text:
# How can the vendor be approved/disapproved based on validation test data?

The power requirement for a sub system is 5V +/- 10% @ 1A of current. A RFQ was issued and a prospect vendor was selected. A first article of inspection (FAIR) has been submitted which includes 15 power supply samples. The samples have been tested with and without a 5 ohm resistor load using two separate calibrated DMM’s . Without the resistor the output voltage is 5V with little variation. Under a 5 ohm load the numbers are not too great. Below are the measurements (V):

4.56, 4.71, 4.95, 5.12, 4.61, 4.83, 4.65, 4.69, 4.79, 4.81, 4.67, 4.74, 4.63, 4.74, 4.81

Should the vendor be approved as a supplier?

• Those numbers are in spec, but if you have other qualms with them, it seems like that's something to discuss internally. And if you want better numbers, revise the tolerance. – Trevor Archibald Feb 1 '15 at 0:34

TL;DR: It depends, but probably not

You might argue that "all 15 samples are within spec, so the supplier should be approved". Not so fast: depending on the parameters of your full production run, the 15 samples may or may not be statistically significant.

There are numerous calculators online to do these calculations; I used this one. A good resource for the actual formulas is online at NIST's Engineering Statistics Handbook or in any undergraduate Statistics textbook.

From the spec, we know that our confidence interval is 10% (two-sided). This means that any parts within 10% of nominal are acceptable.

Next, we need to determine the confidence level. This is usually 95% but sometimes 99%, and represents "how sure" we can be of the result.

The final piece of information we can give if it's known is the population size. In this scenario, this is the total number of parts to be ordered from the vendor over the lifetime of the product/process. Most calculators allow this to be left blank and, if missing, assume a large value, because the effect of the population size decreases as it grows in relation to the sample size.

Assuming a 95% confidence level and our 10% confidence interval, with population left blank, we need a sample size of 96 parts in order to have a statistically significant result. Increasing to a confidence level of 99% requires a sample size of 166 parts.

So, for industry-standard confidence levels, we cannot conclude that the vendor be approved based on the initial sample of 15 parts alone.

Wait a minute—the entire FAIR run was within spec! Well yes, but what's to say the next 15 won't be out of spec? We don't know—that's why we have statistics! :-)

Well, under what circumstances is our sample significant?

Just for illustration, I entered population values until I found a statistically-significant result at $n=15$: for 95% confidence level, $n=15$ samples would be significant only if the population (total production run) is 18 parts! For a confidence level of 99%, the situation is even worse: significance only if the production run is 16 parts!

Other Notes:

The calculations above assume that the process follows a normal distribution and that the sample is representative of the population. In practice, both of these assumptions may be inaccurate.

• it also assumes that resistances of items in the sample are independent, and that may not hold either: items within a batch may be more closely matched than items from separate batches. – EnergyNumbers Feb 1 '15 at 17:08
• @EnergyNumbers this is accounted for in the "representative" assumption: if there is variation across batches, a sample selected from one batch is not representative of the population. – Paul Gessler Feb 1 '15 at 18:10
• Doesn't this also assume we know essentially nothing about the precision of the manufacturing process too though? For example, if I have two groups of parts made to the same design, one forged and one cast, I should need fewer samples of the forged part to get the same confidence interval, because it's accepted that forging is more consistent than casting. – Trevor Archibald Feb 2 '15 at 19:30
• @TrevorArchibald good question, one that I can't immediately think of an answer to. Only that the confidence interval is determined by the spec only - the statistics don't really care about the process used. Though in practice, the precision of the manufacturing process would affect the shape of the distribution. I need to think some more about if using a normalized distribution removes this dependence or not. – Paul Gessler Feb 2 '15 at 19:51
• @PaulGessler I'd think there'd primarily be variation in the standard deviation of different manufacturing processes, but for that to be useful, you'd probably need to have that defined separately. But once you have that, I feel like it shouldn't be hard to alter the calculation of the confidence interval accordingly. – Trevor Archibald Feb 2 '15 at 20:05

The average and standard deviation from the above sample set are:

• Average ($\mu$): 4.754V
• Standard Deviation ($\sigma$): 0.142368
• Mean: 4.74
• USL: 5.5
• LSL: 4.5
• C$_{p}$ : 1.1706
• C$_{pl}$: 0.5947
• C$_{pu}$: 1.7466
• C$_{pk}$: 0.5947

Equations

$\text{Pr}(\mu - \sigma \leq x \leq \mu + \sigma) \approx 0.6827$

$\text{Pr}(\mu - 2\sigma \leq x \leq \mu + 2\sigma) \approx 0.9545$

$\text{Pr}(\mu - 3\sigma \leq x \leq \mu + 3\sigma) \approx 0.9973$

$C_{p} = \frac{(USL - LSL)}{6\sigma}$

$C_{pu} = \frac{USL-\mu}{3*\sigma}$

$C_{pl} = \frac{\mu-LSL}{3*\sigma}$

$C_{pk} = min(C_{pu},C_{pl})$

Based on the above data, it appears that the process is between 1 sigma and 2 sigma capable. Approximately about 10% of the product could fall outside of the requirements.

Also a sample set of 30 is considered a generally acceptable sample size. The sample set in this problem is 15 which is on the low side.

Suggestions:

• Check with the vendor if they sorted the sample parts
• The power supply is not delivering the required 1A of current at full load
• The current carrying conductor might not be capable to delivering the current at full load
• The power supply might have some minor design issues too
• Also check the test system too

The distribution appears to be a normal bell curve. A supplier Quality engineering might be able help address this issue.

