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# Can we supervise on the hidden states of RNN?

I'm trying to generate some history-dependent model with machine learning, whose underline physical model has a clear definition of its "internal state variable" (a state derived from historical inputs) and how this variable interacts with the inputs to get the outputs. Mathematically this reads: $$y_t=f(x_t,h_t)$$, $$h_t=g(x_1, ..., x_{t-1})$$ .

Intuitively, this can be realized by RNN(or LSTM/GRU). Now I wonder if we can add the supervision on state variables to the loss function to improve prediction. For instance, we extract the hidden states $$H$$ output from our RNN and pass them through another branch of feed-forward network to predict the internal state variables $$h$$. The loss function then has two parts, one is for model outputs $$\sum (y-\hat{y})^2$$, and the other is for model internal states $$\sum(h-\hat{h})^2$$.

Any insight toward the following questions are welcomed:

• Will this approach work? (Does it make sense?)
• Potential issues?

I'll also be really grateful if anybody can indicate closely related works in the answers! Thanks in advance!

There are several examples of performing supervision on hidden states for history dependent models.

Latent variables

Loss functions are commonly applied to internal states such as latent variables in variational autoencoders (VAEs).

Hidden states

Supervision directly on hidden states, as opposed to latent states in RNNs, has been performed for imposing sparsity, as in [1]:

In order to impose sparsity, we introduce an L1-norm loss term over the output gate at layer $$l$$ as

$$\mathcal{L}^S(o^l(x,\theta)) = \sum_{1 \leq b \leq B} \sum_{1 \leq > t\leq T} | o_t^l(x^{(b)})|.$$

Autoregressive Neural Networks

Additional, auxilliary latent variables have been used to increase the flexibility of variational inference. An example of this is given in [2] where hidden state $$\mathbf{h}$$ and latent state $$\mathbf{z}_{t-1}$$ are fed into an autoregressive neural network, with final output $$\mathbf{z}_t$$.

# Reference

1 Learning Sparse Hidden States in Long Short-Term Memory, (paper)

2 Improved Variational Inference with Inverse Autoregressive Flow, (arxiv)

Your physical model looks dubious. It is a bit unusual and might not be a perfect match for a RNN/LSTM. I am hoping that your physical model is actually $$y_t=f(x_t,h_t)$$, $$h_t=g(x_{t-1},h_{t-1})$$. If so, then this is a good fit for the kind of supervision on internal states that you plan to do -- and the answer to your question is yes, you can do this, and it is a sensible thing to do.


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# Leave-future-out cross-validation for time-series models

I’m having a hard time making sense of the write-ups on LFO cross-validation found on the Stan website (http://mc-stan.org/loo/articles/loo2-non-factorizable.html) and on arxiv (https://arxiv.org/pdf/1902.06281.pdf), both of which are entitled “Approximate leave-future-out cross-validation for Bayesian time series models.” The Stan-website writeup (in the section “M-step-ahead predictions”) assumes that we have a factorizable model, i.e., the values y[t], 1 <= t <= T, have independent distributions conditional on the model parameters – but this is never true for any but trivial time-series models! The whole point of having a time series model is that past values are informative about future values.

The arxiv writeup correctly notes that “Most time series models have a non-factorizable likelihood”, but then equation (10) treats the likelihood as factorizable anyway. Furthermore, it appears to have the formula for the raw importance ratios inverted – compare equation (10) in the arxiv write-up with the second equation in section “Approximate M-SAP Using Importance Sampling” in the Stan-website writeup.

Can the authors, or anyone else who understands the paper, help me out here?

3 Likes

The link you provided for Stan website has title " Leave-one-out cross-validation for non-factorizable models". Did you mean to link to http://mc-stan.org/loo/articles/loo2-lfo.html which has title " Approximate leave-future-out cross-validation for Bayesian time series models" ?

If we have a model p(y_i|f_i,\phi) and joint time series prior for (f_1,...,f_T) then p(y_i|f_i,\phi) can be considered independent given f_i and \phi and likelihood is factorizable. This is true often and the past values are informative about future values, but conditionally we know f_i, the past values are not providing additional information. This should not be confused with that when we don’t know f_i and integrate over the posterior of (f_1,...,f_T), then y_i are not independent anymore. Also they are not anymore exchangeable as we have the time ordering telling additional information. In cross-validation M-step ahead prediction is more about the usual interest in predicting future and evaluating the time series model for (f_1,...,f_T), but leave-one-out cross-validation is valid for assessing conditional part p(y_i|f_i).

Thanks, we’ll fix this to not say that.

Thanks we need to fix the vignette. We first implemented leave-future-out by fitting the model with all data, and removing one by one observations, but after an anonymous hint, we realized that it’s better to fit first the minimal model and add one by one observations as then the proposal tends to have thicker tails than the target.

Tagging @paul.buerkner so the he also sees what we need to fix.

I hope this helps

2 Likes

Aki: You’re right, I included the wrong link, I meant the one you mention.

If we have a model p(y_i | f_i, ϕ) and joint time series prior for (f_1,\ldots,f_T)

Oh, so for, say, a SSM you’re including the sequence of latent states as part of the parameter vector \theta? I would suggest you make that explicit, as this runs counter to the terminology usually used in the SSM literature. Anyway, thanks, that clarifies things a lot.

Some more comments on approximate LFO CV:

• The autoregressive model discussed as an example for the online writeup (Water Level in Lake Huron) is non-factorizable: there is nothing that plays the role of f_i in your explanation, and y_{i+n} for n>0 is dependent on y_{i-1},\ldots,y_{i-p}.
• Do you really need factorizability if you can efficiently compute p(y_{j+1}\mid y_1,\ldots,y_j,\theta) for all j, as is the case for AR(p) models, and for linear SSMs via the Kalman filter?

Thanks for the clear questions, these will help to improve the paper and the vignette

AR-model can be written with f explicitly shown and this is what is done, e.g., in Stan code produced by brms.

Your questions raises another question that is what does factorizable mean?

If we don’t present f explicitly, AR-model still factorizable for LFO purposes as the joint posterior is
p(y_1|\theta)p(y_2|y_1,\theta)p(y_3|y_2,y_1,\theta),\ldots,p(y_T|y_1,...,Y_{T-1},\theta)
which, e.g., in case of AR(1) reduces to
p(y_1|\theta)p(y_2|y_1,\theta)p(y_3|y_2,\theta),\ldots,p(y_T|Y_{T-1},\theta).
Which are terms we are interested in. This factorization is the reason we can easily use (Pareto smoothed) importance sampling as the contribution of each additional observation is easy to compute. We are restricted to add these contributions in time order, which is not a problem as in LFO we want to do that anyway. For LOO we have a challenge if we can’t represent the model with terms y_i|f_i as discussed in the non-factorizable paper.

That efficient computation Kalman filter is based on that each term in the time ordered factorization is simple. Kalman filter could be used to compute these terms in cases where Kalman filter is applicable. Maybe we should have used non-Gaussian observation model (although then you could ask why we don’t use EKF or UKF, and we would need to again extend our model in a way that is easy in Stan and MCMC, but gets more difficult with *KF type algorithms, and then you could ask why don’t we use particle filters, and then I would say we don’t have those implemented in Stan)

Your questions show that we need to be more careful with the term factorizable and when we have explicit latent values f and when not.

1 Like

Thank you @Kevin_Van_Horn for your really helpful comments! As soon as the LFO paper comes out of review, we will incoroporate your comments in the revised version, in particular clarify what we understand as factorizable and what kind of non-factorizability would be a problem for our method and what would not.

I will update the LFO vignette in the LOO package as soon as we have a final version of the paper, in order to avoid having to update the same things in two places multiple times.

1 Like

This post helps a lot but I am still confused about whether or not it is possible to use LOO-PSIS for time series. In my case, I am interested in model comparison and in the interpretation of the model parameters. As mentioned by @Kevin_Van_Horn, the point-wise log-likelihood can be computed with the Kalman filter for linear Gaussian state space model. Does the time ordered factorization of the likelihood prevent from using LOO-PSIS and we must use LFO-PSIS instead?

Yes, it’s possible. It depends on what are your goals whether LOO or LFO is more useful.

Does this mean that you are not going to use the model to predict future? Which of the model parameters you are interested in?

Yes, and I would recommend it if you are using Kalman filter for linear Gaussian state space models. In this forum I’m a priori assuming people are using MCMC with Stan and write my responses based on that assumption.

No, it doesn’t prevent. It’s valid to use LOO, but it answers a different question than LFO.

I am going to use the model to predict the future only for validation purpose. My main goal is to infer the parameters of a grey-box model to estimate thermal properties of the true physical system.

Sorry I should have clarified this point. I come to the discourse to learn about Bayesian inference in general.

I get that point now, maybe I will try both and compare if the model selection with LFO and LOO leads to different recommendations.

Then LFO would predict what happens in that prediction better.

I try to be more specific. Let’s say you have time t=1,\ldots,T, a conditional model p(y_t|f_t,\phi) and time series model p(f_1,\ldots,f_T|\theta). If you are interested in \phi, LOO is probably enough. If you are interested in \theta, LOO might be enough but LFO would be better. In your model some of the parameters probably belong to the \phi category and some to the \theta category.

No problem, I’m happy to help in general.

My guess is that they give the same order for models, but when comparing models where the difference is in the timser series model part LOO shows smaller differences than LFO.

1 Like

Yes, and I would recommend it if you are using Kalman filter for linear Gaussian state space models. In this forum I’m a priori assuming people are using MCMC with Stan and write my responses based on that assumption.

Those aren’t mutually exclusive. I often use MCMC to estimate the model parameters of a linear Gaussian SSM, and only switch to MAP in situations where I have good reason to believe the parameter uncertainty won’t contribute significantly to the forecast uncertainty, or just can’t afford to do MCMC. (Note that parameter uncertainty can lead to thicker tails, which can be important if you want a high-confidence upper bound forecast.)

Sorry, I’m confused. Can you elaborate what are “Those” you are referring to? I don’t see the connection between what I wrote and what you wrote.

Sorry, I’m confused. Can you elaborate what are “Those” you are referring to? I don’t see the connection between what I wrote and what you wrote.

