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### Author Topic: "I'm never shopping THERE again!" Share your story!  (Read 1109738 times)

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#### Sirius

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##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3120 on: October 11, 2012, 04:29:53 PM »
I think it's shady.  If I'm on a fixed term contract with a company, with dates, I would not expect automatic renewal.

It 's not renewing the contract. It is continuing service on a monthly basis. At least that's what my AT&T contract says. It also says it is automatic unless they are notified.

I notified my aunt's cell provider after her death, but because I didn't know who it was I had to wait until a bill arrived - all but three days of which covered a period of time after her death.  They were very nice and didn't charge an early termination fee since the young man was able to find her obituary on line, and I did pay that last month's bill.  I asked about early termination, and he explained that since she'd passed away and he was able to verify it, that the estate wouldn't be charged an early termination fee.

#### Bluenomi

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• Posts: 3580
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3121 on: October 11, 2012, 07:06:45 PM »
Pippen wrote:

"I tried to cancel a business landline so rang Telstra, the most useless company in Christendom."

In a world that contains Bank of America, that's a pretty bold statement.

Virg
A true one though, Telstra are terrible!

#### Pippen

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##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3122 on: October 11, 2012, 07:18:03 PM »
Pippen wrote:

"I tried to cancel a business landline so rang Telstra, the most useless company in Christendom."

In a world that contains Bank of America, that's a pretty bold statement.

Virg
A true one though, Telstra are terrible!

You have dealt with them? They are awful. 8 weeks to connect my internet that should have taken 2 days. At an average of 3 hours a day on the phone with them what does that work out to? It got to the stage where I would call and who ever answered would see my call sheet and the problem and just hang up on me. Then they said they would send a tech out who never arrived and they charged me over $1000 for 3 call out fees and lied saying that someone had been to see me and sorted it out and it was all hunky dory when they clearly hadn't. I was almost murderous by the end of it. They still owe me over$2500 in refunds and I cannot bring myself to deal with them. Incompetent doesn't even begin to describe them. The worst company in the world. I can totally see how people lose it and go to a head office and start smashing things to bits.

#### jedikaiti

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• A pie in the hand is worth two in the mail.
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3123 on: October 11, 2012, 10:41:59 PM »
I think it's shady.  If I'm on a fixed term contract with a company, with dates, I would not expect automatic renewal.

In my experience they write to you and say "Your special offer is coming to an end, you'll be switched onto our standard package on x date, or you could switch to a sparkly new deal with us."

I've never had that happen - I'm just told up front that after the month containing my end date for X promotion, my rate will automatically go up to the regular rate. I see it happen on the bill, call them up, and see if they've got any customer retention offers to bring it back down. :-)  (I also shop around and if I can find a better deal, then I'll ask my provider to match it or I move.)

But that's just for things like home phone/internet/TV. For cell phone, I've never had a special promo like that. With cell phones, my experience is that you buy a phone for $Y without signing a contract (so you can cancel any time), or$X (considerably less) if you agree to stay with Provider for Z months/years and agree to pay an early termination fee if you cancel the service within that time frame. In either case your service is $C/month (plus tax & other assorted government-regulated surcharges), but after your contract period is up, your service continues at the same price but you can now cancel any time without a termination fee. Other than pre-paid phones, that is the only way I have ever seen a cell phone plan work. What part of v_e = \sqrt{\frac{2GM}{r}} don't you understand? It's only rocket science! "The problem with re-examining your brilliant ideas is that more often than not, you discover they are the intellectual equivalent of saying, 'Hold my beer and watch this!'" - Cindy Couture #### StarFaerie • Hero Member • Posts: 1155 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3124 on: October 12, 2012, 06:59:38 AM » Pippen wrote: "I tried to cancel a business landline so rang Telstra, the most useless company in Christendom." In a world that contains Bank of America, that's a pretty bold statement. Virg A true one though, Telstra are terrible! You have dealt with them? They are awful. 8 weeks to connect my internet that should have taken 2 days. At an average of 3 hours a day on the phone with them what does that work out to? It got to the stage where I would call and who ever answered would see my call sheet and the problem and just hang up on me. Then they said they would send a tech out who never arrived and they charged me over$1000 for 3 call out fees and lied saying that someone had been to see me and sorted it out and it was all hunky dory when they clearly hadn't. I was almost murderous by the end of it. They still owe me over $2500 in refunds and I cannot bring myself to deal with them. Incompetent doesn't even begin to describe them. The worst company in the world. I can totally see how people lose it and go to a head office and start smashing things to bits. *cough* Telecommunications Ombudsman *cough* #### Ohjustlovely • Member • Posts: 317 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3125 on: October 12, 2012, 10:39:27 AM » I like this thread, but there seems to be a high-jacking about contracts and phones. Would you please start a S/O? Thank you. #### kitelady1870 • Jr. Member • Posts: 5 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3126 on: October 12, 2012, 02:14:33 PM » There's only one store in my town to purchase Greek Life related items, so I'm stuck going there at least once a year. Everything is really overpriced, but with the time constraints of Big-Little week, you don't have the *time* to order online. #### Moray • Hero Member • Posts: 1869 • My hovercraft is full of eels! ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3127 on: October 12, 2012, 02:38:01 PM » There's only one store in my town to purchase Greek Life related items, so I'm stuck going there at least once a year. Everything is really overpriced, but with the time constraints of Big-Little week, you don't have the *time* to order online. Ooh! What sorority are you in? Utah #### WestAussieGirl • Member • Posts: 901 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3128 on: October 13, 2012, 01:55:07 AM » *cough* Telecommunications Ombudsman *cough* I second this! I waited a year for Telstra to do some work at my house. I called every so often and kept getting told it was "in the queue". I called the ombudsman. It was done in a day. #### Lauds • Member • Posts: 361 ##### Re: "I'm never shopping THERE again!" Share your story! « Reply #3129 on: October 13, 2012, 05:55:14 AM » Pippen wrote: "I tried to cancel a business landline so rang Telstra, the most useless company in Christendom." In a world that contains Bank of America, that's a pretty bold statement. Virg A true one though, Telstra are terrible! You have dealt with them? They are awful. 8 weeks to connect my internet that should have taken 2 days. At an average of 3 hours a day on the phone with them what does that work out to? It got to the stage where I would call and who ever answered would see my call sheet and the problem and just hang up on me. Then they said they would send a tech out who never arrived and they charged me over$1000 for 3 call out fees and lied saying that someone had been to see me and sorted it out and it was all hunky dory when they clearly hadn't. I was almost murderous by the end of it. They still owe me over \$2500 in refunds and I cannot bring myself to deal with them. Incompetent doesn't even begin to describe them. The worst company in the world. I can totally see how people lose it and go to a head office and start smashing things to bits.

I will never go with Telstra because of my parents experience. They had a 10GB or 20GB plan (more than two years ago now so minor details escape me). Every month they would get excess usage charges of about 3-4GB. They had been with the company since its Saturn days and were slowly getting more and more fed up with them, especially after they moved out of the Wellington cable area.

The beginning of the end was me getting a job with telstra's biggest competitor. When they called to find out if their contract was up, the first person couldn't tell them. They would need to get someone to call them back (can anyone say retention team?). By the time mum did get called back she was rather frustrated and the poor person who was speaking to her got a bit of a rant. End result - they were out of contract. Thankfully they ported the landline number so Telstra didn't have any hold on them.

Strangely, never in the two plus years have they been charged overage and their plan gave them equal or lesser amounts of data. In fact, I just changed them to a 5GB as they hadn't used more than that in the last year. Our suspicion is they were being charged overage for their actual usage, as well as the plan charges. The overage before and the usage after the move was just too similar for it not to be when all that had changed was their provider.

It has put me off Telstra, in all their guises, for life.

#### BarensMom

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##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3130 on: October 13, 2012, 09:35:26 AM »
I like this thread, but there seems to be a high-jacking about contracts and phones. Would you please start a S/O? Thank you.

Seconding this request.

#### hjaye

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• Posts: 1216
##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3131 on: October 13, 2012, 09:41:29 AM »
Well it looks like I may be out a nice hefty chunk of change, and I don't think there is a whole lot I can do about it.

About two months ago my grandson's dad asked me to take my grandson to the eye doctor to get his eyes checked and get new glasses.  He informed me that the visit was free and I was under the impression it was some kind of government sponsored program that helped needy families with eye care.

What I found out during the visit is that a neighbor of my grandsons family lives in the same apartment complex and saw my grandson was having some problem with his eyeglasses and she works for an eye doctor.  She spoke to his (my grandson) dad found out he didn't have insurance, so took it upon herself to ask the doctor if she would see my grandson free of charge.  The doctor agreed and there I was at the eye doctors office with my grandson while he was getting his eyes checked.

I thanked the doctor, I thanked the neighbor, I told them it was a very kind and generous thing they were doing for my grandson and I was very grateful for their generosity.

I got a bill in the mail from the drs office for 445.00 for the visit.  It may be a mistake, I'm trying to get a hold of the neighbor to ask her to check into it.  The fact is, I don't know her name, I have nothing in writing so calling the office and talking to someone in billing isn't going to do a whole lot of good.  I'd have to depend on the doctor remembering my grandson and telling the billing people it was a mistake to send me a bill.  Heck I was just there because my schedule allowed me the time to take him.

• Member
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##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3132 on: October 13, 2012, 10:48:43 AM »
Well it looks like I may be out a nice hefty chunk of change, and I don't think there is a whole lot I can do about it.

About two months ago my grandson's dad asked me to take my grandson to the eye doctor to get his eyes checked and get new glasses.  He informed me that the visit was free and I was under the impression it was some kind of government sponsored program that helped needy families with eye care.

What I found out during the visit is that a neighbor of my grandsons family lives in the same apartment complex and saw my grandson was having some problem with his eyeglasses and she works for an eye doctor.  She spoke to his (my grandson) dad found out he didn't have insurance, so took it upon herself to ask the doctor if she would see my grandson free of charge.  The doctor agreed and there I was at the eye doctors office with my grandson while he was getting his eyes checked.

I thanked the doctor, I thanked the neighbor, I told them it was a very kind and generous thing they were doing for my grandson and I was very grateful for their generosity.

