text:
# Math Help - Quadratic eqn

The roots of the eqn x^2-8x+h
are (alpha) and (alpha+3k)

Express h in terms of k

2. Originally Posted by helloying
The roots of the eqn x^2-8x+h
are (alpha) and (alpha+3k)

Express h in terms of k
For

ax^2 +bx +c =0

Sum of roots of quadratic equation $= \frac{-b}{a}$

Product of roots of equation $= \frac{c}{a}.....(2)$

Hence using (2) we get

$
(\alpha) \times (\alpha + 3k) = h --------(3)$

And

$(\alpha) + (\alpha+3k) = 8$

$\alpha= (8-3k)/2 --------------(1)$

Use one in (3)

$\frac{(8-3k)}{2} \times \frac{8+3k}{2} = h$

$64- 9k^2 = 4h$

3. Originally Posted by helloying
The roots of the eqn x^2-8x+h
are (alpha) and (alpha+3k)

Express h in terms of k
I will guess that the author of the assignment forgot the "equals zero" part of the "equation". (Without the "equals" sign, it's not an "equation"; without the assumption that the other side was "zero", there's no way to find the roots.)

In case you haven't learned about the formulas relating the roots of a quadratic function to the coefficients, you can instead use the relationship between roots and factors to find the factorization:

. . . . . $\mbox{If }\, x\, =\, a\, \mbox{ is a root, then }\, x\, -\, a\, \mbox{ is a factor.}$

In your case, the zeroes are:

. . . . . $x\, =\, \alpha\, \mbox{ and }\, x\, =\, \alpha\, +\, 3k$

Then the factors must have been:

. . . . . $x\, -\, \alpha\, \mbox{ and }\, x\, -\, (\alpha\, +\, 3k)$

Multiplying, we get:

. . . . . $(x\, -\, \alpha)(x\, -\, (\alpha\, +\, 3k))\, =\, x^2\, +\, (-2\alpha\, -\, 3k)x\, +\, (-\alpha^2\, -\, 3\alpha k)$

Comparing coefficients, we see that:

. . . . . $-2\alpha\, -\, 3k\, =\, -8$

Then:

. . . . . $\alpha\, =\, 4\, -\, \frac{3}{2}k$

Comparing coefficients again, we see that:

. . . . . $-\alpha^2\, -\, 3\alpha k\, =\, h$

Substituting from above, we get:

. . . . . $-\left(4\, -\, \frac{3}{2}k\right)^2\, -\, 3\left(4\, -\, \frac{3}{2}k\right)k\, =\, h$

. . . . . $-16\, +\, 12k\, -\, \frac{9}{4}k^2\, -\, 12k\, +\, \frac{9}{2}k^2\, =\, h$

...which simplifies to give a result equivalent to the previous reply.


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## Monday, September 01, 2014 ... /////

### It's always harmful to attach negative adjectives to quantum mechanics

Sabine Hossenfelder attended a global summit dedicated to the question whether it's right for science writers to present quantum mechanics as spooky, strange, and weird. Imagine how much fuel would be saved if the participants could have stayed home and responded with the right answer which is

No.
There is really no room for long yet intelligent debates and diatribes. Quantum mechanics was surprising to the physicists and is still surprising to the beginners who are learning it today. But as Giotis mentioned in the first comment under Sabine's blog post, it is not really surprising that a deep enough theory is surprising because humans and their ancestors have been trained by experience where effective, superficial, and conceptually misleading laws are enough to understand what is going on.

### Production of vacuum cleaners above 1600 watts banned in the EU

In November 2013, I reminded everyone that the EU had an incredibly irresponsible plan to simply ban all vacuum cleaners above 1600 watts of the input power.

(The equivalent figure that Americans would use is 110 times smaller because the voltage is 110 volts and in amps, so 1600 watts is equivalent to 14.55 amps.)

Today, on the 75th anniversary of the outbreak of World War II, the ban came to force. I didn't want to believe that it would ever become valid but it really has. Czech media say that it's still OK to sell them and the retailers have huge inventories, indeed. Some Western European media suggest that it is no longer legal to even sell them – but the sale really seems to continue in Czechia, a country that is telling the EU overlords "screw you, Ken".

## Sunday, August 31, 2014 ... /////

### Science paper: null results should be printed

Null results are 1.4-1.6 times underrepresented in social science literature

Three days ago, Annie Franco, Neil Malhotra, and Gabor Simonovits of Stanford published a paper in Science that discusses an issue that's been often covered on this blog:

Publication bias in the social sciences: Unlocking the file drawer

They looked at some NSF-funded body of social science research – TESS surveys among the U.S. citizens – where one knows how much research was actually performed, how much was written down, and how much was published.

### Czechs, Slovaks won (possible) exemption from new EU sanctions

I urge Moscow to treat EU countries separately, too

A summit in Brussels chose the successor to the EU "president" von Rumpuy. It's no one else than Donald Tusk, center-right prime minister of Poland. He will take the lead in December.

As this 2008 Polish parody "Donald Marží" (Donald Is Dreaming) of a funny 1978 Czech song Joey from the Swamps (the #1 song of your humble correspondent from the kindergarten years) reminds us, Tusk is known as the guy who has ludicrously promised the Poles to completely eradicate corruption and to create another Ireland in Eastern Europe. Well, I still think that the replacement of von Rumpuy by Tusk represents progress in the positive direction.

The Polish coat of arms shows a duck, in order to celebrate Mr Kaczynski and Mr Donald Duck-Tusk.

Radek Sikorski, a former Polish minister of foreign affairs and a staunch Russophobe (not too many other things define his identity), could have been the replacement of Catherine Ashton. But the new EU "minister" of foreign affairs will be Italian social democrat Federica Mogherini who actually has pretty peaceful relations to Russia (and was important in an Italian-US society, too).

## Saturday, August 30, 2014 ... /////

### Don Lincoln's introduction to particle detectors

Fermilab's Don Lincoln has recorded a couple of videos promoting and explaining physics concepts.

In this new 10-minute film, he presents particle detectors as subatomic bomb squads.

## Friday, August 29, 2014 ... /////

### CSTO ready to take over Ukraine

I've spent lots of hours by following the events in Ukraine and worrying about them, too. The Western propagandist outlets have informed their undemanding audiences that Russia has finally invaded Ukraine. Well, it wasn't the first time when we heard such a thing. What strategy did Russia choose to invade Ukraine yesterday?

It has sent about 10 soldiers above. They must have supernatural powers if they're supposed to conquer a country with 40+ million people. Perhaps, it wasn't just the ten folks. There may be a thousand of Russian citizens fighting over there. Soldiers on vacations, retired soldiers, and perhaps other volunteers. What a shock. Do you really expect the Kremlin to exterminate people in their military for these hobbies? The Kremlin understands the emotions behind these hobbies. There is a comparable number of Americans (or at least one hundred) on the other side – who have really no business to be there at all – too.

## Thursday, August 28, 2014 ... /////

### Borexino: photographing the Sun using neutrinos

Technical: Windows 7,8 64-bit users are definitely recommended to download the stable 64-bit version of Chrome which has twice lower instability rate and is faster, not to mention some new abilities. The installer will automatically update your (now 32-bit) Chrome. Perhaps you have to run it twice, closing Chrome perfectly in between. At the end, "About Chrome" should show "Version 37.0.2062.94 unknown-m (64-bit)".
Physics World and most of other MSM science news sources inform us about the observations by Borexino, an Italian experiment located in Gran Sasso (the word could mean "boring former baby") that observed the $pp$ neutrinos, the neutrinos created during$p + p \to \text{d} + e^{+} + \nu_e$ i.e. the fusion of proton pairs in the Sun (which produces about 99% of the solar output), for the first time.

A five-minute 2009 video promoting Borexino. Is her accent Italian?

To achieve the goal that wasn't guaranteed from the beginning, they had to use a thousand of tons of water buried 1,400 meters beneath the surface, in order to shield some of the backgrounds.

## Wednesday, August 27, 2014 ... /////

### Kaggle Higgs contest: the solution file for everyone

Well, an approximate one

Have you ever searched for the solution file for the Higgs Kaggle contest? Have you ever asked why the organizers don't just publish it so that everyone is smarter? ;-)

Did you ever want to be able to estimate your submission's score without sending it to the Kaggle server? Have you ever been confused by the normalization of the weights?

Because I just got a permission from my teammate and it is allowed for the contestants to be generous and share their wisdom with the whole Internet and all the competitors, I just decided to help everyone – and, perhaps, to re-energize the contest a little bit.

### Cristobal, Marie on animated global wind map

Finally, one may see some exciting realtime pictures on the global wind map (click here to get the interactive app; TRF instructions are available, too).

Yes, the hurricane approaching the Bermudas (East of Florida) is called Cristobal. The Pacific hurricane West of California is called Marie.

## Tuesday, August 26, 2014 ... /////

### Visiting Triton

In the most recent week, the American, European, as well as Russian space programs suffered from bummers.