I meant that using the Kalman filter for linear Gaussian SSMs and using MCMC are not mutually exclusive.

So in more layman terms, would it be an accurate assertion to say that LFO is more appropriate if the goal is to see which model is better for out-of-sample predictive accuracy of future points and LOO is more appropriate if the goal is to determine out-of-sample predictive accuracy of unobserved measurements corresponding to existing obs_times?

Also, are there plans to implement LFO in the loo package?

Yes.

Not immediate. It’s likely that refits are needed and refits need knowledge of the data and model structure so it’s not possible to make it generic. This is the same reason why refit for LOO is supported only in rstanarm and brms. There is a vignette (which should be updated) for LFO showing code.

Is it possible to use LFO-CV with Stan directly?

It’s not clear what you mean with Stan directly. 1) We don’t have an example how you could do it using only Stan language. 2) The vignette Approximate leave-future-out cross-validation for Bayesian time series models shows R code that can be modified to be used with models written in Stan language directly (and not via, e.g. brms formulas).

Yes. I was going to say that see ?brms::log_lik for what it computes given the additional asrguments, but I noticed that it doesn’t document oos argument. @paul.buerkner can you clarify that part?
I think oos is passed to brms::prepare_predictions as out-of-sample indices. On second glance, it seems that other parts of the codes do not rely on brms and I just need to sort out log_lik in Stan code. Thanks a lot!

format: plain


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#StackBounty: #pairings Pairing a friendly curve whose group order is a safe prime

Bounty: 100

Are there any pairing-friendly curves whose group order is a safe prime?

That is: the order of the group is $$2q + 1$$ for some prime number $$q$$.

Or, is it impossible to have such groups?

Get this bounty!!!

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# What is the balanced chemical equation of Li(s) + H3PO4(aq) ?

Oct 18, 2015

$6 {\text{Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

#### Explanation:

Lithium metal will react with phosphoric acid to produce lithium phosphate, ${\text{Li"_3"PO}}_{4}$, and hydrogen gas, ${\text{H}}_{2}$.

${\text{Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H}}_{\textrm{2 \left(g\right]}} \uparrow$

To balance this chemical equation, notice that the only atoms that are not balanced are lithium and hydrogen.

More specifically, you have one lithium atom on the reactants' side, and three on the products' side. This means that you need to multiply the former by $3$

$3 {\text{Li"_text((s]) + "H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + "H}}_{\textrm{2 \left(g\right]}} \uparrow$

The only atoms that remain unbalanced are the hydrogen atoms. This time you have three of these on the reactants's side, and only two on the products' side.

This implies that you're going to have to multiply the hydrogen atoms on the reactants' side by $2$ and those on the products' side by $3$ to get a total of six atoms on each side.

$3 {\text{Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> "Li"_3"PO"_text(4(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

Now you've unbalanced the number of phosphorus and oxygen atoms. To fix this, you can treat the phosphate anion, ${\text{PO}}_{4}^{3 -}$, as a unit.

This means that you will need to multiply the lithium phosphate by $2$ to get

$3 {\text{Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

Finally, balance the lithium atoms again by multiplying the lithium metal by $6$ instead of by $3$.

$6 {\text{Li"_text((s]) + 2"H"_3"PO"_text(4(aq]) -> 2"Li"_3"PO"_text(4(aq]) + 3"H}}_{\textrm{2 \left(g\right]}} \uparrow$

And now the chemical equation is balanced.


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Browse Month

# The Problem of the Golden Basket

Somewhat related to the previous post on paydays, I had lunch and then coffee with another friend, and for some reason our discussion turned to personal finance as well. Unlike the last time, I was the one starting with the view that deviates from conventional theory this time.

Suppose you have a choice between two savings accounts. One account pays 2% AER and the other pays 1% AER – so £10000 invested for a year would earn £200 in the first account but only £100 in the second. Should you allocate 100% of your savings to the account that pays 2% AER?

In general, if all other things were held equal, there probably isn’t much of a reason not to allocate everything to the 2% account. However, the standard for all other things has to be high. In practice, I can see quite a few scenarios where there may be legitimate reasons to allocate some money to the 1% account.

One possible reason could be terms of access. Clearly, if the 2% account is a fixed rate bond only allowing access to the money after some amount of time while the 1% is an easy-access account, that provides a clear reason. If one wishes to maintain an emergency fund, for example, the fixed rate bond is probably best avoided even if it pays higher interest. Some savings accounts, while not having a fixed term, place restrictions on withdrawals in terms of frequency or advance notice in exchange for higher rates – again, be careful about using these accounts to park an emergency fund.

Another reason could involve how the accounts compound. The annual equivalent rate (AER) refers to how much money one will have after a year. However, if one wants the funds together with some interest before one full year, then precisely how the accounts pay interest becomes significant. If the 1% account compounds monthly while the 2% account compounds annually only, then between one month and one year after the start date the 1% account has more withdrawable interest. This is a variant of the access problem, though this focuses on access to interest as opposed to the principal. This may seem a little short-term minded, but could be interesting if one is engaging in stoozing or has other pressing financial commitments.

The amount of money one has is also relevant. Financial institutions can fail; in this case, the UK’s Financial Services Compensation Scheme (FSCS) guarantees up to £85,000 per depositor per authorised bank/building society. There is thus certainly a case for keeping not more than that amount with each bank; if one was fortunate enough for one’s savings to exceed £170,000, finding a third bank seems reasonable. I’ve never seen cash as doing the heavy lifting as far as growing my portfolio was concerned. I’d collect interest as available, but would prioritise safety. If the accounts were held with the same provider, of course, then this argument falls down – even if one has multiple accounts, the FSCS limit is on a per-bank basis. In fact, one has to be careful as some bank brands do share authorisations – meaning that an individual will only get the protection once even if several of these banks fail.

In general, the inconvenience that might be caused by failures is something worth considering as well. The FSCS compensation only applies if the bank is suitably authorised; even if one’s balance is fully covered by the FSCS, claims can take one to four weeks to process. I think I’d be much more comfortable having at least a nontrivial amount in a separate account (two to three months’ expenses, ideally) if possible.

Customer service is another factor. I’d probably prioritise that more significantly for more complex products – however, for a savings account it would still be useful.

Furthermore, there are other principles which individuals might find important. MoneySavingExpert on its savings account best buy tables has a section for highly-rated ‘ethical savings accounts’. The criteria Ethical Consumer (which MSE works with) include concerns like tax avoidance and funding of climate change, though I don’t necessarily agree with all of them (“excessive director’s remuneration”, in particular – if someone is that valuable it seems unethical to me to artificially depress their salary). Similarly, Islamic banking is necessary for adherents, since interest is forbidden in Islam.

To conclude, there are quite a number of reasons why one might not actually want to put 100% in the higher interest bearing account. Of course it makes sense ceteris paribus (if all other things were held equal), but that seems unlikely in practice. The standard for all other things being held equal here is high – the access conditions, compounding conventions and bank account provider all need to match (and that isn’t even an exhaustive list).

# Moving Cash Flows

I met a friend for a meal on the weekend, and among other things my friend mentioned that at his company, there was an internal debate over whether payday should be moved forward. This wasn’t a debate on the financial ability or willingness of the company to do this, but was instead focused on individual workers’ preferences.

My initial reaction was a bit of surprise. I wondered why this was even a debate, as I believed the answer should almost always be yes. This reminded me of the standard time-value-of-money question that I wrote about just over a year ago; being paid the same amount of money slightly earlier, mathematically, seems like an outright win. The UK hasn’t had negative interest rates yet – and even in a place like Switzerland where bank rate is negative, this isn’t typically passed on to most depositors.

Cash flow might be a problem if one considers the dollar-today-or-more-tomorrow question; however, this shouldn’t be an issue in this set-up. Valid cash-flow scenarios remain valid even if payday is brought forward. In a sense, an early payday creates additional options; it shouldn’t invalidate any existing ones.

With a bit more discussion and thought, though, we found that there were indeed valid reasons as to why one might not want payday to be shifted forward.

First, although the mathematical argument makes sense, there are some edge cases around tax liability. If one’s salary is close to a marginal rate change and a payment is pushed across a tax year boundary, the amount of tax one pays might change (and can increase).

Also, although we speak of interest as an upside, how much benefit an individual can actually realise may be significantly limited. Some bank accounts pay interest based on the lowest balance on any day in the month, meaning that being paid a few days early yields no benefit. Even if interest is based on the average daily balance, the upside is also in most cases small. For a concrete example, 2017 median UK post-tax earnings would be about £1,884.60 per month. If one was getting paid three days early and storing that into a high-interest current account yielding 5% APR, the additional interest wouldn’t be more than about 30 pence.

Moving away from purely numerical considerations, there are many other plausible reasons too. Clearly, departing from an existing routine may affect one’s own financial tracking. I find this alone to be a little flimsy (surely one’s tracker should be adaptable to variations arising from December and/or weekends?). That said, if one is unlikely to derive much benefit from the money coming early (and it seems like in most cases there indeed wouldn’t be much benefit), the change would likely seem unnecessary.

Another scenario could be if one has many bills or other payments paid by direct debit, and cannot or does not want to pay all of them. In that case, deciding precisely where the money goes could be significant – for example, if one is faced with a decision to lose fuel or premium TV in winter. This is probably not the right system to handle a situation like that, but if one wishes to only make some payments then an unexpected early payday could mess the schedule up. Somewhat related might be joint accounts in households where there are financial disputes.

Taking advantage of an early payday also requires self-control. Consider that if one is living paycheck-to-paycheck, while an early payday might ease financial pressure, it also means that the time to the next paycheck is longer than normal (unless that is also shifted forward). This needs to be dealt with accordingly.

If you’d ask me whether I’d like payday to be shifted forward, I’d almost certainly say yes. Our discussion went to a further hypothetical – would you take a 1% pay cut to have your entire salary for the year paid on January 1st?

