$, So now you can solve for $\lambda$
Bobak | 336 | 1,103 | {"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": 8, "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-2016-50 | longest | en | 0.825201 |
https://ask.sagemath.org/users/26641/tk/?sort=recent | 1,701,496,172,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100327.70/warc/CC-MAIN-20231202042052-20231202072052-00420.warc.gz | 144,921,571 | 8,983 | 2022-03-02 16:49:56 +0100 commented answer lifting modular symbols for newform of level 35 at p = 5, 7 (continued) The reason seems to be that one cannot hash elements of proper extensions of Q_p. How can I resolve this, ho 2022-03-02 16:49:37 +0100 commented answer lifting modular symbols for newform of level 35 at p = 5, 7 I'm using N = 188 and p = 3 now. p does not divide N, so we have to p-stabilize (this is different from your answer wher 2022-03-02 16:49:21 +0100 commented answer lifting modular symbols for newform of level 35 at p = 5, 7 I'm using N = 188 and p = 3 now. p does not divide N, so we have to p-stabilize (this is different from your answer wher 2021-04-29 20:40:10 +0100 received badge ● Notable Question (source) 2020-12-06 14:38:00 +0100 received badge ● Popular Question (source) 2020-04-27 06:24:09 +0100 commented answer PyCharm for SageMath on Linux Ok. Thank you anyway! 2020-04-23 09:11:02 +0100 commented answer PyCharm for SageMath on Linux Only one problem left: How can I use Sage specific syntax (^ instead of **, for example) in PyCharm when not using sage -sh? 2020-04-23 09:07:44 +0100 commented answer lifting modular symbols for newform of level 35 at p = 5, 7 One solution would be to express the K-valued modular symbol (K = NumberField(x²+x-4)) as a K-linear combination of QQ-valued modular symbols and do the procedure for the latter ones. I'm working on this. 2020-04-19 15:16:33 +0100 received badge ● Scholar (source) 2020-04-19 15:16:29 +0100 commented answer PyCharm for SageMath on Linux The problem was that PyCharm used its own python interpreter, not the /usr/bin/python one. I changed it and now it works. 2020-04-19 12:55:46 +0100 commented answer lifting modular symbols for newform of level 35 at p = 5, 7 I think one does not need to $p$-stabilize when p | N: http://math.bu.edu/people/rpollack/Pa... However, only calling phi1.lift instead of phi1.p_stabilize_and_lift also fails. (And sorry for my late response!) 2020-04-19 12:46:19 +0100 received badge ● Editor (source) 2020-04-19 12:45:20 +0100 asked a question PyCharm for SageMath on Linux I am looking for an IDE with syntax highlighting, code completion and debugging support to run SageMath 9.0 with preprocessing on (Arch) Linux. The two most obvious choices seem to be PyCharm and Eclipse, but PyCharm is not able to do from sage.all import * even though my SAGE_ROOT="/usr" and I am running pycharm from a sage -shas described in https://ask.sagemath.org/question/397... or https://ask.sagemath.org/question/387.... 2020-04-19 11:49:01 +0100 received badge ● Supporter (source) 2019-09-25 04:00:26 +0100 received badge ● Good Question (source) 2019-09-25 03:02:28 +0100 received badge ● Nice Question (source) 2019-09-24 09:51:45 +0100 received badge ● Student (source) 2019-09-23 15:58:16 +0100 asked a question lifting modular symbols for newform of level 35 at p = 5, 7 Let $f$ be the unique normalised eigenform in $S_2(\Gamma_0(35))$ of dimension $2$. It has split multiplicative reduction at $p = 5$ ($a_p = +1$) [and non-split multiplicative reduction at $p = 7$ ($a_p = -1$)]. The $p$-adic $L$-function should vanish to the order $1$ at $1$ (because the associated abelian variety has rank $0$). I want to compute the valuation of its leading coefficient using Pollack-Stevens. To do so, I use the following code: from sage.modular.pollack_stevens.space import ps_modsym_from_simple_modsym_space A = ModularSymbols(35,2,1).cuspidal_submodule().new_subspace().decomposition()[1] p = 5 prec = 2 phi = ps_modsym_from_simple_modsym_space(A) ap = phi.Tq_eigenvalue(p,prec) phi1,psi1 = phi.completions(p,prec) phi1p = phi1.p_stabilize_and_lift(p,ap = psi1(ap), M = prec) Unfortunately, the last command fails after a few seconds (also for $p = 7$) with a RuntimeError: maximum recursion depth exceeded while calling a Python object Is there a theoretical problem with computing the $L$-value or is there a problem with the implementation? | 1,208 | 3,969 | {"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} | 3.03125 | 3 | CC-MAIN-2023-50 | latest | en | 0.897646 |
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# D is a point on the side BC of △ABC such that ∠ADC=∠BAC. Prove that CACD=CBCA or CA2=CB×CD
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Solution
## In △ABC and △DAC, we have∠ADC=∠BAC and ∠C=∠C∴ By AA-axiom of similarity△ABC∼△DAC⇒ ABDA=BCAC=ACDC⇒ CBCA=CACD ....Hence, proved
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Tips for solving Linear Equations Quickly
When it comes to solving linear equations quickly, mastering the right techniques can make all the difference. Begin by identifying the form of the equation, whether it’s in standard form, slope-intercept form, or point-slope form. Next, apply the appropriate method, such as the graphical method for visualization, elimination or substitution method for simultaneous equations, cross-multiplication for dealing with fractions, or leveraging matrices and determinants for complex systems.
How to Solve Linear Equation Questions & Definition
• A linear equation is an equation where variable quantities are in the first power only and whose graph is a straight line.
• The above line represent as, y= mx + c
Methods to solve Linear Equation :
There are mainly 6 methods to solve Linear Equation :
1. Graphical Method
2. Elimination Method
3. Substitution Method
4. Cross Multiplication Method
5. Matrix Method
6. Determinants Method
Let’s explain each method in terms of solving linear equations:
1. Graphical Method: The graphical method involves graphing both sides of a linear equation on the same coordinate plane and finding the point(s) where the graphs intersect. The solution to the equation is the x-coordinate (and corresponding y-coordinate) of the point(s) of intersection. In this method, the graphical representation of the equation visually illustrates the solution(s) to the equation.
2. Elimination Method: In the elimination method, you aim to eliminate one variable by adding or subtracting multiple equations to make one variable’s coefficient cancel out. This process leads to an equation with only one variable, which can be easily solved. After finding the value of one variable, you can substitute it back into one of the original equations to find the value of the other variable.
3. Substitution Method: The substitution method involves solving one variable in terms of the other from one of the equations and then substituting this expression into the other equation. By doing so, you reduce the system of equations to a single equation with one variable, which can then be solved easily.
4. Cross Multiplication Method: The cross multiplication method is used to solve equations with fractions. When you have an equation with fractions, you cross-multiply by multiplying the numerator of one fraction with the denominator of the other and vice versa. This process eliminates the fractions and simplifies the equation, making it easier to solve.
5. Matrix Method: In the matrix method, you represent a system of linear equations using matrices and use matrix operations to solve for the variables. This method is particularly useful when dealing with large systems of equations, as it simplifies the process of solving using matrices and their properties.
6. Determinants Method: The determinants method involves using the concept of determinants from linear algebra to solve a system of linear equations. You form a matrix with the coefficients of the variables, a matrix with the constants on the right-hand side, and then find the determinants of these matrices. By using Cramer’s rule or other determinant-based methods, you can solve for the variables.
Type 1: Solve Linear Equations Questions Quickly, Find the value of x or y.
Question 1. If 3a + 7b = 75 and 5a – 5b = 25, what is the value of a + b?
Options:
A. 11
B. 6
C. 5
D. 17
Solution: 3a + 7b = 75 ……(1)
5a – 5b = 25 (divide the equation by 5)
we get, a – b = 5 …….(2)
Now multiplying eq. (2) by 7
and add to eq. (1), we get
3a + 7b = 75
7a – 7b = 35
On solving
10a = 110
a = $\frac{110}{10}$
a = 11
Now put the value of a in eq (2)
11 – b = 5
b = 11 – 5
b = 6
Therefore, a = 11 and b = 6
The value of a + b = 6 + 11 = 17
Correct option: D
Question 2. If 2x + y = 16 and 16x – y = 2, then find the value of x?
Options:
A. $\frac{1}{4}$
B. $\frac{17}{4}$
C. $\frac{17}{8}$
D. 4
Solution: Given, 2x + y = 16
2x + y = 24
x + y = 4….(1)
Now, 16x – y = 2
(24) x – y = 2¹
x – y = $\frac{1}{4}$ ….(2)
On solving equation 1 and 2
We get,
2x = $\frac{17}{4}$
x = $\frac{17}{4 × 2}$ = $\frac{17}{8}$
Correct option: C
Question 3. The system of equations 3a + 5b = 6 and 6a + 10y = 6 has
Options:
A. No solution
B. One solution
C. Two solution
D. Infinite solution
Solution: $\frac{a_{1}}{a_{2}}$ = 3/6 = 1/2
$\frac{b_{1}}{b_{2}}$= 5/10 = 1/2
$\frac{c_{1}}{c_{2}}$ = $\frac{6}{6}$= 1
$\frac{a_{1}}{a_{2}}$ $\frac{b_{1}}{b_{2}}$ $\frac{c_{1}}{c_{2}}$
Therefore, there is no solution
Correct option: A
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Type 2: Solve Quickly Linear Equations Questions, Based on Word problems
Question 1. The difference between the two numbers is 45. The ratio of the two numbers is 8:3. Find the two numbers?
Options:
A. 72 and 27
B. 90 and 45
C. 81 and 36
D. 60 and 15
Solution: Let the first number be 8x
Let the second number be 3x
Now, the difference between the two numbers is 45
Therefore, 8x – 3x = 45
5x = 45
x = $\frac{45}{5}$
x = 9
Now, put the value of x in
8x = 8 × 9 = 72
3x = 3 × 9 = 27
Correct option: A
Question 2. The breadth of a rectangle is twice its length. If the perimeter of the rectangle is 84m. Then, calculate the length and breadth of the rectangle?
Options:
A. L=12 and B= 24
B. L = 14 and B = 28
C. L = 28 and B = 14
D. L = 24 and B = 12
Solution: Perimeter of rectangle = 2 (l+b)
Length of the rectangle = x
Breadth of the rectangle = 2x
Perimeter of the rectangle = 84
2 (x + 2x) = 84
2 (3x) = 84
6x = 84
x = $\frac{84}{6}$
x = 14
Therefore, the Length of the rectangle = 14m
And Breadth of the rectangle = 14 × 2 = 28m
Correct option: B
Question 3. Ajay bought 5 tickets for two concerts A and B and 10 tickets for concert A and C. He paid Rs. 350. Now the total of a ticket for concert A and B and ticket of A and C is Rs. 42, then what is the ticket price for concert A and B?
Options:
A. Rs. 10
B. Rs. 42
C. Rs. 14
D. Rs. 28
Solution: Let the ticket price of concert A and B = a
Let the ticket price of concert A and C = b
According to the question, a + b = 42…… (1)
Ticket bought by Ajay = 5a + 10b = 350
= a + 2b = 70……(2)
Now solve equation 1 and 2
a + b = 42
a + 2b = 70
b = 70 – 42
b = 28
Now put the value of b in equation 1
a+ 28 = 42
a = 42 – 28
a = 14
Hence, the ticket price for concert A and B = Rs. 14
Correct option: C
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Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others | 2,002 | 6,802 | {"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": 0, "wp-katex-eq": 17, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.865711 |
https://www.physicsforums.com/threads/energy-required-to-transfer-an-object-to-a-higher-orbit.370725/ | 1,545,205,590,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831715.98/warc/CC-MAIN-20181219065932-20181219091932-00454.warc.gz | 1,003,455,171 | 13,545 | # Homework Help: Energy required to transfer an object to a higher orbit
1. Jan 19, 2010
1. The problem statement, all variables and given/known data
A spaceship of mass m circles a planet (mass = M) in an orbit of radius R. How much energy is required to transfer the spaceship to a circular orbit of radius 3R?
2. Relevant equations
K=.5mv$$^{2}$$
U$$_{g}$$=-$$\frac{GMm}{r}$$
v=2(pi)r/T
v=(ar)^.5
3. The attempt at a solution
I had no idea how to do this problem. I thought about subtracting the total final energy from the total initial energy to get the energy input required, but I didn't know how to find v. The two equations I could think of for find v are listed above
It would be nice if you guys could at least give me a hint as to how to solve this test problem, since I get to have a second try at it tomorrow.
2. Jan 19, 2010
### D H
Staff Emeritus
What force is needed to keep something in uniform circular motion? How does this relate to the gravitational force that makes a the spaceship orbit the planet?
3. Jan 19, 2010
That would be the centripetal force, which is mv^2/r
And that's equal to the gravitational force, GMm/r^2
Which leads me to v^2 = GM/r
I substituted that into the equations I had and got -GMm/3r for a final answer. Jeez, the solution was that simple??? Thank you very much for helping me! :)
4. Jan 19, 2010
### D H
Staff Emeritus
That simple. However -- are you sure about the sign of your result? Does that make sense?
5. Jan 19, 2010 | 411 | 1,490 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-51 | latest | en | 0.953333 |
http://mathhelpforum.com/differential-geometry/152443-integral-via-residues-2.html | 1,527,380,694,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867949.10/warc/CC-MAIN-20180526225551-20180527005551-00244.warc.gz | 188,438,029 | 12,917 | # Thread: Integral via residues
1. Thank you very much. I suspected an error there, but I didn't see it at all. I'll be careful next time.
2. I've done that error so many times myself!
3. Your poles are outside the contour, because their magnitudes are both greater than pi, which is the radius of the semi-circle.
So, your integral from -pi to pi has become equal to the negative of the integral along the semicircle, since the total integral is, as you've mentioned, zero. So, now you have to compute the negative of the integral along the semicircle. How will you propose to do that?
4. Originally Posted by Ackbeet
Your poles are outside the contour, because their magnitudes are both greater than pi, which is the radius of the semi-circle.
So, your integral from -pi to pi has become equal to the negative of the integral along the semicircle, since the total integral is, as you've mentioned, zero. So, now you have to compute the negative of the integral along the semicircle. How will you propose to do that?
I'm sure I'm not understanding well. Here is what I believe (but I know is false): $\displaystyle \int _{- \pi}^{\pi} \frac{d \theta}{2 \cos (\theta )} = \oint _ C \frac{dz}{1+ 2 \cos (z )}=0$ according to Cauchy's integral theorem. With C being the closed contour of the semi circle going from -pi to pi and being in the 2 first quadrants (upper ones).
5. I think you are making this way too hard and I think the value of your integral is incorrect.
Here is goes First as Olpeg noted is that
$\displaystyle \displaystyle \int_{0}^{\pi}\frac{d\theta}{2+\cos(\theta)}=\frac {1}{2}\int_{0}^{2\pi}\frac{d\theta}{2+\cos(\theta) }$
Now use the substitution $\displaystyle z=e^{i\theta} \implies dz=ie^{i\theta}d\theta \iff d\theta =\frac{dz}{iz}$
Note that $\displaystyle \cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{z+z^{-1}}{2}$
Now we sub all of this into the integral to get
$\displaystyle \displaystyle \int_{0}^{\pi}\frac{d\theta}{2+\cos(\theta)}=\frac {1}{2}\oint_{|z|=1}\frac{1}{(2+\frac{z+z^{-1}}{2})}\frac{dz}{iz}=\frac{1}{i}\oint\frac{dz}{z^ 2+4z+1}=$
$\displaystyle \displaystyle \frac{1}{i}\oint \frac{dz}{(z+2-\sqrt{3})(z+2+\sqrt{3})}$
Now there is only one pole in the interior of the unit circle $\displaystyle z=-2+\sqrt{3}$
So now by the residue theorem we get
$\displaystyle \displaystyle \frac{1}{i}\oint \frac{dz}{(z+2-\sqrt{3})(z+2+\sqrt{3})}=\frac{1}{i}2\pi i \left( \frac{1}{-2+\sqrt{3}+2+\sqrt{3}}\right)=\frac{\pi}{\sqrt{3}}$
6. Thanks TheEmptySet. Could you justify the step $\displaystyle \frac{1}{i}\oint \frac{dz}{(z+2-\sqrt{3})(z+2+\sqrt{3})}=\frac{1}{i}2\pi i \left( \frac{1}{-2+\sqrt{3}+2+\sqrt{3}}\right)$? I know that the line integral is worth $\displaystyle 2 \pi i \sum res (f,-2+\sqrt 3 )$.
I also don't understand well the method. Why did we choose a circle with radius 1? Why not say with radius 0.1 so that we enclose no pole and the integral is worth 0? I never understood the method of complex analysis to calculate real integrals.
7. I don't think we are really choosing a contour. Think of it this way. When you have a line integral with respect to z, to evaluate it you choose a contour (if you are not given one) and then do a substitution(say for a circular contour, it is $\displaystyle z = r\,e^{i\theta}$).
However, in this problem you need to go the other way, since you are already given theta. In essence, the contour has already been chosen for you.
8. Originally Posted by arbolis
Thanks TheEmptySet. Could you justify the step $\displaystyle \frac{1}{i}\oint \frac{dz}{(z+2-\sqrt{3})(z+2+\sqrt{3})}=\frac{1}{i}2\pi i \left( \frac{1}{-2+\sqrt{3}+2+\sqrt{3}}\right)$? I know that the line integral is worth $\displaystyle 2 \pi i \sum res (f,-2+\sqrt 3 )$.
I also don't understand well the method. Why did we choose a circle with radius 1? Why not say with radius 0.1 so that we enclose no pole and the integral is worth 0? I never understood the method of complex analysis to calculate real integrals.
When you have a simple pole at $\displaystyle z=a$. Then the residue at $\displaystyle a$ is
$\displaystyle \displaystyle \lim_{z \to a}(z-a)f(z)$
So in the case above we have
Res$\displaystyle (f,z=-2+\sqrt{3})$$\displaystyle \displaystyle =\lim_{z\to -2+\sqrt{3}}(z+2-\sqrt{3})\frac{1}{(z+2-\sqrt{3})(z+2+\sqrt{3})}=\frac{1}{2\sqrt{3}}$
9. Reply to arbolis at post # 19:
Actually, part of what you wrote down is correct. What you're not seeing is that the integral in the middle of your equation is composed of two integrals, which, when summed, add up to an integral that's equal to zero. Let $\displaystyle C_{1}$ be defined as the segment on the real axis from $\displaystyle -\pi$ to $\displaystyle \pi$. Let $\displaystyle C_{2}$ be the semicircle path from $\displaystyle \pi$ to $\displaystyle -\pi$ all in the upper half plane. Then what you've really got is this:
$\displaystyle \displaystyle{\oint_{C}\frac{dz}{2+\cos(z)}=\int_{ C_{1}}\frac{dz}{2+\cos(z)}+\int_{C_{2}}\frac{dz}{2 +\cos(z)}=0.}$
Now, your original integral is
$\displaystyle \displaystyle{\int_{0}^{\pi}\frac{d\theta}{2+\cos( \theta)}=\frac{1}{2}\int_{-\pi}^{\pi}\frac{dz}{2+\cos(z)}.}$
Hence, from our equation above, we have that
$\displaystyle \displaystyle{\int_{0}^{\pi}\frac{d\theta}{2+\cos( \theta)}=-2\int_{C_{2}}\frac{dz}{2+\cos(z)}.}$
So, in order to finish this problem, you must compute the contour integral on the RHS of this equation. Is this making more sense now?
Incidentally, I think this method of doing the problem is going to end up being about the same amount of work as the other methods described. As a matter of fact, you could use the same substitution Opalg recommended to perform this integral.
Page 2 of 2 First 12 | 1,773 | 5,733 | {"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} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.911999 |
http://www.chegg.com/homework-help/repeat-exercise-data-choice-data-involve-average-monthly-tem-chapter-5.5-problem-110e-solution-9780321559852-exc | 1,475,227,602,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662133.5/warc/CC-MAIN-20160924173742-00233-ip-10-143-35-109.ec2.internal.warc.gz | 381,285,259 | 20,157 | # Algebra and Trigonometry (4th Edition) View more editions Solutions for Chapter 5.5 Problem 110EProblem 110E: Repeat Exercise for data of your choice. The data can involv...
• 9894 step-by-step solutions
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Chapter: Problem:
Repeat Exercise for data of your choice. The data can involve the average monthly temperatures for the region where you live or any data whose scatter plot takes the form of a sinusoidal function.
Exercise
The data show the average monthly temperatures for Washington, D.C.
x (Month) Average Monthly Temperature, °F 1 (January) 34.6 2 (February) 37.5 3 (March) 47.2 4 (April) 56.5 5 (May) 66.4 6 (June) 75.6 7 (July) 80.0 8 (August) 78.5 9 (September) 71.3 10 (October) 59.7 11 (November) 49.8 12 (December) 39.4
a. Use your graphing utility to draw a scatter plot of the data from x = 1 through x = 1 2
b. Use the SINe REGression feature to find the sinusoidal function of the form y = A sin(Bx + C) + D that best fits the data.
c. Use your graphing utility to draw the sinusoidal function of best fi t on the scatter plot.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 4
The data show the average monthly temperature for a Hill Station.
Serial NO Month Average Monthly Temperature F 1 January 33.6 2 February 36.5 3 March 45.2 4 April 55.5 5 May 65.4 6 June 73.3 7 July 79.5 8 August 75.3 9 September 70.3 10 October 57.8 11 November 48.8 12 December 38.4
a)
The display shows the scatter plot of the data from through.
• Chapter , Problem is solved.
Corresponding Textbook
Algebra and Trigonometry | 4th Edition
9780321559852ISBN-13: 0321559851ISBN: Robert F BlitzerAuthors:
Alternate ISBN: 9780131362185, 9780321575067, 9780321575449, 9780321575463, 9780321575470, 9780321575487, 9780321640970, 9780321642417, 9780321655936, 9780321666451, 9780321687364, 9780321691774, 9780321744548, 9780321746214, 9780321848147 | 649 | 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} | 3.859375 | 4 | CC-MAIN-2016-40 | latest | en | 0.834729 |
https://mtp.tools/converters/gibibit-to-kibibyte-calculator | 1,571,646,972,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987763641.74/warc/CC-MAIN-20191021070341-20191021093841-00204.warc.gz | 614,333,696 | 5,491 | # Gibibits (Gib) to Kibibytes (KiB) calculator
Input the amount of gibibits you want to convert to kibibytes in the below input field, and then click in the "Convert" button. But if you want to convert from kibibytes to gibibits, please checkout this tool.