Elon Musk's new SpaceX F9R rocket self-detonated over Texas. Something went wrong and we're told that this rocket's suicide was their mundane Plan B. It's sort of hard to believe that they really planned such a "maneuver" but maybe it's right. One can't get rid of the feeling that these attempts to privatize the space research are perhaps "too cheap" for them to succeed.

Triton, to be discussed later

The second bummer is linked to the unAmerican competitors of the GPS system. A year ago, a Proton-M rocket carrying a GLONASS (Russian GPS) satellite exploded shortly after it was launched.

## Monday, August 25, 2014 ... /////

### A curious traffic accident

Sunday, a day of playing e.g. with my 5-year-old nephew and niece, was probably more likely a day than Monday for such an injury. But I classified an event in which I was injured an hour ago to be a traffic accident.

A curious one, indeed. Some people who drive cars or bikes hit other cars. In 1999, a friend of mine from Prague and Rutgers, mathematician Petr Čížek, was killed by a truck that they had crashed into while his lively Russian female friend was driving on a seemingly empty road in Minnesota (I didn't go to that expedition, mostly due to the qualifying exams).

The underpass going to the soccer stadium after a fan association named Ultraside gave a permission to itself to paint it. ;-) It used to be prettier than at random moments – the city hall should have hired them to paint the underpass regularly.

People may collide with other people of different occupations and races. The individual I hit an hour ago was a boxer. And it's a race, not an occupation. ;-) Well, I know that the native speakers actually call it a "breed", not a "race", but I will use the word "race" because it's the same concept as the human races, isn't it (well, except that the separation to breeds was mostly bred i.e. man-made), and I am Czech who has never attended any dog races so there is no room for confusions.

## Sunday, August 24, 2014 ... /////

### Irradiation of face by LHC beam is better than Botox

Like other trained particle physicists, I have often been asked what would happen if you decided to insert your hand or head or another part of your body right to the LHC beam.

I wasn't sure about the right answer because it's a complicated interdisciplinary question combining particle physics, radiation safety, biology, and some condensed matter physics and thermodynamics. To be sure, I recommended people not to try this experiment because it would almost certainly be lethal.

These charts of Anatoli Bugorski's skull resemble the investigators' map of the JFK assassination scene.

Three weeks ago, Extreme Tech brought us a story that I didn't know and that may actually force us to change our minds. The experiment has already been tried 35 years ago and the guy is still around.

What happened?

## Saturday, August 23, 2014 ... /////

### Europe's support for the Islamic State

"The Islamic State" terrorists are so cruel that at one point, Osama bin Laden had ruled out any collaboration with them.

Off-topic: Czech readers may listen to an episode of Planetarium where they interviewed me about string theory. The program (archive) was aired on Czech Public Radio North on Saturday 18:05 Czech Summer Time.

Volcano: Bárðarbunga began to erupt, so far subglacially, but you may watch the webcam and it may get more interesting.
These überjerks show that when you calibrate your moral sense to the situation of Syria, Bashar Assad is the good guy. The Islamic State is the "real Islam that matters" in the region. Their acts humiliate everyone who has been spreading ludicrous PC fairy-tales about Islam as a religion of peace. There may be people who are peaceful and call themselves Muslims but they are unfortunately irrelevant. What's relevant and "truly Islamic" are groups like the Islamic State – and whether you like it or not, their attitudes are supported by much of the Quran, too. Islam is a form of fascism – and it has been almost nothing else for something like 1300 years.

These monsters are murdering lots of peoples including the Yezidis, an eclectic church sometimes considered as an Islamic sect and sometimes a hybrid of shamanism and Christianity. Those mostly Kurdish people worship Melek Taus, an angel considered to be a "fallen angel" by the Muslims (and probably others) because he refused to lick Adam's aß at one point. Count me as a fan of Melek Taus, too. Adam was just a transitional fossil between mutated apes and homo sapiens – no need to worship this ancestor of ours.

Also, the Islamic State has a P.R. department that is doing pretty much the same thing as MSNBC and other cheap sources of "news" (brainwashing of the undemanding audiences) we know in the West. They use clever tricks to spread their message on the "social media" and they have a superior technology capable of filming hours of professional footage containing dozens of types of murders as well as the warriors' distribution of ice cream among children and they obey many rules of the PC P.R. media as well as Hollywood to impress some people.

## Friday, August 22, 2014 ... /////

### Ashoke Sen: elementary particles are small black holes

Three string theorists added as Dirac Medal winners

On August 8th of every year, the Abdus Salam Institute in Trieste, Italy chooses up to three recipients of the Dirac Medal. (It's the anniversary of Dirac's 1902 birth. There exist three other awards called the "Dirac Medal" which I will ignore because they're less relevant for this blog's audience.)

Of course, the medal tries to decorate deep minds who are doing a similar kind of profound research as Paul Dirac did which is why dozens of string theorists have already won it. The Dirac Medal shows what the Nobel prize would look like if the committee weren't constrained by the required explicit, dynamite-like demonstration of the physical discoveries.

In 2014, i.e. two weeks ago, the Italian institute avoided all experiments and awarded just three string theorists:

Ashoke Sen, Andrew Strominger, Gabriele Veneziano
Congratulations! Of course, Veneziano is the forefather of the whole discipline (the intercourse that has led to the birth was Veneziano's encounter with the Euler Beta function), Andy Strominger is a lot of fun and a perfectly balanced top thinker in one package and I know him the best of all, of course ;-), and Ashoke Sen is among the most brilliant minds, too. He has previously won the Milner award, too.

The Hindu printed a short yet interesting interview with Ashoke Sen yesterday:
‘Elementary particles may be thought of as small black holes’
It's funny – the title is actually a sentence I have included in almost every general physics talk I gave in the last decade, perhaps 30 talks in total. Sometimes I talk about the panic about the LHC-produced black holes and emphasize that only experts may distinguish a small black hole from an elementary particle such as the Higgs boson – and its evaporation from the Higgs decay etc.

It's true that the Hawking radiation of a "larger than minimal" black hole has a higher number of decay products (particles) so it's more uniform but for the truly minimum-size black holes, there's no difference.

### All 365 Sierra Leona ebola casualties due to a herbalist

The Daily Mail and everyone else mentioned the sad and bizarre statistical fact: all 365 people (3/4 of the victims are women because they're more "the social glue" of their communities, caregivers) who died due to ebola (a lethal cousin of flu) in Sierra Leone caught the disease from a herbalist. What an irony.

The mass infection began in Guinea (patient zero has probably been correctly identified) and lots of the infected ones came to the herbalist in Sierra Leone – which used to be a healthy country. She got infected and many others caught the disease from her, many of them during the funeral.

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# An object is pulled along the ground back starting a force of 40 lb on a rope that makes a 15° angle of the ground. How much work is done dragging the

###### Question:

An object is pulled along the ground back starting a force of 40 lb on a rope that makes a 15° angle of the ground. How much work is done dragging the object 47 ft?

### How could you incorporate this Habit of Mind into your lesson plans?

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### Can we talk on a zoom link

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### The length of a rectangle is 5 feet longer than the width. Write a SIMPLIFIED expression to represent

The length of a rectangle is 5 feet longer than the width. Write a SIMPLIFIED expression to represent the area of the rectangle. Use a ^ to denote any exponents....

### Im having to write this problem as a equation then solvethe sum of 4xs a number and 9 is equal to3xs

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### HOW CAN I MAKE MONEY ONLINE AS A 15-YEAR-OLD BADDIE

HOW CAN I MAKE MONEY ONLINE AS A 15-YEAR-OLD BADDIE...

### 'It is better to be feared than loved.''A man forgets more easily the death of his father than the loss ofhis money.''The end justifies the means.'-

"It is better to be feared than loved." "A man forgets more easily the death of his father than the loss of his money." "The end justifies the means." - Niccolo Machiavelli, The Prince (1513) 2. The excerpts above represented a shift in political thinking, which phrase best summarizes this shift? ...

### Question and explain: “do the economic resources reflect a country’s economic success? ” asap

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### King view militancy as

King view militancy as...

### Por que las estrellas polar es usada por los navegantes para orientarse

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### Identify the correct mla style format for citing a book with two authors. a. brown, emma. the twilight

Identify the correct mla style format for citing a book with two authors. a. brown, emma. the twilight letter. dublin: my agent press, 2001. print. b. brown, emma, and linda miller. "the twilight letter." dublin: my agent press, 2001. print. c. brown, emma, and linda miller. the twilight letter...

### How did Perrault change the wolf? Instead of the wolf being a “rapacious beast,” what did he become in Perrault’s

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### Using the portuguese 'fort and factory' model of colonial expansion, became the most prosperous commercial

Using the portuguese "fort and factory" model of colonial expansion, became the most prosperous commercial empire of the seventeenth century....

### The price of a video game was reduced from $70 to$40. by what percent what is the price of the video game reduced.

The price of a video game was reduced from $70 to$40. by what percent what is the price of the video game reduced....

### Which of the following was a direct cause of job specialization in early human civilizations? a. improvement of hunting

Which of the following was a direct cause of job specialization in early human civilizations? a. improvement of hunting methods. b. development of agriculture. c. stratification of society. d. establishment of trade networks...