From a mathematical point of view, you would be comparing a lump sum of $0.99N$ dollars paid now, or (for simplicity) twelve payments of $N/12$ dollars paid $1/12, 2/12, \ldots, 12/12=1$ years from now. Assuming that you can earn interest of $r$% per month and using monthly compounding, after one year we have

$Value_{\text{LumpSum}} = 0.99N (1+r)^{12}$

$Value_{\text{Normal}} = \sum_{i=1}^{12} \left( \frac{N}{12} (1+r)^{12 - i} \right)$

If we set these two to be equal and solve for $r$, we get a break even point of $r = 0.00155$. This computes out to an APR of about $1.87\%$. This is higher than best-buy easy access accounts at time of writing (MoneySavingExpert identifies Marcus at 1.5%). You can beat this with fixed-rate deposits, and probably beat this through P2P loans, REITs and equities – though more risk is involved.

I think I could see that being a yes for me, though I’m not entirely sure I’d have the self control required to manage it properly!

# Chips on the First Floor

Every so often, I spend some time filling out a crossword puzzle. There are quick crosswords where only definitions of terms are given, and cryptic crosswords which include both a definition and some wordplay. Especially for quick crosswords, these clues can be ambiguous – for instance, Fruit (5) could be APPLE, MELON, LEMON, GUAVA or GRAPE; OLIVE if one wants to be technical, and YIELD if one wishes to be indirect.

To resolve this ambiguity, about half of the letters in a quick crossword are checked. This means that their cells are at the intersection of two words, and the corresponding letters must match.

With a recent puzzle I was attempting, I had a clue with a definition for ‘show impatience (5, 2, 3, 3)’. I didn’t get this immediately, but with a few crossing letters in the middle I quickly wrote down CHOMP AT THE BIT. This was fine until I had a down clue with definition ‘problem (7)’ which was D_L_M_O. This should clearly be DILEMMA. It was a cryptic crossword, so I was able to check CHAMP AT THE BIT with the wordplay part, and it made sense. (The clue was “show impatience in talk about politician with silly hat, I bet” – which is CHAT around (an MP and then an anagram of HAT I BET).) The “original” expression is actually CHAMP, though I’ve only heard of the CHOMP version before.

I sometimes have difficulty with crosswords in the UK (and sometimes with crosswords from the US as well) owing to regional variations in English. Singaporean English follows the UK in terms of spelling. However, in terms of definitions, things vary. For example:

• Common with UK usage:
• Tuition refers to additional small-group classes (like in the UK), not the fees one might pay at university (US).
• biscuit is a baked good that’s usually sweet (like in the UK) and probably shouldn’t be eaten with gravy; an American biscuit is a bit more scone-like.
• Common with US usage:
• Chips are thin fried slices of potato (same as US). The word refers to fried strips of potato in the UK (which themselves are fries in both Singapore and the US); the thin slices are called crisps in the UK.
• The first floor of a building is the ground floor (same as US); in the UK that’s the first floor above ground (which is the second floor in Singapore and the US).

Without venturing into Singlish (which incorporates terms from Chinese, Hokkien, Malay and several other languages), there are also terms that aren’t in common with either American or British English. Some of these pertain to local entities. Economy rice is a type of food served in food courts, and the MRT is Singapore’s subway network – though I’ve heard several uses of it as a generic term, much like Xerox for copying.

Others seem a little more random. Sports shoes refer to trainers specifically, and don’t refer to water shoes or hiking boots which are used for sport. The five Cs refer to cash, cars, credit cards, country club memberships and condominiums – five things starting with the letter C that materialistic Singaporeans often chase.

I’ve been resident in the UK for around six years now. This is obviously fewer than the number I’ve spent in Singapore (about 21), though the years in the UK are more recent. I’ve gotten used to the British expressions, especially for life in the UK (I generally like chunky chips more than crisps, and correctly distinguishing the first and ground floors is important for getting around). I don’t think I’ve had too many issues with remembering the correct versions of terms to use when in Singapore or in the US – having had to deal with these inconsistencies has helped here.


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Lemma 57.13.2. Let $k$ be a field. Let $X$, $Y$ be finite type schemes over $k$ with $X$ separated. Let $F : D^ b_{\textit{Coh}}(\mathcal{O}_ X) \to D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$ be a $k$-linear exact functor sending $\textit{Coh}(\mathcal{O}_ X) \subset D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ into $\textit{Coh}(\mathcal{O}_ Y) \subset D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$. Then there exists a Fourier-Mukai functor $F' : D^ b_{\textit{Coh}}(\mathcal{O}_ X) \to D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$ whose kernel is a coherent $\mathcal{O}_{X \times Y}$-module $\mathcal{K}$ flat over $X$ and with support finite over $Y$ which is a sibling of $F$.

Proof. Denote $H : \textit{Coh}(\mathcal{O}_ X) \to \textit{Coh}(\mathcal{O}_ Y)$ the restriction of $F$. Since $F$ is an exact functor of triangulated categories, we see that $H$ is an exact functor of abelian categories. Of course $H$ is $k$-linear as $F$ is. By Functors and Morphisms, Lemma 56.7.5 we obtain a coherent $\mathcal{O}_{X \times Y}$-module $\mathcal{K}$ which is flat over $X$ and has support finite over $Y$. Let $F'$ be the Fourier-Mukai functor defined using $\mathcal{K}$ so that $F'$ restricts to $H$ on $\textit{Coh}(\mathcal{O}_ X)$. The functor $F'$ sends $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ into $D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$ by Lemma 57.9.5. Observe that $F$ and $F'$ satisfy the first and second condition of Lemma 57.11.2 and hence are siblings. $\square$

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# Arc Cosine Function

## Summary

Given a number representing the ratio of a right triangle's adjacent side over its hypotenuse returns the corresponding angle.

### Syntax

acos(n)

### Arguments

Name Description
x A number from -1 to 1 which represents the ratio of the adjacent side over the hypotenuse.


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# Purpose The main objective of this study was to develop and

Purpose The main objective of this study was to develop and evaluate therapeutic effectiveness and safety following systemic delivery of a peptide analgesic into the CNS using an oil-in-water nanoemulsion system. pain induced practical magnetic resonance imaging model in rodents. Initial security evaluations display the nanoemulsion system was well tolerated and did not cause any acute negative effects. Conclusions Overall these results display tremendous chance for the development of revised peptide analgesic-encapsulated nanoemulsion formulations for CNS delivery and restorative efficacy. tests were performed on each voxel of each subject within their unique coordinate system. The baseline threshold was arranged at 2%. BML-277 T-test statistics using a 95% confidence level 2 distributions and heteroscedastic variance assumptions were performed. A statistical composite was created for each treatment group of rats. The composite statistics was built using the inverse transformation matrices. Each composite pixel location (i.e. row column and slice) premultiplied by [Ti]?1 mapped it within a voxel of subject (i). A tri-linear IL12B interpolation of the subject’s voxel ideals (percentage switch) identified the statistical contribution of subject (i) to the composite (row column and slice) location. The use of [Ti]?1 guaranteed that the full volume set of the composite was populated with subject contributions. The average value from all subjects within the group was used to determine the composite value. The average number of activated pixels that experienced the highest composite percent change ideals in a particular ROI was displayed in a composite map. Activated composite pixels were determined as follows:

The composite percent changes for the time history graphs for each region were based on the weighted average of each subject as follows:
$Composite Percent Switch=∑i=1NActivated Pixel Subject(i)×Percent Switch(i)Activated Composite Pixels$

Formulation Safety Evaluations Changes in animal body weight Mice were weighed before dosing (day time 0) followed by dosing a 125μl of saline or DALDA-C8 solution or DALDA-C8 nanoemulsion (3 mg/kg dose) and then weighed continuously at day BML-277 time 1 (24 hours) day time 4 and day time 7. The switch in body weight was identified and compared to the initial body weight (day time 0) of all animals in the three organizations. Changes in liver enzyme and natriuretic peptide levels Aspartate aminotransferase (AST) is found in a variety of cells including liver heart kidney and mind. It is released into the serum when any of these cells is damaged. It is therefore not a highly specific indication of liver injury [25]. Whereas the alanine aminotransferase (ALT) is definitely exclusively found in liver and it is released in plasma in case of liver injury [25]. Therefore it serves as a fairly specific indication of any liver damage. To measure the liver enzyme levels namely ALT and AST plasma from numerous treatment organizations: BML-277 DALDA-C8 remedy or nanoemulsion at 20 min and 180 min was compared to untreated (control) group using the manufacturer’s instructions. Atrial natriuretic peptide (ANP) is definitely a peptide hormone secreted by cardiac myocytes of the.


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# Attempting to find the local minimum of a plane

I'm trying to find the local minimum of a plane. This is being done algebraically because I'm doing it in a computer program for a game I'm making. Sadly, I suck at math and am only in algebra II so I probably screwed a lot up with my minimal knowledge of calculus. When testing, it got a completely random spot and here is the math I used to get there :

$(x,y,z)$=point finding

$(q,r,s)$=player position

$(a,b,c,d)$=plane stuff

$ax+by+cz=d \Longrightarrow z=dx+\frac{-b}{c}y-\frac{a}{c}$

$dist = \sqrt{(x-q)^2+(y-r)^2+(z-s)^2}$

$dist^2 = f(x,y) = (x-q)^2+(y-r)^2+\Bigl((dx+\frac{-b}{c}y+(\frac{a}{c}-s)\Bigr)^2$

$x = 2(x-q)+r^2(dx+\frac{-b}{c}y+(\frac{a}{c}-s))=2x-2q+r^2dx+r^2\frac{-b}{c}y+r^2(\frac{a}{c}-s)$

$x = (2x+r^2dx+r^2\frac{-b}{c}y) + 2q+r^2(\frac{a}{c}-s)$

$x = (2+r^2d)x+(r^2\frac{-b}{c})y + 2q+r^2(\frac{a}{c}-s)$

$x = \frac{r^2\frac{-b}{c}}{-((2+r^2d)-1)}y+2q+r^2\frac{\frac{a}{c}-s}{-((2+r^2d)-1)}$