## Formula
Formula used to convert Gib to KiB:
F(x) = x * 131072
For example, if you want to convert 15 Gib to KiB, just replace x by 15 [Gib]:
15 Gib = 15*131072 = 1966080 KiB
## Steps
1. Multiply the amount of gibibits by 131072.
2. The result will be expressed in kibibytes.
## Gibibit to Kibibyte Conversion Table
The following table will show the most common conversions for Gibibits (Gib) to Kibibytes (KiB):
Gibibits (Gib) Kibibytes (KiB)
0.001 Gib 131.072 KiB
0.01 Gib 1310.72 KiB
0.1 Gib 13107.2 KiB
1 Gib 131072 KiB
2 Gib 262144 KiB
3 Gib 393216 KiB
4 Gib 524288 KiB
5 Gib 655360 KiB
6 Gib 786432 KiB
7 Gib 917504 KiB
8 Gib 1048576 KiB
9 Gib 1179648 KiB
10 Gib 1310720 KiB
20 Gib 2621440 KiB
30 Gib 3932160 KiB
40 Gib 5242880 KiB
50 Gib 6553600 KiB
60 Gib 7864320 KiB
70 Gib 9175040 KiB
80 Gib 10485760 KiB
90 Gib 11796480 KiB
100 Gib 13107200 KiB
A gibibit is a unit of measurement for digital information and computer storage. The binary prefix gibi (which is expressed with the letters Gi) is defined in the International System of Quantities (ISQ) as a multiplier of 2^30. Therefore, 1 gibibit is equal to 1,024 mebibits and equal to 1,073,741,824 bits (around 1.073 gigabits). The symbol commonly used to represent a gibibit is Gib (sometimes as Gibit).
A kibibyte is a unit of measurement for digital information and computer storage. The binary prefix kibi (which is expressed with the letters Ki) is defined in the International System of Quantities (ISQ) as a multiplier of 2^10. Therefore, 1 kibibyte is equal to 1,024 bytes. The symbol used to represent a kibibyte is KiB.
## FAQs for Gibibit to Kibibyte calculator
### What is Gibibit to Kibibyte calculator?
Gibibit to Kibibyte is a free and online calculator that converts Gibibits to Kibibytes.
### How do I use Gibibit to Kibibyte?
You just have to insert the amount of Gibibits you want to convert and press the "Convert" button. The amount of Kibibytes will be outputed in the input field below the button.
### Which browsers are supported?
All mayor web browsers are supported, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari and Opera.
### Which devices does Gibibit to Kibibyte work on?
Gibibit to Kibibyte calculator works in any device that supports any of the browsers mentioned before. It can be a smartphone, desktop computer, notebook, tablet, etc. | 808 | 2,635 | {"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} | 3.265625 | 3 | CC-MAIN-2019-43 | latest | en | 0.714626 |
https://www.convertunits.com/from/myriawatt-hour/to/calorie | 1,627,182,413,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151563.91/warc/CC-MAIN-20210725014052-20210725044052-00259.warc.gz | 703,587,772 | 16,985 | ## ››Convert myriawatt hour to calorie [15 °C]
myriawatt-hour calorie
Did you mean to convert myriawatt-hour to calorie [15 °C] calorie [I.T.] calorie [nutritional] calorie [thermochemical]
How many myriawatt-hour in 1 calorie? The answer is 1.1627222222222E-7.
We assume you are converting between myriawatt hour and calorie [15 °C].
You can view more details on each measurement unit:
myriawatt-hour or calorie
The SI derived unit for energy is the joule.
1 joule is equal to 2.7777777777778E-8 myriawatt-hour, or 0.23890295761862 calorie.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between myriawatt hours and calories.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of myriawatt-hour to calorie
1 myriawatt-hour to calorie = 8600506.47427 calorie
2 myriawatt-hour to calorie = 17201012.94854 calorie
3 myriawatt-hour to calorie = 25801519.42281 calorie
4 myriawatt-hour to calorie = 34402025.89708 calorie
5 myriawatt-hour to calorie = 43002532.37135 calorie
6 myriawatt-hour to calorie = 51603038.84562 calorie
7 myriawatt-hour to calorie = 60203545.31989 calorie
8 myriawatt-hour to calorie = 68804051.79416 calorie
9 myriawatt-hour to calorie = 77404558.26843 calorie
10 myriawatt-hour to calorie = 86005064.7427 calorie
## ››Want other units?
You can do the reverse unit conversion from calorie to myriawatt-hour, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Calorie
15 °C calorie: the amount of energy required to warm 1 g of air-free water from 14.5 °C to 15.5 °C at a constant pressure of 101.325 kPa (1 atm). Experimental values of this calorie ranged from 4.1852 J to 4.1858 J. The CIPM in 1950 published a mean experimental value of 4.1855 J, noting an uncertainty of 0.0005 J.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 668 | 2,342 | {"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-2021-31 | latest | en | 0.673806 |
http://www.askmehelpdesk.com/elementary-school/give-nearest-lenght-1-8-inch-635862.html | 1,369,348,535,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704007597/warc/CC-MAIN-20130516113327-00074-ip-10-60-113-184.ec2.internal.warc.gz | 311,781,751 | 11,556 | # Give the nearest lenght for 1/8 inch?
Asked Feb 13, 2012, 09:47 PM —
Give the nearest lenght for 1/8 inch
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Sarah was asked to round 4 7/8 in to the nearest 1/2 inch and to the nearest inch. Her answer was the same. Explain why Sarah is correct. Would it be 5 inches. :) | 224 | 715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2013-20 | latest | en | 0.928766 |
http://www.slideshare.net/Ashwin12345/lab-report-for-water-experiment | 1,469,494,152,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824499.16/warc/CC-MAIN-20160723071024-00097-ip-10-185-27-174.ec2.internal.warc.gz | 700,765,881 | 27,134 | Upcoming SlideShare
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# Lab report for water experiment
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### Lab report for water experiment
1. 1. Name: ASHWIN BHAKRE Partners: ASHWIN, ALEX DAUGHTRY Date of experiment: 24th and 29th September 2010 Write your Title Here Aim: We are trying to find out if water takes longer time to cool down if the starting temperatures are different. Hypothesis: I think the hotter the water is, the lesser it will cool down in 10 minutes. Variables: Input variable: The starting temperature of the water is our groups input variable. We shall change it by keeping the measured amount of water on the flame for a longer period of time. Output variable: Our output variable is the time the water takes to cool down. We shall measure it by using stopwatches. Control variables: Control variable 1: Our first control variable is the amount of water we are going to use. I plan to keep it constant by using a graduated cylinder to measure it out.
2. 2. (What’s the temperature of the water? You should give more details) Control variable 2: Our second control variable is the room temperature. I plan to keep it constant by keeping an eye on the A/C remote. Control variable 3: Our third control variable is the beaker we are going to use. I plan to keep this one constant by using the same beaker.(How big is the beaker? You have to give a specific number ) Control variable 4: Our fourth control variable is the thermometer we are going to use. I plan to keep this one constant by using the same (red oil) thermometer every time so that there are no errors. Control variable 5: Our fifth control variable is the cooling time. We plan to keep this the same by using a stopwatch to measure out 10 minutes exactly. Materials: • Bunsen Burner • Gauze mat • Tripod • Retort stand • Boss head • 250 ml beaker
3. 3. • Graduated cylinder (100ml) • Thermometer (red oil) • Lighter • Safety glasses/Apron • Metal tongs • Water(how much?) • Bunsen Burner Mat SORRY NO DIAGRAM ! Method: 1. Set up equipment and put on safety equipment. 2. Measure out exactly 100ml of water in a graduated cylinder and pour it into the beaker. 3. Place the beaker (250 ml)on top of the Bunsen burner and with the thermometer inside. 4. Turn the Bunsen burner on and wait till the designated temperature of the water is reached. 5. Using the metal tongs pick up the beaker and place it onto the Bunsen burner mat and start the stopwatch. 6. Observe the temperature drop for 10 minutes and record data into your book for every minute. 7. Repeat the process with the 2 different starting temperatures of the water. Results: STARTING STARTING STARTING TIME IN TEMP. AT TEMP. AT TEMP. AT MINUTES 60 ˚C 70˚C 80˚C 1 minute 58˚C 70˚C 80˚C
4. 4. 2 minute 57˚C 69˚C 75˚C 3 minute 57˚C 67˚C 71˚C 4 minute 56˚C 64˚C 67˚C 5 minute 56˚C 61˚C 64˚C 6 minute 54˚C 59˚C 63˚C 7 minute 53˚C 57˚C 59˚C 8 minute 52˚C 55˚C 57˚C 9 minute 50˚C 54˚C 55˚C 10 minute 50˚C 53˚C 53˚C TOTAL 10˚C in 10 17˚C in 10 27˚C in 10 TEMPERATURE minutes minutes minutes DROP Conclusion: My hypothesis was proved wrong because our results clearly showed that in 10 minutes time the 60˚C water dropped only 10˚C while the 70˚C water dropped 13˚C and
5. 5. 80˚C water dropped a whooping 23˚C. There was one factor that I think must have affected the experiment the most. This experiment was conducted in two different lessons so in the first lesson after I had heated the water up to 60˚C I had just let it stay on the gauze mat and cool. But on the other lesson for the other two tests I had taken the beaker off the gauze mat and put it on the Bunsen burner mat. I think because of leaving the beaker on the gauze mat the water did not cool down as much as the others because the gauze mat must have been hot too which might have made the cooling down process slower. Evaluation: I think my experiment gave accurate results from the second two tests but the first one kind of spoiled it but it still proved my hypothesis wrong so I would call this a successful experiment because there is something that always goes wrong in experiments and we should learn from that mistake. I sure have learnt something from it, which is that you should always do the same thing in an experiment because if you don’t it can change the whole results by a mile. Next time I do this experiment I would try to make my measurements more accurate and keep things constant for each test. I think this experiment was a good one because I learnt something from it. Good report, it is very specific, but it would be better if you have more details. | 1,329 | 4,837 | {"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-2016-30 | latest | en | 0.904199 |
http://weblogs.sqlteam.com/mladenp/archive/2006/06/14/10215.aspx | 1,534,642,301,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221214538.44/warc/CC-MAIN-20180819012213-20180819032213-00459.warc.gz | 446,272,424 | 9,212 | # I want some Moore
Blog about stuff and things and stuff. Mostly about SQL server and .Net
### News
Hi! My name is
Mladen Prajdić I'm from Slovenia and I'm currently working as a .Net (C#) and SQL Server developer.
I also speak at local user group meetings and conferences like SQLBits and NT Conference
Welcome to my blog.
My Books
Users Online:
## The Rozenshtein Method - yet another way to pivot data in SQL Server
I've stumbled accross this post today that explains the Rozenshtein method of pivoting data.
It's quite interesting. I haven't seen the SIGN() function being used at all in SQL server in practice yet so this is kind of cool if you ask me :)
Print | posted on Wednesday, June 14, 2006 11:20 AM | Filed Under [ SQL Server ]
## #re: The Rozenshtein Method - yet another way to pivot data in SQL Server
This is one of those things that when you see it you think why couldnt I think of that...then you appreciate it..think its awesome...and end up still not using it :|.
Great post, very cool trick!
6/14/2006 5:53 PM | Jon
## #re: The Rozenshtein Method - yet another way to pivot data in SQL Server
In SQL 7, I remember writing all kinds of funky expressions like that the old ABS(SIGN(..)) trick to return 1 or 0, since you didn't have CASE available back then. It might also be faster than CASE in some situations, I've never checked.
Even better is to let some data do the crosstabbing .... create a "matrix" table, put in 1's and 0's in the columns, and then just join to that table and multiply.
For example, if you create a "MonthMatrix" table like this:
Month, Jan, Feb, Mar, ..., Dec
1,1,0,0,...
2,0,1,0,...
3,0,0,1, ...
...
12,0,0,0,....,1
Then you could crosstab like this:
Select ID, SUM(Val * Jan) as Jan,
SUM(Val * Feb) as Feb,
SUM(Val * Mar) as Mar,
..
SUM(Val * Dec) as Dec
From
YourData
inner join MonthMatrix
on
YourData.Month = MonthMatrix.Month
group by ID
6/14/2006 7:11 PM | Jeff
## #re: The Rozenshtein Method - yet another way to pivot data in SQL Server
you sure do know your crosstabs jeff :)
Nice!
## #re: The Rozenshtein Method - yet another way to pivot data in SQL Server
It is NOT faster than CASE, and the only reason it was invented is because CASE did not exist at the time. Please do not use that technique. All it does is create unmaintainable garbage code.
6/20/2006 7:41 AM | Adam Machanic
## #re: The Rozenshtein Method - yet another way to pivot data in SQL Server
hey thanx for clearing that for us Adam.
nice to know.
## #re: The Rozenshtein Method - yet another way to pivot data in SQL Server
I don't agree that it makes unmaintainable garbage code. There are lots of people that immediately recognise 1-ABS(SIGN(x-y)).
It's like counting the occurrences of a substring by using replace and comparing the lengths - the first time you see it, you may have to think about what it's doing - but every time after that, you recognise it immediately and understand.
When I see 1-ABS(SIGN(x-y)), I understand that it's generating ones and zeros according to whether x=y or not.
6/30/2006 8:46 AM | Rob Farley
## #T-SQL: The Rozenshtein Method
7/8/2006 6:38 PM | Anatoly Lubarsky
## #re: The Rozenshtein Method - yet another way to pivot data in SQL Server
Greatest method ever, and easy to understand and use. Using it all the time when I need to do report by month, or years, or days of week, or come another criteria. Works GREAT!!!!!
5/20/2008 7:15 PM | Elena
Comments have been closed on this topic. | 956 | 3,492 | {"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} | 2.734375 | 3 | CC-MAIN-2018-34 | latest | en | 0.8941 |
https://docs.sciml.ai/dev/modules/NeuralPDE/pinn/low_level/ | 1,657,012,854,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00639.warc.gz | 267,862,761 | 14,142 | # 1-D Burgers' Equation With Low-Level API
Let's consider the Burgers' equation:
$$$\begin{gather*} ∂_t u + u ∂_x u - (0.01 / \pi) ∂_x^2 u = 0 \, , \quad x \in [-1, 1], t \in [0, 1] \, , \\ u(0, x) = - \sin(\pi x) \, , \\ u(t, -1) = u(t, 1) = 0 \, , \end{gather*}$$$
with Physics-Informed Neural Networks. Here is an example of using the low-level API:
using NeuralPDE, Flux, ModelingToolkit, Optimization, GalacticOptimJL, DiffEqFlux
import ModelingToolkit: Interval, infimum, supremum
@parameters t, x
@variables u(..)
Dt = Differential(t)
Dx = Differential(x)
Dxx = Differential(x)^2
#2D PDE
eq = Dt(u(t,x)) + u(t,x)*Dx(u(t,x)) - (0.01/pi)*Dxx(u(t,x)) ~ 0
# Initial and boundary conditions
bcs = [u(0,x) ~ -sin(pi*x),
u(t,-1) ~ 0.,
u(t,1) ~ 0.,
u(t,-1) ~ u(t,1)]
# Space and time domains
domains = [t ∈ Interval(0.0,1.0),
x ∈ Interval(-1.0,1.0)]
# Discretization
dx = 0.05
# Neural network
chain = FastChain(FastDense(2,16,Flux.σ),FastDense(16,16,Flux.σ),FastDense(16,1))
initθ = Float64.(DiffEqFlux.initial_params(chain))
eltypeθ = eltype(initθ)
parameterless_type_θ = DiffEqBase.parameterless_type(initθ)
strategy = NeuralPDE.GridTraining(dx)
phi = NeuralPDE.get_phi(chain,parameterless_type_θ)
derivative = NeuralPDE.get_numeric_derivative()
indvars = [t,x]
depvars = [u]
_pde_loss_function = NeuralPDE.build_loss_function(eq,indvars,depvars,phi,derivative,nothing,
chain,initθ,strategy)
bc_indvars = NeuralPDE.get_variables(bcs,indvars,depvars)
_bc_loss_functions = [NeuralPDE.build_loss_function(bc,indvars,depvars,
phi,derivative,nothing,chain,initθ,strategy,
bc_indvars = bc_indvar) for (bc,bc_indvar) in zip(bcs,bc_indvars)]
train_sets = NeuralPDE.generate_training_sets(domains,dx,[eq],bcs,eltypeθ,indvars,depvars)
train_domain_set, train_bound_set = train_sets
pde_loss_function = NeuralPDE.get_loss_function(_pde_loss_function,
train_domain_set[1],
eltypeθ,parameterless_type_θ,
strategy)
bc_loss_functions = [NeuralPDE.get_loss_function(loss,set,
eltypeθ, parameterless_type_θ,
strategy) for (loss, set) in zip(_bc_loss_functions,train_bound_set)]
loss_functions = [pde_loss_function; bc_loss_functions]
loss_function__ = θ -> sum(map(l->l(θ) ,loss_functions))
function loss_function_(θ,p)
return loss_function__(θ)
end
f = OptimizationFunction(loss_function_, Optimization.AutoZygote())
prob = Optimization.OptimizationProblem(f, initθ)
cb_ = function (p,l)
println("loss: ", l , "losses: ", map(l -> l(p), loss_functions))
return false
end
# optimizer
opt = BFGS()
res = Optimization.solve(prob, opt; callback = cb_, maxiters=2000)
And some analysis:
using Plots
ts,xs = [infimum(d.domain):dx:supremum(d.domain) for d in domains]
u_predict_contourf = reshape([first(phi([t,x],res.minimizer)) for t in ts for x in xs] ,length(xs),length(ts))
plot(ts, xs, u_predict_contourf, linetype=:contourf,title = "predict")
u_predict = [[first(phi([t,x],res.minimizer)) for x in xs] for t in ts ]
p1= plot(xs, u_predict[3],title = "t = 0.1");
p2= plot(xs, u_predict[11],title = "t = 0.5");
p3= plot(xs, u_predict[end],title = "t = 1");
plot(p1,p2,p3)
See low-level API | 1,026 | 3,099 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-27 | latest | en | 0.522245 |
https://brilliant.org/problems/a-problem-by-syed-baqir-13/ | 1,508,286,170,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822625.57/warc/CC-MAIN-20171017234801-20171018014801-00742.warc.gz | 632,772,132 | 17,801 | # An algebra problem by Syed Baqir
Algebra Level 2
Peter makes a large amount of pink paint by mixing red and white paint in the ratio 2:3. Red paint costs £40 per 5 litres. White paint costs £5 per 10 litres. Peter sells his pink paint in 10-litre tins for £60 per tin. Calculate how much profit he makes for each tin he sells.
× | 94 | 332 | {"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} | 3.078125 | 3 | CC-MAIN-2017-43 | longest | en | 0.938196 |
https://www.acmicpc.net/problem/15466 | 1,709,449,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476205.65/warc/CC-MAIN-20240303043351-20240303073351-00061.warc.gz | 603,868,476 | 8,343 | 시간 제한메모리 제한제출정답맞힌 사람정답 비율
1 초 512 MB60515185.000%
## 문제
The Jaccard similarity coefficient is usually used for measuring the similarity of two sets. Give two sets A and B, the Jaccard similarity coefficient, J(A, B), is defined as the size of the intersection divided by the size of the union of the two sets. That is, J(A, B) = |A∩B|/|A∪B|. For example, if A = {1, 3, 7, 8} and B = {1, 7, 9}, then J(A, B) = |{1,7}|/|{1,3,7,8,9}| = 2/5.
Assume the element i in the set is an integer between 0 to 9 (0 ≤ i ≤ 9) and the size of the set is no larger than 10. Please write a program to compute the Jaccard similarity coefficient of two sets A and B. And output 1 if J(A, B) > 0.5 and 0 if J(A, B) ≤ 0.5.
## 입력
The first line of the input file contains an integer T (T ≤ 25) indicating the number of test cases to follow.
Each test case will consist of three lines. The first line contains two integers m and n (0 < m, n ≤ 10), indicating the number of elements of sets A and B, respectively. The second line contains m integers (the elements of set A) and the third line contains n integers (the elements of set B).
You may assume:
• 1 ≤ T ≤ 25
• m ≤ 10 and n ≤ 10
## 출력
For each test case, output 1 if J(A, B) > 0.5 and 0 if J(A, B) ≤ 0.5.
## 예제 입력 1
3
5 6
0 2 3 5 6
1 2 4 6 7 9
3 2
1 4 6
4 6
7 7
0 1 3 4 6 8 9
0 1 2 3 4 6 7
## 예제 출력 1
0
1
1 | 512 | 1,354 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-10 | latest | en | 0.817238 |
http://www.reddit.com/user/CatManSam | 1,397,972,670,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609538022.19/warc/CC-MAIN-20140416005218-00341-ip-10-147-4-33.ec2.internal.warc.gz | 628,178,787 | 16,818 | reddit is a website about everything
[–] 0 points1 point ago
This is fucked up and you should feel bad
[–] 0 points1 point ago
Alakasakaaaammaaaaa... Alakasiethhhhh. CoughcoughSBF40coughcough
[–] 0 points1 point ago
I don't know how I feel about this. I think WBC are scum, but they have the right to protest.
[–] 0 points1 point ago
So nobody in poverty should have kids? Idiot.
[–] 0 points1 point ago
Thank you Mr. Algebra.
[–] 0 points1 point ago
Remember those slap bracelets you used to get at parties when you were 12 years old? One cut in to my wrist at a Bat Mitzvah and I had to get stitches.
Edit: typo
[–] 3 points4 points ago
What are average GRE math subject scores for various grad schools? (I take it in one week)
[–] 2 points3 points ago
Well fuck you Susan
[–] 0 points1 point ago
Beedrill with anything is absolutely terrifying
[–] 0 points1 point ago
My girlfriend knows two jokes. This is one of them.
[–] 0 points1 point ago
How has no one said Christoph Waltz?
[–] 0 points1 point ago
To prove sin is continuous you must use the formula: sin(a) - sin(b) = 2sin((a-b)/2)cos((a+b)/2)
And note cos(x) <= 1 for all x. The epsilon delta proof falls in to place with those two facts.
[–] 6 points7 points ago
down vote for blatant generalization...