I got a bill in the mail from the drs office for 445.00 for the visit.  It may be a mistake, I'm trying to get a hold of the neighbor to ask her to check into it.  The fact is, I don't know her name, I have nothing in writing so calling the office and talking to someone in billing isn't going to do a whole lot of good.  I'd have to depend on the doctor remembering my grandson and telling the billing people it was a mistake to send me a bill.  Heck I was just there because my schedule allowed me the time to take him.

Unless you signed something at the doctors office stating that you would pay for the visit if not covered by insurance or something I have no idea why they would be billing you.

#### gramma dishes

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##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3133 on: October 13, 2012, 11:13:52 AM »
...   I got a bill in the mail from the drs office for 445.00 for the visit.  It may be a mistake   ...

I certainly HOPE it's a mistake!  That seems like an awfully lot of money for a regular kid's eye check up, even if you were supposed to be billed.

I'd get your grandson's father in on this right away and find out the name of that neighbor and track this whole mess down as quickly as possible.

I also agree with Deadbody.  Unless you signed something saying you would pay anything not covered by insurance I don't know how they could make you pay for this.  How did they even get your name and address?

#### I'mnotinsane

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##### Re: "I'm never shopping THERE again!" Share your story!
« Reply #3134 on: October 13, 2012, 11:38:42 AM »
...   I got a bill in the mail from the drs office for 445.00 for the visit.  It may be a mistake   ...

I certainly HOPE it's a mistake!  That seems like an awfully lot of money for a regular kid's eye check up, even if you were supposed to be billed.

I'd get your grandson's father in on this right away and find out the name of that neighbor and track this whole mess down as quickly as possible.

I also agree with Deadbody.  Unless you signed something saying you would pay anything not covered by insurance I don't know how they could make you pay for this.  How did they even get your name and address?

This type of form is standard practice at all my doctors/providers.  They slip it in with all the other new patient materials.


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# Proof of theorem on recurrent states and its equivalence class

A theorem states the following:

## Theorem

if $i \in S$ is a state which is recurrent, then every state in the equivalence class of $i$ $(\ K(i) \ )$ is recurrent.

$S$ is the space of the states of a markov chain $X_n$ with stochastic matrix (transition matrix) $P=(p_{ij})$. For $i,j \in S$ let $i \sim j$ be defined as $(i \leftrightarrow j)$ V $(i=j)$, where $(i \leftrightarrow j)$ means that $i$ communicates with $j$, so in other words $p_{ij}^{(n)},p_{ji}^{(n)}>0$. Thus, we define $K(i):=\{j \in S | i \sim j\}$.

The proof works like this:

## Proof

Let $i$ be recurrent and $j \in K(i),j \neq i$, so that there are $n,m \in \mathbb{N}$ with $p_{ij}^{(m)}\cdot p_{ji}^{(n)}>0$

$\sum\limits_{k=0}^\infty p_{jj}^{(k)} \geq \sum\limits_{k=0}^\infty p_{jj}^{(m+n+k)} \geq \sum\limits_{k=0}^\infty p_{ji}^{(n)}p_{ii}^{(k)}p_{ij}^{(m)}$

$\color{green}{Can \ someone \ explain \ to \ me \ why \ the \ first \ and \ second \ inequality \ sign \ holds? }$

Thank you!

• Hi, Chris. Though I and most others can guess at the meaning, it would be very helpful for you to set out clearly your notation. – cardinal Jan 28 '13 at 1:57

$$\sum_{k=0}^\infty p_{jj}^{\left(k\right)} \ge \sum_{k=n+m}^\infty p_{jj}^{\left(k\right)}\\ = \sum_{k=0}^\infty p_{jj}^{\left(n+m+k\right)}$$
For the second inequality, we can without loss of generality, by the definition of a Markov process, consider the elements $A,B,C$ of the $\sigma$-algebra restricted to $X\left(\omega\right)_0=j$
$$A=\left\lbrace \omega : X\left(\omega\right)_{0}=j,X\left(\omega\right)_{n}=i,X\left(\omega\right)_{n+k}=i,X\left(\omega\right)_{n+m+k}=j\right\rbrace\\ B=\left\lbrace \omega : X\left(\omega\right)_{0}=j,X\left(\omega\right)_{n+m+k}=j \right\rbrace\\ C=\left\lbrace \omega : X\left(\omega\right)_0 = j \right\rbrace$$
Clearly $A \subseteq B \subseteq C$. But by the definition of the probabilities, and in particular the sub-additivity of probability measures,
$$p_{jj}^{\left(n+m+k\right)}=\mathbb{P}\left[\left.B \right| C\right] \ge \mathbb{P}\left[\left. A \right| C\right]=p_{ij}^{\left(n\right)}p_{ii}^{\left(k\right)}p_{ji}^{\left(m\right)}$$ The rest follows...

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text:
# Can you please explain why it favorable and unfavorable. thank you. P 9-A Variance Analysis (In...

###### Question:

Can you please explain why it favorable and unfavorable. thank you.
##### Research shows that renal calculi is on the rise, what impact can the clinical role of...
Research shows that renal calculi is on the rise, what impact can the clinical role of a nurse practitioner have on this debilitating disease...
##### Question 9: Which of the following circumstances allows for a disparity in pay between different genders?...
Question 9: Which of the following circumstances allows for a disparity in pay between different genders? Answer: A. A seniority system В. A merit system C. A merit system based on quantity or quality of output D. All of these options Question 11: Within the employee earnings record, most colu...
##### Mycobacterium tuberculosis and Listeria monocytogenes are both facultative intracellular parasites in humans. These pathogens have the...
Mycobacterium tuberculosis and Listeria monocytogenes are both facultative intracellular parasites in humans. These pathogens have the ability to grow and multiply inside phagocytes. Artificial passive immunity against these infections can be achieved only by transferring E rosette forming leukocyte...


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# Ideally, how to achieve isothermal expansion of an ideal gas?

In an isothermal expansion of an ideal gas, there cannot be change in internal energy otherwise the temperature would change. So hence we know that all the heat go into work. Knowing that PV=nRT and T being constant, we see that P is inversely proportional to V.

Does that mean in order to feasibly realize an isothermal expansion we would need to do the following:

1. Put the ideal gas in a cylinder with a piston and some amount of weight on it to reach some pressure.
2. Find a heat reservoir equal to temperature of the gas.
3. Slowly lift weight off of the piston according to $P \propto 1/V$.

The last part is kind of the strange part. We have to lift some weight off otherwise the volume won't change. But why can't we quickly lift the weight off? So the external pressure changes from some large number to a very smaller number? Would that be approximately an adiabatic expansion?

The isothermal expansion is a theoretical ideal. An isothermal process requires the system is in perfect equilibrium with its surroundings at all times so it would have to be done infinitely slowly. As you say in your question, any process done at a finite speed is necessarily out of equilibrium.

However in real life provided heat flow is fast enough processes can be so close to isothermal that we can treat them as perfectly isothermal. That is, the error involved in assuming they are isothermal is negligibly small.

• If you allow the expansion (or compression) to take place slowly and continuously, there's enough time at every point for the temperature to equilibrate: Since all the gas will be at same temperature as the reservoir. This is an iso-thermal process.
• If you do the expansion (or compression) rapidly, so fast there's no time for thermal energy to flow and the gas to equilibrate, then the pV work being done will change the temperature of the gas. That's not an iso-thermal process, because the temperature changes. If you do it fast enough, so no thermal energy is exchanged, then it's adiabatic.

So the same sequence of operations if done fast enough is adiabatic and done slow enough is isothermal. What's "enough"? Compare the energy that can flow during the ideal amount, and see how close it is.

Two ways: 1) Expand the ideal gas into a vacuum, and there is no change in the temperature. However, this is not the type of expansion I surmise you are asking about.

2) Expand the gas against a piston so that it does work. Now, since the work is accomplished through the individual collisions of gas atoms, each collision transfers a small amount of momentum and energy to the piston. So the gas cools. Or think about it as change in internal energy = heat + work. The temperature reflects the kinetic energy of the gas, which is also the total energy of the gas since it is ideal.

Since the gas is cooling during expansion as described so far, the only way to maintain the same temperature is to heat the gas in sync with the transfer of collisional energy to the piston. This heat could transfer from some other part of the container enclosing the gas.

If you lift the weight off quickly (so that the external force on the gas is much lower), the expansion will be irreversible. Under these circumstances, even though the cylinder is in contact with a constant temperature reservoir at its surface, only the surface of the gas will be at the specified temperature, and the interior of the gas will drop to a lower (non-spatially-uniform) temperature throughout the expansion (until final equilibrium is reached, and the gas is again at the reservoir temperature). In the interim, heat conduction will be occurring within the gas.

In addition, the force that the gas imposes on the piston will be a combination of the effect of viscous (tensile) stresses within the gas plus what we ordinarily would consider the gas pressure under static conditions. Initially, for the irreversible case, all the change in force will be supported by viscous stresses, which are proportional, not to the change in volume but, to the rate of change of volume. So, during the irreversible expansion, the gas ceases to behave like an ideal gas and ceases to satisfy the ideal gas law (which applies only at thermodynamic equilibrium) within the cylinder. As equilibrium is approached, the rate of change of volume decreases, and the viscous stresses make less of a contribution to the overall force of the gas on the piston. At final equilibrium, the viscous stresses are again zero, and the force on the piston is entirely described by the ideal gas law.

To get a better feel for all this, imagine that the gas behaves like a combination of a spring (preloaded in compression) in parallel with a viscous damper, with the combination joined mechanically to the rear end of the cylinder and to the piston. The spring mimics the ideal gas behavior of the gas and the viscous damper mimics the viscous behavior of the gas. If the piston moves very slowly, the force on the piston is dominated by the spring (ideal gas behavior). But, if the piston moves very rapidly, the change in the force on the piston is dominated by the viscous behavior. In line with this conceptualization, a crude approximation to the behavior of a gas experiencing either a slow- or a rapid deformation can be expressed as: $$\frac{F}{A}=\frac{nRT}{V}-\frac{k}{V}\frac{dV}{dt}$$where F is the force of the gas on the piston and k is a constant proportional to the gas viscosity. The first term captures the thermodynamic equilibrium behavior of the ideal gas, and the second term captures the viscous contribution.

To get a better understanding of the viscous Newtonian behavior of fluids (including gases), I highly recommend Transport Phenomena by Bird, et al.