### When a reader summarizes a theme, the reader should include all the details of the original text. make

When a reader summarizes a theme, the reader should include all the details of the original text. make the summary as long and involved as possible. state personal opinions about the events. include evidence that supports how the theme develops....

### What evidence from the passage best supports the inference that making sugar was difficult? Select two options.

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### The credit remaining on a phone card (in dollars) is a linear function of the total calling time made with the card (in minutes).

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Jot down an example of when this tip may apply. $Jot down an example of when this tip may apply.$...


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# Big bang in Robertson - Walker space - time

## Main Question or Discussion Point

Hey guys. I was looking at the Robertson - Walker metric and went about calculating $$R^{\mu \nu }R_{\mu \nu }$$ as a way of looking at coordinate independent curvature properties through the Ricci Tensor. Considering I did it correctly I ended up with:

$$R^{\mu \nu }R_{\mu \nu } = \frac{1}{R(t)^{2}}[9\ddot{R} + 3\frac{(\ddot{R}R + 2\dot{R}^{2} + 2k)^{2}}{R(t)^{2}}]$$

Where R(t) is the scale factor, k is the spatial curvature index, and dots denote time derivatives. So if R(0) = 0 is taken to be the origin then can it be interpreted as being that moment right at the big bang? I ask this because the big bang is said to be a real, geometric singularity and in this coordinate - invariant expression R(0) = 0 induces infinite curvature in the manifold. Also, what would r = 0 mean in the manifold (r being the radial coordinate as defined by the metric) since it doesn't really show up in this expression (if I did it right of course!)?

Related Special and General Relativity News on Phys.org
fzero
Homework Helper
Gold Member
Since the scale factor evolves with time and curvature invariants (the Ricci scalar would be a more obvious choice) diverge at $$R(t_0)=0$$, it makes sense to treat $$t_0$$ as the origin of the time coordinate. This can always be achieved by a constant shift in time. The singularity at $$r=0$$ is just a coordinate singularity, analogous to the same one in polar coordinates for Cartesian space. Curvature invariants don't do anything special there.

I'm sorry when you say r = 0 do you mean the scale factor or just the radial coordinate?

fzero


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# Glmnet in Matlab

### Description

This is a Matlab port for the extremely efficient procedures for fitting the entire lasso or elastic-net regularization path for linear regression, logistic and multinomial regression models, poisson regression and the Cox model. Two recent additions are the multiresponse gaussian, and the grouped multinomial.

Suppose $X$ is the input matrix and $Y$ the response vector. For the Gaussian family, glmnet solves the penalized residual sum of squares,

$$\min_{(\beta_0, \beta) \in \mathbb{R}^{p+1}}\frac{1}{2N} \sum_{i=1}^N (y_i -\beta_0-x_i^T \beta)^2+\lambda \big[ (1-\alpha)||\beta||_2^2/2 + \alpha||\beta||_1\big],$$

where $\lambda \geq 0$ is a complexity parameter and $0 \leq \alpha \leq 1$ is a compromise between ridge and lasso. Note that it becomes the lasso when $\alpha = 1$ and the ridge regression when $\alpha = 0$.

For other families, glmnet maximizes the appropriate penalized log-likelihood (partial likelihood for the "cox" model), or minimize the penalized negative one. Take the "binomial" model for example, it solves

$$\min_{(\beta_0, \beta) \in \mathbb{R}^{p+1}} -\frac{1}{N} \sum_{i=1}^N y_i \cdot (\beta_0 + x_i^T \beta) + \log (1+e^{(\beta_0+x_i^T \beta)}) + \lambda \big[ (1-\alpha)||\beta||_2^2/2 + \alpha||\beta||_1\big].$$

The algorithm uses cyclical coordinate descent in a pathwise fashion. In addition to basic settings, many more options are available: observation weights, choice of lambda sequence, grouping, etc. For more information, see the reference papers, help file or the documentation (in progress).

The package includes two main functions:

• glmnet.m - the basic function that returns a structure containing all necessary information for further use, like printing, plotting and prediction.
• cvglmnet.m - a more commonly used function that returns a structure after selecting the tuning parameter by cross-validation according to user's requirements.

A basic example is that suppose x is the design matrix and y the response vector, then,

• fit = glmnet(x, y) -- fits the model using all default settings.
• glmnetPrint(fit) -- prints relevant information of the fitted object.
• glmnetPlot(fit) -- plots the coefficients from the fitted object.
• cvfit = cvglmnet(x, y) -- fits the model by cross-validation using all default settings.
• cvglmnetPlot(cvfit) -- plots the cross-validation curve.

For more details and examples, enter help glmnet or help cvglmnet in the Matlab's command window. Other functions are listed below.

 cvglmnet.m Cross-validation for glmnet cvglmnetCoef.m Extract the coefficients from a 'cv.glmnet' object cvglmnetPlot.m Plot the cross-validation curve produced by cvglmnet.m cvglmnetPredict.m Make predictions from a 'cv.glmnet' object glmnet.m Fit a GLM with lasso or elasticnet regularization glmnetCoef.m Extract the coefficients from a 'glmnet' object glmnetControl.m Internal glmnet parameters glmnetPlot.m Plot coefficients from a 'glmnet' object glmnetPredict.m Make predictions from a 'glmnet' object glmnetPrint.m Print a 'glmnet' object

### Authors

Algorithms designed by Jerome Friedman, Trevor Hastie, Noah Simon and Rob Tibshirani. Fortran code was written by Jerome Friedman.
The original MATLAB wrapper was written by Hui Jiang (14 Jul 2009). It was updated (30 Aug 2013) and is maintained by Junyang Qian junyangq@stanford.edu, Department of Statistics, Stanford University, Stanford, California, USA.

The package currently supports the following platforms.

We will add support for other platforms such as Windows 7. Users can also compile the code on their platforms easily. On 32-bit systems, users can simply execute the mex function

mex glmnetMex.F GLMnet.f

On 64-bit systems, compiler options should be used to make it double precision. For example, on the 64-bit Mac OS X, we used

mex FFLAGS='-fdefault-real-8 -ffixed-form' glmnetMex.F GLMnet.f

If the compile is successful, we are ready to go!

### References

Jerome Friedman, Trevor Hastie and Rob Tibshirani. (2008).
Regularization Paths for Generalized Linear Models via Coordinate Descent
Journal of Statistical Software, Vol. 33(1), 1-22 Feb 2010.

Noah Simon, Jerome Friedman, Trevor Hastie and Rob Tibshirani. (2011).
Regularization Paths for Cox's Proportional Hazards Model via Coordinate Descent
Journal of Statistical Software, Vol. 39(5) 1-13.

Robert Tibshirani, Jacob Bien, Jerome Friedman, Trevor Hastie, Noah Simon, Jonathan Taylor, Ryan J. Tibshirani. (2010).
Strong Rules for Discarding Predictors in Lasso-type Problems
Journal of the Royal Statistical Society: Series B (Statistical Methodology), 74(2), 245-266.


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# Equality for floats

Suppose I have two floats (e.g. {x,y} = RandomReal[{0,1},2]). Might there be a situation where all three comparisons evaluate to False (i.e. x > y, x < y and x == y)? I am thinking that the values might not be that stable and the equality might fail. Should something like Around be used instead? I am not fully understanding what is going on behind the hood with == operator and how stable the floats are and how stably they can be carried over.

• see mathematica.stackexchange.com/questions/76896/… in the old days, i.e. when using Fortran, one checks if 2 numbers are the same or not by checking if the absolute difference between them is less than some epsilon (i.e. the machine epsilon) or not. – Nasser Apr 5 '20 at 3:13
• @Nasser, thanks! Is there a way to compare two floats with a margin d? By IntervalIntersection? Or there is a better, more standard way? – Al Guy Apr 5 '20 at 3:21
• From the doc for Equal, "Approximate numbers with machine precision or higher are considered equal if they differ in at most their last seven binary digits (roughly their last two decimal digits)." and "For numbers below machine precision, the required tolerance is reduced in proportion to the precision of the numbers." – Bob Hanlon Apr 5 '20 at 3:32

Both Equal and SameQ compare with tolerance. You can find the tolerances as

Internal$EqualTolerance (* 2.10721 *) Internal$SameQTolerance
(* 0.30103 *)

These values are to be interpreted as decimal digits. The numbers are considered equal if they differ in at most this many digits. Since the numbers are stored in binary, it makes more sense to look at the tolerances in binary digits:

Internal$EqualTolerance Log2[10] (* 7. *) Internal$SameQTolerance Log2[10]
(* 1. *)

Might there be a situation where all three comparisons evaluate to False (i.e. x > y, x < y and x == y)?

I do not think so.

Table[
With[{x = 1 + n \$MachineEpsilon}, {1 < x, 1 == x, 1 > x}],
{n, -100, 100}
]


text:
Difference of squares: a2 - b2 =(a + b)(a - b)

Go !