[–] 0 points1 point ago
Literally every one of these, I think, should not be unsportsmanlike conduct. A little excitement or frustration shown is no grounds for a 15 yd penalty. Ridiculous.
[–] 1 point2 points ago
sorry, this has been archived and can no longer be voted on
This is the best
[–] 2 points3 points ago
sorry, this has been archived and can no longer be voted on
I've never seen a determinant visualized like that! Cool!
[–] 0 points1 point ago
sorry, this has been archived and can no longer be voted on
Token black pug
[–] 7 points8 points ago
sorry, this has been archived and can no longer be voted on
As a Jew.. Not cool.
[–] 0 points1 point ago
sorry, this has been archived and can no longer be voted on
This should be WTF
[–] 0 points1 point ago
sorry, this has been archived and can no longer be voted on
There is no one on earth physically capable of fully and truthfully expressing their thoughts. Even the best art is a mere fraction of the artists true intent. | 643 | 2,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2014-15 | longest | en | 0.954877 |
http://slideplayer.com/slide/3419591/ | 1,571,781,485,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987824701.89/warc/CC-MAIN-20191022205851-20191022233351-00292.warc.gz | 172,805,547 | 19,418 | # Element Loads Strain and Stress 2D Analyses Structural Mechanics Displacement-based Formulations.
## Presentation on theme: "Element Loads Strain and Stress 2D Analyses Structural Mechanics Displacement-based Formulations."— Presentation transcript:
Element Loads Strain and Stress 2D Analyses Structural Mechanics Displacement-based Formulations
Computational Procedure Element Matrices: – Generate characteristic matrices that describe element behavior Assembly: – Generate the structure matrix by connecting elements together Boundary Conditions: – Impose support conditions, nodes with known displacements – Impose loading conditions, nodes with known forces Solution: – Solve system of equations to determine unknown nodal displacements Gradients: – Determine strains and stresses from the nodal displacements
Example B.C.’s Displacements are handled by moving the reaction influences to the right hand side and creation of equations that directly reflect the condition Forces are simply added into the right hand side E1 E2 E3 N1 N2 N3 1000 32426694.11-7680008.00-26666666.670.00-5760027.447680008.00u1 = F1x -7680008.0010239972.560.00 7680008.00-10239972.56v1F1y -26666666.670.0026666666.670.00 u2F2x 0.00 20000000.000.00-20000000.00v2F2y -5760027.447680008.000.00 5760027.44-7680008.00u3F3x 7680008.00-10239972.560.00-20000000.00-7680008.0030239972.56v3F3y 32426694.11-7680008.000.00 7680008.00u1 = 0.00 -7680008.0010239972.560.00 -10239972.56v1-1000 0.00 1.000.00 u20.00 1.000.00 v20.00 1.000.00u30.00 7680008.00-10239972.560.00 30239972.56v30.00 This is it! Solve for the nodal displacements … 32426694.11-7680008.007680008.00u1 = 0 -7680008.0010239972.56-10239972.56v1-1000 7680008.00-10239972.5630239972.56v30 - or - No b.c.’s
Other Loading Conditions Consider the assembled equation system [K] {D} = {F} The only things we can manipulate are: – Terms of the stiffness matrix (element stiffness, connectivity) – The unknown or specified nodal displacement components – The applied nodal force components How do we manage “element” loads? – Self-weight, structural systems where gravity loads are significant – Distributed applied loads, axial, torsional, bending, pressure, etc.
Conversion to Nodal Loads All loads must be converted to nodal loads This is more difficult than it appears It is a place where FEA can go wrong and give you bad results It has consequences for strain and stress calculation q (N/m) F = ? L
You might guess F = qL/2, but why? Setting conc = dist :
Consistent Nodal Loads Consistent nodal loading: – Utilizes the same shape (interpolation) functions (more later) as displacement shape functions for the element – The bar (truss) shape functions specify linear displacement variation between the nodes – We choose a concentrated nodal force that results in an equivalent nodal displacement to the distributed force Question: Are element strain and stress equivalent?
No xx x xx x
Strain and Stress Calculation For bar/truss elements with just nodal boundary conditions: – Find axial elongation L from differences in node displacements – Find axial strain from the normal strain definition – Find axial stress from the stress-strain relationship Even when models become more complicated (higher order displacement/strain relationship, complex constitutive model) this is the general approach
Adjusting Strain and Stress Add analytically-derived fixed-displacement strain and stress This must be done for thermally-induced distributed loading xx x xx x + Note the added constraint …
Mesh Refinement What if we model a bar (truss) or beam element not as a single element, but as many elements? No gain is made in displacement prediction – Holds true for node and element loading Strain and stress prediction improve – Results converge toward the analytical solution even without inclusion of “fixed-displacement analytical stress”
Piece-wise Interpolation If you remember nothing else about FEA, remember this … xx x xx x These are not always flat … 2D/3D elements extend this behavior dimensionally …
To Refine, or Not To Refine … It depends on the purpose of the analysis, the types of elements involved, and what your FEA code does For bar (truss) and beam elements: – Am I after displacements, or strain/stress? – Does my FEA code include analytical strain/stress? – What results does my FEA code produce? – Can I just do my own post-processing? Always refine other element types | 1,177 | 4,457 | {"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-2019-43 | latest | en | 0.635583 |
https://minecraftapp.org/2-d-arrays-part-1-intro-to-java-programming/ | 1,579,922,090,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250628549.43/warc/CC-MAIN-20200125011232-20200125040232-00364.warc.gz | 548,875,551 | 8,250 | # 2-D Arrays Part 1 – Intro to Java Programming
You’ve seen how arrays can store a sequence of values such as this one. Here we have three prices for three different kinds of gasoline. Well, what if we have two gas stations in this case you’ll want to store a two dimensional arrangement with rows and columns, two dimensional arrays can do that. In this example we would have three rows and two columns. In Java we get such a arrangement by saying, give me a new array of doubles, of floating point numbers and I want three rows, two columns. In general, the first number is the number of rows, the second one the number of columns. Note the type, it’s a double brackets brackets meaning, it’s a two dimensional array, of numbers. If you already know which numbers should go into the two dimensional array, then you can supply them like this. Again, here I have a two dimensional array, of prices. And now, you put a, pair of braces, for the entire array, and then for each row you put another pair of braces with the values. So here we have three rows, each of which is enclosed in a pair of braces. Now you know how to declare a two-dimensional array. Next let’s have a look at how we can access its elements. | 270 | 1,213 | {"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} | 2.625 | 3 | CC-MAIN-2020-05 | latest | en | 0.93894 |
https://studyadda.com/question-bank/decimals_q44/2334/207804 | 1,571,244,359,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669057.0/warc/CC-MAIN-20191016163146-20191016190646-00372.warc.gz | 722,197,981 | 19,637 | • # question_answer Suman had to multiply $14.36$ by$23$. Instead, he added the two numbers. What is the difference between his answer and the actual answer? A) $291.92$ B) $292$ C) $293.92$ D) $292.92$
Actual:$14.36\times 23=330.28$ Addition:$14.36+23=37.36$ The difference between his answer and the actual answer $=330.28-37.36=292.92$ | 130 | 439 | {"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} | 3.46875 | 3 | CC-MAIN-2019-43 | latest | en | 0.82898 |
https://web2.0calc.com/questions/geometry_31163 | 1,585,823,268,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506870.41/warc/CC-MAIN-20200402080824-20200402110824-00316.warc.gz | 762,917,315 | 7,558 | +0
# geometry
0
1617
2
Triangle $ABC$ has vertices $A(0,8)$, $B(2,0)$, $C(8,0)$. A vertical line intersects $AC$ at $R$ and $\overline{BC}$ at $S$, forming triangle $RSC$. If the area of $\triangle RSC$ is 12.5, determine the positive difference of the $x$ and $y$ coordinates of point $R$.
Jun 24, 2018
#1
+2346
+1
Constructing a diagram is a crucial step here; digesting all that information visually is extremely difficult without a visual aid. I have concocted one that I am proud of. I have tried to label everything in this diagram for your convenience. This problem most likely has many avenues toward achieving the correct answer, but I will present my approach for you:
In the diagram above, I have added point D, which is the intersection of point A and point C. The point lies on the origin. Let's focus on some characteristics regarding $$\triangle ACD$$ .
Point A and point D have the same x-coordinate, so a vertical segment connects them. Point C and point D share the same y-coordinate, so they must be connected with a horizontal segment. $$\overline{AD}\perp\overline{DC}$$ because horizontal and vertical lines are always perpendicular; thus, $$m\angle ADC=90^{\circ}$$.
$$AD=DC=8$$, so $$\triangle ACD$$ is also an isosceles triangle.
A triangle that is both right and isosceles is also known as a 45-45-90 triangle. It is possible to observe these similarities with $$\triangle RSC$$.
$$\overline{AD}\parallel\overline{RS}$$ because they are both vertical lines. Whenever a parallel line cuts through a triangle, the resulting triangles are similar. In this case, $$\triangle ADC\sim\triangle RSC$$
$$DC=8$$ and $$DS=a$$ , so $$CS=8-a$$ . Because $$\triangle RSC$$ is isosceles, $$RS=CS=8-a$$ . The area of this triangle, according to the original problem, is 12.5. We know the lengths of the side lengths, so we can determine the value of a.
$$12.5=\frac{1}{2}(8-a)(8-a)$$ Multiply by 2 on both sides to solve for a. We are using the area of a triangle to generate this equation. $$25=(8-a)^2$$ Take the square root of both sides. $$5=|8-a|$$
Yes, it is possible to solve for a here, and it is relatively simple from here on out. However, we can take a slight shortcut. The end goal is to determine the difference of x- and y-coordinates, but 8-a represents the height of the triangle. Since 8-a=5, we know that the y-coordinate is 5.
(3,5) is the coordinate of point R. $$|3-5|=2$$ is the positive difference.
Jun 24, 2018
#2
+109345
+1
Thanks, X2 !!!....here's another approach.....
Let SC be the base of triangle RSC.....and let its length = 8 - x
The slope of the line segment AC = (0 - 8) / (8 - 0) = -8/8 = -1
And the equation of the line containing this segment is y = - x + 8 ⇒ y = 8 - x
So...let the height of triangle RSC = SR = 8 - x
And we have that the area of triangle RSC =
(1/2) (8 -x) (8 -x) = 12.5 multiply both sides by 2
(8 - x) ( 8 - x) = 25 simplify
x^2 - 16x + 64 = 25
x^2 - 16x + 39 = 0 factor
(x - 3) ( x - 13) = 0
Setting both factors to 0 and solving for x produces
x = 13 (reject) or
x = 3 (accept)
So.... since R lies on the line y = 8 - x, its coordinates are ( 3, 5)
And the positive difference of its x and y coordinates = 5 - 3 = 2
Jun 24, 2018
edited by CPhill Jun 24, 2018
edited by CPhill Jun 24, 2018 | 1,078 | 3,353 | {"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} | 4.40625 | 4 | CC-MAIN-2020-16 | latest | en | 0.832198 |
https://gotest.pk/short-questions/12th-class-computer-chapter-3-loops-short-questions-answer/ | 1,679,972,630,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00320.warc.gz | 308,399,528 | 23,516 | # 12th Class Computer Chapter 3 Loops Short Questions Answer
## 12th Class Computer Chapter 3 Loops Short Questions Answer
How is the logical flow governed in a structured program?
In a structured program, the logical flow is governed by three control structures.
· Sequential
· Repetition
· Selection
Define sequential stricture?
Sequential structure refers to the execution of statements in the orders in which they appear.
Define a loop?
It is another fundamental idea in programming. It makes possible repeated execution of one or more statements as long as a condition is true.
What is the importance of loop in any program?
We can employ a loop to handle any repetitive task and for most programs loops are essential using a computer to calculate the company payroll, for example would not be practicable without a loop.
Which are called iteration statements?
The statements inside a loop are sometimes called iteration statements.
How many essential elements to a loop?
There are two essential elements to a loop.
What do you know about the loop condition?
A loop condition can take a number of different forms to provide different ways of controlling the loop. We can set the loop condition to suit the circumstances.
What do you know about FOR loop?
The FOR loop is primarily used for executing a statement or block of statements a predetermined number of times. You control for a loop using three expressions.
What do you know about the WHILE loop?
The WHILE loop uses a logical expression to control execution of the loop body. If the condition controlling the loop produces an integer, the loop with continue as long as the value is non-zero. Any non-zero integer means the condition is true and only zero means it is false.
What do you know about the DO WHILE loop?
The DO WHILE loop is similar to the while loop in that the loop continues as long as the specified loop condition remains true.
Write down the general form of the FOR statement.
The general form of the FOR statement is FOR (exp1, exp2, exp3)
{Body of the loop (block of statements)}
Write down the general for of the WHILE statement.
WHILE (expression)
{Body of the loop}
What happens when a while statement is encountered?
When a while statement is encountered expression is evaluated.
Write down the types of characters included in ASCII table.
Following are the types of characters included in ASCII table
0 ~ 31: control codes such as tab, carriage return and bell.
32 ~ 127: Usual printing characters
128 ~ 255: graphics and foreign language characters.
Write down the general form of the DO WHILE loop?
The general form of the DO WHILE loop is do.
DO
{Body of the loop}
WHILE (expression);
What is a nested loop?
A loop within another loop is known as a nested loop.
Wrote down the statements that “C” provides to implement loop structure?
“C” provides three statements to implement loop structure. These three statements are FOR statements. WHILE statements and the DO-WHILE statement.
You Can Learn and Gain more Knowledge through our Online Quiz and Testing system Just Search your desired Preparation subject at Gotest.
error: | 635 | 3,151 | {"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-2023-14 | latest | en | 0.922099 |
http://www.lwfree.cn/shuxue/20170523/7665.html | 1,532,006,692,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590901.10/warc/CC-MAIN-20180719125339-20180719145339-00136.warc.gz | 490,196,069 | 6,549 | ## 曲线积分与积分路径无关性的应用
关键词:曲线积分;全微分;积分
Application Independence of
Curvilinearintegral and the Integration Path
Abstract: This paper introduces the four equivalent conditions plane curve path integral equivalent conditions unrelated to the four points of space curves and path independence, combined with examples of their application: calculated curve points, find the original function, seeking differential equations, find the equations of unknown function, especially in the case of unknown function required to solve a series of use associated with the path integral curves of differential equations seeking independence unknown function problems.
Key word: Curve Points; Fully Differential; Points
1.平面曲线积分与路径无关性的定理及证明 3
1.1利用平面曲线积分与路径无关性求原函数 6
1.2利用平面曲线积分与路径无关性求微分方程的解 7
1.3利用平面曲线积分与路径无关性计算曲线积分的值 9
1.4利用平面曲线积分与路径无关性求未知函数 11
2.空间曲线积分与路径的无关性定理及证明 12
2.1利用空间曲线积分与路径无关性求原函数 15 源自%六[维%论*文\$网.加7位QQ3249^114 www.lwfree.cn
2.2利用空间曲线积分与路径无关性计算曲线积分 16
1.平面曲线积分与路径无关性的定理及证明
首先介绍单连通区域的概念.
若对于平面区域 上任一封闭曲线,皆可不经过 以外的点而连续收缩于属于 的某一点,则称此平面区域为单连通区域,否则称为复连通区域. 曲线积分与积分路径无关性的应用:http://www.lwfree.cn/shuxue/20170523/7665.html
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Archive Language Point 177
# The third conditional
A conditional is used to talk about a possible or imaginary situation (the condition) and the consequences (or the result) of it.
## The third conditional - meaning
We use the third conditional when we want to imagine a different past to that which actually occurred. In this episode of the flatmates, Sophie says the following to Alice. "We wouldn't have had anywhere near such a good Christmas if you hadn't been there to cook the turkey!" The real situation in the past was that Alice cooked the turkey and everybody had a good Christmas. When Sophie speaks, she imagines how the past would have been different, if Alice had not been there and had not cooked the turkey. The third conditional gives the imaginary result, or consequence, of an unreal past.
## The third conditional - form
A basic conditional sentence has two clauses, namely the condition and the consequence. In the following example, the condition is the first clause and the consequence is the second clause. If you hadn't been there, we wouldn't have enjoyed Christmas so much. For the condition: If + past perfect For the consequence: would + have done (the perfect infinitive) For example: If we had left the house earlier, we would have caught the train. If he hadn't been so lazy, he wouldn't have failed his exam. It is possible to put the consequence before the condition. In that case, we do not usually separate the clauses with a comma. We would have caught the train if we had left the house earlier. He wouldn't have failed his exam if he hadn't been so lazy.
## Would or might?
We use would to show that we are certain about the consequence. If we are not certain about the consequence we can use might. If he had studied harder, he would have passed the exam. If he had studied harder, he might have passed the exam.
## If only ...
We often use 'if only + past perfect' to express a strong regret about the past. If only I had studied harder.
## The third conditional - a variation
The basic third conditional expresses an imagined past result of an unreal past condition. If I had studied harder, I would have passed the exam. In that situation, the reality is that the person did not study hard, and he or she failed the exam. However, sometimes the imagined result could be true now, at the moment of speaking. For example, If I hadn't gone to university, I wouldn't be a teacher now. In that situation, the reality is that the person did go to university, and he or she is a teacher now. If the imagined result, or consequence, could be true now, we use 'would do', not 'would have done'. But, because the condition is still an unreal past, we use the past perfect after 'if'. If I hadn't gone to university, I wouldn't be a teacher now. If Alice had stayed in the flat over Christmas, she wouldn't know Sophie.
## Contractions
It is common to use contractions with this language, particularly when it is spoken. If I'd studied harder, I'd have passed the exam If we use the negative forms - hadn't, wouldn't - we do not usually contract 'had' and 'would'.
## Vocabulary:
fireworks explosive devices that create patterns of light in the sky, usually used at celebrations and big parties to get to know someone to meet someone and then become friends with them turkey a large bird that is often eaten for dinner at Christmas in the UK, and at Thanksgiving in the USA
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6 replies [Last post]
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Q1. What was the approximate number of total literate females in MP in the year 2001 ?
a. 99815709
b. 99915709
c. 91951413
d. 81951413
Q2. What was the approximate total population of Karnataka in the year 1981 ?
a. 71191471
b. 51191471
c. 41191471
d. 61191471
Q3. What was the approx difference between the number of females in Kerala in 1991 and that in 2001 ?
a. 10299
b. 11299
c. 12299
d. 14299
Q4. Total life expectancy of MP ( in year at birth ) is what per cent of the average of the total life expectancy of the three given states ?
a. 86.67%
b. 115.36%
c. 96.7%
d. 68.66%
Q5. In 2001 the number of primary schools per thousand persons was the highest for which of the following states ?
a. Can’t say
b. MP
c. Karnataka
d. Kerala
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hey praveen_84 here you didnt
hey praveen_84 here you didnt given any adition information @ population of any state in any year or didnt give total population in any perticular year But dont worry i already read not solved this ppuzzle in BUSINESS & MANAZEMENT magazine
here i complete u r data ok
Total population of MP,Karnataka,Kerala in 2001=191867123
Little Star
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Q2. What was the approximate
Q2. What was the approximate total population of Karnataka in the year 1981 ?
a. 71191471
b. 51191471
c. 41191471
d. 61191471
total population=191867123
so kerala=191867123*0.3790=72717639
now for karnaaka growth from 1991-01 17.3
so in 1991 population 61992872
now growth from 1981 to 1991 21.1 so
ans is 61992872/1.211=51191472
so ans is option(2)
Little Star
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Q4. Total life expectancy of
Q4. Total life expectancy of MP ( in year at birth ) is what per cent of the average of the total life expectancy of the three given states ?
a. 86.67%
b. 115.36%
c. 96.7%
d. 68.66%
average life expectancy of three state to gather=(55.5+63.3+73.3)/3=64.03
so % MP=55.5/64.03=86.67% option(1)
Little STar
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Q5. In 2001 the number of
Q5. In 2001 the number of primary schools per thousand persons was the highest for which of the following states ?
MP population in 2001=191867123*0.4550=87299540
Karnaanka population in 2001=191867123*0.3790=72717639
Kerala population in 2001=191867123*0.1660=31849942
from above data or pie chart data u can say that the population for MP is heighest so per 1000 person
MP has highest school i dont know why number of school given in table
so from my point of view ans is option(2)
Little Star
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working on remaining 2
working on remaining 2 question because now going for GARBA
Little Star
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Q1. What was the approximate
Q1. What was the approximate number of total literate females in MP in the year 2001 ?
a. 99815709
b. 99915709
c. 91951413
d. 81951413
For this question first of all u have to find number of people in MP from pie chart percentage then find the number of female from table
i.e. for MP female in 2001 ,920 compare to 1000 male
so u have to put equation if total MP people xxxxx...
now out of 1920 ,920 female then out of xxxxx... people how many female...
so i get number of female now mutliply that value with 0.503(literacy rate for MP for female in 2001) so u will be got ans
same formula for Question 3 follow it
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n/a | 1,220 | 3,963 | {"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-2018-22 | latest | en | 0.765703 |
bloodredsun.com | 1,519,552,660,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816351.97/warc/CC-MAIN-20180225090753-20180225110753-00631.warc.gz | 40,479,502 | 19,635 | # Why we don’t use Doubles for Financial Calculations
My current client is in the middle of hiring some Java developers and as I mentioned earlier in March (Interviewing – the importance of PASSION!) I’ve been doing some of the interviewing. One of the things we’ve done is to create a technical task to see how the candidates actually code. It’s a simple exercise that requires them to think through some of the basics of financial operations and one thing that has surprised me has been the common use of doubles to represent financial values. It’s been highlighted for some time that this is not a great thing to do but someone actually challenged me to show that it wasn’t.
So here we go…
```package com.bloodredsun;
public class DoubleOperation {
public static void main(String[] args) {
double t1 = 10.266d;
double t2 = 10.0d;
//Outputs 0.266
System.out.println(t1-t2);
double h1 = 100.266d;
double h2 = 100.0d;
//Outputs 0.26600000000000534
System.out.println(h1-h2);
}
}
```
Ouch! That is not what we want but it is the classic behaviour of doubles. The inability to represent some decimals in the IEEE-754 format (as binary fractions) causes this. If we want correct precision the answer is to use BigDecimals but we have to remember to use Strings in the constructors or you end up with the same issues that you were trying to avoid.
