url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
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https://perl6.online/2019/09/09/check-if-an-element-is-in-a-list-in-perl-6/ | 1,568,579,438,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572289.5/warc/CC-MAIN-20190915195146-20190915221146-00275.warc.gz | 616,555,380 | 19,733 | # 📘 Check if an element is in a list in Perl 6
Tell if the given value is in the list.
There are a few approaches to the problem. The most compact one seems to be the use of the smartmatch `~~` operator in combination with the anyfunction:
`my @array = (10, 14, 0, 15, 17, 20, 30, 35);my \$x = 17;say 'In the list' if \$x ~~ any @array;`
To check if a given value is contained among the elements of an array or a list, use the greproutine.
`say 'In the list' if grep \$x, @array;`
The `grep` routine returns a list of all the matched elements. In the Boolean context, the return value is `True` if at least one element was found. In the opposite case, an empty list coerces to `False`. If the element in question is not zero, then the `first` routine may be used instead of `grep`. It returns the first element that matches the search pattern:
`say 'In the list' if first \$x, @array;`
To extend this to zero values, test the values before making a decision:
`say 'In the list' if \$x == first \$x, @array;`
Another solution is to convert the array to a hash and check if there is a key with the given value. This is useful when you need more than one check.
`my %hash = map {\$_ => 1}, @array;say 'In the list' if %hash{\$n};` | 362 | 1,247 | {"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-2019-39 | latest | en | 0.73484 |
https://forum.ansys.com/forums/topic/multiphase-mixture-model-mass-conservation-not-obeyed/ | 1,701,505,186,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100381.14/warc/CC-MAIN-20231202073445-20231202103445-00728.warc.gz | 309,213,971 | 36,044 | Fluids
Topics relate to Fluent, CFX, Turbogrid and more
Multiphase mixture model mass conservation not obeyed
• friday.abolorunke
Subscriber
Hi All,
I am modeling a two-phase problem with interfacial species mass transfer in a closed domain rectangular domain with impermeable BC & zero diffusive flux on all the walls.
Model: Mixture model for two phase. Species mass transfer for mass transfer mechanism.
Issues:
The mass of Vapor co2 = 2.5e-7g and Liquid ethanol = 2.9e-7g. Initial total mass = 5.4e-7g
Vapor co2 converts into liquid co2 via dissolution and goes into the liquid phase. So, in this system the total mass of the liquid increases while the vapor phase decreases. After I ran the simulation to steady-state I calculated the mass of the liquid, vapor and species to see if the mass was conserved. I expect the total mass to be the same (remember domain is closed).
At steady state, Mass of liquid co2 = 1.3e-7g, mass of liquid etoh = 3.45e-7g and mass of vapor co2 is 2.45e-7g. Final total mass = 7.2e-7g. Not equal to 5.47g
Total mass in the system increased. Another strange observation is that the mass of liquid Etoh increased too. I expected that to remain the same because the only mass transfer in the system is due to dissolution of vapor co2.
Please, what could be the reason behind this mass imbalance. Seems my system is gaining mass. Could I be doing something wrong. Any advise? asides from refining the mesh.
Regards,
Friday
• Rob
Forum Moderator
What is the gas density? How good (or likely bad) is the convergence? Why do you need to spam our feeds? We work office hours in various timezones and will get to your thread as we drop into the forum.
• friday.abolorunke
Subscriber
Hi Rob Sorry for spamming your feed and that of the other guys. I did not know it would have that effect.
The initial density of the gas phase is 141.3kg/m^3 and the liquid is 731.6kg/m^3. At the end of the simulation, i.e; at steady-state when the mass fraction & pressure field is uniform, the density of the gas phase becomes 132.45 kg/m^3 and the liquid 772.8 kg/m^3 and the volume of the liquid phase increased too, by about 50%. I am using Peng-Robinson EOS to compute the gas and liquid density as the density is dependent on pressure and composition. The system is isothermal, so the temperature is constant.
As per convergence, I think the convergence is good. The continuity, species & volume fraction residuals are below 1e-7, while velocity residual is below 1e-5.
Thank you for your time.
Regards Abolorunking
• Rob
Forum Moderator
No worries. Using the @ means we get a notification, one or two is fine, but it's become a habit and 15-20 notifications means the panels get overwhelmed. If one of us has missed a reply for a few days it's a different matter, except it's buried in the unnecessary notifications....
Back to the problem.
How are you dissolving the gas? Real gases aren't great around the phase change point. I tend to use ideal gas (if possible) and then set liquid species densities with temperature (liquids don't compress much), the phase density should follow with composition. If you monitor phase mass, species mass and mass transfer during the simulation is it a gradual loss or intermittent?
• friday.abolorunke
Subscriber
The phase change from gas (co2) to liquid (co2) is by concentration driven species mass transfer. The liquid phase has a mixture of ethanol & co2, while the gas phase ethanol & co2 too. These mixtures deviate from ideal gas behavior that's why I used Peng-EOS for both of them. And indeed the Peng-Robinson works well for this system as I have compared it with my experimental density data for ethanol & co2 mixtures. Based on your recommendation, I will set up a another case and use ideal gas for the vapor phase & Peng-Robinson for the liquid phase and see what happens.
To your second question:
It's a gradual gain in mass for the liquid phase & liquid species, but gradual loss in mass for vapor species & phase. Overall the simulation is gaining and not losing mass.
Ethanol gaining species mass gets me really confused because I did not specify any mass transfer for that. I expected the total mass of Ethanol to be the same, while the mass of liquid co2 increases and vapor co2 decreases.
• Rob
Forum Moderator
Check the molecular weights etc. Also how well resolved is the mesh? The fact that all materials are gaining mass tends to suggest it's numerical.
• friday.abolorunke
Subscriber
Molecular weights are good.
Number of elements: 8800. Size of the domain: 22e-5 m by 1e-5 m. Element size: 5e-7 m
I will refine the mesh and change the element size to 1e-7 m and let you know the outcome. Thank you so much.
• Rob
Forum Moderator
You're welcome. Don't forget about adaption, might save you some work.
• friday.abolorunke
Subscriber
Hi Rob I know it took a while to get back. I ran two different cases. Indeed mesh adaptation and increasing the overall mesh density gave me a better result. I was able to match my bench mark with a deviation under ~3%. Meanwhile, I used a different set up for the other two cases. I used UDF to model the interfacial mass transfer for dissolution and evaporation and the VOF model.
Thank you Rob, & DrAmine and Kremella for all the help. This model is the core of what my PhD is about. I started with zero experience in multiphase flow, udfs, interfacial mass transfer and with the help of this forum I am able to bring the project to a close, at least that is what I think haha.
I still have more questions though. Regards Abolorunking
• DrAmine
Ansys Employee
Out of curiosity: Are you using Mixture Model with Slip ( I think so to use species mass transfer) or using your own mass transfer mechanism?
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,117 | 3,458 | {"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-2020-10 | latest | en | 0.894828 |
https://metacpan.org/pod/Math::Inequalities::Parser | 1,718,287,411,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00796.warc.gz | 344,322,447 | 7,542 | The London Perl and Raku Workshop takes place on 26th Oct 2024. If your company depends on Perl, please consider sponsoring and/or attending.
# NAME
Math::Inequalities::Parser - Minimum and maximum values allowed by an inequality.
# SYNOPSIS
`````` use Math::Inequalities::Parser;
my (\$min, \$max) = parse_inequality( ' 10 < n < 20 ' );
# \$min = 11
# \$max = 19``````
# DESCRIPTION
Tiny library for parsing integer maximum and minimum out when given an arbitrary inequality.
Because getting this simple thing right was far harder than it looked, and I never want to have to think about it again.
# FUNCTIONS
## parse_inequality
Parses an inequality string and returns a list of two values, the minimum and the maxium value that string will allow.
# TYPES OF INEQUALITY
## VALUE
The simplest type, a single value, e.g. `42 = Min 42, Max 42`.
## n < VALUE
Maximum is VALUE - 1, Minimum is undefined, e.g. `n < 42 = Min undef, Max 41`.
## n > VALUE
Minimum is VALUE +1, Maximum is undefined, e.g. `n > 42 = Min 43, Max undef`.
## n <= VALUE
Maximum is VALUE, Minimum is undefined, e.g. `n < 42 = Min undef, Max 42`.
## n >= VALUE
Minimum is VALUE, Maximum is undefined, e.g. `n > 42 = Min 42, Max undef`.
## Cases with VALUE, followed by N.
Handled as above, but with minimum and maximum reversed as expected.
## VALUE1 < n < VALUE2
Minimum is VALUE1 + 1, maximum is VALUE2 - 1, e.g `42 < n < 200 = Min 43, Max 199`.
# BUGS
Does not handle `VALUE1 <= n <= VALUE2` or similar. Patches welcome.
Does not complain at impossible `<VALUE1 < n < VALUE 2 `> combinations (e.g. `5 < n < 4`) which result in a higher minumum than the maxiumum. Patches welcome.
Does not work with negative numbers. Patches welcome.
Always uses `n` as the number identifier, this should be configureable at import time.
Uses Exporter (should use Sub::Exporter)
DOES NOT work with floating point numbers. I consider this a feature.
# AUTHORS
Tomas Doran (t0m) `<bobtfish@bobtfish.net>`
Dave Lambley | 559 | 2,003 | {"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-2024-26 | latest | en | 0.794024 |
https://www.physicsforums.com/threads/question-based-on-non-inertial-frame-of-reference.888068/ | 1,532,278,761,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593378.85/warc/CC-MAIN-20180722155052-20180722175052-00474.warc.gz | 960,229,699 | 18,304 | # I Question based on non inertial frame of reference
1. Oct 6, 2016
### parshyaa
Suppose I am observing a object in a noninertial frame from a noninertial frame , then what will happen to the forces acting on a object with respect to both the frames, frame of reference (FOR) moving uniform with inertial FOR are themselves inertial frame , does it follow the same with a FOR moving uniform with a noninertial FOR is itself a non inertial frame.
2. Oct 6, 2016
### drvrm
its best to frame an example and convert non-inertial observers to inertial ones by using 'fictitious forces' generated out of 'non-inertial motion' and then frame a question/problem.
3. Oct 6, 2016
### parshyaa
Suppose a person A moving on a bike which is accelarating ,observes another person B moving on another bike which is also accelarting(with the same accelaration and in same direction) , when person A ask a person B , hey there are you feeling a force , what will he say , here a(AB){accelaration of person A w.r.t B} is 0 , and what will happen if a person standing on ground says , hey there are you feeling a force ,in this case a≠0, then what will be his answer ? , yes you are right we can add fictitious force and make a non-inertial frame of a person A as a inertial frame and as second person is moving uniform w.r.t person A(as a(AB) =0) his frame of reference will also become inertial frame. You mean a noninertial frame moving uniform with a non inertial frame is same as inertial frame moving w.r.t inertial frame after adding fictitious force to any frame .
4. Oct 6, 2016
### A.T.
Fictitious forces cannot be "felt", so the answer about feeling a force is frame independent: Yes, B feels a force forward from the bike. In A's rest frame there is also a fictitious force backwards on B, so coordinate acceleration of B is 0 in that frame.
5. Oct 6, 2016
### parshyaa
Ok then he will give same answer to both the person that yes I am feeling a fictisious force(suppose he is a physicist) , how can you say that force is independent of F.O.R , I belive that yes it is independent of F.O.R, but how can you show this , as F depends on a , and a depends on F.O.R, therefore we can say that F depends on frame of reference , I know its wrong but how can you show that F is independent of F.O.R
6. Oct 6, 2016
### A.T.
The real (interaction) forces are frame independent. The fictitious (inertial) forces are frame dependent.
7. Oct 6, 2016
### parshyaa
[QUOhow c"A.T., post: 5585983, member: 85613"]The real (interaction) forces are frame independent. The fictitious (inertial) forces are frame dependent.[/QUOTE]
Ok, how can you say that real forces are frame independent , there is no gravitational force in rocket in deep space ,but it is there on earth , it means it depends on FOR
8. Oct 6, 2016
### A.T.
Per definition.
A frame of reference is not a location.
9. Oct 6, 2016
### Staff: Mentor
To echo what @A.T. said, real forces are frame independent. They represent some interaction between two objects according to the third law. Fictitious forces only exist in non-inertial frames, and they do not have a corresponding 3rd law pair, also fictitious forces are always proportional to mass and cannot be detected by accelerometers.
Don't worry about gravity at this point. It is complicated, and handled differently by different theories. Stick to bicycles or cars or rotating disks, for now.
10. Oct 6, 2016
### parshyaa
Thanks A.T. and dale
11. Oct 6, 2016
### Khashishi
To answer parshyaa's original question, yes. If frame A is non-inertial, and frame B is moving with uniform velocity with respect to frame A, then frame B is also non-inertial.
Proof is by reversing the statement. Suppose frame B is inertial. If B is moving with uniform velocity wrt A, then A is moving with uniform velocity wrt B.
Uniform velocity is an inertial motion, so it follows that A is inertial. So, for A to be non-inertial, B cannot be inertial.
12. Oct 6, 2016
### parshyaa
Yo I got it , we can also prove it by taking two frame of reference S and S'(S is a non inertial and S' is moving uniform with frame S) , suppose there is a particle 'P' in frame S' , then aPS' =aPS - aS'S, as aS'S is zero , therefore aPS' = aPS , as frame one is non inertial and aPS do not follow newtons first law , therefore aPS' also not follows newtons first law , therefore it is also a non inertial frame, now question arised what if person B is moving nonuniformly with respect to A
13. Oct 9, 2016
### parshyaa
• A frame moving uniform with a inertial frame is itself a inertial frame.
• A frame moving uniform with a non inertial frame is itself a non inertial frame.
• A frame accelerating w.r.t inertial frame is a non inertial frame of reference.
Now question arises, what will happen to a frame accelerating w.r.t a non inertial frame, will it be noninertial or inertial
14. Oct 9, 2016
### parshyaa
• Newton's first and second law acts in both inertial and non inertial frame of reference if we add pseudo forces, but newton's third law acts only in inertial frame.
Is this statement correct?
15. Oct 9, 2016
### Staff: Mentor
Yes.
16. Oct 9, 2016
### parshyaa
Yo mean it will be non-inertial, whats the reason , is it , if we make the first non inertial frame as inertial by adding psuedo forces, and since second frame is accelarting, then as frame accelarting with inertial frame are noninertial, therefore frame accelarting w.r.t non inertial frame will become non inertial
17. Oct 9, 2016
### Staff: Mentor
No, I mean "yes it will be non inertial or inertial."
There is not enough information given.
Last edited: Oct 9, 2016
18. Oct 9, 2016
### Staff: Mentor
I would not say it this way. The third law applies to real forces, both in inertial and in non inertial frames. Fictitious forces violate the third law, and fictitious forces show up only in non inertial frames. But real forces show up in all frames and obey Newton's laws. | 1,581 | 5,972 | {"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} | 3.71875 | 4 | CC-MAIN-2018-30 | latest | en | 0.92702 |
https://stats.stackexchange.com/questions/492480/after-running-a-probit-how-can-i-generate-the-margins-for-the-whole-distributio | 1,627,861,047,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154277.15/warc/CC-MAIN-20210801221329-20210802011329-00122.warc.gz | 547,697,821 | 39,309 | # After running a probit, how can I generate the margins for the whole distribution?
I'm using Stata.
I ran a probit of the form $$\text{outcome}_i = \beta \ f(\text{income}_i) + \gamma\text{ Controls}_i$$
Where $$f(\text{income}_i)$$ is a fractional polynomial. I'm interested on the marginal probability $$\frac{\delta \ P \ (\text{outcome}_i | \text{mean (Controls}_i))}{\delta \ \text{income}_i}$$ for all values of $$i$$. When I run margins, dydx(varname) I only get the average marginal effect, not for the whole distribution. Moreover, I have to calcualte bootstrap standard errors, so I need some way to get the marginal probability that doesn't rely on specifying the functional form beforehand (i.e. _b[var1] + 2*_b[var2]*var2 for a 2nd order polynomial), as $$f(\cdot)$$ will change for each of the 500 bootstrap estimations.
A bit of context in case this is an instance of an xy problem.
I'm implementing an estimator of the form:
$$\beta(\mathbf{x}) = \triangledown \Psi(\mathbf{x}) + \frac{\Psi(\mathbf{x})}{\mathrm{G_M}(\mathbf{x})}\triangledown \mathrm{G_M}(\mathbf{x})$$
Where $$\mathbf{x}$$ is a vector of size $$N$$ (number of observations), $$\Psi(\mathbf{x})$$ is the prediction of the conditional mean, $$\mathrm{G_M}(\mathbf{x})$$ is the prediction of the probability that the outcome is greater than zero, and $$\triangledown \Psi(\mathbf{x})$$, $$\triangledown \mathrm{G_M}(\mathbf{x})$$ refer to their respective marginal effects.
The paper I'm following specifies:
I hope I have been clear enough. Thanks in advance.
• margins has a badly documented gen(me_income) option. Does that do the trick for storing the marginal effects (rather than just their average)? – Dimitriy V. Masterov Oct 18 '20 at 1:12
• That does seem to do the trick regarding storing the marginal effects! I read the documentation for margins all over and couldn't find that! Thanks! – Pablo Derbez Oct 18 '20 at 1:28
• Try help undocumented to see the other goodies. Can I ask how you specify the derivative with respect to the fp-ized variable? – Dimitriy V. Masterov Oct 18 '20 at 1:59
• Thanks! I'm not sure I understood your question but I think I've found a solution to the derivatives issue: After running margins, xb(income_1 income_2) gen(me) it generates me1 and me2. These are the marginal effects wrt the variable transformed according to fp (i.e. raised to some power) so by the chain rule the marginal effect with respect to the non transformed variable should be me1*dydx where dydx is the derivative of the transformed variable wrt to the original. I wrote some code to do that, if you're interested I can post it. – Pablo Derbez Oct 18 '20 at 3:19
• Also you can post your answer and I will mark it as solved, that turned out to be the main issue. – Pablo Derbez Oct 18 '20 at 3:22
First of all, thanks to Dimitriy V. Masterov for pointing me towards the fact that I can save the margins for the whole distribution using margins, dydx(var1 var2), gen(newvar), which will generate newvar1 and newvar2 for var1 and var2 respectively.
Concerning the automation of marginal probability from the fp command, I wrote some code that can handle it, although only for two fp terms.
A fractional polynomial $$x^{f(p)}+x^{f(q)}$$ of $$x$$ is simply $$x^p + x^q$$, with the excepton that $$x^{f(0)}=\ln(x)$$. Also, if any term is repeated, say $$x^{f(p)}+x^{f(p)}$$ then the second term is $$ln(x^{f(p)})$$.
Our outcome is $$y= x^{f(p)}+x^{f(q)}$$. Then, $$\frac{\delta y}{\delta x} = \frac{\delta y}{\delta x^{f(p)}}(\frac{\delta x^{f(p)}}{\delta x})+ \frac{\delta y}{\delta x^{f(q)}}(\frac{\delta x^{f(q)}}{\delta x})$$
Where $$\frac{\delta y}{\delta x^{f(p)}}$$ and $$\frac{\delta y}{\delta x^{f(q)}}$$ are the results from margins, dydx(var1 var2) on the transformed variables,
and $$\frac{\delta x^{f(p)}}{\delta x}$$, $$\frac{\delta x^{f(q)}}{\delta x}$$ are the derivatives of the transformed variables with respect to $$x$$. Thus the code is as follows:
fp <ingpre_noL>, scale(1 1) replace: probit anytransfer <ingpre_noL> \$controls i.id_entidad [pw=factor], robust cluster(upm) iterate(30)
margins, dydx(ingpre_noL_1 ingpre_noL_2) gen(me)
tempvar beta1p beta2p
scalar r = e(fp_fp)[1,1]
if r==0 {
gen beta1p' = 1/(ingpre_noL)
}
else if r!=0 {
gen beta1p' = r*(ingpre_noL)^(r-1)
}
scalar t = e(fp_fp)[1,2]
if r!=t & t==0{
gen beta2p' = 1/(ingpre_noL+1)
}
else if r!=t & t!=0 {
gen beta2p' = t*(ingpre_noL+1)^(t-1)
}
else if r==t & t==0 {
gen beta2p' = 1/(ingpre_noL+1)*ln(ingpre_noL)
}
else if r==t & t!=0{
gen beta2p' = t/(ingpre_noL+1)
}
gen beta_p = me1*beta1p' + me2*beta2p' | 1,410 | 4,616 | {"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": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-31 | latest | en | 0.81958 |
https://ambitiousbaba.com/ibps-po-prelims-exam-analysis-review-questions-asked-3rd-shift-19th-october-2019/ | 1,675,779,142,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500619.96/warc/CC-MAIN-20230207134453-20230207164453-00152.warc.gz | 105,369,550 | 53,179 | # IBPS PO Prelims Exam Analysis, Review & Questions Asked : 3rd Shift 19th October 2019
Table of Contents
## IBPS PO Prelims Exam Analysis, Review & Questions Asked : 3rd Shift
The Institute of Banking Personnel Selection (IBPS) is conducting the IBPS PO 2019 Prelims exam today i.e. 19th October 2019 in 4 shifts. Now the 3rd shift of IBPS PO Prelims exam is over and Here we are come out with detailed exam analysis of the IBPS PO prelims to know the IBPS PO Prelims exam and difficulty level, number of good attempts and questions asked in the exam. The 3rd Shift of 19th October is over now and many aspirants are eagerly waiting for IBPS PO Prelims Exam Analysis and as shared by the candidates, the overall exam was of ‘Moderate’ level. Questions asked in three sections – Reasoning, English Language and Numerical Aptitude were not of difficult level. Candidates were able to attempt around 68-75 questions overall.
1. The sequence of Subjects – Reasoning Ability, Quantitative Aptitude, English Language.
2. Level of Exam– The level of the exam is Moderate.
3. Reasoning Ability Highlights – The topic Puzzles and seating arrangement is asked in the majority.
4. Quantitative Aptitude Highlights – The level of the section was moderate.
## IBPS PO Prelims Exam Analysis 2019: Overall Section-wise Analysis:
Subject Good Attempt Time (in min.) English Language 21-25 20 minutes Reasoning Ability 29-33 20 minutes Quantitative Aptitude 20-24 20 minutes Total 67-73 60 minutes
## IBPS PO Prelims Exam Analysis 2019: Quantitative Aptitude (Moderate):
Two types of DI questions were asked:
• Pie Chart DI
• Tabular DI
Topics No. Of questions Level DI 10 Easy-Moderate Missing Number Series 6 Moderate Quadratic Equation 6 Easy Arithmetic Word Problem 13 Easy-Moderate Total 35 Easy-Moderate
## IBPS PO Prelims Exam Analysis 2019: Reasoning (Easy)
There were 4 puzzles and Seating Arrangement which are given below:-
• Square Seating Arrangement (5 People)
• Box Based Puzzle (6 Boxes, 6 Colours)
• Month Based Puzzle (6 People)
• Vertical Puzzle
Topics No. of Questions Level Puzzles and Seating Arrangement 17 Easy-Moderate Inequality 5 Easy Syllogism 5 Easy-Moderate Coding Decoding 3 Moderate Blood Relation 3 Easy Alphanumeric Series 2 Easy Total 35 Easy-Moderate
## IBPS PO Prelims Exam Analysis 2019: English (Easy-Moderate):
Topics No. of Questions Level Reading Comprehension 10 Easy-Moderate Single Word Filler 5 Easy Error Detection 5 Easy-Moderate Phrase Replacement 5 Easy Miscellaneous 5 Easy-Moderate Total 30 Easy-Moderate
All the best for 3rd Shift
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3 | 840 | 3,296 | {"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-2023-06 | latest | en | 0.86644 |
https://bendwavy.org/klitzing/incmats/f3x5o.htm | 1,643,018,167,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00226.warc.gz | 189,636,246 | 2,022 | Acronym ...
Name f3x5o,
variation of ti,
hull of tigid,
Vertex layers
Layer Symmetry Subsymmetries o3o5o o3o . o . o 1 f3x5o f3x .(f,x)-{6} first f . of-edge first 2 F3f . F . f 3a B3o . x . B 3b B . x 4 x3B . f . C 5a f3B . C . F 5b C3o . o . D 6a o3C . B . B 6b B3f . 7 B3x . D . f 8 o3B . B . B 9a f3F . o . D 9b C . F 10 x3f .opposite (x,f)-{6} f . C 11a B . x 11b x . B 12 F . f 13 f . oopposite f-edge
(F=f+x, B=x+V=2f+x, C=B+x=2f+2x, D=B+f=3f+x)
General of army (is itself convex)
Colonel of regiment (is itself locally convex)
Dihedral angles
• between x5o and f3x: arccos(-sqrt[(5+2 sqrt(5))/15]) = 142.622632°
• between f3x and f3x (at f): arccos(-sqrt(5)/3) = 138.189685°
Confer
uniform variant:
ti
variations:
a3b5c x3f5o (-x)3x5o
Incidence matrix according to Dynkin symbol
f3x5o
. . . | 60 | 1 2 | 2 1
------+----+-------+------
f . . | 2 | 30 * | 2 0
. x . | 2 | * 60 | 1 1
------+----+-------+------
f3x . | 6 | 3 3 | 20 *
. x5o | 5 | 0 5 | * 12 | 496 | 993 | {"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-2022-05 | latest | en | 0.451212 |
https://forums.hackthissite.org/viewtopic.php?f=76&t=52&start=20&view=print | 1,611,503,878,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703549416.62/warc/CC-MAIN-20210124141945-20210124171945-00448.warc.gz | 341,935,340 | 6,140 | Page 3 of 5
### Re: Is there such a thing as truth
Posted: Fri May 30, 2008 7:06 pm
some statements are true like idk SALLYS HAIR IS RED if its truly red then no matter what shade of RED it is its still red
now opinion like GOD IS REAL or something like P.O.D sucks those are opinions just because a lot of people think this that doesnt mean its true
soooo yes there are some truths in this world even though some truths can be broken like
THIS IS A SAFE SITE and a week later a hacker breaks into it so thats and opinion not truth
conclusion: maybe
### Re: Is there such a thing as truth
Posted: Fri May 30, 2008 11:34 pm
but how can sally have red hair if red may or may not be the same 4 all ppl
### Re: Is there such a thing as truth
Posted: Mon Jun 16, 2008 1:47 pm
Jesus_of_Suburbia wrote:Everything I say is a lie.
Did i blow your mind?
Hmmm... I have encountered many people who believe that statement is a paradox. It is not.
My proof:
You have told the truth in the past once before.
Now you say "everything I say is a lie".
That is clearly a a lie, as you have told a truth in the past.
Which has no contradiction to your statement.
This is the correct form of that paradox "This statement is a lie."
### Re: Is there such a thing as truth
Posted: Mon Jun 16, 2008 1:50 pm
Now, on to the subject of the thread.
Yes, truth exists, as it is all around us. Though people may not necessarily agree with something that is true, that makes it no less true. The only real exception to this rule is opinion (and lies of course). If I said "I know a small amount of c++", that is not necessarily a truth. Someone may interpret 'small' as meaning something beyond what is intended. Now this is truth "I know how to declare a variable in c++"
### Re: Is there such a thing as truth
Posted: Mon Jun 16, 2008 3:05 pm
2+2=4, true?
2.49999, rounds down to 2.
1.5 rounds up to 2.
So, 2 could mean 2.4999 just as easily as 1.5 or simply 2.
2 + 2 could mean 1.5 + 1.5 = 3.
The truth is only true when placed in context.
### Re: Is there such a thing as truth
Posted: Wed Jun 18, 2008 1:10 pm
Truth is fundamentally based upon human understanding. "2+2 = 4" is only a true statement in a very specific system. Entirely divorced from context, it has no inherent meaning at all. "My core temps are 58C and 61C" depend on inherently imprecise measurements that we ourselves have defined in the first place. Even "I know how to declare a variable in C++" references a man-made construct - the process of variable declaration is correct because we have deemed it so (and enforce it through vicious-looking compiler errors).
Because human understanding is clearly incomplete, I think it would be premature to conclude that we can declare anything as true with any sort of finality. Of course, applying this line of reasoning would render all debate on anything meaningless ("you don't know that for certain" would be sufficient to end discussion in most cases). The solution, then, would be to further human understanding by applying what we think we know (theory as opposed to "truth"), and revising as necessary.
In short, I would question everything.
### Re: Is there such a thing as truth
Posted: Sun Jul 27, 2008 6:16 am
Is there Truth... maybe...
Would we ever be able to discern if we came across "truth" or not? probably not.
What we know as truth boils down to "intersubjectivity." What is accepted as true by the majority of people is accepted as true. Just look at astronomy or the sciences. It was once believed that the world was flat. This was "true." Then it was discovered that the earth is round and at the center of the universe... so on and so on.
If you want to read a great book on truth and science check out this book by Kuhn, http://search.barnesandnoble.com/Struct ... 083/?itm=4
As he finds, the fundamentals science is based on are no different than, say, religion... and, like I said, truth is basically whatever is generally accepted at the time (intersubjectivity).
The thing is, whether we can discover real Truth or not doesn't matter. What does matter is that we need the appearance of being able to do so.
If you do not have Truth, even if it is just intersubjectivity disguised as Truth, then anything is possible or allowable. There are no guidelines or rules to govern society.
Science, though its basic fundamentals may be no different than religion, has gained the reputation of being Truth, largely due to the quantitative aspect and being able to "prove" something, regardless of whether or not you are really proving anything and whether or not what you proved may be found to be false later on.
Also related to intersubjectivity are differences in personal interpretations of our world.
For those talking about colors, and Red always being Red... that is not necessarily the case.
There is never a one-to-one relationship between the world and how we perceive it. There is always a filter inbetween. Our own personal interpretation/perception of what we experience.
None of our senses are identical. Even the language we speak and our comprehension of it and the meanings of different words, phrases, or sentences differ within a single language. Bring in cross-language differences and you introduce even more problems. Ever heard a foreigner say that they are going to "go get on their car"? It is not because they don't know how to speak english well enough, it is because the relationships between objects differs across cultures.
Back to colors, here in the U.S., we have something like 27 different unique color terms that we can identify. Some cultures have as few as 2 (with the same ocular traits that we have). For them, there may be only light and dark, or white and black. I think one culture, if I remember correctly, has Red and Black. For these cultures, all the other colors terms that we know simply do not exist, they are just shades of their two colors. Then there are all the other cultures somewhere between the min and max.
Then, the best representation of a color also differs between people of the same culture. The "best" representation of Red for one person may very well look like Purple to you. So how do you figure out what the "best" Red is? An average of sorts, Intersubjectivity. What is Red anyways, aside from a human creation to describe something we experience. If cones never developed in the eye would Red still exist? Might there be some evolutionary trait that may change color or light perceptions all-together? Would it be "progress?" Would it be "better?"
Anyways, sorry for the drawn out, wordy response.
Look up that book by Kuhn for more info. That, and Terms of Inquiry by Davis. Maybe read some Popper, Lakatos, or Feyerabend too.
Back to my point, there may very well be Truth, but we have to settle for popular opinion.
### Re: Is there such a thing as truth
Posted: Wed Aug 06, 2008 9:58 pm
There is such a thing as truth to the extent that something is true within our realm and existence. Something that is true to someone, may not be true for someone else. For example: The red chair is red. -- Different people have different definitions of red, as some may be blind, colorblind, or whatnot.So to some people, the red chair could be red, blue, gray, or may not exist at all.
However, this brings it back to my main point,which is that something can be true within our own realm. To me, that chair would be red, and that would be true.
Also keep in mind that anything we talk about being true, has been determined true by ones brain, and ultimately, something determined as true, or fact, is in your own existence, and may or not be true to someone else.
### Re: Is there such a thing as truth
Posted: Tue Aug 12, 2008 10:25 am
Jonas has 9 toes
Poor Jonas.
### Re: Is there such a thing as truth
Posted: Tue Aug 12, 2008 6:02 pm
Going off topic but.... beagle, your sig:
"Our enemies are innovative and resourceful - and so are we, They never stop thinking about new ways to harm our country and our people - and neither do I."
I know he didn't mean to say it like that but if he had he'd have been very clever.
It's not funny at all really, you ought to think like the enemy "how would I attack the US?". | 1,964 | 8,234 | {"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-2021-04 | latest | en | 0.961439 |
https://uscustomwriters.com/2022/08/09/statistics-techniques-management-homework-help/ | 1,669,841,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00790.warc.gz | 623,804,375 | 12,665 | # Statistics techniques | Management homework help
Answer th 5 questions below thoroughly
1) The Appliance Center has six sales representatives at its North Jacksonville outlet.
Listed below is the number of refrigerators sold by each last month.
Sales Rep Number Sold
Zina Craft 54
Woon Junge 50
Ernie DeBrul 52
Jan Niles 48
Molly Camp 50
Rachel Myak 52
a. How many samples of size 2 are possible?
b. Select all possible samples of size 2 and compute the mean number sold.
c. Organize the sample means into a frequency distribution.
d. What is the mean of the population? What is the mean of the sample means?
e. What is the shape of the population distribution?
f. What is the shape of the distribution of the sample mean?
Create the graphs from both distributions?
2) The manufacturer of a new line of ink-jet printers would like to include as part of its advertising the number of pages a user can expect from a print cartridge. A sample of 10 cartridges revealed the following number of pages printed.
(note – Excel’s Data Analysis has an option under Descriptive Statistics that will allow a Confidence Interval to be calculated from a set of raw data)
2,698 2,028 2,474 2,395 2,372 2,475 1,927 3,006 2,334 2,379
a. What is the point estimate of the population mean?
b. Develop a 95% confidence interval for the population mean.
3) Ms. Maria Wilson is considering running for mayor of the town of Bear Gulch, Montana.
Before completing the petitions, she decides to conduct a survey of voters in Bear Gulch. A
sample of 400 voters reveals that 300 would support her in the November election.
Estimate the value of the population proportion.
Develop a 99% confidence interval for the population proportion.
Interpret your findings.
4) Define a hypothesis
Explain the process of testing a hypothesis.
Distinguish between a one-tailed and a two-tailed test of hypothesis.
5) Heinz, a manufacturer of ketchup, uses a particular machine to dispense 16 ounces of its ketchup into containers. From many years of experience with the particular dispensing machine, Heinz knows the amount of product in each container follows a normal distribution with a mean of 16 ounces and a standard deviation of 0.15 ounce. A sample of 50 containers filled last hour revealed the mean amount per container was 16.017 ounces. Does this evidence suggest that the mean amount dispensed is different from 16 ounces? Use the .05 significance level.
(a) State the null hypothesis and the alternate hypothesis.
(b) What is the probability of a Type I error?
(c) Give the formula for the test statistic.
(d) State the decision rule.
(e) Determine the value of the test statistic.
(f) What is your decision regarding the null hypothesis?
(g) Interpret, in a single sentence, the result of the statistical test.
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If you think your paper could be improved, you can request a review. In this case, your paper will be checked by the writer or assigned to an editor. You can use this option as many times as you see fit. This is free because we want you to be completely satisfied with the service offered. | 1,325 | 6,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-49 | latest | en | 0.847009 |
https://www.lmfdb.org/EllipticCurve/Q/3136/p/1 | 1,675,327,986,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499967.46/warc/CC-MAIN-20230202070522-20230202100522-00593.warc.gz | 887,707,030 | 30,857 | # Properties
Label 3136.p1 Conductor $3136$ Discriminant $107943428096$ j-invariant $$\frac{1443468546}{7}$$ CM no Rank $1$ Torsion structure $$\Z/{2}\Z$$
# Related objects
Show commands: Magma / Pari/GP / SageMath
## Simplified equation
$$y^2=x^3-58604x+5460560$$ y^2=x^3-58604x+5460560 (homogenize, simplify) $$y^2z=x^3-58604xz^2+5460560z^3$$ y^2z=x^3-58604xz^2+5460560z^3 (dehomogenize, simplify) $$y^2=x^3-58604x+5460560$$ y^2=x^3-58604x+5460560 (homogenize, minimize)
sage: E = EllipticCurve([0, 0, 0, -58604, 5460560])
gp: E = ellinit([0, 0, 0, -58604, 5460560])
magma: E := EllipticCurve([0, 0, 0, -58604, 5460560]);
sage: E.short_weierstrass_model()
magma: WeierstrassModel(E);
## Mordell-Weil group structure
$$\Z \oplus \Z/{2}\Z$$
### Infinite order Mordell-Weil generator and height
sage: E.gens()
magma: Generators(E);
$P$ = $$\left(28, 1960\right)$$ (28, 1960) $\hat{h}(P)$ ≈ $1.9330572333397874665166854078$
## Torsion generators
sage: E.torsion_subgroup().gens()
gp: elltors(E)
magma: TorsionSubgroup(E);
$$\left(140, 0\right)$$
## Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
$$(28,\pm 1960)$$, $$\left(140, 0\right)$$
## Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$3136$$ = $2^{6} \cdot 7^{2}$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $107943428096$ = $2^{17} \cdot 7^{7}$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$\frac{1443468546}{7}$$ = $2 \cdot 3^{3} \cdot 7^{-1} \cdot 13^{3} \cdot 23^{3}$ Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $1.3174958380090418189147488955\dots$ Stable Faltings height: $-0.63741774231187068864567298162\dots$
## BSD invariants
sage: E.rank() magma: Rank(E); Analytic rank: $1$ sage: E.regulator() magma: Regulator(E); Regulator: $1.9330572333397874665166854078\dots$ sage: E.period_lattice().omega() gp: if(E.disc>0,2,1)*E.omega[1] magma: (Discriminant(E) gt 0 select 2 else 1) * RealPeriod(E); Real period: $0.93493147424956756866377454482\dots$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $8$ = $2\cdot2^{2}$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $2$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $1$ (exact) sage: r = E.rank(); sage: E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: ar = ellanalyticrank(E); gp: ar[2]/factorial(ar[1]) magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12); Special value: $L'(E,1)$ ≈ $3.6145520979503156654594075894$
## Modular invariants
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q + 2 q^{5} - 3 q^{9} - 4 q^{11} + 2 q^{13} + 6 q^{17} - 8 q^{19} + O(q^{20})$$
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 6144 $\Gamma_0(N)$-optimal: no Manin constant: 1
## Local data
This elliptic curve is not semistable. There are 2 primes of bad reduction:
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $2$ $I_{7}^{*}$ Additive -1 6 17 0
$7$ $4$ $I_{1}^{*}$ Additive -1 2 7 1
## Galois representations
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The $\ell$-adic Galois representation has maximal image for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$2$ 2B 8.24.0.111
sage: gens = [[32, 53, 27, 54], [49, 8, 48, 9], [1, 0, 8, 1], [48, 13, 45, 16], [1, 4, 4, 17], [7, 6, 50, 51], [32, 55, 29, 42], [1, 8, 0, 1]]
sage: GL(2,Integers(56)).subgroup(gens)
magma: Gens := [[32, 53, 27, 54], [49, 8, 48, 9], [1, 0, 8, 1], [48, 13, 45, 16], [1, 4, 4, 17], [7, 6, 50, 51], [32, 55, 29, 42], [1, 8, 0, 1]];
magma: sub<GL(2,Integers(56))|Gens>;
The image of the adelic Galois representation has level $56$, index $48$, genus $0$, and generators
$\left(\begin{array}{rr} 32 & 53 \\ 27 & 54 \end{array}\right),\left(\begin{array}{rr} 49 & 8 \\ 48 & 9 \end{array}\right),\left(\begin{array}{rr} 1 & 0 \\ 8 & 1 \end{array}\right),\left(\begin{array}{rr} 48 & 13 \\ 45 & 16 \end{array}\right),\left(\begin{array}{rr} 1 & 4 \\ 4 & 17 \end{array}\right),\left(\begin{array}{rr} 7 & 6 \\ 50 & 51 \end{array}\right),\left(\begin{array}{rr} 32 & 55 \\ 29 & 42 \end{array}\right),\left(\begin{array}{rr} 1 & 8 \\ 0 & 1 \end{array}\right)$
## $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(5,20) if E.conductor().valuation(p)<2]
$p$-adic regulators are not yet computed for curves that are not $\Gamma_0$-optimal.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 add ss ord add ord ord ord ord ss ord ord ord ord ord ord - 7,3 1 - 1 1 1 1 1,1 1 1 1 1 1 1 - 0,0 0 - 0 0 0 0 0,0 0 0 0 0 0 0
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 2 and 4.
Its isogeny class 3136.p consists of 4 curves linked by isogenies of degrees dividing 4.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z$ are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $2$ $$\Q(\sqrt{14})$$ $$\Z/2\Z \oplus \Z/2\Z$$ 2.2.56.1-56.1-b5 $2$ $$\Q(\sqrt{7})$$ $$\Z/4\Z$$ Not in database $2$ $$\Q(\sqrt{2})$$ $$\Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{2}, \sqrt{7})$$ $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $8$ 8.0.493455671296.14 $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $8$ 8.8.40282095616.2 $$\Z/2\Z \oplus \Z/8\Z$$ Not in database $8$ 8.0.1927561216.8 $$\Z/8\Z$$ Not in database $8$ 8.2.16862305517568.25 $$\Z/6\Z$$ Not in database $16$ deg 16 $$\Z/4\Z \oplus \Z/4\Z$$ Not in database $16$ 16.0.59447875862838378496.3 $$\Z/2\Z \oplus \Z/8\Z$$ Not in database $16$ deg 16 $$\Z/2\Z \oplus \Z/6\Z$$ Not in database $16$ deg 16 $$\Z/12\Z$$ Not in database $16$ deg 16 $$\Z/12\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial. | 2,757 | 6,686 | {"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} | 3.171875 | 3 | CC-MAIN-2023-06 | latest | en | 0.262047 |
https://math.answers.com/other-math/What_is_2_over_3_plus_4_over_6_equal_to | 1,721,649,565,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00209.warc.gz | 323,618,184 | 48,020 | 0
# What is 2 over 3 plus 4 over 6 equal to?
Updated: 4/28/2022
Wiki User
11y ago
2/3 + 4/6 = 4/3 or 11/3
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Related questions
### What is 1 over 3 plus 1 over 3 equal?
It is: 1/3 plus 1/3 = 2/3
1
### Whats 2 over 3 plus 3 plus 8?
2/3 plus 3/8 is equal to 25/24.
### What is 2 over 5 plus 1 over 3 plus 5 over 6?
2/5 plus 1/3 plus 5/6 is equal to 1.57 or 1 57/100
2/3+1/4 =11\12
### What does 1 over 7 plus 2 over 3 equal?
1/7 + 2/3 is 17/21
1.75 or 13/4
### What does 24 times 3 plus 6 over 2 equal?
(24 times 3) + (6 over 2) = 75
1 7/15
5/4
### What is 1 over 5 plus 3 over 10 equal?
The answer is 1/5 plus 3/10 = 1/2 in its simplest form
### Two over three plus one over six plus one over six equals?
One over six plus one over six is 2/6. That fraction can be reduced to 1/3. Two over three is written as 2/3. 2/3 plus 1/3 is equal to 1, or a whole. | 410 | 981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-30 | latest | en | 0.861059 |
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Object Oriented Data Structures. Recursion Introduction to Recursion Principles of Recursion Backtracking: Postponing the Work Tree-Structured Programs: Look Ahead in Games. Recursion. Recursion. Recursion. Recursion. Recursion. Recursion. Recursion. Recursion. C. C. C. A. A. A. A.
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Object Oriented Data Structures
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#### Presentation Transcript
Object Oriented Data Structures
RecursionIntroduction to RecursionPrinciples of RecursionBacktracking: Postponing the WorkTree-Structured Programs: Look Ahead in Games
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Recursion
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### Tree of Subprogram Calls
Start
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1 if n = 0
n*(n-1)! if n > 0
n! =
1 if n = 0 and x not 0
x*(xn-1) if n > 0 and x not 0
xn =
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### Designing Recursive Algorithms
• Find the key step
• Find a stopping rule (base case)
• Check termination
• Draw a recursion tree
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### Tail Recursion
• The very last action of a function is a recursive call to itself
• Explicit use of a stack not necessary
• Reassign the calling parameters to the values specified in the recursive call and then repeat the function
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### Backtracking
An algorithm which attempts to complete a search for a solution to a problem by constructing partial solutions, always ensuring that the partial solutions remain consistent with the requirements. The algorithm then attempts to extend a partial solution toward completion, but when an inconsistency with the requirements of the problem occurs, the algorithm backs up (backtracks) by removing the most recently constructed part of the solution and trying another possibility.
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### Knight's Tour
Legal Knight Moves
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### Knight's Tour
Legal Knight Moves
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### Knight's Tour
Legal Knight Moves
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### Knight's Tour
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Legal Knight Moves
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### Application: Depth- And Breadth-First Search
F
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• Hexadecimal – Base 16 numbering system 0-F
• Decimal - Base 10 numbering system 0-9
• Octal - Base 8 numbering system 0-7
• Decimal Digits 0,1,2,3,4,5,6,7,8,9
• Ocal Digits 0,1,2,3,4,5,6,7
• What’s . ?
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• 3210OctDecHex
• 0000000
• 0001111
• 0010222
• 0011333
• 0100444
• 0101555
• 0110666
• 0111777
• 10001088
• 10011199
• 10101210A
• 10111311B
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### Class Cell Maintain:
• A number. This is an integer value used to identify the cell.
• Cells are numbered consecutively from left to right and top to bottom. (Order is important!)
• A list of neighboring cells. Each cell will have an entry in this list for all other neighbor cell that can be reached.
• A Boolean value, named visited, that will be used to mark a cell once it has been visited.
• Traversing a maze often results in dead ends, and the need to back up and start again.
• Marking cells avoids repeating effort and potentially walking around in circles.