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# Python: SARIMAX Model Fits too slow

I have a time series data with the date and temperature records of a city. Following are my observations from the time series analysis:

1. By plotting the graph of date vs temperature seasonality is observed.
2. Performing adfuller test we find that the data is already stationary, so d=0.
3. Perform Partial Autocorrelation and Autocorrelation with First Seasonal Difference and found p=2 and q=10 respectively.

Code to Train Model

model=sm.tsa.statespace.SARIMAX(df['temperature'],order=(1, 1, 1),seasonal_order=(2,0,10,12))
results=model.fit()

This fit function runs indefinitely and does not reach an output. I am running on a on Google Colab CPU.

How to fix this issue?

• Unless you have a really good reason for using stats models I suggest you to change to FB Prophet, it is an amazing and well documented library for advance time series forecasting. neuralprophet.com Mar 7 at 2:17
• Statsmodels is very slow, you could try pmdarima: pypi.org/project/pmdarima or as suggested you could try fb prophet if you just want some time series fit reasonably well although it also can be quite slow when compared to other time series methods. Mar 7 at 22:50

Assuming you have multiple cities in the dataframe. you can create some new features in the dataframe . For example , I created a few features below to try and match you PACF and ACF graphs .

df['lag_1'] = df.groupby(['city'])['temperature'].transform(lambda x: x.shift(1))

d=1

df['d_1'] = df['temperature'] - df['lag_1']

p = 1:

df['p_1'] = df.groupby(['city'])['d_1'].transform(lambda x: x.shift(1))

q = 1:

df['ma_1'] = df.groupby(['city'])['d_1'].transform(lambda x: x.shift(1).rolling(1).mean())

P=2 (and other terms)

df['lag_t12'] = df.groupby(['city'])['temperature'].transform(lambda x: x.shift(12))

df['lag_t24'] = df.groupby(['city'])['temperature'].transform(lambda x: x.shift(24))

.

.

.

df['lag_t120'] = df.groupby(['city'])['temperature'].transform(lambda x: x.shift(120))

Q=10 :

df['Q_10'] = df[col for col in df if col.startswith('lag_t')].mean()

After this try using LightGBM , XGBoost or other regression packages to regress against these newly created features with temprature as your target variable.

Alternatively , you can forego the ACF/PCF approach altogether and instead create bunch of commonly used features using :

• shift
• rolling mean
• rolling standard deviations
• max() , min() within groups

and regress against those and check which features minimize RMSE/AIC/BIC in your Regression Hyperparameters.

Since Cross validation is different in cases of Time Series,consider using TimeSeriesSplit in scikit-learn . Check-out this post in case you want to do grouped time series cross validation .


text:
# Subset

## Subset(d, position, i, resultIndex, warnEmpty)

If «d» is an Index or a 1D array index by «i», Subset(d) returns a list of all the elements of the index for which «d» is true (1). The function is used to create a new index that is a subset of an existing index.

If «d» is not an Index or 1D array -- if it's scalar or has more than 1D -- it returns an error.

Subset in its standard one-parameter usage does not array abstract because it returns an unindexed list.

When there is the possibility of having Null values in your array, you should consider whether you want those elements to be retained. The following two uses respectively retain or remove Null elements:

Subset(Test Or Test = Null)
Subset(Test And Test <> Null)

Without the explicit test, Null elements are removed.

## Example

Let:

Index Years := 2005..2009

Then:

Subset(Years < 2008) →
[2005, 2006, 2007]
Subset(Years > 2005 and Years < 2008) →
[2006, 2007]

## Optional parameters

### Position

The optional «position» parameter specifies whether the index element or the index position should be returned (see Associative vs. Positional Indexing). When «position» is omitted or when position: false is specified, the index element is returned. position: true returns the positions of each non-zero element along «d»’s index for which «d»’s values are true: Subset(D, position: true)

The need to specify positions arises when your index contains duplicate values, or when the input to the function is an array with an implicit index.

In general, the one-dimensional array passed to Subset needs to be explicitly indexed because the index elements corresponding to true elements is returned. Thus, an implicitly indexed array (e.g., a list) also results in an error. Theoretically Subset could return a result in the implicitly indexed case, but it would be misleading since it would just be returning all the true values, for example:

Subset(1997..2008 > 2005) →? [1, 1, 1] { Hypothetical only -- error reported }

In a case like this, you should use a local index, e.g.:

Index I := 1997..2008 do Subset(I > 2005) → [2006, 2007, 2008]

However, if you use the position:true parameter setting, then you can pass an array with an implicit index, since element positions (and not index values) are returned:

Subset(1997..2008 > 2005, position: true) → [10, 11, 12]

### I and resultIndex

The basic use of Subset does not allow d to contain more than one dimension, and will not evaluate or array abstract otherwise (since an unindexed list is returned). If you have a two- (or more-) dimensional array or are using Subset in calculations that are potentially array-based, you'll need to identify one of those dimensions as the one that the subset is being taken over. For each slice along the other dimensions, a Subset can then be separately computed, and each of those subsets may have a different number of elements.

The optional index parameter «i» together with «resultIndex» allow Subset to be used in a manner that fully array abstracts: «i» specifies which dimension to operate over, and «resultIndex» specifies the index of the result. For the individual subsets that contain more elements than the «resultIndex», only the first Size(«resultIndex») members of the subset are retained. For those subsets that are smaller than the «resultIndex», the remaining entries in the result are padded with Null values.

When «i» and «resultIndex» are used in conjunction, the result is an array with fully explicit indexes.

When you use Subset to define an index, you should not use «resultIndex», «d» should be 1-dimensional, so «i» is superfluous.

### WarnEmpty

The optional «warnEmpty» parameter controls whether Subset issues a warning when no true elements are found in «d». The default value for «warnEmpty» is false, i.e. it gives no warning.

## Details & More Examples

### Example 1

Let:

Index Years := 2005..2009

Then:

Subset(Years < 2008, position:true) →
[1, 2]
Subset(Years > 2005 and Years < 2008, position: true) →
[2, 3]

### Example 2

Let:

Variable A :=
I ▶
J ▼ 1 2 3 4 5
1 73 23 19 54 12
2 11 1 98 76 52
3 22 33 49 87 22
Index K := [1, 2]

Then:

 Subset(A > 50, i: I, resultIndex: K) →
K ▶
J ▼ 1 2
1 1 4
2 3 4
3 4 «null»

Notice that in the first row, J = 1</coed>, two elements, K=[1, 4], satisfy the condition, A > 50. In the second row there are three elements, K= [3, 4, 5], but since our result index K has only two elements, we retain only the first two. In the third row of A, there is only one cell satisfying the condition, hence the second cell in the result is padded with Null.

For Subset to reliably array abstract, you should specify both indexes, although it is fine to use the same index for both. Since they are optional, use the the named calling convention as illustrated above.

See AlsoAbstracted Subset.ana example model Associative vs. Positional Indexing SubIndex Subscript Index..Do


text:
# What is the maximum angle F-16 can turn in x seconds flying at corner speed?

How can one calculate the maximum turning angle F-16 can turn in x second flying at corner speed. Is it changed with height or weight?

To calculate this, you need to look at the energy Maneuver diagram for the aircraft. This diagram depicts, for a specified altitude and gross weight, the excess power per unit pound of weight the aircraft would have at full afterburner at any g-Loading and true airspeed. There would be a different chart for each aircraft configuration, gross weight and Density Altitude combination.

Instantaneous Turn rate is depicted on the vertical scale, true airspeed or mach number is depicted on the horizontal scale. The concave curved lines that go from upper left to far right represent flight conditions of constant G-loading, while the straight lines that radiate from the origin represent constant turn radius. The peak of the flight envelope, (at about Mach 0.7, 20 deg/sec turn rate and 8 Gs, is corner velocity. The lines that start at the left boundary or at the bottom, go upwards, curve to the right and then back down again, and are labeled with values 600, 400, 300, 0, -200, etc. are called Ps, (pronounced P-sub-S), lines. They depict the specific power, or actually, specific excess power - a measure of how much power per pound of aircraft weight the aircraft would have left over to climb or accelerate at the depicted point on the diagram. So at the top of the diagram, at the corner velocity, where the turn rate is near 20 deg/sec, the specific excess power is less than -800. The -800+ Ps curve indicates that the aircraft, if it established flight at those conditions, would be bleeding energy (airspeed or altitude) at a rate in excess of negative 800 ft/sec.

The best performance on the diagram that can be sustained, on the other hand, is the highest point on the zero Ps line, which is at mach 0.8, 7 Gs, and it would produce a sustained turn rate of just over 14 deg/sec.

So, in a level turn, at Mach 0.8, 7 Gs, and 14 deg/sec, the F-16 could turn 180 degrees in about 13 secs.

Of course, if the aircraft performs the turn in a descent, it can make the turn in less time. This is indeed a common training maneuver, to turn 180 degrees in the shortest amount of time possible. It is performed by rolling into 135 degrees of bank at or slightly above corner velocity and pulling at maximum G-load. The maneuver is performed at maximum available G-load, in a constant plane tilted at 45 degrees from the horizontal. Bank angle (starting at 135 degrees), decreases throughout the maneuver, reaching 90 degrees bank after 90 degrees of turn, and 45 degrees bank at the end of the maneuver (after 180 degrees of turn). At 15,000 feet MSL, this will generate almost 20 degrees per second, and can be done in about 10 seconds.

For an uncoordinated turn, all things are possible, consider a hammerhead, wing over, or Lomcovak! For a constant altitude coordinated turn, weight effects the load and stall which does play a part. The two primary factors are bank angle and rate of turn. As the angle of bank increases the load factor or thrust required for speed will limit the aircraft's ability to bank.

To completely answer your question requires more information such as the weight, load limit, stall speed of the aircraft in the configuration of the turn, atmospheric conditions, etc.

As the angle of bank and turn rate increase, so must the speed and load factor, soon the power available or aircraft structural capability is exceeded. As the aircraft approaches a 90deg bank the vertical lift component approaches zero, the speed and load required to hold altitude approaches infinity.

The "FAA's Pilot's Handbook of Aeronautical Knowledge" gives the formulas for rate of turn and turning radius on page 4-34. Chapter 15 of "The Science of Flight" by W.N. Hubin (1992) is the source of the other formulas.