$y = (x-q)^2+(y-r)^2+(dx+\frac{-b}{c}y+(\frac{a}{c}-s))^2$

$y = 2(y-r)+q^2(dx+\frac{-b}{c}y+(\frac{a}{c}-s))$

$y = 2y-2r+q^2dx+\frac{-b}{c}q^2y+(\frac{a}{2}-s)q^2$

$y = (2y+q^2dx+\frac{-b}{c}q^2y) + 2r+(\frac{a}{2}-s)q^2$

$y = (2+\frac{-b}{c}q^2)y + q^2dx + 2r+(\frac{a}{2}-s)q^2$

$y = \frac{q^2d}{1+\frac{-b}{c}q^2}x + \frac{2r+(\frac{a}{2}-s)q^2}{1+\frac{-b}{c}q^2}$

--new variables for simplification

$m=2q+r^2\frac{\frac{a}{c}-s}{-((2+r^2d)-1)}$

$g=\frac{r^2\frac{-b}{c}}{-((2+r^2d)-1)}$

$n=\frac{2r+(\frac{a}{2}-s)q^2}{1+\frac{-b}{c}q^2}$

$h=\frac{q^2d}{1+\frac{-b}{c}q^2}$

--substitute

$y = h(gy+m)+n$

$y = hgy+hm+n$

$y = \frac{hm+n}{-(hg-1)}$

$x = gy+m$

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Your first equation, $ax + by + cz = d$, is not equivalent to $z=dx + (-b/c)y - (a/c)$. I'm assuming => was meant to be an implication arrow... –  Arturo Magidin Nov 13 '11 at 8:50
@Arturo Magidin Yeah sorry, I was trying to go off of an example I had seen. –  SDuke Nov 13 '11 at 16:05

I'll assume that Joriki's interpretation is right and you want to find the point on the plane $ax+by+cz=d$ that is closed to $(q,r,s)$. You can do that by minimizing the (squared) distance using calculus, but it is much easier to apply a little geometry.

The line between your target point and $(q,r,s)$ must be perpendicular to the plane; otherwise you could move the target point closer. Therefore what you want to do is draw the line through $(q,r,s)$ that is normal to the plane, and find its intersection with the plane. The equation for the plane directly gives us the direction of the normal, namely $(a,b,c)$. So we can write a parametric equation of the line: $$t\mapsto (q,r,s)+t(a,b,c) = (q+ta, r+tb, s+tc)$$ Finding the intersection is then just a matter of inserting into the plane's equation: $$(q+ta)a + (r+tb)b + (s+tc)c = d$$ and solve for $t$, which easily gives $t=\frac{d-qa-rb-sc}{a^2+b^2+c^2}$. Plug that into your parametric equation, and you have your target point.

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thanks for that, it makes since up until the plugging into the parametric. Once again, algebra II sorry. I understand what a parametric equation is, but that looks like I would only get one number out of it? –  SDuke Nov 13 '11 at 16:38
No, you get $(q+ta, r+tb, s+tc)$, which is a point. (There's nothing left to do except arithmetic because you now know the value of $t$). –  Henning Makholm Nov 13 '11 at 16:42
Ok thank you, I thought that could be it but I wasn't entirely certain. –  SDuke Nov 13 '11 at 16:45

You're not attempting to find the local minimum of a plane; there's no such thing. It seems that you're trying to find the point on a plane at least distance from the player position. You made an error at the very beginning in solving the plane equation for $z$; the result is $z=(d-ax-by)/c$. I suggest you try again using that, and if you don't succeed, present your work in a more systematic form, not just writing lots of equations but explaining what you did, and using $\TeX$ to make the equations more readable (by enclosing them in dollar signs).

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+1 on "not just writing lots of equations but explaining what you did". The TeXing can reasonably be left to editors if the question is otherwise clear; I'd rather see ASCII equations than TeXed math that use $*$ as a multiplication sign. –  Henning Makholm Nov 13 '11 at 15:13
@joriki Thank you, I'm sorry I couldn't get it to look that way, I was trying to but couldn't figure it out. I changed it for you. Also I was going off of some example I had seen, they did it with all numbers and that's what it appeared to be, so that's what I went with. Thank you for the correction. –  SDuke Nov 13 '11 at 16:10


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# ICHEP 2020

28 July 2020 to 6 August 2020
virtual conference
Europe/Prague timezone

## New measurement of the $K^+\to\pi^+\mu^+\mu^-$ decay at NA62

28 Jul 2020, 18:22
15m
virtual conference

#### virtual conference

Talk 05. Quark and Lepton Flavour Physics

### Speaker

Lubos Bician (CERN)

### Description

The flavour-changing neutral current decay $K^+\to\pi^+\mu^+\mu^-$ is induced at the one-loop level in the Standard Model, and is well suited to explore its structure and, possibly, its extensions. The NA62 experiment took data in 2016--2018 with the main goal of measuring the $K^+\to\pi^+\nu\bar\nu$ decay. A scaled down di-muon trigger chain was operating along with the main trigger during the whole data taking period resulting in a large sample of about $3 \times 10^{12}$ kaon decays in the fiducial volume recorded using the di-muon trigger. New results from an analysis of the $K^+\to\pi^+\mu^+\mu^-$ decay using this sample will be presented.

 Secondary track (number) 03


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# Solution for Find the Equations of All Lines of Slope Zero and that Are Tangent to the Curve Y = 1 X 2 − 2 X + 3 ? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Find the equations of all lines of slope zero and that are tangent to the curve $y = \frac{1}{x^2 - 2x + 3}$ ?

#### Solution

Slope of the given tangent is 0.

$\text { Let }\left( x_1 , y_1 \right)\text { be a point where the tangent is drawn to the curve} (1).$

$\text { Since, the point lies on the curve } .$

$\text { Hence }, y_1 = \frac{1}{{x_1}^2 - 2 x_1 + 3} . . . \left( 1 \right)$

$\text { Now,} y = \frac{1}{x^2 - 2x + 3}$

$\Rightarrow \frac{dy}{dx} = \frac{\left( x^2 - 2x + 3 \right)\left( 0 \right) - \left( 2x - 2 \right)1}{\left( x^2 - 2x + 3 \right)^2} = \frac{- 2x + 2}{\left( x^2 - 2x + 3 \right)^2}$

$\text { Slope of tangent }=\frac{- 2 x_1 + 2}{\left( {x_1}^2 - 2 x_1 + 3 \right)^2}$

$\text { Given that }$

$\text { Slope oftangent= slope of the given line }$

$\Rightarrow \frac{- 2 x_1 + 2}{\left( {x_1}^2 - 2 x_1 + 3 \right)^2} = 0$

$\Rightarrow - 2 x_1 + 2 = 0$

$\Rightarrow 2 x_1 = 2$

$\Rightarrow x_1 = 1$

$\text { Now }, y = \frac{1}{1 - 2 + 3} = \frac{1}{2} \left[ \text { From }\left( 1 \right) \right]$

$\therefore \left( x_1 , y_1 \right) = \left( 1, \frac{1}{2} \right)$

$\text { Equation oftangentis},$

$y - y_1 = m \left( x - x_1 \right)$

$\Rightarrow y - \frac{1}{2} = 0 \left( x - 1 \right)$

$\Rightarrow y = \frac{1}{2}$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [3]

Solution Find the Equations of All Lines of Slope Zero and that Are Tangent to the Curve Y = 1 X 2 − 2 X + 3 ? Concept: Tangents and Normals.
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# Is there a “natural” topology on powersets?

Let's say a topology T on a set X is natural if the definition of T refers to properties of (or relationships on) the members of X, and artificial otherwise. For example, the order topology on the set of real numbers is natural, while the discrete topology is artificial.

Suppose X is the powerset of some set Y. We know some things about X, such as that some members of X are subsets of other members of X. This defines an order on the members of X in terms of the subset relationship. But the order is not linear so it does not define an order topology on X.

I haven't found a topology on powersets that I would call natural. Is there one?

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Cofinite topology: the proper closed sets are exactly finite sets. –  user31373 Jul 18 '12 at 5:05
Have you seen this part of the powerset Wikipedia page? –  Zev Chonoles Jul 18 '12 at 5:06
@LeonidKovalev The cofinite topology on what? $X$ or the $\mathscr{P}(X)$. You can define the cofinite topology on $\mathscr{P}(X)$, but it does not really require use anything about $X$. MikeC wants a topology on $\mathscr{P}(X)$ that uses some property of $X$. –  William Jul 18 '12 at 5:09
Perhaps instead of "natural" I should have said "induced". That sounds more precise -- but I don't know how to define "induced"! –  MikeC Jul 18 '12 at 19:45

Natural is far from well-defined. For example, I don’t see why the discrete topology on $\Bbb R$ is any less natural than the order topology; one just happens to make use of less of the structure of $\Bbb R$ than the other.

That said, the Alexandrov topology on $\wp(X)$ seems pretty natural:

$$\tau=\left\{\mathscr{U}\subseteq\wp(X):\forall A,B\subseteq X\big(B\supseteq A\in\mathscr{U}\to B\in\mathscr{U}\big)\right\}\;.$$

In other words, the open sets are the upper sets in the partial order $\langle\wp(X),\subseteq\rangle$.

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Yes, a topology defined in terms of the partial order is exactly the sort of thing I was looking for. Thanks! –  MikeC Jul 18 '12 at 16:47

Note that $\mathcal P(X)$ is in some sense "naturally" equivalent to (and certainly in bijection with) the product space $\prod\limits_{x\in X}\{0,1\}$, since we can identify a subset $S\subseteq X$ with the element of the product space that has a $1$ in the $x$ place if $x\in S$ and a $0$ in the $x$ term otherwise. For example, the set $\{1,3\}\subseteq \{1,2,3\}$ would be identified with the element $(1,0,1)\in \prod\limits_{i=1}^3 \{0,1\}$. Using this, we can give $\mathcal P(X)$ the product topology, taking the discrete topology on $\{0,1\}$.

However, this may not be what you're looking for, since it depends only on the underlying set $X$ and not on its topology.