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### Class Description
classcell
{public: cell(intn) : number(n), visited(false) {}
void addNeighbor(cell * n){neighbors.push_back(n);}voidvisit (deque<cell *> &);
protected:intnumber;boolvisited; list <cell *> neighbors;
};//end class cell
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### Class Maze
classmaze
{public: maze(istream&);voidsolveMaze();
protected: cell * start;boolfinished;deque<cell *> path; // used to hold the path // or paths currently // being traversed
};//end class maze
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maze::maze(istream & infile)// initialize maze by reading from file{intnumRows, numColumns;int counter = 1; cell * current = 0;
infile >> numRows >> numColumns;
vector <cell *> previousRow (numRows, 0);
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for(inti = 0; i < numRows; i++)for(int j=0; j<numColumns; j++)
{ current = new cell(counter++);int walls;infile >> walls;
if((i>0) && ((walls & 0x04)==0))
if((j>0> && ((walls & 0x08) == 0))
}//end maze()
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void maze::solveMaze()// solve the maze puzzle
{ start->visit(path);while ((!finished) && (! path.empty()))
{ cell * current = path.front();path.pop_front(); finished = current->visit(path);}if ( ! finished)cout << “no solution found\n”;}//end solveMaze()
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bool cell::visit(deque<cell *> & path) { //depth firstif(visited) // already been herereturn false; visited = true; // mark as visitedcout << “visiting cell “ << number << endl;if (number == 1)
{cout << “puzzle solved\n”;return true;} list <cell *>:: iterator start, stop; start = neighbors.begin(); stop = neighbors.end();for ( ; start != stop; ++start)if (! (*start)->visited)path.push_front(*start);return false;
}
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bool cell::visit(deque<cell *> & path){// breadth firstif(visited) // already been herereturn false; visited = true; // mark as visitedcout << “visiting cell “ << number << endl;if (number == 1)
{cout << “puzzle solved\n”; return true;} list <cell *>:: iterator start, stop; start = neighbors.begin(); stop = neighbors.end();for ( ; start != stop; ++start)if (! (*start)->visited)path.push_back(*start);return false;
}
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### Depth First vs. Breadth First
• Because all paths of length one are investigated before examining paths of length two, and all paths of length two before examining paths of length three, a breadth-first search is guaranteed to always discover a path from start to goal containing the fewest steps, whenever such a path exists.
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### Depth First vs. Breadth First
• Because one path is investigated before any alternatives are examined, a depth-first search may, if it is lucky, discover a solution more quickly than the equivalent breadth-first algorithm.
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### Depth First vs. Breadth First
• In particular, suppose for a particular problem that some but not all paths are infinite, and at least one path exists from start to goal that is finite.
• A breadth-first search is guaranteed to find a shortest solution.
• A depth-first search may have the unfortunate luck to pursue a never-ending path, and can hence fail to find a solution.
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Chapter 5 Ripples Away
Kruse/Ryba ch05 | 2,623 | 8,311 | {"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} | 3.71875 | 4 | CC-MAIN-2017-17 | longest | en | 0.682874 |
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# operators
nitin sharma
Ranch Hand
Joined: Feb 24, 2001
Posts: 290
public class Q6
{
public static void main(String[] args)
{
int a[]={4,5,9,8,6,6,0};
int i=1;
a[i]=i+=a[i++];
System.out.println(a[i]+" , "+i);
}
}
please explain the way output is coming?
0,6
nitin sharma
Ranch Hand
Joined: Feb 24, 2001
Posts: 290
some body please come up with something.
Cindy Glass
"The Hood"
Sheriff
Joined: Sep 29, 2000
Posts: 8521
i starts at 1
so a[1] gives you 5 (the second element of the array is 5)
i+=5 gives you 6
a[6] gives you 0 (the 7th element of the array is 0).
So the value of i ends up 6, and the valus of a[i] is 0.
"JavaRanch, where the deer and the Certified play" - David O'Meara
Indika Perera
Greenhorn
Joined: Feb 28, 2001
Posts: 8
hi, i got it this way. Hope it's correct.
a[i]=i+=a[i++];
initially i = 1 so, expression become, a[1] = i+ = a[1]
Now, Evaluating from Right to left
first i+=a[1]; i = i + a[1] = 6
then a[1] = 6 Therefore a[6] remains 0 and i is now 6
I agree. Here's the link: http://aspose.com/file-tools
subject: operators | 446 | 1,302 | {"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-2015-48 | latest | en | 0.816505 |
https://rdrr.io/cran/SHT/man/mean2.ttest.html | 1,721,895,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00779.warc.gz | 425,721,547 | 8,027 | # mean2.ttest: Two-sample Student's t-test for Univariate Means In SHT: Statistical Hypothesis Testing Toolbox
mean2.ttest R Documentation
## Two-sample Student's t-test for Univariate Means
### Description
Given two univariate samples x and y, it tests
H_0 : μ_x^2 ≤ft\lbrace =,≥q,≤q \right\rbrace μ_y^2\quad vs\quad H_1 : μ_x^2 ≤ft\lbrace \neq,<,>\right\rbrace μ_y^2
using the procedure by Student (1908) and Welch (1947).
### Usage
mean2.ttest(
x,
y,
alternative = c("two.sided", "less", "greater"),
paired = FALSE,
var.equal = FALSE
)
### Arguments
x a length-n data vector. y a length-m data vector. alternative specifying the alternative hypothesis. paired a logical; whether consider two samples as paired. var.equal a logical; if FALSE, use Welch's correction.
### Value
a (list) object of S3 class htest containing:
statistic
a test statistic.
p.value
p-value under H_0.
alternative
alternative hypothesis.
method
name of the test.
data.name
name(s) of provided sample data.
### References
\insertRef
student_probable_1908SHT
\insertRef
student_probable_1908aSHT
\insertRef
welch_generalization_1947SHT
### Examples
## empirical Type 1 error
niter = 1000
counter = rep(0,niter) # record p-values
for (i in 1:niter){
x = rnorm(57) # sample x from N(0,1)
y = rnorm(89) # sample y from N(0,1)
counter[i] = ifelse(mean2.ttest(x,y)\$p.value < 0.05, 1, 0)
}
## print the result
cat(paste("\n* Example for 'mean2.ttest'\n","*\n",
"* number of rejections : ", sum(counter),"\n",
"* total number of trials : ", niter,"\n",
"* empirical Type 1 error : ",round(sum(counter/niter),5),"\n",sep=""))
SHT documentation built on Nov. 3, 2022, 9:06 a.m. | 518 | 1,692 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-30 | latest | en | 0.63193 |
https://bestbtcxcbdc.netlify.app/taraborelli87443cife/equation-for-rate-of-change-over-time-424.html | 1,722,668,438,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00280.warc.gz | 105,834,886 | 11,288 | ## Equation for rate of change over time
used to evaluate the rate at which the size of the population changes over time. These rates are used by demographers and population ecologists to estimate 4 Aug 2019 Yet, despite the ubiquity of the error and the ramifications for us personally In equation form, the overall percentage change is: Knowingly or not, we deal with multiple percentage changes all the time and when we don't 30 Sep 2019 Churn, expansion revenue, upsells, and changes in growth rate are all To calculate run rate, take your current revenue over a certain time
13 Nov 2019 If your device is not in landscape mode many of the equations will a brief set of examples concentrating on the rate of change application of derivatives. at 35 mph and at the same time Car B starts driving south at 50 mph. Average Rate of Change Formula is one of the integral formulas in algebra. Know more about it and learn how to calculate the average rate of change of a An instantaneous rate of change is equivalent to a derivative. divided by the travel time; An instantaneous rate can be determined is no given formula ( function) for finding the numerator of the ratio from its 25 Jan 2018 Then we can find the distance it covers over any specified time period Let's agree to treat the input x as time in the rate of change formula.
## Average Rate of Change Formula is one of the integral formulas in algebra. Know more about it and learn how to calculate the average rate of change of a
You will generally use the percent change calculation when the order of the numbers does matter; you have starting and ending values or an "old number" and a " Hint: The average rate of change of a function f(t) over the interval [a,b] is f(b)−f(a) b−a. In this problem, you have to calculate the change in displacement. let f(t ) measure the position or displacement of the runner at time t with respect to a Step 1: Calculate the percent change from one period to another using the following formula: Percent Change = 100 × (Present or Future Value – Past or Present Using Formulas for calculating rate of change on a process variable Your browser does not currently recognize any of the video formats available. Click here The percent change formula is used very often in Excel. For example, to Select cell C3. On the Home tab, in the Number group, apply a Percentage format. Percent change calculator - Calculate the percent change between two numbers. any action on the basis of the information provided through this calculator. To calculate a percentage change, you can use this formula: (((y2- y1))/ y1) * 100.
### distance = rate x time When identifying the parts of the word problem, distance is typically given in units of miles, meters, kilometers, or inches. Time is in units of seconds, minutes, hours, or years. Rate is distance per time, so its units could be mph, meters per second, or inches per year.
23 Sep 2014 However, if one wishes to find the average rate of change over an average velocity, or rate of change of distance with respect to time, of 0.1 To calculate this, we shall first calculate the value of the function at these points. As the rate is changing throughout the reaction, we are calculating the average rate over a given time period. For example, the graph below could be used to Substitute using the average rate of change formula. Tap for more steps The average rate of change of a function can be found by calculating the change in y y You will generally use the percent change calculation when the order of the numbers does matter; you have starting and ending values or an "old number" and a " Hint: The average rate of change of a function f(t) over the interval [a,b] is f(b)−f(a) b−a. In this problem, you have to calculate the change in displacement. let f(t ) measure the position or displacement of the runner at time t with respect to a
### Over what period of time has this change occurred? for the same calculation, i.e. a different point for Q, we would get a different average rate of change.
Speed is the rate at which distance changes over time. Students in math and physical science classes are often asked to solve rate problems, the first of which usually deal with speed. Problems may involve calculating speed itself or rearranging the equation for speed to solve for time or distance. Percent change calculation is very easy, if you understand the basics of the calculation method and know how to apply the formula. Therefore the answer is 50%. You can even further calculate percent change over time. Determining the Average Rate from Change in Concentration over a Time Period. We calculate the average rate of a reaction over a time interval by dividing the change in concentration over that time period by the time interval. For the change in concentration of a reactant, the equation, where the brackets mean "concentration of", is To calculate the average rate of change (the average bicycle speed) in Excel, you can easily do as follows: 1. Select the blank cell besides the cell with last distance, in our case select Cell C7, enter the formula =(B7-B2)/((A7-A2)*24) into it and then press the Enter key. distance = rate x time When identifying the parts of the word problem, distance is typically given in units of miles, meters, kilometers, or inches. Time is in units of seconds, minutes, hours, or years. Rate is distance per time, so its units could be mph, meters per second, or inches per year. Rate Of Change - ROC: The rate of change - ROC - is the speed at which a variable changes over a specific period of time. ROC is often used when speaking about momentum, and it can generally be
## If we find the slope we can find the rate of change over that period. This can be By finding the slope of the line, we would be calculating the rate of change.
Determining the Average Rate from Change in Concentration over a Time Period. We calculate the average rate of a reaction over a time interval by dividing the change in concentration over that time period by the time interval. For the change in concentration of a reactant, the equation, where the brackets mean "concentration of", is The rate of reaction is the change of concentration of a substance in a given time. Whether that be reactants disappearing or products appearing; the rate of reaction is affected by the temperature. However, the chemical equation does not tell us how fast things happen, for this we use a rate equation.
What are the underlying principles of how populations change over time? So the division rate could be any number at all and the general equation becomes,. Average Rate of Change Formula The average rate of change is defined as the average rate at which quantity is changing with respect to time or something else that is changing continuously. In other words, the average rate of change is the process of calculating the total amount of change with respect to another. rate of change = change in y change in x = change in distance change in time = 160 − 80 4 − 2 = 80 2 = 40 1 Rate of change is used to mathematically describe the percentage change in value over a defined period of time, and it represents the momentum of a variable. The calculation for ROC is simple in that it takes the current value of a stock or index and divides it by the value from an earlier period. | 1,568 | 7,356 | {"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} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.893857 |
http://www.jiskha.com/display.cgi?id=1329792435 | 1,493,242,469,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121665.69/warc/CC-MAIN-20170423031201-00328-ip-10-145-167-34.ec2.internal.warc.gz | 558,668,538 | 3,784 | # geometry
posted by on .
Joe's living room is a rectangle and measures 15 ft by 27 ft. Using a scale of 1 in. = 3ft. How could I sketch it and show both the actual measurements and the scale measurements
• geometry - ,
Use a scale key: 1 in = 3 ft
Your sketch will be 5 by 9 inches. | 82 | 288 | {"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-2017-17 | latest | en | 0.91966 |
http://www.sciforums.com/threads/why-not-have-new-calender-system.44825/ | 1,531,695,207,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589022.38/warc/CC-MAIN-20180715222830-20180716002830-00180.warc.gz | 544,075,606 | 13,755 | Why Not Have New Calender System ?
Discussion in 'General Science & Technology' started by plakhapate, Feb 10, 2005.
Not open for further replies.
1. plakhapateBannedBanned
Messages:
249
At preesent there are 12 months having different days for each month.
Every year we have to see the calender.
Also date and days are different for every month.
Instead of this consider 7 days /week, 4 weeks/month and 13 months/year
This becomes 364 days.
Every year we shoud have holiday as year ending day as extra day.
During leap year we can have extra holiday after year ending day.
Thus 1st of every month will be always on Monday.
Since there are 4 weeks/month, every month will be identical.
13th Month can be named as Thirteenember
Year Ending Day can be named as YEDAY (Date will be 29th Thirteenember)
Leap Year Day can be named as LEDAY((Date will be 30th Thirteenember)
By using this calender our measuring system becomes very simple.
Can anybody answer why are we not implementing this ?
P.J.LAKHAPATE
pjlakhap@bechtel.com
3. WingsRegistered Senior Member
Messages:
89
Remember the Millenium virus scare? There was a glitch in the computer stuff? Imagine all the reprogramming that would have to be done just to fix the computers. Then there's changes to national holidays, religious holidays (Catholic Easter and American Thanksgiving would be particularly tricky). People would forget your birthday, and that would be sad. Voting systems would change. I cant imagine the world community is ready to take on that much effort. The financial costs would absurd. Those countries that couldn't afford it would be in a fix.
5. IrisRegistered Senior Member
Messages:
107
Sheesh, we can't even finish switching over to the metric system--how we gonna implement an entirely new calendar?
7. GondolinHell hath no fury like squidRegistered Senior Member
Messages:
709
I also have an idea. We can read time differently. I have created a series of dots, letters, numbers, exclamation points, and animals that will help us read time easier.
Example: 1:30pm would be ..sw42!23..goose...fj44:
Easy huh?
Messages:
4,969
9. plakhapateBannedBanned
Messages:
249
Can anybody tell me the name and address (phone/fax no/ email ) of the committee or panel who decides about the accptance of new calender system ?
P.J.LAKHAPATE
pjlakhap@bechtel.com
10. Jolly RodgerBannedBanned
Messages:
460
yeah lets do it, lets change the world, maybe if we get enopugh people, just us oin the forum to change our calander, everyone will follow after all, we are the cool nerds!!
I mean i dont leave my computer much but if you guys did i am sure we could start a tread, because that would make us cool tready nerds
cool idea
11. vslayerRegistered Senior Member
Messages:
4,969
lets just start with time for now, we can write a proram for a metric clock
12. Avatarsmoking revolverValued Senior Member
Messages:
19,083
I count time in ATU's (Avatar Time Units),
everything befory my this-life is b.a.c.e. (Before Avatar's Current Era).
Last edited: Feb 14, 2005
13. Jolly RodgerBannedBanned
Messages:
460
how does that work?
14. Avatarsmoking revolverValued Senior Member
Messages:
19,083
Well, since my position in timespace continum varies I can not use a time counting system based on full moons, etc, because at one place there may be no moon at all, or be some 15 of them, so 850 carbon-based life cycles ago I thought that I need to count my being in something that is always where I am, that is myself. Because I'm not always carbon based my prime time measuring unit is Avatar Time Unit (as told). An ATU is basically the time it takes for 1 RIGT (or Random Idea Generation Time) to occur in particulary my mind.
The only problem is that it is rather difficult for other entities to accept time measuring in ATU's, and they always annoy me and convert ATU's in moons, many summers, seconds, cesium decay times, etc.
15. catoless hate, more scienceRegistered Senior Member
Messages:
2,959
lol @ Gondolin
16. melodicbardRegistered Senior Member
Messages:
208
plakhapate,
Maybe the ISO.
The week-based ISO 8601 system is used in commerce and industry.
17. 1100fBannedRegistered Senior Member
Messages:
807
In England, they have succesfully switched to the metric system.
Now they are trying to switch to drive on the right side of the roads. They are preparing to prepare an experiment where all cars in England will be driven on the right side of the roads. If, after on month, this experiment succeed, they will let trucks and buses drive on the right side of the roads also.
Last edited: Feb 18, 2005
18. vslayerRegistered Senior Member
Messages:
4,969
why drive on the right? the left works just fine. despite the avoidance of crashes caused by ignorant americans on the wrong side of the road, there are no advantages, it wolud jsut cost money and during the first few years ther wolud be a limbo state where no-one knows where to go, and crashes are 10x more often
19. jennyRaterLuck B me 2niteRegistered Senior Member
Messages:
401
we should have -
10 long months of 36 days each (have a global vote on which 2 monthnames to drop).
5 special days not part of any month, 1 fitted in between evry 2 months. these would be public holidays - the 1st would come before january, and thatd be New Years day. The 2nd special day could be Easter, so we wouldnt have to keep changing its date every year. Thhe 5th special day could be Halloween, and the other 2... we could invent names.
Christmas + Indepndence day would stay inside their months, of course. For leap years, just make the New year holiday 2 days long before starting January.
20. Avatarsmoking revolverValued Senior Member
Messages:
19,083
p.s. possible only in the USA
21. OphioliteValued Senior Member
Messages:
9,232
I am quite happy with the current system. Do we really want to have the same date always falling on the same weekday? Boring, in my view. Stay with what we have until tidal forces have slowed us down to another day or two per year.
If you want a serious calendar discussion then have a look at what we will need for Mars.
Messages:
19,083 | 1,512 | 6,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-30 | latest | en | 0.952801 |
https://www.expertsmind.com/library/how-nonverbal-factors-can-cause-conflict-51069383.aspx | 1,719,329,968,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00343.warc.gz | 654,545,290 | 14,783 | ### How nonverbal factors can cause conflict
Assignment Help HR Management
##### Reference no: EM131069383
Give a specific example of how nonverbal factors can cause conflict in intercultural communication, or if you feel comfortable doing so, give a personal example of a conflict caused by your perceptions of nonverbal styles.
#### Questions Cloud
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Why should the auditor review transactions and events that occurred after the balance sheet date? Does this responsibility extend indefinitely? What are the types of subsequent events that must be evaluated by the auditor and what actions should b.. | 1,039 | 5,207 | {"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-2024-26 | latest | en | 0.889414 |
https://docs.sympy.org/0.7.2-py3k/_modules/sympy/logic/inference.html | 1,550,436,224,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247482478.14/warc/CC-MAIN-20190217192932-20190217214932-00325.warc.gz | 560,642,734 | 4,234 | /
# Source code for sympy.logic.inference
"""Inference in propositional logic"""
from sympy.logic.boolalg import And, Or, Not, Implies, Equivalent, \
conjuncts, to_cnf
from sympy.core.basic import C
from sympy.core.sympify import sympify
def literal_symbol(literal):
"""
The symbol in this literal (without the negation).
Examples:
>>> from sympy import Symbol
>>> from sympy.abc import A
>>> from sympy.logic.inference import literal_symbol
>>> literal_symbol(A)
A
>>> literal_symbol(~A)
A
"""
if literal.func is Not:
return literal.args[0]
else:
return literal
[docs]def satisfiable(expr, algorithm="dpll2"):
"""
Check satisfiability of a propositional sentence.
Returns a model when it succeeds
Examples:
>>> from sympy.abc import A, B
>>> from sympy.logic.inference import satisfiable
>>> satisfiable(A & ~B)
{A: True, B: False}
>>> satisfiable(A & ~A)
False
"""
expr = to_cnf(expr)
if algorithm == "dpll":
from sympy.logic.algorithms.dpll import dpll_satisfiable
return dpll_satisfiable(expr)
elif algorithm == "dpll2":
from sympy.logic.algorithms.dpll2 import dpll_satisfiable
return dpll_satisfiable(expr)
raise NotImplementedError
def pl_true(expr, model={}):
"""
Return True if the propositional logic expression is true in the model,
and False if it is false. If the model does not specify the value for
every proposition, this may return None to indicate 'not obvious';
this may happen even when the expression is tautological.
The model is implemented as a dict containing the pair symbol, boolean value.
Examples
========
>>> from sympy.abc import A, B
>>> from sympy.logic.inference import pl_true
>>> pl_true( A & B, {A: True, B : True})
True
"""
if isinstance(expr, bool):
return expr
expr = sympify(expr)
if expr.is_Symbol:
return model.get(expr)
args = expr.args
func = expr.func
if func is Not:
p = pl_true(args[0], model)
if p is None: return None
else: return not p
elif func is Or:
result = False
for arg in args:
p = pl_true(arg, model)
if p == True: return True
if p == None: result = None
return result
elif func is And:
result = True
for arg in args:
p = pl_true(arg, model)
if p == False: return False
if p == None: result = None
return result
elif func is Implies:
p, q = args
return pl_true(Or(Not(p), q), model)
elif func is Equivalent:
p, q = args
pt = pl_true(p, model)
if pt == None:
return None
qt = pl_true(q, model)
if qt == None:
return None
return pt == qt
else:
raise ValueError("Illegal operator in logic expression" + str(expr))
class KB(object):
"""Base class for all knowledge bases"""
def __init__(self, sentence=None):
self.clauses = []
if sentence:
self.tell(sentence)
def tell(self, sentence):
raise NotImplementedError
raise NotImplementedError
def retract(self, sentence):
raise NotImplementedError
class PropKB(KB):
"""A KB for Propositional Logic. Inefficient, with no indexing."""
def tell(self, sentence):
"""Add the sentence's clauses to the KB
Examples
========
>>> from sympy.logic.inference import PropKB
>>> from sympy.abc import x, y
>>> l = PropKB()
>>> l.clauses
[]
>>> l.tell(x | y)
>>> l.clauses
[Or(x, y)]
>>> l.tell(y)
>>> l.clauses
[Or(x, y), y]
"""
for c in conjuncts(to_cnf(sentence)):
if not c in self.clauses: self.clauses.append(c)
"""Checks if the query is true given the set of clauses.
Examples
========
>>> from sympy.logic.inference import PropKB
>>> from sympy.abc import x, y
>>> l = PropKB()
>>> l.tell(x & ~y)
True
False
"""
if len(self.clauses) == 0: return False
from sympy.logic.algorithms.dpll import dpll
query_conjuncts = self.clauses[:]
query_conjuncts.extend(conjuncts(to_cnf(query)))
s = set()
for q in query_conjuncts:
s = s.union(q.atoms(C.Symbol))
return bool(dpll(query_conjuncts, list(s), {}))
def retract(self, sentence):
"""Remove the sentence's clauses from the KB
Examples
========
>>> from sympy.logic.inference import PropKB
>>> from sympy.abc import x, y
>>> l = PropKB()
>>> l.clauses
[]
>>> l.tell(x | y)
>>> l.clauses
[Or(x, y)]
>>> l.retract(x | y)
>>> l.clauses
[]
"""
for c in conjuncts(to_cnf(sentence)):
if c in self.clauses:
self.clauses.remove(c) | 1,128 | 4,099 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-09 | latest | en | 0.518441 |
http://www.eiwrqb.tw/contents/14/1735.html | 1,591,063,814,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347422065.56/warc/CC-MAIN-20200602002343-20200602032343-00332.warc.gz | 164,032,461 | 5,518 | 您當前位置：首頁 >> OS作業OS作業
日期：2018-09-24 01:45
[HW-01] Square root approximation
1. Description
The square root of a number N can be approximated by repeated calculation using the formula:
NG = 0.5(LG + N/LG)
where
NG = next guess
LG = last guess
Write a function that calculates the square root of a number using this method. The initial
guess will be the starting value of LG = 1.0. The program needs to calculate a value of NG using
the given formula. It is considered to be the root is found when |NG-LG| < 0.005. In this case
NG is accepted as the square root: otherwise, the new guess (NG) becomes the last guess (LG)
and the process is repeated.
2. Output
Check the validity of all input values as shown in <Input handling example 1>.
※ note: Gray colored background area is where the user needs to input
<Input handling example 1>
Enter a number to find the square root value (0.01 ~ 1000000.00) : 4.0
Square root of 4.00 is : 2.000
Enter C/c for continue or Q/q for quit : C
Enter a number to find the square root value (0.01 ~ 1000000.00) : 5
Square root of 5.00 is : 2.336
Enter C/c for continue or Q/q for quit : Q
Program Terminating. Bye!
<Input handling example 2>
Enter a number to find the square root value (0.01 ~ 1000000.00) : 0
Not a valid number. Try again.
Enter a number to find the square root value (0.01 ~ 1000000.00) : 0.75
Square root of 0.75 is : 0.866
Enter C/c for continue or Q/q for quit : Q
Program Terminating. Bye!
2 / 6
3. Output Guidelines
1) Display a friendly message to explain the program.
2) Display a friendly message to receive input from user.
3) Display a friendly message for the invalid inputs.
4) Define macros as needed
5) Leave a space line after receiving the inputs and before the result is displayed.
6) Follow the format of the outputs shown in <example 1> and <example 2>.
<Supporting information>
When the inputs needs to be read several times in the program, there could be a input buffer issue that will
cause some trouble reading the input.
Use fflush() in the <stdio.h> or define a macro FLUSH in your program as below and use this before a new
#define FLUSH while (getchar() != ‘\n’)
fflush() or FLUSH macro will empty the previous input line so that another input can read from a new input
line.
3 / 6
[HW-02] Prime Fibonacci number
1. Description
Write a program that checks whether the nth Fibonacci number is prime. Main function should call two subfunctions
to accomplish this task: one to find nth Fibonacci number and another one to find whether the
number is prime number.
2. Output
Check the validity of all input values as shown in <Input handling example 1>.
※ note: Gray colored background area is where the user needs to input
<Input handling example 1>
Enter a number to find whether nth Fibonacci number is prime (3 ~ 20) : 3
Fibonacci (3) = 3 is a prime number.
Enter C/c for continue or Q/q for quit : C
Enter a number to find whether nth Fibonacci number is prime (3 ~ 20) : 5
Fibonacci (5) = 8 is not a prime number.
Enter C/c for continue or Q/q for quit : Q
Program Terminating
<Input handling example 2>
Enter a number to find whether nth Fibonacci number is prime (3 ~ 20) : 2
Not a valid number. Try again.
Enter a number to find whether nth Fibonacci number is prime (3 ~ 20) : 5
Fibonacci (5) = 8 is not a prime number.
Enter C/c for continue or Q/q for quit : Q
Program Terminating
3. Output Guidelines
1) Display a friendly message to explain the program.
2) Display a friendly message to receive input from user.
4 / 6
3) Display a friendly message for the invalid inputs.
4) Leave a space line after receiving the inputs and before the result is displayed.
5) Follow the format of the outputs shown in <example 1> and <example 2>.
5 / 6
[HW-03] N Locker door problem
1. Description
N students are assigned lockers 1 through N. The student assigned to locker number 1 closes all N
lockers. The student assigned to locker number 2 then opens all lockers whose numbers are
multiples of 2. The student assigned to locker number 3 changes the status of all lockers whose
numbers are multiples of 3 (e.g. locker number 3, which is open gets closed, locker number 6, which
is closed, gets opened). The student assigned to locker number 4 changes the status of all locker
whose numbers are multiples of 4, and so on for all N lockers.
Write a program that read input N, process door alteration as described, and display the results as
shown in the output examples.
2. Output
Check the validity of the input value as shown in <Input handling example 1>.
※ note: Gray colored background area is where the user needs to input
<Input handling example 1>
Enter a number of lockers (1 ~ 1000): 5
Door 1 closed
Door 4 closed
There are 3 lockers open and 2 lockers closed.
Enter C/c for continue or Q/q for quit : C
Enter a number of lockers (1 ~ 1000): 10
Door 1 closed
Door 4 closed
Door 9 closed
There are 7 lockers open and 3 lockers closed.
Enter C/c for continue or Q/q for quit : Q
Program Terminating
<Input handling example 2>
6 / 6
Enter a number of lockers (1 ~ 1000): 0
Invalid input. Try again.
Enter a number of lockers (1 ~ 1000): 5
Door 1 closed
Door 4 closed
There are 3 lockers open and 2 lockers closed.
Enter C/c for continue or Q/q for quit : Q
Program Terminating
3. Output Guidelines
1) Display a friendly message to explain the program.
2) Display a friendly message to receive input from user.
3) Display a friendly message for the invalid inputs.
4) Leave a space line after receiving the inputs and before the result is displayed.
5) Follow the format of the outputs shown in <example 1> and <example 2> | 1,507 | 5,722 | {"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-2020-24 | latest | en | 0.747282 |
https://mathfilefoldergames.com/2023/09/13/decimal-dungeon-an-engaging-place-value-card-game-for-your-math-classroom/ | 1,695,635,662,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508959.20/warc/CC-MAIN-20230925083430-20230925113430-00339.warc.gz | 422,066,429 | 23,691 | # Decimal Dungeon: An Engaging Place Value Card Game for Your Math Classroom
Greetings fellow math educators! If you’ve been scouring the internet for an exciting new way to help your students master the seemingly tricky topic of decimal place values, particularly focusing on tenths and hundredths, you’re in luck! I have a fun, engaging, and entirely original card game that will not only reinforce your students’ understanding of decimals but also foster a sense of camaraderie and healthy competition in your classroom. Let’s delve into the game I affectionately call “Mastering the Decimal Dungeon”.
## Why “Mastering the Decimal Dungeon”?
I believe that learning can be enjoyable, engaging, and still be effective. Gone are the days of rote memorization and dry drills. It’s time to introduce fun into our math classrooms and replace the fear and anxiety associated with learning new concepts with excitement and curiosity. “Mastering the Decimal Dungeon” has been designed with this philosophy in mind. In this captivating game, students are “dungeon explorers,” fighting their way through the “decimal dungeon” by battling decimal monsters using their knowledge of place values.
## Who is it for?
“Mastering the Decimal Dungeon” is a fantastic math activity for 3rd and 4th graders who are ready to conquer the world of decimals. However, it can also be easily adapted for other grade levels or for differentiated instruction, a topic we’ll cover later in this post.
## What you’ll need:
• A standard deck of playing cards (only Ace, 2-9 are needed per suit, with Aces representing 1, face cards are not used)
• Dice or coins to decide who goes first
## Gameplay Instructions:
Setup:
1. Each student will need a deck of nine playing cards (Aces to Nines) from a suit. In a classroom setting, you can divide a standard card deck among four students.
2. Set the boundaries of the “Decimal Dungeon”. This can be an area on a desk or a section of the classroom floor.
3. The teacher, or “Dungeon Master”, shuffles the cards and lays out a 7×4 grid face-down in the “Decimal Dungeon”. The grid represents different decimal monsters lurking in the dungeon, each with a different strength (decimal value).
Playing the Game:
1. Determine who goes first by rolling a dice or flipping a coin.
2. The first player, or “Dungeon Explorer”, chooses a card from their hand (A-9) and places it face up in front of them, representing their strength for the round.
3. The Dungeon Master flips the top card of the Decimal Dungeon to reveal the strength (decimal value) of the decimal monster.
4. The player must then determine if their card represents a decimal value that is less than, greater than, or equal to the decimal monster. For example, a ‘3’ card would represent ‘0.30’ or ‘0.03’, depending on whether the tenths or hundredths place is being considered.
5. If the player’s decimal is greater than the monster’s decimal, they “defeat” the monster and take the card, adding it to their score pile. If it’s less than or equal to, the monster “defeats” the player, and the card is discarded.
6. The next player then takes their turn, following the same process.
7. The game continues until all cards in the Decimal Dungeon have been battled.
8. The player with the most cards in their score pile at the end of the game is the winner!
## Accommodations and Modifications:
Every classroom is unique, and every student has a different style and pace of learning. This game can be easily modified to fit the needs of your students and align with their individual learning goals.
For advanced learners: You can include the face cards in the deck with Jacks representing ‘0.11’, Queens ‘0.12’, and Kings ‘0.13’. This will provide a greater range of numbers for the students to work with and challenge their understanding of place values beyond the tenths and hundredths.
For struggling learners: You can introduce the game with only the tenths place. Once students are comfortable with this concept, you can add the hundredths place to increase the difficulty gradually.
For students with special needs: You can use manipulatives like base ten blocks or unifix cubes to visually represent the decimal values. This can help students who struggle with abstract concepts to better understand and engage with the game.
## Example Gameplay Scenario:
To help visualize how “Mastering the Decimal Dungeon” works, let’s walk through a round.
Player 1, Annie, draws a ‘4’ from her deck, which she decides to use as ‘0.40’. The Dungeon Master flips a card from the Decimal Dungeon, revealing ‘0.06’. Since Annie’s decimal is greater, she defeats the decimal monster and takes the card, adding it to her score pile.
Next, Player 2, Ben, draws a ‘7’ and decides to use it as ‘0.07’. The Dungeon Master flips the next card, showing ‘0.5’. Unfortunately, Ben’s decimal is less than the monster’s, so his card is discarded, and the monster card is removed from the game.
The game continues in this way, promoting the understanding of decimal place values while engaging students in a fun, competitive activity.
## Link to Common Core State Standards (CCSS):
“Mastering the Decimal Dungeon” aligns with several CCSS, particularly for grades 3 and 4. These include:
• CCSS.MATH.CONTENT.3.NBT.A.1: Use place value understanding to round whole numbers to the nearest 10 or 100.
• CCSS.MATH.CONTENT.4.NF.C.7: Compare two decimals to hundredths by reasoning about their size.
By engaging in this game, students are developing their understanding of decimal place values, a skill essential for these standards and future mathematical concepts.
“Mastering the Decimal Dungeon” is a fantastic resource to engage students in understanding and practicing decimal place values. It offers an excellent blend of fun and learning, reinforcing concepts in a relaxed and enjoyable environment. I encourage you to try it in your classrooms and share your experiences. Happy teaching!
## Transform Your Child into a Math Genius Overnight with the Ultimate Place Value Learning Bundle – Guaranteed Fun & Success!
Master the world of place value with the Ultimate Focus on Place Value Bundle! Perfect for kids in the 2nd to 6th grades, this bundle includes 28 stimulating games, assorted place value cards, and 132 decimal place value mazes/worksheets. Transform math struggles into victories by building a solid foundation of place value understanding. Enjoy long-term use with activities designed to grow with your child’s learning level over five years. Games like “Speed” Place Value Yahtzee and Place Value Battleship turn learning into exciting challenges. Don’t wait! Enhance your child’s math journey today. Click ‘Buy Now’ and make math fun with the Ultimate Focus on Place Value Bundle! | 1,445 | 6,779 | {"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-2023-40 | latest | en | 0.921585 |
https://www.physicsforums.com/threads/estimation-of-central-temperature-of-a-star.580360/ | 1,550,757,940,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247504790.66/warc/CC-MAIN-20190221132217-20190221154217-00224.warc.gz | 920,536,726 | 12,650 | # Estimation of central temperature of a star
1. Feb 22, 2012
### Darkkk
Problem:
Relatively low mass stars can be shown to have their radii dependent on their mass as;
R$\propto$M1/2
From this, show that the temperature in kelvin at the centre of such a star can be estimated to be
$\approx$24$\times$106(M/Msolar)1/2
Relevant Equations:
Tc=(2/3ℝ)$\times$Pc$\times$(R3/M)
Attempt at a Solution:
I tried to sub in the relation between R and M of the star, but to be quite honest I didn't see myself getting very far...Any pointers at all would be much appreciated! Not too sure on this one.
2. Feb 23, 2012
### zhermes
What did you get when you plugged in?
Do you know $P_c$ and $\mathcal{R}$
3. Feb 24, 2012
### Darkkk
I do know R=8.314 joules per Kelvin but Pc is unknown, Im assuming the proportion relation will be used at the end of the problem to get the temperature in terms of the solar mass. Im just not quite sure what the first step is. Perhaps I ought to sub the Tc given in the question into the central pressure equation Pc. What do you think? sounds like circular logic to me...
4. Feb 24, 2012
### zhermes
The first step is to plug in the mass-radius relation. That way you eliminate the radius from the equation. You know R, so all you need to find is the central pressure to solve for the temperature.
5. Feb 25, 2012
### Darkkk
Ah yes I see! I have it now. Thank you for the advice :) | 396 | 1,426 | {"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} | 3.3125 | 3 | CC-MAIN-2019-09 | latest | en | 0.919059 |
https://www.jiskha.com/display.cgi?id=1454624014 | 1,529,478,235,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863489.85/warc/CC-MAIN-20180620065936-20180620085936-00244.warc.gz | 822,869,111 | 4,051 | # Algebra
posted by Carmen
Suppose the growth of a population (in thousands) is represented by the expression 32 * 2.1^x where x represents the number of years since 2010. Evaluate the expression for they year 2015. Round to the nearest whole number.
A. 41
B. 1,307
C. 67
D. 622
Is it B or D??
1. Steve
what, no calculator?
Since 2015 is 5 years from 2010, you want to evaluate
32 * 2.1^5 = 1306.9
D should have been clearly wrong, since 2^5=32, so 32*32 = 1024 would be a good guess, though somewhat low.
2. Yosabeth
-3x+9=-16+2x
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https://cs.stackexchange.com/questions/106492/how-to-prove-space-tmsat-is-pspace-hard | 1,718,356,925,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00872.warc.gz | 167,313,338 | 39,321 | # How to prove SPACE-TMSAT is PSPACE-hard?
I understand that the language:
$$\operatorname{SPACE-TMSAT} = \{⟨M, w, 1^n⟩ : \text{DTM M accepts w in space n}\}$$
is in PSPACE since it doesn't use more than $$n$$ space. But to prove that it is PSPACE-complete, we need to do the reduction from any other PSPACE language to it. If we assume the language to run on a DTM using $$O(n^k)$$ space, how could we reduce it to SPACE-TMSAT?
Suppose $$L$$ is accepted by some machine $$M$$ using up to $$p(n)$$ space, for some polynomial $$p(n)$$. We map $$w$$ to $$\langle M, w, 1^{p(|w|)} \rangle$$.
• No, $n$ is completely arbitrary. Commented Apr 5, 2019 at 4:03 | 212 | 657 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.902403 |
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`f(x) = sin(x), [0,2pi]` Determine whether Rolle’s Theorem can be applied to `f` on the closed interval `[a, b]`. If Rolle’s Theorem can be applied, find all values of `c` in the open interval
### Textbook Question
Chapter 3, 3.2 - Problem 17 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook. | 162 | 531 | {"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-2017-13 | longest | en | 0.58036 |
https://ask.sagemath.org/question/8503/is-there-a-way-to-block-diagonalize-a-matrix/?answer=12955 | 1,603,237,952,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874340.10/warc/CC-MAIN-20201020221156-20201021011156-00198.warc.gz | 223,563,025 | 14,615 | # Is there a way to block diagonalize a matrix?
I am trying to block diagonalize a four by four symbolic matrix in to two matrices of dimension two by two matrices. Is there a simple way to do it in sage?
edit retag close merge delete
Sort by » oldest newest most voted
How do you know that this is possible to do?
Some 4x4 matrices are not block diagonalizable into 2x2 blocks. For example a nilpotent matrix with a singe Jordan block. If you know for some reason that your symbolic matrix is diagonalizable into 2x2 blocks then probably there is a way to do this, but I don't think possible to write an algorithm that can decide if a symbolic matrix is block diagonalizable.
more
I know that my matrix block diagonalizable from physics arguments. So can you recommend some place where I can read about it?
( 2011-11-30 19:57:23 -0500 )edit
Can you give me an example of a symbolic matrix that is block diagonalizable? Unless the matrix is of a very special form it must depend heavily on assumptions about the domain of the symbolic entries.
( 2011-12-11 09:05:13 -0500 )edit
The matrix I am trying to block diagonalise is [[Cos(theta),Sin(theta),0,mu],[-Sin(theta),Cos(theta),mu,0],[0,mu,Cos(theta),Sin(theta)],[mu,0,-Sin(theta),Cos(theta)]]. And I am trying to get rid of the mu in the off-diagonal block. I know it is possible because it is a Hamiltonian and I can always go to a basis in which the two systems decouple.
( 2011-12-12 05:35:31 -0500 )edit
The inverse of a matrix isn't guaranteed to exists, but there is a function for it anyway.
You can use SciLab's bdiag (numeric). http://help.scilab.org/docs/5.3.2/en_...
more
( 2011-12-11 09:03:00 -0500 )edit | 449 | 1,684 | {"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-2020-45 | latest | en | 0.90407 |
https://cpt.hitbullseye.com/CoCubes/Co-Cubes-Technical-Questions.php | 1,675,533,105,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500151.93/warc/CC-MAIN-20230204173912-20230204203912-00259.warc.gz | 193,777,767 | 39,483 | # Technical Questions asked in Co-Cubes
Views:85358
1. What is a the output signal when the input to NOT gate is A=11100?
1. 00000
2. 00011
3. 11111
4. None of these
1. What is function of logic gate?
1. It makes logic decisions
2. It works on binary algebra
3. It alternates between 0 and 1 values
4. None of these
2. What is output of division of (100011)2 by (101)2?
1. 1002
2. 10102
3. 1112
4. 1012 | 144 | 403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-06 | longest | en | 0.839894 |
https://fr.slideserve.com/amory/ch-7-the-capital-asset-pricing-model-another-view-about-risk | 1,709,366,932,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475757.50/warc/CC-MAIN-20240302052634-20240302082634-00407.warc.gz | 264,011,663 | 22,828 | 1 / 38
# Ch.7 The Capital Asset Pricing Model: Another View About Risk
Ch.7 The Capital Asset Pricing Model: Another View About Risk. 7.1 Introduction 7.2 The Traditional Approach to the CAPM 7.3 Using the CAPM to value risky cash flows 7.4 The Mathematics of the Portfolio Frontier: Many Risky Assets and No Risk-Free Asset
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## Ch.7 The Capital Asset Pricing Model: Another View About Risk
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### Presentation Transcript
1. Ch.7 The Capital Asset Pricing Model: Another View About Risk • 7.1 Introduction • 7.2 The Traditional Approach to the CAPM • 7.3 Using the CAPM to value risky cash flows • 7.4 The Mathematics of the Portfolio Frontier: Many Risky Assets and No Risk-Free Asset • 7.5 Characterizing Efficient Portfolios -- (No Risk-Free Assets) • 7.6 Background for Deriving the Zero-Beta CAPM: Notion of a Zero Covariance Portfolio • 7.7 The Zero-Beta Capital Asset Pricing Model • 7.8 The Standard CAPM • 7.9 Conclusion
2. 7.1 Introduction • Equilibrium theory (in search of appropriate risk premium) • Exchange economy • Supply = demand: for all asset j, • Implications for returns
3. 7.2 The Traditional Approach • All agents are mean-variance maximizers • Same beliefs (expected returns and covariance matrix) • There exists a risk free asset • Common linear efficient frontier • Separation/ Two fund theorem • T = M
4. (7.1) 7.2 The Traditional Approach to the CAPM • a. The market portfolio is efficient since it is on the efficient frontier. • b. All individual optimal portfolios are located on the half-line originating at point . • c. The slope of the CML is .
5. Figure 7-1 The Capital Market Line
6. Proof of the CAPM relationship (Appendix 7.1) • Refer to Figure 7.1. Consider a portfolio with a fraction 1- a of wealth invested in an arbitrary security j and a fraction a in the market portfolio As a varies we trace a locus which - passes through M (- and through j) - cannot cross the CML (why?) - hence must be tangent to the CML at M Tangency = = slope of the locus at M = slope of CML =
7. (7.2) (7.3) Define Only a portion of total risk is remunerated = Systematic risk
8. (7.4) (7.5) A portfolio is efficient iff no diversifiable risk (7.6) b is the only factor; SML is linear
9. Figure 7-2 The Security Market Line
10. 7.3 Using the CAPM to value risky cash flows (i)
11. Using the CAPM to value risky cash flows (ii)
12. 7.4 The Mathematics of the Portfolio Frontier: Many Risky Assets and No Risk-Free Asset Goal: understand better what the CAPM is really about – In the process: generalize. • No risk free asset • Vector of expected returns e • Returns are linearly independent • Vij = sij
13. Definition 7.1: A frontier portfolio is one which displays minimum variance among all feasible portfolios with the same . since
14. (vector) (vector) l (scalar) g (scalar) (7.15) (vector) (vector) (scalar) If E = 0, wp = g If E = 1, wp = g + h
15. Proposition 7.1: The entire set of frontier portfolios can be generated by ("are convex combinations" of) g and g+h. • Proof: To see this let q be an arbitrary frontier portfolio with as its expected return. Consider portfolio weights (proportions) ; then, as asserted,
16. . • Proposition 7.2. The portfolio frontier can be described as convex combinations of anytwo frontier portfolios, not just the frontier portfolios g and g+h. • Proof: To see this, let be any two distinct frontier portfolios; since the frontier portfolios are different, Let q be an arbitrary frontier portfolio, with expected return equal to . Since , there must exist a unique number such that (7.16) Now consider a portfolio of with weights a, 1- a, respectively, as determined by (7.16). We must show that .
17. (7.17) • For any portfolio on the frontier, , with g and h as defined earlier. Multiplying all this out yields:
18. (i) the expected return of the minimum variance portfolio is A/C; • (ii) the variance of the minimum variance portfolio is given by 1/C; • (iii) equation (7.17) is the equation of a parabola with vertex (1/C, A/C) in the expected return/variance space and of a hyperbola in the expected return/standard deviation space. See Figures 7.3 and 7.4.