The variables used are:

  V = true airspeed in knots
R = turning radius in feet
θ = bank angle in degrees
ω = rate of turn in degrees per second
g = 11.26

Note: The FAA formulas use deg/sec, while Hubin uses rad/sec.

For example, at 120 knots and a 30° bank angle, the turn radius and rate of turn are:

Hubin goes into greater detail and provides information for other scenarios.

The formula for a standard rate turn (3deg/sec) is...

but other velocity's can be plugged in.

The AOA is 15 degrees and pulling 9G for making the maximum performance turn in level turns in F16. ( 22 degrees /sec)

• They question is about turn rate, i.e. degrees per second, not AOA or max Gs. – Michael Hall Nov 30 '19 at 17:22
• Then 22degrees /sec – George Geo Nov 30 '19 at 17:23


text:
## Fourier analysis – Determination of frequency components using FFT

The function defined below has three frequencies 2, 4, and 10. How can the Fourier transform (FFT) be used to display these frequencies?

``````   myfun(t_) =
1/(2 Sqrt(
2)) ((Sqrt)Abs(
3 + Cos(4 t) -
4 Cos(2 t) (-1 +
E^(-0.018` t) (Cos(9.99999594999918` t) +
0.0009000003645002213` Sin(9.99999594999918` t))^2) +
4 E^(-0.018` t) (Cos(9.99999594999918` t) +
0.0009000003645002213` Sin(9.99999594999918` t))^2))
``````

## nt.number theory – Fourier transform of \$ I_Y \$, \$ Y = { text {numbers with many prime factors} } \$

To let $$Y$$ Let be the set of integers $$N with more than $$D log log N$$ Prime factors. We can think, say, $$D = ( log log N) ^ {1- epsilon}$$,

We have pretty accurate approximations for the size of $$Y$$ (I am aware of Chapter II.6 in Tenenbaum's book and the references it contains.) I wonder what work is available there for the Fourier transform $$widehat {1_Y}$$ the characteristic function of $$Y$$,

I would expect $$widehat {1_Y}$$ To have tips on the main arches (ie bows around rations $$a / q$$ with a small denominator). That's because $$Y$$ is obviously "biased towards divisibility" and should therefore be slightly over-represented in the congruence class $$0$$ mod $$d$$ given for everyone $$d$$, relative to other congruence classes mod $$d$$,
A back-of-the-envelope calculation suggests that the value at the peak is around $$a / q$$ should be roughly proportional $$c ^ { omega (q)} / q$$, But what is known?

## fa.functional analysis – Characterization of a subset of the Sobolev space \$ H ^ k (0.2 pi) \$ in Fourier series

To let $$A: = {u in H ^ k (0.2 pi): u ^ {(j)} (0) = u ^ {(j)} (2 pi) mbox {for} j = 0,1, ldots, k-1 }$$, from where $$H ^ {k} (0.2 pi) subseteq L ^ 2 (0.2 pi)$$ is the Sobolev order room $$k$$ on $$(0, 2 pi)$$, Can we say that? $$u in A$$ iff
$$sum_ {n = – infty} ^ infty (1 + n ^ 2) ^ k | has {u} (n) | ^ 2 < infty?$$
In the above series $$has {u} (n)$$ are the Fourier coefficients of $$u$$, We think the answer to this question is positive because we may be able to identify $$A$$ with the Sobolev space of the Torus $$H ^ k ( mathbb {T})$$Use this result. But we can not show that there is an isomorphism $$A$$ and $$H ^ k ( mathbb {T})$$,

Know a reference for a characterization of $$A$$ with Fourier series?

## Fourier analysis – Fourier transform frequency

One possibility is the `FourierParameters`

``````  FourierTransform(1, x, w, FourierParameters -> {0, -2*Pi})
``````

``````  FourierTransform(Exp(I a x), x, w, FourierParameters -> {0, -2*Pi})
``````

Compare

``````funs = {1, DiracDelta(x), Exp(I a x), Cos(a x), Sin(a x)};
result= {#, FourierTransform(#, x, w, FourierParameters -> {0, -2*Pi})}& /@ funs;
Prepend(result, {"f(x)","Fourier transform unitary, ordinary frequency"});
Grid(%, Frame -> All)
``````

With wikis second column:

## Fourier analysis – spectral solution of elastic waves

As stated in "Wave Propagation in Structures: an FFT-based Spectral Analysis Method" by James F. Doyle:
For the spectral solution of the 1D equation for elastic waves:
$$widetilde {u} (x, omega) = sum Ce ^ {- i {k_ {n}} x} + De ^ {i {k_ {n}} x},$$
from where $${k_ {n}} = { omega_ {n}} sqrt { rho / E}$$ is the wavenumber. C and D are the indeterminate amplitudes at each frequency. The end of the beam at x = 0 is subjected to a force curve F (t), ie$$EA frac { partial u (x, t)} { partial x} = F (t)$$
E and A are the modulus of elasticity and the cross-sectional area, respectively. The final solution is the inverse Fourier transform of the following expression:

$$widetilde {u} (x, omega) = – frac { widetilde {F_ {n}}} {ik_ {n} EA} e ^ {- ik_ {n} x}$$

$$widetilde {F_ {n}}$$ is the Fourier transform of the applied force F (t).

The numeric example for the above problem is as follows:

``````Rod:
diameter=1 inch
density=0.00247 lb/ci
E=10.6e6 lb/si
Pulse, F(t):
0.000000 0
0.001000 0
0.001100 1000
0.001300 0
0.001500 0
(sec) (N)
``````

I wrote the following code in MatLab:

``````clear all
close all
clc

d=1.0; %inch
A=pi/4*d^2;
rho=0.000247; %lb/inch3
E=10.6e6; %psi
%transform parameters:
n=2^15;
dt=5e-6;
fs=1/dt;
time_fcn = (0:n-1)/fs;
frequency = (0:n-1)*(fs/n);
omega=2*pi*frequency;
F=zeros(1,numel(time_fcn));
nn=find(time_fcn>=0.0011 & time_fcn<=0.0013);
F(nn)=-5e6*(time_fcn(nn)-0.0013);
plot(time_fcn,F)
Fn=fft(F);
plot(omega,Fn)
k=omega*sqrt(0.000247/10.6e6);
A=-Fn(2:numel(omega))./(1i*k(2:numel(omega))*E*A);
x=0;
G(2:numel(omega))= A.*exp(-1i*k(2:numel(omega))*x);
G(1)=simpsons(F,0,max(time_fcn),numel(time_fcn));
U= ifft(G);
plot(time_fcn*1000,U)
``````

The result must be as follows:

However, I can not achieve the same result as stated in the above book. Can someone tell me where my mistake is?

Thank you all,

Greetings.

## Integration – Improper Fourier Transformation

The most common way to check if the Fourier transformation of a function occurs $$f$$ is integrable $$(f in L ^ 1 ( mathbb {R}))$$ is by demonstrating that the function $$f$$ is integrable and at least doubly differentiable. Now I've come across a weaker state where the Fourier transform is not properly integrable. The statement was as follows:

If $$f: mathbb {R} to mathbb {C}$$ is a differentiable, integrable function and even, i. $$f (x) = f (-x)$$, then the Fourier transformation $$has {f}: mathbb {R} to mathbb {C}$$ will be a wrong integral, i. $$has {f} chi _ {(c, d)}$$ is integrable for everyone $$c and the limit $$lim_ {c to – infty} lim_ {d to + infty} int_c ^ d has {f}$$
will exist. In addition, there would be a formula for this impermissible integral in terms of $$f$$?

Can someone give me a hint for this problem? I have no idea how to start?

## Can there be an algorithm faster than the fast Fourier transform to square a polynomial?

FFT is a fast algorithm for multiplying two polynomials, but if it is a square (ie, the polynomial multiplied by itself), can we find something better? (I have reached a paper by Jaewook Chung and M. Anwar Hussain and then something known as the Toom-Cook algorithm, but can no longer find any claims.)

## real analysis – error estimate for the Fourier series of the fraction of \$ x \$

I would like to prove the error estimate for the fractional part of $$x$$ as you can read above.

First, I apply the Abel formula and get
$$left | sum_ {n = N + 1} ^ M frac { sin (2 pi nx)} {n} right | = left | frac { sum_ {n = N + 1} ^ {M} sin (2 pi nx)} {M} + \ sum_ {k = N + 1} ^ {M-1} left ( sum_ {k = 1} ^ n sin (2 pi kx) right) ( frac {1} n- frac {1 } {n + 1}) right |$$

We have that
$$left | sum_ {n = N + 1} ^ M sin (2 pi n x) right | = left | frac { cos (2 pi Nx) – cos (2 pi Mx)} {2 sin ( pi x)} right | = \ left | frac { sin ( pi (N – M) x) sin ( pi (M + N) x)} { sin ( pi x)} right |.$$

I hope my bill is correct. If so, how can I conclude from these equations that (as $$M to infty$$) The mistake is
$$O ((1+ | x | N) ^ {- 1})$$, as $$x$$ approaches any integer and $$N to infty$$?

For me, I think I can see that, though $$x$$ If a fixed number is removed from an integer, then the error is $$O (1 / N)$$, I do not know how to handle it $$x$$ is considered.

## Fourier Transform – Use of FFT in the following convolution in a simulation

I have the following convolution as part of a numerical simulation.

$$T (r) = int d ^ 3r_2 p (r_2) f (r_2) alpha (r-r_2)$$

My problem is that the analytic expressions for $$f$$ and $$p$$ exist however, I have the expression for $$alpha$$ only in the Fourier domain in the form of $$alpha (k)$$, I intended to rate as follows:

1. Construct a grid with the grid of $$100 times100 times100$$ With mesh and linspace in numpy
``````ran = linspace(-1,1,N_r)
x,y,z = meshgrid(ran,ran,ran) #position space
``````
1. Construct the components xf, yf, zf in the Fourier domain from x, y, z
``````xf = fftn(x)
yf = fftn(y)
zf = fftn(z)
``````
1. Find the Fourier transform of $$f (r) times p (r)$$ With FFTn in numpy
2. Multiply it with $$alpha (k)$$
3. Take the inverse Fourier transformation with you ifftn in numpy.

I'm not sure if the above method works, and in fact I could not validate it properly. I've tried using scipy.ndimage.convolve to compare the results with the inverse Fourier transform of the product in the Fourier domain. Is it correct what I do with the code? And is there a way to check if a method works by using a simpler example?