## Calculate and simplify:

 $(a^2-b^2)(a^2+b^2) =$ $(xy+3)(xy-3)(\frac{x^2y^2+9}{2}) =$

## Factorize:

 $x^6-y^2 =$ $(x^3+y)(x^3-y)$ $a^3-ab^2 =$


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# 03.08.2022 - Reading Notes / Forced Synchronous Layout and Layout Thrashing

When a DOM node is written (change of style, attribute, or inner content), and the action triggers a layout change, the browser must do a layout recalculation (reflow). The browser usually waits for the end of a frame to perform this calculation.

However, if we try to read the geometrics information of the DOM node, the browser is forced to perform layout recalculation early, during JavaScript execution. So it could have the latest information to give back to you. This forced behavior is called forced synchronous layout.

For example:

// read geometry. current layout information will be used.
let width = element.clientWidth;

// write geometry. if we stop here, reflow will be scheduled
// at the end of the frame
element.style.width = '600px';

// read geometry again. this force a reflow right in JS execution
let width = element.clientWidth;

Layout thrashing is when many force synchronous layouts happen in a short amount of time (for example, in a loop), making a janky user experience.

For example:

for (let i = 0; i < 1_000_000; i++) {
element.style.width = element.clientWidth + 10;
}

In the above example, we force reflow by setting a new value for element.style.wdith, and read the value immediately with element.clientWidth. To make it worse, we do it in a loop.

In Chrome Dev Tools, we can use the Performance tool to record and detect force reflow, it is usually shown as a purple bar with a red corner like this:

The Google Sample repository provided an example to help you see layout thrashing in action, you can give it a try at https://googlesamples.github.io/web-fundamentals/tools/chrome-devtools/rendering-tools/forcedsync.html

Also, see Paul Irish’s What forces layout/reflow note for a list of what action triggers reflow, so you can avoid them or use them better.

References


text:
# cirq.linalg.AxisAngleDecomposition

Represents a unitary operation as an axis, angle, and global phase.

The unitary $U$ is decomposed as follows:

$$U = g e^{-i heta/2 (xX + yY + zZ)}$$

where heta is the rotation angle, (x, y, z) is a unit vector along the rotation axis, and g is the global phase.

## Methods

### canonicalize

View source

Returns a standardized AxisAngleDecomposition with the same unitary.

Ensures the axis (x, y, z) satisfies x+y+z >= 0. Ensures the angle theta satisfies -pi + atol < theta <= pi + atol.

Args
atol Absolute tolerance for errors in the representation and the canonicalization. Determines how much larger a value needs to be than pi before it wraps into the negative range (so that approximation errors less than the tolerance do not cause sign instabilities).

Returns
The canonicalized AxisAngleDecomposition.

View source

View source


text:
# How can we measure absolute time?

In existence of mass, spacetime gets warped, this means time speeds up near the mass, so how can we measure the actual rate of flow of time?


text:
2. Economics
3. at rams college the tuition fee for the domestic canadian...

Question: at rams college the tuition fee for the domestic canadian...

Question details
1. At rams College, the tuition fee for the domestic (Canadian) students is $285.00 per credit for the University Transfer and Associate Degree courses, while the tuition fee for the international students is$550.00 per credit Yet most students at the college are the international students. Explain why. For your explanation,

1. Draw the demand (for the courses at AC) and supply diagrams for two markets, (1) domestic students and (2) international students, that show the difference.

1. Explain, in words, WHY the demand (or supply) is greater in one market than in the other.


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In a contest, a bowl contains 10 keys, one of which will open a treasu

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In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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22 Apr 2015, 02:14
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In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?

A. 1/10
B. 1/9
C. 1/5
D. 19/90
E. 2/9

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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22 Apr 2015, 02:23
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KUDOS
Bunuel wrote:
In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?

A. 1/10
B. 1/9
C. 1/5
D. 19/90
E. 2/9

Kudos for a correct solution.

The total possibilities existing while withdrawing two keys are: 10C2 = 45

For Anna to win, 9 combinations exist assuming Key 1 to unlock the treasure E.g(1,2), (1,3)....(1,9)

P = 9/45 = 1/5

Option C
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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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22 Apr 2015, 21:44
Can i please have the solution to this?
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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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23 Apr 2015, 03:20
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jayanthjanardhan wrote:
Can i please have the solution to this?

Probability of having the right key = Favorable Cases/Total Cases = (Cases in which the right key will be picked)/(No of ways in which you can pick two keys)

No of ways in which you can pick two keys out of 10 = 10C2 = 10*9/2 = 45

No of cases in which the right key will be picked = 1*9C1 (you pick the correct key and one other key) = 9

Probability of winning = 9/45 = 1/5

You can also do the question with arrangement since you need probability.

Pick the first key in 10 ways and second in 9 ways. Total = 10*9 = 90 ways
Pick the first correct key in 1 way and second key in 9 ways. Total = 1*9 = 9 ways
Pick the first key in 9 ways and second correct key in 1 way. Total = 9*1 = 9 ways

Probability = (9+9)/90 = 1/5

Another Method is using probabilities
She can win the contents by picking the first key correctly or second key correctly.
Probability of picking the first key correctly = 1/10
Probability of picking the first key incorrectly and second key correctly = (9/10)*(1/9) = 1/10

Probability of winning = 1/10 + 1/10 = 1/5
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11025 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink] Show Tags 24 Apr 2015, 10:15 2 This post received KUDOS Expert's post Hi All, When dealing with probability questions, there are 2 outcomes that you can calculate: what you WANT to have happen and what you DON'T WANT to have happen. When you add these two outcomes together, the result will equal 1. Here, it's actually pretty easy to calculate the outcome that you DON'T WANT: NOT getting the winning key. Then we can subtract that fraction from 1 to get the probability of what we do WANT... There are 10 keys and 1 of them is the 'winning' key. Probability of NOT getting the winning key on the first grab = 9/10 Probability of NOT getting the winning key on the second grab = 8/9 (9/10)(8/9) = 8/10 = 4/5 So, the probability of NOT getting the winning key = 4/5 The probability that we DO get the winning key = 1 - 4/5 = 1/5 Final Answer: [Reveal] Spoiler: C GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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27 Apr 2015, 03:06
Bunuel wrote:
In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?

A. 1/10
B. 1/9
C. 1/5
D. 19/90
E. 2/9

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

When Anna reveals her two keys, there are two possible sequences that allow her to win:

1) The first key is the winner, the second is not

2) The first key is not the winner, but the second one is

And you can calculate those probabilities:

1) 1/10 * 9/9 = 1/10

2) 9/10 * 1/9 = 1/10

Add those two favorable probabilities and you have 2/10 which reduces to 1/5.
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In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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since the two keys have been drawn simultaneously(and not one by one) and hence the no of cases must be 10 c2 not 10 *9. Now-

one key is correct one and second key incorrect one= 1* 9c1

there is no logic of first and second key as the draw was simutaneous. so probability =9c1 /10c2=1/5
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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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05 Jun 2016, 20:02
vishnu440 wrote:
since the two keys have been drawn simultaneously(and not one by one) and hence the no of cases must be 10 c2 not 10 *9. Now-

one key is correct one and second key incorrect one= 1* 9c1

there is no logic of first and second key as the draw was simutaneous. so probability =9c1 /10c2=1/5

Note that considering first and second key in a valid method even though we talk of simultaneous draws because when you pick out two, you will touch one of them first. They are both drawn out simultaneously but not picked simultaneously. Of course when you consider that first key is A and second key is B, you have to consider the other side too where the first key is B and second key is A.

The answer will be the same no matter which method you use - combinations or probability.
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In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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05 Jun 2016, 20:24
Bunuel wrote:
In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?

A. 1/10
B. 1/9
C. 1/5
D. 19/90
E. 2/9

Kudos for a correct solution.

There are two ways in which the right key is taken out

(1st key is the right one & second wrong one) or (1st key is the wrong one & 2nd one is right)

Since it is "or" hence we will add the probabilities.

($$\frac{1}{10}$$ * $$\frac{9}{9}$$) + ($$\frac{9}{10}$$ * $$\frac{1}{9}$$)

$$\frac{1}{10}$$ = Only one key is the right key. Total options are 10

$$\frac{9}{9}$$ = After first key is taken out, there r 9 favorable options (all keys r wrong). Total options are also 9

$$\frac{9}{10}$$ = For first key to be the wrong one, favorable options are nine. Total options are 10.

$$\frac{1}{9}$$ = For second key to be correct one, only one favorable option is there out of available 9 options.

$$\frac{9}{90}$$ + $$\frac{9}{90}$$

$$\frac{18}{90}$$ = $$\frac{1}{5}$$

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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu [#permalink]

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21 Sep 2017, 14:02
Bunuel wrote:
In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?

A. 1/10
B. 1/9
C. 1/5
D. 19/90
E. 2/9

P(Anna wins) = 1 - P(Anna does not win)

Let’s determine P(Anna does not win). Note that the probability that the first key doesn’t open the treasure chest is 9/10 and the probability that the second key doesn’t open the treasure chest (given that the first key didn’t work) is 8/9.

P(Anna does not win) = 9/10 x 8/9 = 4/5.

Thus, the probability that she wins is 1 - ⅘ = 1/5.