19. Figure 7-4 The Set of Frontier Portfolios: Mean/SD Space
20. 7.4 Characterizing Efficient Portfolios -- (No Risk-Free Assets) • Definition 7.2: Efficient portfolios are those frontier portfolios for which the expected return exceeds A/C, the expected return of the minimum variance portfolio.
21. Proposition 7.3 : Any convex combination of frontier portfolios is also a frontier portfolio. • Proof : Let , define N frontier portfolios ( represents the vector defining the composition of the ith portfolios) and let , i = 1, ..., N be real numbers such that . Lastly, let denote the expected return of portfolio with weights . The weights corresponding to a linear combination of the above N portfolios are : Thus is a frontier portfolio with .
22. Proposition 7.4 : The set of efficient portfolios is a convex set . • This does not mean, however, that the frontier of this set is convex-shaped in the risk-return space. • Proof : Suppose each of the N portfolios considered above was efficient; then , for every portfolio i. However ; thus, the convex combination is efficient as well. So the set of efficient porfolios, as characterized by their portfolio weights, is a convex set.
23. 7.5 Background for Deriving the Zero-Beta CAPM: Notion of a Zero Covariance Portfolio • Proposition 7.5: For any frontier portfolio p, except the minimum variance portfolio, there exists a unique frontier portfolio with which p has zero covariance. We will call this portfolio the "zero covariance portfolio relative to p", and denote its vector of portfolio weights by . • Proof: by construction.
24. E ( r ) p A/C MVP ZC(p) E[r ( ZC(p)] (1/C) Var ( r ) Figure 7-6 The Set of Frontier Portfolios: Location of the Zero-Covariance Portfolio
25. Let q be any portfolio (not necessary on the frontier) and let p be any frontier portfolio. (7.27)
26. 7.6 The Zero-Beta Capital Asset Pricing Model (Equilibrium) (i) agents maximize expected utility with increasing and strictly concave utility of money functions and asset returns are multivariate normally distributed, or (ii) each agent chooses a portfolio with the objective of maximizing a derived utility function of the form , W concave. (iii) common time horizon, (iv) homogeneous beliefs about e and V
27. (7.28) (7.29) • All investors hold mean-variance efficient portfolios • the market portfolio is convex combination of efficient portfolios is efficient. • (7.21) .
28. (7.30) n x 1 n x n a number n x 1 7.7 The Standard CAPM Solving this problem gives
29. (7.31) (7.32) (7.33)
30. (7.34) or (7.35) or
31. What have we accomplished? • The pure mathematics of the mean-variance portfolio frontier goes a long way • In particular in producing a SML-like relationship where any frontier portfolio and its zero-covariance kin are the heroes! • The CAPM = a set of hypotheses guaranteeing that the efficient frontier is relevant (mean-variance optimizing) and the same for everyone (homogeneous expectations and identical horizons)
32. What have we accomplished? • The implication: every investor holds a mean-variance efficient portfolio • Since the efficient frontier is a convex set, this implies that the market portfolio is efficient. This is the key lesson of the CAPM. It does not rely on the existence of a risk free asset. • The mathematics of the efficient frontier then produces the SML
33. What have we accomplished? • In the process, we have obtained easily workable formulas permitting to compute efficient portfolio weights with or without risk-free asset
34. Conclusions • The asset management implications of the CAPM • The testability of the CAPM: what is M? the fragility of betas to its definition (The Roll critique) • The market may not be the only factor (Fama-French) • Remains: do not bear diversifiable risk!
More Related | 1,956 | 7,995 | {"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-2024-10 | latest | en | 0.712122 |
http://navalfacilities.tpub.com/mo221/mo2210129.htm | 1,539,732,311,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510893.26/warc/CC-MAIN-20181016221847-20181017003347-00119.warc.gz | 251,074,663 | 7,898 | Section 2. Concepts of Metering Electricity
Custom Search
CONCEPTS OF METERING ELECTRICITY
Section 2.
1. INTRODUCTION. To effectively manage or participate in electric energy
management programs, a basic knowledge of electricity and meters is highly
beneficial. A course in electricity and meters is beyond the scope of this
manual. The material presented here acquaints individuals with fundamental
electrical units and phenomena specifically related to meters. The major
emphasis is to show how information obtained from electrical meters is
applicable to management programs. There is some mathematical and technical
discussion, but it has been kept to a basic level.
2. COMMON METERING TERMS. The following definitions provide the means for a
basic application of metering data.
Ampere (I):
Unit of current or rate of flow of electricity.
Blondel's Theorem: The power/energy in a circuit of N lines can be
metered by N single-phase watt/watthour meters with the potential circuits
connected from each line to any common point. If the common point is on one
of the lines, the power/energy can be metered by N-1 watt/watthour meters.
Demand Analysis: Demand analysis is the critical examination of demand
profiles obtained from a power survey recorder (PSR), or compiled from data
collected using kilowatthour meters, rate schedules, demand registers, and
other instruments. One of its purposes is to locate peaks and valleys in a
demand profile to evaluate the possibility of load shifting to level out the
profile and to reduce demand charges.
Demand Factor: Ratio of maximum demand to the total connected load.
Demand Peaks: Although demand billing procedures are not uniform,
time-dependent demand charges are typical. One method uses onpeak, midpeak,
and offpeak hours to assess demand charges. Although hours may vary between
summer and winter months, a typical summer schedule is: onpeak-demand--l:00
PM to 7:00 PM, weekdays; midpeak-demand--9:00AM to 1:00 PM and 7:00 PM to
11:00 PM, weekdays; offpeak-demand--all other hours including holidays.
Highest charges are assessed for onpeak-demand with proportional cost
reductions for midpeak and offpeak periods.
Demand Profile: A demand profile is a graphic representation of
electrical demand made on a system during a given period. The profile
illustrates the relationship between demand, in kilowatt, and time of day
demand occurred. When place demand on y-axis and time on x-axis.
Diversity Factor: Ratio of the sum of individual maximum demands of
various subdivisions of a system to the maximum demand of the whole system.
Horsepower (hp): Measure of time rate of doing work; equivalent to
raising 33,000 lbs, one ft in one minute; 746 watts.
9-3 | 639 | 2,730 | {"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} | 3.328125 | 3 | CC-MAIN-2018-43 | latest | en | 0.885154 |
https://numbermatics.com/n/86220288/ | 1,597,061,143,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738674.42/warc/CC-MAIN-20200810102345-20200810132345-00364.warc.gz | 432,586,620 | 9,457 | # 86220288
## 86,220,288 is an even composite number composed of four prime numbers multiplied together.
What does the number 86220288 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 320 divisors.
86220288 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of three hundred twenty divisors.
## Prime factorization of 86220288:
### 29 × 37 × 7 × 11
(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 7 × 11)
See below for interesting mathematical facts about the number 86220288 from the Numbermatics database.
### Names of 86220288
• Cardinal: 86220288 can be written as Eighty-six million, two hundred twenty thousand, two hundred eighty-eight.
### Scientific notation
• Scientific notation: 8.6220288 × 107
### Factors of 86220288
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 18
• Sum of prime factors: 23
### Divisors of 86220288
• Number of divisors d(n): 320
• Complete list of divisors:
• Sum of all divisors σ(n): 322122240
• Sum of proper divisors (its aliquot sum) s(n): 235901952
• 86220288 is an abundant number, because the sum of its proper divisors (235901952) is greater than itself. Its abundance is 149681664
### Bases of 86220288
• Binary: 1010010001110011110000000002
• Base-36: 1FC000
### Squares and roots of 86220288
• 86220288 squared (862202882) is 7433938062802944
• 86220288 cubed (862202883) is 640956280749031918927872
• The square root of 86220288 is 9285.4880324085
• The cube root of 86220288 is 441.7770556557
### Scales and comparisons
How big is 86220288?
• 86,220,288 seconds is equal to 2 years, 38 weeks, 3 days, 22 hours, 4 minutes, 48 seconds.
• To count from 1 to 86,220,288 would take you about four years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 86220288 cubic inches would be around 36.8 feet tall.
### Recreational maths with 86220288
• 86220288 backwards is 88202268
• 86220288 is a Harshad number.
• The number of decimal digits it has is: 8
• The sum of 86220288's digits is 36
• More coming soon! | 749 | 2,542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-34 | latest | en | 0.823262 |
https://boredofstudies.org/threads/short-answer.188507/ | 1,643,339,741,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305341.76/warc/CC-MAIN-20220128013529-20220128043529-00280.warc.gz | 206,105,687 | 16,064 | #### JosephW
##### Member
How did everyone find short answer? I didnt find any of them that hard.
Was the cheese question, you had to use cosine rule to get how wide the box is?
#### soap~
##### Legal Eagle
Ayup I used cosine rule.
I have to say the last simmo question got me. Also I kinda didn't go over home loan repayments so I lost out on that one )=
Not too difficult otherwise.
#### risole91
##### I'm Coming Home
JosephW said:
How did everyone find short answer? I didnt find any of them that hard.
Was the cheese question, you had to use cosine rule to get how wide the box is?
I did; fuck 27 and 28 were hard
#### bling05
##### Member
What did you need cosine rule for??
#### JosephW
##### Member
The loan repayents, all u needed to do was the present value formula, and use algebra to get the M contributions. And the second question of it, u need to add the 250 to the other contribution and make it 20 terms instead of 30. In the end, she was short by a few thousand
#### lind27
##### Member
Wat a fucken stupid test barely any algebra or trig and i swear there was more than 30% yr 11 stuff, which sucks coz i didn't do any yr 11 general. Shouldn't have dropped 2 unit
#### davohan123
##### Member
lind27 said:
Wat a fucken stupid test barely any algebra or trig and i swear there was more than 30% yr 11 stuff, which sucks coz i didn't do any yr 11 general. Shouldn't have dropped 2 unit
hey lins didnt like the test hey
#### samjohnno91
##### Member
wasnt the cheese question arc length formula, and then you times your answer by four to get the amount going across?
#### JosephW
##### Member
samjohnno91 said:
wasnt the cheese question arc length formula, and then you times your answer by four to get the amount going across?
No cause that will give a greater length. U had to use cosine to get a straight length.
#### soap~
##### Legal Eagle
samjohnno91 said:
wasnt the cheese question arc length formula, and then you times your answer by four to get the amount going across?
Not really. The arc length will be longer than required because it includes the curve. You needed to use cosine to actually work out how wide it was.
The diagram underneath that bit was a damn good hint.
##### New Member
JosephW said:
No cause that will give a greater length. U had to use cosine to get a straight length.
i put a line down he middle of the triangle to spilt it up into 2 right angle trianges, then used sin20 = x/15 , n doubled that answer.
would that work?
#### samjohnno91
##### Member
im a bit confused, looking at the diagram, it shows the cheese with the curve at the bottom being taken into account and that each ends of the cheeses touch the side of the box. All cheese blocks at their respective ends just touch, so still dont see why arc length wouldnt work.
and then to find the 3 going across i just did 7x3
#### samjohnno91
##### Member
yea, but it says packed into a rectangular box with the arches at the start. it is only until the second question that they cut the ends off for straight line to calculate the remaining voludem taken up by the 12 slices.
#### daveyvc
##### New Member
arc length to get the length, i just assumed length of side things would be the same if its a cylinder.. so radius was still 15. meh. we will see.
#### Cerry
##### Member
The arc length isn't actually the width - the width of the cheese is the same as the base when the arc is cut off.
The only way I can think off to demonstrate is make up a model using string, with a seperate piece for each side. Then straighten out the arc, and you'll see that it's longer than the bottom of the triangle that the other two bits of string make.
#### JosephW
##### Member
samjohnno91 said:
im a bit confused, looking at the diagram, it shows the cheese with the curve at the bottom being taken into account and that each ends of the cheeses touch the side of the box. All cheese blocks at their respective ends just touch, so still dont see why arc length wouldnt work.
and then to find the 3 going across i just did 7x3
I can either go around a football field to point A or go straight through it. Is it the same distance walked? | 1,001 | 4,170 | {"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-2022-05 | latest | en | 0.961105 |
https://flylib.com/books/en/3.315.1.175/1/ | 1,566,775,067,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330907.46/warc/CC-MAIN-20190825215958-20190826001958-00148.warc.gz | 466,051,529 | 6,261 | 19.3 SceneNode Properties and Methods
19.3 SceneNode Properties and Methods
Aside from sustaining a scene hierarchy and containing other nodes, SceneNodes can affect the objects they contain. SceneNodes can be moved, scaled, and rotated, which affects all attached child Entities. Likewise, SceneNodes can have their visibility status set to true or false, effectively hiding and unhiding all attached Entities. Following is a list of some of the methods of SceneNode that allow transformation.
• void scale (const Vector3 &scale)
Scales the node, combining its current scale with the passed-in scaling factor.
• void scale (Real x, Real y, Real z)
Scales the node, combining its current scale with the passed-in scaling factor.
• void translate (const Vector3 &d, TransformSpace relativeTo=TS_PARENT)
Moves the node along the Cartesian axes.
• void translate (Real x, Real y, Real z, TransformSpace relativeTo=TS_PARENT)
Moves the node along the Cartesian axes.
• void translate (const Matrix3 &axes, const Vector3 &move, TransformSpace relativeTo=TS_PARENT)
Moves the node along arbitrary axes.
• void translate (const Matrix3 &axes, Real x, Real y, Real z, TransformSpace relativeTo=TS_PARENT)
Moves the node along arbitrary axes.
• void roll (const Radian &angle, TransformSpace relativeTo =TS_LOCAL)
Rotates the node around the z-axis.
• void pitch (const Radian &angle, TransformSpace relativeTo=TS_LOCAL)
Rotates the node around the x-axis.
• Void rotate (const Vector3 &axis, const Radian &angle, TransformSpace relativeTo=TS_LOCAL)
Rotates the node around an arbitrary axis.
• Void rotate (const Quaternion &q, TransformSpace relativeTo=TS_LOCAL)
Rotates the node around an arbitrary axis using a quaternion.
Introduction to Game Programming with C++ (Wordware Game Developers Library)
ISBN: 1598220322
EAN: 2147483647
Year: 2007
Pages: 225
Authors: Alan Thorn
Similar book on Amazon | 439 | 1,916 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-35 | latest | en | 0.632278 |
https://www.physicsforums.com/threads/integration-problems.763203/ | 1,519,194,176,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813431.5/warc/CC-MAIN-20180221044156-20180221064156-00637.warc.gz | 949,311,258 | 14,254 | # Integration problems
1. Jul 24, 2014
### eric_999
Hey!
I have this integral: ∫((1/2)/(2x-1))dx.
The first time, I did like this: ∫((1/2)/(2x-1))dx = (1/2)∫(1/(2x-1))dx. If I set u = 2x-1, then du = 2dx, so I can rewrite (1/2)∫(1/(2x-1))dx as (1/2)*(1/2)∫(1/u)du = 1/4∫(1/u)du = 1/4ln|u| = 1/4ln|2x-1|.
But when I do like this (I cannot see what im doing wrong) ∫((1/2)/(2x-1))dx = ∫(1/(4x-2)dx and set u = 4x-2, so du = 4dx, I can rewrite it as 1/4∫(1/u)du = 1/4ln|u| = 1/4ln|4x-2|. What am I doing wrong??
2. Jul 24, 2014
### da_nang
The error you made is that you forgot the constant of integration. Include that, and recall that $\ln{ab} = \ln{a} + \ln{b}$ for positive, non-zero a and b, you should be able to see that the methods are equivalent.
3. Jul 24, 2014
### jbunniii
Note that
\begin{align} \frac{1}{4}\ln|4x-2| &= \frac{1}{4}\ln(2|2x-1|) \\ &= \frac{1}{4}(\ln(2) + \ln|2x-1|)\\ &= \frac{1}{4}\ln|2x-1| + \frac{\ln(2)}{4} \end{align}
which differs from your other answer $\frac{1}{4}|2x-1|$ by a constant. Since indefinite integrals can differ by a constant ($+C$), both answers are equally valid.
4. Jul 24, 2014
### eric_999
Thanks a lot, i understand now! | 520 | 1,188 | {"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} | 4.15625 | 4 | CC-MAIN-2018-09 | longest | en | 0.826494 |
https://pbaumgarten.com/java/05-comparing.md | 1,722,957,409,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00710.warc.gz | 350,476,184 | 3,678 | # Paul Baumgarten
Computer Scientist, teacher, freelance programmer and self confessed geek
# Java: Comparing values
The power of programming comes from letting the computer do work for us. To do that it needs to make decisions. We can have Java make decisions on the basis of comparing one value to another. These comparisons should always generate a boolean result, that is true or false. We can then instruct Java to execute code based on whether a comparison is true or false.
## Numeric comparisons
To compare the values of any numeric values or variables, the following operators exist:
• `(a == b)` ... is equal to
• `(a != b)` ... is not equal to
• `(a > b)` ... greater than
• `(a >= b)` ... greater than or equal to
• `(a < b)` ... less than
• `(a <= b)` ... less than or equal to
Note that a and b can either be a value or a variable, and that the surrouding set of parenthesis is required.
For example
``````boolean result;
int a = 10, b = 3;
result = ( a == b );
System.out.println( result );
result = ( a != b );
System.out.println( result );
result = ( a > b );
System.out.println( result );
result = ( a < b );
System.out.println( result );
``````
Take careful note of the difference in punctuation between setting a value to a variable, and comparing two values! Setting a variable to a given value uses the single equal sign `=` whereas comparing two values or variables uses the double equal sign `==`.
## String comparisons
There are a variety of functions suitable for comparing the values of strings as the following illustrates:
``````java.util.Scanner keyb = new java.util.Scanner(System.in);
System.out.print("Type string 1: ");
String s1 = keyb.nextLine();
System.out.print("Type string 2: ");
String s2 = keyb.nextLine();
System.out.println( s1.equals(s2) ); // also Objects.equals(s1, s2)
System.out.println( s1.compareTo(s2) );
System.out.println( s1.contains(s2) );
System.out.println( s1.endsWith(s2) );
System.out.println( s1.startsWith(s2) );
System.out.println( s1.isEmpty() );
``````
You will notice they all return boolean results except `.compareTo()` which returns an integer result indicating how closely the two strings compare or differ. Specifically, it will return:
• zero when string values match
• negative when the parent object (s1) is alphabetically before the parameter (s2)
• positive when the parent object (s1) is alphabetically after the parameter (s2)
• the size of the positive or negative number indicates the value of number of characters different. For example `"a".compareTo("d")` will return -3 because the letter a is three values before the letter d.
## Multiple comparisons
Boolean logic can be used to diasy chain multiple comparisons into one instruction.
• The AND operator is the double ampersand `&&`.
• The OR operator is the double pipe `||`.
• The NOT operator is the exclaimation `!`.
The following is a valid example:
``````int a = 13;
int b = 4;
int c = 10;
boolnea result;
result = ( (a > b) && (a < c) );
result = ( (a > b) || (a < c) );
result = ( (a != b) && (a < c) || (b == c) );
``````
Where you are uncertain about order of precedence, it is recommended to use an additional set of parenthesis to enforce your intended outcome.
## Ternary operator
You may see a this used when searching online documentation, so I wanted to briefly show it here. It is known as the ternary operator.
The ternary operator allows you to set a resulting value based on whether or not a comparison condition is true. For example, if we didn't have the `Math.max()` function, we could use the tenary operator to easily identify the larger of two numbers...
``````int largerOfTheTwo = (a>b) ? a : b;
``````
The general rule for the ternary operator is:
``````result = (condition) ? result_if_true : result_if_false ;
``````
## Problem set
The following problems all use the following code:
``````int i = 10
int j = 3;
boolean true_false;
``````
What will be the value of `true_false` after each comparison?
1. true_false = ( j > i );
2. true_false = ( i > j );
3. true_false = ( i= = j );
4. true_false = ( (j<=i)||(j>=i) );
5. true_false = ( (i>j)&&(j==0) );
6. true_false = ( (j<50)||(j!=33) );
7. true_false = ( !(j>=0)||(i<=50) );
8. true_false = ( !(! (!true)) );
9. true_false = ( 5 < = 5 );
10. true_false = ( j != i );
11. What is returned by the following expression? (Recall that the precedence order of logical operators is `!`, `&&`, and finally `||`.)
`!( (2>3) | | (5= =5) && (7>1) && (4<15) | | (35<=36) && (89!=34) )`
(Exercises adapted from Blue Pelican Java) | 1,186 | 4,571 | {"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-2024-33 | latest | en | 0.703368 |
http://docplayer.net/20722477-Boundary-between-ccm-and-dcm-in-dc-dc-pwm-converters.html | 1,544,814,036,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826306.47/warc/CC-MAIN-20181214184754-20181214210754-00474.warc.gz | 79,663,267 | 27,840 | # Boundary between CCM and DCM in DC/DC PWM Converters
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1 Boundary between CCM and DCM in DC/DC PWM Converters ELENA NICULESCU and E. P. IANCU Dept. of Electronics and Instrumentation, and Automation University of Craiova ROMANIA Abstract: - It is presented a modality to compute the boundary between the discontinuous (DCM) and continuous (CCM) operating modes in the DC/DC fourth-order PWM converters with parasitic resistances. To reach this goal, an averaged model of the converter has been used. The boundary between DCM and CCM, considered to be the critical load resistance R crit, or, equivalently, the critical value of the inductor conduction parameter k ctrit, is found as the positive root of a second-degree equation with unknown R crit or k ctrit. The algorithm carried out and implemented with MATLAB environment allows studying the effect of the parasitic resistances and the coupling of inductors over the boundary R crit or k tcrit, the steady-state properties and the external characteristics of the converter. The simulation results show that the parasitic resistances and the inductor coupling change the boundary between the discontinuous and continuous conduction modes of a fourth-order PWM converter. Key-Words: - DC/DC fourth-order PWM converters, boundary between CCM and DCM 1 Introduction In the elementary PWM converters (buck, boost, buck-boost), which are second-order converters, the discontinuous conduction mode (DCM) appears at the moment when the diode current becomes zero. This will occur at a specific operating point (duty ratio D and load resistance R). The current through the inductor becomes discontinuous and would stay discontinuous for the third switched interval D T s. Here, the boundary between CCM and DCM is defined by the critical value of the inductor conduction parameter (k crit ) or by the critical value of the load resistance (R crit ). Considering the converter to be ideal, this boundary has been calculated as the positive root of a second-degree equation [1. Including the parasitic resistances of the circuit, the degree of the equation with the unknown R crit or k crit remains unchanged [. A fourth-order converter consists of two inductors, two capacitors, and a single switch realized by transistor and diode combination, such as the non isolated Cuk, Sepic and Zeta converters and the ripple-free input-current PWM boost converter. In these converters, the two inductors conduct some constant current I, even for the third switched interval D T s. In this third switched interval D T s, the sum i 1 +i of two inductors currents that traverses the diode is zero [ [ 5. Using the individual inductor voltage and current waveforms and the volt-secondbalance law on the inductors, for the assumed 100% efficiency, the boundary between CCM and DCM has been easily found as the positive root of a second-degree equation, for a non-isolated Cuk converter with DCM and a ripple-free input-current boost converter with DCM [, [ 4. Boundary between CCM and DCM It is known that the boundary between CCM and DCM is established by the specific value of the parameter D, namely D =1-D, regardless of the order (two or four) of PWM switching converters. So, a PWM converter operates with CCM when D >1-D and with DCM when D <1-D [1, [. For the elementary or second-order PWM converters (buck, boost, buck-boost) no parasitic included, the boundary between CCM and DCM is given either as the critical value of the inductor conduction parameter k, written k crit, or as the critical value of the load resistance R, written R crit. It is natural to express the boundary in terms of the load resistance R, rather than the dimensionless parameter k. The parameter k is defined as Lf R k = s = nom (1) R R
2 where is a design parameter called nominal load resistance of the converter and f s is the constant switching frequency. Considering (1) for k crit, the load boundary R crit results as Ripple-free Input-current Boost D 1 ( D) 1 R crit =. () kcrit Similarly, the boundary between CCM and DCM for the fourth-order converters (non-isolated Cuk, Sepic, Zeta and ripple-free input-current boost converters etc.) is given either as the critical value of conduction parameter k tcrit (k t= L e f s /R) through an equivalent inductor L e (L e =L 1 //L ) or as the critical value of the load resistance R crit [-[5. Like the second-order converters with ideal circuit components, these critical values have been found as the positive root of a second-degree equation with unknown k tcrit or R crit. The expressions of k tcrit as functions on duty ratio and dc voltage conversion ratio are shown in Table 1, for a non-isolated Cuk PWM converter and a ripple-free input-current boost PWM converter [, [4. The dc voltage conversion ratio of the converter with DCM is written. Table 1: k tcrit for a non-isolated Cuk PWM converter and a ripple-free input-current boost converter (no parasitic case) Converter k tcrit = f ( D) k = f ( ) Cuk Ripple-free Input-current Boost ( 1 D) ( 1 D) D tcrit 1 ( 1 + ) MD 1 The functions R crit = f (D) and R crit = f ( ) are given in Table, for the same ideal converters [, 4. Table : R crit for a non-isolated Cuk PWM converter and a ripple-free input-current boost converter (no parasitic case) Converter R crit = f ( D) R crit = f ( ) Cuk ( 1 + M ) ( 1 D) D Equation of the Boundary between CCM and DCM for Nonideal Converters Despite its complexity, the state-space averaging (SSA) method is the most popular approach used for the modelling of the dc-dc switching converters. This method allows including all the parasitic resistances of the circuit even in the initial stage and supplies both the static and dynamic models of the converter. The following parasitic resistances of the converter (written p i with i = 1, 6 ) have been considered here: the loss resistance of the inductors (r 1 and r ), the equivalent-series resistance of the capacitors (r and r 4 ), the conducting-state resistance of the transistor (r 5 ) and the diode (r 6 ). Using the SSA method, a fourth-degree equation in unknown D has been carried out for the fourthorder converters with parasitic included [8: 4 m4 D + m D + md + m1d + m0 = 0. () In order to obtain the equation (), the matrices A and B that describe the averaged steady-state model of the fourth-order PWM converter with DCM have been set in the form: A = Ae + DA B = Be + DB where Ae = DA1 + D A = [ ae( m, n) Be = DB1 + D B = [ be( m) A = A A = [ a( m, n) B = B B = [ b( m) D = 1 D with m = n = 1, 4. Generally, the coefficients m i with i = 1, 4, from the equation (), are functions on the load resistance R. At the boundary between the two operating modes, D =1-D and R becomes R crit. In order to distinguish this last element, the components of the matrices A e and A, which are functions on R, have been split in two terms: one term is independent of R and another that is function on R. For example, it writes: ae ( m, n) = ae0( m, n) + r1 ( m, n) f1( R)
3 ( m, n) a ( m, n) r ( m, n) f ( R) a = 0 +. Replacing D with 1-D and R with R crit into the equation (), a second-degree equation in unknown R crit results for the fourth-order PWM converter with parasitic included, like as for the second-order PWM converters [: n Rcrit + n1r + n0 = 0. (4) crit The coefficients n, n 1 and n 0 from the equation (4) are functions of the duty ratio D, all the parameters and parasitic resistances of the circuit, the switching frequency and the coupling coefficient k c of the inductors: R crit =f (D, L 1, L, C 1, C, p i, f s, k c ). The general form of the coefficients from the equation (4), as well as those from the equation (), allows studying four PWM converter configurations, namely: the non isolated Cuk, Sepic and Zeta converters, and the ripple-free input current boost converter. A second-degree equation in unknown k tcrit is obtained if the equation () is introduced into (4): n 0 k tcrit 1 nom tcrit nom = + n R k + n R 0. (5) Like as R crit, the coefficients of this last equation depend on the duty ratio D, all the parameters and parasitic resistances of the circuit, the switching frequency and the coupling coefficient k c of the inductors: k tcrit =f (D, L 1, L, C 1, C, p i, f s, k c ). The same second-degree equation with the unknown R crit or k tcrit can be obtained by equaling the expressions of M C and, deduced from the averaged steady-state models corresponding to DCM and CCM, and replacing D with 1-D in. The dc characteristics of the converter as the average values of the currents that traverse the inductors, the voltage drops across the capacitors and the dc voltage conversion ratio (M=V O /V I ) of converter are obtained through the SSA method, for both DCM and CCM. The two dc voltage conversion ratio M C for CCM and for DCM have the same value (M C = ) on the boundary between CCM and DCM. Consequently, plotting the external characteristics of the converter for both continuous and discontinuous operating modes, that is M C =f (D, R) and =f (D, R), the boundary R crit between CCM and DCM could be found as the intersection of these two external characteristic families of converter: R crit = R. The value k Mc = tcrit can be MD calculated with the formula (). But this way of determination of the boundary between DCM and CCM is not recommended because this is a graphical method that cannot prove a satisfactory accuracy. 4 Simulation results MATLAB environment offers a very simple implementation of this algorithm carried out for establishing the boundary between DCM and CCM in the fourth-order PWM converter. The analytical way for establishing R crit or k tcrit in the fourth-order PWM converters with parasitic resistances has been studied on a ripple-free inputcurrent boost converter with the specifications: L 1 =5.1 µh, L =0.7 µh, C 1 =18 µf, C =1000 µf and f s =00 khz [4. The following parasitics have been considered here: r 1 =r =r 4 =r 5 =r 6 =0.1 Ω and r =0.01 Ω. For this converter, the effect of the inductor coupling and the parasitic resistances of the circuit over R crit can be seen in Figures 1 and. In Fig. 1, the function R crit =f (D, k c ) is plotted for several values of D and k c, for an ideal converter (p i =0). As it can be seen from this figure, the increasing of the value of k c causes the decreasing of R crit for all the values of D. Fig. 1: Plots of R crit as function on the duty ratio and the coupling coefficient of inductors The function R crit =f (D, k c, p i ) is plotted in Fig. for the same values of D and k c as in Fig. 1. For the converter with parasitics, the effect of the value of k c over R crit is changed in function on the value of the duty ratio. So, the increasing of the value of the coupling coefficient causes the increasing of R crit at small values of D and the decreasing of R crit at big values of D. As k c and D tend to 1, R crit tends to zero.
4 Fig. : Plots of R crit as function on the duty ratio, the coupling coefficient of inductors and the parasitic resistances of converter The effect of the coupling coefficient k c of the inductors and the parasitic resistances of the circuit over k tcrit is shown in Fig. and 4. Here, the function k tcrit =f (D, k c ) is plotted for the same values of D and k c as for R crit, and for the two cases: no parasitic included (Fig. ) and parasitic included (Fig. 4). As it can be seen from Fig., k tcrit is a continuous function on the duty ratio D for all the values of the coupling coefficient k c from k c =0 (separate inductors) to k c =0.87. This function has a maximum at D=0.5 that decreases with the increasing of k c in the ideal converter case. The parasitic resistances of the circuit cause to move and to decrease the maximum of the function k tcrit with the increasing of k c from zero to 0.8 (Fig. 4). This maximum increases again if k c surpasses 0.8. So, k tcrit.max =0.17 at D=0.5 for k c =0 (separate inductors), k tcrit.max =0.107 at D=0.5 for k c =0.8 and k tcrit.max =0.155 at D=0.7 for k c =0.87. All these results differ from that obtained for an ideal converter. Using the formula given by the Table 1 for a ripple-free input-current boost converter, it finds k tcrit.max =0.148 at D=0.. Hence, if k t is greater than 0.148, then the converter operates in the continuous conduction mode for all D. The simulation results disable this general conclusion. After establishing R crit, the steady-state model of converter with coupled or separate inductors and all the parasitic included and its external characteristics M = f ( R / ) can be computed and plotted for both DCM (R>R crit ) and CCM (R<R crit ) [6. Fig. : Plots of k tcrit as function on the duty ratio and the coupling coefficient of inductors Fig. 4: Plots of k tcrit as function on the duty ratio, the coupling coefficient of inductors and the parasitic resistances of converter 5 Conclusion The order of the equation of the boundary between DCM and CCM, with unknown R crit or k tcrit, is an invariant of the PWM converters. It is two, regardless of the order of converter (two or four), coupled or separate inductors and the including or non-including of the parasitics. Both the parasitic resistances and the inductor coupling change the boundary between the discontinuous and continuous conduction modes of a fourth-order PWM converter. The algorithm carried out for establishing the boundary between DCM and CCM in a fourth-order PWM converter and implemented with MATLAB environment allows studying the effect of the parasitic resistances and the coupling of inductors over the boundary R crit or k tcrit, the steady-state
5 properties and the external characteristics of converter. References [1 S. Cuk and M. D. Middlebrook, A General Unified Approach to Modelling Switching DCto-DC Converters in Discontinuous Conduction Mode, PESC 77 Record, 1977, pp.6-57 [ E. Niculescu and A. Cirstea, Computation of Averaged Characteristic Coefficients for Elementary PWM Converters, COM.P.EL. 98 Proceedings, IEEE Catalog 98TH858, 1998, pp [ S. Cuk, Discontinuous Inductor Current Mode in the optimum topology switching converter, PESC 78 Record, 1978, pp [4 J. Wang, W.G. Dunford and K. Mauch, Analysis of a Ripple-Free Input-Current Boost Converter with Discontinuous Conduction Characteristics, IEEE Transactions on Power Electronics, Vol. 1, No. 4, 1997, pp [5 R. W. Erickson, Fundamentals of Power Electronics, Chapman & Hall (ed.), International Thomson Publishing, 1997 [6 E. Niculescu and E. P. Iancu, Modeling and Analysis of the DC/DC Fourth-Order PWM Converters, COM.P.EL. 000 Proceedings, IEEE Catalog 00TH855, 000, pp [7 E. Niculescu and E. P. Iancu, State Sensitivity of the Fourth-Order PWM Converters, CIFA 000 Proceedings, 000, pp [8 E. Niculescu and E. P. Iancu, Decay Interval of the Inductor Currents in PWM Converters, Annals of the University of Craiova, Year 4, No. 4, 000, pp
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Topic: Uncountability of the Real Numbers Without Decimals
Replies: 110 Last Post: Dec 10, 2013 4:33 AM
Messages: [ Previous | Next ]
gnasher729 Posts: 419 Registered: 10/7/06
Re: Uncountability of the Real Numbers Without Decimals
Posted: Dec 3, 2013 11:53 AM
On Tuesday, December 3, 2013 11:19:51 AM UTC, WM wrote:
> It seems you have not yet understood the main point. To contradict a nonsensical definition does not prove anything: There is no sequence containing "all" rational numbers.
It seems you have some kind of obsession going on.
Zeit Geist's post didn't say anything about any sequences of rational numbers.
And a proof that the rational numbers are countable is really simple.
And if there is no sequence containing "all" rational numbers, as you incorrectly claim, then the real numbers, which are a superset of the rational numbers, are clearly uncountable, as was claimed. So what is your point?
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1. ## Programming the sin function
If I were to write my own sine function, in some programming language, how would I do it?
I know if given an angle, you can simply substitute that into a right triangle with hypotenuse 1 and determine the triangle. But to get to return the result, I keep coming back to using the sine function, which is what I'm trying to program.
So I'm just curious as to how the sine function works to a computer.
Thanks everyone.
2. Originally Posted by lucius
If I were to write my own sine function, in some programming language, how would I do it?
I know if given an angle, you can simply substitute that into a right triangle with hypotenuse 1 and determine the triangle. But to get to return the result, I keep coming back to using the sine function, which is what I'm trying to program.
So I'm just curious as to how the sine function works to a computer.
Thanks everyone.
First because $\sin(x)$ is periodic with period $2\pi$ and has other symmetries we need only calculate it for $x \in [0,\pi/2]$
Then:
$\sin(x)= x(1-x^2/6)+\varepsilon \ \ \ \ \ (1)$
where
$|\varepsilon|\le |x|^5/120\ \ \ \ \ \ \ \ (2)$
Also:
$\sin(x)=2\sin^2(x/2)-1 \ \ \ \ \ \ (3)$
So if $x$ is small enough so that the error in $(2)$ is acceptable then use $(1)$. If the error is too big use $(3)$ and $\sin (x/2^n)$ in $(1)$ with $n$ choosen to give acceptable error.
In practice a better polynomial approximation is used as are neater ways of doing the calculations in general.
3. You can also use a Taylor Series Approximation, which requires some degree of iteration to get a good approximation. The formula for sine is: Summation from n = 0 to infinity of (-1^n)/(2n+1)! * x^(2n+1).
4. Hello, lucius!
. . . So I'm just curious as to how the sine function works to a computer.
I would use the infinite series for $\sin x$, where $x$ is in radians.
. . $\sin x \;=\;x - \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} - \frac{x^9}{9!} - \frac{x^{11}}{11!} +\; \hdots$
5. Originally Posted by macosxnerd101
You can also use a Taylor Series Approximation, which requires some degree of iteration to get a good approximation. The formula for sine is: Summation from n = 0 to infinity of (-1^n)/(2n+1)! * x^(2n+1).
Originally Posted by Soroban
Hello, lucius!
I would use the infinite series for $\sin x$, where $x$ is in radians.
. . $\sin x \;=\;x - \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} - \frac{x^9}{9!} - \frac{x^{11}}{11!} +\; \hdots$
In my earlier post you will note that I used a Taylor/McLaurin series truncated at the $x^3$ term for my small angle approximation, and provided a bound for the error. Almost universally a long Taylor series is never used. The general approach is to use a polynomial approximation for small angles and then the properties of the trig functions to reduce the required angle to something in the valid range of the approximation.
The fact that I used a Taylor approximation is of course misleading, in practice an optimal polynomial for the required angle range is used.
Note in the approximation that I used the maximum error $<10^{-12}$ when $|x|<0.01$ radian. | 890 | 3,180 | {"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": 22, "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} | 4.15625 | 4 | CC-MAIN-2018-09 | latest | en | 0.88601 |
https://blog.hellholestudios.top/archives/141 | 1,653,066,564,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00212.warc.gz | 202,417,014 | 18,278 | # Rougelike Game
You are given a graph of N*M grids. Each grid can be a wall 'X',a path'.' or unknown '?'
One grid is the start point 'S' and one grid is the end point 'E'
Now we change the '?' randomly to 'X' or '.' and run shortest path from S to E. (4 directions walking)
What's the expected length of route from S to E if we ignore the unreachable situation.
If impossible, print -1.
Examples:
3 3
S..
.X.
??E
Output:
3
Explain:
Could be -
S..
.X.
..E 3
S..
.X.
.XE 3
S..
.X.
X.E 3
S..
.X.
XXE 3
Constriants: | 167 | 513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-21 | longest | en | 0.653103 |
https://proofwiki.org/wiki/Row_Operation_to_Clear_First_Column_of_Matrix/Examples/Arbitrary_Matrix_1 | 1,627,171,254,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00149.warc.gz | 476,680,451 | 10,667 | # Row Operation to Clear First Column of Matrix/Examples/Arbitrary Matrix 1
## Example of Use of Row Operation to Clear First Column of Matrix
Let $\mathbf A$ be the matrix:
$\mathbf A = \begin {pmatrix} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$
The matrix $\mathbf R$ corresponding to the row operation to clear the first column is:
$\mathbf R = \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end {pmatrix}$
and the matrix $\mathbf B$ which results from the row operation $\mathbf R$ is:
$\mathbf B = \begin {pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end {pmatrix}$
## Proof
We use Row Operation to Clear First Column of Matrix as follows:
$(1): \quad$ Apply the elementary row operation $r_1 \leftrightarrow r_2$ to exchange row $1$ with row $2$.
From Elementary Matrix corresponding to Elementary Row Operation: Exchange Rows, this is accomplished by pre-multiplying by the matrix:
$\mathbf E_1 := \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A$ to $\mathbf A_1$ as follows:
$\ds \mathbf A_1$ $=$ $\ds \mathbf E_1 \mathbf A$ $\ds$ $=$ $\ds \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix} \begin {pmatrix} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end {pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end {pmatrix}$
$(2): \quad$ Apply the elementary row operation $r_3 \to r_3 + \paren {-1} r_1$ to add $-1$ times row $1$ to row $3$.
From Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add, this is accomplished by pre-multiplying by the matrix:
$\mathbf E_2 := \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end {pmatrix}$
which converts $\mathbf A_1$ to $\mathbf A_2$ as follows:
$\ds \mathbf A_2$ $=$ $\ds \mathbf E_2 \mathbf A_1$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end {pmatrix} \begin {pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end {pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & 1 \end {pmatrix}$
The matrix is in the correct form, and so:
$\mathbf B = \mathbf A_2$
$\Box$
Hence we can calculate $\mathbf R$:
$\ds \mathbf R$ $=$ $\ds \mathbf E_2 \mathbf E_1$ $\ds$ $=$ $\ds \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end {pmatrix} \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end {pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & -1 & 1 \end {pmatrix}$
$\blacksquare$ | 994 | 2,469 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2021-31 | latest | en | 0.491217 |
https://metanumbers.com/54429 | 1,638,496,218,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00021.warc.gz | 454,681,786 | 7,355 | # 54429 (number)
54,429 (fifty-four thousand four hundred twenty-nine) is an odd five-digits composite number following 54428 and preceding 54430. In scientific notation, it is written as 5.4429 × 104. The sum of its digits is 24. It has a total of 2 prime factors and 4 positive divisors. There are 36,284 positive integers (up to 54429) that are relatively prime to 54429.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 24
• Digital Root 6
## Name
Short name 54 thousand 429 fifty-four thousand four hundred twenty-nine
## Notation
Scientific notation 5.4429 × 104 54.429 × 103
## Prime Factorization of 54429
Prime Factorization 3 × 18143
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 54429 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 54,429 is 3 × 18143. Since it has a total of 2 prime factors, 54,429 is a composite number.
## Divisors of 54429
1, 3, 18143, 54429
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 72576 Sum of all the positive divisors of n s(n) 18147 Sum of the proper positive divisors of n A(n) 18144 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 233.3 Returns the nth root of the product of n divisors H(n) 2.99983 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 54,429 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 54,429) is 72,576, the average is 18,144.
## Other Arithmetic Functions (n = 54429)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 36284 Total number of positive integers not greater than n that are coprime to n λ(n) 18142 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5537 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 36,284 positive integers (less than 54,429) that are coprime with 54,429. And there are approximately 5,537 prime numbers less than or equal to 54,429.
## Divisibility of 54429
m n mod m 2 3 4 5 6 7 8 9 1 0 1 4 3 4 5 6
The number 54,429 is divisible by 3.
## Classification of 54429
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (54429)
Base System Value
2 Binary 1101010010011101
3 Ternary 2202122220
4 Quaternary 31102131
5 Quinary 3220204
6 Senary 1055553
8 Octal 152235
10 Decimal 54429
12 Duodecimal 275b9
20 Vigesimal 6g19
36 Base36 15zx
## Basic calculations (n = 54429)
### Multiplication
n×y
n×2 108858 163287 217716 272145
### Division
n÷y
n÷2 27214.5 18143 13607.2 10885.8
### Exponentiation
ny
n2 2962516041 161246785595589 8776501293182313681 477696188886620151343149
### Nth Root
y√n
2√n 233.3 37.8975 15.2742 8.85454
## 54429 as geometric shapes
### Circle
Diameter 108858 341987 9.30702e+09
### Sphere
Volume 6.75429e+14 3.72281e+10 341987
### Square
Length = n
Perimeter 217716 2.96252e+09 76974.2
### Cube
Length = n
Surface area 1.77751e+10 1.61247e+14 94273.8
### Equilateral Triangle
Length = n
Perimeter 163287 1.28281e+09 47136.9
### Triangular Pyramid
Length = n
Surface area 5.13123e+09 1.90031e+13 44441.1
## Cryptographic Hash Functions
md5 d1d07e2687728b478177313c1e95020d c680cf04f1f69679a71b85cd18a5be6a09d1414b 06d67ab1d48ac0fb351317a5bc21a060a3f39a4100365a992178603fa1ada59e 73ebe4d5c80ee12ddbb1b76aec688d3894e35739dd1b2b4cecbb2a86a0e9c43e6a09f53a33304ee697c8ce00b370da61508a58edaf22be296bb7aaf20911b9fc 15fe2e8c71817267e42ed7e58c2f62cd6404cb3d | 1,474 | 4,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-49 | latest | en | 0.813495 |
https://www.convertunits.com/forum/message/94/Re-converting-square-feet-to-acres | 1,596,848,944,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737233.51/warc/CC-MAIN-20200807231820-20200808021820-00348.warc.gz | 615,742,069 | 3,610 | # ››Re: converting square feet to acres (11/05/2007, 1:34 PM)
Use this page first:
http://www.convertunits.com/from/0.219+acres/to/square+feet
0.219 acres = 9539.64 square feet
So if one side is 144 feet, the other side can be computed using this formula:
9539.64 square feet / 144 feet = 66.2475 feet
So the other side is 66.2475 feet in length.
Current Thread... 11/05/2007, 5:10 AM converting square feet to acres 11/05/2007, 1:34 PM Re: converting square feet to acres 11/06/2007, 4:16 AM Re: Re: converting square feet to acres | 179 | 542 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-34 | latest | en | 0.692363 |
https://wisdomessays.com/answers-only-statistics/ | 1,702,013,743,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00576.warc.gz | 703,632,982 | 16,094 | Rate this post
See attachment:
74. Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity
Race/Ethnicity AP Population Overall Student Population Survey Frequency
Asian, Asian American 10.2% 5.4% 113
Black or African-American 8.2% 14.5% 94
Hispanic or Latino 15.5% 15.9% 136
American Indian 0.6% 1.2% 10
White 59.4% 61.6% 604
Not reported/other 6.1% 1.4% 43
102. Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in Table 11.55. Conduct a test for homogeneity at a 5% level of significance
French Toast Pancakes Waffles Omelettes
Men 47 35 28 53
Women 65 59 55 b 60
114. chi-square test statistic = ________ (look below)
116. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the p-value
Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes.