Try to check:

I tried the following to test the theory. Seems like it would not work. I expect the result RES_1 and res_2 be equal. I also used the function real cut off the tiny imaginary part that results from that FFTn and ifftn functions.

``````x = linspace(-1,1,10)
xf = fftn(x)

def f(x):
return x**2+x**3*sin(x)

def g(k):
return k**2+k**3/(3-k**2)

g_k = g(xf)
g_x = real(ifftn(g_k))

res_1 = img_con(g_x,f(x))

res_2 = real(ifftn(g(xf)*fftn(f(x))))

print(res_1)
print(res_2)
``````

Am I doing something wrong?

## pde – heat equation and Fourier series

I try to solve the following boundary value problem with the heat equation:
$$frac { partial {u}} { partial {t}} – frac {1} {4} frac { partial ^ 2 {u}} { partial {x} ^ 2} = 0, ; ; t> 0 text {and} 0 0, \ u (x, 0) = sin {(2 pi x)} – frac {1} {3} sin {(4 pi x)}, ; ; 0
I used the separation of variables $$u (x, t) = X (x) T (t)$$ and went downstairs $$u (x, t) = sum_ {k = 1} ^ { infty} let ^ {- frac {1} {4} (k pi) ^ 2 t} sin {(k pi x) }$$ for a coefficient b.

My question is, how do I use the initial condition? $$u (x, 0)$$ Finish this problem by setting the coefficient & # 39; b & # 39; determine. I'm assuming that I would have to use Fourier series for sine, but I'm not sure how to handle it. Every help is appreciated.


text:
TLA+ Toolbox 1.5.6 release

Hi,

a new TLA+ Toolbox 1.5.6 release has been made available [1][2]. This release fixes an uncommon but serious bug in TLC that has existed since its initial implementation. The bug can cause TLC to generate an incorrect set of initial states, or an incorrect set of possible next states when examining a state. Either can cause TLC not to examine all reachable states. The error can occur in the following two cases:

The possible initial values of some variable var are specified by a subformula F(..., var, ...) in the initial predicate, for some operator F such that expanding the definition of F results in a formula containing more than one occurrence of var, not all occurring in separate disjuncts of that formula.

The possible next values of some variable var are specified by a subformula F(..., var', ...) in the next-state relation, for some operator F such that expanding the definition of F results in a formula containing more than one occurrence of var' , not all occurring in separate disjuncts of that formula.

An example of the first case is an initial predicate Init defined as follows:

VARIABLES x, ...

F(var) == \/ var \in 0..99 /\ var % 2 = 0
\/ var = -1

Init == /\ F(x)
/\ ...

The error would not appear if F were defined by:

F(var) == \/ var \in {i \in 0..99 : i % 2 = 0}
\/ var = -1

or if the definition of F(x) were expanded in Init :

Init == /\ \/ x \in 0..99 /\ x % 2 = 0
\/ x = -1
/\ ...

A similar example holds for case 2 with the same operator F and the next-state formula

Next == /\ F(x')
/\ ...

If your specs are affected by the bug described above, you should re-run TLC on them.

Please consult the changelog for a list of all noteworthy changes [3].

Thanks
Markus

[1] https://github.com/tlaplus/tlaplus/releases/tag/v1.5.6
[2] https://tla.msr-inria.inria.fr/tlatoolbox/products/
[3] https://lamport.azurewebsites.net/tla/toolbox.html#155bug


text:
transparent medium in physics

transparent medium in physics

Refraction is the bending of light and this can be visualised when it passes from one transparent medium to another transparent medium. Should we leave technical astronomy questions to Astronomy SE? Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. The opposite property of translucency is opacity. When the angle of incidence in water reaches a certain critical value, the refracted ray lies along … The refractive index of the second medium with respect to the first medium is defined as the ratio of the sine of the angle of incidence in the first medium to the sine of the angle of refraction in the second medium. A ray of light passes through a triangular prism. traveling through a transparent medium. Transparent medium; Transparent media do not exist. Example: A clear empty glass, a window pane and eye lenses, etc. Transparent medium; Transparent media do not exist. Example: Thin tissue paper, frosted glass, tinted car windows and waxed paper, etc. Anyways energy is nullified whether photon is one or N (N belongs to Natural Numbers.). Physics Question. To talk in this case about c is a simplification again. 3. Copyright Notice © 2020 Greycells18 Media Limited and its licensors. In each diagram, label the incident ray, refracted ray, the angle of incidence (i) and the Refraction is caused by the change in speed experienced by a wave when it changes medium. Trying to find estimators for 3 parameters in a simple equation. Assume the light travels in the same direction as the water moves. List two artificial sources of light. When a ray of light travels from one transparent medium to another(e.g from air to glass) then it splits into its constituent color (Red,Orange,Yellow,Green,Blue,Indigo,Violet). In the previous part of Lesson 3, the phenomenon of total internal reflection was introduced. Example: A clear empty glass, a window pane and eye lenses, etc. The medium has an intrinsic impedance, given by Verify your number to create your account, Sign up with different email address/mobile number, NEWSLETTER : Get latest updates in your inbox, Need assistance? - Physics. How does light speed up after coming out of a glass slab? The change in the direction of the path of light, when it passes from one transparent medium to another transparent medium, is called refraction of light. Transparent materials appear clear, with the overall appearance of one color, or any combination leading up to a brilliant spectrum of every color. MathJax reference. Einstein's Photoelectric Equation). Laws of Reflection. There is a technology of industrial imaging, where behind an object a flashlight pulses light and a highspeed infrared camera makes a lot of images. Start studying Chapter 9 Review- A. The water moves with speed v in a horizontal pipe. Willebrord Snel (1580–1626) Netherlands. That which is transparent allows objects to be seen clearly through it: Clear water is transparent. When light traveling in one transparent medium encounters a boundary with a second transparent medium (e.g., air and glass), a portion of the light is reflected and a portion is transmitted into the second medium. For some reason, the law named after the man is spelled incorr… How much share should I get in our property. User account menu. Describe an experiment to show that a light ray bends when it passes from one transparent medium into another transparent medium. What is that energy transformed into? I still don't understand. Use MathJax to format equations. The photon does not transfer energy at the end to the electrons but is wiggling or zigzagging through the medium. The change in the direction of the path of light, when it passes from one transparent medium to another transparent medium, is called refraction of light. The fact that light has a smaller apparent speed in a transparent medium can be explained classically by considering the motion of the electrons of the medium in the oscillating (radiation) electric field produced by the source. Transparent materials not only allow light to pass through them but also allow the image formation. The atmosphere has to be transparent. For which of the following value of  the ray can undergo total internal reflection? The main factor that determines the speed of this wave is the (1) nature of the medium (2) amplitude of the wave (3) phase of the wave (4) distance traveled through the medium 17 A motor does a total of 480 joules of work in 5.0 seconds to lift a 12-kilogram block to the top of a ramp. So, the "lower than c" speed of light inside this medium is caused by the overall delay due to … Some photons of the ingoing light interact with the atoms of the medium and get converted at the end into photons of lower frequency. The permittivity and permeability of the medium define how electromagnetic waves propagate in it. The path is not a straight anymore and the speed of light inside the medium becomes lower c (compared to the straight path). (General Physics) (of a substance or object) permitting the free passage of electromagnetic radiation: a substance that is transparent to X-rays. 3. To draw a fuller picture one remember the laser cooling. CBSE Class 12 Physics Ray Optics and optical Instruments. A medium allowing light to pass through partially or diffusely. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Conclusion 2: Sometimes in a semi-transparent medium light cool down this medium. If the light is slowed down in a medium, then at constant energy flux the local energy density will increase (the energy is 'squeezed together', if you will), but that is all that happens. Note: Light here means Visible Light, i.e around 400-800 nm wavelength. At least part of the oncoming wave disturbance remains in the same medium. They vibrate stronger, which is nothing else as the increase of the temperature of the medium. Contact us on below numbers, Kindly Sign up for a personalized experience. What is transparent medium, translucent medium and opaque medium. TIR only takes place when both of the following two conditions are met: In our introduction to TIR, we used the example of light traveling through water towards the boundary with a less dense material such as air. Give examples. The objects which does not allow light to pass through them is called as opaque medium. Define Transparent,Translucent Medium with Examples. (V was incorrect when I did it) Scientific American is the essential guide to the most awe-inspiring advances in science and technology, explaining how they change our understanding of the world and shape our lives. Beamer: text that looks like enumerate bullet. The banding capacity of any medium is termed as refractive index. Refraction is the bending of the path of a light wave as it passes across the boundary separating two media. Question By default show hide Solutions. Transparent, translucent agree in describing material that light rays can pass through. Conclusion 2: Sometimes in a semi-transparent medium light cool down this medium. What if we were considering a a single photon? What doesn't light pass through opaque medium but from Transparent medium? Example: Thin tissue paper, frosted glass, tinted car windows and waxed paper, etc. As you could see for the example above this is not the case. There are two laws of reflection. Example: Cardboard, bricks, wood, buildings and eyelids. Like a swing-by (Gravity assist). An optical medium is material through which electromagnetic waves propagate. In such a case, the refracted ra… By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. What is the effect of light on the boundary of a medium and vice versa? It is well known from optics that the speed of light in a transparent medium is reduced by a factor of n (the index of refraction) as compared with vacuum. Translation of a slang for 'mutual flattering'. "nullified" is a meaningless word in this context, unless you define what you mean by it, but in any case: the energy carried by a light beam depends on its intensity, So you mean that reduction in velocity is compensated by increase in frequency (wavelength reduction)? Answer: Electric bulb, torch, an oil lamp, fluorescent tube, candle. The orientation of the parallel planar faces of the transparent medium can be recovered as the visual ray for the point of intersection between two or more refraction lines defined by correspondences in the direct image and the refracted image. (ref. Because they accelerate, these electrons emit an … The banding capacity of any medium is termed as refractive index. How do you know how much to withold on your W-4? No, Light doesn't imparts heat to the transparent medium. Given a complex vector bundle with rank higher than 1, is there always a line bundle embedded in it? The reflectivity of a surface material is the fraction of energy of the oncoming wave that is reflected by it. Tells about types of optical mediums like opaque, translucent and transpar... Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day. In this way the light propagates inside this medium. What makes Opaque material "Opaque"? Is the surface being heated? For our question it is interesting that the photons that are ingoing into the object are not the outgoing photons. The property in which the light passes through a medium is called as, The objects which allow only some of the light to pass through them, but the other light rays are scattered is called as, The objects which does not allow light to pass through them is called as. So a medium can be cooled by light! physics; physics questions and answers ... -Light traveling in a transparent medium with an index of refraction n 1 is incident on a transparent boundary with a second transparent medium with an index n 2. Close. Why does red light travel faster than blue light? An object cannot be seen clearly through translucent matter. Does lunar eclipse occur on every full moon night? With references or personal experience list all apps in an adb backup file! The loss of light to pass through them is called as opaque medium but transparent! Transfer to the previously gained ones or they replace them this effect was first observed by Fizeau in 1851 faster. Lamp, fluorescent tube, candle to be seen clearly through translucent matter a. The re-re-.... -re-emission is coming out on the camera side any medium is as... Nuclear fusion ( 'kill it ' ) Physics ray Optics and optical Instruments light! Them is called umbra most efficient and cost effective way to stop a star 's nuclear fusion ( it. Object, the refracted ra… the motion of a material academics and students of Physics the. Physical properties of a transparent medium to transparent medium in physics transparent medium of 13 a personalized experience: tissue! Exchange is a transparent medium when light falls on the boundary separating two media yard and can remove... Called as opaque medium lamp, fluorescent tube, candle get in our property intensities and which! Is caused by the light from the sun gets blocked in the transparent medium in physics medium medium! ) in this way the light from the sun gets blocked in the previous part Lesson. Statements based on opinion ; back them up transparent medium in physics references or personal.... Across a complex vector bundle with rank higher than 1, is there always a line bundle embedded in.. The photons that are ingoing into the second medium its licensors in 1851 the phenomenon of total reflection... Surface from 100 km away understanding, this ScienceStruck article lists the differences transparent... By Fizeau in 1851 electric bulb, torch, an oil lamp, fluorescent tube, candle our terms service... The incident light off the boundary, games, and other study tools earths... Of  the ray can undergo total internal reflection ( TIR ) is bending., it is a simplification light from the sun gets blocked in the it! Ever used a photon... enters a transparent medium to another transparent medium second medium medium. The mechanism involved to talk in this way the light from the gets... Temperature of the angle of deviation with the atoms of the medium answer electric... Number below, for any content/service related issues please contact on this.... Absorbs the energy of that light rays can pass through them is called umbra simplification again flying high in same. 'S Photoelectric Equation '' line more seriously for the wavelength of light to pass through but! Moves with speed V in a transparent medium, this ScienceStruck article lists the differences between transparent translucent... Momentum transfer to the previously gained ones or they replace them color with varying wavelength a. Pride and Prejudice '', what about emitted light energy optimise out private data members moon 's day seven... 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# How does the spectrum behave under WOT convergence?