```package com.bloodredsun;
import java.math.BigDecimal;
public class BigDecimalOperation {
public static void main(String[] args) {
BigDecimal t1 = new BigDecimal("10.266");
BigDecimal t2 = new BigDecimal("10.0");
//Outputs 0.266
System.out.println(t1.subtract(t2));
BigDecimal h1 = new BigDecimal("100.266");
BigDecimal h2 = new BigDecimal("100.0");
//Outputs 0.266
System.out.println(h1.subtract(h2));
}
}
```
That’s great but wouldn’t it be nice to use the normal operators rather than the overly-verbose method calls for the mathematical operations.
Now there is no way that we can do this in Java but if we let ourselves use another language on the JVM…
```package com.bloodredsun
object ScalaBigDecimalOperation {
def main (args: Array[String]) {
var t1 = BigDecimal("10.266")
var t2 = BigDecimal("10.0")
//Outputs 0.266
println(t1 - t2)
var h1 = BigDecimal("100.266")
var h2 = BigDecimal("100.0")
//Outputs 0.266
println(h1 - h2)
}
}
```
Scala FTW!
PS if you want to know more about floating point operations have a read of What Every Computer Scientist Should Know About Floating-Point Arithmetic
## 34 thoughts on “Why we don’t use Doubles for Financial Calculations”
1. Ed
Are you sure doubles are so bad in finance?? Really?
Each time I do alpha + beta I get a value signma + an error term epsilon. Provided epsilon is sufficiently less than the precision, you’re ok because you can just round the value to the appropriate accuracy (less than half epsilon I guess)
What’s the performance difference between big decimal and the scala version?
1. Martin
As you say, providing epislon is small enough it’s not a problem and in many cases we can live with it. The trouble arises when you have billions of calculations that sum up to something significant or when you are taking leveraged positions that can massively exacerbate any imprecision.
As for the performance of the Java and Scala version, they are identical since they both compile down to byte code. The difference is purely a syntactical one.
1. Martin Post author
My point was that despite all the focus given to this topic, there are still developers who do not know about the imprecision of floating point values.
Like most rules, knowing this gives you the freedom to break it when you are able to, such as when the error is acceptably small and the performance benefit is a great enough advantage.
2. stratton
Hey Ed,
The problem with doubles as he shows is you can’t represent the numbers exactly. Imagine adding a whole bunch of numbers that should have added up to an exact value. It’s hard to verify your books if cents are gone here and there. When working with money, you need your numbers to be exact.
“That’s great but wouldn’t it be nice to use the normal operators rather than the overly-verbose method calls for the mathematical operations.”
As the author noted, with Scala BigDecimal has its operators overloaded so you can use normal + and – operators, while with java because BigDecimal isn’t a primitive, you need to use “.subtract()” and such.
3. Rich
Doubles are bad because they do not mean what naive programmers expect them to mean. Though in this case I think a workable solution is to pick a suitable precision and just use longs.
4. Tristan
In the 238k range a float will(may) be off by a penny. This may not be an issue for day to day banking, Depending on the value of the whole part the precision of the fractional goes down. It turns out that at about 238000 the precision for the fractional goes down to ~0.005 which will impact financial transactions.
besides this is *your* money, is “shouldn’t be a problem” ok?
5. Chris Fairhall
Floating point in finance becomes a real problem when it comes to rounding.
In general the finance sector uses “half even” rounding.
If your calculation is supposed to end up as 36.295 and then rounded to 2DP it should go to 36.30.
If it actually ends up being 36.294999999999995, that rounds down to 36.29
2. John Haugeland
Are … you retarded?
The *correct* answer is to do integer math on cents. Unambiguous, correct, primitive, fast, and no need to switch languages.
Why are Java people unable to cope with basic software topics?
1. Wouter Lievens
What if you need sub-cent accuracy? You could just move the imaginary decimal point, sure, but what if you don’t know the location of said point?
2. Gavin
Wow, such ignorance….
Were you writing a calculator to use at the till in your toy shop?
Cent accuracy is far too coarse in finance, what are you going to do, just round to the nearest cent every time you need to find a fraction of a value?
Also, who are Java people? Java is just a language, stop with all the high school fan boy crap.
3. Martin Post author
Insulting, smug and wrong in one post – surely this is the ultimate internet post trifecta!
Integers are performant but all well and good until your precision changes and you realise you’ve coded yourself into a corner. And as Gavin mentioned, who says that cents are good enough? Or better yet if you’re dealing with massive numbers your chosen solution causes an overflow.
The reality is that you choose the right solution to the problem. If you can live with the imprecision, use a double. If you need specified precision (and don’t want to roll your own rounding), use something like a BigDecimal or create your own data object that can handle both sides of the decimal point independently. What you don’t do is present a solution as simplistic as yours and then get to claim that everyone else is an idiot.
4. Ed
Sure, that works great if you are adding and subtracting monetary amounts. But that’s Accounting. We’re talking Finance here. What happens when interest rates come into the picture?
Consider:
import java.math.BigDecimal;
import java.math.RoundingMode;
public class Main{
private static Transformer floor = new BigDecToDiscreteViaFloor();
private static Transformer ceil = new BigDecToDiscreteViaCeil();
private static Transformer round = new BigDecToDiscreteViaRound();
private static Transformer i = new BigDecIdentity();
public static void main(String[] args){
BigDecimal initialInvestment = new BigDecimal(“1000000”);
BigDecimal annualRate = new BigDecimal(“0.07125”);
int compoundingPeriodsPerYear = 4;
int years = 10;
BigDecimal test = new BigDecimal(“10.12345”);
BigDecimal testA = new BigDecimal(“10.105”);
BigDecimal testB = new BigDecimal(“10.115”);
System.out.println(“To get some confidence in our Transformers:”);
System.out.println(test + ” via Floor: ” + floor.transform(test));
System.out.println(test + ” via Ceil: ” + ceil.transform(test));
System.out.println(test + ” via Round: ” + round.transform(test));
System.out.println(testA + ” via Round: ” + round.transform(testA));
System.out.println(testB + ” via Round: ” + round.transform(testB));
System.out.println(test + ” via Identity: ” + i.transform(test));
System.out.println();
BigDecimal correct = futureValue(initialInvestment, annualRate, compoundingPeriodsPerYear, years, i);
BigDecimal viaFloor = futureValue(initialInvestment, annualRate, compoundingPeriodsPerYear, years, floor);
BigDecimal viaCeil = futureValue(initialInvestment, annualRate, compoundingPeriodsPerYear, years, ceil);
BigDecimal viaRound = futureValue(initialInvestment, annualRate, compoundingPeriodsPerYear, years, round);
System.out.println( “What is the future value of \$” + initialInvestment +
” invested at an annual rate of ” + annualRate +
” compounded ” + compoundingPeriodsPerYear + ” time(s) per year for ” +
years + ” years?\n”);
//Consider \$1M invested for 10 years at 7.125 % interest,
//compounded quarterly calculated with decimals
System.out.println( “‘Correct’ value (to two decimal places):” + round.transform(correct) + “\n”);
//Consider the same using integers (where integers are obtained
//via the floor function on decimals)
System.out.println(“Using floor (to two decimal places):” + round.transform(viaFloor) + “\n”);
//Consider the same using integers (where integers are obtained
//via the ceiling function on decimals)
System.out.println(“Using ceiling (to two decimal places):” + round.transform(viaCeil) + “\n”);
//Consider the same using integers (where integers are obtained
//via the rounding)
System.out.println(“Using round (to two decimal places):” + round.transform(viaRound) + “\n”);
}
public static BigDecimal futureValue( BigDecimal initialInvestment,
BigDecimal annualRate,
int compoundingPeriodsPerYear,
int years,
Transformer t) {
return t.transform(initialInvestment).
multiply(
BigDecimal.ONE.
t.transform(annualRate).
divide(
new BigDecimal(compoundingPeriodsPerYear)
)
).pow( (compoundingPeriodsPerYear * years ) )
);
}
private static interface Transformer {
public T transform(T t);
}
private static class BigDecToDiscreteViaFloor implements Transformer {
public BigDecimal transform(BigDecimal d) {
return d.divide(BigDecimal.ONE, 2, RoundingMode.FLOOR);
}
}
private static class BigDecToDiscreteViaCeil implements Transformer {
public BigDecimal transform(BigDecimal d) {
return d.divide(BigDecimal.ONE, 2, RoundingMode.CEILING);
}
}
private static class BigDecToDiscreteViaRound implements Transformer {
public BigDecimal transform(BigDecimal d) {
return d.divide(BigDecimal.ONE, 2, RoundingMode.HALF_EVEN);
}
}
private static class BigDecIdentity implements Transformer {
public BigDecimal transform(BigDecimal d) {
return d;
}
}
}
Which outputs:
To get some confidence in our Transformers:
10.12345 via Floor: 10.12
10.12345 via Ceil: 10.13
10.12345 via Round: 10.12
10.105 via Round: 10.10
10.115 via Round: 10.12
10.12345 via Identity: 10.12345
What is the future value of \$1000000 invested at an annual rate of 0.07125 compounded 4 time(s) per year for 10 years?
‘Correct’ value (to two decimal places):2026334.83
Using floor (to two decimal places):2001597.34
Using ceiling (to two decimal places):2208039.66
Using round (to two decimal places):2001597.34
There’s a lot of disparity there, no matter how you choose to convert Decimals to “Dollars and Cents”.
3. Chris
Why not just impose a furthest place value — say, thousandths — and use ints (or longs if you’re worried about overflow)?
``` int t1 = 10266; int t2 = 10000d; ```
``` //Outputs 266 System.out.println(t1-t2); int h1 = 100266d; int h2 = 100000d; ```
```//Outputs 266 System.out.println(h1-h2); ```
Just make sure to divide by 1000 before displaying the results to the user.
1. Rich
Eeek… nice comment until the division bit. The dollars and cents thing is a presentational issue, not an arithmetic one.
4. jmalcolm
Scala is pretty cool but this is not the best advertisement for it. After all, the following has worked in C# since version 1.0 (released in 2002)
``` using System;```
``` class MainClass { public static void Main (string[] args) { var t1 = 10.266m; var t2 = 10.0m; // Outputs 0.266 Console.WriteLine(t1-t2); var h1 = 100.266m; var h2 = 100.0m; ```
``` // Outputs 0.266 Console.WriteLine(h1-h2); } } ```
Actually, I guess you would need to replace `var` with explicit `decimal` back in the 1.0 days.
I am not trashing on Scala though. As I said, Scala is pretty cool.
1. Martin Post author
As I understand it, the m datatype is just an alias for System.Decimal so it actually amounts to the same thing 🙂
All I was trying to do was to have a little bit of fun with Scala to highlight that the implicit typing is a far better solution than Java’s, nothing more.
5. Gorokon
This is not a big deal. Any developer unaware of the problems with floating point arithmetic is not worth feeding. Fixed point integers are good and they are faster too.
6. Derek D.
Integers are your friends. They’ll never leave you for more exciting values. They’ll stay the same forever (unless you spend too much time adding to them, that is).
<3
Also, dear God, floats belong in the graphics department, keep them out of anything serious.
They're floaty like a viscous liquid, can't be trusted.
7. francis
you can also use Decimal in C#
so money variables would be declared like this; 2.50m
or you can just create a fixed point number class that will allow one to change the precision
8. Aprogrammer
If you are doing simple retail like calcs it makes sense. But try using big decimals is an iterative calculation thousands of times (say like a Monte Carlo simulation) in a HFT business. You’ll be killed on latency and resources as you churn a truck load more objects causing high levels of gc whilst the Market moves is against you. Using doubles has it’s place, primitive longs better if you keep hold of the decimal and calculate that at the end.
If that was your criteria for hiring people then I despair……
1. Martin Post author
You’re right about each approach, doubles/longs/BigDecimals, having their place. What I meant to highlight, and that particular point of the hiring process, was that each approach has its shortfall. As long as you are aware of the failings of each approach then that is all that matters but the candidates routinely don’t have this awareness.
Like many rules, it is lies-to-children. When you can appreciate the complexity of reality you are then able to break it for the right reason.
9. Patrice
Martin,
This is a good post, you’re taking a lot of unjustified heat here IMO.
It’s a simple, general rule, which like some many others says “unless you know for a fact that you have a better answer, you should always use [BigDecimal] for [currency arithmetics] rather than double/long”.
A simple rule that keeps you out-of-trouble 99% of the time. It is intellectually honest, unlike some of the (expected) comments.
Yes there are other ways to do it, if you really need to (e.g. raw performance) and if you absolutely know what you’re doing, but you never claimed otherwise.
Your post made the programming masses a bit better, and it’s well worth a thank you. Keep ’em coming.
1. Martin Post author
Thanks for the kind words Patrice 🙂
Quite a few people seemed to have missed the fact that I never said that doubles should never be used. I say that they are “not a great thing to do” but each data type has their own strengths and weaknesses and should be used when appropriate. The trick is, as you mention, having sufficient understanding of the complexities to know when to break the rule. My fault for a contentious title I guess!
10. Ara
The problem with using BigDecimal is the overhead in performance. An arithmetic operation on a double (in most CPUs) is typically a one or two-instruction process. An arithmetic operation on a BigDecimal involves many many more instructions (check out the source for subtract() at http://www.docjar.com/html/api/java/math/BigDecimal.java.html).
By carefully controlling the rounding of doubles, it is possible to get accurate financial results without incurring the performance overhead. I know this for a fact because, 20 years ago, I used to write FORTRAN programs for scientific calculations that used FLOAT and DOUBLE PRECISION data types and produced accurate (within epsilon) results. BigDecimal may still be necessary, but only when adding really large numbers.
11. Keith Thompson
If all you’re doing is adding and subtracting amounts of money, scaled integers (where 1 represents \$0.01 and 100 represents \$1.00) are good enough.
But as Ed points out, as soon as you start dealing with interest calculations, you have to deal with fractional cents, and you have to get the answers right.
As I understand it, there are regulations that specify exactly how these calculations must be done, with exact rules for rounding vs. truncating.
You shouldn’t even begin to write code that deals with interest calculations until you understand these regulations. Don’t assume that money can be expressed in whole numbers of cents, or even in real numbers of dollars. Don’t assume that the regulations match what your intuition tells you about how money *should* work. If your calculation yields a result that’s mathematically perfect, but it doesn’t match what the regulations require, your calculation is wrong.
(Disclaimer: I have little or no idea what these regulations actually say, or how to find them. It’s entirely possible that I’ve misunderstood the situation myself.)
12. Joe
Partical answer: go to a clearing house and ask them WHY they don’t want you to use float/double to perform their calculations. They don’t care about speed or efficiency, what they care about is accuracy. So fractional operations are performed on integer values and results are later scaled to obtain floating representation. When you’re shuffling millions for one account to another a few hundred times every night, a tiny rounding error is not so tiny. | 4,176 | 18,089 | {"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-2018-09 | longest | en | 0.842243 |
https://ir.nctu.edu.tw/handle/11536/222 | 1,503,297,752,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107720.63/warc/CC-MAIN-20170821060924-20170821080924-00293.warc.gz | 790,203,707 | 8,772 | 標題: Parallelogram-free distance-regular graphs 作者: Liang, YJWeng, CW交大名義發表應用數學系National Chiao Tung UniversityDepartment of Applied Mathematics 公開日期: 1-十一月-1997 摘要: Let Gamma=(X, R) denote a distance-regular graph with distance function partial derivative and diameter d greater than or equal to 4. By a parallelogram of length i (2 less than or equal to i less than or equal to d), we mean a 4-tuple xyzu of vertices in X such that partial derivative(x, y) = partial derivative(z, u) = 1, partial derivative(x,u) = i, and partial derivative(x,z) = partial derivative(y, z) = partial derivative(y, u)=i-1. We prove the following theorem. THEOREM. Let Gamma denote a distance-regular graph with diameter d greater than or equal to 4, and intersection numbers a(1) = 0, a(2) not equal 0. Suppose Delta is Q-polynomial and contains no parallelograms of length 3 and no parallelograms of length 4. Then Gamma has classical parameters (d, b, alpha, beta) with b < -1. By including results in [3]; [9], we have the following corollary. COROLLARY. Let Gamma denote a distance-regular graph with the Q-polynomial property. Suppose the diameter d greater than or equal to 4. Then the following (i)-(ii) are equivalent. (i) Gamma contains no parallelograms of any length. (ii) One of the following (iia)-(iic) holds. (iia) Gamma is bipartite. (iib) Gamma is a generalized odd graph. (iic) Gamma has classical parameters (d b, alpha, beta) and either b < -1 or Gamma is a Hamming graph or a dual polar graph. (C) 1997 Academic Press. URI: http://dx.doi.org/10.1006/jctb.1997.1787http://hdl.handle.net/11536/222 ISSN: 0095-8956 DOI: 10.1006/jctb.1997.1787 期刊: JOURNAL OF COMBINATORIAL THEORY SERIES B Volume: 71 Issue: 2 起始頁: 231 結束頁: 243 顯示於類別： 期刊論文 | 527 | 1,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-34 | longest | en | 0.757915 |
http://mizar.uwb.edu.pl/version/current/html/proofs/waybel35/41 | 1,571,262,463,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-43/segments/1570986670928.29/warc/CC-MAIN-20191016213112-20191017000612-00442.warc.gz | 145,434,076 | 1,886 | let L be reflexive transitive RelStr ; :: thesis: for R being auxiliary(ii) Relation of L
for C being Subset of L
for x, y being Element of L st x <= y holds
SetBelow (R,C,x) c= SetBelow (R,C,y)
let R be auxiliary(ii) Relation of L; :: thesis: for C being Subset of L
for x, y being Element of L st x <= y holds
SetBelow (R,C,x) c= SetBelow (R,C,y)
let C be Subset of L; :: thesis: for x, y being Element of L st x <= y holds
SetBelow (R,C,x) c= SetBelow (R,C,y)
let x, y be Element of L; :: thesis: ( x <= y implies SetBelow (R,C,x) c= SetBelow (R,C,y) )
assume A1: x <= y ; :: thesis: SetBelow (R,C,x) c= SetBelow (R,C,y)
let a be object ; :: according to TARSKI:def 3 :: thesis: ( not a in SetBelow (R,C,x) or a in SetBelow (R,C,y) )
assume A2: a in SetBelow (R,C,x) ; :: thesis: a in SetBelow (R,C,y)
then reconsider L = L as non empty reflexive RelStr ;
reconsider a = a as Element of L by A2;
A3: a in C by ;
A4: a <= a ;
[a,x] in R by ;
then [a,y] in R by ;
hence a in SetBelow (R,C,y) by ; :: thesis: verum | 351 | 1,017 | {"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-2019-43 | latest | en | 0.802203 |
https://maas.museum/observations/2016/01/22/energy-mass-the-velocity-of-light-and-cake/ | 1,532,019,674,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591150.71/warc/CC-MAIN-20180719164439-20180719184439-00106.warc.gz | 695,461,760 | 10,401 | Observations
# Energy, Mass, the Velocity of Light and Cake
Einstein’s Relativity is a little over a century old, and is still our best description of space and time. But trying to explain this unintuitive theory of distorted space and time, without using mathematics, has always been a challenge. I recently referred back to one of Einstein’s early papers on relativity, dated September 1905, to try and find a clear and concise way to explain the derivation of one of the most important equations in physics, E=mc2.. The first thing that hits you about this paper is its brevity, not quite two and a half short pages. So I certainly had a concise explanation in my hands, but was it going to be clear. The title did not instill confidence
“Does the inertia of a body depend upon its energy-content?”
With Einstein framing the title as a question, and not a statement, you are left thinking that this man, who was happy to distort space and time, was having a hard time with this concept. In fact he wrote to a friend saying “I cannot know whether the dear Lord doesn’t laugh about this and has played trick on me” when he first came up with his derivation. So after having read many texts explaining E=mc2 I was looking for new inspiration. And indeed it came. Not immediately after reading the paper, but 3am the next morning in a bolt of clarity that woke me, and kept me awake for fear I would lose it. It was clear to me that any explanation I was to give needed a simple prop. It had to be a cake. A cake would provide me with all the qualities I needed to transport my class from the familiar to the wonderful realm of abstract higher physics. Consuming the cake after would be the proverbial icing. Perfect!
The essence of the explanation is that a stationary cake, if it were to radiate light uniformly, remains stationary. Yet when the same thing is viewed from a moving vantage point the cake has kinetic energy, and some of that energy reveals itself in the radiated light. For the cake to radiate kinetic energy away without changing speed means the light must be carrying away inertia.
Now if the reason for my rapture at this realisation is lost on you, do not be disappointed.
I would love to be able to give you a clearer explanation of where E=mc2 came from. Sadly I cannot in these short paragraphs. The full explanation that I gave my class came after exposing them to 3 evenings of relativity at the observatory. But I encourage you to download Einstein’s paper, which is freely available, to experience the work of a genius. And I invite you to come along to the observatory and have some cake with me.
Paul Payne conducts evening adult education courses at Sydney Observatory.
Understanding Relativity, 6 Tuesday evenings, commences 8th March
Astronomical Concepts, 8 Thursday evenings, commencing 25th February | 612 | 2,841 | {"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} | 2.96875 | 3 | CC-MAIN-2018-30 | latest | en | 0.977242 |
http://www.jiskha.com/display.cgi?id=1325914554 | 1,498,283,178,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320226.61/warc/CC-MAIN-20170624050312-20170624070312-00165.warc.gz | 562,080,412 | 3,632 | # Physics
posted by .
Two forces act on a 5-kg object sitting on a frictionaless horizontal surface. One force is 30N in the +x- direction and the other 35N in the -x-direction. What is the acceleration of the object?
• Physics -
The net force is
F = -5 N which means 5 Newtons in the -x direction.