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"Natural" very much depends on context. The product topology of discrete spaces described in this answer is natural in descriptive set theory, even if the factors (or exponents) carry a topology, other than the discrete one. (For example in the context of determinacy, where products of the form $\kappa^{\aleph_0}$ are common for large uncountable cardinals $\kappa$, or in the context of Borel equivalence relations, where one favors Polish topologies.) –  Andres Caicedo Jul 18 '12 at 21:21

Your question is vague. You want a topology on $\mathscr{P}(X)$ that uses some property of $X$. If $X$ is any arbitrary set, you don't exactly know if $X$ satisfies any particular property.

However, there is a possible topology you can put on $\mathscr{P}(X)$ if $X$ is a topological space. I am not entirely sure that this is what you are looking for since a topology on $X$ is not a property of $X$, although often times a property of $X$ (such order as mentioned) can be used to define topologies.

Anyway, suppose $X$ has a topology. You can define the topological space generated by the sub basis

$\{K \in \mathscr{P}(X) : K \subseteq U\}$

$\{K \in \mathscr{P}(X) : K \cap U \neq \emptyset\}$

where $U$ ranges over all open subsets of $X$.

This is something artificial I made up, but if you replace $\mathscr{P}(X)$ with the set of all compact subset of a topological space $X$, then you obtain the Vietoris topology of the set of all compact subsets of a topological space $X$. The Vietoris Topology is actually a useful topology.

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What about the following: A set $S\subset P(X)$ is open if for any $x\in S$ it also contains all $y\subset x$.

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I wonder if this is the same as the Alexandrov topology mentioned by Brian M. Scott. It's not obvious to me one way or the other. –  MikeC Jul 18 '12 at 19:43
@MikeC: If I understand it correctly, in the example of Brian M. Scott for every set in $S$ all the supersets are also in $S$. Which if I'm not mistaken would mean the open sets in that topology are exactly the complements of the open sets in my example. However from the link I gather that the term "Alexandrov topology" is more general, and I think my example should also be one. –  celtschk Jul 18 '12 at 20:19

The question may be more relevant to your motivation if you suppose $Y$ is a topological space and you ask for a "reasonable" topology on the power set of $Y$. This is quite tricky but there are topologies on the family of closed subsets, or of open subsets, see for example

R. Brown and A.M. Abd-Allah, A compact-open topology on partial maps with open domain'', J. London Math Soc. (2) 21 (1980) 480-486.

The idea is to use the Sierpinski space $S= \{0,1\}$ in which $1$ is open but not closed. Then the set $\mathcal O(Y)$ of open sets of $Y$ is given by a corresponding set $\mathcal{CO}(Y)$ of characteristic maps $Y \to S$; this set can be given the compact-open topology. This topology is non-Hausdorff of course.

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text: Edit: perhaps the best way to clarify what I mean when I refer to the `small quantum group' is to give the definition: take the $\mathbb{C}$-algebra (other fields will do) generated by (for $n=3$) $E_1, E_2, F_1, F_2, K_1, K_2$ subject to the following relations.


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# Hexagon grid coordinate system

If I had a grid of squares, they can be labeled with Cartesian coordinates such that given square $(x,y)$, you know it shares a boundary with squares $(x+1,y),(x-1,y),(x,y+1),(x,y-1)$.

Is there a way of labeling a tessellated hexagon grid, so that given any hexagon label you can work out it's neighbors?

• Everything you want to know about hex grids is at redblobgames.com/grids/hexagons . In particular see the sections on "coordinate systems" and "neighbors". – Michael Lugo Apr 27 '17 at 13:41

In the interest of closing the question I've posted this answer, but full credit for the answer goes to Michael Lugo for his comment on my initial Post.

All information comes from this link. It goes into far more detail than this answer.

There are multiple approaches to labeling Hexagon grids, each with advantages and disadvantages.

## Offset coordinates

The most common approach is to offset every other column or row. Columns are named col or q. Rows are named row or r. You can either offset the odd or the even column/rows, so the horizontal and vertical hexagons each have two variants.

## Cube coordinates

Another way to look at hexagonal grids is to see that there are three primary axes, unlike the two we have for square grids. There’s an elegant symmetry with these.

Instead of having two values per coodinate, representing how far North and South the hexagon is, you can use three dimensions.

## Axial Coordinates

The axial coordinate system, sometimes called “trapezoidal”, is built by taking two of the three coordinates from a cube coordinate system. Since we have a constraint such as x + y + z = 0, the third coordinate is redundant. Axial coordinates are useful for map storage and for displaying coordinates to the user. Like cube coordinates, you can use the standard add, subtract, multiply, divide operations from cartesian coordinates.

So for my purposes, the most useful map system was cube coordinates, as they have the simplest and most consistent rules for changing coordinates based movement between cells.

Create the grid by drawing horizontal lines thru ALL hexagon centers and drawing vertical lines thru ALL hexagon centers. The neighbors are:

(+1,-1), (+2,0), (+1,+1), (-1,+1), (-2,0), (-1,-1)

Note: the x-axis is scaled by a factor of sqrt(3). This is ok. Cartesian coordinates can have an axis that is scaled.

Note: the x,y does not match the hexagon label. This is ok.

If we set the upper left hexagon with a coordinate of (0,0), the hexagon label is: Math.floor(x/2)+1 plus y+1

(0,0) => 0101 (2,0) => 0201 (4,0) => 0301 (1,1) => 0102 (3,1) => 0202 (5,1) => 0302


text:
# Buffon type problems - Introduction and simulations

Goal of problem - estimate $$\pi$$. Part of the Virtual Laboratories in Probability and Statistics. This link is to the extensive expository material which in turn links to associated applet material. In the applet simulation, the length of the needle is a parameter. Data collected includes: the angle of the needle relative to the crack in floorboards, the distance of the center to the floor board is collected, and whether or not the needle crosses. $$\Pi$$ is estimated with each update of the simulation. The number of crack crossings is explained to be a binomial distribution, with parameters equal the number of tosses and two times the length of the needle divided by $$\pi$$.
Identifier:
http://www.math.uah.edu/stat/buffon/Buffon.html
Rating:
Creator(s):
Kyle Siegrist
Cataloger:
Carolyn Cuff
Publisher:
MathDL
Rights:
Creative Commons


text:
# Ratio between planet-size and terrain elevation?

Let's imagine we have an algorithm that produce an elevation-map for a sphere. I wonder if the ratio between the planet radius and the delta between the highest and lowest altitude is a constant or can be guessed depending a few factors (main chemical components of the planet, atmosphere thickness, ...). Of course, I speak about telluric planets.

For example, Earth has a delta of, approximately, 20 km (Mount Everest in Nepal is 8,848 m hight and Mariana trench is 10,911 m deep in Pacific Ocean). See Wikipedia for more details. And, radius is about 6300 km. So, the final ratio is 0,003 (radius/delta).

On Mars (see here), the highest point is the peak of Olympus Mons at 21,229 m, and the deepest is in the Hellas Impact Crater which is 8,200 m deep. So, the total delta is about 29 km. Then, Mars radius is about 3400 km, which makes a ratio of 0.008.

As you can see, the variation of this ratio between these two planets are quite different.

So, I would like to have some way of "guessing" this ratio (maybe I am missing a few factors that I did not take into account, the radius is probably not enough). My point is to be able to make a map-making algorithm that will stay within realistic elevations when computing the points.

It can also be that I am totally wrong and such ratio do not exist (or has absolutely no sense at all), but, then, I would like to have a few arguments about it.

EDIT

Just to make it clear, what I am looking for is an equation providing the delta of the elevation map (highest and deepest points) according to several parameters such as planet density and planet size (radius) and others...

Something like:

$$\Delta \text{(meter)} = \text{constant(m}^3\text{/kg)} \times \text{planet radius(meters)} \times \text{planet density(kg/m}^3\text{)}$$

EDIT 2

I have collected a few samples to illustrate the formula that I am looking for. I recall that I am looking for the elevation delta based on various physical parameters which are only linked to the physics and NOT evolution of the landscape (no tectonic activity, no erosion, ...).

         delta     radius    density        surface gravity
Earth    20 km     6300 km   5.51 g/cm^3    1g
Mars     29 km     3400 km   3.93 g/cm^3    .376 g
Mercury  30 km     2439 km   5.43 g/cm^3    .38 g
Moon     18 km     1700 km   3.34 g/cm^3    .16 g

Somehow, I suspect that the planet radius and the surface gravity are involved in the formula but I don't see quite well how they interact right now. And, I suspect that I am still missing one parameter.

• If I have time later, I'll find the references, but the highest peak a planet can support is based on the mass of the planet, that's why Mars has such higher peaks, less gravity pulling them down. – bowlturner Sep 21 '14 at 15:19
• Gravity (or Mass)... Of course ! This was my missing parameter... I guess that computing this delta will involve: planet average density, radius and some magic constants. If you can find the precise equation it would be extremely useful ! Thanks ! – perror Sep 21 '14 at 15:21
• You state the ratio is radius/delta, but the figures you have calculated appear to be delta/radius. Either is fine as long as they are consistent, so I don't know which one to correct... – trichoplax Sep 21 '14 at 15:46
• There are many factors to consider. An equation can only give rough upper and lower bounds, but I imagine such bounds would be useful. The exact figure will depend on the depth and density of atmosphere, the rate of incident space debris, the level of tectonic activity, the presence or absence of weathering and water erosion, and lots of things I can't think of... Also the answer will vary widely depending on whether you measure altitude from an idealised spherical planet surface, or from "sea level" (the oblate spheroid associated with rotation, regardless of whether there is a sea). – trichoplax Sep 21 '14 at 15:57
• What I am looking for is a way to compute this delta (in order to give limits to the algorithm building the elevation map). As I am looking for an equation, parameters can move from left to right and reverse... – perror Sep 21 '14 at 17:02

The main factors that would be involved are:

1. Surface gravity - an effect of diameter and density
2. Tectonic activity levels
3. Erosion rate.

Mars has lower surface gravity, and due to its thinner atmosphere, a lower erosion rate. This means that there is less gravity to prevent taller mountains and less weather to wear them down.

So, unfortunately there isn't a single factor.