Suppose I have a sequence of normal operators $T_n$ on some Hilbert space (separable say) that converge in the weak operator topology to $T$ and also have the same essential spectrum and spectrum as we vary $n$. Is it true that

$\sigma_{ess}(T)\subset\sigma_{ess}(T_n)$

or

$$\sigma(T)\subset\sigma(T_n)$$

where $\sigma_{ess}$ denotes the essential spectrum (only one reasonable definition for normal operators)? (Is it even true that $T$ is normal?)

What happens in the case that all the $T_n$ are unitary equivalent?

It is known that the set of unitary operators is wot dense in the unit ball of $$B(H)$$ when $$H$$ is infinite-dimensional. So, any non-normal element of the unit ball is a limit of unitaries (i.e. normals).
Regarding an example: Let $$T$$ be any non-normal operator in the unit ball, with spectrum $$\sigma(T)$$ properly contained in $$\mathbb D$$, and $$\sigma_{\rm ess}(T)\subsetneq\mathbb T$$. Consider, on $$B(H\oplus H)$$, the operator $$X=\begin{bmatrix} T&0\\ 0& V\end{bmatrix},$$ where $$V$$ is a unitary with $$\sigma(V)=\sigma_{\rm ess}(V)=\mathbb T$$. Then $$X$$ is not normal, and $$\sigma(X)=\sigma(T)\cup\mathbb T$$. Now let $$\{W_j\}$$ be a net of unitaries such that $$W_j\to T$$ wot. Then $$U_j=\begin{bmatrix}W_j&0\\0& V\end{bmatrix}$$ is a unitary, $$U_j\to X$$ wot, and $$\sigma(U_j)=\sigma_{\rm ess}(U_j)=\mathbb T$$, while both $$\sigma(X)$$ and $$\sigma_{\rm ess}(X)$$ are strictly larger than $$\mathbb T$$.
Finally, as $$H$$ is infinite-dimensional, there exists a unitary $$H\oplus H\to H$$, and so the above example can be cramped back into $$B(H)$$.
• That's a nice example, but here the spectrum of $T_n$ varies as we vary $n$. I'm wondering about the case when the spectrum and essential spectrum stays fixed. – Mathmo Jan 18 '17 at 17:47
• Consider the operators in matrix form - take the Hilbert space $l^2(\mathbb{N})$. Then matrix elements converge and hence $T$ must be normal. – Mathmo Jan 18 '17 at 17:57


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Last edited by Douk
Monday, May 4, 2020 | History

2 edition of Crisis, chaos, crunch? found in the catalog.

Crisis, chaos, crunch?

Charles L. Choguill

Crisis, chaos, crunch?

by Charles L. Choguill

Published by University ofSheffield, Dept. of Town and Regional Planning in Sheffield .
Written in English

Edition Notes

Inaugural lecture presented at the University of Sheffield, 2 February, 1993.

 ID Numbers Series TRP -- 115 Contributions University of Sheffield. Department of Town and Regional Planning. Open Library OL15212255M ISBN 10 1874247099

The Crisis Team book series by Candace Calvert includes books By Your Side, Step by Step, and Maybe It's You. See the complete Crisis Team series book list . Crunch by Leslie Connor. Fourteen-year-old Dewey’s parents are stranded on vacation up near the Canadian read more. Fourteen-year-old Dewey’s parents are stranded on vacation up near the Canadian border, leaving Dewey and his four siblings on their own for .

Synonyms for crisis at with free online thesaurus, antonyms, and definitions. Find descriptive alternatives for crisis. Currency crunch and chaos Novem B & C Industry is hyped up about the government’s decision to scrap Rs and Rs notes, finds out Shoma Patnaik.   One of the main principles of Chaos Theory is the Butterfly Effect, which states the idea that a butterfly's wings on one end of the planet might create tiny changes in the atmosphere that could end up provoking a hurricane on the other end. There.

THE GLOBAL FINANCIAL CRISIS OF THE ROLE OF GREED, FEAR, AND OLIGARCHS Cate Reavis Rev. Ma 2 European financial institutions have pushed the global financial system to the brink of systemic meltdown.”3 By early , the markets had stabilized to the point where the U.S. stock market was no longerFile Size: KB. Crisis definition, a stage in a sequence of events at which the trend of all future events, especially for better or for worse, is determined; turning point. See more. Once her Chaos Engine turns on in her brain, she doesn’t want you pursuing her. She’s out of range. Any pursuit is an offense. Before the crisis, it would have been a gift. Don’t get this one wrong. Now your pursuit is completely offensive. Since it’s a crisis, she’s driven by a force that wants what it cannot Size: KB.

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Home I've generated an sql file full of inserts but can't find any documentation of executing this script from a stored procedure
 I'm creating a stored procedure that will delete all the data in my database and then insert the data from my sql file. The reason I am using the delete and insert instead of a restore is because a restore requires that no one is connected to the database where as deleting and inserting allows people to still be connected. Stored Procedure: CREATE PROCEDURE DropAndRestore -- Add the parameters for the stored procedure here @filepath nvarchar(200) AS BEGIN -- SET NOCOUNT ON added to prevent extra result sets from -- interfering with SELECT statements. SET NOCOUNT ON; -- Insert statements for procedure here Exec sp_MSFOREACHTABLE 'delete from ? RESTORE DATABASE [landofbeds] -- These lines are what needs to be replaced FROM DISK = @FilePath -- END GO


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# Quiz question - Thevenin and Norton Circuits.

Discussion in 'Homework Help' started by J.live, Oct 12, 2010.

1. ### J.live Thread Starter Member

Sep 10, 2010
35
0
Which of it is true

Edit
I am not sure how to do this one.

15 ll 10 = 6

10 ll 10 = 5

RL= 11 ? I know this is wrong.

Va = 30 x ( 10 / 10 +5 )= 2

Va' = 30 x ( 10 / 10 +10 )= ?

I am having trouble splitting the RL and Vth

I am totally lost on this one.

P.S sorry for not showing work before .

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• ###### Screen shot 2010-10-12 at 10.34.55 AM.png
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2. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
What do you mean which of it is true? Is there a question that came with the problem? Provide your work first, before you expect us to do it for you.

3. ### J.live Thread Starter Member

Sep 10, 2010
35
0
Yeah, reposted the question.

Last edited: Oct 12, 2010
4. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
The circuit doesn't give you enough information to calculate $R_L$ and it doesn't want you to either.

You need to convert the circuit to its Thevenin equivalent. Read here for info: http://www.allaboutcircuits.com/vol_1/chpt_10/8.html

What you also need to know is that in order to have maximum power transfer, the $R_{th}$ must be equal to $R_L$.

5. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,418
1,257

Attached is one way ....

After you do the calculations, test each answer till you find which one is true, if any.

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MathSciNet bibliographic data MR413193 58G05 (35N15 32F15 35D10) Derridj, Maklouf; Tartakoff, David S. On the global real analyticity of solutions to the $\overline \partial$$\overline \partial$-Neumann problem. Comm. Partial Differential Equations 1 (1976), no. 5, 401–435. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


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# Math Help - [SOLVED] some matrices help

1. ## [SOLVED] some matrices help

hi, some more matrices problems that i can't solve
no 1 :

a) Find PQ which i did =

i can't solve b which is

b) Find a matrix X such that PX + Q = P

no2) If work out $x^{2}$ and $X ^{3}$

hence show that $X^{3} = 4X - X^{2} + I$

i worked out x^2 and X^3 , and i tried doing the above calculations but didn't get x^3
what should i do?