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Re: In a contest, a bowl contains 10 keys, one of which will open a treasu   [#permalink] 21 Sep 2017, 14:02
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Fundamentals of Fluid Mechanics B Questions and Answers
 Question by Student AME536B Concerning question 4, the velocity at the wake was obtained by assuming that $x$ goes to infinity. Why we can't say that $\frac{u_w}{u_{inf}}=0.99999$ at the edge of the wake in order to find its height, as this strategy causes problems only close to the cylinder.
 04.30.20
It's because far away from the cylinder, $u_w$ will be almost exactly equal to $u_\infty$ everywhere within the profile. As $x \rightarrow \infty$, $u_w \rightarrow u_\infty$. Thus, the point $u_w=0.999999 u_\infty$ won't lie anywhere on the profile! The way you determine the height must work equally well close to the cylinder and when $x$ is very large. Think about this more.
 Question by Student AME536B For Q#2,I am able to reach this point: $$\frac{\partial}{\partial t}\int_{C\forall}\rho V_{i}d\forall + \oint_{CS}\left(V_{C\forall}\cdot\hat{n}\right)\rho V_{i}dS = -\oint_{CS}\left(P\cdot\hat{n}_i\right)dS + \oint_{CS}\tau_{ij}\hat{n}_{j} dS + \int_{C\forall}\vec{f}_{i}d\forall$$ But I am not sure about the 2nd term on the LHS of the equation, would you provide a hint for this please? $$\oint_{CS}\rho V_{i}\left(V_{C\forall}\cdot\hat{n}\right)dS \stackrel{?}{=} \oint_{CS}\rho V_{i} \left(V - V_{C\forall}\right)\cdot\hat{n}dS$$ Note: I started by contracting the momentum eq in conservative form to its vector form $$\frac{D}{Dt}\left(\rho\vec{V}\right) = -\nabla\cdot \vec{P} + \nabla\cdot\vec{\tau} + \vec{f}_b$$ I applied the Reynolds transport on the LHS and the Divergence theorem to the pressure gradient term and the shear gradient
 05.02.20
I do not agree with your last equation. How can you have a pressure vector? Pressure is a scalar. Also, you shouldn't transform the momentum equation convection terms as a substantial derivative. I don't see how this can help you. Simply substitute the time derivative of the momentum equation within the first term on the RHS of the Reynolds transport theorem listed on the last page of the tables. Then, there'll be several integrals you need to rewrite using Gauss's theorem. The ones involving the velocity are easy. The ones involving pressure and the shear stress are more challenging.
 Question by Student AME536B Is it possible to give us a hint on how to find the wake height on question 4?
Well I think I gave you many examples of what not to do in class. This is sufficient. You need to formulate the definition of the wake height in a way that is fool proof, that will work equally well in the near or the far field.
 Question by Student AME536B Per university policy, I am not sure if a homework applies Under, Policy Memo: Final Examination Regulations and Information 4. It is Faculty Senate policy that all forms of exams (quizzes, take homes, etc.) are prohibited on any scheduled class or reading day during the calendar week in which regularly scheduled final exams begin. Specific exceptions for certain courses may be made with approval from the appropriate academic unit head and academic dean. Students shall be informed of any such exceptions in the class syllabus. https://www.registrar.arizona.edu/cours ... pring-2020
 05.04.20
Assignment 10 definitely does not qualify as an exam. You won't receive a grade for it and it doesn't count towards the final score. The purpose of the assignments in this course is to help you learn the material and give you feedback. The material for the final exam includes the last lecture hence the need for assignment 10. The submission of the assignment 10 can take place anytime between Wednesday 6 pm and Sunday 11 am. If you don't submit it, you don't lose any point. If you do, I'll give you feedback. It's up to you.
 Question by AME536A Student I have a problem with question 3. As we did in class, I found the velocity $u$ from the streamfunction through: $$u=\frac{\partial \Psi}{\partial y}$$ Then I integrated u over y in the limits from 0 to $x^{\frac{2}{3}}$. The second x-term cancels out, but I still end up with an expression that contains $\frac{1}{x}$, which I think should not be the case. Is there something wrong with this approach?
This is possible. Then if this happens, you need to explain what this entails. Answer the other parts of the Q3 carefully. You need to think about this.
 Question by AME536A Student I believe you mentioned this in class, but I wanted to double check — Is the final exam open note? If not, will we be expected to do proofs of equations for the final exam? Are there any that might be excluded from being on the exam?
 05.07.20
Yes it's open notes, and the exam will be designed in consequence. I could ask you part of a proof that I didn't finish in class. No assignment problem is excluded. You should make sure you understand and remember the proofs and other problems in the assignments. I may ask you the same problem as in the assignment or a very similar problem that builds upon the assignment problem. Thus, I expect you to have finished all your assignment problems correctly. I recommend not to open your notes during the exam — most likely this will end up being detrimental. You should have all the theory and assignments well understood and inside your brains.
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 [2015] Spiderweb theory tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/29/2015, 06:25 PM (This post was last modified: 03/29/2015, 09:44 PM by tommy1729.) After thinking about tetration and fake function theory it is time for the 3rd topic which I started years ago at sci.math. Chapter 3 : Spiderweb theory. It all started - like most things started on sci.math - with a flame war. And maybe my knowledge is not big enough , its possible. But they did not answer my questions either. It started by talking about Taylor series. They said they only converge within a radius. Just like with the axiom of choice , I had some objections ... well conditional objections. Using summability methods it is possible for some Taylor series with a nonzero radius to be valid outside the expansion point. Here is a simple example : f(x) = 0 + 2^2^1 x + 2^2^1 x^2 + 2^2^2 x^3 + 2^2^2 x^4 + ... Now this has a radius 0 at x = 0. but for x = -1 this function gives f(-1) = 0. SO it converges outside its radius. More complicated examples can converge in many places. Many summabilty methods exist. Im not sure how it all works though. But assume it converges in a spiderweb shape. That is to say the points where it converges are dense on a spiderweb shape. ( like the roots of unity are dense on the unit circle ) What is the theory behind that ? I believe a nonzero radius Taylor series cannot converge in a region with a positive area that is connected to the expansion point unless that is part of the boundary of a radius of another expansion point. Also if for instance every VALID (leading to convergeance) point q gives f(q) = 0 then we say the analytic continuation is f(z) = 0. That crazy idea is part 1 of the spiderweb theory. part 2 : An example : $T(x) = \frac {(exp x - 1 + 0.5 x^2)^{[1/2]}} {(exp x - 1)^{[1/2]}}$ I think T(x) is analytic or it converges in a spiderweb. edit : another example : $R(x) = \frac {(x + x^2 + x^3)^{[1/2]}} {(x + x^3)^{[1/2]}}$ I think R(x) is analytic or it converges in a spiderweb. regards tommy1729 « Next Oldest | Next Newest »

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An Algebraic-Combinatorial Proof Technique for the GM-MDS Conjecture

# An Algebraic-Combinatorial Proof Technique for the GM-MDS Conjecture

Texas A&M University, College Station, TX 77843 USA
###### Abstract

This paper considers the problem of designing maximum distance separable (MDS) codes over small fields with constraints on the support of their generator matrices. For any given binary matrix , the GM-MDS conjecture, due to Dau et al., states that if satisfies the so-called MDS condition, then for any field of size , there exists an MDS code whose generator matrix , with entries in , fits (i.e., is the support matrix of ). Despite all the attempts by the coding theory community, this conjecture remains still open in general. It was shown, independently by Yan et al. and Dau et al., that the GM-MDS conjecture holds if the following conjecture, referred to as the TM-MDS conjecture, holds: if satisfies the MDS condition, then the determinant of a transformation matrix , such that fits , is not identically zero, where is a Vandermonde matrix with distinct parameters. In this work, we generalize the TM-MDS conjecture, and present an algebraic-combinatorial approach based on polynomial-degree reduction for proving this conjecture. Our proof technique’s strength is based primarily on reducing inherent combinatorics in the proof. We demonstrate the strength of our technique by proving the TM-MDS conjecture for the cases where the number of rows () of is upper bounded by . For this class of special cases of where the only additional constraint is on , only cases with were previously proven theoretically, and the previously used proof techniques are not applicable to cases with .

## I Introduction

In recent years, there has been a growing interest in designing maximum distance separable (MDS) codes with constraints on the support of the codes’ generator matrix [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. Such constraints arise in wireless network coding and distributed storage scenarios where each user/server has access to only a subset of the information symbols. Two examples of such scenarios are cooperative data exchange in the presence of an eavesdropper [1, 5], and simple multiple access networks with link/relay errors [6, 8].

Given an binary (support) matrix and a field of size , the problem is to design an MDS code with a generator matrix , (i.e., all sub-matrices of are full-rank) fitting (i.e., if , then ). Note that for some , there exists no MDS code whose generator matrix fits (i.e., is not completable to MDS). Nevertheless, there is a simple condition, known as the MDS condition, which characterizes all matrices that are completable to MDS for sufficiently large fields [4, 5]. A matrix satisfies the MDS condition if:

 |∪i∈Isupp(Mi)|≥n−m+|I|,∀I⊆{1,…,m},I≠∅,

where is the th row of , and is the support of . The existence of MDS codes (over sufficiently large fields) whose generator matrix’s support satisfies the MDS condition was shown, e.g., in [8], via Edmonds matrix and Hall’s marriage theorem. The following conjecture, due to Dau et al. [4], aims for generalizing this result for small fields.