• Those factors are connected, lower gravity tends to lead to less atmosphere. – Tim B Sep 21 '14 at 16:17
• In fact, I would like to leave the factors such as tectonic activity and erosion rate to the algorithm computing the elevation map. On the other hand, the effect of gravity and/or planet mass cannot be workaround by the algorithm and are constants... and this is precisely what I would like to know (all the effects that cannot be changed in the algorithm). – perror Sep 21 '14 at 17:08
• I think this is a good answer, but I would add vulcanism as a separate item, as it can be caused by forces not related to tectonics. The Hawaiian islands were caused by an interior hot spot. My recollection is that Mars has exhibited very little, if any tectonic activity, so Olympus Mons was probably created the same way. – Donald.McLean Sep 25 '14 at 13:15
• There is evidence that suggests that Mars once had a significant amount of surface water, and there is evidence of plate tectonics there: news.discovery.com/space/… – Monty Wild Sep 25 '14 at 14:51
• @Donald.McLean: In fact, the reason Olympus Mons is so tall is because Mars has no tectonic activity. Where Hawaii was created as the plate moved across the hotspot, Olympus Mons just stayed in one place, so all the eruptions built on the previous ones, creating one super-mountain instead of a series of small mountains. – mao47 Oct 16 '14 at 18:11

The tallest mountains possible depend on the strength of the rocks, which are fluid under high pressures and geologic time. They will sag even as their being pushed up. Mountains on the moon are very tall because the rocks are dry. Mountains on Venus are stubby because of the heat.

So it depends greatly on the composition and gravity. A larger planet will have stronger surface gravity and thus shorter mountains, which is scaling the opposite direction you were supposing.

You could look up the details for various rocky worlds and check for yourself.


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# Determine the set of points at which the function is continuous.

This question aims to find the set of points at which the function is continuous if the points ( x, y ) of the given function are not equal to ( 0, 0 ).

A function is defined as the expression which gives an output of the given input such that if we put values of x in the equation, it will give exactly one value of y. For example:

$y = x ^ 4 + 1$

This expression can be written in the form of function as:

$f ( y ) = x ^ 4 + 1$

The given function is $f ( x, y) = \frac { x ^ 2 y ^ 3 } { 2 x ^ 2 + y ^ 2}$. The function f ( x ) is a rational function and every point in its domain makes it a continuous function. We have to check the continuity of function f ( x, y ) at the origin. We will limit the function as:

$Lim _ { ( x, y ) \implies ( 0, 0 ) } f ( x, y ) = f ( 0, 0 )$

We have to check along the line  by putting the value of y = 0 in the function:

$Lim _ { x \implies 0 } = \frac { x ^ 2 ( 0 ) ^ 3 } { 2 x ^ 2 + ( 0 ) ^ 2 }$

$Lim _ { x \implies 0 } = 0$

This means that the function f ( x, y ) must be zero when its limit is such that ( x, y ) equals ( 0, 0 ). The value of f ( 0, 0 )
does not satisfy this condition. Hence, a function is said to be continuous if the set of points makes it continuous at the origin.

## Numerical Results

The given function $f ( x, y) \frac { x ^ 2 y ^ 3 } { 2 x ^ 2 + y ^ 2}$ is not a continuous function.

## Example

Determine the set of points at which the function is continuous when the function is given as:

$f ( x , y ) = \frac { y ^ 2 x ^ 3 } { 3 y ^ 3 + ( y ) ^ 2 }$

We have to check the continuity of function f ( x ) at the origin. We will limit the function as:

$Lim _ { ( x, y ) \implies ( 0, 0 ) } f ( x, y ) = f ( 0, 0 )$

$Lim _ { x \implies 0 } = \frac { y ^ 2 x ^ 3 } { 3 y ^ 3 + y ^ 2 }$

We have to check along the line  by putting the value of y = 0 in the function:

$f ( 0, 0) = \frac { 0^ 2 x ^ 3 } { 3 (0) ^ 3 + ( 0 ) ^ 2 }$

$Lim _ { x \implies 0 } = 0$

This means that the function f ( x, y ) must be zero when its limit is such that ( x, y ) equals ( 0, 0 ). The value of f ( 0, 0 ) does not satisfy this condition. The given function is not continuous at the origin.

Image/Mathematical drawings are created in Geogebra.


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# Lagrangian for a relativistic free particle

The Lagrangian for a relativistic free particle is of the form $L(\vec{v}^2)$ due to the definition of an inertial frame.

We must also demand invariance under Lorentz transformations:

v’=\frac{v+V}{1+\frac{vV}{c^2}}

Keeping to first order in $V$ and using $(1+x)^n \approx 1+nx, x \ll 1$:

v’^2=(v+V)^2\left(1+\frac{vV}{c^2}\right)^{-2}=(v^2+2vV+V^2)\left(1-\frac{2vV}{c^2}+…\right)=v^2+2vV\left(1-\frac{v^2}{c^2}\right)

The new Lagrangian $L(v’^2)$ is given by:

L(v’^2)=L(v^2)+\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)

Applying the Lorentz transformation for new time $t’$:

t'(x,t)=\frac{t-\frac{Vx}{c^2}}{\sqrt{1-\frac{V^2}{c^2}}}

\frac{dt’}{dt}=\frac{\partial t’}{\partial x}\frac{dx}{dt}+\frac{\partial t’}{\partial t}=\left(-\frac{\gamma V}{c^2}\right)v+\gamma=\gamma\left(1-\frac{vV}{c^2}\right)=\frac{1-\frac{vV}{c^2}}{\sqrt{1-\frac{V^2}{c^2}}}

Note that $\frac{dx}{dt}=v$. Keeping to first order in V:

\frac{dt’}{dt}=\left(1-\frac{vV}{c^2}\right)\left(1-\frac{V^2}{c^2}\right)^{-1/2}=1-\frac{vV}{c^2}

The new action is given by:

S’=\int L(v’^2)\,dt’=\int \left[L(v^2)+\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)\right]\left(1-\frac{vV}{c^2}\right)\,dt

Keeping to first order in V:

S’=\int\left[L(v^2)\left(1-\frac{vV}{c^2}\right)+\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)\right]\,dt

The change in the integrand of the action should be a total derivative:

\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)-L(v^2)\frac{vV}{c^2}=\frac{d}{dt}f(x,t)

Factoring out $v=\frac{dx}{dt}$, this suggests setting the rest to a constant. Furthermore, the $1-\frac{v^2}{c^2}$ term needs to be cancelled out which indicates $C=0$.

\frac{\partial L}{\partial (v^2)}2V\left(1-\frac{v^2}{c^2}\right)-L(v^2)\frac{V}{c^2}=0

Simplifying:

\frac{\partial L}{\partial (v^2)}\left(1-\frac{v^2}{c^2}\right)-\frac{L(v^2)}{2c^2}=0

The only nontrivial solution is:

L(v^2)=C\sqrt{1-\frac{v^2}{c^2}}

where C is a constant.

We can deduce what $C$ is because in the limit $v \ll c$, we should extract the Lagrangian for a nonrelativistic free particle.

L_r(v^2)=C\left(1-\frac{v^2}{c^2}\right)^{-1/2} \approx C\left(1-\frac{v^2}{2c^2}\right)=C-\frac{Cv^2}{2c^2}

Ignoring the additive constant which doesn’t affect the motion and setting the right term equal to the nonrelativistic Lagrangian, we obtain:

-\frac{Cv^2}{2c^2}=L_{nr}(v^2)=\frac{1}{2}mv^2

C=-mc^2

Thus the form of $L(v^2)$ for a relativistic particle is given by:

L=-mc^2\sqrt{1-\frac{v^2}{c^2}}


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# What is the square root of 156.25?

Sep 18, 2015

$\sqrt{156.25} = \frac{25}{2}$

#### Explanation:

$15625 = {5}^{6}$

So $156.25 = \frac{{5}^{6}}{100} = \frac{{5}^{6}}{{2}^{2} \cdot {5}^{2}} = {5}^{4} / {2}^{2}$

If $a , b \ge 0$, then $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

If $a , b , c > 0$ then ${a}^{b c} = {\left({a}^{b}\right)}^{c}$

So $\sqrt{156.25} = \sqrt{{5}^{4} / {2}^{2}} = \frac{\sqrt{{5}^{4}}}{\sqrt{{2}^{2}}} = \frac{\sqrt{{\left({5}^{2}\right)}^{2}}}{\sqrt{{2}^{2}}} = {5}^{2} / 2 = \frac{25}{2}$


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HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# Solution for The probability mass function for X = number of major defects in a randomly selected appliance of a certain type is - HSC Science (Electronics) 12th Board Exam - Mathematics and Statistics

ConceptVariance of Binomial Distribution (P.M.F.)

#### Question

The probability mass function for X = number of major defects in a randomly selected
appliance of a certain type is

 X = x 0 1 2 3 4 P(X = x) 0.08 0.15 0.45 0.27 0.05

Find the expected value and variance of X.

#### Solution

E(X)=sumx_i.P(x_i)

=0(0.08)+1(0.15)+2(0.45)+3(0.27)+4(0.05)

=0+0.15+0.9+0.81+0.2=2.06

E(X^2)=sumx_i^2.P(x_i)

=0(0.08)+1^2(0.15)+2^2(0.45)+3^2(0.27)+4^2(0.05)

=0 + 0.15 + 1.8 + 2.43 + 0.8=5.18

Var(X)=E(X^2)-[E(X)]^2

=5.18-(2.06)^2

=5.18-4.2436

=0.9364

Is there an error in this question or solution?

#### APPEARS IN

2015-2016 (July) (with solutions)
Question 5.2.2 | 4 marks
2016-2017 (July) (with solutions)
Question 5.1.2 | 3 marks
Solution for question: The probability mass function for X = number of major defects in a randomly selected appliance of a certain type is concept: Variance of Binomial Distribution (P.M.F.). For the courses HSC Science (Electronics), HSC Science (Computer Science), HSC Arts, HSC Science (General)
S


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Previous issue ·  Next issue ·  Most recent issue · All issues

# Journal of Operator Theory

Volume 50, Issue 1, Summer 2003  pp. 157-167.