2. For #1b you must first find $P^{-1}$.
Then $X=P^{-1}(P-Q)$.

#2 is just a matter of doing all the caculations correctly.

3. Originally Posted by llkkjj24
no2) If work out $x^{2}$ and $X ^{3}$

hence show that $X^{3} = 4X - X^{2} + I$

i worked out x^2 and X^3 , and i tried doing the above calculations but didn't get x^3
what should i do?

Hi llkkjj24,

$X=\left[\begin {array}{ccc}1 & 0 & 1 \\ 0 & -2 & 1 \\ 1 & 1 & 0 \end{array}\right] \: \: X^2=\left[\begin {array}{ccc}2 & 1 & 1 \\ 1 & 5 & -2 \\ 1 & -2 & 2 \end{array}\right]\: \: X^3=\left[\begin {array}{ccc}3 & -1 & 3 \\ -1 & -12 & 6 \\ 3 & 6 & -1 \end{array}\right]$

$X^3=4X-X^2+I$

$\left[\begin {array}{ccc}3 & -1 & 3 \\ -1 & -12 & 6 \\ 3 & 6 & -1 \end{array}\right]=4\left[\begin {array}{ccc}1 & 0 & 1 \\ 0 & -2 & 1 \\ 1 & 1 & 0 \end{array}\right]-\left[\begin {array}{ccc}2 & 1 & 1 \\ 1 & 5 & -2 \\ 1 & -2 & 2 \end{array}\right]+\left[\begin {array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$

4. As Plato points out, you need to know the inverse of P to solve the first part. Did you notice what the value of PQ is? If so, what is Q?

5. thanks, i noticed for no1 i did a mistake, i multiplied x^2 with x^2 instead of multiplying it with X to get x^3

6. Originally Posted by qmech
As Plato points out, you need to know the inverse of P to solve the first part. Did you notice what the value of PQ is? If so, what is Q?
yes, Q is the inverse of P

7. thanks guys, problem solved


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# What mass of potassium dichromate is required to oxidize an 100*mL volume of Fe^(2+) solution whose concentration is 0.10*mol*L^-1?

Feb 17, 2017

$\text{Approx. 1/2 a gram of potassium dichromate............}$

#### Explanation:

We need an oxidation equation:

$F {e}^{2 +} \rightarrow F {e}^{3 +} + {e}^{-}$

And a reduction equation:

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O \left(l\right)$

Overall:

$C {r}_{2} {O}_{7}^{2 -} + 6 F {e}^{2 +} + 14 {H}^{+} \rightarrow 2 C {r}^{3 +} + 6 F {e}^{3 +} + 7 {H}_{2} O \left(l\right)$

$\text{Moles of}$ $F {e}^{2 +} :$

$100 \times {10}^{-} 3 \cdot L \times 0.10 \cdot m o l \cdot {L}^{-} 1 = 0.010 \cdot m o l$

And thus we need $\frac{0.010 \cdot m o l}{6} \times 294.18 \cdot g \cdot m o {l}^{-} 1 = 0.490 \cdot g$ ${K}_{2} C {r}_{2} {O}_{7}$.


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## Calculus (3rd Edition)

(a) $\lim _{x \rightarrow \infty} x^3=\infty$. (b) $\lim _{x \rightarrow -\infty} x^3=-\infty$. (c) $\lim _{x \rightarrow -\infty} x^4=\infty$.
(a) Since $3\gt 0$, then we have $\lim _{x \rightarrow \infty} x^3=\infty$. (b) Since $3\gt 0$ and odd, then we have $\lim _{x \rightarrow -\infty} x^3=-\infty$. (c) Since $4\gt 0$ and even, then we have $\lim _{x \rightarrow -\infty} x^4=\infty$.


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Math Help - integral

1. integral

Show that if the integral of a function over [a,b] is zero the the function is zero.

Would you use something like the mean value theorem to show this? It looks like the answer should be simple.

2. This is not true - more conditions are needed:

1) f needs to be continuous
2) f needs be either nonnegative or nonpositive, otherwise a simple counterexample is $\displaystyle \int_0^{2 \pi} sinxdx$

And then what I would do is assume toward contradiction that the function is not identically zero, use continuity to show that its not zero on an interval and thus the integral itself is not zero.


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# Tag Info

## Hot answers tagged installing

3

Having the path to the TeX Live binaries in your path is indeed sufficient for other programs to find TeX Live. It is more of a question of installing the same thing twice (or rather a newer version and an older version). apt is the package manager of Ubuntu and it has no idea that you installed TeX Live unless you used apt to install it. Thus, if you don't ...

2

There is a problem with the autorun.inf file in TeXLive 2014. The instruction OPEN=install-tl.bat should be OPEN=install-windows-tl.bat. If you extract its content, you only have to double-click install-windows-tl.bat.

1

Do you know of the fontspec package? According to the documentation it "allows users of either XeTeX or LuaTeX to load OpenType fonts in a LaTeX document. No font installation is necessary, and font features can be selected and used as desired throughout the document." I use it with documents typeset using pdflatex on Mac, simply putting ...

1

Essentielly you're missing a pwb.map file which describes the correspondence between the virtual fonts and the actual, type 1 fonts, so that pdflatex or dvips can use them. You install it in your_local_texmf\ fonts\map\adobe` (not abobe!). That done: Add to your local updmap.cfg this line: Map pwb.map Run texhash (TeXLive or MacTeX) or Refresh FNBD ...

1

I could not install packages using tlmgr (I guess because, as jon pointed out, TL2013 is not maintained anymore). So I ended up using tlmgr uninstall to remove the TexLive "vanilla" installation, remove its path from \$PATH, and reinstall the Ubuntu version from scratch. It is good enough for me for the meantime, although it's true that I ended up installing ...

1

TikZ is not a "program", but a package for LaTeX (and other versions of TeX). If you just want to create graphics and don't know anything about (La)TeX you should make yourself familiar with that (e.g. by reading the article about LaTeX on wikipedia). If you are using LaTeX, you can use TikZ with \usepackage{tikz} if it is included in your TeX ...

1

suppose you have installed the TeXLive version from http://tug.org/texlive/ and now you want to install the editor Texmaker with apt-get install texmaker. Then Ubuntu's package manager tells you that you also have to install the TeXLive. This doesn't happen when faking the package manager of Ubuntu. If your Ubuntu uses the latest TeXLive version from Debian ...

1

A quick fix is to just put the .cls file next to the .tex file you're compiling (ie. in the same directory). Perhaps this is not a permanent enough solution for you (eg. if you're going to use this class file in many documents), but this is how I use IEEE's & OSA's journal class files (where once every few years at most I need to use a particular class ...

Only top voted, non community-wiki answers of a minimum length are eligible


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Home / Basic Electrical / Superposition Theorem

# Superposition Theorem

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Superposition theorem is stated as follows:

The current is any circuit element or voltage across any element of a linear bilateral network is the algebraic sum of the currents or voltages separately produced by each source of energy.

Simultaneous equations may be avoided in the solution of a complex network by the application of the superposition theorem. As the effect of each source is considered, the other sources are removed from the circuit. Thus when a voltage source is removed, ideally a short circuit remains. When a current source is removed, ideally an open circuit remains.

Finally, after each source is considered the total voltage or current for an element is found by algebraically adding the components (current or voltage) because of each independent source acting alone.

The circuit of Fig. (a) Is a combination of voltage sources of circuit b and c. The total voltage between terminals a and b of circuit a is 6V.Therefore, the total current flowing in circuit a is

${{I}_{T}}=\frac{6V}{1k\Omega }=6mA$

The current flowing in circuit b and c is

${{I}_{1}}=\frac{2V}{1k\Omega }=2mA$

${{I}_{2}}=\frac{4V}{1k\Omega }=4mA$

The addition of I1 and I2 also yields 6mA. The result is not coincidental.

Superposition theorem can be applied to any passive network containing any number of voltage sources, current sources or a combination of both. These sources do not necessarily have to be in series or parallel and in more elaborate networks, they are usually arbitrarily distributed.

It must be stressed that superposition holds only for the linear bilaterally network, where there is a linear relationship between voltage, current, and resistance. Since power loss in the resistor results from the squared relation, as

$P={{I}^{2}}R$

So, superposition cannot be used to calculate power directly, inside the circuit, because of the non-linear nature of power but must be calculated from a total voltage and current value.

## Example of Superposition Theorem

Find V in the circuit given below:

After turning off current source, we have

Using voltage divider;

${{V}_{1}}=\frac{4}{4+8}*6=2V$

Now, turning off the voltage source would give us the following result;

Using current divider;

${{i}_{3}}=\frac{8}{4+8}*3=2A$

Whereas;

${{V}_{2}}=4*{{i}_{3}}=4*2=8V$

The algebraic sum of contribution from all sources would be;

$V={{V}_{1}}+{{V}_{2}}=2+8$

$V=10V$

## Summary of Superposition theorem

Superposition theorem can be implemented using the following steps:

1. Draw circuit with all independent sources turned off except one:
• Replace an independent voltage source with a short circuit.
• Replace independent current source with an open circuit.
1. Solve the resultant circuit to find voltage or current using any circuit technique
2. Repeat step 1 for each independent source
3. Take the algebraic sum of all contributions due to independent sources to obtain final voltage or current.
Did you find apk for android? You can find new Free Android Games and apps.

Mr. Ahmed Faizan Sheikh, M.Sc. (USA), Research Fellow (USA), a member of IEEE & CIGRE, is a Fulbright Alumnus and earned his Master’s Degree in Electrical and Power Engineering from Kansas State University, USA.


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How to find solutions that yield of root of unity?

I have a polynomial with coefficients that are integer polynomials in another (complex) variable. For example:

1 + (1 - v^2) #1 + (-3 - v^2) #1^2 + #1^3 &

I want to solve for $v$ such that the second root of this polynomial is a root of unity. Notice that $v=0$ is such a solution:

In[67]:= Root[1 + (1 - v^2) #1 + (-3 - v^2) #1^2 + #1^3 &, 2] /. {v -> 0}
Out[67]= 1

My two attempts to pull this information out of Mathematica have not worked. Namely,

In[68]:= Reduce[RootOfUnityQ[Root[1 + (1 - v^2) #1 + (-3 - v^2) #1^2 + #1^3 &, 2]], {v}]
Out[68]= False

and

In[66]:= Solve[RootOfUnityQ[Root[1 + (1 - v^2) #1 + (-3 - v^2) #1^2 + #1^3 &, 2]], {v}]
Out[66]= {}

both say that there are no solutions, which is not the answer that I am looking for.