The acceleration is a = F/m
You know the mass, m. Do the calculation. | 102 | 375 | {"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-2017-26 | latest | en | 0.89333 |
https://oeis.org/A125618 | 1,638,927,937,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363420.81/warc/CC-MAIN-20211207232140-20211208022140-00225.warc.gz | 489,785,436 | 4,268 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A125618 (Sum of the squares of the quadratic nonresidues of prime(n)) / prime(n). 6
10, 23, 23, 40, 65, 117, 127, 199, 209, 254, 319, 474, 441, 654, 583, 765, 1071, 826, 1218, 1252, 1246, 1476, 1637, 2000, 2042, 1899, 2028, 2974, 3155, 2998, 3394, 3593, 4291, 3983, 4469, 5525, 4867, 5743, 5301, 7274, 5964, 6321, 7446, 7684, 9013, 9099 (list; graph; refs; listen; history; text; internal format)
OFFSET 4,1 COMMENTS Always an integer for primes > 5. REFERENCES D. M. Burton, Elementary Number Theory, McGraw-Hill, Sixth Edition (2007), p. 185. LINKS N. Hobson, Table of n, a(n) for n = 4..1000 N. Hobson, Home page (listed in lieu of email address) FORMULA a(n) = A125617(n)/prime(n). EXAMPLE The quadratic nonresidues of 7=prime(4) are 3, 5 and 6. Hence a(4) = (3^2 + 5^2 + 6^2)/7 = 10. PROG (PARI) vector(46, m, p=prime(m+3); t=1; for(i=2, (p-1)/2, t+=((i^2)%p)^2); (p-1)*(2*p-1)/6-t/p) CROSSREFS Cf. A076409, A076410, A125613-A125618. Sequence in context: A333104 A214959 A054095 * A341261 A154033 A275238 Adjacent sequences: A125615 A125616 A125617 * A125619 A125620 A125621 KEYWORD easy,nonn AUTHOR Nick Hobson, Nov 30 2006 STATUS approved
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Last modified December 7 20:40 EST 2021. Contains 349589 sequences. (Running on oeis4.) | 695 | 1,944 | {"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} | 3.75 | 4 | CC-MAIN-2021-49 | latest | en | 0.654086 |
https://my-assignmentexpert.com/2022/08/17/%E9%87%91%E8%9E%8D%E4%BB%A3%E5%86%99%E5%88%A9%E7%8E%87%E7%90%86%E8%AE%BA%E4%BB%A3%E5%86%99portfolio-theory%E4%BB%A3%E8%80%83fin586-estimating-models-with-excess-returns/ | 1,696,011,136,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510520.98/warc/CC-MAIN-20230929154432-20230929184432-00890.warc.gz | 437,432,084 | 46,412 | # 金融代写|利率理论代写Portfolio Theory代考|FIN586 ESTIMATING MODELS WITH EXCESS RETURNS
my-assignmentexpert™提供最专业的一站式服务：Essay代写，Dissertation代写，Assignment代写，Paper代写，Proposal代写，Proposal代写，Literature Review代写，Online Course，Exam代考等等。my-assignmentexpert™专注为留学生提供Essay代写服务，拥有各个专业的博硕教师团队帮您代写，免费修改及辅导，保证成果完成的效率和质量。同时有多家检测平台帐号，包括Turnitin高级账户，检测论文不会留痕，写好后检测修改，放心可靠，经得起任何考验！
## 金融代写|利率理论代写Portfolio Theory代考|ESTIMATING MODELS WITH EXCESS RETURNS
When excess returns $\left(R^{c}\right)$ are used to estimate and test asset pricing models, the moment conditions (pricing equations) are
$$E\left(m R^{e}\right)=0_{N} .$$
Let $m=\theta_{0}-\left(\theta_{1} f_{1}+\cdots+\theta_{K} f_{K}\right)$. In this case, the mean of the SDF cannot be identified or, equivalently, the parameters $\theta_{0}$ and $\left(\theta_{1}, \ldots, \theta_{K}\right)$ cannot be identified separately. This requires a particular choice of normalization. One popular normalization is to set $\theta_{0}=1$, in which case $m=1-\left(\theta_{1} f_{1}+\cdots+\theta_{K} f_{K}\right)$. An alternative (preferred) normalization is to set $\theta_{0}=1+\theta_{1} E\left(f_{1}\right)+\cdots+\theta_{K} E\left(f_{K}\right)$, in which case $m=1-\theta_{1}\left[f_{1}-E\left(f_{1}\right)\right]-\cdots-\theta_{K}\left[f_{K}-E\left(f_{K}\right)\right]$ with $E(m)=1$. These two normalizations can give rise to very different results (see Kan and Robotti, 2008; Burnside, 2010)
Kan and Robotti (2008) argue that when the model is misspecified, the first (raw) and the second (de-meaned) normalizations of the SDF produce different GMM estimates that minimize the quadratic form of the pricing errors. Hence, the pricing errors and the $p$-values of the specification tests are not identical under these two normalizations. Moreover, the second (de-meaned) specification imposes the constraint $E(m)=1$ and, as a result, the pricing errors and the HJ-distances are invariant to affine transformations of the factors. This is important because in the first normalization, the outcome of the model specification test can be easily manipulated by simple scaling of factors and changing the mean of the SDF. This problem is not only a characteristic of linear SDFs but also arises in nonlinear models. The analysis in Burnside (2010) further confirms these findings and links the properties of the different normalizations to possible model misspecification and identification problems discussed in the previous two subsections.
## 金融代写|利率理论代写Portfolio Theory代考|CONDITIONAL MODELS WITH HIGHLY PERSISTENT PREDICTORS
The usefulness of the conditional asset pricing models crucially depends on the existence of some predictive ability of the conditioning variables for future stock returns. While a large number of studies report statistically significant coefficients for various financial and macro variables in in-sample linear predictive regressions of stock returns, several papers raise the concern that some of these regressions may be spurious. For example, Ferson, Sarkissian, and Simin (2003) call into question the predictive power of some widely used predictors, such as the term spread, the book-to-market ratio, and the dividend yield. Spurious results arise when the predictors are strongly persistent (near unit root processes) and their innovations are highly correlated with the predictive regression errors. In this case, the estimated slope coefficients in the predictive regression are biased and have a nonstandard (nonnormal) asymptotic distribution (Elliott and Stock, 1994; Cavanagh, Elliott, and Stock, 1995; Stambaugh, 1999). As a result, $t$-tests for statistical significance of individual predictors based on standard normal critical values could reject the null hypothesis of no predictability too frequently and falsely signal that these predictors have predictive power for future stock returns. Campbell and Yogo (2006) and Torous, Valkanov, and Yan (2004) develop valid testing procedures when the predictors are highly persistent and revisit the evidence on the predictability of stock returns.
Spuriously significant results and nonstandard sampling distributions also tend to arise in long-horizon predictive regressions, where the regressors and/or the returns are accumulated over $r$ time periods so that two or more consecutive observations are overlapping. The time overlap increases the persistence of the variables and renders the sampling distribution theory of the slope coefficients, $t$-tests and $R^{2}$ coefficients, nonstandard. Campbell (2001) and Valkanov (2003) point out several problems that emerge in long-horizon regressions with highly persistent regressors. First, the $R^{2}$ coefficients and $t$-statistics tend to increase with the horizon, even under the null of no predictability, and the $R^{2}$ is an unreliable measure of goodness of fit in this situation. Furthermore, the $t$-statistics do not converge asymptotically to well-defined distributions and need to be rescaled to ensure valid inference. Finally, the estimates of the slope coefficients are biased and, in some cases, not consistently estimable. All these statistical problems provide a warning to applied researchers and indicate that the selection of conditioning variables for predicting stock returns should be performed with extreme caution.
# 利率理论代写
## 金融代写|利率理论代写PORTFOLIO THEORY代 考|ESTIMATING MODELS WITH EXCESS RETURNS
$$E\left(m R^{e}\right)=0_{N}$$
## Matlab代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 1,730 | 6,023 | {"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} | 3.03125 | 3 | CC-MAIN-2023-40 | longest | en | 0.637134 |
https://www.nagwa.com/en/videos/917127295931/ | 1,701,388,073,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100258.29/warc/CC-MAIN-20231130225634-20231201015634-00873.warc.gz | 1,013,180,746 | 9,829 | # Question Video: Applying the Right-Angled Triangle Altitude Theorem and the Pythagorean Theorem Mathematics
From the figure, determine the length of the line π΅π·. If necessary, round your answer to the nearest hundredth.
02:26
### Video Transcript
From the figure, determine the length of the line π΅π·. If necessary, round your answer to the nearest hundredth.
Weβre asked in this example to find the length of the line π΅π·, which we can see is the perpendicular projection to π· on the hypotenuse of the right triangle π΄π΅πΆ from the right angle at π΅. To find π΅π·, our first step will be to use the Pythagorean theorem to find the length of the hypotenuse π΄πΆ of triangle π΄π΅πΆ. We can then use this in one part of the right triangle altitude theorem to find the side length πΆπ· and then use this value in the Pythagorean theorem in right triangle πΆπ·π΅ to find the side length we want, which is π΅π·.
So, letβs start with the Pythagorean theorem applied to triangle π΄π΅πΆ to find the length of π΄πΆ. We have π΄πΆ squared equal to πΆπ΅ squared plus π΄π΅ squared. And substituting our two known side lengths, this gives 15 squared plus eight squared on the right-hand side. This evaluates to 289. And taking the positive square root on both sides, positive since lengths are positive, we have π΄πΆ equals the square root of 289, which is 17 centimeters.
So, now marking this on the diagram and clearing some space, next we can use the right triangle altitude theorem to find side length πΆπ·. We know that πΆπ΅ is 15 centimeters. And weβve just found that π΄πΆ is 17 centimeters. And so we have 15 squared equals πΆπ· multiplied by 17. Now, dividing through by 17 and evaluating 15 squared, we have πΆπ· equal to 225 over 17.
Next, applying the Pythagorean theorem to triangle πΆπ·π΅, we have π΅π· squared equals π΅πΆ squared minus πΆπ· squared. Thatβs 15 squared minus 225 over 17 all squared. This is 49.826 and so on. And taking the square root on both sides, we have π΅π· equal to 7.058 and so on. Finally, rounding to the nearest hundredth, which is to two decimal places, we have that π΅π· is equal to 7.06 centimeters. | 726 | 2,276 | {"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} | 4.78125 | 5 | CC-MAIN-2023-50 | latest | en | 0.818626 |
www.financegecko.com | 1,718,981,654,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00384.warc.gz | 684,761,633 | 9,824 | Home » Loans » Signature Loans » Signature Loan Calculator
# Signature Loan Calculator
If you are looking for a no collateral loan a signature loan calculator might be something for you to investigate. Basically a signature loan calculator can help you to determine how much money you can borrow and how much your payments for the loan might be. The following article explains what signature loan calculators are and also looks at the the benefits they offer.
## What Is A Signature Loan Calculator?
Signature loans have become one of the most popular types of loan today. However, before one applies for such a loan, you need to learn more about them so that you may understand the important details them. A signature loan is a form of unsecured personal loan which can be used by borrowers for whatever purpose they want. When we say ‘unsecured’ in this context it means the borrower need not present collateral to have the signature loan application granted. Depending on the borrower’s credit score and ability to pay, creditors usually offer signature loana from as low as \$500 to as high as \$25,000.
What about the interest raates charged on this type of loan? A signature loan calculator can help you to map out the interest rate variances that are charged by a given loans provider, since each signature loan may vary significantly. For those applicants who are considered as low risk, lenders may lend them with prime rates and zero fees. However, if the borrower is someone who falls under the category of high risk borrower, they may get a loan with an interest rate that is triple the prime rate with fees included to it to boot. In these scenarios, signature loans calulators can be extremely effective to help you understand what you are likely to pay.
## The Advantages Of Using Signature Loans Calculators
To determine the exact amount that you may have to pay for a signature loan you can use a signature loan calculator. Most signature loan calculators commonly used these days require you to enter certain loan information needed by the calculator to present you a profile of the loan you might get. Say for example, your term is 4 years, the interest rate is 12 percent, and the loan that you are looking for is \$10,000. After you have inputt the figures into the signature loan calculator, your monthly payments may be expressed. So, using the figures above as an example, we will get a result of \$263.33 as your monthly payment.
If you want to see the differences of increased payments over time, a signature loan calculator should be able to show you what the impact of overpayment will be, giving you a flexible and powerful tool to help you understand how you can save money on a signature loan.
## Where Can You Find A Signature Loan Calculator?
Many websites offer free signature loans calculators that might usually be in the form of a downloadable Microsoft Excel spreadsheet with included formulas to predict your loan payments. Oftentimes, lenders themselves offer a signature loan calculator on their websites to help you see how their loans work before you apply. | 608 | 3,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.972603 |
https://fiunz.447.in.net/san-andreas-pelicula-completa-en-espa%C3%B1ol-latino.html | 1,582,470,564,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145774.75/warc/CC-MAIN-20200223123852-20200223153852-00138.warc.gz | 376,639,652 | 4,930 | Jul 01, 2011 · This Excel tutorial shows how to calculate the principal and interest payments of a loan. Watch more at http://www.lynda.com/Excel-2010-tutorials/Financial-F... The Excel PMT Function (payment function) is a really simple to use but highly useful Financial Function used to calculate the repayment amount on a loan. This function assumes that payments are made consistently (repayment frequency and amount remain constant) at a constant interest rate. Schedule of Loan Amortization in Excel – Step by Step. Let us take home loan example for preparing a schedule of Loan Amortization in Excel. Let us assume that a home loan is issued at the beginning of month 1. The principal is \$1,500,000 the interest rate is 1% per month and the term is 60 months. Repayments are to be made at the end of each ...
Sep 18, 2018 · Calculate loan repayments with Excel. Photo by rawpixel on Unsplash. Check out the PMT tutorial before starting this one to create the initial data table.. Of course, making loan calculations can be done easily using free online loan calculators, but that’s not as fun as doing it yourself in Excel and this way you can link your information into your budget to run some dynamic financial ... Dec 27, 2018 · You can quickly create a spreadsheet in Microsoft Excel to perform the calculation for you--and, in the process, gain a greater understanding of just how a mortgage loan works. Launch Microsoft Excel. However, the loan you're probably most familiar with, is the amortizing loan, where the principal of the loan is paid back over the life of the loan, typically through a set of equal payments. In this lesson, I'm going to show you how to calculate the repayments on an amortizing loan, using the example of a home mortgage. The loan payment formula can be used to calculate any type of conventional loan including mortgage, consumer, and business loans. The formula does not differ based on what the money is spent on, but only when the terms of repayment deviate from a standard fixed amortization. Jan 09, 2020 · Schedule Loan Repayments With Excel Formulas Understanding Your Mortgage. Using Excel, you can get a better understanding... Calculate the Monthly Payment. First, here's how to calculate the monthly payment for a mortgage. Calculate the Annual Interest Rate. We have seen how to set up the ...
Mar 29, 2019 · Steps 1. Launch Microsoft Excel and open a new workbook. 2. Save the workbook file with an appropriate and descriptive name. 3. Create labels in cells A1 down to A4 for the variables and result of your monthly payment... 4. Enter the variables for your loan or credit card account in the cells ... Oct 21, 2019 · Interest - The amount of the total paid that is interest. Principal - The amount of the total paid that is not interest (e.g., loan payment). Extra Payment - The dollar amount of any extra payments you make. Loan - The amount of your loan that remains after a payment. However, the loan you're probably most familiar with, is the amortizing loan, where the principal of the loan is paid back over the life of the loan, typically through a set of equal payments. In this lesson, I'm going to show you how to calculate the repayments on an amortizing loan, using the example of a home mortgage. Re: Loan drawdown and repayment - how to automate Thank You for the feedback and for marking the thread as 'Solved' (if problems arise you can reverse that action using the 'Thread Tools' link). A good way to see what a formula is doing is to select one of the cells using the formula then select/run the 'Evaluate Formula' tool on the 'Formulas ... Mar 07, 2018 · Loan Amortization Schedule – Excel Loan End Date: – To calculate Loan End Date, enter EDATE function and take EMI Start Date as Start Date and then Monthly Installment in second argument and subtract 1 from it to get the Loan End Date. Excel Amortization Schedule. With the last step, half of the work for this template has been finished. Loan amortization schedule Create a loan summary and payment schedule based on the terms of the initial loan with this loan amortization schedule template.This is an accessible template. Excel
Mar 07, 2018 · Loan Amortization Schedule – Excel Loan End Date: – To calculate Loan End Date, enter EDATE function and take EMI Start Date as Start Date and then Monthly Installment in second argument and subtract 1 from it to get the Loan End Date. Excel Amortization Schedule. With the last step, half of the work for this template has been finished. The loan payment formula can be used to calculate any type of conventional loan including mortgage, consumer, and business loans. The formula does not differ based on what the money is spent on, but only when the terms of repayment deviate from a standard fixed amortization. Find out about compound interest and how to use the compounding interest formula in Microsoft Excel to calculate the compound interest on a loan. If you're starting to shop around for student loans, you may want a general picture of how much you're going to pay. If you're refinancing existing debt, you may want a tool to compare your options based on how far you've already come with repayment. Either way, check out these simple Excel formulas to compare different student loan options. Total Amount of a Loan Using Excel. Microsoft Excel can be easily used to calculate the total amount of a loan to be repaid. The total amount to be repaid in a loan is a combination of the initial amount borrowed and the total amount of interest to be added, excel can very easily calculate the total amount to be paid using the PMT function. Jan 09, 2020 · Schedule Loan Repayments With Excel Formulas Understanding Your Mortgage. Using Excel, you can get a better understanding... Calculate the Monthly Payment. First, here's how to calculate the monthly payment for a mortgage. Calculate the Annual Interest Rate. We have seen how to set up the ...
PMT, one of the financial functions, calculates the payment for a loan based on constant payments and a constant interest rate. Use the Excel Formula Coach to figure out a monthly loan payment. At the same time, you'll learn how to use the PMT function in a formula. Syntax. PMT(rate, nper, pv, [fv], [type]) However, the loan you're probably most familiar with, is the amortizing loan, where the principal of the loan is paid back over the life of the loan, typically through a set of equal payments. In this lesson, I'm going to show you how to calculate the repayments on an amortizing loan, using the example of a home mortgage. The loan payment formula can be used to calculate any type of conventional loan including mortgage, consumer, and business loans. The formula does not differ based on what the money is spent on, but only when the terms of repayment deviate from a standard fixed amortization. Oct 30, 2019 · Build a spreadsheet: You can also build advanced spreadsheets in programs like Google Sheets and Microsoft Excel. Those sheets complete calculations and show you how the loan works year-by-year. See more details about using a spreadsheet for standard amortizing loans (including auto loans, home loans, and many personal loans). Mar 07, 2018 · Loan Amortization Schedule – Excel Loan End Date: – To calculate Loan End Date, enter EDATE function and take EMI Start Date as Start Date and then Monthly Installment in second argument and subtract 1 from it to get the Loan End Date. Excel Amortization Schedule. With the last step, half of the work for this template has been finished. Mar 29, 2019 · Steps 1. Launch Microsoft Excel and open a new workbook. 2. Save the workbook file with an appropriate and descriptive name. 3. Create labels in cells A1 down to A4 for the variables and result of your monthly payment... 4. Enter the variables for your loan or credit card account in the cells ...
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Sep 18, 2018 · The other thing that the repayment amount doesn’t take into account is monthly bank fees, so if you expect to have to pay fees, you might like to factor these into your calculations. how to calculate loan repayments in excel. First you need to enter your data. Principle = the amount you want to borrow. Dec 27, 2018 · You can quickly create a spreadsheet in Microsoft Excel to perform the calculation for you--and, in the process, gain a greater understanding of just how a mortgage loan works. Launch Microsoft Excel. Re: Loan drawdown and repayment - how to automate Thank You for the feedback and for marking the thread as 'Solved' (if problems arise you can reverse that action using the 'Thread Tools' link). A good way to see what a formula is doing is to select one of the cells using the formula then select/run the 'Evaluate Formula' tool on the 'Formulas ... Interest calculation formula for loans in Excel The calculation of interest on the loan in Excel and calculate the effective interest rate, with the following information on the bank offers credit: We carry out interest calculation on loan and calculate the effective interest rate with the following information on the bank offers credit.
# Repayment loan formula excel
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Schedule of Loan Amortization in Excel – Step by Step. Let us take home loan example for preparing a schedule of Loan Amortization in Excel. Let us assume that a home loan is issued at the beginning of month 1. The principal is \$1,500,000 the interest rate is 1% per month and the term is 60 months. Repayments are to be made at the end of each ... Schedule of Loan Amortization in Excel – Step by Step. Let us take home loan example for preparing a schedule of Loan Amortization in Excel. Let us assume that a home loan is issued at the beginning of month 1. The principal is \$1,500,000 the interest rate is 1% per month and the term is 60 months. Repayments are to be made at the end of each ... Excel does have PMT function Hello, Exel does have the formula you are lloking for and it is: =PMT(rate,nper,pv,fv,type) Where rate is the rate, nper is number of periods, pv is the amount of mortgage/loan FV and type are optional. This article considers how to model debt repayment calculations from a practical perspective. It addresses three common calculations using Excel’s financial functions for the last item. For completeness, my examples include the seemingly more convoluted mathematical formulas that arrive at the same answer. Find out about compound interest and how to use the compounding interest formula in Microsoft Excel to calculate the compound interest on a loan. How to calculate total interest paid on a loan in Excel? Let’s say you have bought a house with a bank loan, and you need to pay the bank every month in coming years. Do you know how much interest you will pay on the loan? Actually, you can apply the CUMIPMT function to figure it out easily in Excel. Calculate total interest paid on a loan in ... Find out about compound interest and how to use the compounding interest formula in Microsoft Excel to calculate the compound interest on a loan. | 2,413 | 11,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-10 | latest | en | 0.949436 |
https://byjus.com/question-answer/why-is-a-first-degree-polynomial-in-two-variables-ax-by-c-0-called-a-1/ | 1,708,776,785,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474533.12/warc/CC-MAIN-20240224112548-20240224142548-00802.warc.gz | 148,772,720 | 21,605 | Question
# Why is a first degree polynomial in two variables ax + by + c = 0, called a linear equation?
A
It has infinitely many solutions
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B
The geometrical representation is a straight line
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C
It has two variables
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D
None of the above
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Solution
## The correct option is B The geometrical representation is a straight line The degree (the highest power on any variable) of the given equation is 1. Hence, its geometrical representation is a straight line.
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Join BYJU'S Learning Program | 231 | 891 | {"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-2024-10 | latest | en | 0.90912 |
https://goprep.co/ex-4.1-q1-the-cost-of-a-notebook-is-twice-the-cost-of-a-pen-i-1njggz | 1,618,485,569,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00334.warc.gz | 384,227,650 | 28,808 | # The cost of a not
Let the cost of pen be y and the cost of notebook be x
As per the question,
Cost of a notebook = Twice the cost of pen = 2y
So,
2y = x
x – 2y = 0
This is a linear equation in two variables to represent this statement.