Convex trace functions of several variables on $C^*$-algebras

Authors Gert K. Pedersen
Author institution: Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK-2100 Copenhagen, Denmark

Summary:  For each trace $\tau$ on a $C^*$-algebra $A$ generated by mutually commuting $C^*$-subalgebras $A_1,A_2, \ldots, A_n$ and every convex function $f$ of $n$ variables we show that the function $(x_1, x_2, \ldots , x_n) \to \tau (f(x_1, x_2, \ldots , x_n))$ is convex on the space $\bigoplus\limits_{k=1}^{ n} ((A_k)_{\sa})$.

Contents    Full-Text PDF


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How to adjust SSE or RMSE for the number of free parameters in the model?

How do I adjust SSE (sum of squared errors) or RMSE (root-mean-square errors) for the number of free parameters in the model?

Is there an "adjusted" RMSD metric similar to the adjusted r-squared metric?

• please don't assume that we know what RMSD stands for. The more effort you invest in carefully phrasing your question, the more motivated people will feel to answer it. – Artem Kaznatcheev Feb 8 '13 at 5:04
• This sounds more like a question better suited for stats.stackexchange.com . Is there an aspect of this that relates particularly to cognitive modelling? – Jeromy Anglim Feb 8 '13 at 5:13
• Not sure to understand what you mean by "adjusted" here ? Do you mean adjusted as for "goodness of fit" or as correction in regards to some parameters ? – Cheatboy2 Feb 8 '13 at 14:30
• @Cheatboy2 I believe he is referring to a correction in the number of parameters, as in adjusted $r^2$ – Jeff Feb 8 '13 at 23:17

To my knowledge, there is no adjusted RMSD.

RMSD, unlike $R^2$, isn't typically used to compare models across the literature. $R^2$ represents the proportion of variance explained by the model, a construct which translates well across different experimental designs. Adjusted $R^2$ distorts this by accounting for the number of parameters in your model, but is a better estimate of the proportion of variance in the population explained by your model.

RMSD is in whatever units you happen to be using. It should be interpreted against some benchmark for what you think a good model should be, or whatever has been deemed an "acceptable error rate". It's not clear to me that adjusting for the number of parameters in RMSD would provide a number that is inherently meaningful, but I'm not sure if this is true.

If you would like to select among several model candidates, you may want to compute AIC (Akaike Information Criterion) or BIC (Bayesian Information Criterion), both of which are attempts to maximize the likelihood of your model while minimizing the number of parameters. Both metrics are commonly used in cognitive modeling, and I think this may be what you are looking for.

While I think this question is on-topic here, you will likely get better answers on stats.stackexchange.com (as Jeromy points out) because your question is not really specific to cognitive modeling.

There are at least 3 ways to discount SSE (or RMSE) by the number of free params:

$$\text{adjusted RMSE} = \sqrt{\frac{SSE}{n - k}}$$

$$AIC = n \times ln\left(\frac{SSE}{n}\right) - k \times ln(n)$$

$$BIC = n \times ln\left(\frac{SSE}{n}\right) - 2 \times k$$

or in computer code style:

k = number of free params
n = number of DV's
SSE = sum of squared errors

adjusted RMSE = sqrt ( SSE / n - k )
AIC = n * ln(SSE/n) - k * ln(n)
BIC = n * ln(SSE/n) - 2 * k


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Expressing a particle's matter wave in terms of its momentum

I'm following Zettili's QM book and on p. 39 the following manipulation is done,

Given a localized wave function (called a wave packet), it can be expressed as $$\psi(x,t) = \frac{1}{\sqrt{ 2 \pi}} \int_{-\infty}^{\infty} \phi(k) e^{i(kx-\omega t)} dk$$ Now use the de broglie relations: $p = \hbar k$ and $E = \hbar \omega$ and define $\tilde{\phi}(p) = \phi(\frac{k}{\hbar})$.

This should yield $$\psi(x,t) = \frac{1}{\sqrt{ 2 \pi \hbar}}\int_{-\infty}^{\infty} \tilde{\phi}(p) e^{i(px-E t)/ \hbar} dp$$ but I get $$\psi(x,t) = \frac{1}{\sqrt{ 2 \pi} \hbar}\int_{-\infty}^{\infty} \phi\biggl(\frac{p}{h}\biggr) e^{i(px-E t)/ \hbar} dp$$ when I make the change-of-variable $k=\frac{p}{\hbar}$. What am I missing?

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Are you just talking about the factor of $\sqrt{\hbar}$? –  Colin K Jan 24 '12 at 22:01
It's correct as written. So no, $\tilde{\phi}$ does not align either. –  StuartHa Jan 24 '12 at 23:54
Sorry, this is a really weak question. It's a straightforward and trivial algebra to get the right power of $\hbar$. Moreover, mature physicists - probably including the author of the textbook you mention - use units with $\hbar=1$, so missing powers of $\hbar$ are "not really mistakes". On the other hand, you have real mistakes in your text, e.g. the claim $\tilde \phi(p)=\phi(k/\hbar)$ at the top. You meant $p/\hbar$ on the right hand side, right? When you make this fix, you do see that the "obtained" and "desired" expressions only differ by $\sqrt{\hbar}$, don't you? –  Luboš Motl Jan 25 '12 at 6:30
I'm sorry the author of an introductory textbook doesn't use your convention for mature physicists. And no, the text states $k/\hbar$, not $p/\hbar$. This is not in the errata either. –  StuartHa Jan 25 '12 at 15:58
I haven't found this wrong formula in this version of the book. But on the page 40 in the paragraph just before eq.(1.98) there is another one: $\tilde\phi(p) = \phi(k)/\sqrt{\hbar}$ –  Maksim Zholudev Jan 25 '12 at 16:56

In bra-ket notation your formulas should look as follows $$\left<x\right.\left|\psi\right> = \int_{-\infty}^\infty \left<x\right.\left|k\right> \left<k\right.\left|\psi\right> dk = \int_{-\infty}^\infty \left<x\right.\left|p\right> \left<p\right.\left|\psi\right> dp$$ where
$\left<k\right.\left|\psi\right> = \phi(k)\exp(-i\omega t)$ and $\left<p\right.\left|\psi\right> = \tilde\phi(p)\exp(-i\omega t)$
are the wavefunctions in terms of $k$ and $p$;
$\left<x\right.\left|k\right>$ and $\left<x\right.\left|p\right>$ are the eigenfunctions of the operators $\hat{k}$ and $\hat{p}$ respectively.

These eigenfunctions should be normalized and the usual normalization for continuous quantum numbers ($k$ and $p$) is the one with delta function: $$\left<k'\right.\left|k\right> = \int_{-\infty}^\infty \left<k'\right.\left|x\right> \left<x\right.\left|k\right> dx = \delta(k'-k)$$ The eigenfunctions normalized like this are $$\left<x\right.\left|k\right> = \frac{1}{\sqrt{2\pi}} e^{ikx}$$ and $$\left<x\right.\left|p\right> = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}$$ So here is the lost $\sqrt{\hbar}$, in the normalization coefficient.

Edit: version without bra-kets

$$\psi(x, t) = \int_{-\infty}^\infty \xi_k(x)\varphi(k,t) dk = \int_{-\infty}^\infty \tilde\xi_p(x)\tilde\varphi(p,t) dp$$ where
$$\varphi(k,t) = \phi(k)\exp(-i\omega t) = \int_{-\infty}^\infty \xi_k^*(x)\psi(x, t) dx$$ and $$\tilde\varphi(p,t) = \tilde\phi(p)\exp(-i\omega t) = \int_{-\infty}^\infty \tilde\xi_p^*(x)\psi(x, t) dx$$
are the wavefunctions in terms of $k$ and $p$;
$\xi_k(x)$ and $\tilde\xi_p(x)$ are the eigenfunctions of the operators $\hat{k}$ and $\hat{p}$ respectively.

These eigenfunctions should be normalized and the usual normalization for continuous quantum numbers ($k$ and $p$) is the one with delta function: $$\int_{-\infty}^\infty \xi_{k'}^*(x)\xi_k(x) dx = \delta(k'-k)$$ The eigenfunctions normalized like this are $$\xi_k(x) = \frac{1}{\sqrt{2\pi}} e^{ikx}$$ and $$\tilde\xi_p(x) = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}$$ So here is the lost $\sqrt{\hbar}$, in the normalization coefficient.

-
I haven't studied bra-kets yet so I will have to save this to read later. But you are basically saying to just re-normalize the wave function? That makes notation tricky because then the $\psi(x,t)$ are no longer equal to themselves. –  StuartHa Jan 25 '12 at 16:06
@user19192, nothing happens to $\psi(x,t)$ here. You have wrong assumption that functions $\tilde\phi(p)$ and $\phi(p/\hbar)$ are equal. They differ by $\sqrt{\hbar}$ because the basis functions are also different. See the 2nd and 3rd formulas in the updated part of my answer. –  Maksim Zholudev Jan 25 '12 at 16:45

The culprit seems to be a typo

$$\widetilde{\phi}(p) ~=~ \phi(\frac{k}{\hbar}) \qquad\qquad ({\rm Wrong!})$$

right above eq. (1.98) in the 1st edition of N. Zettili, Quantum Mechanics: Concepts and Applications, which is corrected to

$$\widetilde{\phi}(p) ~=~ \frac{\phi(k)}{\sqrt{\hbar}}$$

in the 2nd edition, see also a comment from Maksim Zholudev.

-


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# Tag Info

Let us begin with the following principle: the CC ansatz always returns an appropriate electronic wavefunction (meaning a combination of Slater determinants). Still, at least in a non-relativistic approach, the Hamiltonian does commute with the Spin operator $$S^{2}=S_{+}S_{-}-S_{z}+S_{z}^{2}$$ (I'm using atomic units here) and ...


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## Cryptology ePrint Archive: Report 2012/222

A secret sharing scheme of prime numbers based on hardness of factorization

Kai-Yuen Cheong

Abstract: Secret sharing schemes usually have perfect information theoretic security. This implies that each share must be as large as the secret. In this work we propose a scheme where the shares are smaller, while the security becomes computational. The computational security assumption is hardness of factorization, which is a simple and rather standard assumption in cryptography. In our scheme, the shared secret can only be a set of prime numbers.