76. The following table shows data on average per capita wine consumption and heart disease rate in a random sample of 10 countries.
Yearly wine consumption in liters 2.5 3.9 2.9 2.4 2.9 0.8 9.1 2.7 0.8 0.7
Death from heart diseases 221 167 131 191 220 297 71 172 211 300
a. Enter the data into your calculator and make a scatter plot.
b. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a.
c. Explain in words what the slope and y-intercept of the regression line tell us.
d. How well does the regression line fit the data? Explain your response.
e. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.
f. Do the data provide convincing evidence that there is a linear relationship between the amount of alcohol consumed and the heart disease death rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question.
82.
Size (ounces) Cost (\$) Cost per ounce
16 3.99
32 4.99
64 5.99
200 10.99
a. Using “size” as the independent variable and “cost” as the dependent variable, draw a scatter plot.
b. Does it appear from inspection that there is a relationship between the variables? Why or why not?
c. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
d. Find the correlation coefficient. Is it significant?
e. If the laundry detergent were sold in a 40-ounce size, find the estimated cost.
f. If the laundry detergent were sold in a 90-ounce size, find the estimated cost.
g. Does it appear that a line is the best way to fit the data? Why or why not?
h. Are there any outliers in the given data?
i. Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would you cost? Why or why not?
j. What is the slope of the least-squares (best-fit) line? Interpret the slope. | 1,008 | 4,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-50 | latest | en | 0.67895 |
https://www.semesprit.com/33885/triangle-congruence-practice-worksheet/ | 1,719,005,726,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00151.warc.gz | 858,310,109 | 18,736 | # Triangle Congruence Practice Worksheet
If you are thinking about learning about the Triangle Congruence, this short tutorial will tell you a bit more about it. The Triangle Congruence is a law of attraction that gives you the power to attract anything and everything into your life.
I know this sounds very “religious” and I do not want to sound like an “anglican priest” but the idea behind the Triangle Congruence is actually very scientific. All things are combinations of a right angle. That is, a very rare and unique alignment of three things to create a more valuable end result. When the right angle is properly aligned, the laws of physics come into play.
Just like our human mind is capable of combining a far more diverse variety of items to form the perfect “triangle,” when we get it right, our mind can do so much more. We have the power to change and transform any given situation. Sometimes the real value of the Triangle Congruence practice worksheet is that it gets us to be in a comfortable mindset that helps us to see the good stuff more clearly.
This is a great solution for anyone who is feeling defeated by a bad situation. It is also a great way to take control of the situation, rather than letting it control you. The Triangle Congruence practice worksheet is all about having a very positive attitude towards the things that surround us, like our family, friends, and even other people.
The great thing about the Triangle Congruence is that it does not only apply to our daily lives. It applies to every part of our existence as well. In fact, if we do not understand this fact we could become enslaved to a negative outlook on our world and everything that surrounds us.
So, if you are currently trapped in a situation where you are under stress, a Triangle Congruence exercise could be just what you need to get out of that poor state you are in. When I say this I am referring to the free space where you are stuck. Let’s put it in simple terms for those of you who are unfamiliar with the concept.
When you have negative energy around you, such as a lack of balance, wealth, and prosperity, and you feel as though the situation will never change for the better, or that there is nothing you can do about it, this is not good for your overall well-being. When this happens, that is when you have a positive attitude towards life and everything it has to offer.
Most people get stuck in a downward spiral, spiraling down into misery, neglect, and possibly even depression. If they are not careful, this spiral will take them down a road of devastation and despair. The best way to eliminate this roadblock is to focus on how wonderful life is right now, and how you can make it better. | 568 | 2,725 | {"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-2024-26 | latest | en | 0.960742 |
https://www.visitmilitarybases.com/interesting/what-time-is-1900-military-time.html | 1,618,164,913,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064898.14/warc/CC-MAIN-20210411174053-20210411204053-00231.warc.gz | 1,201,645,452 | 19,970 | # What time is 1900 military time
7:00 p.m.
11:00 PM
## What are military hours?
Military time is the clock format in which the day lasts from midnight to midnight . At midnight , the hours begin at 00 and go up to 23.
## How do you use military time in a sentence?
Military time typically does not use a colon (:) to separate the hours and minutes; they are just written together. The first two numbers in military time tell you the hour while the last two numbers tell you the minutes. So, if you want to use 10:35 A.M., you would write 1035 and say ten thirty five.
## What is 11 am military time?
9:00 p.m. — 2100 hrs. 10:00 p.m. — 2200 hrs. 11 :00 p.m. — 2300 hrs.
## Is it 2400 or 0000?
If it is the beginning of the day, activity, or event then use 0000 (Pronounced: “Zero Hundred Hours” or “Midnight”). If it is the end of the day, activity or event then use 2400 (Pronounced: “Zero Hundred Hours,” “Twenty Four Hundred Hours,” or “Midnight”). Refer to our military time chart for easy reference.
4:00 pm
## What time is 3 30 pm in military time?
Military time 1530 is: 03: 30 PM using 12-hour clock notation.
## What time is 4 30 pm in military time?
Military Time Conversion Chart
Standard Time Military Time Equivalent Standard Time
4:15 a.m. 04:15 4:15 p.m.
4:30 a.m. 04:30 4:30 p.m.
4:45 a.m. 04:45 4:45 p.m.
5:00 a.m. 05:00 5:00 p.m.
You might be interested: How to become a nurse in the military
## How do I quickly tell military time?
The hours from 1am to noon are the same as in civilian time -keeping. For hours below 10, you just add a zero in front of it. So 9:00am becomes 0900. For a military time that’s 1300 or larger, simply subtract 1200 to get the standard time .
## What is 8 o’clock in military time?
24-hour clock
24-hour clock 12-hour clock
08:00 8 :00 a.m.
09:00 9:00 a.m.
10:00 10:00 a.m.
11:00 11:00 a.m.
## How do you say 2000 hours?
“Two zero hundred” formally, “Twenty hundred” informally. But I would write it as 2000 , not 20:00, so I guess I would be more likely to read the version with the colon as twenty hundred. Never “twenty hundred hours “! Why never twenty hundred hours ?
## How do you say 0015 in military time?
Military Time 0015 is: 12:15 AM using 12-hour clock notation, 00:15 using 24-hour clock notation. See, what time is in the other military time zones at 0015Z (Zulu Time Zone) .
## How do you say 1730 in military time?
Military Time 1730 is: 05:30 PM using 12-hour clock notation, 17:30 using 24-hour clock notation.
## How do you say 1230 in military time?
Making the conversion to military time from 12:30 standard time can seem confusing at first! You are likely here to get a quick answer, then you should know that 12:30 AM military time is equivalent to 0030 military time and 12:30 PM is equivalent to 1230 military time . | 823 | 2,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-17 | longest | en | 0.905505 |
http://www.mathworks.com/matlabcentral/fileexchange/6110-toolbox-fast-marching/content/toolbox_fast_marching/divgrad.m | 1,427,581,823,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131297831.4/warc/CC-MAIN-20150323172137-00210-ip-10-168-14-71.ec2.internal.warc.gz | 657,956,634 | 10,073 | Code covered by the BSD License
# Toolbox Fast Marching
### Gabriel Peyre (view profile)
24 Oct 2004 (Updated )
A toolbox for the computation of the Fast Marching algorithm in 2D and 3D.
%
%
% if M is a 2D array, compute gradient,
% if M is a 3D array, compute divergence.
% Use centered finite differences.
%
% Copyright (c) 2007 Gabriel Peyre
options.null = 0;
if size(M,3)==2
G = mydiv(M,options);
else
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function d = mydiv(g,options)
bound = getoptions(options, 'bound', 'sym');
n = size(g,1);
d = zeros(n);
if strcmp(bound,'sym')
d(2:end-1,:) = d(2:end-1,:) + ( g(3:end,:,1)-g(1:end-2,:,1) )/2;
d(1,:) = d(1,:) + g(2,:,1)-g(1,:,1);
d(end,:) = d(end,:) + g(end,:,1)-g(end-1,:,1);
d(:,2:end-1) = d(:,2:end-1) + ( g(:,3:end,2)-g(:,1:end-2,2) )/2;
d(:,1) = d(:,1) + g(:,2,2)-g(:,1,2);
d(:,end) = d(:,end) + g(:,end,2)-g(:,end-1,2);
else
sel1 = [2:n 1];
sel2 = [n 1:n-1];
d = g(sel1,:,1)-g(sel2,:,1) + g(:,sel1,2)-g(:,sel2,2);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
bound = getoptions(options, 'bound', 'sym');
n = size(M,1);
g = zeros(n,n,2);
if strcmp(bound,'sym')
% on x
g(2:end-1,:,1) = ( M(3:end,:)-M(1:end-2,:) )/2;
g(1,:,1) = M(2,:)-M(1,:);
g(end,:,1) = M(end,:)-M(end-1,:);
% on y
g(:,2:end-1,2) = ( M(:,3:end)-M(:,1:end-2,:) )/2;
g(:,1,1) = M(2,:)-M(1,:);
g(:,end,1) = M(:,end)-M(:,end-1);
else
sel1 = [2:n 1];
sel2 = [n 1:n-1];
g = cat( 3, M(sel1,:)-M(sel2,:), M(:,sel1)-M(:,sel2) )/2;
end | 612 | 1,689 | {"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-2015-14 | longest | en | 0.724573 |
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# Jamie Moffa
##### Last seen: 28 jours ago
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Finding Perfect Squares
Given a vector of numbers, return true if one of the numbers is a square of one of the other numbers. Otherwise return false. E...
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Make a random, non-repeating vector.
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Check if number exists in vector
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Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
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Return the first and last character of a string
Return the first and last character of a string, concatenated together. If there is only one character in the string, the functi...
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Roll the Dice!
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Length of the hypotenuse
Given short sides of lengths a and b, calculate the length c of the hypotenuse of the right-angled triangle. <<http://upload....
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Select every other element of a vector
Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...
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Triangle Numbers
Triangle numbers are the sums of successive integers. So 6 is a triangle number because 6 = 1 + 2 + 3 which can be displa...
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Sum all integers from 1 to 2^n
Given the number x, y must be the summation of all integers from 1 to 2^x. For instance if x=2 then y must be 1+2+3+4=10.
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Return area of square
Side of square=input=a Area=output=b
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Maximum value in a matrix
Find the maximum value in the given matrix. For example, if A = [1 2 3; 4 7 8; 0 9 1]; then the answer is 9.
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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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Solved | 711 | 2,562 | {"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} | 3.53125 | 4 | CC-MAIN-2021-04 | latest | en | 0.722505 |
http://openstudy.com/updates/50ec566ee4b07cd2b648eaf0 | 1,448,631,721,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398449160.83/warc/CC-MAIN-20151124205409-00176-ip-10-71-132-137.ec2.internal.warc.gz | 172,113,404 | 10,428 | ## ms.hi 2 years ago 108 is 36% of what number? Write and solve a proportion to solve the problem
1. Lilmoney
108 is 36 of what number. 108 divided by 0.36 = 300
2. ms.hi
thanks :)
3. Lilmoney
Welcome
4. mathstudent55
100% corresponds to the number 36% corresponds to a part which is 108 100% is to x as 36% is to 108 100/x = 36/108 That's the proportion. Now cross multiply and solve for x
5. ms.hi
360?
6. Lilmoney
Wait what.
7. ms.hi
?
8. mathstudent55
100/x = 36/108 Cross multiply: 36x = 100(108) 36x = 18,800 x = 300 Which is what Lilmoney got above.
9. ms.hi
thanks :
10. mathstudent55
You're welcome | 216 | 627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2015-48 | longest | en | 0.886956 |
https://blogs.solidworks.com/teacher/author/mgruber | 1,721,867,157,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00395.warc.gz | 118,577,868 | 18,965 | ## Aero Design Series – Airplane Stability Playlist
Seven videos are up on airplane stability! Ever wonder why aiplanes have stabilizers on the aft of the tail? How do they work? Why are they there? Where does the center of gravity of the aircraft relative to the wings
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## Aero Design Series – Airplane Stability – Part 1: Introduction
In this 1st video in a series on airplane stability, we introduce the concepts that we’ll cover. We continue where we left off in the series on airfoils. Using SolidWorks for model demonstrations, we get a feel for the forces
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## Aero Design Series – Airfoils – Part 4: Physical Interpretation of Coefficient of Moment
In this 4th video in a series of 4 about airfoils, we search for a physical understanding and interpretation of the coefficient of moment “Cm” of airfoils. This parameter is often confusing to grasp at first, but a physical understanding
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## Aero Design Series – Airfoils – Part 3: Interpreting Cl and Cd
In the first 2 videos about airfoils we have imported airfoil curves into SolidWorks from airfoiltools.com. In the 3rd and 4th videos in the series of 4 we delve into physical interpretations of the performance parameters of airfoils. In this
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## Aero Design Series – Airfoils – Part 2: Importing Airfoils Into Solidworks
In this 2nd video in a series of 4 about airfoils, we cover various methods to import airfoils into SolidWorks in order to design the parts and assemblies of your aircraft.
... Continued
## Aero Design Series – Airfoils – Part 1: Introduction – What is an Airfoil?
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... Continued
## Aero Design Series- Stock Wheel – Part 9: Relative References
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## Aero Design Series – Stock Wheel – Part 5: Fillets and Re-Do Rim Wall
In this tutorial, we will learn to focus on the fillets of the wheel as well as going back up the tree and fixing previous mistakes, for example in the rim wall.
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## Aero Design Series – Stock Wheel – Part 4: Right Plane Cutouts
In this Aero Design Video Series tutorial, we will learn how to do right plane cutouts that form the spokes of the wheel.
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## Aero Design Series – Stock Wheel – Part 3: Saving and Renaming
Welcome to the first in a line of Aero Design Video Series created by Matthew Gruber in collaboration with SOLIDWORKS and SolidXPerts! This first series is for beginners using SOLIDWORKS for the SAE Aero Design and Design/Build/Fly competitions. In this tutorial, we will learn focus on saving and renaming parts.
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## Aero Design Series- Stock Wheel – Part 2: Front Plane Sketch and Revolve
Welcome to the #Aero Design Video Series created by Matthew Gruber! This is the second tutorial in the 10 part series where we will go from the initial mockup to a detailed front plane sketch and revolve feature.
... Continued | 726 | 3,175 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.89793 |
https://us.metamath.org/mpeuni/cardlim.html | 1,726,052,890,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00722.warc.gz | 550,831,756 | 7,804 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > cardlim Structured version Visualization version GIF version
Theorem cardlim 9434
Description: An infinite cardinal is a limit ordinal. Equivalent to Exercise 4 of [TakeutiZaring] p. 91. (Contributed by Mario Carneiro, 13-Jan-2013.)
Assertion
Ref Expression
cardlim (ω ⊆ (card‘𝐴) ↔ Lim (card‘𝐴))
Proof of Theorem cardlim
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 sseq2 3918 . . . . . . . . . . 11 ((card‘𝐴) = suc 𝑥 → (ω ⊆ (card‘𝐴) ↔ ω ⊆ suc 𝑥))
21biimpd 232 . . . . . . . . . 10 ((card‘𝐴) = suc 𝑥 → (ω ⊆ (card‘𝐴) → ω ⊆ suc 𝑥))
3 limom 7594 . . . . . . . . . . . 12 Lim ω
4 limsssuc 7564 . . . . . . . . . . . 12 (Lim ω → (ω ⊆ 𝑥 ↔ ω ⊆ suc 𝑥))
53, 4ax-mp 5 . . . . . . . . . . 11 (ω ⊆ 𝑥 ↔ ω ⊆ suc 𝑥)
6 infensuc 8717 . . . . . . . . . . . 12 ((𝑥 ∈ On ∧ ω ⊆ 𝑥) → 𝑥 ≈ suc 𝑥)
76ex 416 . . . . . . . . . . 11 (𝑥 ∈ On → (ω ⊆ 𝑥𝑥 ≈ suc 𝑥))
85, 7syl5bir 246 . . . . . . . . . 10 (𝑥 ∈ On → (ω ⊆ suc 𝑥𝑥 ≈ suc 𝑥))
92, 8sylan9r 512 . . . . . . . . 9 ((𝑥 ∈ On ∧ (card‘𝐴) = suc 𝑥) → (ω ⊆ (card‘𝐴) → 𝑥 ≈ suc 𝑥))
10 breq2 5036 . . . . . . . . . 10 ((card‘𝐴) = suc 𝑥 → (𝑥 ≈ (card‘𝐴) ↔ 𝑥 ≈ suc 𝑥))
1110adantl 485 . . . . . . . . 9 ((𝑥 ∈ On ∧ (card‘𝐴) = suc 𝑥) → (𝑥 ≈ (card‘𝐴) ↔ 𝑥 ≈ suc 𝑥))
129, 11sylibrd 262 . . . . . . . 8 ((𝑥 ∈ On ∧ (card‘𝐴) = suc 𝑥) → (ω ⊆ (card‘𝐴) → 𝑥 ≈ (card‘𝐴)))
1312ex 416 . . . . . . 7 (𝑥 ∈ On → ((card‘𝐴) = suc 𝑥 → (ω ⊆ (card‘𝐴) → 𝑥 ≈ (card‘𝐴))))
1413com3r 87 . . . . . 6 (ω ⊆ (card‘𝐴) → (𝑥 ∈ On → ((card‘𝐴) = suc 𝑥𝑥 ≈ (card‘𝐴))))
1514imp 410 . . . . 5 ((ω ⊆ (card‘𝐴) ∧ 𝑥 ∈ On) → ((card‘𝐴) = suc 𝑥𝑥 ≈ (card‘𝐴)))
16 vex 3413 . . . . . . . . . 10 𝑥 ∈ V
1716sucid 6248 . . . . . . . . 9 𝑥 ∈ suc 𝑥
18 eleq2 2840 . . . . . . . . 9 ((card‘𝐴) = suc 𝑥 → (𝑥 ∈ (card‘𝐴) ↔ 𝑥 ∈ suc 𝑥))
1917, 18mpbiri 261 . . . . . . . 8 ((card‘𝐴) = suc 𝑥𝑥 ∈ (card‘𝐴))
20 cardidm 9421 . . . . . . . 8 (card‘(card‘𝐴)) = (card‘𝐴)
2119, 20eleqtrrdi 2863 . . . . . . 7 ((card‘𝐴) = suc 𝑥𝑥 ∈ (card‘(card‘𝐴)))
22 cardne 9427 . . . . . . 7 (𝑥 ∈ (card‘(card‘𝐴)) → ¬ 𝑥 ≈ (card‘𝐴))
2321, 22syl 17 . . . . . 6 ((card‘𝐴) = suc 𝑥 → ¬ 𝑥 ≈ (card‘𝐴))
2423a1i 11 . . . . 5 ((ω ⊆ (card‘𝐴) ∧ 𝑥 ∈ On) → ((card‘𝐴) = suc 𝑥 → ¬ 𝑥 ≈ (card‘𝐴)))
2515, 24pm2.65d 199 . . . 4 ((ω ⊆ (card‘𝐴) ∧ 𝑥 ∈ On) → ¬ (card‘𝐴) = suc 𝑥)
2625nrexdv 3194 . . 3 (ω ⊆ (card‘𝐴) → ¬ ∃𝑥 ∈ On (card‘𝐴) = suc 𝑥)
27 peano1 7600 . . . . . 6 ∅ ∈ ω
28 ssel 3885 . . . . . 6 (ω ⊆ (card‘𝐴) → (∅ ∈ ω → ∅ ∈ (card‘𝐴)))
2927, 28mpi 20 . . . . 5 (ω ⊆ (card‘𝐴) → ∅ ∈ (card‘𝐴))
30 n0i 4232 . . . . 5 (∅ ∈ (card‘𝐴) → ¬ (card‘𝐴) = ∅)
31 cardon 9406 . . . . . . . . 9 (card‘𝐴) ∈ On
3231onordi 6274 . . . . . . . 8 Ord (card‘𝐴)
33 ordzsl 7559 . . . . . . . 8 (Ord (card‘𝐴) ↔ ((card‘𝐴) = ∅ ∨ ∃𝑥 ∈ On (card‘𝐴) = suc 𝑥 ∨ Lim (card‘𝐴)))
3432, 33mpbi 233 . . . . . . 7 ((card‘𝐴) = ∅ ∨ ∃𝑥 ∈ On (card‘𝐴) = suc 𝑥 ∨ Lim (card‘𝐴))
35 3orass 1087 . . . . . . 7 (((card‘𝐴) = ∅ ∨ ∃𝑥 ∈ On (card‘𝐴) = suc 𝑥 ∨ Lim (card‘𝐴)) ↔ ((card‘𝐴) = ∅ ∨ (∃𝑥 ∈ On (card‘𝐴) = suc 𝑥 ∨ Lim (card‘𝐴))))
3634, 35mpbi 233 . . . . . 6 ((card‘𝐴) = ∅ ∨ (∃𝑥 ∈ On (card‘𝐴) = suc 𝑥 ∨ Lim (card‘𝐴)))
3736ori 858 . . . . 5 (¬ (card‘𝐴) = ∅ → (∃𝑥 ∈ On (card‘𝐴) = suc 𝑥 ∨ Lim (card‘𝐴)))
3829, 30, 373syl 18 . . . 4 (ω ⊆ (card‘𝐴) → (∃𝑥 ∈ On (card‘𝐴) = suc 𝑥 ∨ Lim (card‘𝐴)))
3938ord 861 . . 3 (ω ⊆ (card‘𝐴) → (¬ ∃𝑥 ∈ On (card‘𝐴) = suc 𝑥 → Lim (card‘𝐴)))
4026, 39mpd 15 . 2 (ω ⊆ (card‘𝐴) → Lim (card‘𝐴))
41 limomss 7584 . 2 (Lim (card‘𝐴) → ω ⊆ (card‘𝐴))
4240, 41impbii 212 1 (ω ⊆ (card‘𝐴) ↔ Lim (card‘𝐴))
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 → wi 4 ↔ wb 209 ∧ wa 399 ∨ wo 844 ∨ w3o 1083 = wceq 1538 ∈ wcel 2111 ∃wrex 3071 ⊆ wss 3858 ∅c0 4225 class class class wbr 5032 Ord word 6168 Oncon0 6169 Lim wlim 6170 suc csuc 6171 ‘cfv 6335 ωcom 7579 ≈ cen 8524 cardccrd 9397 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2729 ax-sep 5169 ax-nul 5176 ax-pow 5234 ax-pr 5298 ax-un 7459 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2557 df-eu 2588 df-clab 2736 df-cleq 2750 df-clel 2830 df-nfc 2901 df-ne 2952 df-ral 3075 df-rex 3076 df-rab 3079 df-v 3411 df-sbc 3697 df-csb 3806 df-dif 3861 df-un 3863 df-in 3865 df-ss 3875 df-pss 3877 df-nul 4226 df-if 4421 df-pw 4496 df-sn 4523 df-pr 4525 df-tp 4527 df-op 4529 df-uni 4799 df-int 4839 df-br 5033 df-opab 5095 df-mpt 5113 df-tr 5139 df-id 5430 df-eprel 5435 df-po 5443 df-so 5444 df-fr 5483 df-we 5485 df-xp 5530 df-rel 5531 df-cnv 5532 df-co 5533 df-dm 5534 df-rn 5535 df-res 5536 df-ima 5537 df-ord 6172 df-on 6173 df-lim 6174 df-suc 6175 df-iota 6294 df-fun 6337 df-fn 6338 df-f 6339 df-f1 6340 df-fo 6341 df-f1o 6342 df-fv 6343 df-om 7580 df-er 8299 df-en 8528 df-dom 8529 df-card 9401 This theorem is referenced by: infxpenlem 9473 alephislim 9543 cflim2 9723 winalim 10155 gruina 10278
Copyright terms: Public domain W3C validator | 2,997 | 5,187 | {"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} | 3.484375 | 3 | CC-MAIN-2024-38 | latest | en | 0.242325 |
https://ubraintv-jp.com/dividing-a-trinomial-by-a-binomial/ | 1,638,262,901,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00639.warc.gz | 675,379,561 | 5,570 | act Math assist » Algebra » Variables » Polynomials » Trinomials » exactly how to divide trinomials
You are watching: Dividing a trinomial by a binomial
Simplify the expression:
Explanation:
Once simplified, (x+1) appears on both the numerator and also denominator, an interpretation we can cancel out both of them.
Which offers us:
Explanation:
can be divided using lengthy division.
The set up would certainly look very comparable to the division of actual numbers, together as when we desire to divide 10 by 2 and the price is 5.
The first step after setting up the "division house" is to watch what the very first term in the outer trinomial requirements to multiply by to match the
in the house. In this case, it"s
will it is in multiplied throughout the other two state in the outer trinomial and the product will be subtracted indigenous the expression inside the department house. The adhering to steps will certainly take ar in the exact same way.
While we could proceed to divide, the would need the use of portion exponents that would make the answer much more complicated. Therefore, the hatchet in red will certainly be the remainder. Due to the fact that this remainder is still topic to it is in being separated by the trinomial external of the department house, we will make the remainder component of the last answer by creating it in fraction form:
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your Infringement notice may it is in forwarded to the party that made the content accessible or to 3rd parties such together ChillingEffects.org. | 943 | 4,406 | {"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-2021-49 | latest | en | 0.914796 |
https://electronics.stackexchange.com/questions/153911/single-supply-op-amp-audio-amplifier/153916#153916 | 1,642,592,860,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301309.22/warc/CC-MAIN-20220119094810-20220119124810-00237.warc.gz | 298,676,925 | 38,378 | # Single supply op-amp audio amplifier
I am trying to create an op-amp amplifier that would work from single 5V supply, and would be able to amplify -100mV to +100mV audio signal to around a 1V peak-peak or so. I've came across this circuit from this article, that could seem to work, but am having trouble calculating the actual values:
simulate this circuit – Schematic created using CircuitLab
From the article I read that R1 and R2 should both be the same and around 42kOhm for 5V power supply. R4 should be R3+(0.5*R1) and thats about it...
So how would I go about actually calculating the capacitor, resistor values needed for a varying frequency signal with maximum frequency at around 20kHz and gain of about 5?
Thank you for helping me!
EDIT:
In the article the author wrote by the ground symbol: "*STAR GROUND". Is it really imporant that I combine all ground trances in the schematic to one point, or can I use a ground plane across the whole circuit?
• What is the load on the op-amp output? Feb 12 '15 at 19:12
• @Andyaka I am not quite sure, this will get connected to an actual audio amplifier. Is there any way I could measure the input impedence of the amplifier? Feb 12 '15 at 19:16
• If it's going to an audio amp it'll be fine. Just wanted to make sure you weren't driving a speaker or headphones. Star grounds are probably best for audio rather than groundplanes but a combo of both (providing you know what you are doing) is better. Feb 12 '15 at 19:27
• @Andyaka Haven't really read that much about this subject. I guess I'll read up some articles and go with star grounding. Thanks! Feb 12 '15 at 19:42
• Good going. You've effectively created a "virtual earth" at the junction of R1, R2, R3, C2, getting around the single supply rail problem. It may be best to connect the "grounded" end of C3 to that point rather than the power supply ground, to prevent noise in the power supply being injected there. Feb 13 '15 at 11:43
You seemed to have actually found a reasonable circuit on the internet. I heard there was out there somewhere.
The equations you cite are overly strict. Instead of just telling you the values, it's better to explain what each part does.
R1 and R2 are a voltage divider to make 1/2 the supply voltage. This will be the DC bias the opamp will operate at. C2 low pass filters the output of that voltage divider. This is to squash glitches, power supply ripple, and other noise on the 5 V supply so they don't end up in your signal. R3 is needed only because C2 is there. If R3 weren't there, C2 would squash your input signal too, not just the noise on the power supply. Ultimately, the right end of R3 is intended to deliver a clean 1/2 supply signal with high impedance. The high impedance is so that it doesn't interfere with your desired signal coming thru C1.
C1 is a DC blocking cap. It decouples the DC level at IN from the DC level the opamp is biased at.
R4 and R5 form a voltage divider from the output back to the negative input. This is the negative feedback path, and the overall circuit gain is the inverse of the voltage divider gain. You want a gain of 10, so the R4-R5 divider should have a gain of 1/10. C3 blocks DC so that the divider only works on your AC signal, not the DC bias point. The divider will pass all DC, so the DC gain from the + input of the opamp to its output will be 1.
C4 is another DC blocking cap, this time decoupling the opamp DC bias level from the output. With the two DC blocking caps (C1, C4), the overall amplifier works on AC and whatever DC biases may be at IN and OUT are irrelevant (within the voltage rating of C1 and C4).
Now for some values. The MCP6022 is a CMOS input opamp, so it has very high input impedance. Even a MΩ is small compared to its input impedance. The other thing to consider is the range of frequencies you want this amplifier to work over. You said the signal is audio, so we'll assume anything below 20 Hz or above 20 kHz is signal you don't care about. In fact, it's a good idea to squash unwanted frequencies.
R1 and R2 only need to be equal to make 1/2 the supply voltage. You mention no special requirement, like battery operation where minimizing current is of high importance. Given that, I'd make R1 and R2 10 kΩ each, although there is large leeway here. If this were battery operated, I'd probably make them 100 kΩ each and not feel bad about it. With R1 and R2 10 kΩ, the output impedance of the divider is 5 kΩ. You don't really want any relevant signal on the output of that divider, so let's start by seeing what capacitance is needed to filter down to 20 Hz. 1.6 µF. The common value of 2 µF would be fine. Higher works too, except that if you go too high, the startup time becomes significant on a human scale. For example, 10 µF would work to filter noise nicely. It has a 500 ms time constant with the 5 kΩ impedance, so would take a few seconds to stabilize after being turned on.
R3 should be larger than the output of R1-R2, which is 5 kΩ. I'd pick a few 100 kΩ at least. The input impedance of the opamp is high, so lets use 1 MΩ.
C1 with R3 form a high pass filter that needs to pass at least 20 Hz. The impedance seen looking into the right end of R3 is a bit over 1 MΩ. 20 Hz with 1 MΩ requires 8 nF, so 10 nF it is. This is a place you don't want to use a ceramic cap, so lower values are quite useful. A mylar cap, for example, would be good here and 10 nF is within the available range.
Again, the overall impedance of the R4-R5 divider doesn't matter much, so lets arbitrarily set R4 to 100 kΩ and work out the other values from there. R5 must be R4/9 for a overall amplifier gain of 10, so 11 kΩ works out. C3 and R5 form a filter that has to roll off at 20 Hz or below. C3 must be 720 nF or more, so 1 µF.
Note one issue with this topology. Frequency-wise, C3 is acting with R5, but the DC level that C3 will eventually stabilize at is filtered by R4+R5 and C3. That is a filter at 1.4 Hz, which means this circuit will take a few seconds to stabilize after power is applied.
C4 forms a high pass filter with whatever impedance will be connected to OUT. Since you may not know, you want to make it reasonably large. Let's pick 10 µF since that's readily available. That rolls off at 20 Hz with 8 kΩ. This amp will therefore function as specified as long as OUT is not loaded with less than 8 kΩ.
• Wow, I've learned so much from reading your response, thanks! But I still have some questions: Why does R3 has to be larger that R1-R2? And why is a ceramic capacitor not OK to use as input decoupling ? Feb 12 '15 at 18:41
• "You seemed to have actually found a reasonable circuit on the internet." :) Good one. Feb 12 '15 at 18:53
• @Gola: A resistor divider isn't perfect. It's output will change depending on how it is loaded. By making R3 large compared to the parallel combination of R1 and R2, loading effects are minimized. Note that this is really to keep the frequency response flat of the desired 20 Hz to 20 kHz range. Due to C1, the divider can't be loaded with DC. Feb 12 '15 at 19:53
• If you wanted unipolar output (let's say, to drive an ADC that expected 0-5V input) would you just leave off C4? Feb 12 '15 at 22:43
• @Russ: Yes. --- Feb 12 '15 at 22:49
There is no need to make R4 that particular value since this is a CMOS op-amp (no matching of input bias currents).
Capacitor values are determined by the desired lower corner frequency. C = $\frac{1}{2\pi f_c R}$
So if fc = 20Hz, and R1, R2 = 39K. Let's arbitrarily make R3 100K. Then C = 100nF is about right.
C2 depends on what's on the power supply that you're trying to attenuate, but let's say 1uF for that.
Let's pick R4 = 100K just to keep two resistors the same. R5 will then be 11K for a gain of +10.1
Finally, C3 can be calculated from R5 to be about 1uF (using the above equation).
That's it! | 2,062 | 7,854 | {"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} | 2.921875 | 3 | CC-MAIN-2022-05 | latest | en | 0.958448 |
https://web2.0calc.com/questions/help-pls_10089 | 1,717,042,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059418.31/warc/CC-MAIN-20240530021529-20240530051529-00143.warc.gz | 514,851,079 | 5,522 | +0
# Help pls!
0
94
3
A spinner has 10 equally sized sections, 8 of which are gray and 2 of which are blue. The spinner is spun twice. What is the probability that the first spin lands on blue and the second spin lands on gray ?
Apr 14, 2023
#1
0
The probability is 2/10*8/9 = 8/45.
Apr 14, 2023
#2
0
Follow this reasoning through, and presumeably after ten spins the spinner will not stop, unable to land on either grey or blue.
Guest Apr 14, 2023
#3
0
uhh it said it was wrong but Ty ig
Guest Apr 14, 2023 | 166 | 518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-22 | latest | en | 0.941292 |
https://socratic.org/questions/how-do-you-use-the-binomial-series-to-expand-1-1-2x-2 | 1,632,529,484,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057584.91/warc/CC-MAIN-20210924231621-20210925021621-00366.warc.gz | 557,754,493 | 5,865 | # How do you use the binomial series to expand 1 / (1 + 2x)^2?
Oct 23, 2016
the binomial expansion is $1 - 4 x + 12 {x}^{2} - 32 {x}^{3} + 80 {x}^{4} - 192 {x}^{5} + \ldots .$
#### Explanation:
We must write in the form${\left(1 + b\right)}^{n}$
The development of ${\left(1 + b\right)}^{n} = 1 + \frac{n}{1} \left(b\right) + \frac{n \left(n - 1\right)}{1 \cdot 2} {b}^{2} \cdot \ldots \ldots$
n can be positive or negative
So we apply this and we get
$\frac{1}{1 + 2 x} ^ 2 = {\left(1 + 2 x\right)}^{- 2} = 1 + \frac{- 2}{1} \left(2 x\right) + \frac{- 2 \cdot - 3}{1 \cdot 2} {\left(2 x\right)}^{2} + \frac{- 2 \cdot - 3 \cdot - 4}{1 \cdot 2 \cdot 3} {\left(2 x\right)}^{3} + \frac{- 2 \cdot - 3 \cdot - 4 \cdot - 5}{1 \cdot 2 \cdot 3 \cdot 4} {\left(2 x\right)}^{4} + \frac{- 2 \cdot - 3 \cdot - 4 \cdot - 5 \cdot - 6}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} {\left(2 x\right)}^{5} + \ldots$
$= 1 - 4 x + 12 {x}^{2} - 32 {x}^{3} + 80 {x}^{4} - 192 {x}^{5} + \ldots \ldots .$ | 467 | 977 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 4.71875 | 5 | CC-MAIN-2021-39 | latest | en | 0.216043 |
www.fiat-squadra.pl | 1,638,605,759,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362952.24/warc/CC-MAIN-20211204063651-20211204093651-00173.warc.gz | 767,500,761 | 9,449 | The goods are equal to the sample two piece fitting 1 1 2
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The p-value is found by (P(X^2>22.152)) with degrees of freedom =((2-1)(3-1) = 2). Minitab calculates this p-value to be less than 0.001 and reports it as 0.000. Given this p-value of 0.000 is less than the alpha of 0.05, we reject the null hypothesis that political affiliation and their opinion on … | 2,880 | 10,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-49 | latest | en | 0.883208 |
http://scipy.github.io/devdocs/generated/scipy.ndimage.percentile_filter.html | 1,579,819,751,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00158.warc.gz | 146,565,799 | 4,224 | # scipy.ndimage.percentile_filter¶
scipy.ndimage.percentile_filter(input, percentile, size=None, footprint=None, output=None, mode='reflect', cval=0.0, origin=0)[source]
Calculate a multidimensional percentile filter.
Parameters
inputarray_like
The input array.
percentilescalar
The percentile parameter may be less then zero, i.e., percentile = -20 equals percentile = 80
sizescalar or tuple, optional
See footprint, below. Ignored if footprint is given.
footprintarray, optional
Either size or footprint must be defined. size gives the shape that is taken from the input array, at every element position, to define the input to the filter function. footprint is a boolean array that specifies (implicitly) a shape, but also which of the elements within this shape will get passed to the filter function. Thus size=(n,m) is equivalent to footprint=np.ones((n,m)). We adjust size to the number of dimensions of the input array, so that, if the input array is shape (10,10,10), and size is 2, then the actual size used is (2,2,2). When footprint is given, size is ignored.
outputarray or dtype, optional
The array in which to place the output, or the dtype of the returned array. By default an array of the same dtype as input will be created.
modestr or sequence, optional
The mode parameter determines how the input array is extended when the filter overlaps a border. By passing a sequence of modes with length equal to the number of dimensions of the input array, different modes can be specified along each axis. Default value is ‘reflect’. The valid values and their behavior is as follows:
‘reflect’ (d c b a | a b c d | d c b a)
The input is extended by reflecting about the edge of the last pixel.
‘constant’ (k k k k | a b c d | k k k k)
The input is extended by filling all values beyond the edge with the same constant value, defined by the cval parameter.
‘nearest’ (a a a a | a b c d | d d d d)
The input is extended by replicating the last pixel.
‘mirror’ (d c b | a b c d | c b a)
The input is extended by reflecting about the center of the last pixel.
‘wrap’ (a b c d | a b c d | a b c d)
The input is extended by wrapping around to the opposite edge.
cvalscalar, optional
Value to fill past edges of input if mode is ‘constant’. Default is 0.0.
originint or sequence, optional
Controls the placement of the filter on the input array’s pixels. A value of 0 (the default) centers the filter over the pixel, with positive values shifting the filter to the left, and negative ones to the right. By passing a sequence of origins with length equal to the number of dimensions of the input array, different shifts can be specified along each axis.
Returns
percentile_filterndarray
Filtered array. Has the same shape as input.
Examples
>>> from scipy import ndimage, misc
>>> import matplotlib.pyplot as plt
>>> fig = plt.figure()
>>> plt.gray() # show the filtered result in grayscale
>>> ax1 = fig.add_subplot(121) # left side
>>> ax2 = fig.add_subplot(122) # right side
>>> ascent = misc.ascent()
>>> result = ndimage.percentile_filter(ascent, percentile=20, size=20)
>>> ax1.imshow(ascent)
>>> ax2.imshow(result)
>>> plt.show()
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# Find max absolute of two numbers and retain it's original sign.
Experts,
I am trying to find the absolute highest value of two columns and store it in the third column. The only catch here is the value stored in the third column should retain it's original sign.
For example, for the values -0.1430000060 and -0.1410000030 when I run =MAX(ABS(A1:B1)), it returns 0.1430000060 without the - sign. Any suggestions?
Sample.xlsx
0
sukhoi35
• 2
• 2
1 Solution
Commented:
Are the numbers always negative as in your example? If so, you're not looking for the highest, but the lowest and you can just use the MIN function.
=MIN(A1:B1)
0
Author Commented:
The numbers can be either positive or negative. Sorry, the sample I chose contains only negative, it is a pretty huge file running into several hundred thousands of rows.
0
Commented:
OK, as you're only comparing 2 columns, you can use a simple IF statement to compare the absolute values...
=IF(ABS(A1)>ABS(B1), A1, B1)
0
Author Commented:
Thank you, Wayne!
0
Question has a verified solution.
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Have a better answer? Share it in a comment. | 327 | 1,264 | {"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-2018-26 | longest | en | 0.900013 |
https://www.lmfdb.org/Character/Dirichlet/25/c?number_field= | 1,702,013,198,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00841.warc.gz | 948,831,582 | 5,545 | # Properties
Label 25.c Modulus $25$ Conductor $5$ Order $4$ Real no Primitive no Minimal no Parity odd
# Related objects
Show commands: PariGP / SageMath
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(25, base_ring=CyclotomicField(4))
M = H._module
chi = DirichletCharacter(H, M([1]))
chi.galois_orbit()
[g,chi] = znchar(Mod(7,25))
order = charorder(g,chi)
[ charpow(g,chi, k % order) | k <-[1..order-1], gcd(k,order)==1 ]
## Basic properties
Modulus: $$25$$ Conductor: $$5$$ sage: chi.conductor() pari: znconreyconductor(g,chi) Order: $$4$$ sage: chi.multiplicative_order() pari: charorder(g,chi) Real: no Primitive: no, induced from 5.c sage: chi.is_primitive() pari: #znconreyconductor(g,chi)==1 Minimal: no Parity: odd sage: chi.is_odd() pari: zncharisodd(g,chi)
## Related number fields
Field of values: $$\Q(\sqrt{-1})$$ Fixed field: $$\Q(\zeta_{5})$$
## Characters in Galois orbit
Character $$-1$$ $$1$$ $$2$$ $$3$$ $$4$$ $$6$$ $$7$$ $$8$$ $$9$$ $$11$$ $$12$$ $$13$$
$$\chi_{25}(7,\cdot)$$ $$-1$$ $$1$$ $$i$$ $$-i$$ $$-1$$ $$1$$ $$i$$ $$-i$$ $$-1$$ $$1$$ $$i$$ $$-i$$
$$\chi_{25}(18,\cdot)$$ $$-1$$ $$1$$ $$-i$$ $$i$$ $$-1$$ $$1$$ $$-i$$ $$i$$ $$-1$$ $$1$$ $$-i$$ $$i$$ | 457 | 1,226 | {"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} | 2.9375 | 3 | CC-MAIN-2023-50 | latest | en | 0.28899 |
http://algebra-test.com/algebra-help/powers/ti-83-plus-graphing-calculator.html | 1,721,764,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00003.warc.gz | 1,651,102 | 5,104 | Do more than just homework: Do the Math! Algebra Master shows you all the steps to solving every problem – from Pre-Algebra straight through to Advanced College level.
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More Testimonials >> | 811 | 3,441 | {"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-2024-30 | latest | en | 0.922039 |
https://stackoverflow.com/questions/58964734/implement-simple-quiescence-search-algorithm | 1,580,220,676,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251778272.69/warc/CC-MAIN-20200128122813-20200128152813-00530.warc.gz | 666,320,373 | 28,532 | # Implement simple quiescence search algorithm
I'm trying to implement a simple quiescence search for my negamax AI (not chess - 10-10-5 mnk or gomoku style game called Take 5). It already has a transposition table. The main negamax algorithm looks like this:
``````def negamax(board, depth, alpha, beta, player, ttable):
if board in ttable:
value, vtype = ttable[board]
if vtype == 0:
return value
elif vtype == 1:
beta = min(alpha, value)
else:
alpha = max(beta, value)
score = board.getValue()
if abs(score) == 1000000000000:
return player * score
if depth == 0:
return player * score
oldAlpha = alpha
possibleMoves = board.getBestNextMoves()
value = -1000000000000
for move in possibleMoves:
#child = board#Board(board)
board.place(move, player)
value = max(value, -negamax(board, depth-1, -beta, -alpha,
-player, ttable))
board.place(move, 0)
alpha = max(alpha, value)
if alpha >= beta:
break # alpha beta prune
if value <= oldAlpha:
vtype = 1
elif value >= beta:
vtype = -1
else:
vtype = 0
ttable[board] = (value, vtype)
return value
``````
I've looked online for some quiescence search algorithms, but they seem to not work properly, but this is what I've come up with:
``````def quiesce(board, alpha, beta, player, ttable):
if board in ttable:
value, vtype = ttable[board]
if vtype == 0:
score = value
else:
score = player * board.getValue()
ttable[board] = (score, 0)
else:
score = board.getValue()
ttable[board] = (score, 0)
if score >= beta:
return score
if alpha < score:
alpha = score
for move in board.getForcedMoves(player):
board.place(move, player)
score = -quiesce(board, -beta, -alpha, -player, ttable)
board.place(move, 0)
if score >= beta:
return score
if score > alpha:
alpha = score
return alpha
``````
Can you check that I have implemented properly a quiescence search for this algorithm?
• accidentally posted the question before I was done making it, it is ready now – micsthepick Nov 20 '19 at 23:00
• It seems you know the algorithm and you implement it by yourself. So can not you run it on your laptop to see if it works or not ? – codrelphi Nov 20 '19 at 23:13
• @codrelphi it appears to work, however that does not mean that it works 100% correctly – micsthepick Nov 20 '19 at 23:15
• for example, min-max may try to minimize the score for the wrong player, but if when you search through and pick the move with the best evaluation, it will appear to play good moves – micsthepick Nov 20 '19 at 23:16
• Ok. It is better to post your code in the code review section: codereview.stackexchange.com – codrelphi Nov 20 '19 at 23:29 | 738 | 2,564 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-05 | latest | en | 0.729562 |
https://svn.rot13.org/cgi-bin/viewvc.cgi/webpac/branches/ffzg/doc/exact_match.txt?sortby=date&r1=303&r2=289&pathrev=360 | 1,669,886,591,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710808.72/warc/CC-MAIN-20221201085558-20221201115558-00300.warc.gz | 598,575,197 | 3,005 | This is repository of my old source code which isn't updated any more. Go to git.rot13.org for current projects!