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## Newton First Law MCQs: Questions and Answers
MCQ 1:
Reference frame in which Newton's law holds is termed as
1. second reference frame
2. inertial reference frame
3. net reference frame
4. external reference frame
MCQ 2:
When two or more forces acts on a body, by adding the individual forces vectorially, we get
1. resultant force
2. initial force
3. origin force
4. kinetic force
MCQ 3:
Fact that states that a single force that has magnitude and direction of the net force has the same effect on the body as all the individual forces together is known as
1. principle of superposition for forces
2. principle of addition of forces
3. principle of initialization of forces
4. principle of superposition of velocities
MCQ 4:
A body will keep moving with a constant velocity, if
1. no force acts on it
2. external force acts on it
3. centripetal force acts on it
4. acceleration is variable
MCQ 5:
If no force acts on a body, the body's velocity cannot change, that is the body cannot accelerate, is the Newton's
1. 1st Law
2. 2nd Law
3. 3rd Law
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# Harry Schwester
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Was ist denn los?
Nur für deutschsprechende Leute! Wie geht's? ...also gut, bis bald!
5 years ago | 1 | 76 solvers
Solved
What number has this problem?
This problem is added because it is problem number *???* in the "Community" problems section. <http://www.mathworks.de/matlab...
5 years ago
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only input
Return the output without writing any code into the function.
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Beginner's Problem - Squaring
Try out this test problem first. Given the variable x as your input, square it by two and put the result in y. Examples: ...
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Given an input variable x, output a variable y that is 7 greater than x. Example: Input x = 1 Output y is 8 Input ...
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Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
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Pizza!
Given a circular pizza with radius _z_ and thickness _a_, return the pizza's volume. [ _z_ is first input argument.] Non-scor...
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Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
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Is my wife right?
Regardless of input, output the string 'yes'.
5 years ago
Solved | 400 | 1,457 | {"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} | 3.078125 | 3 | CC-MAIN-2020-50 | longest | en | 0.723886 |
https://discuss.codechef.com/t/your-approach-to-solve-sandwich/14618?page=2 | 1,621,032,976,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991829.45/warc/CC-MAIN-20210514214157-20210515004157-00174.warc.gz | 248,934,303 | 6,587 | # Your approach to solve SANDWICH?
1 Like
Given N,K
Now, minimum number of partitions you can have is P = ceil(N/K),
Now what is the maximum size of sandwich can be made with P partitions ?
That is P*K, suppose initially all partitions are full, for example if N=10 and K=4, we have P=3
and partitions are like this :
|…|…|…| now we have 3 groups each is fully filled.
Now coming to the question,
we need to delete some ‘.’ from these groups in order to get a valid configuration in which total size of sandwich is N, suppose if we add a ‘X’ to any group we reduce its size by 1. Now the current size is `P*K` and required size is `N`, so we need to distribute `(P * K - N)` ‘X’ (s) to these P groups, if you observe `R = (P * K - N) = (K - N % K)`, and now our problem is reduced to finding ways of distributing R things in P groups ( any group can get 0 ‘X’ ) for which you can compute the binomial coefficient `C(P+R-1,P-1)`.
Hi guys!
Here is a video editorial by @jtnydv25 on the problem.
It uses Number Theory and the Chinese Remainder Theorem to get to the solution. Feel free to leave your doubts and suggestion in the video comments.
Video Editorial - SANDWICH
3 Likes
There was similar problem on hackerank. And the link is editorial to that problem which gives detailed explanation about how to implement ncr % m, whether m is prime or composite;
Ratings haven’t updated.
1 Like
But CRT only works for square free numbers right?
Ik, just in case he read that after the ratings were updated, it would’ve been weird.
I am not quite sure , but i think CRT works fine if N and m are co- prime.
Here’s a quote from that link:
“You can find the result of nCr % m for each m = 27, 11, 13, 37. Once you have the answers, you can reconstruct the answer modulo 142857 using Chinese Remainder Theorem. These answers can be found by Naive Methods since, m is small.”
how do u calculate ans modulo all prime powers in M .??
Since prime power can be non prime ( i am assuming prime power means p^q)
So denominator and prime power can be non co prime too right ??
So how do you calculate modular inverse !!
1 Like
@sanket407 nCr % p^q has to be calculated using Lucas theorem for prime powers
Yes N and K are not 10^18 in that testcase.
how ?? p^q is non prime right ?? lucas theorem works for only prime modulus right ?? Correct me if wrong !!
so constraints were n,k <= 10^18 or the same as of subtask ,2??
How did you identify that (nk−n%k+knk)(nk−n%k+knk) is the solution? I would appreciate if you can explain that as well
1 Like
Thank you ymondelo20
Thanks abdullah786
@c0d3h4ck3d Updated the answer with resources for both.
@vijju123 It was more of an observation, after writing a brute force algorithm.
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Thanks, mathecodecian
it is, but in a paper of Andrew Granville, it has been extended to prime powers, and after that u need to use chinese remainder theorem | 757 | 2,903 | {"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-2021-21 | latest | en | 0.931084 |
http://regency-group.com/karutha-pakshikal-rit/71ebcc-track-modulus-is-defined-as-formula | 1,632,492,814,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00120.warc.gz | 45,656,090 | 6,317 | It may be calculated using the formula: Example: 100 mod 9 equals 1 Because 100/9 = 11 with a remainder of 1 Another example: 14 mod 12 equals 2 Because 14/12 = 1 with a remainder of 2 σ is the Stress, and ε denotes strain. This is accomplished by the rail acting as a continuously supported beam, distributing the load across several ties. Section Modulus calculator uses Section Modulus=(Breadth of the Section*Height of the Section^2)/6 to calculate the Section Modulus, The Section Modulus formula is defined as a geometric property of the cross section used for designing beams and flexural members. Bulk Modulus Formulas . You next response, understandably, might be, “That doesn’t clarify anything,” so let’s take a closer look: How It Works. (a) Vertical loads consisting of dead loads, dynamic augment of loads including the effect of speed, the hammer blow effect, the inertia of reciprocating masses, etc. The bulk modulus of a material may be measured by powder diffraction, using x-rays, neutrons, or electrons targeting a powdered or microcrystalline sample. A rail is subjected to heavy stresses due to the following types of forces. A 1 meter length of rubber with a Young's modulus of 0.01 GPa, a circular cross-section, and a radius of 0.001 m is subjected to a force of 1,000 N. Fritz Birmann, in Railroad Track Mechanics and Technology, 1978. (b) Lateral forces due to the movement of live loads, eccentric vertical loading, shunting of locomotives, etc. Dividing this equation by tensile strain, we obtain the expression for Young’s modulus: . Definition of . Publisher Summary. In this article, we will discuss its concept and Young’s Modulus Formula with examples. Let us learn the interesting concept! The evaluation of the effects of surcharge loads on buried pipes can be addressed using the Iowa formula. function of the track support, which is generally defined in terms of the vertical track modulus (lb/in/in). Young’s modulus $$Y$$ is the elastic modulus when deformation is caused by either tensile or compressive stress, and is defined by Equation \ref{12.33}. We can write the expression for Modulus of Elasticity using the above equation as, E = (F*L) / (A * δL) So we can define modulus of Elasticity as the ratio of normal stress to longitudinal strain. Young’s Modulus or Elastic Modulus or Tensile Modulus, is the measurement of mechanical properties of linear elastic solids like rods, wires, etc. This formula, over 100 years old, matches the results of state-of-the-art finite element analysis with pipe-soil interaction. At most tracks, a white line traces the exterior of curbs in corners; that white line represents the track limit. The Modulo Operation Expressed As a Formula; Use Cases for the Modulo Operation. In the formula as mentioned above, “E” is termed as Modulus of Elasticity. The modulus operator, written in most programming languages as % or mod, performs what is known as the modulo operation. Modulo Operation. The modulo (or "modulus" or "mod") is the remainder after dividing one number by another. Given an integer n > 1, called a modulus, two integers are said to be congruent modulo n, if n is a divisor of their difference (i.e., if there is an integer k such that a − b = kn).. Congruence modulo n is a congruence relation, meaning that it is an equivalence relation that is compatible with the operations of addition, subtraction, and multiplication. Forces Acting on the Track . 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As a continuously supported beam, distributing the load across several ties to high speeds most,. Discuss its concept and Young ’ s modulus formula with examples in most programming languages as % or mod performs! To high speeds elasticity of a track in ballast with regard to high speeds the exterior of curbs in ;!, and ε denotes strain remainder after dividing one number by another matches the results of state-of-the-art finite analysis! The effects of surcharge loads on buried pipes can be addressed using formula... In ballast with regard to high speeds Mechanics and Technology, 1978 Use Cases for modulo. This formula, over 100 years old, matches the results of state-of-the-art finite element with. Formula: track limits are essentially defined based on the intended layout the. The intended layout of the track support, which is generally defined in terms of the dynamic modulus elasticity. Σ is the remainder after dividing one number by another ) Lateral forces due to the following of. | 2,489 | 12,008 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 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} | 3.78125 | 4 | CC-MAIN-2021-39 | latest | en | 0.884045 |
https://m.scirp.org/papers/63271 | 1,590,901,822,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347410745.37/warc/CC-MAIN-20200531023023-20200531053023-00570.warc.gz | 418,038,604 | 16,699 | On Irresolute Topological Vector Spaces
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Abstract: In this paper, our focus is to investigate the notion of irresolute topological vector spaces. Irresolute topological vector spaces are defined by using semi open sets and irresolute mappings. The notion of irresolute topological vector spaces is analog to the notion of topological vector spaces, but mathematically it behaves differently. An example is given to show that an irresolute topological vector space is not a topological vector space. It is proved that: 1) Irresolute topological vector spaces possess open hereditary property; 2) A homomorphism of irresolute topological vector spaces is irresolute if and only if it is irresolute at identity element; 3) In irresolute topological vector spaces, the scalar multiple of semi compact set is semi compact; 4) In irresolute topological vector spaces, every semi open set is translationally invariant.
Received 22 December 2015; accepted 25 January 2016; published 29 January 2016
1. Introduction
If a set is endowed with algebraic and topological structures, then by means of a mathematical phenomenon, we can construct a new structure, on the bases of an old structure which is well known. This is the case we have introduced and discussed for beautiful interaction between linearity and topology in this paper. Although the new notion is similar to the notion of topological vector spaces, mathematically it behaves differently. To define irresolute topological vector space, we keep the algebraic and topological structures unaltered on a set but continuity conditions of vector addition and scalar multiplication are replaced by one of the characterizations of irresolute mappings.
A topological vector space [1] is a structure in topology in which a vector space X over a topological field F(R or C) is endowed with a topology such that the vector space operations are continuous with respect to.
The axioms for a space to become a topological vector space or linear topological space have been given and studied by Kolmogroff [2] in 1934 and von Neumann [3] in 1935. The relation between the axioms of topological vector space has been discussed by Wehausen [4] in 1938 and Hyers [5] in 1939. Also, Kelly [6] has done classical work on topological vector spaces. In the last decade, we can see the work of Chen [7] , on fixed points of convex maps in topological vector spaces. Bosi et al. [8] and Clark [9] have researched on conics in topological vector spaces. More work, in recent years, has been done by Drewnowski [10] , Alsulami and Khan [11] and Kocinac et al. [12] . In 2015, Moiz and Azam [13] defined and investigated s-topological vector spaces, which is a generalization of topological vector spaces.
The motivation behind the study of this paper is to investigate such structures in which the topology is endowed upon a vector space which fails to satisfy the continuity condition for vector addition and scalar multiplication or either. We are interested to study such structures for irresolute mappings in the sense of Levine. The concept of irresolute was introduced by Crossely and Hildebrand in 1972 as a consequence of the study of semi open sets and semi continuity in topological spaces, defined by Levine [14] . In this paper, several new facts concerning topologies of irresolute topological vector spaces are established.
2. Preliminaries
Throughout in this paper, X and Y are always representing topological spaces on which separation axioms are not considered until and unless stated. We will represent field by F and the set of all real numbers by. and are assumed negligible small but positive real numbers.
Semi open sets in topological spaces were firstly appeared in 1963 in the paper of N. Levine [14] . With invent of semi open sets and semi continuity, many interesting concepts in topology were further generalized and investigated by number of mathematicians. A subset A of a topological space X is said to be semi open if, and only if, there exists an open set O in X such that, or equivalently if. denotes the collection of all semi open sets in the topological space. The complement of a semi open set is said to be semi closed; the semi closure of, denoted by, is the intersection of all semi closed subsets of X containing A [15] . It is known that if, and only if, for any semi open set U containing , is non-empty. Every open set is semi open and every closed set is semi closed. It is known that union of any collection of semi open sets is semi open set, while the intersection of two semi open sets need not be semi open. The intersection of an open set and a semi open set is semi open set. A subset A of a topological space X is said to be semi compact if for every cover of A by semi open sets of X, there exists a finite sub cover.
If is a vector space then e denotes its identity element, and for a fixed, , and, , denote the left and the right translation by x, respectively. The addition mapping is defined by, and the scalar multiplication mapping is defined by.
Definition 1. Let be single valued function between topological spaces (continuity not assumed). Then:
1) is termed as semi continuous [14] , if and only if, for each V open in Y, there exists.
2) is termed as irresolute [15] , if, and only if, for each, there exists . Note that the function is irresolute at, if for each semi open set V in containing, there exists a semi open set U in X containing x such that.
Recall that a topological vector space is a vector space over a topological field F (most often the
real or complex numbers with their standard topologies) that is endowed with a topology such that:
1) Addition mapping defined by is continuous function.
2) Multiplication mapping defined by. is continuous function (where the domains of these functions are endowed with product topologies).
Equivalently, we have a topological vector space X over a topological field F (most often the real or complex numbers with their standard topologies) that is endowed with a topology such that:
1) for each, and for each open neighbourhood W of in X, there exist neighbourhoods U and V of x and y respectively in X, such that.
2) for each and for each open neighbourhood W in X containing, there exist neighbourhoods U of in F and V of x in X such that. Or equivalently, we have: topological Vector Space X over the field with a topology on X such that is a topological group and is a continuous mapping.
3. Irresolute Topological Vector Spaces
In this section we will define and investigate basic properties of irresolute topological vector spaces. Examples are given to show that topological vector spaces are independent of irresolute topological vector spaces in general.
Definition 2. A space is said to be an irresolute topological vector space over the field F if the
following two conditions are satisfied:
1) for each and for each semi open neighbourhood W of in X, there exist semi open neighbourhoods U and V in X of x and y respectively, such that.
2) for each and for each semi open neighbourhood W of in X, there exist semi open neigh- bourhoods U of in F and V of x in X, such that.
Remark 1. Topological vector spaces are independent of irresolute topological vector spaces.
The following example shows that is neither a topological vector space nor an irresolute topological vector space.
Example 1. Consider the vector space R(R) endowed with the lower limit topology on, generated by
the base, then is neither a topological vector space nor an irre-
solute topological vector space.
Example 2. Let be a topology on generated by the base, then
is a topological vector space as well as irresolute topological vector space over the field.
The next example shows that is an irresolute topological vector space which fails to be a topological
vector space.
Example 3. Consider the field with standard topology on F. Let, where topology defined on
X be generated by the base. Then is not a topological vector
space, because for but, if we choose an open neighbourhood of in X, then, there does not exist any open neighbourhoods U and V of x and y respectively in X, which satisfy the relation.
Now, we show that is an irresolute topological vector space. To verify the first condition, let
.
Case I: Let Consider a semi open neighbourhood (or, ) of in X. Then, for the selection of semi open neighbourhoods (resp.) and (resp.) of x and y respectively in X, we have for each.
Case II: Let, for or. Consider a semi open neighbourhood (or,) of in X. Then, for the selection of semi open neighbourhoods (resp.) and (resp.) of x and y respectively in X, we have for each.
Now, we have to verify the second condition. For this we have four cases,
Case I: Let and. Then for each semi open neighbourhood (or,) of in X, we can choose semi open neighbourhoods (resp.) and (resp.) containing and x in F and X respectively. Then, for every (resp.).
Case II: Let, and. Then for each semi open neighbourhood (or,) of in X, we can choose semi open neighbourhoods (resp.) and, (resp.) containing and x in F and X respectively. Then,
for every (resp.).
Case III: Let and. Then for each semi open neighbourhood (or,) of in X, we can choose semi open neighbourhoods (resp.) and (resp.) containing and x in F and X respectively. Then,
for every (resp.).
Case IV: Let and. Then for each semi open neighbourhood (or,) of in X, we can choose semi open neighbourhoods (resp.) and (resp.) containing and x in F and X respectively. Then, for every (resp.).
Since, both conditions for irresolute topological vector spaces are satisfied, therefore, is an irreso-
lute topological vector space.
Theorem 1. Let be an irresolute topological vector space. Then:
1) The (left) right translation defined by; for all, is irresolute.
2) The translation, defined by; for all, is irresolute.
Proof. 1. Let W be a semi open neighbourhood of. Then by definition, there exist semi open neighbourhoods U and V in X containing y and x respectively, such that. Or . This proves that, is irresolute mapping.
2. Let, then. Let W be any semi open neighbourhood of, then by definition, there exist semi open neighbourhoods U in F of and V in X of x, such that. This gives that
. This proves that is an irresolute mapping.
Remark 2. In topological vector spaces, every open set is translationally invariant whereas in irresolute topological vector spaces, every semi open set is translationally invariant.
Theorem 2. Let be an irresolute topological vector space. If, then:
1) for every.
2) for every.
Proof 1. Let, and let, then we have to prove that z is a semi-interior point of. Now, , where x is some point in A. We can write. By the right translation
, we have. Since, X is irresolute topological vector space and is
irresolute, by Theorem 1 , we have for any semi open neighbourhood A containing, there exists
semi open neighbourhood of z such that, that is. Thus for any
, we can find a semi open neighbourhood such that. Hence.
2. Let and. This means, for some, so we can write
and. Then we can define mapping by. Since, X is an irresolute
topological vector space and by Theorem 1(2), is irresolute mapping, so, we have for any semi open neighbourhood containing, there exists semi open neighbourhood of z such that
. This gives. That is, for any, we can find a semi open neighbourhood
, such that. Hence.
Theorem 3. Let be an irresolute topological vector space. If and B is any subset of
X, then is semi open in X.
Proof. Suppose and. Then, for each and by Theorem 2 (1), We have
. Now, for each. Because arbitrary union of
semi open sets is semi open, therefore is semi open in X.
Corollary 1. Suppose is an irresolute topological vector space. If, then the set
is semi open in X.
Theorem 4. Let be an irresolute topological vector space. Then is an irresolute mapping.
Proof. Let and. The. Let W be a semi open neighbourhood of in X.
Since is an irresolute topological vector space, therefore there exist semi open neighbourhoods U of
in F and V of x in X such that,. Or. Since, and, therefore,. This proves that is an irresolute mapping.
Theorem 5. Let be an irresolute topological vector space. The defined by
is an irresolute mapping.
Proof. Let and. Let W be a semi open neighbourhood of in X. Since is an Irresolute topological vector space, therefore, there exist semi open neighbourhoods U of x and
V of y in X such that,. Or. Since, and
, therefore,. This proves that is an irresolute mapping.
Let A be semi open in X. Then, by Theorem 3, and. Similarly, we can prove that each set is semi open in X. Thus the set is semi open in X.
Definition 3. A mapping f form a topological space to itself is called irresolute-homeomorphism [15] , if it is bijective, irresolute and pre-semi open.
Theorem 6. Let be an irresolute topological vector space. For given and with
, each translation mapping and multiplication mapping, where is irresolute homeomorphism onto itself.
Proof. First, we show that is an irresolute homeomorphism. It is obviously bijective. By Theorem 1, is irresolute. Moreover, is pre-semi open because for any semi open set U, by Theorem 2 (1), is semi open.
Similarly, we can prove that is an irresolute homeomorphism.
Definition 4. An irresolute topological vector space is said to be irresolute homogenous space, if
for each, there exists irresolute homeomorphism f of the space X onto itself such that.
Theorem 7. Every irresolute topological vector space is an irresolute homogenous space.
Proof. Let be an irresolute topological vector space. Take, put. Define,
by. By Theorem 6, is irresolute homeomorphism, therefore
is an irresolute homogenous space.
Theorem 8. Suppose that is an irresolute topological vector space and S is a subspace of X. If S
contains a non-empty semi open subset of X, then S is semi open in.
Proof. Suppose U is a non-empty semi open subset in X, such that. By Theorem 2 (1),
is semi open subset of X for each. Thus is semi open in X being union of semi open
sets.
In general, intersection of two semi open sets is not semi open; however we have the following lemma.
Lemma 1. [17] Let be a topological space and. If A is open and U is semi open, then.
Lemma 2. [17] Suppose is a topological space., where, then if, and only if,.
Theorem 9. Every open subspace S of an irresolute topological vector space is also an irresolute topological vector space.
Proof. Suppose is an irresolute topological vector space and S is an open subspace of X. Then, it
satisfies the following properties.
1) For all, we have.
2) For any and, we have. We define topology on S as,. We show that is itself an irresolute topological vector space.
Now, let, and W be any semi open neighbourhood of in S, then W is semi open neighbor-
hood of in X. As is an irresolute topological vector space, therefore, there exist semi open
neighbourhoods of x and of y such that. Now, the sets and are semi open in X containing x and y respectively. By Lemma, 2, , and.
Again, for, and, let W be a semi open neighbourhood of in S and hence semi open in X.
As is an irresolute topological vector space, therefore there exist semi open neighbourhoods A of in F and B of in X such that. Now, the sets and are semi open in F and
X respectively. Since, S is open, therefore by Lemma 2, V is semi open in S. Hence for each semi open neighbourhood W of in S, there exist semi open neighbourhoods U in F of and V in S of x such that
. This proves that is an irresolute topological vector space.
Theorem 10. In irresolute topological vector spaces, for any semi open neighbourhood U of 0, there exists a semi open neighbourhood V of 0 such that.
Proof. The proof is trivial, therefore omitted.
Theorem 11. Let A and B be subsets of an irresolute topological vector space. Then .
Proof. Let and, and let W be a semi open neighbourhood of. Then there exist semi open neighbourhoods U and V of x and y respectively, such that. Since, , , there are and. Then,. This implies. That is,.
Theorem 12. Let be an irresolute topological vector space, then every semi open subspace of X is
semi closed in X.
Proof. Let H be a semi open subspace of X. As right translation is irresolute homeomorphism,
therefore, is semi open. Then, is also semi open. Hence, is semic-
losed.