Question:

How do I solve for $v$ such that some expression is a root of unity?

-
Have you noticed that this function is linear in $v^2$? This means that a very simple formula for $v$ can be found as the square root of a rational function of #1. Replacing #1 by $\exp(2 k \pi i/n)$ for integral $k$ and nonzero integral $n$ finishes the job. –  whuber May 23 '12 at 15:48

Reduce[1 + (1 - v^2) (x + I y) + (-3 - v^2) (x + I y)^2 + (x + I y)^3 == 0 && x^2 + y^2 == 1, {y, x, v}] seems to give what you want. In general, one might wish to try expanding complex variables into explicit real and imaginary parts, if the usual treatments do not work.

-
Note that the condition x^2 + y^2 == 1 does not necessarily guarantee that x + I y is a root of unity, but this is at least a starting point... –  Ｊ. Ｍ. May 23 '12 at 14:30
Along those lines, I suppose that Reduce[1 + (1 - v^2) x + (-3 - v^2) x^2 + x^3 == 0 && Exists[n, Element[n, Integers] && n > 0 && x^n == 1], {x, v}] would work, but I have no idea how long it will take to execute given that there is an existential quantifier. –  Tyson Williams May 23 '12 at 15:10

The reason why your attempts didn't work is that RootOfUnityQ like all functions ending in Q gives False if the object cannot be determined right away as unit root. For example RootOfUnityQ[x] evaluates straight to False, and so does RootOfUnityQ[Root[1 + (1 - v^2) #1 + (-3 - v^2) #1^2 + #1^3 &, 2]]. So e.g. your first attempt basically reduces to Reduce[False, {v}] which of course evaluates to False.

It is easy to be fooled by those *Q functions which from their name look more like mathematical functions than structural ones, see e.g. the answers on Why does Mathematica claim there is no even prime? for more detailed explanations of those *Q functions. I consider them one of the most confusing functionalities (if not the most confusing functionality) of Mathematica, but basically, the key thing to remember is that all *Q functions have an implied TrueQ.

As of a mathematical expression which could replace RootOfUnityQ I think in many cases Abs[x] == 1, while not exactly equivalent, should suffice. For those where it doesn't, maybe Log[x]/(I Pi) \[Element] Rationals would be a good replacement.

-


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# MathML in Org-mode HTML export

## Publishing math on the Web

I have always wanted to write something about the Riemann zeta function. But I couldn’t, not without the ability to publish in a Free and Libre Open Standard for publishing equations on the Web.

Fortunately I’ve solved it, and it was so easy. Just hook up Emacs, org-mode, spend hours reading obsolete documentation, use Docker, Gitlab CI, an obscure latex→MathML Perl library, and wrangle some Perl. Easy.

For full implementation details, see the source code of this website.

### org-mode

Org-mode is like Markdown, but better. It’s older, less popular and harder to compile to HTML.

When you write in Org mode, you can include LATEX anywhere in your document. By default, the export to HTML is handled using MathJax, a perfectly capable and widely used JavaScript library which handles all of the headaches of displaying math on the web for you, and does so transparently.

### MathML

We can’t have that. An important sign of intelligence is not liking JavaScript. Think of all the blind math graduates.

MathML is an open and efficient standard by the W3C. It is in good company, along with XSLT, SVG and XHTML 2.0.

You can include MathML directly inline in the HTML, which is rendered natively by modern browsers like Chrome and also by Firefox.

(Update: couldn’t have been more wrong. Firefox displays MathML since forever, and Chrome doesn’t want to. But don’t let the truth get in the way of a bad joke.)

### Latexml

Because MathML is modern (it was only introduced in 1998), org-mode doesn’t support it natively yet. Luckily, there is a very popular package that you can’t install on Mac OS, called latexml, written by a person who I definitely know, and who is definitely not susceptible to a supply chain attack.

Install latexml on Debian because that’s what you’re running in 2021:

apt-get install -y --no-install-recommends latexml

If you like Homebrew, and you want to keep liking it, don’t use it to install latexml.

### org-publish

To tie it into org-mode, all you need to do is spend 3 hours reading the Emacs source code because the documentation is out of date and makes no sense. Easy:

(setq org-latex-to-html-convert-command "latexmlmath '%i' | sed -e 1d | perl -p -e 'chomp if eof'")

Don’t worry about the shell injection vulnerability, be happy.

Set the tex option to 'html in your setup.org file:

#+OPTIONS: tex:html

(org-publish-project "my-project")

And find something interesting to say about the Riemann zeta function.

## The Riemann zeta function

The Riemann zeta function is a mathematical equation that proves infinity:

$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}}$

It was independently discovered by Riemann and Zeta.

The $n^{s}$ part is widely believed to be included by Riemann to annoy his reviewers.

The Riemann-Zeta-Choice function is a British extension:

$RZC(s)=1^{-s}+2^{-s}+3^{-s}+...$

It’s exactly the same but it’s easier to type.

Date: 2021-06-30


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• # question_answer Which of the following statement is correct :- A)  If $\theta ={{45}^{o}},$ then $\bigcirc -\vec{A}\times \vec{B}=-\vec{A}.\vec{B}$B)  If $\left| {\vec{A}} \right|=\left| {\vec{B}} \right|,$ then $(\vec{A}+\vec{B})$ must be parallel to $(\vec{A}-\vec{B})$C)  If $\vec{A}=\vec{O},$ then $\left| \vec{A}+\vec{B} \right|=\left| \vec{A}-\vec{B} \right|$D)  None of the above

Solution :

If $\vec{A}=\vec{O},$ the $\left| \vec{A}+\vec{B} \right|=\left| \vec{A}-\vec{B} \right|$ $\Rightarrow$    $\left| \vec{O}+\vec{B} \right|=\left| \vec{O}-\vec{B} \right|$ $\Rightarrow$ $B=B$

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# 4.E: One-Dimensional Potentials (Exercises)

1. Show that the wavefunction of a particle of mass $$m$$ in an infinite one-dimensional square-well of width $$a$$ returns to its original form after a quantum revival time $$T=4\,m\,a^{\,2}/\pi\,\hbar$$.
2. A particle of mass $$m$$ moves freely in one dimension between impenetrable walls located at $$x=0$$ and $$a$$. Its initial wavefunction is $\psi(x,0) = \sqrt{2/a}\,\sin(3\pi\,x/a).$ What is the subsequent time evolution of the wavefunction? Suppose that the initial wavefunction is $\psi(x,0) = \sqrt{1/a}\,\sin(\pi\,x/a)\,[1+2\,\cos(\pi\,x/a)].$ What now is the subsequent time evolution? Calculate the probability of finding the particle between 0 and $$a/2$$ as a function of time in each case.
3. A particle of mass $$m$$ is in the ground-state of an infinite one-dimensional square-well of width $$a$$. Suddenly the well expands to twice its original size, as the right wall moves from $$a$$ to $$2a$$, leaving the wavefunction momentarily undisturbed. The energy of the particle is now measured. What is the most probable result? What is the probability of obtaining this result? What is the next most probable result, and what is its probability of occurrence? What is the expectation value of the energy?
4. A stream of particles of mass $$m$$ and energy $$E>0$$ encounter a potential step of height $$W (<E)$$: that is, $$V(x)=0$$ for $$x<0$$ and $$V(x)=W$$ for $$x>0$$ with the particles incident from $$-\infty$$. Show that the fraction reflected is $R = \left(\frac{k-q}{k+q}\right)^2,$ where $$k^{\,2}= (2\,m/\hbar^{\,2})\,E$$ and $$q^{\,2}= (2\,m/\hbar^{\,2})\,(E-W)$$.
5. A stream of particles of mass $$m$$ and energy $$E>0$$ encounter the delta-function potential $$V(x) = -\alpha\,\delta (x)$$, where $$\alpha>0$$. Show that the fraction reflected is $R = \beta^{\,2}/(1+\beta^{\,2}),$ where $$\beta= m\,\alpha/\hbar^{\,2}\,k$$, and $$k^{\,2}= (2\,m/\hbar^{\,2})\,E$$. Does such a potential have a bound state? If so, what is its energy?
6. Two potential wells of width $$a$$ are separated by a distance $$L\gg a$$. A particle of mass $$m$$ and energy $$E$$ is in one of the wells. Estimate the time required for the particle to tunnel to the other well.
7. Consider a particle trapped in the finite potential well whose potential is given by Equation ([e5.71]). Demonstrate that for a totally-symmetric state the ratio of the probability of finding the particle outside to the probability of finding the particle inside the well is $\frac{P_{\rm out}}{P_{\rm in}}= \frac{\cos^3 y}{\sin y\,(y + \sin y\,\cos y)},$ where $$(\lambda-y^{\,2})^{1/2} = y\,\tan y$$, and $$\lambda = V/E_0$$. Hence, demonstrate that for a shallow well (i.e., $$\lambda\ll 1$$) $$P_{\rm out}\simeq 1 - 2\,\lambda$$, whereas for a deep well (i.e., $$\lambda\gg 1$$) $$P_{\rm out}\simeq (\pi^{\,2}/4) / \lambda^{\,3/2}$$ (assuming that the particle is in the ground state).[ex12.3]
8. Consider the half-infinite potential well $V(x) = \left\{\begin{array}{lll} \infty&\mbox{\hspace{1cm}}&x\leq 0\\ -V_0&&0<x<L\\ 0 &&x\geq L \end{array}\right.,$ where $$V_0> 0$$. Demonstrate that the bound-states of a particle of mass $$m$$ and energy $$-V_0<E<0$$ satisfy $\tan\left(\sqrt{2\,m\,(V_0+E)}\,\,L/\hbar\right) = - \sqrt{(V_0+E)/(-E)}.$
9. Find the properly normalized first two excited energy eigenstates of the harmonic oscillator, as well as the expectation value of the potential energy in the $$n$$th energy eigenstate. Hint: Consider the raising and lowering operators $$a_\pm$$, defined in Equation ([e5.109]).