###### Conjecture 1 (GM-MDS Conjecture)

If the matrix satisfies the MDS condition, then for any field of size , there exists an MDS code whose generator matrix with entries in fits the matrix .

Notwithstanding all the efforts by the coding theory community, the GM-MDS conjecture remains still open in general. The GM-MDS conjecture and a simplified version of this conjecture where the supports of rows of have the same size were shown in [8] to be equivalent (using a generalized version of Hall’s theorem). Despite this simplification, there are only three classes of special cases for which this conjecture is theoretically proven: (i) the rows of are divided into three groups, and the rows in each group have the same support [6]; (ii) the size of intersection of the supports of every two rows of is upper bounded by [8]; and (iii) the number of rows of is upper bounded by [11]. More importantly, the previously used proof techniques are not applicable to more general cases due to the combinatorial explosion.

One possible approach to find a completion of to MDS is to leverage the structure of Generalized Reed-Solomon (GRS) codes [5, 4] which are known to be MDS. Let be the set of independent indeterminates . Let be an binary matrix whose rows’ supports have the same size, and let be a generic Vandermonde matrix with parameters . Let be a generic transformation matrix such that fits , and let . If the evaluations of are distinct, then every sub-matrix of is non-singular (i.e., is a generator matrix of a GRS code with evaluation points ) so long as is non-singular. That is, if is not generically singular (i.e., the determinant of as a multivariate polynomial in variables is not identically zero), then for any field of size , there exists such that is a generator matrix of an MDS code, and fits . Thus, the GM-MDS conjecture holds if the following conjecture, proposed independently by Dau et al. [4] and Yan et al. [5], holds:

###### Conjecture 2 (TM-MDS Conjecture)

If satisfies the MDS condition, then is not generically singular.

The contributions of this work are as follows. First, we present a generalization of the TM-MDS conjecture for the cases where the supports of rows of have arbitrary sizes. Then, we present an algebraic-combinatorial approach based on polynomial-degree reduction for proving this conjecture. Our technique’s strength is primarily due to reducing the inherent combinatorics in the proof. Specifically, we demonstrate this strength by proving the TM-MDS conjecture for the cases where the number of rows of is upper bounded by .

## Ii Basic Notations and Definitions

Let be a field. For , let be independent indeterminates. Let be a ring of multivariate polynomials in variables with coefficients in , and let be a module of univariate polynomials in variable with coefficients in . Fix and such that . For , denote by . Define polynomials of degrees in :

 Pi(α)≜∏γ∈Ni(α−γ),∀i∈[m], (1)

where , the set of roots of , is a (proper) subset of of size . (Note that the roots of are indeterminates.) For and , .

Note that , where are polynomials in . Define . Note that is a polynomial in .

###### Definition 1

A polynomial in is identically zero, denoted by , if the coefficients of all monomials in the polynomial expansion of are zero.

###### Definition 2

A set of polynomials of degree has rectangular property (RP) if, for some , there exist at least polynomials in with at least common roots. Otherwise, has non-rectangular property (NRP).

###### Definition 3

A set of polynomials of degrees has generalized RP (GRP) if, for some and , there exist at least polynomials of degrees at most in with at least common roots. Otherwise, has generalized NRP (GNRP).

## Iii Main Conjectures and Theorems

The following conjecture is equivalent to the TM-MDS conjecture (Conjecture 2).

###### Conjecture 3

Let be polynomials of degree in . If , then has RP.

The sketch of proof of the equivalency between the TM-MDS conjecture and Conjecture 3 follows. Let be an binary matrix. Let be the th row of and let be the support of . Let , where are independent indeterminates. Suppose that all have the same size. Defining () as in (1), it follows that the matrix satisfies the MDS condition iff has NRP (refer to this as Fact 1). Let be a generic generator matrix of a Generalized Reed-Solomon (GRS) code with evaluation points such that fits (i.e., if , then ). Let be a generic Vandermonde matrix with parameters , and let be a generic transformation matrix such that . Taking and , where , it follows that (for more details, see [5]). Since , then (i.e., is not generically singular) iff (refer to this as Fact 2). By Facts 1 and 2, the TM-MDS conjecture and Conjecture 3 are equivalent.

In the following, we propose a new conjecture which generalizes Conjecture 3 for the cases where the degrees of polynomials are arbitrary. (Conjecture 4 is equivalent to a generalized version of the TM-MDS conjecture where the supports of rows of have arbitrary sizes.)

###### Conjecture 4

Let be polynomials of arbitrary degrees in . If , then has GRP.

If for all , then Conjecture 4 holds trivially: (i) since for all , and (ii) has GRP since for and , there exist polynomials of degrees at most in with at least common roots. Hereafter, w.l.o.g., we assume for some .

The following theorems, which are our main results, prove the GM-MDS conjecture for . More specifically, Theorems 12, and 3 settle Conjecture 4 (and so Conjecture 3) for , and Theorem 4 settles Conjecture 3 for .

###### Theorem 1

For any such that , if , then has GRP.

###### Theorem 2

For any such that , if , then has GRP.

###### Theorem 3

For any such that , if , then has GRP.

###### Theorem 4

For any such that , if , then has GRP.

## Iv Main Ideas and Lemmas

In this section, we explain the main ideas and state the useful lemmas for the proofs of our main results.

Consider an arbitrary set () of polynomials (with the sets of roots ) such that for all , and for some . Define a class of reduction processes over , where any process in this class is associated with a unique reduction set , and it reduces to . Let . Note that . Restrict your attention to those reduction sets such that for some , and for all . Such are referred to as acceptable. For any acceptable reduction set, w.l.o.g., assume that for all and (and so, since ). For any (acceptable) reduction set, the following result holds.

If , then .

###### Proof 1

Consider the resulting from for an arbitrary (acceptable) reduction set . Let . For any , let be the number of polynomials in such that , and let . Let be the resulting polynomial from by taking derivative times with respect to each variable . (Since , and is the sum of monomials for some (depending on ), then the derivatives of with respect to any variable are independent of .) Note that (by using the Leibniz formula for determinant), and (since and for all , and and , where ). Since (by assumption), then . Thus, .

Lemma 1 enables us to use an inductive argument towards the proof of Conjecture 4 as follows. Suppose that Conjecture 4 holds for any , i.e., for any such that for all and for some , if , then has GRP. We need to prove that for any such that for all and for some , if , then has GRP. The proof follows by contradiction. Assume that and does not have GRP. Consider the resulting from for an (acceptable) reduction set such that has GNRP. By definition, for all . Since (by assumption), then (by Lemma 1), and so, has GRP (by the induction hypothesis), yielding a contradiction. Our goal is thus to devise an (acceptable) reduction process such that if has GNRP, then so does . The problem of designing such a process is still open in general. In the following, we propose a simple yet powerful reduction process which solves this problem for and for all , and for and for all .

From now on, we assume that () is a set of polynomials (with the sets of roots ) such that for all , and .

###### Definition 4

A subset is an -subset in a subset of if belongs to polynomials in (i.e., there exist polynomials in such that ), and . Moreover, an -subset has higher order than an -subset if , or and .

The following lemma gives the intuition behind the definition of -subsets.

###### Lemma 2

If has GNRP, then there exists no -subset in such that .

###### Proof 2

The proof is straightforward and follows from the definitions (and hence omitted).

Intuitively, for any (acceptable) reduction set , any highest-order -subset , if not broken (i.e., ), is the most likely to cause rectangularity in for any with non-rectangular property. This is the main idea of the proposed reduction process.

###### Definition 5

An element of a subset is removable if is a root of some but not all polynomials of degree .

###### Definition 6

A subset is weakly reducible if belongs to a polynomial of degree , and has a removable element.

###### Definition 7

A weakly reducible -subset is strongly reducible if no other weakly reducible -subset has higher order than .

### Proposed Reduction Process

Given , choose an arbitrary strongly reducible subset in , and choose an arbitrary removable element of , say , such that no other removable element of , when compared to , belongs to more polynomials of degree in . Break via removing from the sets of roots of all polynomials , and update all polynomials via replacing by . Repeat this process (in rounds) over the resulting if there exist more than one polynomial of degree . Otherwise, terminate the process, and return the resulting denoted by .

Note that if has GNRP initially, then (i) in each round of the process, such exists, and (ii) the process terminates eventually. Otherwise, there must exist two (or more) identical polynomials of degree in (and hence has GRP), which is a contradiction.

Consider an arbitrary run of the reduction process over and its corresponding . Let be the set of the roots that the reduction process removes over the rounds. (Note that, due to the arbitrary choices in the reduction process, may or may not be unique.) Hereafter, for any such , assume, w.l.o.g., that the (initial) indexing of polynomials in is such that () and , and denote by .