# Diff of /branches/ffzg/doc/exact_match.txt
revision 289 by dpavlin, Sun Mar 14 20:08:29 2004 UTC revision 303 by dpavlin, Sun Apr 4 22:11:13 2004 UTC
# Line 27 on TitleAndResponsibility, you would use Line 27 on TitleAndResponsibility, you would use
27 <input type="text" name="v1"> <input type="text" name="v1">
28 <input type="hidden" name="e1" value="5"> <input type="hidden" name="e1" value="5">
29
30
31 What are bit-masks?
32
33 Bit-mask is usage of one byte (8 bits) as 8 separate bits with it's own
34 meaning (this is simplification, but bear with me for now).
35
36 So, 1 = 2^0, thus it's bit 1. With analogy, 2=2^1 and 3=2^0+2^1.
37 So, for 1-3 we use two bits and have:
38
39 number bits
40 1 01 (just begin bit set)
41 2 10 (just end bit set)
42 3 11 (begin and end bit set)
43
44 Thus, with two bits (and values 1-3) we can express should we exact match from
45 beginning, end or both. For wild-card match, we use additional bit 3 (2^3 = 4)
46 so we have:
47
48 number bits exact match
49 1 001 begin
50 2 010 end
51 3 011 begin+end
52 4 100 (not used)
53 5 101 (4+1) begin+wild-card
54 6 110 (4+2) end+wild-card
55 7 111 (4+3) begin+end+wild-card
56
Legend:
Removed from v.289 changed lines Added in v.303 | 549 | 1,513 | {"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-2022-49 | latest | en | 0.543298 |
https://www.numbersaplenty.com/5724 | 1,718,434,973,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861584.65/warc/CC-MAIN-20240615062230-20240615092230-00556.warc.gz | 819,873,099 | 3,510 | Search a number
5724 = 223353
BaseRepresentation
bin1011001011100
321212000
41121130
5140344
642300
722455
oct13134
97760
105724
114334
123390
1327b4
14212c
151a69
hex165c
5724 has 24 divisors (see below), whose sum is σ = 15120. Its totient is φ = 1872.
The previous prime is 5717. The next prime is 5737. The reversal of 5724 is 4275.
It is a happy number.
5724 is nontrivially palindromic in base 11.
It is a Harshad number since it is a multiple of its sum of digits (18).
It is a plaindrome in base 7.
It is a nialpdrome in base 9.
It is a junction number, because it is equal to n+sod(n) for n = 5697 and 5706.
It is a congruent number.
It is an unprimeable number.
5724 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 82 + ... + 134.
It is an arithmetic number, because the mean of its divisors is an integer number (630).
25724 is an apocalyptic number.
5724 is a gapful number since it is divisible by the number (54) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 5724, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (7560).
5724 is an abundant number, since it is smaller than the sum of its proper divisors (9396).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
5724 is a wasteful number, since it uses less digits than its factorization.
5724 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 66 (or 58 counting only the distinct ones).
The product of its digits is 280, while the sum is 18.
The square root of 5724 is about 75.6571212775. The cubic root of 5724 is about 17.8881958732. Note that the first 3 decimals are identical.
Adding to 5724 its reverse (4275), we get a palindrome (9999).
It can be divided in two parts, 572 and 4, that added together give a square (576 = 242).
The spelling of 5724 in words is "five thousand, seven hundred twenty-four". | 650 | 2,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-26 | latest | en | 0.918942 |
http://mathlesstraveled.com/2009/01/16/predicting-pi-pretty-graphs-and-convergents/ | 1,455,444,165,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701174607.44/warc/CC-MAIN-20160205193934-00008-ip-10-236-182-209.ec2.internal.warc.gz | 142,841,316 | 17,873 | ## Predicting pi: pretty graphs and convergents
Recall the challenge I posed in a previous post: given the sequence of integers $\lfloor \pi \rfloor, \lfloor 2\pi \rfloor, \lfloor 3\pi \rfloor, \ldots$, what can you learn about $\pi$ (assuming you didn’t know anything about it before)? The answer, as explained in another post, is that you can learn $\pi$ to whatever precision you like, if you wait long enough. Not only that, but you can do better than is initially obvious. You can learn n digits of the decimal expansion of $\pi$ if you wait for $\lfloor 10^n \pi \rfloor$, but this is throwing away too much information: instead, noting that $\lfloor n \pi \rfloor \leq n\pi < \lfloor n \pi \rfloor + 1$, we get upper and lower bounds on $\pi$ from each term in the sequence:
$\displaystyle\frac{\lfloor n\pi \rfloor}{n} \leq \pi < \frac{\lfloor n\pi \rfloor + 1}{n}.$
Let’s graph these upper and lower bounds for various values of n:
Neato! We can view this in a slightly different way by graphing the error (the amount under or over $\pi$) of the approximations instead of the approximations themselves:
So, that looks pretty cool… but the best approximations quickly get so close to $\pi$ that it’s hard to see what’s really going on. So now we’re going to (a) zoom way out—the graph will go to n = 50000 instead of just 200, and (b) use a log-log scale—that is, we’ll graph the logarithm of both the error and n.
Cool! Now we can see some more interesting structure at the bottom of the graph (in the previous graph, the approximations quickly got so close to the axis that it was impossible to tell what was going on; this is why plotting the logarithm of the error helps). We can see that most of the approximations are in that big mass sloping gently down and to the right. This represents about how well we would do at approximating pi if we just looked at the $10^n$th elements of the sequence. But the interesting thing is the approximations which are not part of that big mass. At several places—most notably at around n=5, n=100, and n=30000—there are downward spikes, representing approximations which jump out as being way better than most of the other ones at that point. These are precisely the convergents of $\pi$, which are (in a specific technical sense) the best rational approximations to $\pi$. This is why, by looking at all the approximations and choosing the best ones, we can do much better than just looking at the $10^n$th elements of the sequence (imagine a line drawn through the bottommost points in the graph; it’s much steeper than the average slope of the big mass of approximations above it). In a future post I hope to write a bit more about these so-called “convergents” and where they come from; I’ve actually written a bit about them before, but didn’t point it out at the time!
Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
This entry was posted in convergence, famous numbers, pattern, sequences and tagged , , , . Bookmark the permalink.
### 5 Responses to Predicting pi: pretty graphs and convergents
1. Michael Lugo says:
What’s more, the downward spikes (in general) converge about twice as fast as the big mass of approximations. That is, the best approximation of a real number x by rationals with denominator less than n is good to about 1/n^2. (This result is known as “X’s Theorem” for some name X that I can’t remember now.)
2. Brent says:
Yes, X = Dirichlet if I recall correctly. =)
3. Michael Lugo says:
Armed with that knowledge, I poked around a bit. Dirichlet’s theorem says the exponent is at least 2; the Thue-Siegel-Roth theorem says it’s at most 2.
4. Brent says:
Ah, thanks, I hadn’t remembered the Thue-Siegel-Roth results, although now that you mention it I think it was mentioned (but not proved) in my undergrad number theory course. | 974 | 3,911 | {"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": 14, "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} | 3.953125 | 4 | CC-MAIN-2016-07 | latest | en | 0.928515 |
https://aenhancers.com/viewtopic.php?f=11&t=2231 | 1,603,695,412,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890586.57/warc/CC-MAIN-20201026061044-20201026091044-00477.warc.gz | 214,357,954 | 10,332 | ## Lighten layers based on distance from camera. Fog?
What type of scripts do you need?
Moderator: byronnash
esbowman
Posts: 10
Joined: February 25th, 2009, 1:34 pm
So what I'm hoping to achieve here is a simple way to automatically lighten background 3d layers based on their distance from the 3D after effects camera. My goal is to have a way to mimic something like this photo.
Hoping to achieve an atmospheric effect in which subjects far from the camera/eye are gradually lightened - perhaps even desaturated a bit. I'm sure there's a better way to describe this but the picture above should help. I can obviously do this by hand, but with 30-50 3d layers in my comp I'm hoping to apply an expression that automates this a bit more. I think using the hue/saturation effect or maybe just a levels effect will get me the results I'm looking for on a per/layer level.
I think it would be pretty amazing to have this included with a 3D camera in much the same way that depth of field works. It would be incredibly useful to just have a checkbox for this. In Maya I suppose this would be a volumetric effect or a way to add distance based fog.
Any suggestions? Thank you!!!
drockne
Posts: 13
Joined: March 2nd, 2012, 10:10 am
Sounds like you need an expression that calculates the distance in 3d space from the camera to the layer in question. Hopefully this equation helps you:
Define your two points. Point 1 at (x1, y1, z1) and Point 2 at (x2, y2, z2).
xd = x2-x1
yd = y2-y1
zd = z2-z1
Distance = SquareRoot(xd*xd + yd*yd + zd*zd)
Once you have that you can just paste whatever effect you want to use for the lighting and add the expression to alter its value as a function of distance. You'd need some way of setting your min and max distance values, kind of like setting a focal length for depth of field. Lastly when using coordinates to calculate distance make sure to use their world position, otherwise things like parenting will throw it off. Here's a link for how to get those values in expressions: http://forums.creativecow.net/thread/227/3790
P.S. One you have figured out your effect and expression, there's nothing stopping you from applying it to all selected layers via a script. Best of luck!
esbowman
Posts: 10
Joined: February 25th, 2009, 1:34 pm
Thanks so much for the info/advice and the expression. I'm going to take a crack at integrating this into my comp soon. Since a pesky deadline was forcing my hand, I just created two semi-transparent white 3D layers separated in z space to do a quick and dirty fog. Works for now, but I can see how this could be used for many things in the future. Thanks so much! | 662 | 2,639 | {"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-2020-45 | latest | en | 0.948283 |
http://nrich.maths.org/5516/solution | 1,503,073,998,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104704.64/warc/CC-MAIN-20170818160227-20170818180227-00224.warc.gz | 310,214,199 | 5,661 | ### Roll These Dice
Roll two red dice and a green dice. Add the two numbers on the red dice and take away the number on the green. What are all the different possibilities that could come up?
### Stop or Dare
All you need for this game is a pack of cards. While you play the game, think about strategies that will increase your chances of winning.
### Game of PIG - Sixes
Can you beat Piggy in this simple dice game? Can you figure out Piggy's strategy, and is there a better one?
# It's a Tie
##### Stage: 2 Challenge Level:
Jannis from Northcross Intermediate said:
"I think Kaia was right. The ties come out of the closet randomly. We know each tie has 1 eighth of a chance to be pulled out, so its quite likely that the father will not wear 1 tie twices in the week. But its random so there is a chance of him wearing a tie twice in the week."
Making the point that they're probabilities so you can't be sure.
Matthew and James from Stradbroke Primary did an experiment, and found that they picked the same tie twice in 9 out of 10 weeks. Ege and Onur from FMV Ozel Erenkoy Isik Primary School, Turkey, also did experiments, and you can find their results here and here .
Some people have tried to calculate the probability of picking the same tie twice, but none correctly yet. Here is what happened when we tried the experiment:
We found that there were:
6 weeks out of 20 when he wore a different tie every day (green), so the probability of this is 6/20,
12 weeks out of 20 when he wore the same tie twice in a week (blue), so the probability is 12/20, and
2 weeks out of 20 when he wore the same tie more than twice, so the probability of this is 2/20.
These are only approximate probabilities, and each time you do the experiment you will get a different answer. You can improve the approximation by increasing the number of times you do the experiment and adding up all the results as this will be more accurate.
Here are some some photos of children working on this problem:
Preparing materials
Drawing ties
Completing the table
Collating results | 478 | 2,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-34 | latest | en | 0.963537 |
https://gsebsolutions.com/gseb-solutions-class-7-maths-chapter-1-ex-1-4/ | 1,701,227,938,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00148.warc.gz | 347,379,751 | 53,020 | # GSEB Solutions Class 7 Maths Chapter 1 Integers Ex 1.4
Ā Ā Ā
Gujarat BoardĀ GSEB Textbook Solutions Class 7 Maths Chapter 1 Integers Ex 1.4 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 7 Maths Chapter 1 Integers Ex 1.4
Question 1.
Evaluate each of the following:
(a) (- 30) Ć· 10
(b) 50 Ć· (- 5)
(c) (- 36) Ć· (- 9)
(d) (- 49) Ć· (49)
(e) 13 Ć· [(- 2) +1]
(f) 0 Ć· (- 12)
(g) (- 31) Ć· [(- 30) + (- 1)]
(h) [(- 36) + 12] + 3
(i) [(- 6) + 5] Ć· [(- 2) + 1]
Solution:
(a) (- 30) Ć· 10 = $$\frac { – 30 }{ 10 }$$
(b) 50 Ć· (- 5) = $$\frac { 50 }{ (- 5) }$$
(c) (- 36) Ć· (- 9) = $$\frac { – 36 }{ (- 9) }$$
= 4
(d) (- 49) Ć· 49 = $$\frac { ( – 49) }{ 49 }$$
= 1
(e) 13 Ć· [(- 2) + 1] = 13 Ć· [- 1]
= $$\frac { 13 }{ (- 1) }$$ = – 13
(f) 0 Ć· (- 12) = 0
(g) (- 31) Ć· [(- 30) + (- 1)] = (- 31) Ć· (- 31)
= $$\frac { – 31 }{ (- 31) }$$ = 1
(h) [(- 36) Ć· 12] Ć· 3 = [ $$\frac { (- 36) }{ 12 }$$ ] + 3
= [- 3] Ć· 3 = – 1
(i) [ (- 6) + 5] Ć· [(- 2) + 1] = [- 1] + [- 1]
= $$\frac { (- 1) }{ (- 1) }$$ = 1
Question 2.
Verify that a Ć· (6 + c) ā (a Ć· b) + (a Ć· c) for each of the following values of a, b and c.
(a) a = 12, b = – 4, c = 2
(b) a = – 10, b = 1, c = 1
Solution:
(a) Given that a = 12, b = – 4 and c = 2
L.H.S. = a Ć· (b + c) = 12 + (- 4 + 2)
= 12 Ć· (- 2) = – 6
R.H.S. = (a Ć· b) + (a Ć· c)
= [12 Ć· (- 4)] + (12 Ć· 2)
= [- 3] + 6 = 3
āµ – 6 ā 3
ā“ L.H.S. ā R.H.S.
Hence, a Ć· (b + c) ā (a Ć· b) + (a Ć· c)
(b) Given that a = – 10, b = 1, c = 1
L.H.S. = a + (b + c)
= (- 10) + (1 + 1) = (- 10) Ć· 2 = – 5
R.H.S. = (a Ć· b) + (a Ć· c)
= [(- 10) Ć· 1] + [(- 10) Ć· 1]
= (- 10) + (- 10) = – 20
āµ – 5 ā 20
ā“ L.H.S. ā R.H.S.
Hence, a Ć· (b + c) ā (a Ć· b) + (a Ć· c)
Question 3.
Fill in the blanks:
(a) 369 Ć· ____= 369
(b) (- 75)Ć· ____ = – 1
(c) (- 206) Ć· ____ = 1
(d) – 87 Ć· ____ = 87
(e) ____Ć· 1 = – 87
(f) ____ Ć· 48 = – 1
(g) 20 Ć· ____ = – 2
(h) ____ Ć· 4 = – 3
Solution:
(a) āµ a + 1 Ć· aĀ Ā Ā Ā Ā Ā Ā Ā ā“ 369 + 1 = 369
(b) āµ (- a) Ć· a = – 1Ā Ā Ā Ā Ā ā“ (- 75) Ć· 75 = – 1
(c) āµ (- a) Ć· (- a) = 1Ā Ā Ā Ā ā“ (- 206) Ć· (- 206)= 1
(d) āµ (- a) Ć· (- 1) = aĀ Ā Ā Ā ā“ (- 87) Ć· (- 1) = 87
(e) āµ (- a) Ć· 1 = – aĀ Ā Ā Ā Ā ā“ (- 87) Ć· 1 = – 87
(f) āµ (- a) Ć· a = – 1Ā Ā Ā Ā Ā ā“ (- 48) Ć· 48 = – 1
(g) āµ 20 Ć· 10 = 2Ā Ā Ā Ā Ā Ā Ā ā“ 20 Ć· (-10) = -2
(h) āµ 12 Ć· 4 = 3Ā Ā Ā Ā Ā Ā Ā Ā ā“ (- 12) Ć· 4 = – 3
Question 4.
Write five pairs of integers (a, b) such that a Ć· b = – 3. One such pair is (6, – 2) because 6 Ć· (- 2) = (- 3).
Solution:
I. Since, 3 Ć· 1 = 3
3 + (- 1) = – 3
Compairing it with a Ć· b = – 3 we have a = 3 and b = (- 1)
ā“ The required pair of integers = (3, – 1)
II. Since, 3 Ć· 1 = 3
ā“ (- 3) – 1 = – 3
Compairing with a + b = – 3 we have a = (- 3) and b = 1
ā“ The required pair of integers = (- 3, 1)
III. Since, 9 Ć· 3 = 3
ā“ 9 Ć· (- 3) = – 3
Compairing it with a + b = – 3 we have a = 9 and b = – 3
ā“ The required pair of integers = (9, – 3)
IV. Since, 12 Ć· 4 = 3
ā“ (- 12) Ć· 4 = – 3
Comparing it with a Ć· b = – 3 we have a = – 12 and b = 4
ā“ The required pair of integers = [- 12, 4].
V. Since, 12 Ć· 4 = 3
ā“ 12 + (- 4) = – 3
Comparing it with a + b = – 3 we have
a = – 12 and b = – 4
ā“ The required pair of integers = [12, (- 4)]
Question 5.
The temperature at 12 noon was 10Ā°C above zero. If it decreases at the rate of 2Ā°C per hour , until mid-night, at what time would the temperature be 8Ā°C below zero? What would be the temperature at mid-night?
Solution:
Temperature at 12 noon = +10Ā°C Rate of change in temperature = – 2Ā°C per hour
Number of hours from 12 noon to mid- night =12
ā“ Change in temperature in 12 hours = 12Ā° x (- 2) = – 24Ā°C
ā“ Temperature at mid- night (i.e. 12 hours after 12 noon) = + 10Ā°C + (- 24Ā°C) = – 14Ā°C
Thus, temprature at mid- night = – 14Ā°C
Now, temperature difference between + 10Ā°C and – 8Ā°C = +10Ā°C – (- 8Ā°C) = 18Ā°C
āµ $$\frac { 18 }{ 2 }$$ = 9
ā“ Temperature change of 18Ā°C will take place in 9 hours from 12 noon. Time after 9 hours from 12 noon = 9 p.m. Thus, the temperature 8Ā°C below 0Ā° (- 8Ā°C) would be at 9 p.m.
Question 6.
In a class test (+3) marks are given for every correct answer and (- 2) marks are given for every incorrect answer and no marks for not attempting any Question uestion.
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
(ii) Mohini scores – 5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution:
Marks for every correct answer = +3
Marks for every incorrect answer = – 2
(i) Total marks scored by Radhika = 20
Marks scored for correct answers = 12 x 3 = 36
ā“ Marks given for incorrect answers = 20 – 36 = – 16
Number of incorrect answers = (- 16) Ć· (- 2) = 8
Thus, Radhika attempted 8 questions incorrectly.
(ii) Marks obtained by Mohini = – 5
Marks obtained for 7 correct answers = 7 x 3 = 21
ā“ Marks obtained for incorrect answers = – 5 – 21 = – 26
ā“ Number of incorrect answers = – 26 Ć· (- 2) = 13
Question 7.
An elevator descends into a mine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach – 350 ml
Solution:
Present position of the elevator is at 10 m above the ground level.
Distance to be moved by the elevator below the ground level = 350 m
ā“ Total distance to be moved by the elevator = 350 m + 10 m = 360 m
The rate of descent = 6 m/min | 2,631 | 5,536 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-50 | longest | en | 0.333077 |
https://www.gradesaver.com/textbooks/science/chemistry/chemistry-the-central-science-13th-edition/chapter-6-electronic-structure-of-atoms-additional-exercises-page-254/6-95 | 1,532,319,699,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594886.67/warc/CC-MAIN-20180723032237-20180723052237-00519.warc.gz | 896,959,206 | 12,547 | ## Chemistry: The Central Science (13th Edition)
The wavelength of the electron is $\lambda\approx10.59pm$
The kinetic energy of the electron is $E_K=13.4keV=1.34\times10^4eV$ $1eV\approx1.602\times10^{-19}J$ So, $E_K=(1.34\times10^4eV)\times(1.602\times10^{-19}J)\approx2.147\times10^{-15}J$ 1) Calculate the velocity of the electron The velocity of the electron is calculated according to the formula $$E_K=\frac{1}{2}mv^2$$$$v^2=\frac{2E_K}{m}$$ - Mass of the electron $m\approx9.109\times10^{-31}kg$ So, $$v^2=\frac{2\times(2.147\times10^{-15}J)}{9.109\times10^{-31}kg}\approx4.714\times10^{15}m/s$$$$v=6.866\times10^7m/s$$ 2) Find the wavelength of the electron $(\lambda)$ The wavelength of the electron (matter) is calculated as follows $$\lambda=\frac{h}{mv}$$$$\lambda=\frac{6.626\times10^{-34}J.s}{(9.109\times10^{-31}kg)(6.866\times10^7m/s)}$$$$\lambda\approx1.059\times10^{-11}m\approx10.59pm$$ | 341 | 907 | {"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} | 3.609375 | 4 | CC-MAIN-2018-30 | longest | en | 0.555026 |
https://www.systutorials.com/docs/linux/man/3-zla_hercond_x.f/ | 1,695,668,609,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510085.26/warc/CC-MAIN-20230925183615-20230925213615-00236.warc.gz | 1,125,030,106 | 9,013 | # zla_hercond_x.f (3) - Linux Manuals
## NAME
zla_hercond_x.f -
## SYNOPSIS
### Functions/Subroutines
DOUBLE PRECISION function zla_hercond_x (UPLO, N, A, LDA, AF, LDAF, IPIV, X, INFO, WORK, RWORK)
ZLA_HERCOND_X computes the infinity norm condition number of op(A)*diag(x) for Hermitian indefinite matrices.
## Function/Subroutine Documentation
### DOUBLE PRECISION function zla_hercond_x (characterUPLO, integerN, complex*16, dimension( lda, * )A, integerLDA, complex*16, dimension( ldaf, * )AF, integerLDAF, integer, dimension( * )IPIV, complex*16, dimension( * )X, integerINFO, complex*16, dimension( * )WORK, double precision, dimension( * )RWORK)
ZLA_HERCOND_X computes the infinity norm condition number of op(A)*diag(x) for Hermitian indefinite matrices.
Purpose:
``` ZLA_HERCOND_X computes the infinity norm condition number of
op(A) * diag(X) where X is a COMPLEX*16 vector.
```
Parameters:
UPLO
``` UPLO is CHARACTER*1
= 'U': Upper triangle of A is stored;
= 'L': Lower triangle of A is stored.
```
N
``` N is INTEGER
The number of linear equations, i.e., the order of the
matrix A. N >= 0.
```
A
``` A is COMPLEX*16 array, dimension (LDA,N)
On entry, the N-by-N matrix A.
```
LDA
``` LDA is INTEGER
The leading dimension of the array A. LDA >= max(1,N).
```
AF
``` AF is COMPLEX*16 array, dimension (LDAF,N)
The block diagonal matrix D and the multipliers used to
obtain the factor U or L as computed by ZHETRF.
```
LDAF
``` LDAF is INTEGER
The leading dimension of the array AF. LDAF >= max(1,N).
```
IPIV
``` IPIV is INTEGER array, dimension (N)
Details of the interchanges and the block structure of D
as determined by CHETRF.
```
X
``` X is COMPLEX*16 array, dimension (N)
The vector X in the formula op(A) * diag(X).
```
INFO
``` INFO is INTEGER
= 0: Successful exit.
i > 0: The ith argument is invalid.
```
WORK
``` WORK is COMPLEX*16 array, dimension (2*N).
Workspace.
```
RWORK
``` RWORK is DOUBLE PRECISION array, dimension (N).
Workspace.
```
Author:
Univ. of Tennessee
Univ. of California Berkeley | 635 | 2,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-40 | longest | en | 0.685993 |
https://books.google.com.jm/books?qtid=347e78f0&lr=&id=lAUAAAAAMAAJ&sa=N&start=110 | 1,685,674,739,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00323.warc.gz | 179,453,769 | 6,278 | Search Images Maps Play YouTube News Gmail Drive More »
Books Books
A Circle is a plane figure bounded by a curved line every point of which is equally distant from a point within called the center.
Cyclopedia of Architecture: Architectural drawing. Lettering - Page 43
1912
...well chosen to express fashion's frivolities The circle is a plane figure bounded by a curved line, every point of which is equally distant from a point within called the center. An arc in this bounding line is the most monotonous curve we have. Wherever it is taken, however great...
## Everyday Arithmetic: Book One-three, Book 3
Franklin Sherman Hoyt, Harriet E. Peet - Arithmetic - 1915 - 378 pages
...bank to pay a specified sum of money. Circle. A circle is a plane figure bounded by a curved line, every point of which is equally distant from a point within called the center. Circumference. A circumference is the boundary of a circle. Collateral note. A collateral note is a...
## Advertising, Its Principles and Practice
...well chosen to express fashion's frivolities The circle is a plane figure bounded by a curved line, every point of which is equally distant from a point within called the center. An arc in this bounding line is the most monotonous curve we have. Wherever it is taken, however great...
## Practical Applied Mathematics
Joseph Woodwell Ledwidge Hale - Mathematics - 1915 - 232 pages
...FIG. 150. 71. The Sphere.— The sphere, Fig. 150, is a solid bounded by a uniformly curved surface every point of which is equally distant from a point within called the center, as 0. The area or the surface of a sphere equals the square of the diameter multiplied by Pi (IT),...
## Buker-Felter Arithmetics, Book 3, Parts 1-2
Eva F. Buker - 1915 - 436 pages
...Circles—Relations between Dimensions 1. A portion of surface, bounded by a curved line, every part of which is equally distant from a point within (called the center) is a circle. The bounding line of a circle is its circumference. A line from the center of a circle...
## Gunners' Instruction: 1916-1917 (mine Companies).
Gunnery - 1916 - 176 pages
...joins them will lie wholly in the surface. A circle is a plane figure bounded by a curve (called a circumference), every point of which is equally distant from a point within, called the center. A circumference is frequently called a circle. Any line passing through the center and terminating...
## Mechanical Drawing: Instruction Paper, Part 2
Ervin Kenison - Mechanical drawing - 1916 - 44 pages
...perpendicular distance between the bases. SPHERES A sphere is a solid bounded by a curved surface, every point of which is equally distant from a point within called the center. The diameter is a straight line drawn through the center and having its extremities in the curved surface....
## Manual of Natural Education
Winifred d'Estcourte Sackville Stoner - Education - 1916 - 246 pages
...quadrilateral with two sides parallel. A circle is a plane figure whose perimeter is a curved line, every point of which is equally distant from a point within called the center. The diameter is a straight line passing through the center and having both ends in the circumference.... | 728 | 3,202 | {"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-2023-23 | latest | en | 0.93466 |
https://365datascience.com/question/different-from-dates/ | 1,716,049,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057440.7/warc/CC-MAIN-20240518152242-20240518182242-00144.warc.gz | 63,584,508 | 41,647 | 21 Dec 2022
Posted on:
19 Nov 2022
0
# Different from dates
1. what is the difference between from date and hire date.
2. Why do i get a different answer when i use from_date in the salaries table instead of from_date from the dept_employees table?
SELECT
dept_name, d.dept_no, AVG(salary) AS salary
FROM
t_salaries t
JOIN
t_dept_emp de ON t.emp_no = de.emp_no
JOIN
t_departments d ON d.dept_no = de.dept_no
WHERE
t.from_date BETWEEN '1999-01-01' AND '1999-12-31'
GROUP BY dept_no
ORDER BY dept_no
;
Instructor
Posted on:
21 Dec 2022
1
Hi Pius!
Thanks for reaching out.
1. The hire_date column indicates the date that the employee has started working. The from_date and to_date columns indicate the start date and the end date for the current work that the employee has worked. Let's check for emp_no=10001. If we run the following query:
`SELECT `
`* `
`FROM `
`employees `
`WHERE emp_no = 10001; `
we see Georgi Facello has started working at 1986-06-26. This is his career start for the company. But in this company he has worked at different positions during different periods. If we run the following query:
2. `SELECT `
`*`
`FROM`
`salaries `
`WHERE emp_no = 10001; `
we see all different periods of work with all salaries and the columns from_date and to_date for all his previous jobs. If he is currently working at a particular position the date is set to '9999-01-01'.
3. The different answer comes from the different records that each table contains for this particular column. You can check this with the SELECT statement comparing the result sets of the queries which select this column from these tables.
Hope this helps.
Best,
Tsvetelin | 431 | 1,664 | {"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-2024-22 | longest | en | 0.843651 |
https://aisky.men/problem/leetcode/LeetCode-166.%C2%A0%E5%88%86%E6%95%B0%E5%88%B0%E5%B0%8F%E6%95%B0/ | 1,571,771,888,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987823061.83/warc/CC-MAIN-20191022182744-20191022210244-00039.warc.gz | 374,169,447 | 27,612 | # LeetCode: 166. 分数到小数¶
## 1、题目描述¶
输入: numerator = 1, denominator = 2
输入: numerator = 2, denominator = 1
输入: numerator = 2, denominator = 3
## 2、解题思路¶
再找思路的过程中，反向思考，如果是无限循环小数，如何变成分数呢？
步骤1.将无限循环小数分为2个部分,0.3454545...45为例,将其分0.3+0.04545...45这2个部分.
0.0454545...45的划分方法.先设它为a,那么就有：
10a=0.454545...45
1000a=45.4545.45
1000a-10a=45
990a=45
a=45/990=1/22
3/10+1/22=66/220+10/220=76/220=19/55
这个题目可以这样做，首先将数据分成两部分，整数部分和小数部分，
• 建立一个哈希，缓存余数
• 对小数部分进行判断，是不是循环的，如果余数不为0，表示要继续计算下去
• 如果余数出现过了，表示这个就是循环的，当前的数字不需要加入结果集中
class Solution:
def fractionToDecimal(self, numerator, denominator):
"""
:type numerator: int
:type denominator: int
:rtype: str
"""
sign = 1
if (numerator > 0 and denominator < 0) or (numerator < 0 and denominator > 0):
sign = -1
numerator = abs(numerator)
denominator = abs(denominator)
integer_part = numerator // denominator
remainder = numerator % denominator
if sign == 1:
result = ""
else:
result = "-"
if remainder == 0:
return result + str(integer_part)
remainder_hash = {}
decimal_cache = []
is_cycle = False
cycle_index = 0
while remainder != 0:
temp = remainder = remainder
cur_num = remainder * 10 // denominator
remainder = remainder * 10 % denominator
if remainder_hash.get(temp) is None:
remainder_hash[temp] = cycle_index
decimal_cache.append(cur_num)
cycle_index += 1
else:
cycle_index = remainder_hash[temp]
is_cycle = True
break
if not is_cycle:
result += str(integer_part) + "." + "".join([str(i) for i in decimal_cache])
else:
result += str(integer_part) + "." + "".join([str(i) for i in decimal_cache[:cycle_index]]) + "(" + "".join(
[str(i) for i in decimal_cache[cycle_index:]]) + ")"
return result | 617 | 1,660 | {"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} | 4 | 4 | CC-MAIN-2019-43 | latest | en | 0.287417 |
https://www.physicsforums.com/threads/what-are-the-basic-mathematical-objects-in-qft.1045305/ | 1,723,710,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00085.warc.gz | 721,040,938 | 20,916 | What are the basic mathematical objects in QFT?
• snoopies622
In summary: These functional fields are what we want to describe as our wavefunction. The key point is that the wavefunction is not a function at a single point in space, it is a function at every point in space. The wavefunction is given by a functional field equation in which the coefficients depend on the position and momentum of every particle in the system. In summary, in order to be truly compatible with special relativity, we need to discard the notion that \phi and \psi in the Klein-Gordon and Dirac equations respectively describe single particle states. In their place, we propose the following new ideas: — The wave functions \phi and \psi are not wave functions at all, instead they are
snoopies622
TL;DR Summary
How are phi and psi (solutions to the Klein-Gordon and Dirac equations) expressed mathematically in quantum field theory?
I found a copy of David McMahon's "Quantum Field Theory Demystified" and I'm already confused on page 4 where he says, " . . in order to be truly compatible with special relativity, we need to discard the notion that $\phi$ and $\psi$ in the Klein-Gordon and Dirac equations respectively describe single particle states. In their place, we propose the following new ideas:
— The wave functions $\phi$ and $\psi$ are not wave functions at all, instead they are fields.
— The fields are operators that can create new particles and destroy particles."
As i understand things,
— the $\psi$ in the Schrodinger equation represents a complex number at every point in space and time, while in the Dirac equation represents four complex numbers at every point in space and time. (I don't know what the $\phi$ in the Klein-Gordon equation represents, but I'm guessing something similar.)
— an operator is something that changes a function into a different function. One way to think about it is - if a function is a vertical list of n complex numbers, then an operator is an nxn matrix that can be multiplied by the column of n numbers to produce a different column of n numbers.
In quantum field theory, what exists at every point in space and time? A matrix? More than one matrix?
Last edited:
snoopies622 said:
I found a copy of David McMahon's "Quantum Field Theory Demystified"
Get a real QFT book.
The "basic objects" in QFT are operator-valued distributions
aaroman and dextercioby
strangerep, PeroK, malawi_glenn and 3 others
vanhees71
Last edited:
vanhees71, strangerep and topsquark
vanhees71 said:
https://www.amazon.com/dp/9814635502/?tag=pfamazon01-20
It's a gem! It's the best book, which makes relativistic QFT "as simple as possible but not simpler".
Hmm. Although I've only just now skimmed the first lecture, I will say that I like his style.
vanhees71
Sidney was a genius, and there's a reason why his students populate the theoretical physics departments of so many universities. But I don't think this is the place to start for someone who is just starting out, especially with gaps.
vanhees71 and malawi_glenn
In one particle quantum mechanics we have a system described by a state space and a number of observables like ##\hat{q}##, ##\hat{p}##, ##\hat{S_z}## etc. At any point in time the system is in some state ##|\psi\rangle## and the wavefunction is given by ##\langle q|\psi\rangle## where ##|q\rangle## are the position eigenstates of ##\hat{q}##.
In quantum field theory our system is again described by a state space however now there are observables ##\hat{\phi}(\mathbf{x})##, ##\hat{\pi}(\mathbf{x})## for every point in space ##\mathbf{x}##. These field observables admit eigenstates ##|f(\mathbf{x})\rangle## for each c-number function of spacetime ##f(\mathbf{x})##, so evidently ##\langle f(\mathbf{x}) | \psi \rangle## is not a function but a functional i.e. a function of functions.
topsquark, vanhees71, PeroK and 1 other person
1. What is QFT?
QFT, or quantum field theory, is a theoretical framework used to describe the behavior of particles at the subatomic level. It combines elements of quantum mechanics and special relativity to explain the interactions between particles and their corresponding fields.
2. What are mathematical objects in QFT?
The basic mathematical objects in QFT are fields, operators, and states. Fields describe the properties of particles, operators represent physical observables, and states represent the possible configurations of a system.
3. How are fields represented in QFT?
In QFT, fields are represented by mathematical functions that describe the values of a particle's properties at different points in space and time. These functions are called field operators.
4. What is the significance of operators in QFT?
Operators in QFT are used to calculate the values of physical observables, such as energy or momentum, for a given state of a system. They also play a crucial role in the mathematical formulation of quantum mechanics and are essential for understanding the behavior of particles at the subatomic level.
5. How do states relate to mathematical objects in QFT?
States in QFT represent the possible configurations of a system, including the arrangement of particles and their corresponding fields. They are described by mathematical objects called wave functions, which are used to calculate the probability of a particle having a particular property at a given time and location.
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2K | 1,365 | 5,780 | {"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} | 3.21875 | 3 | CC-MAIN-2024-33 | latest | en | 0.916075 |
https://achyuts.wordpress.com/2016/09/14/floating-point-numbers/ | 1,632,111,124,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057018.8/warc/CC-MAIN-20210920040604-20210920070604-00423.warc.gz | 132,218,731 | 20,263 | ## Rounding off double to specified number of digits
You can’t set the precision of a double (or Double) to a specified number of decimal digits, because floating-point values don’t have decimal digits. They have binary digits.
You will have to convert into a decimal radix, either via BigDecimal or DecimalFormat, depending on what you want to do with the value later.
``````DecimalFormat df = new DecimalFormat("#.####");
df.setRoundingMode(RoundingMode.CEILING);
for (Number n : Arrays.asList(12, 123.12345, 0.23, 0.1, 2341234.212431324)) {
Double d = n.doubleValue();
System.out.println(df.format(d));
}``````
```You can try BIGDECIMAL for this purpose
Double toBeTruncated = new Double("3.5789055");
Double truncatedDouble = new BigDecimal(toBeTruncated)
.setScale(3, BigDecimal.ROUND_HALF_UP)
.doubleValue();
``` | 207 | 824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-39 | latest | en | 0.54001 |
https://stats.stackexchange.com/questions/102759/differences-between-logistic-regression-and-perceptrons/146911 | 1,560,653,311,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997533.62/warc/CC-MAIN-20190616022644-20190616044644-00374.warc.gz | 601,325,516 | 36,802 | # Differences between logistic regression and perceptrons
As I understand, a perceptron/single-layer artificial neural network with a logistic sigmoid activation function is the same model as logistic regression. Both models are given by the equation:
$F(x) = \frac{1}{1-e^{-\beta X}}$
The perceptron learning algorithm is online and error-driven, whereas the parameters for logistic regression could be learned using a variety of batch algorithms, including gradient descent and Limited-memory BFGS, or an online algorithm, like stochastic gradient descent. Are there any other differences between logistic regression and a sigmoid perceptron? Should the results of a logistic regressor trained with stochastic gradient descent be expected to be similar to the perceptron?
• Looks like this question is similar, and it seems to contain better responses :) – Ralph Tigoumo May 7 '16 at 9:34
You mentioned already the important differences. So the results should not differ that much.
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Xi'an Mar 1 '15 at 17:14
• Actually I tried to answer both questions: 1) "Are there any other differences between logistic regression and a sigmoid perceptron?" and 2) "Should the results of a logistic regressor trained with stochastic gradient descent be expected to be similar to the perceptron?" – Michael Dorner Mar 1 '15 at 17:27
• That's a reasonable position, @MichaelDorner. Would you mind expanding your answer a little to make that clearer? – gung Mar 1 '15 at 17:44
I believe one difference you're missing is the fact that logistic regression returns a principled classification probability whereas perceptrons classify with a hard boundary.
This is mentioned in the Wiki article on Multinomial logistic regression.
There is actually a big substantial difference, which is related to the technical differences that you mentioned. Logistic regression models a function of the mean of a Bernoulli distribution as a linear equation (the mean being equal to the probability p of a Bernoulli event). By using the logit link as a function of the mean (p), the logarithm of the odds (log-odds) can be derived analytically and used as the response of a so-called generalised linear model. Parameter estimation on this GLM is then a statistical process which yields p-values and confidence intervals for model parameters. On top of prediction, this allows you to interpret the model in causal inference. This is something that you cannot achieve with a linear Perceptron.
The Perceptron is a reverse engineering process of logistic regression: Instead of taking the logit of y, it takes the inverse logit (logistic) function of wx, and doesn't use probabilistic assumptions for neither the model nor its parameter estimation. Online training will give you exactly the same estimates for the model weights/parameters, but you won't be able to interpret them in causal inference due to the lack of p-values, confidence intervals, and well, an underlying probability model.
Long story short, logistic regression is a GLM which can perform prediction and inference, whereas the linear Perceptron can only achieve prediction (in which case it will perform the same as logistic regression). The difference between the two is also the fundamental difference between statistical modelling and machine learning. | 701 | 3,439 | {"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} | 2.6875 | 3 | CC-MAIN-2019-26 | longest | en | 0.90338 |
http://s3-us-west-1.amazonaws.com/lamecool/confidence-level-bounds.html | 1,576,075,082,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540531917.10/warc/CC-MAIN-20191211131640-20191211155640-00182.warc.gz | 125,362,593 | 9,703 | # Law of attraction movie rotten tomatoes 7500
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## Confidence level bounds,best mate is dating my ex,loving kindness meditation mantra download,top ways to be successful in life pdf - And More
23.01.2014
Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Given a 95% confidence level, how do I demonstrate 95% of the intervals actually contain the population mean?
Let me say first off that I'm consistently impressed with the quality of responses on this site. My basic goal is to prove that confidence intervals actually work, and by work I mean that that a certain percentage of them, say 95%, actually include the population mean (assuming that we know the population mean in the first place).
A) I’ve generated a data set in Excel consisting of 20 samples with 64 observations each (i.e. C) Below is the best Excel formula I found that can generate something approaching a normal distribution. As you might have already guessed, here’s the issue: theoretically if 95% of these intervals contained the population mean, then only 1 interval out of 20 would have failed to include it.
In case someone happens to see this thread, here are screen shots of the 2 main excel spreadsheets; one version with just the values, the other version with just the formulas.
Compared to your population standard deviation of 12, the width of your confidence intervals looks too narrow in many cases. Use meaningful names for ranges and variables rather than cell references wherever possible.
To illustrate, let me share a spreadsheet I created long ago for exactly this purpose: to show, via simulation, how confidence intervals work. Finally, these 100 columns drive a graphic that shows all 100 confidence intervals relative to the specified mean Mu and also visually indicates (via the spikes at the bottom) which intervals fail to cover the mean. In any case, you have to simulate an infinite number of samples to get the result you want. Not the answer you're looking for?Browse other questions tagged normal-distribution confidence-interval binomial excel random-generation or ask your own question.
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Yes, the methods seem very simple to me and are easily applied to all as they are universal but the crux is that they have to be applied to real people. 3: Make companies understand how everyone works differently and stop focusing on the difference. Almost 6 years on I have a set of methods to apply universally to the whole world to do just this - but it'll never be about me just talking about them.
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You folks have done a better job of explaining difficult concepts than most instructors or textbooks I've encountered. I couldn’t even find agreement on whether the truncation disqualifies it as normal in the strict sense.
Is Excel simply incapable of generating random normally distributed data with the level of accuracy needed to make this example work?
If you do \$n\$ CI's, the number that cover the true mean will follow a binomial distribution \$(n,p)\$, with \$p=0.95\$.
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I only know that RandBetween produces a uniform distribution so it’s useless for these purposes. Also, I tried a Normal Quantile Plot Test for Normality that I found online on the 128.1k data point example and I did find a very slight amount of leptokurtosis for anyone who knows what to make of that. Note that I also tried Excel’s data analysis->random number generation tool and got virtually identical results as the Excel formula I included above.
Here were the results: at the 90% CL only 60 out of 100 had at least 18 intervals with the pop. Below is a comparison of the original 100 iterations I recorded compared to the binomial distribution per Excel.
I was told recently that "you have made me look differently at my employees and I can see how they learn now" and this was from working with someone's child. If I regenerate the prices repeatedly using F9, it only meets the 19 or greater threshold around 70% of the time with some iterations producing 20 out 20 intervals that contain the mean to as low as only 16 out of 20. Your intervals are much narrower, which leads me to suspect you are calculating something wrong. As far as choosing a value of 12, the logic came from trying out various std devs that would result in a data set that was fairly consistently bound between 25 and 100 give or take.
Please note, though--in response to your title--that indeed these CIs are provably correct. I can now demonstrate that out of 20 confidence intervals, the highest probability belongs to scenarios where exactly 2 do not contain the population mean at the 90% confidence level. But by understanding their own child in a different way now from experience they see their own employees differently.
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### Comments to «Confidence level bounds»
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Been in a LENGTHY interval of being very concern. | 2,446 | 11,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2019-51 | latest | en | 0.906937 |
http://mathoverflow.net/questions/16010/do-separable-and-normal-have-topological-meanings-for-fields | 1,469,681,562,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00116-ip-10-185-27-174.ec2.internal.warc.gz | 160,668,926 | 15,425 | # Do separable and normal have topological meanings for fields?
The terminology would suggest that a separable field extension is so because the resulting field extension has some sort of separable topology, and that a normal extension corresponds to one with a normal topology.
I imagine this is true, or else they wouldn't have named them in such a way.
Also, I'm not sure what subfield this falls under, so if you could suggest additional tags, that would be great as well.
-
Separable has a topological meaning that is not T_2. See also: mathoverflow.net/questions/7389/… – Jonas Meyer Feb 22 '10 at 2:02
I see your edit has made part of my comment irrelevant, but the link is intended to get across the same point as Gerald Edgar's answer. – Jonas Meyer Feb 22 '10 at 2:36
Then I misunderstood your comment. After reading it I thought there still was some connection, and mining the link for information came back with nothing, so I was still in a state of confusion. Simply saying "This is incorrect." answers my question better. – Andrew Homan Feb 22 '10 at 2:53
The "normal" in "normal extension" comes from normal subgroups, or maybe the other way around. The "separable" in "separable extension" comes from separable polynomials, i.e. polynomials whose roots are distinct (hence can be "separated"). – Qiaochu Yuan Feb 22 '10 at 3:53
## 1 Answer
This is incorrect. Words like "separable" and "normal" occur in unrelated ways in various parts of mathematics. (Normal subgroup, separable differential equations...) Other words, too. Like "regular", "perfect" ...
"Separable" in toplogy... Does it mean something can be "separated"? What? I believe it goes back to Fréchet, but what did he mean by it?
-
So there really is no connection. Hmph. – Andrew Homan Feb 22 '10 at 2:55
My guess is that the connection between "separable" in topology and separation is that the fact that Q is dense in R (so R is separable) means that any two elements of R can be "separated" by an element of Q. – Qiaochu Yuan Feb 22 '10 at 3:50
I can "guess" things, too, but does anyone actually know? – Gerald Edgar Mar 5 '10 at 2:46 | 543 | 2,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2016-30 | latest | en | 0.951206 |
https://www.alpertron.com.ar/MODFERM.HTM | 1,713,369,509,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817158.8/warc/CC-MAIN-20240417142102-20240417172102-00804.warc.gz | 569,863,814 | 10,809 | # Factors of Generalized Fermat Numbers
1. Alpertron
2. Math problems
3. Factors of Generalized Fermat Numbers
These numbers have the form 23n + 1 and 23n - 1.