Category / Keywords: foundations / secret sharing

Date: received 17 Apr 2012, last revised 20 Aug 2012

Contact author: kaiyuen at jaist ac jp

Available format(s): PDF | BibTeX Citation


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Ergodicity

# Ergodicity

Discussion

Encyclopedia
In mathematics
Mathematics
Mathematics is the study of quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity...

, the term ergodic is used to describe a dynamical system
Dynamical system (definition)
The dynamical system concept is a mathematical formalization for any fixed "rule" which describes the time dependence of a point's position in its ambient space...

which, broadly speaking, has the same behavior averaged over time as averaged over space. In physics
Physics
Physics is a natural science that involves the study of matter and its motion through spacetime, along with related concepts such as energy and force. More broadly, it is the general analysis of nature, conducted in order to understand how the universe behaves.Physics is one of the oldest academic...

the term is used to imply that a system satisfies the ergodic hypothesis
Ergodic hypothesis
In physics and thermodynamics, the ergodic hypothesis says that, over long periods of time, the time spent by a particle in some region of the phase space of microstates with the same energy is proportional to the volume of this region, i.e., that all accessible microstates are equiprobable over a...

of thermodynamics
Thermodynamics
Thermodynamics is a physical science that studies the effects on material bodies, and on radiation in regions of space, of transfer of heat and of work done on or by the bodies or radiation...

.

## Etymology

The word ergodic is derived from the Greek words έργον and οδός, work and path. This was chosen by Boltzmann while working on a problem in statistical mechanics.

## Formal definition

Let be a probability space
Probability space
In probability theory, a probability space or a probability triple is a mathematical construct that models a real-world process consisting of states that occur randomly. A probability space is constructed with a specific kind of situation or experiment in mind...

, and be a measure-preserving transformation. We say that T is ergodic with respect to (or alternatively that is ergodic with respect to T) if one of the following equivalent definitions is true:
• for every with either or .
• for every with either or (where denotes the symmetric difference
Symmetric difference
In mathematics, the symmetric difference of two sets is the set of elements which are in either of the sets and not in their intersection. The symmetric difference of the sets A and B is commonly denoted by A\,\Delta\,B\,orA \ominus B....

).
• for every with positive measure we have .
• for every two sets E and H of positive measure, there exists an n > 0 such that .

### Measurable flows

These definitions have natural analogues for the case of measurable flows and, more generally, measure-preserving semigroup actions. Let {Tt} be a measurable flow on (X, Σ, μ). An element A of Σ is invariant mod 0 under {Tt} if

for each tR. Measurable sets invariant mod 0 under a flow or a semigroup action form the invariant subalgebra of Σ, and the corresponding measure-preserving dynamical system
Measure-preserving dynamical system
In mathematics, a measure-preserving dynamical system is an object of study in the abstract formulation of dynamical systems, and ergodic theory in particular.-Definition:...

is ergodic if the invariant subalgebra is the trivial σ-algebra consisting of the sets of measure 0 and their complements in X.

## Markov chains

In a Markov chain
Markov chain
A Markov chain, named after Andrey Markov, is a mathematical system that undergoes transitions from one state to another, between a finite or countable number of possible states. It is a random process characterized as memoryless: the next state depends only on the current state and not on the...

, a state is said to be ergodic if it is aperiodic and positive recurrent. If all states in a Markov chain are ergodic, then the chain is said to be ergodic.

## Ergodic decomposition

Conceptually, ergodicity of a dynamical system is a certain irreducibility property, akin to the notions of irreducible representation in algebra and prime number
Prime number
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is called a composite number. For example 5 is prime, as only 1 and 5 divide it, whereas 6 is composite, since it has the divisors 2...

in arithmetic. A general measure-preserving transformation or flow on a Lebesgue space admits a canonical decomposition into its ergodic components, each of which is ergodic.


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How to reconcile the definitions of nilpotence class of groups using upper and lower central series?

This is part of a theorem in Dummit & Foote, but the proof is omitted. I want to prove it by myself but I'm starting to give up:

Def. $Z^0(G)=1$, $Z^1(G)=Z(G)$, ..., $Z^{i+1}(G)/Z^i(G)=Z(G/Z^i(G))$. The series $Z^0(G)\le Z^1(G)\le\dots$ is the upper central series of $G$. $G$ is of nilpotence class $c$ iff. $Z^c(G)=G$ but $Z^{c-1}(G)<G$.

Def. $\gamma_0(G)=G$, $\gamma_1(G)=[G,G]$, ..., $\gamma_{i+1}(G)=[G,\gamma_i(G)]$. The series $\dots\le\gamma_1(G)\le\gamma_0(G)$ is the lower central series of $G$.

Thm. $G$ is of nilpotence class $c$ iff. $\gamma_c(G)=1$ but $\gamma_{c-1}(G)>1$. If $G$ is of nilpotence class $c$, then $$Z^0(G)\le\gamma_{c-1}(G)\le Z^1(G)\le\gamma_{c-2}(G)\le Z^2(G)\le\dots\le Z^{c-1}(G)\le \gamma_0(G)\le Z^c(G)$$

I tried to show that the upper central series is the fastest ascending central series of $G$. Suppose $1=H_0\le H_1\le\dots\le\ H_n=G$ is a central series of $G$. By def., central series is a normal series.

I want to show that $H_i\le Z^i(G)$ for all $i$. Obviously the result is true for $i=0,1$. Suppose $H_k\le Z^k(G)$.

I'll show that $H_{k+1}\le Z^{k+1}(G)$, i.e. that $H_{k+1}\le\pi^{-1}(Z(G/Z^k(G)))$, i.e. $$H_{k+1}Z^k(G)/Z^k(G)\le Z(G/Z^k(G))-(*)$$ where $\pi$ is the natural projection homomorphism $G\rightarrow G/Z^k(G)$.

Now I only know that $[G,H_{k+1}]\le H_k\le Z^k(G)$, how to continue from here? Or is it even possible to do so?

I also tried to show that the lower central series is the fastest descending central series of $G$. Suppose $1=K_n\le\dots\le K_1\le K_0=G$ is a central series of $G$. Then i show that $\gamma_i(G)\le K_i$ for all $i$. Obviously the result is true for $i=0,1$. Suppose $\gamma_k(G)\le K_k$.

I'll show that $\gamma_{k+1}(G)\le K_{k+1}$. This is way easier: $\gamma_{k+1}(G)=[G,\gamma_k(G)]\le [G,K_k]\le K_{k+1}$. The last inclusion is due to an equiv. def. of a central series which I know how to prove.

But then how to conclude that the two series are of the same length?

For the 'alternating-term' part of the theorem, I have no idea. I can only prove that the result is true for $c=0,1,2$. And I can prove that $\gamma_{c-i}\le Z^i(G)$. We do a backward induction: The reselt is true for $i=c$. Suppose $\gamma_{c-k}\le Z^k(G)$, then $\gamma_{c-(k-1)}(G)=\gamma_{c-k+1}(G)=[G,\gamma_{c-k}(G)]\le [G,Z^k(G)]\le Z^{k-1}(G)$. The last inclusion is direct from how we define the centers of $G$. How to prove that $Z^i(G)\le\gamma_{c-i-1}(G)$??

P.S. As Derek Holt pointed out in the comments, $Z^i(G)\le\gamma_{c-i-1}(G)$ is false in general: let $H$ be a nilpotent group with nilpotence class $3$. Let $G=H\times\Bbb Z_p$. Then $G$ is also nilpotent of class $3$. $Z^1(G)=Z(H)\times\Bbb Z_p$, but $\gamma_1(G)=[G,G]=[H,H]\times[\Bbb Z_p,\Bbb Z_p]=[H,H]\times1\neq Z^1(G)$

• It seems to me that you have done all the hard work and then for some reason lost your nerve just before the finishing line! $[G,H_{k+1}] \le Z^k(G)$ implies $[G,H_{k+1}Z^k(G)] \le Z^k(G)$, which implies $(*)$. The fact that the upper central series is the fastest ascending shows that no other central series can be longer. Similarly for the descending central series, so they must both have the same length. – Derek Holt Sep 16 '17 at 9:04
• @DerekHolt I know how $[G,H_{k+1}Z^k(G)]\le Z^k(G)$ implies $(*)$ now. The '$[G,H_{k+1}]$ implies $[G,H_{k+1}Z^k(G)]\le Z^k(G)$' part was not apparent to me though until after you've said so. It uses the fact that $Z^k(G)\unlhd G$. By the way, why is $Z^i(G)\le\gamma_{c-i-1}(G)$? I have absolutely no clue. – user441558 Sep 16 '17 at 9:20
• @DerekHolt Plus, why upper central series being the fastest ascending shows that no other central series can be longer? Isn't it 'shorter' not 'longer'? And for the descending central series, it is also the 'shortest' one, and in fact since both the upper central series and the lower central series are the 'shortest' central series, they must be of the same length, right? – user441558 Sep 16 '17 at 9:43
• Yes sorry, I wrote "longer " instead of "shorter" (so i would not do so well in an exam these days), but I think you understand that argument now. – Derek Holt Sep 16 '17 at 10:20
• $Z^i(G) \le \gamma_{c-i-1}(G)$ is not true in general. Where did you get that from? For example if $H$ is a nilpotent group of class $3$ and $G = H \times C_p$, then $Z_1(G)$ contains the direct factor $C_p$, but that is not contained in $[G,G]$. – Derek Holt Sep 16 '17 at 10:23


text:
# A first-order ODE solvable in a case

The ODE

$$y'+\left(-1+\frac{a-1}{x}\right)y+b x -c=0$$

is solved by $$y=bx$$ in the particular case $$c=ab$$.

How would you proceed to find a more general solution?

The equation $$y'+\left(-1+\frac{a-1}{x}\right)y+b x -c=0$$ is linear.
You have $$y= y_h + y_p$$ where $$y_h$$ is the solution to the homogeneous equation, $$y'+\left(-1+\frac{a-1}{x}\right)y=0$$ and $$:y_p$$. is a particular solution to the in-homogeneous equation.
Since this equation can be put in the form $$y'+a(x)y=b(x)$$, you can solve by using the integrating factor $$e^{\int a(x)\operatorname dx}$$.