The proofs of our main theorems rely on the following properties of the proposed reduction process.

###### Lemma 3

If has GNRP, then any -subset in such that belongs to a polynomial in such that .

###### Proof 3

Let be an arbitrary -subset in such that . Let be the set of all polynomials in such that . Note that is a set of at most polynomials of degrees at most . Since has GNRP (by assumption), then (and hence ) has GNRP. Thus, there exists no -subset in such that (by Lemma 2). Suppose that belongs to no polynomial in . Then, is an -subset in such that . This is, however, a contradiction. Thus, belongs to a polynomial in .

###### Lemma 4

If has GNRP, then any -subset in such that is weakly reducible.

###### Proof 4

Let be an arbitrary -subset in such that . Note that belongs to a polynomial in (and so ) of degree (by Lemma 3). Note, also, that has degree . Thus, if there exists such that , then is weakly reducible since is removable (by definition). Otherwise, if , then is an -subset in . Since , then has GRP (by Lemma 2), yielding a contradiction.

###### Lemma 5

If has GNRP, then the strongly reducible -subsets in such that belong to disjoint subsets of .

###### Proof 5

Let and be two arbitrary strongly reducible -subsets in such that . Let or be the set of polynomials in such that or belongs to , respectively. Note that . First, suppose that . Then, is an -subset in . Since , then has GRP (by Lemma 2), and hence a contradiction. Next, suppose that . We consider two cases. First, suppose that . Then, is a -subset in . Let and . Since , then has GRP (by Lemma 2), which is a contradiction. Next, suppose that . Then, is a -subset in . Let and . Note that . If , then has GRP (by Lemma 2), and hence a contradiction. If , then is weakly reducible (by Lemma 4), and has higher order than and since and . This is also a contradiction since and are strongly reducible (by assumption). Thus, .

###### Lemma 6

For and for all , and for and for all , if has GNRP, then the reduction process breaks any strongly reducible -subset in such that .

###### Proof 6

Let be an arbitrary strongly reducible -subset in such that . Since has GNRP, belongs to a polynomial in of degree (by Lemma 3) and no other strongly reducible -subset in belongs to (by Lemma 5). Moreover, for any and any , there exists no other -subset in such that (otherwise, has GRP). Thus must be broken to reduce .

For and , is either a - or - or -subset in . First, suppose that is a - or -subset in . Since there exists no other - or -subset in (otherwise, has GRP), then must be broken to reduce . Next, suppose that is a -subset in . Let be the set of two polynomials in , say and , such that belongs to both and . Let be an arbitrary (if any) strongly reducible -subset in . If does not exist, then must be broken to reduce both and . If exists, no element of is a common root of both and (otherwise, there exists a strongly reducible -subset in , which is a contradiction since is a strongly reducible -subset). Since has GNRP, then there exists no other strongly reducible -subset in (by Lemma 5), and breaking cannot reduce both and simultaneously. Thus, must be broken to reduce or (or both).

## V Proofs of Main Theorems

In this section, we prove our main theorems. For simplicity, we denote the degree-set of polynomials and by and , respectively.

###### Proof 7 (Proof of Theorem 1)

Assume that . If , then , which is a contradiction. If , then . Thus, , i.e., has RP (and hence GRP).

###### Proof 8 (Proof of Theorem 2)

The proof follows by contradiction. Assume that , and has GNRP. If , then (since the reduction process either reduces both and simultaneously, or it first reduces one, and then reduces the other one). Since (by assumption), then (by Lemma 1). Thus, has GRP (by Theorem 1), i.e., there exists a -subset in . Thus, is a strongly reducible -subset in such that (by Lemmas 4 and 2), and it must have been broken (by Lemma 6), yielding a contradiction.

If , then (since the reduction process must reduce , and reducing may or may not reduce ). If , then has GRP, yielding a contradiction as before. If , then . Thus, (by Lemma 1), which is again a contradiction.

If , then (since the reduction process must reduce , and reducing does not reduce ). Since , then (by Lemma 1), yielding a contradiction. If , then (since the reduction process does not reduce and ). Thus, () has GRP (by the same argument as before), which is a contradiction. If , then , yielding a contradiction. If , then . Thus, has GRP, again a contradiction.

###### Proof 9 (Proof of Theorem 3)

Due to the lack of space, we only give the proofs for the cases of . (The proofs of the rest of the cases follow the exact same lines.) The proof is by way of contradiction. Assume that , and has GNRP. Since (by Lemma 1), then has GRP (by Theorem 2).

First, consider . By the procedure of the reduction process, . Since has GRP, either there exists a -subset , or if does not exist, there exists a -subset , in . Since (or ) is a strongly reducible -subset in such that (by Lemmas 4 and 2), (or ) must have been broken (by Lemma 6), which is a contradiction.

Second, consider . Then, . For any of these cases, by the same arguments as for the previous case, we arrive at a contradiction.

Next, consider . Then, . For the cases of , similar to the previous cases, we reach a contradiction. For the case of


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# Multiplying fractions and whole numbers

### Multiplying fractions and whole numbers

We learned previously that whole numbers can be written We learned previously that whole numbers can be written as fractions with 1 as the denominator and the whole number as the numerator. To make the calculation easier, we can first make the whole numbers into fraction when we multiply whole numbers with fractions. By doing so, we turn the questions into multiplying fractions only.


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# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
 any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Stability of stochastic delay neural networks. (English) Zbl 0991.93120
The stochastically perturbed network with delays $$dx(t)= \bigl[ -Bx(t)+ Ag\bigl(x_\tau (t)\bigr) \biggr]dt +\sigma\bigl( x(t),x_\tau (t), t\bigr)dw(t), t\ge 0;\ x(s)=\xi(s),\ -\tau\le s\le 0;\tag 1$$ is considered. Here $w(t)$ is an $m$-dimensional Brownian motion, $\sigma(x,y,t)$ is locally Lipschitz continuous and satisfies the linear growth conditions. Several sufficient criteria are established for almost sure exponential stability of (1).

##### MSC:
 93E15 Stochastic stability 92B20 General theory of neural networks (mathematical biology) 93C23 Systems governed by functional-differential equations
Full Text:
##### References:
 [1] Coben, M. A.; Crosshery, S.: Absolute stability and global pattern formation and patrolled memory storage by competitive neural networks. IEEE trans. Systems man cybernet. 13, 815-826 (1983) [2] Hopfield, J. J.: Neural networks and physical systems with emergent collect computational abilities. Proc. natl. Acad. sci. USA 79, 2554-2558 (1982) [3] Hopfield, J. J.: Neurons with graded response have collective computational properties like those of two-state neurons. Proc. natl. Acad. sci. USA 81, 3088-3092 (1984) [4] Hopfield, J. J.; Tank, D. W.: Computing with neural circuits. Model sci. 233, 3088-3092 (1986) [5] Liao, X. X.: Absolute stability of nonlinear control systems. (1993) · Zbl 0817.93002 [6] Quezz, A.; Protoposecu, V.; Barben, J.: On the stability storage capacity and design of nonlinear continuous neural networks. IEEE trans. Systems man cybernet. 18, 80-87 (1983) [7] Marcus, C. M.; Westervelt, R. M.: Stability of analog networks with delay. Physical review A 39, No. 1, 347-359 (1989) [8] Haykin, S.: Neural networks. (1994) · Zbl 0828.68103 [9] Le Cun, Y.; Galland, C. C.; Hinton, G. E.: GEMINI: gradient estimation through matrix inversion after noise injection. Advances in neural information processing systems, vol. I, 141-148 (1989) [10] Mao, X.: Exponential stability of stochastic differential equations. (1994) · Zbl 0806.60044 [11] Mao, X.: Stochastic differential equations and applications. (1997) · Zbl 0892.60057 [12] Mohammed, S. -E.A.: Stochastic functional differential equations. (1986) [13] Arnold, L.: Stochastic differential equations: theory and applications. (1972) · Zbl 0216.45001 [14] Friedman, A.: Stochastic differential equations and applications. (1976) · Zbl 0323.60057 [15] Has’minskii, R. Z.: Stochastic stability of differential equations. (1981) [16] Kolmanovskii, V. B.; Myshkis, A.: Applied theory of functional differential equations. (1992) · Zbl 0917.34001 [17] Liao, X. X.; Mao, X.: Exponential stability and instability of stochastic neural networks. Stochast. anal. Appl. 14, No. 2, 165-185 (1996) · Zbl 0848.60058 [18] Liao, X. X.; Mao, X.: Stability of stochastic neural networks. Neural, parallel sci. Comput. 4, No. 2, 205-224 (1996) · Zbl 1060.92502 [19] Liptser, R. Sh.; Shiryayev, A. N.: Theory of martingales. (1986)


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# How do I ensure a single character for \ldots

How do I ensure that \ldots comes out in the PDF as a single character?

I'm formatting a document for a publisher, who has given this condition in its formatting rules, but I do not know how I can do this in LaTeX.

I'm using MiKTeX 2.9

EDIT: Yes the question could be phrased as: is it possible for \ldots to produce a single "triple point glyph" rather than 3 separate dots (like "..."), since that's what it currently does.