The problem here is to find factors of these numbers.
Since 23n+1 + 1 = (23n + 1) (43n - 23n + 1)
the first factor of the RHS has the same form of the LHS, so we are interested on factors of the second expression.
In the same way 23n+1 - 1 = (23n - 1) (43n + 23n + 1)
Notice that 43n - 23n + 1 is always multiple of 3, because 43n = 13n = 1 (mod 3), 23n = 2odd = 2 (mod 3), so 1 - 2 + 1 = 0 (mod 3).
The numbers 43n + 23n + 1 are known as Generalized Fermat Number Fn,2.
Prime factors of (43n - 23n + 1) / 3
ExponentPrime factorsDiscoverer
0No prime factors
12*32+1 = 19
23230*33+1 = 87211
32*34+1 = 163
1672*34+1 = 135433
3358160*34+1 = 272010961
46*35+1 = 1459
574*35+1 = 139483
4291937 2147783270*35+1 = 1042940743 1911334611
377829239 3426209464*35+1 = 9 1812505160 2568899753
5312568*36+1 = 227862073
4267065754*36+1 = 3110690934667
2 9752059431 0685684488*36+1 = 2168 9251325248 9863991753
15 1179583137 9924450952*36+1 = 11020 9916107596 4924744009
Prime cofactor with 78 digits
68*37+1 = 17497
2478*37+1 = 5419387
2766*37+1 = 6049243
11338334*37+1 = 2 4796936459
84279294*37+1 = 1 84318815979
504889554 2903149065 7884451200 5439490118*37+1 = 110 4193455232 9187006879 3294775589 6164888067Eric Jeancolas (14 Nov 2022)
2892851800 1521723769 6354515718 1899996240*37+1 = 632 6666886932 8009884192 7325875681 5291776881Eric Jeancolas (14 Nov 2022)
78*38+1 = 52489
4410*38+1 = 28934011
80183520*38+1 = 526084074721
440 5633237922*38+1 = 2890535 9674006243Philippe Strohl (7/2003)
1107 8656127098*38+1 = 7268706 2849889979Philippe Strohl (7/2003)
862*39+1 = 1220347
141 0521316064*39+1 = 2776329 1064087713Philippe Strohl (7/2003)
9400455322*310+1 = 2364 6486308779Dario Alpern
13760*314+1 = 3635056441Dario Alpern
33493592*314+1 = 16019 8812234649Dario Alpern
145659504816*315+1 = 8120770 8270836113Dario Alpern (10/2005)
27622252160*315+1 = 39634912 7374389121Dario Alpern (12/2005)
152*316+1 = 86093443Dario Alpern
1638480104*317+1 = 4969326902816953Dario Alpern
1764074*318+1 = 24823580412187Dario Alpern
1833158976*319+1 = 3853940 0089977793Dario Alpern
134770686*319+1 = 15663877 5218956363Dario Alpern
2040749616*321+1 = 42625537 6246620049Dario Alpern
23228978058*324+1 = 6467016678 5259533899Dario Alpern (10/2005)
27578221330*328+1 = 1 322784935944 1514518131Dario Alpern (10/2005)
286*329+1 = 41178 2264189299Dario Alpern
101256*329+1 = 694923749 0458593049Dario Alpern
291373422682*330+1 = 2827 7555084144 9107428619Dario Alpern (10/2005)
303028366*331+1 = 18 7054111241 0831440603Dario Alpern (10/2005)
3132*332+1 = 5929664 6043258913Dario Alpern
3200*332+1 = 592966460 4325891201Dario Alpern
326912*333+1 = 3842422663 6031774977Dario Alpern
5 3257453584*333+1 = 2960614 1009397550 8931344433Dario Alpern (1/2006)
33234402*334+1 = 39 0916474476 5243106739Dario Alpern
3915098*340+1 = 1835 5643310084 1511037499Dario Alpern
42110*343+1 = 361 0826641339 9078538971Dario Alpern
436152*344+1 = 60583 1059023357 6304683913Dario Alpern
4696*347+1 = 25525 2617845992 0315627553Dario Alpern
4761406648*348+1 = 4 8981898922 3354715405 5937023929Dario Alpern (10/2005)
5054*351+1 = 1162994 7400608011 9380780339Dario Alpern
518*352+1 = 516886 5511381338 6391457929Dario Alpern
5312202*354+1 = 7095430909 1109480834 2140842139Dario Alpern
542544*355+1 = 4437987928 0720173555 7057769809Dario Alpern
55211435592*356+1 = 110654 3166109387 2738317683 8841455433Dario Alpern (10/2005)
5614*357+1 = 219806005 8714914256 2967483883Dario Alpern
24*357+1 = 376810295 7796995867 9372829513Dario Alpern
574663458*358+1 = 21965 4873542025 7844537383 2096296563Dario Alpern (10/2005)
586*359+1 = 847823165 5043240702 8588866403Dario Alpern
5962080*360+1 (34 digits)Dario Alpern
3587982080*360+1 (39 digits)Dario Alpern (10/2005)
65860704*366+1 (38 digits)Dario Alpern
758*376+1 (38 digits)Dario Alpern
84203214*385+1 (46 digits)Dario Alpern
8560037080*386+1 (49 digits)Dario Alpern (10/2005)
8762192*388+1 (47 digits)Dario Alpern
9319388858*394+1 (53 digits)Dario Alpern (10/2005)
9423922928*395+1 (53 digits)Dario Alpern (10/2005)
99260280*3100+1 (54 digits)Dario Alpern
11030*3111+1 (55 digits)Dario Alpern
11132*3112+1 (55 digits)Dario Alpern
11892774*3119+1 (62 digits)Dario Alpern
1258800130*3126+1 (68 digits)Dario Alpern (10/2005)
1306*3131+1 (64 digits)Dario Alpern
1361190881376*3137+1 (75 digits)Dario Alpern (3/2006)
14414*3145+1 (71 digits)Dario Alpern
161616400864*3162+1 (87 digits)Dario Alpern (5/2006)
163104620466*3164+1 (87 digits)Dario Alpern (11/2005)
166150*3167+1 (82 digits)Dario Alpern
16756*3168+1 (82 digits)Dario Alpern
175407078202*3176+1 (93 digits)Dario Alpern (6/2006)
17654*3177+1 (87 digits)Dario Alpern
1846222*3185+1 (93 digits)Dario Alpern
20296*3203+1 (99 digits)Dario Alpern
20682880766*3207+1 (107 digits)Dario Alpern (8/2006)
2084104*3209+1 (104 digits)Dario Alpern
2148256*3209+1 (107 digits)Dario Alpern (10/2005)
21111128*3212+1 (106 digits)Dario Alpern
2141062270*3215+1 (109 digits)Dario Alpern (10/2005)
2212208*3222+1 (110 digits)Dario Alpern
22278*3223+1 (109 digits)Dario Alpern
238968*3239+1 (118 digits)Dario Alpern
281723538*3282+1 (141 digits)Dario Alpern (10/2005)
2859960*3286+1 (141 digits)Dario Alpern
2861127030*3287+1 (143 digits)Dario Alpern (10/2005)
30881816*3309+1 (153 digits)Dario Alpern
3145826134*3315+1 (158 digits)Dario Alpern (10/2005)
3192*3320+1 (153 digits)Dario Alpern
32516950610*3326+1 (163 digits)Dario Alpern (1/2006)
33024*3331+1 (160 digits)Dario Alpern
3333108970*3334+1 (166 digits)Dario Alpern (1/2006)
362109064*3363+1 (179 digits)Dario Alpern
398528*3399+1 (194 digits)Dario Alpern
405413610*3406+1 (200 digits)Dario Alpern (10/2005)
499316072*3500+1 (245 digits)Dario Alpern (10/2005)
525168074*3526+1 (257 digits)Dario Alpern
5603522382*3561+1 (275 digits)Eric Jeancolas (9 Dec 2022)
67914922*3680+1 (329 digits)Dario Alpern (10/2005)
6946*3695+1 (333 digits)Dario Alpern
7061974806*3707+1 (344 digits)Dario Alpern (1/2006)
72438102*3725+1 (351 digits)Dario Alpern (10/2005)
7812*3782+1 (374 digits)Dario Alpern
7974650*3798+1 (385 digits)Dario Alpern (10/2005)
8206*3821+1 (393 digits)Dario Alpern
9367262*3937+1 (451 digits)Dario Alpern (10/2005)
9798*3980+1 (469 digits)Dario Alpern
122326*31224+1 (586 digits)Dario Alpern
12512*31252+1 (598 digits)Dario Alpern
136614*31367+1 (654 digits)Dario Alpern
140748*31408+1 (674 digits)Dario Alpern
14532*31454+1 (695 digits)Dario Alpern
1530558*31531+1 (734 digits)Dario Alpern (10/2005)
187029758*31871+1 (898 digits)Donovan Johnson (9 Nov 2007)
20561296*32057+1 (985 digits)Dario Alpern (10/2005)
2183458*32184+1 (1045 digits)Dario Alpern (10/2005)
22522512*32253+1 (1079 digits)Dario Alpern (11/2005)
240814*32409+1 (1151 digits)Dario Alpern
245148*32452+1 (1172 digits)Dario Alpern
262332*32624+1 (1254 digits)Dario Alpern
2703210*32704+1 (1293 digits)Dario Alpern (11/2005)
2789210*32790+1 (1334 digits)Dario Alpern (11/2005)
40473872*34048+1 (1935 digits)Eric Jeancolas (22 Nov 2022)
45998*34600+1 (2196 digits)Dario Alpern (11/2005)
54792*35480+1 (2615 digits)Dario Alpern (12/2005)
58281606*35829+1 (2785 digits)Eric Jeancolas (3 Dec 2022)
6313362*36314+1 (3016 digits)Eric Jeancolas (22 Nov 2022)
78406*37841+1 (3742 digits)Dario Alpern (12/2005)
8851338*38852+1 (4227 digits)Eric Jeancolas (22 Nov 2022)
9409160*39410+1 (4492 digits)Dario Alpern (1/2006)
Exponents 0 to 5 are completely factored.
For exponent 6 to 9 I ran 300 ECM curves with B1 = 50000 and 700 with B1 = 250000.
For exponents 10 and 11 I ran 250 ECM curves with B1 = 50000.
All cofactors for exponents 6 to 12 were found composite by Donovan Johnson on 9 November 2007.
The limits in the search for factors are:
• Exponents 11 to 37: 2*1011
• Exponents 38 to 64: 2*1010
• Exponents 65 to 200: 5*109
• Exponents 201 to 283: 2*108
• Exponents 284 to 400: 2*107
• Exponents 401 to 801: 2000000
• Exponents 802 to 1400: 200000
• Exponents 1401 to 2900: 20000
• Exponents 2901 to 5000: 2000
• Exponents 5001 to 10000: 200
Factors found so far: 144
Prime factors of 43n + 23n + 1
ExponentPrime factorsDiscoverer
02*31+1 = 7
18*32+1 = 73
29728*33+1 = 262657
332*34+1 = 2593
878*34+1 = 71119
1205998*34+1 = 97685839
42*35+1 = 487
68 945611600*35+1 = 1675 3783618801
794 122245632*35+1 = 19297 1705688577
1527979 4910498594*35+1 = 371299016 3251158343
5110*36+1 = 80191
134*36+1 = 97687
520*36+1 = 379081
911835396 9088590993 8096624124 8655269152*36+1Robert Silverman
Prime cofactor with 90 digitsRobert Silverman
618*37+1 = 39367
3477936*37+1 = 7606246033
120345960*37+1 = 26 3196614521
241912920*37+1 = 52 9063556041
732*38+1 = 209953
198*38+1 = 1299079
1067874826 9716005072*38+1 = 700 6326739760 6709277393
87192*39+1 = 141560137
298 0726001840*39+1 = 5866962 9894216721
14 8184916806 7520103946 5100468042*39+1Dario Alpern (9/2005)
98*310+1 = 472393
686184*310+1 = 40518479017
1151958*310+1 = 6 8021967943Yannick Saouter (1995)
2819752928*310+1 = 16650 3590645473Donovan Johnson (9 Nov 2007)
106175 5910828066*311+1 = 1093987433 5459407703Donovan Johnson (9 Nov 2007)
11166*312+1 = 88219207Yannick Saouter (1995)
122735946*313+1 = 436 1981634559Yannick Saouter (1995)
133 5198840558*314+1 = 16835496 3224856703Dario Alpern (11/2005)
1418*315+1 = 258280327Yannick Saouter (1995)
1626*317+1 = 3357644239Yannick Saouter (1995)
1698664*317+1 = 21936 5745842233Yannick Saouter (1995)
1731078*318+1 = 1204 0253957143Yannick Saouter (1995)
19190*320+1 = 66 2489036191Yannick Saouter (1995)
1677086*320+1 = 584763 7303935487Yannick Saouter (1995)
218*322+1 = 25 1048476873Yannick Saouter (1995)
23126*324+1 = 3558 6121596607Yannick Saouter (1995)
261386*327+1 = 1056907 8114191983Yannick Saouter (1995)
292483478*330+1 = 5 1132609695 2154709223Yannick Saouter (1995)
1 3802075504*330+1 = 28417 2495077438 3172378097Dario Alpern (10/2005)
5 7791187848*330+1 = 118986 9309111924 2074625353Dario Alpern (1/2006)
30298*331+1 = 18406667 2092616207Yannick Saouter (1995)
33288*334+1 = 480302832 9503971873Yannick Saouter (1995)
3419 5296351960*335+1 = 97709782 4075685982 2208875721Dario Alpern (1/2006)
39128*340+1 = 15 5618117875 9286886529Yannick Saouter (1995)
15152*340+1 = 1842 1294703563 0585192753Yannick Saouter (1995)
4288*343+1 = 288 8661313071 9262831177Dario Alpern
44287122*345+1 = 8482481 7293028847 3221775447Dario Alpern
45539382*346+1 = 47805092 8885440534 6124375879Dario Alpern
4640160*347+1 = 10678067 8465573333 2037525921Dario Alpern
476288*348+1 = 5015713 9406737434 2020813969Dario Alpern
501289566968*351+1 = 2777 3325939632 5098361020 2554605097Dario Alpern (10/2005)
556*356+1 = 31400857 9816416322 3281069127Dario Alpern
572304*358+1 = 10 8521365184 5534809965 9374899457Dario Alpern
642*365+1 = 20 6021029217 5507490794 7094535687Dario Alpern
602*365+1 = 6201 2329794482 7754729207 5455241487Dario Alpern (10/2005)
75923680232*376+1 (46 digits)Dario Alpern (11/2005)
7626*377+1 (39 digits)Dario Alpern
8159425822*382+1 (47 digits)Dario Alpern (10/2005)
871140517862*388+1 (52 digits)Dario Alpern (1/2006)
9426*395+1 (47 digits)Dario Alpern
9522*396+1 (48 digits)Dario Alpern
9914008*3100+1 (52 digits)Dario Alpern
1023883390160*3103+1 (59 digits)Dario Alpern (2/2006)
103161129448*3104+1 (58 digits)Dario Alpern (10/2005)
115537432*3116+1 (62 digits)Dario Alpern (10/2005)
11814010*3119+1 (61 digits)Dario Alpern
120354*3121+1 (61 digits)Dario Alpern
121762630*3122+1 (65 digits)Dario Alpern (10/2005)
1298*3130+1 (63 digits)Dario Alpern
136379406664*3137+1 (74 digits)Dario Alpern (3/2006)
1582266*3159+1 (80 digits)Dario Alpern
16170*3162+1 (80 digits)Dario Alpern
2041362*3205+1 (101 digits)Dario Alpern
282154*3205+1 (104 digits)Dario Alpern (10/2005)
22154264*3222+1 (111 digits)Dario Alpern
225247968*3226+1 (114 digits)Dario Alpern (10/2005)
227192*3228+1 (112 digits)Dario Alpern
2461223920*3247+1 (124 digits)Dario Alpern (10/2005)
27310152*3274+1 (135 digits)Dario Alpern
28810*3289+1 (139 digits)Dario Alpern
315626280*3316+1 (157 digits)Dario Alpern (10/2005)
32924*3330+1 (159 digits)Dario Alpern
36332*3364+1 (176 digits)Dario Alpern
4307136*3431+1 (210 digits)Dario Alpern
43231890*3433+1 (212 digits)Dario Alpern
57132*3572+1 (275 digits)Dario Alpern
5752272*3576+1 (279 digits)Dario Alpern
626766514*3627+1 (306 digits)Dario Alpern (1/2006)
63050*3631+1 (303 digits)Dario Alpern
653224*3654+1 (315 digits)Dario Alpern
6662408*3667+1 (322 digits)Dario Alpern
56976*3667+1 (323 digits)Dario Alpern (10/2005)
76130*3762+1 (366 digits)Dario Alpern
78032*3781+1 (375 digits)Dario Alpern
7881069946*3789+1 (383 digits)Dario Alpern (1/2006)
8956*3896+1 (429 digits)Dario Alpern
100034*31001+1 (480 digits)Dario Alpern
10051422*31006+1 (484 digits)Dario Alpern (10/2005)
106826*31069+1 (512 digits)Dario Alpern
11542544*31155+1 (555 digits)Dario Alpern (10/2005)
120696*31207+1 (578 digits)Dario Alpern
1257387478*31258+1 (606 digits)Eric Jeancolas (6 Dec 2022)
128024*31281+1 (613 digits)Dario Alpern
1556267322*31557+1 (749 digits)Eric Jeancolas (6 Dec 2022)
15584336*31559+1 (748 digits)Dario Alpern (10/2005)
16548*31655+1 (791 digits)Dario Alpern
1734178*31735+1 (831 digits)Dario Alpern
212026*32121+1 (1014 digits)Dario Alpern
23436822*32344+1 (1123 digits)Dario Alpern (11/2005)
25058934*32506+1 (1200 digits)Dario Alpern (11/2005)
28124616*32813+1 (1346 digits)Dario Alpern (11/2005)
3709302*33710+1 (1773 digits)Dario Alpern (11/2005)
42162*34217+1 (2012 digits)Dario Alpern (11/2005)
5088274*35089+1 (2431 digits)Eric Jeancolas (22 Nov 2022)
527132*35272+1 (2517 digits)Dario Alpern (12/2005)
571970*35720+1 (2731 digits)Dario Alpern (12/2005)
622218*36223+1 (2971 digits)Dario Alpern (12/2005)
8757198*38758+1 (4181 digits)Dario Alpern (1/2006)
Exponents 0 to 5 are completely factored.
For exponent 6 to 9 I ran 300 ECM curves with B1 = 50000 and 700 with B1 = 250000.
For exponent 10 I ran 300 ECM curves with B1 = 50000.
All cofactors for exponents 6 to 12 were found composite by Donovan Johnson on 9 November 2007.
The limits in the search for factors are:
• Exponents 11 to 37: 2*1011
• Exponents 38 to 64: 2*1010
• Exponents 65 to 200: 5*109
• Exponents 201 to 283: 2*108
• Exponents 284 to 400: 2*107
• Exponents 401 to 801: 2000000
• Exponents 802 to 1400: 200000
• Exponents 1401 to 2900: 20000
• Exponents 2901 to 5000: 2000
• Exponents 5001 to 10000: 200
Factors found so far: 123
If you have a number that can be in these tables or you have any comment please fill the form.
In October 2005 I wrote a program to find factorizations of numbers of the form (2^pe+1)-1)/(2^pe-1) and (2^pe+1+1)/(2^pe+1). The numbers in this page can be factored using p=3. In December 2022 I changed it to execute on 64-bit processors, dropping support for 32-bit operating systems. You can download it here. It includes both the source code and the Windows executable. The source code can be compiled in Linux also by typing:
`gcc -O2 genferm.c montmult.s -o genferm -lm`
Last updated on December 16th, 2022. | 6,648 | 15,024 | {"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} | 4.09375 | 4 | CC-MAIN-2024-18 | latest | en | 0.41763 |
https://www.physicsforums.com/threads/can-someone-explain-to-me-what-a-function-arrow-is.564138/ | 1,542,348,055,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742978.60/warc/CC-MAIN-20181116045735-20181116071106-00061.warc.gz | 968,538,208 | 14,180 | # Homework Help: Can someone explain to me what a function arrow is?
1. Dec 31, 2011
### IntegrateMe
Can someone explain to me what a "function arrow" is?
In particular, what this may mean:
I = [a, b]
f : I --> R
I'm just trying to understand what the arrow means.
Or if we have R --> R what that would mean.
Thank you!
2. Jan 1, 2012
### JHamm
Re: Can someone explain to me what a "function arrow" is?
The thing to the left of the arrow is the domain of the map f and the thing to the right is the range.
As an example a map which returns the square of all real numbers would be
$$f: \mathbb{R} \longrightarrow \mathbb{R}$$
A map which multiplies natural numbers by $\frac{12}{7}$ would be
$$f: \mathbb{N} \longrightarrow \mathbb{R}$$
And a map which takes a number from 1 to 4 and returns a complex number would be
$$f: [1, 4] \longrightarrow \mathbb{R}^2$$
3. Jan 1, 2012
### DavidAlan
Re: Can someone explain to me what a "function arrow" is?
It is just notation. It means that the map entails some correspondence between sets. In English, the arrow points to the possible set of outputs of the map. If your map takes as its domain some set and "points" to the set of complex numbers, then your map is "complex valued," meaning that it takes inputs that are elements of the set at the beginning of the arrow, and outputs elements of the complex numbers (i.e. numbers of the form a+ib).
For example, if f: R^2 --> C is defined by f(x,y)=x+iy then f turns elements of R^2 into "high school" complex numbers.
4. Jan 4, 2012
### IntegrateMe
Re: Can someone explain to me what a "function arrow" is?
JHamm,
If we use the example of:
f: R --> R
You said that it would return the square of all real numbers. But wouldn't it just return all real numbers, since the range, which is to the right, is R? The second one which involves 12/7 is also confusing me. Can you, or anyone for that matter, please clarify this for me.
Thanks for the help, guys, greatly appreciated.
Edit: Is this how you're describing the function f? For example, could I have said that f(x) = x^3 so that f: R --> R would actually return the cube of all real numbers?
Last edited: Jan 4, 2012
5. Jan 4, 2012
### Deveno
Re: Can someone explain to me what a "function arrow" is?
the usual notation is:
$$f: A \to B$$
what is meant by this is:
for any element a in the set A, f(a) is defined, and is an element of the set B. B is called the "co-domain" of f, and is not to be confused with the "range" of f (or the image set), which is f(A). f(A) is always a subset of B, it may not be all of B itself.
for example, if we have:
$$f:\mathbb{R} \to \mathbb{R}, f(x) = x^2$$
then we are saying f (the "squaring function") is defined for every real number x, and that x2 is also a real number. as it turns out, x2 is always non-negative, so in this case, the range of f is considerably smaller than the co-domain.
in general, you can always make the domain (the set A) smaller (this is called a restriction), and the co-domain bigger. in calculus, it is common to choose the entire real numbers as the co-domain, rather than trying to figure out ahead of time what the actual range is. often, we are only concerned with the behavior of f on a particular kind of subset of R, an interval (usally denoted I, or [a,b] or (a,b) if we wish to specify whether the interval includes the boundary points).
often, A is called "the source (or start) set", and B the "target set" (and the arrow is suggestive of some kind of motion, or action, or transformation).
6. Jan 5, 2012
### JHamm
Re: Can someone explain to me what a "function arrow" is?
Exactly, the bit with the arrow doesn't tell you what the function really does, it just tells you the set of its domain and the set of its range. | 1,034 | 3,784 | {"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} | 3.953125 | 4 | CC-MAIN-2018-47 | latest | en | 0.917552 |
http://www.happylearnhaskelltutorial.com/1/skwak_the_squirrel.html | 1,701,625,478,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00476.warc.gz | 75,138,336 | 12,117 | # Happy Learn Haskell Tutorial Vol 1
Written and illustrated by GetContented.
Published on 2017-07-08.
## Contents
Previous chapter: 18. Times-Table Train of Terror
19. Skwak the Squirrel
... 19.1. More on the (\$) Function
... 19.2. Mid-Lesson Homework
... 19.3. Continuing On
... 19.4. Problems and Homework
Next chapter: 20. Basic Input
# 19. Skwak the Squirrel 🔗
Games! We saw the Fridge game. It took place in one single “room” which made it very limited. Then, the Train game didn’t really give you any freedom, but at least it had more rooms.
Next we’ll see a game that lets us imagine that we’re a squirrel and we live in a tree. Our new game will only have two areas, but it will provide more capability than before. And, in the process, we’ll get some more practice with all the things we’ve seen so far.
If we wanted to make a game that is a tiny bit more like a real text adventure game, it’d have to let the player move around between its game areas.
To keep it ultra-simple, let’s say we (our squirrel) could be able to go between only the inside and outside of the tree, perhaps. The program we’ll be discovering in this section is a lot more complicated than the Fridge game, but it’s about one of the most simple, basic text adventures possible.
Let’s look at the types first:
``````
data GameObject = Player
| Acorn
deriving (Eq, Show)
data Room =
Room Description [GameObject]
deriving (Show)
type Description = String
type Inventory = [GameObject]
type GameMap = [Room]
type GameState = (GameMap, Inventory)
``````
We have a `GameObject` type whose values can be either `Player` or `Acorn`. Fairly straightforward, this is just a sum type like we’ve seen before. What about `deriving (Eq, Show)`, though? Well, this is a way to make a type become an instance of these typeclasses without having to write the code for it manually ourselves. Being an instance of `Eq` means we can use `(==)` and other comparison functions on values of this type, and being an instance of `Show` means it can be converted to strings with the `show` function.
Now, the `Room` type is a product type and it has a type constructor called `Room` as well as a value constructor of the same name, which is used to make values of the `Room` type. It has fields of `Description`, and a list of type `GameObject` which is used to hold the contents of the `Room`. So, because of the way the `GameObject` type is defined, a `Room` can have one or more `Player` or `Acorn` values in it. Our game will only ever have one of each in the whole game.
Next are a bunch of type synonyms which should be pretty easy to understand by now, possibly with the exception of `GameState` which is a 2-Tuple (otherwise known as a pair) of `GameMap`, and `Inventory`, which is a list of `GameObject`. `GameMap` is a list of `Room`.
The `GameState` type will be what we use to store all the changing pieces of our game as the player plays it. The `GameMap` will be all the rooms in the game, and the `Inventory` is what the player is holding moment by moment.
Let’s see a definiton for the initial state of the game and the `main` function, that will start the game.
``````
initialState :: GameState
initialState =
( [ Room "You are inside a tree." [Player]
, Room "You are outside of a tree." [Acorn]]
, [] )
main :: IO ()
main = do
putStrLn "Welcome to Skwak the Squirrel."
putStrLn "You are a squirrel."
gameLoop initialState
``````
Now we see the `main` function, which as we know is the `IO` action that will be executed by Haskell when we compile and run our program.
All it does it announce the game, then runs a function called `gameLoop` using the `intialState` which is the state that the game should start with. The `gameLoop` contains the bulk of the program, and we’ll see it in a moment.
First, Let’s look at `initialState`. This is a `GameState`, which as we know from the types is a 2-Tuple that has a list of `Room` then a list of `GameObject`. We start our game with two rooms... we use the `Room` constructor to build the rooms. We have the outside and the inside of the tree. The `Player` is inside, and the `Acorn` is outside.
Let’s look at the `gameLoop` function now:
``````
gameLoop :: GameState -> IO ()
gameLoop (rooms, currentInv) = do
let currentRoom =
case findRoomWithPlayer rooms of
Just r -> r
Nothing -> error \$ "Somehow the player "
++ "ended up outside the map!"
possibleCmds =
validCommands currentRoom currentInv
if playerWon (rooms, currentInv)
then gameOverRestart
else do
describeWorld currentRoom currentInv possibleCmds
takeActionThenLoop
currentRoom currentInv possibleCmds rooms
``````
This function takes a `GameState` and returns an `IO` action. It’s the `main` functionality of the game. Each move the player makes goes through the game loop once, which is why it’s called a loop, because it’s like a circle.
We see that we can have a `let` expression in our `do` block which essentially sets up temporary variables in the `do` block from that point onwards.
We’re finding and grabbing the current room from the list of rooms, or throwing up our hands if it can’t find the `Player`.
Then we work out which commands are valid for the current room and the current inventory. Some player commands are only possible with some combination of `Acorn` being in the room with the player, or in the inventory.
Once that is done, we check if the player has won with an `if` expression, and if so, we tell the player they’ve won and offer to play again. If they didn’t win, then we describe the world to them at the room they’re in, and let them take an action by grabbing their command, then go back again to the start (using the `takeActionThenLoop` function).
Now we’ll go through the remaining functions. Note, though, that we can use any expressions (such as `if`) with no trouble in a `do` block for `IO`, even other nested `do` blocks, so long as those expressions result in an `IO` value of some kind. We’ll see in a moment that all of those functions we just discussed yield `IO` values of some kind.
``````
findRoomWithPlayer :: [Room] -> Maybe Room
findRoomWithPlayer rooms =
L.find (\(Room _ obs) ->
any (== Player) obs)
rooms
``````
We can tell by the name that `findRoomWithPlayer` should return the `Room` whose objects include the `Player`. We’re returning a `Maybe Room` because it’s technically possible that the `Player` might not be in any `Room`. If that’s the case, there’s a problem, because the game won’t work. That’s why there’s an application of the `error` function in `gameLoop` if `findRoomWithPlayer` can’t find a `Player`.
We’re using a lambda as our Room-testing function here. The way find works is it looks through the list passed into it, applies the predicate section function `(== Player)` to each item, and if it finds one that returns `True`, it returns that one, wrapped in the `Just` data constructor from the `Maybe` type. If it doesn’t, it returns the `Nothing` value (also from the `Maybe` type.
In depth, our lambda takes a `Room`, pulls it apart using pattern-matching to get at the objects (which is a list of `GameObject`), names that `obs`, and then passes that to `any`, which will check to see if any of them return `True` for the function `(== Player)` which checks to see if something is equal to the `Player` value.
Next we’ll see the function that crafts the valid commands for a particular room and inventory combination:
``````
validCommands :: Room -> Inventory -> [String]
validCommands (Room _ gameObjs) invItems =
["go"] ++ takeCommandList
++ dropCommandList ++ ["quit"]
where
takeCommandList =
if somethingToTake gameObjs
then ["take"]
else []
dropCommandList =
if length invItems > 0
then ["put"]
else []
``````
This function takes a `Room` and an `Inventory` and returns a list of `String` values that are all the commands that a player can type in depending on the current data: whether there is anything to take (in the room), or drop (from inventory).
It’s using the list concatenation operator `(++)` to join the `String` lists together into one list to return, and two definitions with `if` expressions to ensure we only put “take” or “put” on the command list if it’s appropriate.
``````
somethingToTake :: [GameObject] -> Bool
somethingToTake objs =
any (/= Player) objs
``````
Of course, we can eta-reduce this function as we know by now (that is, get rid of the repeated trailing arguments):
``````
somethingToTake :: [GameObject] -> Bool
somethingToTake = any (/= Player)
``````
Here we take a list of `GameObject` which is used by `validCommands` as the list of objects in the room that can be taken. We’re using it to work out if there is anything that is not a `Player` in the list. The `any` function takes a predicate function, evaluates it on each item of a list, and returns `True` if any of the evaluations return `True`, otherwise `False`. The function `(/=)` means not equal to, and comes from the `Eq` typeclass. The end effect answers the question “Are there any non-player objects in this room?”.
As we’ve seen before, the use of parentheses around `(/= Player)` makes it into what’s called a section: it creates a function of one argument, having applied one of the 2-arguments required by the operator. Here we’re creating a function that takes a `GameObject` and returns a `Bool` that indicates whether the `GameObject` was `Player`.
``````
playerWon :: GameState -> Bool
playerWon (rooms, currentInv) =
any hasAcornAndInside rooms
where hasAcornAndInside (Room desc objs) =
desc == "You are inside a tree."
&& any (==Acorn) objs
``````
This function answers the question “Has the player won yet?” The player has won if the acorn is in the tree (and not in the player’s inventory). Pay careful attention to the indentation (where the lines start). This matters!
We test for this by pulling the `GameState` tuple apart into a variable each for rooms and current inventory. We then go through all the rooms, and using the `any` function again, check if there’s one which has the description “You are inside a tree.” that also has the Acorn in its objects list.
Note the use of the `&&` operator, which takes two `Bool} expressions and returns \{True` if they’re both `True`. This is called the logical and operator. It’s one way we can connect up logic expressions into larger chunks of meaning. There is also an `(||)` operator that is logical or, which will return `True` if either of its inputs evaluate to `True`, and a `not` function that takes only one `Bool` expression and returns the logical inverse of it (that is, if it’s `True`, it returns `False` and vice-versa).
Our `playerWon` function is a not the best way we could write it, because we’re relying on the description of the room to check if it’s the inside room. This gives us a lot of room for errors to creep in by accident. We did this to keep the types a bit simpler as you’re learning. It’s bad because what if we changed the description in the actual `Room` data, but then forgot to change this check `String`. It would mean it’d be impossible to win the game.
A better way would be to pull this description into its own definition that is defined in only one place. The best way, though, would be to have an identifier in the `Room` data type representing the type of `Room` it is, and to check against this to see if this `Room` was the winning goal `Room`.
``````
gameOverRestart :: IO ()
gameOverRestart = do
putStrLn \$ "You won!"
++ "You have successfully stored the acorn"
++ " for winter. Well done!"
putStrLn "Do you want to play again? y = yes"
playAgain <- getLine
if playAgain == "y"
then gameLoop initialState
else putStrLn "Thanks for playing!"
``````
Our function to check if the game is now over uses a `do` block to print some messages congratulating the player, then asks if they want to play again, collects a line of input from them, and restarts the game from the beginning by using `initialState` if their input equals `"y"`. Very straight-forward!
``````
getCommand :: IO String
getCommand = do
putStrLn "What do you want to do?"
getLine
``````
`getCommand` is a simple little action that prints a query to the player, gets that response and returns it. The type is `IO String`, which is the same type as `getLine`. Notice that as `getLine` is the last line of the function, we don’t need to use anything to pull the value out here, because it has the return type already. We’ll see how it’s used now in `takeActionThenLoop`:
``````
takeActionThenLoop :: Room ->
Inventory ->
[String] ->
[Room] ->
IO ()
takeActionThenLoop currentRoom
currentInv
possibleCmds
rooms =
do
command <- getCommand
if any (==command) possibleCmds
then case command of
"go" ->
do
putStrLn "You go..."
gameLoop \$ movePlayer (rooms, currentInv)
"take" ->
do
putStrLn "You take the acorn..."
gameLoop \$
moveAcornToInventory (rooms, currentInv)
"put" ->
do
putStrLn "You put the acorn down..."
gameLoop \$
moveAcornFromInventory (rooms, currentInv)
"quit" ->
putStrLn \$ "You decide to give up."
++ " Thanks for playing."
_ ->
do
putStrLn "That is not a command."
gameLoop (rooms, currentInv)
else do
putStrLn \$ "Command not possible here,"
++ " or that is not a command."
gameLoop (rooms, currentInv)
``````
Even though this is one of the largest functions in the program, it’s reasonably simple. It’s structured by firstly taking a command from the user, then checking if the command is one of the valid ones for the player right now. If not it complains that it can’t understand and just goes back to the `gameLoop` with the current data. However, if it does actually match against a valid command, it runs through the case expression trying to find a match.
Each of the commands has a corresponding function which results in a modified `GameState`, which is then passed back into the `gameLoop` function for a further turn. We’ll see each of these functions momentarily.
Notice that we use the `(\$)` function all over the place, such as to apply `gameLoop` to the result of the evaluation of the command functions, such as `movePlayer`.
Also notice that there is a pattern match for “`_`”, which for completeness will match on any other values for the `command`. In our case there aren’t actually any other commands, but not having this is a potential source of problems in the future as things change, and Haskell will complain if we leave it off. Take a moment and think what would happen if you added a new command to the `validCommands` function, but didn’t add it to the `takeActionThenLoop` function. If we keep this underscore catch-all, then play the game, the game will tell us that our new command is not recognised as a command. If we leave it off, though, our program will crash. That’s why it’s better to have total functions.
Having said this, a better solution would be to represent our commands with an algebraic data type. For now and to keep things slightly more simple, we’ll stick to the `String` type.
These next three functions are to adjust the game state when the player does something.
``````
movePlayer :: GameState -> GameState
movePlayer (rooms, inv) =
(newRooms, inv)
where
newRooms =
if any (==Player) objs
then (Room d (filter (/=Player) objs))
else (Room d (Player : objs))
moveAcornToInventory :: GameState -> GameState
moveAcornToInventory (rooms, inv) =
(newRooms, newInv)
where
newInv =
Acorn : inv
newRooms =
Room d (filter (/=Acorn) objs)
moveAcornFromInventory :: GameState -> GameState
moveAcornFromInventory (rooms, inv) =
(newRooms, newInv)
where
newInv =
filter (/=Acorn) inv
newRooms =
if any (==Player) objs
then Room d (Acorn : objs)
else Room d objs
``````
Our game only has two rooms, so the `movePlayer` function just goes over “all” two rooms using the `map` function, and applies a function that takes the `Player` out if it’s there using `filter`, or puts it in if it’s not there, using the list-prepend function `(:)`. It does this by pattern-matching the `Room` into its pieces, then reconstructing it with a new `GameObject` list.
The `moveAcornToInventory` function returns a new 2-tuple (a `GameState`) that is comprised of an adjusted set of rooms by mapping a deconstructing-reconstructing function that uses `filter` to remove all acorns from the rooms’ `GameObject` list (there should only be one anyway), and an adjusted inventory by prepending an `Acorn` to it using `(:)`. This “moves” the `Acorn` into inventory.
The `moveAcornFromInventory` does the reverse of this. The mapping function that puts the `Acorn` into the room where the player is is quite interesting because it has an `if` expression to detect if a particular room’s game objects contains the `Player`, and if so, it prepends an `Acorn`, otherwise it just re-builds the passed in room as it was before.
``````
describeWorld :: Room ->
Inventory ->
[String] ->
IO ()
describeWorld currentRoom
currentInv
possibleCmds =
do
putStrLn \$ describeRoom currentRoom
putStrLn \$ describeInventory currentInv
putStrLn \$ describeCommands possibleCmds
``````
This is pretty straightforward, it just uses other functions to describe to the player which room they’re in, what they’re carrying and also which commands they can use here.
Notice, again, that we’re using another `do` block to make a single `IO` action out of a bunch of `putStrLn` function applications.
## 19.1. More on the (\$) Function 🔗
Also we see the function `(\$)` again at work, applying the function on the left of it to the result of evaluating everything on the right of it. The function application operator is often used as above, to avoid having to type brackets around expressions, as you probably know by now. It has the lowest possible precedence, which means it’s evaluated last, after all the other parts of an expression.
The type signature of this is `(\$) :: (a -> b) -> a -> b` which means it takes a function from `a` to `b` (on the left of it) and an `a` (on the right of it), and applies the function to the “a” which gives a `b`. Obviously this is an extremely general function because `(\$)` can apply to any function, and any value as long as they fit together.
Again, it’s important to realise that even though the type signature of `(\$)` says it takes a function from `a` types to `b` types, it doesn’t have to be different types, but they can be if you like. So, `(+3)` fits the description even though its type is `Num a => a -> a` and this is because `(a -> b)` goes from one “any type” to another “any type”, so not necessarily the same type. `(a -> b)` is actually the type that fits any type of function at all!
There is no difference in result between writing `(+3) \$ 7` and writing `(+3) 7`, but when you start adding more arguments it starts to matter, like in `(+3) \$ 7 * 5` which gives `38`, a different answer than `(+3) 7 * 5`, which is 50. From this we can see that `(\$)` changes the order that things are evaluated in, just like putting parentheses around some things (so, `(+3) \$ 7 * 5` is equivalent to `(+3) (7 * 5)`).
## 19.2. Mid-Lesson Homework 🔗
Your homework (yes, mid-lesson homework) is to experiment with the `(\$)` function, and explicit bracketing, and explore the differences and similarities if you can.
## 19.3. Continuing On 🔗
Next, the three functions that `describeWorld` uses:
``````
describeRoom :: Room -> String
describeRoom (Room desc objs) =
desc ++ if any (==Acorn) objs
then " There is an acorn here"
else ""
describeInventory :: Inventory -> String
describeInventory [] =
"You are holding nothing"
describeInventory inv =
"You are holding: " ++
(concat \$ map show inv)
describeCommands :: [String] -> String
describeCommands commands =
"Commands: "
++ (L.intercalate ", " commands)
``````
The `describeRoom` function takes a single `Room` and pattern-matches it into the `desc` and `objs` variables, which are the description of the `Room`, and the objects that the `Room` contains respectively. Notice that we’re using `(++)` to append the result of an `if` expression to the `Room`’s description here. The `if` expression uses a section. Remember that a section is effectively a partially applied operator. In this case, `(==Acorn) :: GameObject -> Bool` will check if something is an `Acorn`, and return a `Boolean` value for that (`True`/`False`).
Let’s think some more about the function `any :: Foldable t => (a -> Bool) -> t a -> Bool` for a moment. This is a function that takes a predicate from types `a` to `Bool`, a `Foldable t a` and then returns a `Bool`. This function will tell us if any item in the list “satisfies” the predicate. That is, if any item in the list returns `True` for our predicate function.
Or, partially applied to the `(==Acorn)` section as in `any (==Acorn)`, it will tell us if there’s an `Acorn` in a supplied list! If there is an `Acorn`, it’ll say so. If not, just an empty `String`.
Next we’ll look at the `describeInventory` function. There are two definitions on this function. The first is for when there is nothing in the passed-in `Inventory`, so we’re matching on an empty list. This is simple.
The second definition is where the meat is. If the match with the empty list failed, that means there must be something in the `Inventory`, which we now pattern-match to the “inv” variable.
We then use the list concatenation operator (++) to append the result of an expression to the end of the beginning of a description of what they player is holding. Of course, this works because `String` is `[Char]`, so therefore a list.
The expression we’re appending is `(concat \$ map show inv)`. We can break this apart into its pieces. Firstly, we now know about the `(\$)` function, and can think about it like it’s bracketing like this: `(concat (map show inv))`.
Thinking about the expression `map show inv`, we can see that this takes the `inv` list and returns a new list that it builds by applying the `show` function to each `GameObject` in it, turning it into a list of strings. We can see this clearly if we check its type: `map show :: Show a => [a] -> [String]`.
The function `concat :: Foldable t -> t [a] -> [a]` takes a `Foldable`-wrapped list of type `a`, and returns a list of type `a`. So in this case, because we’re passing it a `[String]`, which as you know is the same as a `[[Char]]`. So that means the `Foldable t` that `concat` needs will be the list type, so a list of list of `Char`. The `concat` function will concatenate (join) that `[String]` to be a single `String` (or turn the `[[Char]]` into a `[Char]`, same thing), by joining all the strings into one `String`.
The end result is that `describeInventory` will either tell the player they’re holding nothing, or that they’re holding a list of what they’re holding. The fact that there will only ever be one item in the list means we don’t need to comma-separate it or anything.
Next is the `describeCommands` function. It takes a list of commands, which are just `String` values, and calls the `intercalate` function from `Data.List` on that it. This is a pretty neat function. Its type is `[a] -> [[a]] -> [a]`. The first argument is a separator, and the second is a list of items of the same type. It then joins them all together into one big list after putting the separator between each of the items. So, here we’re using it to separate all the commands by comma-space so that it reads nicely.
## 19.4. Problems and Homework 🔗
While this game is designed to be small and simple for teaching purposes, it’s not hard to see that it would be pretty easy to adjust it to have a larger map, or more items. There are some problems, though.
Your homework is to go back through the chapter and think about what the problems of this program are as it’s built. For example, what would happen to the program if you added an additional room? Why not try it out and see.
It can be very useful to think about all the possible ways that programs could be changed, and how important it is to design programs in ways that make extending them easier, and also to make them resilient to change in as many ways as possible. This game is very easy to extend in some ways, but it’s not written to be extended at all in other ways.
Ideally when we write programs, we make it very easy to adjust them. We can do this by building them up from smaller pieces that closely match the type of data and functionality we want. For example, perhaps a `List` isn't a very good data type to use for our game map, and perhaps when we wrote our map movement functions, we could have written them in a way that meant they would work on any size `List`, not just a two-element one.
We can see from the language Haskell itself that its designers have followed this pattern because this language is very easy to extend, and encourages a style of programming where we build the language up to the problem we’re trying to solve. | 6,076 | 24,841 | {"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} | 2.8125 | 3 | CC-MAIN-2023-50 | latest | en | 0.911011 |
https://mathoverflow.net/questions/332152/if-a-distance-metric-on-a-connected-riemannian-manifold-locally-agrees-with-th | 1,561,496,397,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999946.25/warc/CC-MAIN-20190625192953-20190625214953-00166.warc.gz | 521,090,886 | 33,426 | If a (distance) metric on a connected Riemannian manifold locally agrees with the Riemannian metric, is it equal to the induced metric?
Let $$(M,g)$$ be a connected Riemannian manifold. Let $$d_g$$ be the induced distance metric of $$g$$. Now let $$d$$ be some other metric on $$M$$.
Suppose that for each $$x \in M$$, there is a neighborhood $$U$$ of $$x$$ so that $$d = d_g$$ on $$U \times U$$.
Question: Does this imply that $$d = d_g$$ on $$M \times M$$?
I am not assuming that $$d$$ is the metric of some Riemannian metric; I know the answer is yes in that case.