Theorem 13. Let be a homomorphism of irresolute topological vector spaces. f is
irresolute on X if it is irresolute at.
Proof. Let. Suppose W is semi open neighbourhood of in Y. Since, is irresolute, therefore, there is a semi open neighbourhood V of 0 such that. Now, since f is irresolute at, there exists semi open neighbourhood U of 0 in X such that. Since is irresolute, therefore, is semi open neighbourhood of x. Thus, . This proves that, f is irresolute at x and hence on X.
Theorem 14. Let be an irresolute topological vector space and are subsets of X. If B is
semi open, then for any set A, we have.
Proof. As we know that so. Conversely, let and write where and. There exists a semi open neighbourhood V of zero such that. Now, V is semi open neighbourhood of 0 in X, this gives that is also semi open neighbourhood of 0 in X. Since, , so,. We know that . Therefore,. Hence,.
Theorem 15. Let be an irresolute topological vector space. Then the scalar multiple of semi closedset is semi closed.
Proof. Let, then.. Therefore,.
Theorem 16. Let be an irresolute topological vector space. Then scalar multiple of semi-compact
set is semi-compact.
Proof. Let A be a semi-compact subsets of X. Let be a semi open cover of for some non
zero, then. This gives. Since, and is an irresolute topological vector space, therefore, for each αÎ. Since, A is semi-compact therefore, there exist a finite subset of such that This implies that. Hence is semi-compact in X.
Definition 5. [18] A space is said to be P-regular, if for each semi closed set F and, there exist disjoint open sets U and V such that and.
Theorem 17. Let be a P-regular and irresolute topological vector space. Then the algebraic sum of
a semi-compact set A and semi-closed set B is semi-closed.
Proof. Let, then for some,. Since, the translation mapping is irresolute homeomorphism so, where is semi closed. Since X is P-regular space, therefore, there exist
open sets and such that and. Also is semi open and contains a. Hence,. Since, A is semi-compact, therefore there exists a finite subset of elements of A, such that. Let, then U is a neighbourhood
of x. We claim that. If not, then, then for some i and, which is contradiction to the fact that.
Cite this paper: Khan, M. and Iqbal, M. (2016) On Irresolute Topological Vector Spaces. Advances in Pure Mathematics, 6, 105-112. doi: 10.4236/apm.2016.62009.
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[6] Kelly, J.L. (1955) General Topology. Van Nastrand, New York.
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[8] Bosi, G., Candeal, J.C., Indurain, E. and Zudaire, M. (2005) Existence of Homogenous Representations of Interval Orders on a Cone in Topological Vector Space. Social Choice and Welfare, 24, 45-61.
http://dx.doi.org/10.1007/s00355-003-0290-2
[9] Clark, S. T. (2004) A Tangent Cone Analysis of Smooth Preferences on a Topological Vector Space. Economic Theory, 23, 337-352.
http://dx.doi.org/10.1007/s00199-003-0366-3
[10] Drewnowski, L. (2007) Resolution of Topological Linear Spaces and Continuity of Linear Maps. Journal of Mathematical Analysis and Applications, 335, 1177-1194.
http://dx.doi.org/10.1016/j.jmaa.2007.02.032
[11] Alsulami, S.M. and Khan, L.A. (2013) Weakly Almost Periodic Functions in Topologicl Vector Spaces. African Diaspora Journal of Mathematics, 15, 76-86.
[12] Kocinac, L.D.R. and Zabeti, O. (2015) A Few Remarks on Bounded Operators on Topological Vector Spaces.
http://arxiv.org/abs/1410.6299
[13] Khan, M.D. and Azam, S. (2015) S-Topological Vector Spaces. Jr. of Linear and Topological Algebra, 4, 153-158.
[14] Levine, N. (1963) Semi-Open Sets and Semi-Continuity in Topological Spaces. The American Mathematical Monthly, 70, 36-41.
http://dx.doi.org/10.2307/2312781
[15] Crossley, S.G. and Hildebrand, S.K. (1972) Semi-Topological Properties. Fundamenta Mathematicae, 74, 233-254.
[16] Crossley, S.G. and Hildebrand, S.K. (1971) Semi-Closure. Texas Journal of Science, 22, 99-112.
[17] Noiri, T. (1973) Semi-Continuous Mappings. Accad. Nazionale Dei Lincei, LIV.
[18] Khan, M. and Ahmad, B. (1995) On P-Regular Spaces. Math. Today, XIII, 51-56.
Top | 5,339 | 21,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-24 | longest | en | 0.93224 |
https://qazmath.com/blog-math/the-logic-of-mathematical-proofs | 1,696,471,214,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00775.warc.gz | 512,124,281 | 22,030 | # The Logic of Mathematical Proofs
Title: The Logic of Mathematical Proofs: A Comprehensive Guide
Subtitle: How to Master the Art of Proving Your Mathematical Theorems
Introduction
Mathematics is a field of study that has fascinated people for centuries. It is the language of science and technology, the foundation of engineering, and the key to unlocking the mysteries of the universe. But beneath all the complexity and beauty of mathematics lies a fundamental logic that can be used to prove theorems and solve problems. This logic is known as mathematical proof, and it is the cornerstone of mathematics.
In this article, we’ll explore the logic of mathematical proofs. We’ll discuss what a proof is, how to construct one, and the various types of proofs. We’ll also provide examples of proofs and answer some frequently asked questions. By the end of this article, you’ll have a better understanding of how to prove mathematical theorems and solve problems.
Body
A mathematical proof is a logical argument that establishes the truth of a mathematical statement. It is a way of demonstrating that a statement is true, and it is an essential part of mathematics. To prove a statement, you must start with a set of assumptions or axioms, which are accepted as true. You then use logic and reasoning to derive a conclusion from these assumptions.
There are several types of proofs, each with its own set of rules and techniques. The most common type of proof is direct proof, which is a straightforward argument that starts with the assumptions and ends with the conclusion. Another type of proof is proof by contradiction, which starts with the assumption that the statement is false and then uses logic to show that this assumption leads to a contradiction. Other types of proofs include proof by induction, proof by construction, and proof by exhaustion.
Examples
Let’s look at an example of a direct proof. Consider the statement “For every integer n, n2 + 2n is even.” To prove this statement, we start by assuming that n is an integer. We then use algebra to show that n2 + 2n is always divisible by 2, which means it is even.
Here is an example of a proof by contradiction. Consider the statement “There are no real numbers x and y such that x2 + y2 = -1.” To prove this statement, we assume that there are real numbers x and y such that x2 + y2 = -1. We then use algebra to show that this assumption leads to a contradiction, which means the statement is true.
FAQ Section
Q: What is a mathematical proof?
A: A mathematical proof is a logical argument that establishes the truth of a mathematical statement. It is a way of demonstrating that a statement is true, and it is an essential part of mathematics.
Q: How do you construct a proof?
A: To construct a proof, you must start with a set of assumptions or axioms, which are accepted as true. You then use logic and reasoning to derive a conclusion from these assumptions.
Q: What are the different types of proofs?
A: The most common type of proof is direct proof, which is a straightforward argument that starts with the assumptions and ends with the conclusion. Other types of proofs include proof by contradiction, proof by induction, proof by construction, and proof by exhaustion.
Summary
In this article, we explored the logic of mathematical proofs. We discussed what a proof is, how to construct one, and the various types of proofs. We also provided examples of proofs and answered some frequently asked questions. By the end of this article, you should have a better understanding of how to prove mathematical theorems and solve problems.
Conclusion
Mathematical proofs are the cornerstone of mathematics. They are the logical arguments that establish the truth of a mathematical statement. Understanding the logic of mathematical proofs is essential for anyone who wants to master the art of proving their mathematical theorems. With practice and dedication, you can learn to construct proofs and solve problems with confidence.
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June 21, 2007, 04:50 torque calculations #1 Mechstud Guest Posts: n/a Hi .. I am working on a simulation of a crop harvesting machine and I want to draw a contour plot of the torque acting on the blade due to fluid. For this I need magnitude of the torque acting on different points of the blade surface. I used the torque function in CFX post.. torque_z@Rot Blade 1 But this function calculates the total magnitude of the torque acting on the whole blade ..not on discrete infinitesimal points of the blade..please help me to calculate torque on various points of the blade. Thanks in advance Mechstud
June 21, 2007, 18:44 Re: torque calculations #2 Glenn Horrocks Guest Posts: n/a Hi, You could write a CEL function which takes the torque component out of the pressure and wall shear variables, and does a cross product with the rotation axis. Or you could just do contour plots of pressure and/or wall shear on the blade. No need for CEL functions then. Glenn Horrocks
June 22, 2007, 02:40 Re: torque calculations #3 Mechstud Guest Posts: n/a Thank you very much Glenn... Although I have also figured out the solution to this problem 2 hurs later I posted this topic. I defined variable which calculate torque on a point using forces in x and y direction multiplying with y and x respectively. and then doing vector addition. ..Well in your solution ..what do you mean by wall shear variables ??..I am a bit new to CFX .. Thanks a lot for your help once again Best Wishes ...Mechstud
June 24, 2007, 21:50 What is wall shear? #4 Glenn Horrocks Guest Posts: n/a Hi, Fluid transfers momentum to a surface by two means. The first is by pressure and the second is by friction, that is shear in the fluid boundary layer. In CFX they are expressed as two separate variables, pressure and wall shear. Glenn Horrocks
June 26, 2007, 08:53 Re: What is wall shear? #5 Mechstud Guest Posts: n/a thanx....its all clear now... mechstud
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Contact Us - CFD Online - Top | 692 | 2,729 | {"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-2016-44 | latest | en | 0.889184 |
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The question may sound silly. If it is I'm sorry for it but I just couldn't find an answer anywhere else.
I have just learned about vector spaces and their properties and on the other hand have also started with Lagrangian mechanics. The author writes: "The configuration of the system of N particles, moving freely in space, may be represented by the position of a single point in 3N dimensional space, which is called the configuration space of the system."
My questions here are how is it possible for us to visualize the position of say 700 particles using just a single point in 2100 dimensional space? I cannot make any sense of it. Since there are no constraints, for every particle we are adding up 3 dimensions; What is the advantage in doing so?
And is this in anyway related to vector spaces?
• What about e.g. the Wikipedia article is unclear to you? – ACuriousMind Aug 27 '15 at 15:16
• I believe the answer is no, because it doesn't make sense to add two different configurations to each other, and addition of elements is necessary for the space to be considered a vector space. – march Aug 27 '15 at 16:28
My questions here are how is it possible for us to visualize the position of say 700 particles using just a single point in 2100 dimensional space?
Think of it as a map. You can read a map and loom around and you can learn how to relate the common points. So of you have 700 particles there are 2100 scalars. So have a giant 2100 dimensional vector space. You can place the first three to be the xyz of the first particle then the next three for the xyz of the second particle and so on.
Just think of it as a map given one you find the other one by finding the correspondence. So once you get hood at moving between the two you are fine. You can start with a simpler example. Three particles on a line. Then you find that the plane x=y is two of them touching, as are the planes x=z and y=z and the line x=y=z is all three touching.
What is the advantage in doing so?
Lagrangian mechanics would be the biggest reason.
And is this in anyway related to vector spaces?
Yes. It doesn't have to be for instance if there are constraints. But unlike one of the comments you can consider the configuration at two times and subtract them yo get a displacement in configuration space and then you can scale that by the time interval and thus get a derivative. Derivatives often are related to vector space structures.
• You mean to say that configuration do behave as vector spaces? If so what will adding of two configurations give me if subtraction gives displacement? And are inner products associated with them? – Weezy Aug 30 '15 at 12:52
• @eoshah Some configuration spaces behave as vector spaces. Adding them isn't as physical as subtracting because when you add them the answer depends on which configuration is the origin configuration If you treat the configuration as a displacement from the origin then the sum is the sum of the displacements. And yes there can be an inner product again that helps for taking the limit to get that derivative. – Timaeus Aug 30 '15 at 13:56
They are not related structurally: Configuration space is a manifold which in general has no vector space structure. For example $\mathbb{R}$, the configuration space of a free particle moving on a line can be viewed as a vector space (you can sensibly "add" two configurations to get a new one and so on), but if you constrain it to move on a circle this structure is lost.
There is an interesting link between the two though which appears in quantum mechanics. One can naturally associate a vector space to a classical configuration space: If your system has configuration space $M$, one can take the vector space to be $L^2(M)$, the space of square integrable functions on $M$. This is the natural setting of the quantum version of this classical system.
(Thanks to ACuriousMind for pointing out the very fundamental flaw in the first version of this answer!)
• Wrong, the symplectic manifold you are talking about is the phase space, which is the cotangent bundle of the configuration space. Quantization takes place on the phase space, but indeed obtains the wavefunctions, broadly speaking, as the square-integrable functions on the original configuration space. – ACuriousMind Aug 27 '15 at 19:46
• Ah I see! I didn't think about this carefully enough. Since the relation between position and momentum plays an important role in quantum mechanics, phase space (the total space of $M$ and $T^*M$) is what's quantized... I'll edit the answer. – childofsaturn Aug 27 '15 at 19:51
• Okay so what I could understand from your answer was that configuration spaces can behave like vector spaces but this is not the general case.? Also I couldn't understand the 2nd paragraph and the comments to this answer since I'm relatively new to quantum mechanics. – Weezy Aug 30 '15 at 12:57
• @eoshah Right, they can be vector spaces, but they don't have to be (for example when there are constraints). It is probably better to just learn what they are for many examples. It's also like $\mathbb R^2$ if you use polar coordinates of isn't as easy to add vectors though it still is a vector space. Sometimes it is about what is easier rather than what is possible. – Timaeus Aug 30 '15 at 14:02 | 1,202 | 5,359 | {"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} | 3.53125 | 4 | CC-MAIN-2019-43 | latest | en | 0.951817 |
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# 3 - laplace_disc_steps - Step functions Recall the...
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Step functions April 9, 2011 Recall the Heaviside step function (or unit-step function) is defined as u c ( t ) = ( 0 x < c 1 x > c (1) Also recall the “t-shifting” theorem (as opposed to the “s-shifting” theorem): L { u c ( t ) f ( t - c ) } = e - cs L { f ( t ) } . (2) In order to solve differential equations with discontinuous forcing functions, we express them in terms of the Heaviside functions, and use above theorem to take their Laplace transforms. Problem statement: (“Forward” problem) Find the Laplace transform of g ( t ) = u c ( t ) h ( t ) (3) Steps: 1. Identify f ( t - c ) . This function is what happens to be multiplied by u c ( t ) . In problem statement above, f ( t - c ) = h ( t ) . 2. Evaluate f ( t ) from Step 1. This means replacing t with t + c in the identified function. 3. From Eq. ( 2 ), L { g ( t ) } = e - cs L { f ( t ) } , where f ( t ) is the func- tion determined in Step 2. Problem statement:
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differentiate and do not simplify: Y=e^ -sq.rt x + e^3 - 2^(3x-1)
91. ### Calculus
I'm trying to find out what the graph of: y= |x+1| + |x+2| looks like. could someone help me out?
92. ### calculus
find the domain of f(x)=5/¡Ì(x2-4)
93. ### Calculus
Derivative of (Sec(z))^2
94. ### Calculus
Find dy/dx y=e^(4xtan2x)
95. ### Calculus
derivative of (4Cos^2(3x)+3)^2
96. ### calculus
Differentiate. sqrt((6x^2 + 5x + 1)^3)
97. ### Calculus
derivative of cot(6/x)
98. ### Calculus
Find lim [f(x+h)-f(x)]/h h->0 f(x) = 2 - 5x + x^2
99. ### calculus
Calculate R8 for f(x)=8−x over [3,5]. R8=
100. ### calculus
Calculate R8 for f(x)=8−x over [3,5]. R8= | 2,688 | 6,439 | {"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} | 3.8125 | 4 | CC-MAIN-2018-30 | latest | en | 0.590958 |
http://sportdocbox.com/Sailing/69776631-C-miles-per-hour-d-all-of-the-above-2-when-you-look-at-the-speedometer-in-a-moving-car-you-can-see-the-car-s.html | 1,560,879,242,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998808.17/warc/CC-MAIN-20190618163443-20190618185443-00185.warc.gz | 166,768,140 | 25,437 | C) miles per hour. D) all of the above. 2) When you look at the speedometer in a moving car, you can see the car's
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1 Practice Kinematics Questions (Answers are at the end ) 1) One possible unit of speed is. A) light years per century. B) kilometers per hour. C) miles per hour. D) all of the above.. 2) When you look at the speedometer in a moving car, you can see the car's A) average acceleration. B) instantaneous acceleration. C) average distance traveled. D) instantaneous speed. E) average speed. 3) Acceleration is defined as the CHANGE in. A) velocity divided by the time interval. B) time it takes to move from one speed to another speed. C) velocity of an object. D) time it takes to move from one place to another place. E) distance divided by the time interval. 4) Suppose you are in a car that is going around a curve. The speedometer reads a constant 30 miles per hour. Which of the following is NOT true? A) Your velocity is constant. B) Your acceleration is constant. C) Your speed is constant. D) You and the car are accelerating. E) Your direction is constantly changin 5) An object travels 8 meters in the first second of travel, 8 meters again during the second second of travel, and 8 meters again during the third second. Its acceleration is. A) 8 m/s to power of (2). B) 0 m/s to power of (2). C) 16 m/s to power of (2). D) 32 m/s to power of (2). 6) Ten seconds after starting from rest, a car is moving at 40 m/s. What is the car's average acceleration? A) 2.5 m/s to power of (2) B) 4.0 m/s to power of (2) C) 40 m/s to power of (2) D) 10 m/s to power of (2) E) 0.25 m/s to power of (2) 7) As an object falls freely in a vacuum, its. A) velocity increases. B) acceleration increases. C) both A and B D) none of the above
2 8) In the absence of air resistance, objects fall at constant. A) distances each successive second. B) acceleration. C) speed. D) velocity. E) all of the above 9) Speed is. A) a measure of how fast something is moving. B) always measured in terms of a unit of distance divided by a unit of time. C) the distance covered per unit time. D) all of the above.. 10) A ball is thrown upwards and caught when it comes back down. In the absence of air resistance, the speed of the ball when caught would be. A) less than the speed it had when thrown upwards. B) more than the speed it had when thrown upwards. C) the same as the speed it had when thrown upwards. 11) Suppose an object is in free fall. Each second the object falls. A) a larger distance than in the second before. B) with the same average speed. C) with the same instantaneous speed. D) the same distance as in the second before. 12) If you drop a feather and a coin at the same time in a tube filled with air, which will reach the bottom of the tube first? A) Neither-they will both reach the bottom at the same time. B) The coin C) The feather 13) Consider drops of water leaking from a water faucet. As the drops fall they. A) remain at a relatively fixed distance from each other. B) get closer together. C) get farther apart. 14) A ball tossed vertically upward rises, reaches its highest point, and then falls back to its starting point. During this time the acceleration of the ball is always. A) directed downward. B) directed upward. C) in the direction of motion. D) opposite its velocity.
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NEWTONS 1st LAW 1. A 10.00 kg mass is tied to a string with a maximum strength of 100 N. A second string of equal strength is tied to the bottom of the mass. a) If the bottom string is pulled with a jerk
Chapter 3: Two-Dimensional Motion and Vectors
Assumption College English Program Mr. Stephen Dobosh s EP- M 4 P h y s i c s C l a s s w o r k / H o m e w o r k P a c k e t Chapter 3: Two-Dimensional Motion and Vectors Section 1: Introduction to Vectors
ACTIVITY THE MOTION OF PROJECTILES
Name (printed) ACTIVITY THE MOTION OF PROJECTILES First Day Stamp INTRODUCTION In this activity you will begin to understand the nature of projectiles by mapping out the paths of two projectiles over time;
Unit 3 ~ Learning Guide Name:
Unit 3 ~ Learning Guide Name: Instructions: Using a pencil, complete the following notes as you work through the related lessons. Show ALL work as is explained in the lessons. You are required to have
6 Motion in Two Dimensions BIGIDEA Write the Big Idea for this chapter.
6 Motion in Two Dimensions BIGIDEA Write the Big Idea for this chapter. Use the What I Know column to list the things you know about the Big Idea. Then list the questions you have about the Big Idea in
The speed of an inline skater is usually described in meters per second. The speed of a car is usually described in kilometers per hour.
The speed of an inline skater is usually described in meters per second. The speed of a car is usually described in kilometers per hour. Speed How are instantaneous speed and average speed different? Average | 7,641 | 30,089 | {"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} | 3.75 | 4 | CC-MAIN-2019-26 | latest | en | 0.885026 |
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# Problem on Natural Numbers | TIFR B 2010 | Problem 4
Try this problem of TIFR GS-2010 using your concepts of number theory and congruence based on natural numbers.
## Problem on Natural Numbers | TIFR 201O| PART B | PROBLEM 4
Which of the following statements is false?
• There exists a natural number which when divided by $3$ leaves remainder $1$ and when divided by $4$ leaves remainder $0$
• There exists a natural number which when divided by $6$ leaves remainder $2$ and when divided by $9$ leaves remainder $1$
• There exists a natural number which when divided by $7$ leaves remainder $1$ and when divided by $11$ leaves remainder $3$
• There exists a natural number which when divided by $12$ leaves remainder $7$ and when divided by $8$ leaves remainder $3$
### Key Concepts
NUMBER THEORY
CONGRUENCE
CHINESE REMAINDER THEOREM
Answer:There exists a natural number which when divided by $6$ leaves remainder $2$ and when divided by $9$ leaves remainder $1$
TIFR 2010|PART B |PROBLEM 12
ELEMENTARY NUMBER THEORY DAVID M.BURTON
## Try with Hints
Let us take the equations $x\equiv1(mod 3)$ and $x\equiv0(mod 4)$
Now we will apply Chinese remainder theorem to get the value of $x$
Since $3$,$4$ are relatively prime,gcd($3$,$4$)$=1$. Let $m=3\times4=12$
Then $M_1=4$,$M_2=3$.
Then gcd($M_1$,$3$)$=1$,gcd($M_2$,$4$)$=1$
Since gcd($4$,$3$)$=1$,therefore the linear congruence equation $4x\equiv1(mod 3)$ has a unique solution and $x\equiv1(mod 3)$ is the solution.
Since gcd($3$,$4$)$=1$,therefore the linear congruence equation $3x\equiv0(mod 4)$ has a solution and $x\equiv4(mod 4)$ is the solution.