Following the suggestion of kahen, I tried with XeLaTeX and the ellipsis character "…" as well as the \ldots command. Both worked and looked the same.

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I don't understand your request. Could you please explain in more detail? –  Gonzalo Medina Oct 26 '11 at 18:11
I think that the question is: "Does \ldots produce the single character (tripple point glyph?) or three periods?" –  Yossi Farjoun Oct 26 '11 at 18:13
If you use XeTeX, you can enter the ellipsis character directly into your document, but it will not look the same as \ldots. I don't think there is a way to get \ldots as a single glyph unless your font has a properly spaced out ellipsis in it. –  kahen Oct 26 '11 at 18:23
Check out this thread on the XeTeX mailing list as well: tug.org/pipermail/xetex/2010-December/019576.html –  kahen Oct 26 '11 at 18:41

The command \ldots, which works both in text mode and in math mode, will create a single glyph in xelatex and lualatex, assuming you're using a font that knows how to set character U+2026, "Horizontal Ellipsis."

In an earlier version of this answer, I mistakenly referred to what's created in pdflatex by the \ldots command as a "single glyph". In fact, it's composed of three separate dots, concatenated by non-breaking whitespace. Thus, while it's not a single glyph in the narrow, technical sense, it behaves like one for typesetting purposes. For instance, it won't ever get broken up across two lines.

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\ldots does not produce a single glyph in the resulting PDF file. It's produced by three peroids with manual space between them. Hence if you copy to a plain text file, you'll get . . . or something similar. –  kahen Oct 26 '11 at 18:26
Is this true? I'm able to select each period/dot in \ldots when I view a PDF. \ldots defaults to either \mathellipsis or \textellipsis, depending on whether you're in math mode or not, both of which are comprised of three individual dots. –  Werner Oct 26 '11 at 18:28
The publisher gave no other rules in regard to this. But with Xelatex and \ldots it works to produce a single glyph. –  user8868 Oct 26 '11 at 18:59
Glad that xelatex is working for you. I know there's a unicode code for "textellipsis", so that's why I was thinking that it was a single glyph. –  Mico Oct 26 '11 at 19:01
@Werner: I strongly suspect the publisher cares mostly about the dots not getting broken up across lines. The OP mentioned that ellipses have to be a "single character" -- this would be satisfied by a single glyph, but also by any construct that will not ever get broken up (in the pdf file, of course). –  Mico Oct 26 '11 at 19:43

You could use the dots family from the MnSymbol package; in the following example I used \hdots to define a new command \Ldots to produce the same result as \hdots, but using lowered instead of centered dots (I used \mathop in my definition to give the symbol the spacing corresponding to a standard operator, but you could replace it with \mathord to get the same spacing as \hdots):

\documentclass{article}
\usepackage{MnSymbol}

\newcommand\Ldots{\mathop{\lower.5ex\hbox{$\hdots$}}}

\begin{document}

$a\hdots b$

$a\Ldots b$

\end{document}

This remark is on page 64 of the The Comprehensive LaTeX Symbol List :

Also, unlike most of the other dot symbols in this document, MnSymbol’s dots are defined as single characters instead of as composites of multiple single-dot characters.

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Unlike \ldots, this creates three dots which are centered vertically. –  diabonas Oct 26 '11 at 18:35
@diabonas: good point. I've updated my answer accordingly. –  Gonzalo Medina Oct 26 '11 at 18:46
Some quick tests shows that using \smash{\raisebox{-.5ex}{$\hdots$}} pushes it down to the baseline. –  Werner Oct 26 '11 at 18:47


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# What is center of inertia?

1. Dec 28, 2014

### Chemist@

What is the center of inertia? Is it the same as the center of mass?

2. Dec 28, 2014

### Doug Huffman

What other component of inertia can have space-like - center - dimensions?

3. Dec 28, 2014

### Chemist@

I don't understand.

4. Dec 28, 2014

### mathman

"A https://www.physicsforums.com/wiki/point [Broken], near or https://www.physicsforums.com/wiki/within [Broken] a https://www.physicsforums.com/wiki/body [Broken], at which the body's https://www.physicsforums.com/wiki/mass [Broken] can be https://www.physicsforums.com/wiki/assumed [Broken] to be https://www.physicsforums.com/wiki/concentrated [Broken] when considering its rotational motion and https://www.physicsforums.com/wiki/moment_of_inertia [Broken]. This may be different from its https://www.physicsforums.com/wiki/centre_of_mass [Broken] which is the equivalent for linear motion."

Above from wikidictionary.

Last edited by a moderator: May 7, 2017
5. Dec 28, 2014

### Staff: Mentor

Mathman, all of the links in the body of your text point to nonexistent pages. PF doesn't have a wiki.

6. Dec 29, 2014

### Chemist@

How to calculate it?

7. Dec 29, 2014

### mathman

8. Dec 29, 2014

### FactChecker

Can anyone give an example where it is easy to see that the center of mass is not the center of inertia?

9. Dec 30, 2014

### jbriggs444

As I understand the concept from the rather terse dictionary definition...

Suppose that you have a turntable. On this turntable you have an object. The "center of inertia" of the object is where you could place its entire mass and wind up with the same moment of inertia as the original object.

Suppose, for instance that the object is a thin hoop of mass m, radius r placed flat on the turntable with its center R units from the center of the turntable. The moment of inertia of this hoop with respect to the center of the turntable is $mr^2 + mR^2$.

Its "center of inertia", C, is at distance $\sqrt{r^2+R^2}$ from the center of the turntable so that the moment of inertia works out to $mC^2 = mr^2 + mR^2$

10. Jan 26, 2015

### Chemist@

Shouldn't moment of inertia be used instead of the masses?

11. Jan 27, 2015

### dean barry

Have a look at radius of gyration (mechanics), it assumes all the mass is concentrated at a single point and radius.
For instance a cylinder with a mass of 10 kg and a radius of 0.1 metres, rotating about its longitudinal axis has a moment of inertia of
0.04 kg - m², the radius of gyration = 0.0632 metres

12. Feb 8, 2015

### shiromani

Is center of inertia a valid concept? I think inertia is quantitative concept.


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Domain wall fluctuations in ferroelectrics coupled to strain    [PDF]

R. T. Brierley, P. B. Littlewood
Using a Ginzburg--Landau--Devonshire model that includes the coupling of polarization to strain, we calculate the fluctuation spectra of ferroelectric domain walls. The influence of the strain coupling differs between 180 degree and 90 degree walls due to the different strain profiles of the two configurations. The finite speed of acoustic phonons, $v_s$, retards the response of the strain to polarization fluctuations, and the results depend on $v_s$. For $v_s \to \infty$, the strain mediates an instantaneous electrostrictive interaction, which is long-range in the 90 degree wall case. For finite $v_s$, acoustic phonons damp the wall excitations, producing a continuum in the spectral function. As $v_s\ to 0$, a gapped mode emerges, which corresponds to the polarization oscillating in a fixed strain potential.
View original: http://arxiv.org/abs/1307.7886


text:
• # question_answer 45) A current of $(1/4\pi )$ampere is flowing in a long straight conductor. The line integral of magnetic field around a closed path enclosing the current carrying conductor is: A) ${{10}^{-7}}$ weber per meterB) $4\pi \times {{10}^{-7}}$ weber per meterC) $16{{\pi }^{2}}\times {{10}^{-7}}$weber per meterD)  0

By Amperes Law, we get, $\vec{B}.\overrightarrow{dl}={{\mu }_{0}}I=4\pi \times {{10}^{-7}}\times \frac{1}{4\pi }={{10}^{-7}}\,w/{{m}^{2}}$ Hence, the correction option is [a].


text:
1. ## differentiation

i know this is a simple differentiation, but i just can't remember how to do it!

((x^3)+3(x^2)+1) ^(1/3)

also how can i write x^2 on this forum so it looks like how x^2 should look, with a superscript 2.

2. Hello, mathmonster!

You need the Chain Rule . . .

$f(x) \;=\;\left(x^3+3x^2+1\right)^{\frac{1}{3}}$

$f'(x)\;=\;\frac{1}{3}\left(x^3+3x^2+1\right)^{-\frac{2}{3}}(3x^2 +\, 6x) \;=\;\frac{3x^2+6x}{3(x^3+3x^2+1)^{\frac{2}{3}}} \;=\;\frac{x^2+2x}{(x^3+3x+1)^{\frac{2}{3}}}$

3. Originally Posted by mathmonster
also how can i write x^2 on this forum so it looks like how x^2 should look, with a superscript 2.
Left-click on the script that Soroban wrote in his answer. That will show you how to code it.

There is a thread on how to use LaTeX in one of the subforums. (Or you could do a web search. Tips on how to code are all over it.)

-Dan

4. $
\log_{5} 25=2
$

just checking now to see how this works

5. $
f(x) = (x^3+3x^2+1)^{\frac{1}{3}}
$

6. Originally Posted by mathmonster
$
f(x) = (x^3+3x^2+1)^{\frac{1}{3}}
$
Looks like you've got it!

-Dan