Remarks:
1. It's clear to me that $$d = d_g$$ on a neighborhood of the diagonal in $$M \times M$$; this essentially is just a rephrasing of the hypothesis.
2. Connectedness is necessary to avoid stupid counter examples. For example, we could take two disjoint points $$\{x,y\}$$. Then $$d_g$$ between them is infinity, but we can take $$d(x,y) = 1$$. They both agree in a neighborhood of $$x$$, namely $$x$$. (We can build similar examples by taking disjoint unions of non-zero dimensional manifolds.)
3. Maybe the 'right' question is actually: let X be a (path) connected topological space and consider two metrics $$g$$ and $$h$$ on X, compatible with the topology on X. If these metrics are locally equal in the sense of the question (i.e. agree on a neighborhood of the diagonal), are they equal? (I feel like this either has an obvious proof or some terrible counter example. It appears to be true on graphs metrized by assigning edge lengths - oops, this is wrong by the cut off metric example.)
4. I (think) I can prove a special case. See motivation section below.
Motivation: Let $$(N,h)$$ be a connected Riemannian manifold, with geodesic distance function $$d_N$$. Let $$G$$ be the group of a covering space action by isometries on $$N$$, with quotient map $$\pi$$, with $$G$$ finite. Then $$N / G$$ inherits a Riemannian metric $$g$$ by using local trivializations. $$N / G$$ also inherits a metric, defined by $$d([x], [y]) = \inf_{g \in G} d_N(gx, y)$$. Locally $$d = d_g$$ because of local trivializations. I want to know if $$d = d_g$$ on all of $$N / G$$.
I think this is true if the geodesic distance between any two points in $$N$$ and $$N / G$$ is always realized by some shortest path, by using the covering space path lifting + being a geodesic is a local condition + isometry of covering actions in order to relate shortest paths in $$N$$ to those in $$N / G$$. That is, if $$g$$ minimizes $$d(gx, y)$$ let $$\gamma : [0,1] \to N$$ be a shortest path in $$N$$ between $$gx$$ and $$y$$. Then $$\pi( \gamma)$$ is a (potentially self intersecting) geodesic, of the same length as $$\gamma$$, connecting $$[x]$$ and $$[y]$$. Hence $$d_g \leq d$$. On the other hand, if we have a geodesic path $$\gamma$$ from $$[x]$$ to $$[y]$$, we can lift it to a geodesic from $$x$$ to $$hy$$ of the same length, for some $$h \in G$$, and hence $$d_g \geq d$$.
I would like to be able to drop this condition about there always being shortest path witnesses to shortest distances (I mean that the inf in geodesic distance is achieved). I think that maybe in this covering space case one can achieve this by taking a sequence of paths approximating the geodesic distance, but the argument starts to get a lot fuzzier, and I am already outside my comfort zone and procrastinating on my actual work as it is...
I would appreciate a reference for either the main question or the motivation.
• You might be looking for the notion of “length metric”. See Bridson—Haefliger’s book Metric spaces of non-positive curvature for details. – HJRW May 22 at 5:55
The answer is no. Take any non-convex region in the plane, and let the Riemannian metric be the ordinary Euclidean metric $$ds^2=dx^2+dy^2$$. Then define the new metric as the infimum of Euclidean diameters of curves connecting $$a$$ to $$b$$. This new metric coincides with the Riemannian metric locally, but does not coincide globally. This metric even has a name: Mazurkiewicz metric.
One can also construct a compact example by taking $$M$$ to be a ramified covering of the sphere, and pullback of the spherical metric. This Riemannian metric in general does not coincide with the Mazurkiewicz metric corresponding to it. (Diameter used in the definition is the diameter of the projection of the curve on the sphere. Projection of the Riemannian-shortest path on $$M$$ can have small diameter but large length).
In general, any distance $$d$$ which locally coincides with the Riemannian distance $$d_R$$ must satisfy $$d\leq d_R$$. Just break the curve on which the Riemannian distance is (almost) achieved into small pieces, and use the triangle inequality for $$d$$.
• Thank you! What about the special case described in the motivation section? (The counter example you suggests that something about the convexity like condition I impose is necessary.) – Lorenzo May 22 at 2:23
• There is no convexity in my second example. – Alexandre Eremenko May 22 at 2:31
• Thanks, I really like your examples! They definitely answer my question. In the motivation section I'm looking at a particular metric, which is just induced by the quotient action (without the Mazurkiewicz metric idea). Since the more general question has a negative answer, I'd be interested in knowing if the case I'm considering there is correct. (I'm sure I can write out a detailed proof if it's necessary at some point, this question just came up talking to a friend about their research. Mostly I'm concerned about missing some key counter example, like the one you shared.) – Lorenzo May 22 at 2:46
• Never mind -- this seems to be answered by Nik Weavers comment below. Thanks a lot for your help! – Lorenzo May 22 at 17:07
Take $$d(x,y) = {\rm min}(|x-y|, 1)$$ on $$\mathbb{R}$$,
• Thanks! I definitely should have thought of this, because I've seen this trick recently in the construction of a metric on path space $C([0,\infty), \mathbb{R})$. As a follow up -- do you know if what I was sketching in the 'motivation section' (the particular case when the Riemannian metric and the metric arise as the quotient of a group action) is correct? – Lorenzo May 22 at 4:31
• Yes, more generally the quotient of any length space by a proper isometric action is again a length space, which implies that the two metrics agree globally. See Proposition 4.1 of these nice notes. – Nik Weaver May 22 at 15:14
• Great - thank you! – Lorenzo May 22 at 17:08
• You are welcome! – Nik Weaver May 22 at 19:26 | 1,716 | 6,511 | {"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": 68, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-26 | longest | en | 0.901574 |
http://dict.youdao.com/w/eng/fitting_a_curve/ | 1,590,409,903,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388427.15/warc/CC-MAIN-20200525095005-20200525125005-00297.warc.gz | 32,138,595 | 8,546 | go top
您要找的是不是:
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曲线拟合
fitting of a curve 曲线拟合 ; 翻译
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文章提出曲线拟合极小极大法，通过研究极小极大剩余向量计算方法，将问题转化为求解相容方程组问题
• The problem of quadrature formula, normalization, symmetrization and peak spreading correction for experimental GPC chromatograms after partial smoothing or fitting a curve is discussed.
最后讨论了公式归一化对称GPC实验加宽改正问题
• Fitting a curve for an experimental energy spectrum of plasma particles with the least square method, a square error sum between the fitted curve and experimental spectrum data is usually minimized.
最小二乘法等离子体粒子能谱实验数据时，通常使函数实验能谱数据之间误差平方和极小化。
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- 来自原声例句 | 316 | 909 | {"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-2020-24 | latest | en | 0.468422 |
https://stats.stackexchange.com/questions/32662/question-about-calculating-mutual-information | 1,560,733,293,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998339.2/warc/CC-MAIN-20190617002911-20190617024911-00006.warc.gz | 612,415,394 | 37,146 | # Question about calculating mutual information
I have two vectors $x,y \in \mathbb{R}.$ Based on the count and vector length I can compute $p(x)$ and $p(y)$ but I have no information on the joint density. How can I calculate mutual information in this case?
• What do you mean by $p(x)$ and $p(y)$? What are you counting? How exactly did these vectors arise--what are they modeling? – whuber Jul 20 '12 at 12:41
• Imagine I have two vectors $x = \begin{pmatrix} -1 \\ -2 \\ -3 \\ 1 \\ 2 \\ 3 \end{pmatrix}$ and $y = \begin{pmatrix} 1 \\ 4 \\ 9 \\ 1 \\ 4 \\ 9 \end{pmatrix}.$ If I were to compute the correlation I need to evaluate $\frac{E\left[(x-\mu_x)(y-\mu_y)\right]}{\sigma_x \sigma_y}$ which I can readily compute from vectors $x,y$ to obtain $corr(x,y) = 0.$ However, for mutual information I need $p(x)$ which I have (p(x=-1) = 1/6, p(x=-2)= 1/6, p(x=-3)= 1/6, p(x=1)= 1/6, p(x=2)= 1/6, p(x=3) = 1/6 and $p(y)$ which I also have p(y=1)=1/3, p(y=4)=1/3, p(y=9)=1/3... – user1137731 Jul 20 '12 at 13:22
• and inferred based on the empirical frequencies counting occurrences and taking the vector length into account. However, then I have no information on the joint density $p(x,y).$ How could I calculate mutual information then? – user1137731 Jul 20 '12 at 13:22
• You seem to be confusing vectors, random variables, and paired observations. Mathematically they may look similar, but they function quite differently. In this case it seems you intend your "vectors" really to be paired observations, whence your "$p$" is really an observed frequency, not a probability. But because you have paired data, you can just as readily estimate the joint density: it's the empirical joint density of $(x,y)$. – whuber Jul 20 '12 at 13:25
• Indeed, I would like to use the observed frequencies to model $p(x)$ and $p(y).$ For my example above, could you please give me pointers on how to compute it as example? – user1137731 Jul 20 '12 at 13:37
The Mutual Information of two discrete random variables $X$ and $Y$ taking values in $R_X$ and $R_Y$ respectively is the difference between the expectation of $\log p(x,y)$ (the logarithm of the joint probability of $(X,Y)$) and the expectation of $\log\left(p(x)p(y)\right)$ (which would be the joint probability for independent variables having marginal probabilities of $p(x)$ and $p(y)$).
When the vectors constitute an iid sample of $(X,Y)$, we can compute the mutual information of their joint empirical density. This is just the observed frequency: if a particular combination of values $(x,y)$ occurs $k(x,y)$ times in the dataset out of $n$ total occurrences, the empirical density $\hat{p}(x,y)$ is just the ratio $k(x,y)/n$.
To compute expectations with respect to the empirical density, let's introduce some notation. Let $R$ ("rows") be the set of distinct observed values of $X$ and $C$ ("columns") the set of distinct observed values of $Y$. For $x\in R$ and $y\in C$, $k(x,*) = \sum_{y\in C}k(x,y)$ is the row sum, counting all elements of the dataset whose first component is $x$. Likewise, $k(*,y) = \sum_{x\in R}k(x,y)$ is the column sum. These determine the marginal densities. Notice that the sum of all the $k(x,y)$, the sum of all the $k(x,*)$, and the sum of all the $k(*,y)$ all count the elements of the dataset, whence they are all equal to $n$. The mutual information equals
\eqalign{ &\sum_{x\in R, y\in C} \frac{k(x,y)}{n} \left(\log \frac{k(x,y)}{n} - \log \left(\frac{k(x,*)}{n} \frac{k(*,y)}{n}\right)\right)\\ =&\frac{1}{n}\sum_{x\in R, y\in C}k(x,y)\left(\log k(x,y) - \log(n)\right)\\ &- \frac{1}{n}\sum_{x\in R}k(x,*)\left(\log k(x,*) - \log(n)\right) \\ &- \frac{1}{n}\sum_{y\in C}k(*,y)\left(\log k(*,y) - \log(n)\right) \\ =&\log(n) + \\ & \frac{1}{n} \left( \sum_{x\in R, y\in C}k(x,y)\log k(x,y) - \sum_{x\in R}k(x,*)\log k(x,*) - \sum_{y\in C}k(*,y)\log k(*,y) \right). }
The first equality is just exploiting properties of logarithms while the last equality is due to the sum-to-$n$ properties of the $k(,)$.
In the example, $n=6$, $R=\{-1,-2,-3,1,2,3\}$, $C=\{1,4,9\}$, all the $k(x,*)=1$, all the $k(*,y)=2$, and all the $k(x,y)$ are either $0$ or $1$. The mutual information equals
$$\log(6) + \frac{1}{6}\left(6 \times (1\times \log(1)) - 6\times(1\times \log(1)) - 3\times(2\times \log(2)) \right) = \log(3).$$
• (+1) It seems the important point for addressing the OP's concern may be to emphasize the necessity of observing $(X,Y)$ pairs. Observing only marginal counts tells you next to nothing about the (empirical) mutual information. – cardinal Jul 20 '12 at 14:34
• Indeed, based on his answer to my question on how to infer $p(x,y)$ we look at counts of $(x_i,y_i) \in \mathbb{R}^2, i \in \{1,\ldots,n\},$ i.e. the joint density is inferred the same way we count event occurrences for the marginal densities $p(x), p(y).$ Thanks whuber and cardinal! :) – user1137731 Jul 20 '12 at 20:39
• Note that the plugin estimator (given by whuber) can be severely biased. It is asymptotically unbiased, and it is the ML estimate, but it can be quite off when only a handful of observations are given. – Memming Jul 21 '12 at 14:51
• User1137731, I would caution you, adding my voice to that of @Memming, not to use this as an estimator of mutual information unless you have a lot of data, and even then you probably should do some careful smoothing and binning first. As far as being a computation of MI of the data goes, there's nothing in my analysis that required integer values: it works regardless. – whuber Jul 23 '12 at 14:17
• There exists a paper by Strimmer 'Entropy Inference and the James-Stein Estimator, with Application to Nonlinear Gene Association Networks' jmlr.csail.mit.edu/papers/volume10/hausser09a/hausser09a.pdf with an associated package 'entropy' cran.r-project.org/web/packages/entropy/index.html. But this approach seems to expect count data and not continuous data for vectors $x,y.$ – user1137731 Jul 26 '12 at 15:15 | 1,839 | 5,947 | {"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} | 3.9375 | 4 | CC-MAIN-2019-26 | longest | en | 0.914043 |
http://math.stackexchange.com/questions/tagged/quadratic-programming+matlab | 1,408,710,868,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500823598.56/warc/CC-MAIN-20140820021343-00459-ip-10-180-136-8.ec2.internal.warc.gz | 127,193,520 | 11,883 | # Tagged Questions
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https://irfpy.irf.se/projects/util/_modules/irfpy/util/ballistic.html | 1,708,468,822,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473347.0/warc/CC-MAIN-20240220211055-20240221001055-00129.warc.gz | 343,899,224 | 8,948 | # Source code for irfpy.util.ballistic
"""Ballistic trajectory related module
- :mod:irfpy.util.keplernumeric
"""
import numpy as np
[docs]class Trajectory:
def __init__(self, r0, th0, vr0, om0, mass, g=6.674e-11, m=1):
r""" Initialize the trajectory.
:param r0: Initial position, :math:r_0.
:param th0: Initial phase, :math:\theta_0. Radians.
:param vr0: Initial radial velocity, :math:v_0=dr/dt(t=0).
:param om0: Initial angular veloicty, :math:\omega_0=d\theta/dt(t=0).
:param mass: Mass of the central body (or effective mass of the center of gravity).
:keyword g: Gravity of constant, :math:G. Default value is for MKSA unit value.
:keyword m: Mass of the moving object. It DOES NOT concern the trajectory,
but some physical quantities do.
*Theory*: From the initial conditions, the trajectory is expressed in the parametric equation
:math:r = LL / (1 + e \cos (\theta + \Phi). Here, r is the distance from the
central body, e is eccentricity, Phi represents the phase shift.
"""
# From the initial condition, the trajectory is expressed in the parametric
# elipse equation: r = LL / (1 + e cos (_theta + Phi) ).
# LL, e, and Phi are calculated analytically below.
self._r0 = r0
self._th0 = th0
self._vr0 = vr0
self._om0 = om0
self._M = mass
self._GM = mass * g
self._m = m
self._L = m * r0 * r0 * om0 # Angular moment. Constant over the trajectory.
self._LL = ((self._L / self._m) ** 2) / self._GM # Parametric ellipse equation: LL/(1+e cos theta)
# The phase at the initial time
self._Phi = np.arctan2(self._vr0 * self._LL / (self._r0 ** 2) / self._om0, self._LL / self._r0 - 1) - self._th0
phi0 = self._Phi + self._th0
sin = np.sin(phi0)
cos = np.cos(phi0)
self._es = self._vr0 * self._LL / (self._r0 * self._r0 * self._om0 * sin)
self._ec = ((self._LL / self._r0) - 1) / cos
if np.abs(sin) > np.abs(cos):
### Use sin conterpart of formulation
self._e = self._es
else:
### Use cos conterpart of formulation
self._e = self._ec
[docs] def print_values(self):
print('r0:', self._r0)
print('th0:', self._th0)
print('v0:', self._vr0)
print('om0:', self._om0)
print('M:', self._M)
print('GM:', self._GM)
print('m:', self._m)
print('L:', self._L)
print('LL:', self._LL)
print('Phi:', self._Phi)
print('es:', self._es)
print('ec:', self._ec)
print('e:', self._e)
[docs] def angular_momentum(self):
r""" Return the angular momentum.
The angular momentum is constant.
It is defined by the initial condition: L=m r0^2 omega0.
.. math::
L = m r_0^2 \omega_0
"""
return self._L
[docs] def eccentricity(self):
return self._e
[docs] def r(self, th):
r""" Return the distance r as a function of th
:param th: The phase angle in radians.
*Theory*: From LL, e, and Phi, the radial distance is
:math:LL / (1 + e \cos(\theta+\Phi)).
"""
r = self._LL / (1 + self._e * np.cos(th + self._Phi))
return r
[docs] def omega(self, th):
r""" Return :math:omega as a function of th.
:param th: The phase angle in radians.
:returns: The angular speed, omega.
*Theory*: The angular momentum (L see :meth:angular_momentum) is conserved, so that the
angular speed omega can be retrieved via the equation
.. math::
L = m r(\theta)^2 \omega(\theta).
The distance r is retrieved from :meth:r
"""
r = self.r(th)
L = self._L
m = self._m
return L / m / r / r
[docs] def v(self, th):
""" Return the radial velocity as a function of th.
Deprecated. Use :meth:vr.
"""
return self.vr(th)
[docs] def vr(self, th):
r""" Return the radial velocity as a function of th
:param th: The phase angle in radians.
*Theory:* We derived the vr as follows.
.. math::
r = \frac{LL}{1+e\cos(\theta+\Phi)}.
By taking time derivative, we obtain
.. math::
v_r = \frac{dr}{dt} = e \sin(\theta+\Phi)\frac{LL}{(1+e\cos(\theta+\Phi))^2}\frac{d\theta}{dt}
By simplifying the equation, we obtain
.. math::
v_r = \frac{e l}{m LL}\sin{\theta + \Phi}.
"""
L = self._L
m = self._m
e = self._e
LL = self._LL
phi = self._Phi
v = L / m * e * np.sin(th + phi) / LL
return v
[docs] def integ_distance_num(self, end_phase, ndiv=36000):
r""" Integrate the distance flown from initial phase until the given end_phase.
The distance along the trajectory is integrated along the orbit.
Simple daikei-kousiki (?what is the English?) is used.
:param end_phase: The end of the phase (:math:\theta_1). Radians.
:param ndiv: Division used for the separation.
:return: The distance.
*Sample*
Let's see a lunar orbiter. The 100 km circular orbit (distance of 1838 km) is considered.
It is easier to use the helper function :func:speed_from_distances to get the
velocity for such orbiter.
>>> vper, vapo = speeds_from_distances(1838e3, 1838e3, 7.342e22)
>>> print('{:.1f} m/s'.format(vper))
1632.8 m/s
Create the :class:Trajectory object.
>>> orbiter = Trajectory(1838e3, 0, 0, 1632.8 / 1838e3, 7.342e22)
>>> print('{:.2f}'.format(orbiter._e)) # Eccentrycity. This is (almost) circle.
0.00
Calculate the full length of the trajectory. It is :math:2\pi R where R is
the radius of the orbiter (1838 km). Thus, 11548.5 km.
>>> trajectory_length = orbiter.integ_distance_num(2 * np.pi)
>>> print('{:.3e} m'.format(trajectory_length))
1.155e+07 m
"""
# separate the phase angle by the division
thetalist = np.linspace(self._th0, end_phase, ndiv)
thetalist_c = (thetalist[1:] + thetalist[:-1]) / 2.
d_thetalist = thetalist[1:] - thetalist[:-1]
rlist = self.r(thetalist_c) # r(th)
distance_list = rlist * d_thetalist # r dth
distance = distance_list.sum()
return distance
[docs] def integ_time_num(self, end_phase, ndiv=36000):
r""" Calcuate the time from the initial time to the end at the end_phase.
The calculation uses numerical calculation
.. math::
t = \int_{\theta_0}^{\theta_1} \frac{d\theta}{\omega}
:param end_phase: The end of the phase (:math:\theta_1). Radians.
:param ndiv: Division used for the separation.
:return: The distance.
*Sample*
Let's see a lunar orbiter. The 100 km circular orbit (distance of 1838 km) is considered.
Create the :class:Trajectory object, in the same manner as :meth:integ_distance_num.
>>> orbiter = Trajectory(1838e3, 0, 0, 1632.8 / 1838e3, 7.342e22)
>>> print('{:.2f}'.format(orbiter._e)) # Eccentrycity. This is (almost) circle.
0.00
Calculate the circular period. In this case, the total distance flying over is
11548.5 km (see :meth:integ_distance_num).
The speed of the orbiter is 1632.8 m/s.
This resuts in 7072.8 s.
>>> period = orbiter.integ_time_num(2 * np.pi)
>>> print('{:.0f} s'.format(period))
7073 s
"""
# separate the phase angle by the division
thetalist = np.linspace(self._th0, end_phase, ndiv)
thetalist_c = (thetalist[1:] + thetalist[:-1]) / 2.
d_thetalist = thetalist[1:] - thetalist[:-1]
wlist = self.omega(thetalist_c) # omega(th)
dt_list = d_thetalist / wlist
t = dt_list.sum()
return t
[docs]def speeds_from_distances(dperi, dapo, mass, g=6.674e-11):
r""" Calculate the speeds at peri- and apo-centers from the distances
It is common request to calulate the trajectory for, say, spacecraft with
specification of the heights.
:param dperi: Pericenter distance from the center
:param dapo: Apocenter distance from the center
:param mass: Mass of the central body
:keyword g: Constant of gravity, :math:G
:returns: (vper, vapo), where vper is the pericenter spped and vapo for apocenter.
The formulation is simple: Using the energy and angular momentum conversations.
Derivation is based on MKSA system.
.. math::
v_p = \frac{1}{d_p}\sqrt{\frac{2GM}{\frac{1}{d_a}+\frac{1}{d_p}}} \\
v_a = \frac{1}{d_a}\sqrt{\frac{2GM}{\frac{1}{d_a}+\frac{1}{d_p}}} \\
Example below is an orbiter around the Moon. The pericenter altitude 30 km and
apocenter altitude of 270 km are considered. Moon radius of 1738 km is taken.
>>> dp = (30 + 1738) * 1000 # in meter
>>> da = (270 + 1738) * 1000 # in meter
>>> from irfpy.util.planets import moon
>>> v_p, v_a = speeds_from_distances(dp, da, moon['mass'])
>>> print('{:.3f}'.format(v_p))
1717.547
>>> print('{:.3f}'.format(v_a))
1512.262
The results say the pericenter velocity of 1.718 km/s and apocenter velocity of 1.512 km/s
"""
sq = np.sqrt(2 * g * mass / (1 / dapo + 1 / dperi))
return sq / dperi, sq / dapo
[docs]class Trajectory3D:
r""" 3-D trajectory of ballistic motion.
*Theory*
The class represents the ballistic motion of a mass *m* under a huge gravity star of mass *M*.
(*M >> m* is assumed, but theoretically, if you use "equivalent mass" then one can extend this to more generic problems)
Consider the mass *M* at origin, and stationary.
The mass *m* moves around under Kepler motion.
Let's take the initial position :math:\vec{r_0} with the velocity :math:\vec{v_0}.
:math:r_0 = |\vec{r_0}| and :math:v_0 = |\vec{v_0}| should not be zero.
Also they should not be parallel, i.e. :math:\vec{r_0}\times\vec{v_0}\ne 0.
This is a 2-D problem, which can be represented by :class:Trajectory.
This, a conversion between the 3-D coordinates to the 2-D in :class:Trajectory is the key to solve.
Conversion:
Let's take the new axis as follows.
- :math:\hat{s}: Along the positon vector, i.e., :math:\hat{s} = \frac{\vec{r_0}}{r_0}.
- :math:\hat{t}: :math:\hat{s}-\hat{t} plane to include the velocity vector, with :math:v_t>0.
This means that :math:\hat{t} // (\vec{r_0}\times\vec{v_0})\times\vec{r_0}.
*Usage*
Let's try the following example.
- An orbiter around the Moon (mass=7.342e22 kg). In inertia frame
- The position of orbiter at t=0 is 1850 km at the equator, longitude at 45.
>>> r0 = 1850e3 * np.array([np.cos(np.pi / 4), np.sin(np.pi / 4), 0])
>>> print(r0) # doctest: +NORMALIZE_WHITESPACE
[1308147.54519511 1308147.54519511 0. ]
- The initial velocity of the orbiter at t=0 is (0, 0.01, 1.65) km/s.
>>> v0 = 1000 * np.array([0, 0.01, 1.65])
>>> print(v0) # doctest: +NORMALIZE_WHITESPACE
[ 0. 10. 1650.]
Now it is time to make the Trajectory3D object.
>>> traj = Trajectory3D(r0, v0, 7.342e22)
>>> print(traj)
Trajectory3D: s=(0.707,0.707,0) t=(-0.00303,0.00303,1) r0=1.85e+06 vs0=7.07 w0=0.000892
The initial position, r0, should be converted to the original position in xyz.
>>> print(traj.st2xyz(traj.s(0), traj.t(0))) # doctest: +NORMALIZE_WHITESPACE
[1308147.54519511 1308147.54519511 0. ]
The initial velocity, v0, should be converted to the original velocity in xyz.
>>> v0recal = traj.st2xyz(traj.vs(0), traj.vt(0))
>>> print('{v[0]:.3f} {v[1]:.3f} {v[2]:.3f}'.format(v=v0recal))
0.000 10.000 1650.000
Later, the velocity can be retrieved by vxyz method.
>>> vxyz1 = traj.vxyz(1) # Velocity at phase = 1 radian
>>> print('{v[0]:.3f} {v[1]:.3f} {v[2]:.3f}'.format(v=vxyz1)) # It is in m/s
-952.900 -947.372 912.080
>>> traj.print_values() # doctest: +SKIP
--- 2d ---
r0: 1850000.0
th0: 0.0
v0: 7.07106781187
om0: 0.00086487331077
...
The period of such orbit is ~7450 sec.
>>> print('{:.3f}'.format(traj.integ_time_num(np.pi * 2)))
7452.076
*Notes*
All the quantities are derived via the parameter :math:\theta. This is frequently a difficulty, because
it is more natural to use the time *t* as a parameter. However, the theory of two-body collision is built
such that. I do not want to dig into more in details now.
Of course practically, it could be worthwhile to get a function such like :math:\theta=\theta(t)
by inverting :math:t=t(\theta).
"""
def __init__(self, r0vec, v0vec, mass, g=6.674e-11, m=1):
r""" Initialize the trajectory.
:param r0vec: Initial position, :math:\vec{r_0}.
:param v0vec: Initial radial velocity, :math:\vec{v_0}.
:param mass: Mass of the central body (or effective mass of the center of gravity).
:keyword g: Gravity of constant, :math:G. Default value is for MKSA unit value.
:keyword m: Mass of the moving object. It DOES NOT concern the trajectory,
but some deriving physical quantities do.
"""
### New axis
self.r0vec = np.array(r0vec)
self.r0 = np.sqrt((self.r0vec ** 2).sum())
if self.r0vec.shape != (3,) or self.r0 == 0:
raise ValueError('Position should be non-zero 3-D vector, but {}'.format(str(self.r0vec)))
self.v0vec = np.array(v0vec)
self.v0 = np.linalg.norm(self.v0vec)
if self.v0vec.shape != (3,) or self.v0 == 0:
raise ValueError('Velocity should be non-zero 3-D vector, but {}'.format(str(self.v0vec)))
self.hats = self.r0vec / self.r0
self.hatt = np.cross(self.r0vec, self.v0vec)
if (self.hatt == 0).all():
raise ValueError('Position and Velocity should note be paralell, but {} and {}.'.format(str(self.r0vec), str(self.v0vec)))
self.hatt = np.cross(self.hatt, self.r0vec)
self.hatt = self.hatt / np.linalg.norm(self.hatt)
self.th0 = 0. # Initial phase is always zero in s-t plane.
cosalpha = np.clip(np.dot(self.r0vec, self.v0vec) / self.r0 / self.v0, -1, 1)
self.alpha = np.arccos(cosalpha)
sinalpha = np.sqrt(1 - cosalpha ** 2)
self.vs0 = self.v0 * cosalpha
self.om0 = self.v0 * sinalpha / self.r0
self._2dtraj = Trajectory(self.r0, self.th0, self.vs0, self.om0, mass, g=g, m=m)
[docs] def st2xyz(self, s, t):
"""
"""
return s * self.hats + t * self.hatt
[docs] def r(self, th):
""" Return r from the phase angle as a parameter.
"""
return self._2dtraj.r(th)
[docs] def omega(self, th):
""" Return the tangential angular velocity
"""
return self._2dtraj.omega(th)
[docs] def s(self, th):
""" Return the s value
"""
return self.r(th) * np.cos(th)
[docs] def t(self, th):
""" Return the t value
"""
return self.r(th) * np.sin(th)
[docs] def v(self, th):
"""
return self._2dtraj.v(th)
[docs] def vr(self, th):
""" Return the radial velociyt. Same as :meth:v
"""
return self.v(th)
[docs] def vth(self, th):
""" Return the phase velocity."""
return self.omega(th) * self.r(th)
[docs] def vs(self, th):
""" Return the s-direction velocity
"""
return self.v(th) * np.cos(th) - self.vth(th) * np.sin(th)
[docs] def vt(self, th):
""" Return the tangential velocity
"""
return self.v(th) * np.sin(th) + self.vth(th) * np.cos(th)
[docs] def xyz(self, th):
""" Return the (x, y, z) coordinates
"""
s = self.s(th)
t = self.t(th)
return self.st2xyz(s, t)
[docs] def longitude(self, th):
"""Return the longitude in degrees at the phase angle |theta|.
:param th: Phase angle |theta|
:return: The longitude in degrees.
"""
xyz = self.xyz(th)
return lon
[docs] def latitude(self, th):
""" Return the Latitude in degrees at the phase angle |theta|.
:param th: The phase angle |theta|
:return: The latitude in degrees
"""
r = self.r(th)
z = self.xyz(th)[2]
cos = np.clip(z / r, -1, 1)
[docs] def vxyz(self, th):
""" Return (vx, vy, vz)
"""
vs = self.vs(th)
vt = self.vt(th)
return self.st2xyz(vs, vt)
[docs] def integ_time_num(self, end_phase, ndiv=36000):
return self._2dtraj.integ_time_num(end_phase, ndiv=ndiv)
[docs] def integ_distance_num(self, end_phase, ndiv=36000):
return self._2dtraj.integ_distance_num(end_phase, ndiv=ndiv)
[docs] def print_values(self):
print('--- 2d ---')
print('r0:', self._2dtraj._r0)
print('th0:', self._2dtraj._th0)
print('v0:', self._2dtraj._vr0)
print('om0:', self._2dtraj._om0)
print('M:', self._2dtraj._M)
print('GM:', self._2dtraj._GM)
print('m:', self._2dtraj._m)
print('L:', self._2dtraj._L)
print('LL:', self._2dtraj._LL)
print('Phi:', self._2dtraj._Phi)
print('es:', self._2dtraj._es)
print('ec:', self._2dtraj._ec)
print('e:', self._2dtraj._e)
def __str__(self):
s = 'Trajectory3D: s=({s[0]:.3g},{s[1]:.3g},{s[2]:.3g}) t=({t[0]:.3g},{t[1]:.3g},{t[2]:.3g}) '.format(s=self.hats, t=self.hatt)
s = s + 'r0={:.3g} vs0={:.3g} w0={:.3g}'.format( self.r0, self.vs0, self.om0)
return s | 5,103 | 15,490 | {"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": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-10 | latest | en | 0.5374 |
http://www.goodmath.org/blog/2015/01/28/the-abc-conjecture-aka-the-soap-opera-of-the-math-world/?replytocom=120064 | 1,695,444,184,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00882.warc.gz | 63,717,206 | 24,134 | # The ABC conjecture – aka the soap opera of the math world.
Sorry for the silence of this blog for the last few months. This spring, my mother died, and I was very depressed about it. Depression is a difficult thing, and it left me without the energy or drive to do the difficult work of writing this kind of material. I’m trying to get back into the cycle of writing. I’m trying to make some progress in writing about type theory, but I’m starting with a couple of easier posts.
In the time when I was silent, I had a couple of people write to me to ask me to explain something called the ABC conjecture.
The ABC conjecture is a mathematical question about number theory that was proposed in the 1980s – so it’s relatively new as number theory problems go. It’s gotten a lot of attention recently, due to an almost soap-operatic series of events.
It’s a very hard problem, and no one had made any significant progress on it until about five years ago, when a well respected Japanese mathematician named Shinichi Mochizucki published a series of papers containing a proof of the conjecture.
Normally, when a proof of a hard problem gets published, mathematicians go nuts! Everyone starts poring over it, trying to figure it out, and see if it’s valid. That’s what happened the previous time someone thought they’d prooved it. But this time, no one has been able to make sense out of the proof!
The problem is that in order to build his proof, professor Mochizucki created a whole new mathematical theory, called inter-universal Teichmüller theory. The entire ABC conjecture proof is built in this new theory, and no one other than professor Mochizucki himself understands Teichmüller theory. Before anyone else can actually follow the proof, they need to understand the theory. Professor Mochizucki is a bit of a recluse – he has declined to travel anywhere to teach his new mathematical system. So in the five years since he first published it, no one has been able to understand it well enough to determine whether or not the proof is correct. One error in it was found, but corrected, and the whole proof remains in question.
Exactly why the proof remains unchecked after five years is a point of contention. Lots of mathematicians are angry at Professor Mochizucki for not being willing to explain or teach his theory. A common statement among critics is that if you create a new mathematical theory, you need to be willing to actually explain it to people: work with a group of mathematicians to teach it to them, so that they’ll be able to use it to verify the proof. But Professor Mochizuchki’s response has been that he has explained it: he’s published a series of papers describing the theory. He doesn’t want to travel and take time away from his work for people who haven’t been willing to take the time to read what’s he’s written. He’s angry that after five years, no one has bothered to actually figure out his proof.
I’m obviously not going to attempt to weigh in on whether or not Professor Mochizuki’s proof is correct or not. That’s so far beyond the ability of my puny little brain that I’d need to be a hundred times smarter before it would even be laughable! Nor am I going to take sides about whether or not the Professor should be travelling to teach other mathematicians his theory. But what I can do is explain a little bit about what the ABC conjecture is, and why people care so much about it.
It’s a conjecture in number theory. Number theorists tend to be obsessed with prime numbers, because the structure of the prime numbers is a huge and fundamental part of the structure and behavior of numbers as a whole. The ABC conjecture tries to describe one property of the structure of the set of prime numbers within the system of the natural numbers. Mathematicians would love to have a proof for it, because of what it would tell them about the prime numbers.
Before I can explain the problem, there’s a bit of background that we need to go through.
1. Any non-prime number N is the product of some set of prime numbers. Those numbers are called the prime factors of N. For example, 8 is 2×2×2 – so the set of prime factors of 8 is {2}. 28 is 2×2×7, so the prime factors of 28 are {2, 7}. 360 = 8 × 45 = 2×2×2×(9×5) = 2×2×2×3×3×5, so the prime factors of 360 are {2, 3, 5}.
3. Given two positive integers N and M, N and M are coprime if they have no common prime factors. A tiny bit more formally, if pf(N) is the set of prime factors of N, and M and N are coprime if and only if pf(N) ∩ pu(M) = ∅. (Also, if M and N are coprime, then rad(M×N) = ram(M)×rad(N).)
The simplest way of saying the ABC conjecture is that for the vast majority of integers A, B, and C, where A + B = C and A and B are coprime, C must be smaller than rad(A*B).
Of course, that’s hopelessly imprecise for mathematicians! What does “the vast majority” mean?
The usual method at times like these is to find some way of characterizing the size of the relative sizes of the set where the statement is true and where the statement is false. For most mathematicians, the sizes of sets that are interesting are basically 0, 1, finite, countably infinite, and uncountably infinite. For the statement of the ABC conjecture, they claim that the set of values for which the statement is true is infinite, but that the set of values for which it is false are finite. Specifically, they want to be able to show that the set of numbers for which rad(A*B)>C is finite.
To do that, they pull out a standard trick. Sadly, I don’t recall the proper formal term, but I’ll call it epsilon bounding. The idea is that you’ve got a statement S about a number (or region of numbers) N. You can’t prove your statement about N specifically – so you prove it about regions around N.
As usual, it’s clearest with an example. We want to say that C > rad(A*B) for most values of A and B. The way we can show that is by saying that for any value ε, the set of values (A, B, C) where A and B are coprime, and A + B = C, and rad(A*B) > C + ε is finite.
What this formulation does is give us a formal idea of how rare this is. It’s possible that there are some values for A and B where rad(A*B) is bigger that 1,000,000,000,000,000,000 + C. But the number of places where that’s true is finite. Since the full system of numbers is infinite, that means that in the overwhelming majority of cases, rad(A*B) < C. The size of the set of numbers where that's not true is so small that it might at well be 0 in comparison to the size of the set of numbers where it is true. Ultimately, it seems almost trivial once you understand what the conjecture is. It's nothing more that the hypothesis that that if A + B = C, then most of the time, pf(A)*pf(B) < C. Once you've got that down, the question is, what's the big deal? Professor Mochuzuki developed five hundred pages of theory for this? People have spent more than five years trying to work through his proof just to see if it’s correct for a statement like this? Why does anybody care so much?
One answer is: mathematicians are crazy people!
The better answer is that simple statements like this end up telling us very profound things about the deep structure of numbers. The statements reduce to something remarkably simple, but the meaning underneath it is far more complex than it appears.
Just to give you one example of what this means: If the conjecture is true, then there’s a three-line proof of Fermat’s last theorem. (The current proof of Fermat’s last theorem, by Andrew Wiles, is over 150 pages of dense mathematics.) There’s quite a number of things that number theoreticians care about that would fall out of a successful proof.
## 14 thoughts on “The ABC conjecture – aka the soap opera of the math world.”
1. NickF
Seconded. Welcome back, and thanks for taking the time to write a little about mathematics!
2. John Armstrong
Nitpick: plenty of people understand Teichmüller theory; it’s Mochizuki’s version that parameterizes number field structures on elliptic curves (AIUI) that nobody but him understands well enough yet.
3. mike miller
I don’t think your assessment that nobody understands IUTeich, and that Mochizuki is unwilling to help teach it, is entirely valid. The report published maybe a month ago here: http://www.kurims.kyoto-u.ac.jp/~motizuki/IUTeich%20Verification%20Report%202014-12.pdf shows that he does indeed work to help (quite a small audience that is willing to spend their time understanding the conjecture) promote understanding, and that some well-regarded mathematicians are working to understand his work.
4. Barry Leiba
On the “vast majority” thing: the term I’d always heard used for this was “almost all”, which meant “all but a finite number in an infinite domain.” I always liked the term because it was well defined, but sounded like it wasn’t.
1. Chai
I smiled when my RSS feed for this blog was in bold for the first time in ages. Really good to see you posting again.
1. John Armstrong
First of all, you’re looking for rad(A*B) B = 1
E = 2 => B = 3
E = 3 => B = 7
E = 4 => B = 3*5
E = 5 => B = 31
E = 6 => B = 3^2*7
E = 7 => B = 127
E = 8 => B = 3*5*17
E = 9 => B = 7*73
E = 10 => B = 3*11*31
As we see, we only get a repeated prime for E = 6 so far. For the rest we’re going to get a much larger value of rad(A*B).
5. Lisa R.
Really glad to see you’re feeling up to posting again! And sorry you had to go through so much to get to the other side.
6. Luca
I am curious about the 3 lines proof of Fermat’s last theoreme, can you post it? Could it be the same proof that Fermat’s could not write on his notes?
1. Bob Silverman
The assertion that ABC yields a 3 line proof of FLT is false.
What is true is that ABC yields a 3-line proof of FLT for all sufficiently
large exponents. However, the proof does not yield an actual value
for what “sufficiently large” means. | 2,424 | 9,878 | {"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-2023-40 | latest | en | 0.970432 |
http://www.instructables.com/id/Double-Makalu/ | 1,419,196,071,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802772398.133/warc/CC-MAIN-20141217075252-00175-ip-10-231-17-201.ec2.internal.warc.gz | 580,066,727 | 35,345 | Double Makalu - Origami Scupture: 12 woven pentagons (no glue)
Hi and welcome to a small part of my world :)
I decided that MAKALU wasn't enough of a challenge, so I made it double!
Double the units, double the difficulty, double the fun :)
Makalu is one of the 5 Himalayan peaks of origami folding, and this double model adds a little spice to the mix!
This is an icosidodecahedron (origami is a great way of learning about geometry and polyhedra).
Ok lets get started.
What you will need for this instructable:
1) Time.
2) 15 squares of paper.
3) Patience.
4) Good music.
5) Mild OCD (not essential, but it helps)
Remove these ads by Signing Up
Step 1: The paper!
Firstly, good music! Easily embarrassed is great origami music: https://soundcloud.com/easilyembarrassed
Take your 15 squares, and cut them in half.
Accurately.
Now take your 30 2x1 rectangles, and cut them in half again. Accurately.
You should have 60 4x1 rectangles now.
Step 2: The folding begins!
Start with the white side up (if you are using coloured paper).
Valley fold your paper in half lengthwise, and then into quarters.
This can be difficult to do accurately, so fold little segments of the fold at a time to make sure that everything is lined up.
Step 3:
Valley fold the top right hand corner down to the left edge, and the mirror on the bottom.
Don't crease heavily, these are just guideline folds.
Step 4:
Valley fold the bottom right hand corner to the first crease (as if you were going to fold the quarter in half again), and do the same with the top left corner.
Unfold.
Fold the bottom edge up to the 2 folds you just made. Repeat at the top. Check photos.
If you get stuck, the assembly of the models is identical to that of Makalu, the paper dimensions just differ slightly.
Step 5:
Valley fold the edges to the middle.
Open up the top right edge.
Reverse the fold inside there, and push flat.
Repeat at the bottom.
Step 6:
Valley fold in half again.
Bend the top flap down along the edge, and flatten.
Repeat on the bottom.
torstenp161 year ago
I know I keep saying this, but I will eventually get around to making one of these. These'd be pretty sweet as a stone or metal sculpture too!
joettle (author) torstenp161 year ago
I have been wanting to make metal sculptures for a while now...
I'll post an instructable if I ever get it done... | 591 | 2,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2014-52 | longest | en | 0.915377 |
https://books.google.com.jm/books?qtid=357e7974&lr=&id=xjcPAAAAYAAJ&sa=N&start=0 | 1,725,868,331,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00780.warc.gz | 123,861,217 | 5,924 | Books Books
All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Elements of Geometry: Containing the First Six Books of Euclid, with a ... - Page 292
by Euclid, John Playfair - 1846 - 317 pages
## The Young Surveyor's Guide: Or, A New Introduction to the Whole Art of ...
Edward Laurence - Surveying - 1716 - 408 pages
...external < a, is equal to t wo right Angles (by tbe^tb /)confequently all the internal and external Angles are equal to twice as many right Angles as the Figure has fides. But all its internal Angles are equal to twiee as many right Angles eicept 4 as it has fides...
## The Young Gentleman's Arithmetick, and Geometry: Containing Such Elements of ...
Edward Wells - Arithmetic - 1723 - 358 pages
...the Sum of all the Angles in all the Tri~ angles, into which the Figure is divided, will together be equal to twice as ma-ny right Angles, as the Figure has Sides. But the Angles about P, the inward Point of each Figure, wherein all the Triangles concur, are (by...
## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...
Robert Simson - Trigonometry - 1762 - 488 pages
...gether with four right angles. Therefore all the angles of the figure^ together with four right angles, are equal to twice as many right angles as the figure has fides. C o R. 2 . All the exterior angles of any rectilineal figure are together equal to four right...
## Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books
Euclid - Geometry - 1765 - 492 pages
...taken together) therefore .all the angles of a right-lined figure, together with four right angles, are equal to twice as many right angles as the figure has fides. And taking away four right angles from each, there will remain all the angles of the figure...
## A Royal Road to Geometry: Or, an Easy and Familiar Introduction to the ...
Thomas Malton - 1774 - 484 pages
...by the Sides. ie equal to four Right Angles. And, all the internal Angles of any Right-lined Figure are equal to twice as many Right Angles as the Figure has Sides, wanting four, (Th. i. i0. i.) confequently, the external Angles being equal to thofe four (Th. 2. of the fame) are...
## The Elements of Euclid, Viz: The Errors, by which Theon, Or Others, Have ...
Robert Simson - Trigonometry - 1775 - 534 pages
...together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has fides. CoR. 2. All the exterior angles of any re&ilineal figure, are together equal to four right angles....
## The First Six Books: Together with the Eleventh and Twelfth
Euclid - 1781 - 552 pages
...together with four right angles. Thprefpre all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has fides. CoR. 2. All the exterior angles of any rectilineal figure, arc together equal to four right...
## A Complete Treatise on Practical Mathematics: Including the Nature and Use ...
John McGregor (teacher of mathematics.) - Mathematics - 1792 - 558 pages
...ft, I. 32. Euclid. All the anterior angles of any reoilineal figure, together with four right angles, are equal to twice as many right angles as the figure has fides. Hence the following rule. RULÉ. From double thé number of fides f übt vail: 4, and the remainder... | 881 | 3,465 | {"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-2024-38 | latest | en | 0.893594 |
https://goprep.co/ex-16.2-q30-a-number-lock-on-a-suitcase-has-3-wheels-each-i-1nl5r8 | 1,628,061,163,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154796.71/warc/CC-MAIN-20210804045226-20210804075226-00023.warc.gz | 271,689,229 | 39,902 | # <span lang="EN-US
In the question, we have to find the possible number of 3 digit numbers formed by the numbers 0 to 9 when repetition of digits is not allowed.