Therefore,$x=1\times4\times1 +0\times3\times4=4$ is a solution.
The solution of the given system is $x\equiv4(mod 12)$
So we have used the Chinese Remainder Theorem to check the statements, you may use it to check for other options. | 577 | 1,929 | {"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} | 4.65625 | 5 | CC-MAIN-2021-39 | latest | en | 0.840339 |
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Oracle9i SQL Reference
Release 2 (9.2)
Part Number A96540-01
Home Book List Contents Index Master Index Feedback
Functions, 72 of 177
## MOD
#### Syntax
mod::=
Text description of mod
#### Purpose
`MOD` returns the remainder of `m` divided by `n`. Returns `m` if n is 0.
#### Examples
The following example returns the remainder of 11 divided by 4:
```SELECT MOD(11,4) "Modulus" FROM DUAL;
Modulus
----------
3
```
This function behaves differently from the classical mathematical modulus function when m is negative. The classical modulus can be expressed using the `MOD` function with this formula:
```m - n * FLOOR(m/n)
```
The following table illustrates the difference between the `MOD` function and the classical modulus:
m n MOD(m,n) Classical Modulus
`11`
`4`
`3`
`3`
`11`
`-4`
`3`
`-1`
`-11`
`4`
`-3`
`1`
`-11`
`-4`
`-3`
`-3`
See Also:
Copyright © 1996, 2002 Oracle Corporation.
All Rights Reserved.
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# SEMESTER SPRING 2014 Financial Statement analysis (FIN621) ASSIGNMENT NO. 02 DUE DATE: August 04, 2014 Total Marks: 10 Marks
SEMESTER SPRING 2014
Financial Statement analysis (FIN621)
ASSIGNMENT NO. 02
DUE DATE: August 04, 2014
Total Marks: 10 Marks
LEARNING OBJECTIVES:
After attempting this assignment whole-heartedly, the students will be able to:
Calculate the accounting ratios based on missing data
Evaluate and analyze the company’s strategies for improving cash funds.
Question #1:
ABC Techno Labs is engaged in manufacturing fertilizer, chemicals and pesticides
products in Pakistan. The company is known for cost-effective products and has gained a
tremendous growth during last 15 years. Due to its research and devolvement, patents and
dozens of successful products available in the market; the profit has increased 20% or
more annually. The company has never paid any dividends but it has a higher price per
share in the market. The company occasionally suffered from cash shortage due to heavy
research expenditures and rapid growth. In order to cope with this problem of cash
shortage, the company has decided to (1) reduce the credit period to 20 days instead of 45
days to its debtors (2) reduce its research and development expenditure by 30% annually.
Required:
In the light of above decided actions; evaluate and analyze (in detail) the above case in
perspective of:
1. Short term creditors of the company.
2. Credit customers of the company.
Discuss with logical arguments.
Question: # 2
If sales are 8 times of closing stock; Gross Profit Ratio is 30% and Net Profit is 25% of
gross profit and closing Stock is 0.4 million then prepare the Income Statement of ABC
company as below (also show complete working):
ABC Company Limited
Income Statement
For the year Ended 20XXX
Sales
Cost of sales
Gross Profit
Expenses
Net Profit
IMPORTANT:
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uploading difficulties. This extra time should only be used to meet the
emergencies and above mentioned due dates should always be treated as final
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IMPORTANT INSTRUCTIONS:
Complete calculations are required for every part of the problem.
All the students are recommended to attempt the assignment their own.
OTHER IMPORTANT INSTRUCTIONS:
Make sure to upload the solution file before the due date on VULMS.
Any submission made via email after the due date will not be accepted.
FORMATTING GUIDELINES:
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You may also compose your assignment in Open Office format.
Use black and blue font colors only.
Dear students!
As you know that Post Mid-Term semester activities have started and load
shedding problem is also prevailing in our country. Keeping in view the fact, you
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VULMS for the current semester, therefore no excuse will be entertained after
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guyzzz me ne dosra question solve kiya hai
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ans btao ayesha
mera tu net profit 340000
first question ke kuch samj ni a rhe
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# If S is a sequence of consecutive multiples of 3
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17 Jun 2014, 14:24
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If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?
(1) There are 15 terms in S.
(2) The greatest term of S is 126.
I don't agree with the official answer. Any thoughts on this question?
[Reveal] Spoiler: OA
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17 Jun 2014, 14:52
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If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?
(1) There are 15 terms in S.
In every set of 3 consecutive multiples of 3, there will be one multiple of 9:
{-9, -6, -3}
{-6, -3, 0}
{-3, 0, 3}
{0, 3, 6}
{3, 6, 9}
...
15 consecutive multiples of 3 could be break down into 5 different such sets, each of which will contain 1 multiple of 9. Therefore, there are 5 multiples of 9 in the set. Sufficient.
(2) The greatest term of S is 126. Clearly insufficient.
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]
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17 Jun 2014, 15:05
ajithkumar wrote:
If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?
(1) There are 15 terms in S.
(2) The greatest term of S is 126.
I don't agree with the official answer. Any thoughts on this question?
Indeed the correct answer is (A) because whatever sequence with 15 terms you pick you will always have 5 multiples of 9, whether you start the sequence with a multiple of 9 or not (there's a pattern: every 3 numbers there's a multiple of 9 and in a 15 terms there're 5 sets of 3 terms )
Hope that helps
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]
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17 Jun 2014, 16:31
Bunuel wrote:
If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?
(1) There are 15 terms in S.
In every set of 3 consecutive multiples of 3, there will be one multiple of 9:
{-9, -6, -3}
{-6, -3, 0}
{-3, 0, 3}
{0, 3, 6}
{3, 6, 9}
...
15 consecutive multiples of 3 could be break down into 5 different such sets, each of which will contain 1 multiple of 9. Therefore, there are 5 multiples of 9 in the set. Sufficient.
(2) The greatest term of S is 126. Clearly insufficient.
{-6, -3, 0} {-3,0,3} in these sets there are no multiples of 9..
So the answer is C isn't it?
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]
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17 Jun 2014, 16:32
clipea12 wrote:
ajithkumar wrote:
If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?
(1) There are 15 terms in S.
(2) The greatest term of S is 126.
I don't agree with the official answer. Any thoughts on this question?
Indeed the correct answer is (A) because whatever sequence with 15 terms you pick you will always have 5 multiples of 9, whether you start the sequence with a multiple of 9 or not (there's a pattern: every 3 numbers there's a multiple of 9 and in a 15 terms there're 5 sets of 3 terms )
Hope that helps
you are not considering 0, and negative multiples of 3. So the answer should be C
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]
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17 Jun 2014, 16:38
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ajithkumar wrote:
Bunuel wrote:
If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?
(1) There are 15 terms in S.
In every set of 3 consecutive multiples of 3, there will be one multiple of 9:
{-9, -6, -3}
{-6, -3, 0}
{-3, 0, 3}
{0, 3, 6}
{3, 6, 9}
...
15 consecutive multiples of 3 could be break down into 5 different such sets, each of which will contain 1 multiple of 9. Therefore, there are 5 multiples of 9 in the set. Sufficient.
(2) The greatest term of S is 126. Clearly insufficient.
{-6, -3, 0} {-3,0,3} in these sets there are no multiples of 9..
So the answer is C isn't it?
0 is divisible by EVERY integer except 0 itself, (or, which is the same, zero is a multiple of every integer except zero itself).
Check here for more: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1371030
Hope it helps.
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]
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17 Jun 2014, 16:52
errrrrr... how did I miss it???
Thanks...
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]
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18 Jun 2014, 02:25
For the second statement, 126 terms can be divided into 42 sets isn't it?
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]
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18 Jun 2014, 02:27
pretzel wrote:
For the second statement, 126 terms can be divided into 42 sets isn't it?
The second statement does not say that there are 126 terms in the set. It says that the greatest term of S is 126, there can be any number of elements.
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If S is a sequence of consecutive multiples [#permalink]
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19 Aug 2014, 04:59
If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?
(1) There are 15 terms in S.
(2) The greatest term of S is 126.
1) statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked;
2) statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked;
3) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked;
5) statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]
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19 Aug 2014, 05:01
Saulius1 wrote:
If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?
(1) There are 15 terms in S.
(2) The greatest term of S is 126.
1) statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked;
2) statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked;
3) BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked;
5) statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
Merging similar topics. Please refer to the discussion above.
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Re: If S is a sequence of consecutive multiples of 3 [#permalink]
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15 Jul 2015, 22:16
Bunuel wrote:
pretzel wrote:
For the second statement, 126 terms can be divided into 42 sets isn't it?
The second statement does not say that there are 126 terms in the set. It says that the greatest term of S is 126, there can be any number of elements.
To make it more clear,
S could be {120,123,126} which contains 1 multiple of 9 (126)
or
S could be {111,114,117,120,123,126} which contains 2 multiples of 9 (117 & 126)
So clearly insufficient.
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Display posts from previous: Sort by | 3,317 | 11,025 | {"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-2016-36 | longest | en | 0.894895 |
https://stats.stackexchange.com/questions/531273/how-to-compute-the-sandwich-variance-ml-estimator-in-r | 1,726,855,734,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00106.warc.gz | 498,099,769 | 42,077 | # How to compute the sandwich variance ML estimator in R
I'm currently estimating a DCC-type model by maximum likelihood. Im using the command solnp and it return an object where I can compute the Hessian H evaluated at the optimal values.
If I want to find the covariance matrix of the estimators, following Davidson and MacKinnon (2004), I can compute $$\hat{Var} ( \hat{\theta} ) = -H^{-1} ( \hat{\theta} )$$ However, there is a robust estimation for distribution miss-especification: the sandwich estimator or the quasi-maximum likelihood estimator (QMLE). It is as follows: $$\hat{Var} ( \hat{\theta} ) = H^{-1} ( \hat{\theta} ) G^{T} ( \hat{\theta} ) G ( \hat{\theta} ) H^{-1} ( \hat{\theta} )$$ where G is the gradient.
As solnp does not compute the gradient, I am using the command gHgenb to compute it at the optimal values of the parameters.
The problem is that when I compute the non-robust covariance matrix I get the following results:
Mean SE t Pvalue
[Joint]dcca1 0.0087397376 9.267334e-03 0.9430692 3.461028e-01
[Joint]dccb1 0.9653333580 1.255250e-02 76.9036529 1.723509e-278
0.0001513922 4.115874e-05 3.6782522 2.604074e-04
And when I compute the sandwich estimators I get the following results:
Mean SE t Pvalue
[Joint]dcca1 0.0087397376 6.622702e-09 1319663.4 0
[Joint]dccb1 0.9653333580 6.706874e-08 14393193.4 0
0.0001513922 1.538495e-10 984028.1 0
where the standard errors are extremely low.
The questions is: what should I do? Should I multiplicate any of the matrix by the number of observations (which one?)?, as Bollerslev and Wooldridge (1992) state that the non-robust matrix is: $$\hat{Var} ( \hat{\theta} ) = -H^{-1} ( \hat{\theta} ) / T$$
Thank you so much.
References:
Bollerslev, T., & Wooldridge, J. M. (1992). Quasi-maximum likelihood estimation and inference in dynamic models with time-varying covariances. Econometric reviews, 11(2), 143-172.
Davidson, R., & MacKinnon, J. G. (2004). Econometric theory and methods (Vol. 5). New York: Oxford University Press.
I just found the answer to my problem by comparing the manual method versus the QMLE robust standard errors for a GARCH using the package fGarch. The solution is as follows: Let hessian be the hessian computed by the optimization at the optimal values. Now, I computed a $$N\times K$$ matrix (where $$N$$ is the number of observations and $$K$$ the number of parameters). Using the command grad, from the package numDeriv, I computed the gradient for every observation in the optimal parameters and with those values I filled the matrix $$N\times K$$, which I called hessian. Then, the Information matrix and the QMLE robust standard errors are computed as follows:
meat<-t(gradient)%*%gradient
Info<-solve(hessian)%*%meat%*%solve(hessian)
SE_robust <- sqrt(diag(Info))
The sandwich matrix is $$-H^{-1}VH^{-1}$$ for $$V$$ an estimate of the expectation of $$G(\theta_0)^TG(\theta_0)$$, where $$G$$ is a row vector (following the poster's notation). Using $$G(\hat{\theta})^TG(\hat{\theta})$$ to approximate this expectation will not provide a good answer, because $$G(\hat{\theta}) = 0$$ by definition of $$\hat{\theta}$$!
To provide slightly more detail: if $$\hat{\theta}$$ solves $$\sum_{i=1}^{n}\psi_i(\theta) = 0$$ where $$\psi_i$$ are independent functions of the response $$Y_i$$, and $$\theta_0$$ solves $$\Psi(\theta) = 0$$ where $$\Psi = E(\psi_1)$$, then $$\sqrt{n}(\hat{\theta} - \theta_0)$$ is asymptotically $$N(0,V)$$ where $$V = H^{-1}SH^{-1}$$. These matrices are $$H = E\nabla_\theta\psi_1(\theta_0)$$ and $$S = E\psi_1(\theta_0)\psi_1(\theta_0)^T$$. We approximate $$H$$ by $$\tilde{H} = \sum_{i=1}^{n}\psi^{\prime}_{i}(\hat{\theta})$$ and $$V$$ by $$\tilde{V} = \sum_{i=1}^{n}\psi_i(\hat{\theta})\psi_i(\hat{\theta})^T$$. This last step is the part that's different from the question: $$\sum_{i=1}^{n}\psi_i(\hat{\theta}) = 0$$ so $$\left(\sum_{i=1}^{n}\psi_i(\hat{\theta})\right)\left(\sum_{i=1}^{n}\psi_i(\hat{\theta})\right)^T = 0$$, but $$\sum_{i=1}^{n}\psi_i(\hat{\theta})\psi_i(\hat{\theta})^T \neq 0$$.
Here is a useful reference: https://www.stat.berkeley.edu/~census/mlesan.pdf
My explanation is a poor recreation of material from van der Vaart's Asymptotic Statistics, Section 5.3, page 51/52; that's where to look for the good stuff! | 1,387 | 4,394 | {"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": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-38 | latest | en | 0.68634 |
https://polytope.miraheze.org/wiki/Pentagonal_gyrocupolarotunda | 1,686,310,547,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00415.warc.gz | 495,268,398 | 13,440 | # Pentagonal gyrocupolarotunda
Pentagonal gyrocupolarotunda
Rank3
TypeCRF
SpaceSpherical
Notation
Bowers style acronymPegycuro
Coxeter diagramxoxo5ofxx&#xt
Elements
Faces5+5+5 triangles, 5 squares, 1+1+5 pentagons
Edges5+5+5+5+10+10+10
Vertices5+5+5+10
Vertex figures5 isosceles trapezoids, edge lengths 1, 2, (1+5}0/2, 2
10 rectangles, edge lengths 1 and (1+5)/2
10 irregular tetragons, edge lengths 1, 1, 2, (1+5)/2
Measures (edge length 1)
Volume${\displaystyle 5\frac{11+5\sqrt5}{12} ≈ 9.24181}$
Dihedral angles3–4: ${\displaystyle \arccos\left(-\frac{\sqrt3+\sqrt{15}}{6}\right) ≈ 159.09484°}$
4–5 cupolaic: ${\displaystyle \arccos\left(-\sqrt{\frac{5+\sqrt5}{10}}\right) ≈ 148.28253°}$
3–5: ${\displaystyle \arccos\left(-\sqrt{\frac{5+2\sqrt5}{15}}\right) ≈ 142.62263°}$
3–3: ${\displaystyle \arccos\left(-\frac{\sqrt5}{5}\right) ≈ 116.56505°}$
4–5 join: ${\displaystyle \arccos\left(-\sqrt{\frac{25-11\sqrt5}{50}}\right) ≈ 95.15242°}$
Central density1
Related polytopes
ArmyPegycuro
RegimentPegycuro
Abstract & topological properties
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryH2×I, order 10
ConvexYes
NatureTame
The pentagonal gyrocupolarotunda is one of the 92 Johnson solids (J33). It consists of 5+5+5 triangles, 5 squares, and 1+1+5 pentagons. It can be constructed by attaching a pentagonal cupola and a pentagonal rotunda at their decagonal bases, such that the two pentagonal bases are rotated 36° with respect to each other.
If the cupola and rotunda are joined such that the bases are in the same orientation, the result is the pentagonal orthocupolarotunda.
## Vertex coordinates
A pentagonal gyrocupolarotunda of edge length 1 has vertices given by the following coordinates:
• ${\displaystyle \left(0,\,\sqrt{\frac{5+\sqrt5}{10}},\,\sqrt{\frac{5+2\sqrt5}{5}}\right),}$
• ${\displaystyle \left(±\frac12,\,-\sqrt{\frac{5+2\sqrt5}{20}},\,\sqrt{\frac{5+2\sqrt5}{5}}\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{5-\sqrt5}{40}},\,\sqrt{\frac{5+2\sqrt5}{5}}\right),}$
• ${\displaystyle \left(0,\,-\sqrt{\frac{5+2\sqrt5}{5}},\,\sqrt{\frac{5+\sqrt5}{10}}\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}{4},\,\sqrt{\frac{25+11\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}\right),}$
• ${\displaystyle \left(±\frac{3+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{40}},\,\sqrt{\frac{5+\sqrt5}{10}}\right),}$
• ${\displaystyle \left(±\frac12,\,±\frac{\sqrt{5+2\sqrt5}}{2},\,0\right),}$
• ${\displaystyle \left(±\frac{3+\sqrt5}{4},\,±\sqrt{\frac{5+\sqrt5}{8}},\,0\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}{2},\,0,\,0\right),}$
• ${\displaystyle \left(±\frac12,\,\sqrt{\frac{5+2\sqrt5}{20}},\,-\sqrt{\frac{5-\sqrt5}{10}}\right),}$
• ${\displaystyle \left(±\frac{1+\sqrt5}{4},\,-\sqrt{\frac{5+\sqrt5}{40}},\,-\sqrt{\frac{5-\sqrt5}{10}}\right),}$
• ${\displaystyle \left(0,\,-\sqrt{\frac{5+\sqrt5}{10}},\,-\sqrt{\frac{5-\sqrt5}{10}}\right).}$
## Related polyhedra
A decagonal prism can be inserted between the two halves of the pentagonal gyrocupolarotunda to produce the elongated pentagonal gyrocupolarotunda. | 1,198 | 3,075 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 18, "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 | 0.581655 |
https://www.glassdoor.com/Interview/If-we-were-paid-5-every-time-someone-Liked-our-clients-page-and-30-of-people-that-click-on-the-ad-like-the-page-what-QTN_511729.htm | 1,566,542,909,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318011.89/warc/CC-MAIN-20190823062005-20190823084005-00162.warc.gz | 821,690,236 | 50,023 | Ampush Interview Question: If we were paid \$5 every time... | Glassdoor
## Interview Question
Media Analyst Interview(Student Candidate) San Francisco, CA
# If we were paid \$5 every time someone "Liked" our clients
page, and 30% of people that click on the ad like the page, what is the maximum we can pay per click to have a 20% profit margin?
0
1.47?
Anonymous on Mar 31, 2014
28
1.20?
Anonymous on Apr 3, 2014
1
\$1.20
Anonymous on May 7, 2014
4
1.25
Anonymous on Oct 2, 2014
0
expected return is 1.50 (5*0.3) for person. Profit margin = (1.50-X)/1.50= 20%. X = 1.20\$
Anonymous on Mar 27, 2015 | 203 | 614 | {"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} | 2.953125 | 3 | CC-MAIN-2019-35 | latest | en | 0.913041 |
https://goprep.co/ex-5.c-q30-find-the-values-of-a-and-b-for-which-cc-a-b-a-2b-i-1nlmn9 | 1,603,677,960,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890108.60/warc/CC-MAIN-20201026002022-20201026032022-00484.warc.gz | 337,689,633 | 45,068 | # Find the values of a and b for which
Given :
To find : a and b
Formula used :
Where cij = ai1b1j + ai2b2j + ai3b3j + ……………… + ainbnj
If A is a matrix of order a b and B is a matrix of order c d ,then matrix AB exists and is of order a d ,if and only if b = c
=
= = =
=
Equating similar terms,
2a – b = 5
-2a – 2b = 4
Adding the above two equations,we get
-3b = 9
b = = -3
b = -3
substituting b = -3 in 2a – b = 5,we get
2a + 3 = 5
2a = 5 – 3 = 2
a = 1
a = 1 and b = -3
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This is a tutorial/explanation on how DFS and BFS can be used for the following problem: 1141A - Game 23
#### Statement
We are given two integers $n$ and $m$ and are asked to determine in how many steps can we tranform $n$ to $m$ with the following steps:
• multiply $n$ by $2$
• multiply $n$ by $3$
If $n$ can not be made equal to $m$ using the transformations mentioned above then the answer show be $-1$
#### Solution
Let's try and visualize the following example: $120, 51840$
The highlighted path is the one from $n$ to $m$, so we can see that it takes $7$ transformations
Observations:
• every number generated is the form of $n \cdot 2^{k_1} \cdot 3^{k_2}$
• distance from $n$ to $n \cdot 2^{k_1} \cdot 3^{k_2}$ is $k_1 + k_2$
• the tree follows this rule:
• there is one particular case where $n = m$, then the answer is $0$
Now that we know what rule it follows we can generate the tree itself. But since we want to know if a path exists from $n$ to $m$ we can use any graph treversal algorithm, in this blog im going to show you how it can be done using DFS and BFS.
So that we dont generate a number multiple times we will use a frequency array. The problem is that we haven’t got enough memory to store for each number up to $10^{18}$. As mentioned before all numbers generated are in the form of $n \cdot 2^{k_1} \cdot 3^{k_2}$ so we can use a frequency matrix in which $viz[i][j]$ is equal to $1$ if $n \cdot 2^{i} \cdot 3^{j}$ has already been generated, or $0$ it that numbers has not been generated yet
#### Implementation
DFS implementation
BFS implementation
• 0
» 3 days ago, # | 0 multiply n by 2 multiply m by 3 That m should be n according to the problem statement. Regardless of that, your approach is interesting.
• » » 3 days ago, # ^ | 0 Thanks for pointing that out. I have now changed it | 525 | 1,857 | {"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} | 4.6875 | 5 | CC-MAIN-2022-21 | latest | en | 0.883556 |
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