We will use the concept of multiplication because there are three sub jobs dependent on each other because a number appearing on any one place will not appear in any other place.
In the first position from left we will have ten choices, in the second position we will have nine choices, and in the third position, we will have eight choices because repetition is not allowed and one digit is occupied in each position.
The number of ways in which we can form three - digit numbers with the help of given data is 10 × 9 × 8 = 720
There will be only one correct combination out of these so the incorrect combinations will be 720 - 1 = 719
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If 20 lines are dMathematics - Exemplar | 375 | 1,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-31 | latest | en | 0.871482 |
https://www.gradesaver.com/textbooks/science/chemistry/general-chemistry-principles-and-modern-applications-10th-edition/chapter-15-priciples-of-chemical-equilibrium-exercises-writing-equilibrium-contants-expressions-page-688/8 | 1,701,639,086,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00165.warc.gz | 890,302,401 | 14,117 | ## General Chemistry: Principles and Modern Applications (10th Edition)
(a) $$K_p = 0.114$$ (b) $$K_p =4.26 \times 10^3$$ (c) $$K_p = 3.45 \times 10^{-4}$$
(a) T = 25 + 273.15 = 298.15 K $$K_p = K_c(RT)^{\Delta n} = 4.61 \times 10^{-3} (0.08314 \times 298.15)^{2 - 1} = 0.114$$ (b) $$K_p = K_c(RT)^{\Delta n} =0.154 (0.08314 \times 2000)^{4 - 2} = 4.26 \times 10^3$$ (c) $$K_p = K_c(RT)^{\Delta n} =(5.27 \times 10^{-8}) (0.08314 \times 973)^{5 - 3} = 3.45 \times 10^{-4}$$ | 233 | 474 | {"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} | 2.546875 | 3 | CC-MAIN-2023-50 | latest | en | 0.531467 |
https://discuss.leetcode.com/topic/90103/java-greedy-algorithm-with-merge-sort-source-code-and-proof | 1,516,660,509,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891543.65/warc/CC-MAIN-20180122213051-20180122233051-00786.warc.gz | 648,140,367 | 8,658 | # Java greedy algorithm with merge sort source code and proof
• ``````public class Solution {
//Use greedy algorithm. Sort the given array in ascending order and group 2 integers from beginning to end.
//Poof: If there exists any reverse in optimal solution: (ai,bj), (aj, bi) (i<j);
//Then the sum of those two pairs is ai+bi. The orginal is ai+aj, which is bigger than ai+bi (bi<aj). Contradiction.
public int arrayPairSum(int[] nums) {
mergeSort(nums);
int sum = 0;
for(int i = 0; i < nums.length; i=i+2){
sum += nums[i];
}
return sum;
}
public void mergeSort(int[] nums) {
if(nums.length>1){
//merge
int p = nums.length/2;
int[] arr1 = new int[p];
int[] arr2 = new int[nums.length-p];
for (int i = 0; i < p; i++) {
arr1[i] = nums[i];
arr1[i] = nums[i];
}
for (int i = p; i < nums.length; i++) {
arr2[i-p]=nums[i];
arr2[i-p]=nums[i];
}
mergeSort(arr1);
mergeSort(arr2);
//sort
int m = 0,n=0;
for (int i = 0; i < nums.length; i++) {
if (m<arr1.length&&n<arr2.length&&arr1[m]<arr2[n]) {
nums[i]=arr1[m];
nums[i]=arr1[m];
m++;
}
else if (m<arr1.length&&n<arr2.length&&arr1[m]>=arr2[n]) {
nums[i]=arr2[n];
nums[i]=arr2[n];
n++;
}
else if (m>=arr1.length&&n<=arr2.length) {
nums[i]=arr2[n];
nums[i]=arr2[n];
n++;
}
else if (n>=arr2.length&&m<=arr1.length){
nums[i]=arr1[m];
nums[i]=arr1[m];
m++;
}
}
return;
}
else {
return;
}
}
}
``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 483 | 1,435 | {"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} | 3.203125 | 3 | CC-MAIN-2018-05 | latest | en | 0.465635 |
https://javascriptio.com/view/180895/how-to-draw-the-three-triangle-in-single-page-using-canvas | 1,607,215,155,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141753148.92/warc/CC-MAIN-20201206002041-20201206032041-00047.warc.gz | 359,745,289 | 6,860 | # How to draw the three triangle in single page using canvas?
``````function areaval() {
var equation = \$('#number').val();
var secod = \$('#acure').val();
var thrd = \$('#pmet').val();
var frd = \$('#cmet').val();
var a = Math.abs(parseInt(equation * secod) );
var d = Math.abs(a/2);
document.getElementById("result").innerHTML = "Area=(1/2)*b*h<br><br>A="+d+"cm2";
document.getElementById("step4.1").innerHTML = "step1=(1/2)*"+equation+"*"+secod;
var f= Math.abs(parseInt(equation) + parseInt(thrd) + parseInt(frd));
document.getElementById("result2.2").innerHTML = "Perimeter=a+b+c<br><br>P="+f+"cm2";
document.getElementById("step4.6").innerHTML = "step2="+equation+"+"+thrd+"+"+frd;
\$('input').val('');
return false;
}
function rightang() {
var givenone = \$('#oppsite').val();
var giventwo = \$('#adjacent').val();
var giventhree = \$('#hyper').val();
var givenhig = \$('#hidgt').val();
var agive = Math.abs(parseInt(givenone * givenhig) );
var ragive = Math.abs((agive)/2);
document.getElementById("raitresult").innerHTML = "Area=(1/2)*b*h<br><br>A="+ragive+"cm2";
document.getElementById("step5.1").innerHTML = "step1=(1/2)*"+givenone+"*"+givenhig;
var subperi = Math.abs((2*givenone)+(2*giventwo));
var rtsub = Math.sqrt(subperi);
var srtadd = Math.floor(parseInt(givenone)+parseInt(giventwo));
var finres = Math.floor(parseInt(rtsub)+parseInt(srtadd));
document.getElementById("periresult2").innerHTML = "Perimeter=a+b+c<br><br>P="+finres+"cm2";
document.getElementById("step5.5").innerHTML = "step2="+givenone+"+"+giventwo+"+"+giventhree;
var hysid = Math.floor(parseInt(finres)/parseInt(giventwo));
var hyang = Math.abs(2*parseInt(hysid));
document.getElementById("periresult3").innerHTML = "Angle of a=A*2/b<br><br>angle="+hyang+"degree";
document.getElementById("step5.8").innerHTML = "step3="+hysid+"*"+"2"+"/"+givenone;
\$('input').val('');
return false;
}
function obtuseang() {
var oppavall = \$('#oppsite6').val();
var oppbval = \$('#oppsite7').val();
var obtont = \$('#oppsite1').val();
var obttwo = \$('#oppsite2').val();
var obttriagle = Math.abs(parseInt(obtont * obttwo ) );
var obtval = Math.abs((obttriagle)/2);
document.getElementById("raitresult43").innerHTML = "Area=(1/2)*b*h<br><br>A="+obtval+"cm2";
document.getElementById("step6.1").innerHTML = "step1=(1/2)*"+obtont+"*"+obttwo;
var obtperi = Math.abs(parseInt(oppavall)+parseInt(oppbval)+parseInt(obtont));
document.getElementById("obtuseresult").innerHTML = "Perimeter=a+b+c<br><br>P="+obtperi+"cm2";
document.getElementById("stepfine").innerHTML = "step2="+oppavall+"+"+oppbval+"+"+obtont;
\$('input').val('');
return false;
}``````
``````#equilateral-try {
width: 0;
height: 0;
border-left: 128px solid transparent;
border-right: 48px solid transparent;
border-bottom: 95px solid black;
position:absolute;
top:100px;
left:100px;
}
#coverbox {
position:absolute;
top:calc(8%);
left:calc(3%);
width:300px;
height:500px;
}
#equilateral-try:after {
width: 100%;
height: 100%;
border-left: 110px solid transparent;
border-right: 40px solid transparent;
content: "";
border-bottom: 80px solid #fff;
position:absolute;
top:10px;
left:-110px;
z-index:1;
}
#strightline {
position:absolute;
left:66%;
top:22%;
z-index:2;
}
#right-try {
width: 0;
height: 0;
border-bottom: 100px solid black;
border-right: 100px solid transparent;
position:absolute;
top:16%;
left:45%;
}
#coverbox11 {
position:absolute;
top:calc(8% );
left:calc(30% );
width:300px;
height:500px;
}
#right-try:after {
width: 100%;
height: 100%;
content: "";
border-bottom: 85px solid #fff;
border-right: 85px solid transparent;
position:absolute;
top:12px;
left:6px;
z-index:6;
}
#obtuse-try {
width: 0;
height: 0;
border-bottom: 100px solid black;
border-left: 140px solid transparent;
position:absolute;
top:17%;
left:20%;
}
#coverbox12 {
position:absolute;
top:calc(8% );
left:calc(65% );
width:300px;
height:500px;
}
#obtuse-try:after {
width: 100%;
height: 100%;
content: "";
border-bottom: 80px solid #fff;
border-left: 120px solid transparent;
position:absolute;
top:14px;
left:-125px;
z-index:8;
}
#strightline3 {
position:absolute;
left:68.8%;
top:16%;
z-index:9;
}``````
``````<html>
<head>
<script src="angleoftriangle.js"></script>
<link rel="stylesheet" type="text/css" href="angleoftriangle.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>
</head>
<body>
<div id="coverbox">
<div id="equilateral-try">
</div>
<form id="text" onsubmit="return areaval()">
<p id="heading1"style="position:absolute;top:8%;left:20%;font-size:22px;color:black;">Acute Triangle</p>
<p id="heading5"style="position:absolute;top:42%;left:20%;font-size:20px;color:black;">abc<90deg.</p>
<input type="text" value ="" id="number" style="position:absolute;width:8%;height:4%;left:60%;top:40%;outline:none;border:none;font-size:25px;background-color:transparent;" placeholder="b"required>
<input type="text" value ="" id="acure" style="position:absolute;width:8%;height:4%;left:78%;top:31%;outline:none;border:none;font-size:25px;z-index:4;background-color:transparent;" placeholder="h"required>
<input type="text" value ="" id="pmet" style="position:absolute;width:8%;height:4%;left:42%;top:25%;outline:none;border:none;font-size:25px;z-index:6;background-color:transparent;" placeholder="a"required>
<input type="text" value ="" id="cmet" style="position:absolute;width:8%;height:4%;left:88%;top:25%;outline:none;border:none;font-size:25px;z-index:5;background-color:transparent;" placeholder="c"required>
<button style="position:absolute;height:20px;left:70%;top:41%;">sol</button>
</form>
<div id="result"style="position:absolute;top:55%;left:20%;font-size:20px;color:#878787;"></div>
<div id="step4.1"style="position:absolute;top:60%;left:20%;font-size:20px;color:#878787;"></div>
<div id="result2.2"style="position:absolute;top:68%;left:20%;font-size:20px;color:#878787;"></div>
<div id="step4.6"style="position:absolute;top:73%;left:20%;font-size:20px;color:#878787;"></div>
<svg height="16%" width="14%"id="strightline">
<line x1="30" y1="0" x2="30" y2="105" style="stroke:black;stroke-width:2" />
</svg>
</div>
<div id="coverbox11">
<div id="right-try">
</div>
<form id="text1" onsubmit="return rightang()">
<p id="heading1"style="position:absolute;top:7%;left:32%;font-size:22px;color:black;">Rightangle Triangle</p>
<p id="heading5"style="position:absolute;top:42%;left:43%;font-size:20px;color:black;"> a=90degree.</p>
<input type="text" value ="" id="oppsite" style="position:absolute;width:8%;height:4%;left:38%;top:23%;outline:none;border:none;font-size:25px;background-color:transparent;" placeholder="a"required>
<input type="text" value ="" id="hidgt" style="position:absolute;width:8%;height:4%;left:48%;top:23%;outline:none;border:none;font-size:25px;z-index:20;background-color:transparent;" placeholder="h"required>
<input type="text" value ="" id="adjacent" style="position:absolute;width:8%;height:4%;left:53%;top:37%;outline:none;border:none;font-size:25px;z-index:4;background-color:transparent;" placeholder="b"required>
<input type="text" value ="" id="hyper" style="position:absolute;width:8%;height:4%;left:68%;top:23%;outline:none;border:none;font-size:25px;z-index:6;background-color:transparent;" placeholder="c"required>
<button style="position:absolute;height:20px;left:64%;top:38%;">sol</button>
</form>
<div id="raitresult"style="position:absolute;top:55%;left:43%;font-size:20px;color:#878787;"></div>
<div id="step5.1"style="position:absolute;top:60%;left:43%;font-size:20px;color:#878787;"></div>
<div id="periresult2"style="position:absolute;top:70%;left:43%;font-size:20px;color:#878787;"></div>
<div id="step5.5"style="position:absolute;top:75%;left:43%;font-size:20px;color:#878787;"></div>
<div id="periresult3"style="position:absolute;top:85%;left:43%;font-size:20px;color:#878787;"></div>
<div id="step5.8"style="position:absolute;top:90%;left:43%;font-size:20px;color:#878787;"></div>
</div>
<div id="coverbox12">
<div id="obtuse-try">
</div>
<svg height="15%" width="15%"id="strightline3">
<path id="lineAB" d="M 100 150 l 70 -200" stroke="black" stroke-width="2" fill="none" />
</svg>
<form id="text2" onsubmit="return obtuseang()">
<p id="heading16"style="position:absolute;top:8%;left:20%;font-size:22px;color:black;">Obtuse Triangle</p>
<p id="heading17"style="position:absolute;top:42%;left:30%;font-size:20px;color:black;">a>90degree.</p>
<input type="text" value ="" id="oppsite1" style="position:absolute;width:8%;height:4%;left:48%;top:38%;outline:none;border:none;font-size:25px;background-color:transparent;" placeholder="b"required>
<input type="text" value ="" id="oppsite2" style="position:absolute;width:8%;height:4%;left:58%;top:28%;outline:none;border:none;font-size:25px;z-index:16;background-color:transparent;" placeholder="h"required>
<input type="text" value ="" id="oppsite6" style="position:absolute;width:8%;height:4%;left:71%;top:23%;outline:none;border:none;font-size:25px;z-index:16;background-color:transparent;" placeholder="c"required>
<input type="text" value ="" id="oppsite7" style="position:absolute;width:8%;height:4%;left:36%;top:23%;outline:none;border:none;font-size:25px;z-index:16;background-color:transparent;" placeholder="a"required>
<button style="position:absolute;height:20px;left:58%;top:39%;">sol</button>
</form>
<div id="raitresult43"style="position:absolute;top:55%;left:30%;font-size:20px;color:#878787;"></div>
<div id="step6.1"style="position:absolute;top:60%;left:30%;font-size:20px;color:#878787;"></div>
<div id="obtuseresult"style="position:absolute;top:70%;left:30%;font-size:20px;color:#878787;"></div>
<div id="stepfine"style="position:absolute;top:75%;left:30%;font-size:20px;color:#878787;"></div>
</div>
<p id="heading19"style="position:absolute;top:-1%;left:30%;font-size:28px;color:blue;">Different types of triangle in angle method.</p>
</body>
</html>``````
I have some code in triangle shape calculator that generates CSS code used to draw the triangle shape. But I want to draw the different shapes in the triangle using canvas method. How can I change the programming code in CSS to canvas?
## Answers:
``````function draw() {
var canvas = document.getElementById('canvas');
if (canvas.getContext){
var ctx = canvas.getContext('2d');
ctx.beginPath();
ctx.moveTo(75,50);
ctx.lineTo(100,75);
ctx.lineTo(100,25);
ctx.fill();
}
}``````
``````<html>
<body onload="draw();">
<canvas id="canvas" width="100" height="100"></canvas>
</body>
</html>``````
You can reffer following artice-Drawing shapes in canvas.
`
``````<!DOCTYPE html>
<html>
<head>
<title>Triangle Canvas Example</title>
</head>
<body>
<canvas id="myCanvas" width="500" height="500"></canvas>
<script>
var canvasElement = document.querySelector("#myCanvas");
var context = canvasElement.getContext("2d");
// the triangle
context.beginPath();
context.moveTo(200, 100);
context.lineTo(100, 300);
context.lineTo(300, 300);
context.closePath();
// the outline
context.lineWidth = 10;
context.strokeStyle = '#666666';
context.stroke();
// the fill color
context.fillStyle = "#FFCC00";
context.fill();
</script>
</body>
</html>``````
`Try this
Html :
``````<!DOCTYPE html>
<html>
<head>
<title>Triangle Canvas Example</title>
</head>
<body>
<canvas id="myCanvas" width="500" height="500"></canvas>
</body>
</html>
``````
javascript :
`````` var canvasElement = document.querySelector("#myCanvas");
var context = canvasElement.getContext("2d");
// the triangle
context.beginPath();
context.moveTo(100, 100);
context.lineTo(100, 300);
context.lineTo(300, 300);
context.closePath();
// the outline
context.lineWidth = 10;
context.strokeStyle = '#666666';
context.stroke();
// the fill color
context.fillStyle = "#FFCC00";
context.fill();
``````
``````<!DOCTYPE HTML>
<html>
<head>
<style>
#test {
width: 100px;
height:100px;
margin: 0px auto;
}
</style>
<script type="text/javascript">
function drawShape(){
var canvas = document.getElementById('mycanvas');
if (canvas.getContext){
var ctx = canvas.getContext('2d');
ctx.beginPath();
ctx.moveTo(25,25);
ctx.lineTo(105,25);
ctx.lineTo(25,105);
ctx.fill();
ctx.beginPath();
ctx.moveTo(125,125);
ctx.lineTo(125,45);
ctx.lineTo(45,125);
ctx.closePath();
ctx.stroke();
}
}
</script>
</head>
<body id="test" onload="drawShape();">
<canvas id="mycanvas"></canvas>
</body>
</html>``````
## Top Questions
©2020 All rights reserved. | 3,639 | 12,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.124492 |
https://www.coursehero.com/file/6171178/Exam1D-FA02/ | 1,498,666,158,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323711.85/warc/CC-MAIN-20170628153051-20170628173051-00508.warc.gz | 876,080,796 | 149,463 | Exam1D. FA02
# Exam1D. FA02 - COURSE MSCI 3710 EXAM 1 Version D S.S Print...
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1 Use the information given below to answer the next 2 questions. The CEO of Mean Green Cab Company believes there is a difference between the durability of two brands of tires, Minus and Rub . The CEO decides to test her belief by running an experiment. To conduct the experiment, her R & D department uses a machine with a metallic device to wear down the tires and records the tread depth on a random sample of 10 tires of each brand after new tires are spun for an equivalent of 10,000 miles. Excel output for the test, on the sample data, conducted at the 5% significance level is shown in the tables below. t-Test: Two-Sample Assuming Equal Variances Rub Minus Mean 0.2415 0.2538 Variance 0.0004 0.0009 Observations 10 10 Pooled Variance 0.0007 Hypothesized Mean Difference 0 df 18 t Stat -1.0183 P(T<=t) one-tail 0.161 t Critical one-tail 1.734 P(T<=t) two-tail 0.322 t Critical two-tail 2.101 1. To test the claim that there is a difference between the durability of Rub and Minus brands (μ R – μ M ), which of the following is the critical value of the test statistic at the 5% level of significance? A. 1.734 B. 2.101* C. 1.330 D. -2.113 E. -1.018 2. What are the decision and conclusion for the test, if the test is conducted at the 5% significance level? A. Reject the null hypothesis, conclude there is evidence of a significant difference between the durability of the Minus and Rub brands. B. Reject the null hypothesis, conclude there is insufficient evidence that there is no significant difference between the durability of the Minus and Rub brands. C.
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## This note was uploaded on 03/09/2011 for the course DSC 3710 taught by Professor Staff during the Spring '09 term at North Texas.
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# first attempt GMAT score 760
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Re: first attempt GMAT score 760 [#permalink]
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08 Oct 2011, 14:01
congrats mate
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22 Oct 2011, 23:17
Good Job!
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01 Nov 2011, 15:18
Nice score!
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Re: first attempt GMAT score 760 [#permalink]
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08 Jul 2012, 14:56
great score mate! well done
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Re: first attempt GMAT score 760 [#permalink]
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22 Nov 2016, 20:03
Can you share the pdf of the error log that you made? I've just started preparing for GMAT and I need help!
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Re: first attempt GMAT score 760 [#permalink] 22 Nov 2016, 20:03
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# Grouping of Dataset based on then sum of a measure
I have a requirement to group orders based on the order amount (an order can have multiple lines). I want to group orders based on the order size.
Example of the Raw Data
Example of Final View
Is the 'Final view' achievable based on the raw data and using functions available in Tableau. If not, any other pointer towards implementation will be good.
• ###### 1. Re: Grouping of Dataset based on then sum of a measure
Yes. Very achievable.
It is called histogram. The order categories are bins. CountD(Order Number) is the count.
• ###### 2. Re: Grouping of Dataset based on then sum of a measure
can you supply a sample workbook?
• ###### 3. Re: Grouping of Dataset based on then sum of a measure
Hi Jeffrey,
I have attached the sample workbook.
-Shijin
• ###### 4. Re: Grouping of Dataset based on then sum of a measure
Hi Alexander
Please correct me if wrong. It is achievable easily if an order has only one line in the data set and if the bins are equal size.
In my case the data set is large and an order can have multiple lines.
Below is what i get If I use the bin concept.
-Shijin
• ###### 5. Re: Grouping of Dataset based on then sum of a measure
Hi Shijin,
You can achieve final you by creating calculated filed and place in row shlef
Cal Filed:
if [Amount \$]=<10 then "0-10\$"
ELSEIF [Amount \$]>11 and [Amount \$]<20 then "11-20\$"
ELSEIF [Amount \$]>21 and [Amount \$]<30 then "21-30\$" end
Use count as your measure aggregation.
For better solution, can you share workbook with sample data or data file
-
Sai
• ###### 6. Re: Grouping of Dataset based on then sum of a measure
Hi Sai,
Below is what I get if I use your logic
Also attached sample workbook.
-Shijin
• ###### 7. Re: Grouping of Dataset based on then sum of a measure
try something like this.
• ###### 8. Re: Grouping of Dataset based on then sum of a measure
Hi Jeffrey,
Thanks for trying to help but unfortunately that's not what I want. Thanks for your effort though.
-Shijin
• ###### 9. Re: Grouping of Dataset based on then sum of a measure
so, what is it that you want?
• ###### 10. Re: Grouping of Dataset based on then sum of a measure
I want the final view to be like below based on the data that I shared with you.
• ###### 11. Re: Grouping of Dataset based on then sum of a measure
Here is my solution workbook.
https://public.tableau.com/views/Sample1_81/Sheet1?:embed=y&:showTabs=y&:display_count=yes
See if it works for you.
On Tue, Jun 2, 2015 at 10:18 AM, Shijin Mathew <
1 of 1 people found this helpful
• ###### 12. Re: Grouping of Dataset based on then sum of a measure
it is like your screenshot, it's just based of the data that you supplied in your sample workbook.
• ###### 13. Re: Grouping of Dataset based on then sum of a measure
Hi Alex,
Thanks for the sample solution. The solution partly works for me.
The second column "size" should be the count of orders which has the amount sum in that range.
In the example that you provided the size refers to the number of rows in the final table.
To make my request more clearer I added one more row to the raw data as below
suppose the raw data was
Order Number Amount(\$) Product Category 100 10 Low End 101 5 High End 102 3 High End 100 10 High End 103 7 Low End 102 17 High End 104 15 High End
I would like the end result to be
Order Category(\$) Count of Order High End(\$) Low End(\$) 0-10 (\$) 2 5 7 10-20 (\$) 3 45 10 20-30 (\$) 0 0 0
You can see the column 'Count of order' column corresponding to "20-30 (\$)" has count 3 now because I added another order with amount 15.
-Shijin
• ###### 14. Re: Grouping of Dataset based on then sum of a measure
Hi Jeff,
The second column "count of orders" should be the count of orders which has the amount sum in that range.
In the below example "count of orders" is 2 for each row since there are 2 orders whose sum of amount comes in each of that range - that is the part which I am having trouble implementing.
1 2 Previous Next | 1,102 | 4,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-24 | latest | en | 0.903025 |
http://fora.xkcd.com/viewtopic.php?f=7&t=101043&start=5800&view=print | 1,566,519,083,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317688.48/warc/CC-MAIN-20190822235908-20190823021908-00001.warc.gz | 75,119,101 | 5,920 | Page 146 of 2688
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:01 pm UTC
fweeeee
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:01 pm UTC
Damn. I hope the sand castle survives the bombardment.
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:01 pm UTC
We have a lot of information about the trajectory there, can someone predict where it will hit?
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:01 pm UTC
Meaux_Pas wrote:Randy doesn't read the forums. Like ever. Seriously.
trust me on this one.
Well, if not, he's missing out on this thread.
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:02 pm UTC
INCOMING!
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:02 pm UTC
Meaux_Pas wrote:Randy doesn't read the forums. Like ever. Seriously.
trust me on this one.
Well, if not, he be missing out on this small sequence of programmed instructions.
Yes. He misses every thread. On purpose.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:03 pm UTC
speising wrote:for all wondering how much Things that are on my side for 600, Alex! passes between frames: this animation moves at the speed of plot.
Quite right. But should passing intervals be measured in plots (or plotlets)? soon we'll be discussing dekaplots and kiloplots.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:03 pm UTC
Meaux_Pas wrote:Randy doesn't read the forums. Like ever. Seriously.
trust me on this one.
Well, if not, he be missing out on this small sequence of programmed instructions.
How is it possible to enjoy Time without reading?
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:03 pm UTC
Sappharos wrote:We have a lot of information about the trajectory there, tin someone predict where it woll hit?
It looks like it will hit close to Cueball.
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:03 pm UTC
I'm not sure what happens when a mini-trebuchet misfires, but I'm betting it involves missing fingers.
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:03 pm UTC
I'm guessing it is going to hit the second flag from the right.
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:04 pm UTC
Definitely becoming more & more of a solid metaphor for the building/repairing/destruction of a relationship. Cueball seems to be doing a lot of work to keep the "castle" together.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:04 pm UTC
mybrainhurts wrote:C'mooon mods - still no replacement for TREBUCHET!?!
And this is the best way to make sure we never will.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:04 pm UTC
peewee_RotA wrote:I reckon she built a catapult!
It's a trebuchet, but I still win the internet!
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:05 pm UTC
So, is this flying object going to destroy the two prideful towers? T:me will say...
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:06 pm UTC
Azrael wrote:
mybrainhurts wrote:C'mooon mods - still no replacement for TREBUCHET!?!
And this be the goodest way to make sure we never woll.
Perfect
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:06 pm UTC
Corrado wrote:Definitely becoming more & more of a solid metaphor for the building/repairing/destruction of a relationship. Cueball seems to be doing a lot of work to keep the "castle" together.
His fingertips are holding onto the cracks in their Magic Kingdom.
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:06 pm UTC
I did in fact say "who uses a trebuchet for such a small projectile?" - well we now know it's someone who has a small trebuchet. I'm just glad Megan is back. Quick Cueball! Tackle her before she launches another volley!
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:06 pm UTC
Sappharos wrote:We have a lot of information about the trajectory there, tin someone predict where it woll hit?
I call middle of the big tower to the right of the two arches.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:07 pm UTC
Foarm wrote:
Meaux_Pas wrote:Randy doesn't read the forums. Like ever. Seriously.
trust me on this one.
Well, if not, he be missing out on this small sequence of programmed instructions.
How be it possible to enjoy Things that are on my side for 600, Alex! without reading?
it actually makes it more awesome. His imagination is not tainted by ours. He comes up with original content by himself
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:07 pm UTC
Kyrie wrote:So, be this flying object going to destroy the two prideful towers? T:me woll say...
I think it will hit the arches.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:08 pm UTC
beargarden wrote:
peewee_RotA wrote:
trusstopher wrote:Not really a fan of the transformation this place has undergone since I departed yesterday. I Guess I shall resort to just watching progress with no commentary until completion. I have so many more free moments in my day now that I shant have to keep up with countless postings.
But we be going to elect a new pope.‽‽ eventually. When I get around to continue posting it.
tin it wait about twelve timeframes?
I need to sleep, but as Cardinal I reckon I ought to be present for the vote. hwæt to do!
I'll postpone the vote
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:08 pm UTC
I call an eagle grabbing the projectile in mid-air.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:10 pm UTC
Foarm wrote:
Kyrie wrote:So, be this flying object going to destroy the two prideful towers? T:me woll say...
I reckon it woll hit the arches.
It looks like she moved the trebuchet a metre right, so I expect the ball will hit a metre right of the previous one.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:10 pm UTC
Foarm wrote:
Kyrie wrote:So, be this flying object going to destroy the two prideful towers? T:me woll say...
I reckon it woll hit the arches.
Yep, I'm thinking arches as well. And he just built those!
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:11 pm UTC
pdg wrote:
Sappharos wrote:We have a lot of information about the trajectory there, tin someone predict where it woll hit?
I call middle of the big tower to the right of the two arches.
Arc is higher which means steeper trajectory. She's aiming for Cueball, me thinks.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:11 pm UTC
pdg wrote:
Opiboble wrote:
Dictator wrote:
Montov wrote:There be no clarification yet on the yukky substance in Megan's mouth, but it be the same colour as the trebuchet. So be the trebuchet made of coffee? Coffee be black too.
Backshadowing. Everything tin be black. Keeps the options open; more fun.
Some bodily fluids tin’t be black... Some be quite clear...
Well, they tin sometimes be black, but that be usually a sign you need to see a doctor.
I stand corrected, maybe that is where they went off scene after the swim.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:11 pm UTC
Lord Cathbad wrote:I call an eagle grabbing the projectile in mid-air.
I was going to say Cueball catches it (or it hits him).
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:11 pm UTC
It`s gonna be Cueball's HEAD that`s gonna be hit!
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:15 pm UTC
mybrainhurts wrote:It`s gonna be Cueball's HEAD that`s gonna be hit!
Ohh, and he's going fall, causing more damage! Smart girl!
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:15 pm UTC
Wow. Sure missed a tone while I was gone.
Also, I am glad to see I'm not the only one who seriously doubts Randall would change the course of it all based on this board. I mean, this is fun and all, but he clearly planned this out in advance. The presence of a trebuchet should not mean he's listening to it, since most of the fans here who suggested a trebuchet before it appeared probably found their appreciation and passion for trebuchets from other earlier xkcd comics to begin with. He liked the idea long before a bunch of people he's never met and who he DEFINITELY doesn't live for the approval or acceptance of brought it up in connection with this comic. Similarly, if we end up seeing raptors, that shouldn't be cited as evidence that he's "obeying the forum requests" or any such nonsense. Randall is the reason why most of you guys like raptors to begin with (well, Randall and Jurassic Park...), and so if they show up, or the eternity bird, or the Waterman map, it will be because Randall liked them first and planned them into things, not because we have any measure of control over the experience.
The discussion board here is amusing, but it is a corollary to the experience, not the experience itself.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:16 pm UTC
Foarm wrote:
Meaux_Pas wrote:Randy doesn't read the forums. Like ever. Seriously.
trust me on this one.
Well, if not, he be missing out on this small sequence of programmed instructions.
How be it possible to enjoy Things that be on my side for 600, Alex! without reading?
it actually makes it more awesome. His imagination be not tainted by ours. He comes up with original content by himself
He is great at being original.
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:19 pm UTC
Randall loves raptors cause he loves Calvin&Hobbes!
That said, the actual situation reminds me a lot of the snowball fights of the latter...
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:20 pm UTC
Watching it backwards is fun, but it doesn't make quite as much sense now that we have a trebuchet.
### Re: 1190: "Things that are on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:22 pm UTC
I felt inspired.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:22 pm UTC
Valarya wrote:Christ.. I sleep for 16NPs and have a 4NP meeting at work, and the small sequence of programmed instructions be now 45 more pages deep and has a new title.
I seem to have missed a lot. Why has the thread title changed? And why isn't the Time wiki being updated?
And what are these small sequences of programmed instructions all about?
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:23 pm UTC
bouer wrote:Azrael just told us he be not reading this. Seriously.
I missed that post some how... But seriously, this comic (if we can even call it that) is awesome (in the traditional sense, not the new watered down sense).
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:25 pm UTC
SecondTalon wrote:
Kyrie wrote:hwæt the... hwæt be this hodor thingy? When did it happen?
I replaced every word on the forum - literally everything that could be replaced - with the word HODOR as a Song of Ice and beflame reference.
As it made everything unusable, I turned it off after a minute.
This, and all the other fun turn-loose-the-mods on the forums ... I love this place. I'm moving in. You don't mind, do you? I could even help build the Monastery of Time.
### Re: 1190: "Things that be on my side for 600, Alex!"
Posted: Fri Mar 29, 2013 5:26 pm UTC
kenmelken wrote:Wow. Sure missed a tone while I be-did go-done.
Also, I be glad to see I be not the only one who seriously doubts Randall woll-did change the course of it all based on this board. I mean, this be fun and all, but he clearly planned this out in advance. The presence of a trebuchet should not mean he be listening to it, since most of the fans here who suggested a trebuchet before it appeared probably found their appreciation and passion for trebuchets from other earlier xkcd comics to begin with. He liked the idea long before a bunch of people he be never met and who he DEFINITELY doesn't live for the approval or acceptance of brought it up in connection with this comic. Similarly, if we end up seeing raptors, that shouldn't be cited as evidence that he be "obeying the forum requests" or any such nonsense. Randall be the reason why most of you guys like raptors to begin with (well, Randall and Jurassic Park...), and so if they show up, or the eternity bird, or the Waterman map, it woll be because Randall liked them first and planned them into things, not because we have any measure of control over the experience.
The discussion board here be amusing, but it be a corollary to the experience, not the experience itself.
Spoken like an outsider. Heresy!
### Re: 1190: "Time"
Posted: Fri Mar 29, 2013 5:27 pm UTC
KarMann wrote:
KuwabatakeSanjuro wrote:I've been thinking on how to hide that comic's link in my class I teach, so my students tin find it. Maybe as an extra-credit activity or something. Or perhaps I should just open up next semester with it in my online welcome message. (Yes, I quite enjoy that comic.)
Hrm. Probably Goode Homolosine. (Though I admit, when I be-did at a Buckminster Fuller exhibit in some museum and there be-did a Dymaxion map... geekness ensued. But by description in the comic, Goode Homolosine.)
High five, fellow Goode Homolosine! | 3,902 | 14,026 | {"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-2019-35 | latest | en | 0.914458 |
https://socratic.org/questions/5634bfd511ef6b4a8a99b912 | 1,580,125,919,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251700675.78/warc/CC-MAIN-20200127112805-20200127142805-00229.warc.gz | 656,733,889 | 7,316 | # Question 9b912
Oct 31, 2015
${\text{ZnI}}_{2}$
#### Explanation:
The idea here is that you have a sample of zinc and a sample of iodine that combine to form, like the problem tells you, a compound of zinc and iodine.
Now, if you assume that all the zinc and all the iodine that took part in the reaction are now part of the compound, you can say that the mass of the resulting compound will be
$\text{ "m_"compound" = m_"zinc" + m_"iodine}$
$\text{ "m_"compound" = "2.50 g" + "9.70 g" = "12.20 g}$
To find the chemical formula of this compound, you need to figure out how many moles of each element you get in that sample.
Basically, the same principle applies - all the moles of zinc and all the moles of iodine are now part of the compound.
So, how many moles of each did you have in the initial samples?
To find that out, use the molar amsses of two elements
$\text{ "2.50color(red)(cancel(color(black)("g zinc"))) * "1 mole zinc"/(65.38color(red)(cancel(color(black)("g zinc")))) = "0.03824 moles zinc}$
and
$\text{ "9.70color(red)(cancel(color(black)("g iodine"))) * "1 mole zinc"/(126.9color(red)(cancel(color(black)("g iodine")))) = "0.07644 moles iodine}$
To find the mole ratio that exists between the two elements in the compound, divide both values by the smallest one
" " "For zinc: " (0.03824color(red)(cancel(color(black)("moles"))))/(0.03944color(red)(cancel(color(black)("moles")))) = 1
" " "For iodine: " (0.07644color(red)(cancel(color(black)("moles"))))/(0.03824color(red)(cancel(color(black)("moles")))) = 1.999 ~~ 2#
This means that the empirical formula of the compound will be
${\text{ " "Zn"_1"I"_2 implies "ZnI}}_{2}$
As it turns out, this is also the formula unit of the compound.
For ionic compounds, the empirical formula, which tells you what the smallest whole number ratio between the elements that make up that compound is, is also the formula unit.
SInce you can't have a whole number that's smaller than $1$ (and different from zero, of course), it follows that $1 : 2$ is the smallest whole number ratio in which the two elements can be combined.
You can double-check the result by using the molar mass of zinc iodide, ${\text{ZnI}}_{2}$, and the number of moles you have.
Since you have
${\text{ ""Zn" + "I"_2 -> "ZnI}}_{2}$
you can say that every mole of zinc will produce one mole of zinc iodide. If this is the case, the you can expect the mass of the resulting compound to be
${\text{ " 0.03824color(red)(cancel(color(black)("moles"))) * ("319.19 g ZnI"_2)/(1color(red)(cancel(color(black)("mole")))) = "12.2 g ZnI}}_{2}$ | 761 | 2,590 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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} | 3.90625 | 4 | CC-MAIN-2020-05 | longest | en | 0.908546 |
http://nrich.maths.org/10746/submitsolution | 1,502,963,809,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103167.97/warc/CC-MAIN-20170817092444-20170817112444-00505.warc.gz | 301,276,887 | 5,159 | ### Pair Sums
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
### Summing Consecutive Numbers
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Big Powers
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
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Country: | 308 | 1,368 | {"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-2017-34 | latest | en | 0.943521 |
https://metacpan.org/pod/release/KRYDE/Math-NumSeq-72/lib/Math/NumSeq/FibonacciWord.pm | 1,534,705,335,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215284.54/warc/CC-MAIN-20180819184710-20180819204710-00417.warc.gz | 700,173,986 | 11,309 | ++ed by:
Kevin Ryde
and 1 contributors
# NAME
Math::NumSeq::FibonacciWord -- 0/1 related to Fibonacci numbers
# SYNOPSIS
`````` use Math::NumSeq::FibonacciWord;
my \$seq = Math::NumSeq::FibonacciWord->new;
my (\$i, \$value) = \$seq->next;``````
# DESCRIPTION
This is a sequence of 0s and 1s formed from the Fibonacci numbers.
`````` 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, ...
starting i=0``````
The initial values are 0,1. Then Fibonacci number F(k) many values are copied from the start to extend, repeatedly.
`````` 0,1 initial
0,1,0 append 1 value
0,1,0,0,1 append 2 values
0,1,0,0,1,0,1,0 append 3 values
0,1,0,0,1,0,1,0,0,1,0,0,1 append 5 values
0,1,0,0,1,0,1,0,0,1,0,0,1,0,1,0,0,1,0,1,0 append 8 values
etc``````
## Morphism
The same sequence is had by starting with 0 and then repeatedly expanding
`````` 0 -> 0,1
1 -> 0``````
## Fibbinary and Zeckendorf
The result is also the Fibbinary numbers modulo 2, which is the least significant bit of the Zeckendorf base representation of i.
The Zeckendorf base breakdown subtracts Fibonacci numbers F(k) until reaching 0 or 1. This effectively undoes the above append expansion procedure. (See "Zeckendorf Base" in Math::NumSeq::Fibbinary.)
`````` start at i
until i=0 or i=1 do
subtract from i the largest Fibonacci number <= i
final resulting i=0 or i=1 is Fibonacci word value``````
For example i=11 has largest Fibonacci<=11 is 8, subtract that to leave 3. From 3 the largest Fibonacci<=3 is 3 itself, subtract that to leave 0 which is the Fibonacci word value for i=11.
## Dense Fibonacci Word
Option `fibonacci_word_type => "dense"` selects the dense Fibonacci word
`````` 1,0,2,2,1,0,2,2,1,1,0,2,1,1,...
starting i=0``````
This is the above plain word with each two values (not overlapping) encoded in a binary style as
`````` plain pair dense value
---------- -----------
0,0 0
0,1 1
1,0 2``````
For example the Fibonacci word starts 0,1 so the dense form starts 1. A pair 1,1 never occurs in the plain Fibonacci word so there's no value 3 in the dense form.
# FUNCTIONS
See "FUNCTIONS" in Math::NumSeq for behaviour common to all sequence classes.
`\$seq = Math::NumSeq::FibonacciWord->new ()`
`\$seq = Math::NumSeq::FibonacciWord->new (fibonacci_word_type => \$str)`
Create and return a new sequence object. The `fibonacci_word_type` option (a string) can be either
`````` "plain" (the default)
"dense"``````
## Iterating
`\$seq->seek_to_i(\$i)`
Move the current i so `next()` will return `\$i` (and corresponding value) on the next call.
## Random Access
`\$value = \$seq->ith(\$i)`
Return the `\$i`'th value in the sequence. The first value is at i=0.
`\$bool = \$seq->pred(\$value)`
Return true if `\$value` occurs in the sequence. This simply means 0 or 1, or for the dense Fibonacci word 0, 1 or 2.
Math::PlanePath::FibonacciWordFractal | 968 | 3,066 | {"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} | 3.71875 | 4 | CC-MAIN-2018-34 | latest | en | 0.796007 |
https://studymaterialcenter.in/question/when-m-1-gram-of-ice-at-10-circ-c-specific-heat-0-5-cal-g-1-circ-c-1-is-added-to-m-2-gram-of-water-at-50-circ-c-finally-no-ice-is-left-and-the-water-is-at-0-circ-c-the-value-of-latent-heat-of-ice-in-c/ | 1,675,912,974,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00325.warc.gz | 565,683,475 | 24,851 | # When $\mathrm{M}_{1}$ gram of ice at $-10^{\circ} \mathrm{C}$ (Specific heat $=0.5 \mathrm{cal} g^{-1}{ }^{\circ} \mathrm{C}^{-1}$ ) is added to $\mathrm{M}_{2}$ gram of water at $50^{\circ} \mathrm{C}$, finally no ice is left and the water is at $0^{\circ} \mathrm{C}$. The value of latent heat of ice, in cal $g^{-1}$ is:
Question:
When $\mathrm{M}_{1}$ gram of ice at $-10^{\circ} \mathrm{C}$ (Specific heat $=0.5 \mathrm{cal} g^{-1}{ }^{\circ} \mathrm{C}^{-1}$ ) is added to $\mathrm{M}_{2}$ gram of water at $50^{\circ} \mathrm{C}$, finally no ice is left and the water is at $0^{\circ} \mathrm{C}$. The value of latent heat of ice, in cal $g^{-1}$ is:
1. $\frac{50 M_{2}}{M_{1}}-5$
2. $\frac{5 M_{1}}{M_{2}}-50$
3. $\frac{50 M_{2}}{M_{1}}$
4. $\frac{5 M_{2}}{M_{1}}-5$
JEE Main Previous Year Single Correct Question of JEE Main from Physics Termal Properties of Matter chapter.
JEE Main Previous Year 12 April 2019 I
Correct Option: 1
Solution:
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Download App | 1,346 | 4,658 | {"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-2023-06 | latest | en | 0.752579 |
https://www.usingenglish.com/forum/threads/36215-Scale-Unit-Degree | 1,477,169,201,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719041.14/warc/CC-MAIN-20161020183839-00182-ip-10-171-6-4.ec2.internal.warc.gz | 1,023,485,022 | 15,499 | # Thread: Scale Unit / Degree
1. ## Scale Unit / Degree
Hallo,
I am a little bit confused right now. This morning I listened to an English radio sender, and as they talked about the weather here in Germany they said that we had 42 degrees . I don't know, maybe they measured in the desert instead of Germany, but on my thermometer there were only 6 degrees.
Do you have other scale units?
I am also confused about your other measure units like your weight or your size But in the meantime I know that when someone said that they have a weigt of 200 that this are not kilogram
2. ## Re: Scale Unit / Degree
Originally Posted by Dany
Hallo,
I am a little bit confused right now. This morning I listened to an English radio sender, and as they talked about the weather here in Germany they said that we had 42 degrees . I don't know, maybe they measured in the desert instead of Germany, but on my thermometer there were only 6 degrees.
Do you have other scale units?
I am also confused about your other measure units like your weight or your size But in the meantime I know that when someone said that they have a weigt of 200 that this are not kilogram
The term degree is used in several scales of temperature. The symbol ° is usually used, followed by the initial letter of the unit, for example °C for degree(s) Celsius. (For temperature differences, the usage is sometimes reversed; then 100 C°, or "100 Celsius degrees", is a temperature difference, while 100 °C, or "100 degrees Celsius", is an actual temperature.) Scales of temperature include:
degree Celsius (°C)
degree Delisle (°De)
degree Fahrenheit (°F)
degree Newton (°N)
degree Rankine (°R or °Ra)
degree Réaumur (°R)
degree Rømer (°Rø)
degree Kelvin (°K)
This degree Kelvin (°K) is a former name for the SI unit of temperature on the thermodynamic (absolute) temperature scale. Since 1967 it has been known simply as the kelvin, with symbol K.
degree absolute (°A) is obsolete terminology, often referring specifically to the kelvin but sometimes the degree Rankine as well
Degree symbol
Degree (temperature - Wikipedia, the free encyclopedia)
3. ## Re: Scale Unit / Degree
Thanks a lot for your detailed explanation
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• | 545 | 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} | 3.203125 | 3 | CC-MAIN-2016-44 | longest | en | 0.950948 |
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