I agree with What Trevor said in a comment. You specified 5V ±10% (4.5-5.5 V) and that's what you got. I don't see the problem here. All the samples you got are within spec. There may be other reasons to not use this vendor, but samples failing to meet specs is not one of them.

However, first article samples don't tell you that the vendor is OK, they only tell you whether the vendor is definitely not OK. In the real world, you don't just qualify a vendor and then blindly buy from them after that. It is a ongoing relationship. The samples were all within spec, but then again, unless the vendor is really stupid, of course they were. They can afford to manually check and hand-pick 15 items to send you.

Especially since one out of 15 was quite close to the edge, you should have the vendor explain to you how their design or calibration procedure guarantees all the units will be in spec, how you know they will stay in spec over the expected lifetime, etc. In short, it's still your job to have some confidence that these sample are representative and that you know there is a system in place you believe in to deliver future product in spec.

If you are really worried about this, then set up your own incoming inspection. You could possibly do this with a random sampling from every lot received, or every unit, depending on how critical this is to you.

Of course if you're really this worried, then your original specs are probably too loose. Allowing the bare power supply to be ±10% sounds unnecessarily high in the first place. That's not pushing any limits, so tighter requirements like ±5% shouldn't cost much more, if any. Some electronic parts are specified for 5V ±10%, so you can't have the bare supply eat up the entire error budget. The voltage ulitimately delivered where it is used will have a wider tolerance than whatever it is at the supply.


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## Buy and sell currencies – Debates and Help

To buy and sell currencies, a Forex trader needs a good means of trading and the broker is the medium in exchange operations. Before selecting our agent, we should check several things from that agent to obtain adequate facilities and security from them. FXdirects is an honest and safe operations broker in this challenging market and they have been operating in this market for many years. I never face any re-quote or slip problem during my operation.

## of currencies – Obtaining cash in USD in Munich in cards of the United Kingdom

There are at least ATMs at the airport where you can receive cash payments in US dollars and I'm relatively sure that ReiseBank has an ATM in the main train station, which can also deliver you notes in US dollars.

However, I do not assume that this is the cheapest solution. The German ATM operators noticeably hide all kinds of strange fees and I would expect that if you try to make a payment in US dollars in a German ATM from a UK card, the ATM operator will convert the amount in US dollars to Euros at a low exchange rate. and then load your card in EUR and not directly in US \$.

The change of currency in regular banks is often only provided to bank customers and is also usually quite expensive. The most prominent exchange offices, eg. Western Union or ReiseBank do not offer very good exchange rates either. I guess the best solution is to get EUR with your card from an ATM and then go to any of the smaller independent exchange offices. Many of them are located in the area south of the main train station between Bayerstraße and Schwanthaler Straße.

## probability – What is the correct mathematical answer for a cryptographic hash function like in digital currencies?

Most tutorials on digital currencies (such as BitCoin) say that a hash function is a cryptographic function that takes an input and always returns a binary string of length $$256$$ in such a way that a small change in the input would result in a totally different hash.

Obviously, in the simplest interpretation, the hash function must be injective. However, since the length of the incoming messages is unlimited while there is only $$2 ^ {256}$$ binary strings, it is obvious that an injective function can not exist.

So, what is the correct interpretation? I tried to formulate a possible interpretation and I looked it up in Google to see if there is a real mathematical definition of the hash function as in the BitCoin tutorials, but I had no luck.

Anyway, here is my interpretation:

Leave $$mathcal {M}$$ denotes the set of messages. We can think of this set as binary strings of unlimited length, that is, $$mathcal {M} = Large { large (a_1, a_2, a_3, cdots, a_n, cdots) large mid a_i = 0,1 ( forall i in mathbb {N}) Large }$$ We can define a distance in this set as follows:

$$d (m_1, m_2) = sum_ {i = 1} ^ { infty} frac {a_i} {2 ^ i}$$

Then the statement becomes something like this:

$$forall m_1 in mathcal {M}: , , mathbb {P} ( {{f (m_2) | f (m_2) neq f (m_1) }}) = 1 – frac {c} {2 ^ {256}}$$

Where $$mathbb {P}$$ It is the uniform distribution in the set of binary chains of length. $$256$$ Y $$c$$ it's a constant that $$c ll 2 ^ {256}$$. For example, $$c leq 2 ^ {10}$$.

Are there alternative interpretations? A formal definition maybe?

## A Gdax and Bitmex Bot for daily trading in CryptoCurrency currencies. – Advertising, Offers.

By

blockchainerz

Facebook, is a social network based on Web2 that has 2.4 billion active users, has recently announced a change to Web3. The medium "Facebook launching a cryptocurrency" is being introduced. The reality is more complex. Facebook plans to launch its own distributed ledger, token, Libra, and Calibra's wallet manages a native token. Libra It is assumed that through the back door of this portfolio, the company is making a move to govern the future of providing global sovereign digital identities. Facebook users will be able to use the Libra tokens for online purchases and P2P remittances through that wallet for "low to none" rates. The technical document indicates that Libra, a stable token, is backed by several fiduciary currencies. It would be competition for state-of-the-art technology providers and credit card companies. It could be a token that charges merchants as much as 2.5 percent or more in purchases. In addition, it could compete with money transfer companies that many expatriates around the world use to send funds to friends and family in their home countries.

Key elements of the Libra Network:

Infrastructure

Libra runs in a federated authorized and distributed ledger, as a security mechanism. That does not use a "chain of blocks". Security is established through legal contracts between known actors of that federated network. However, a time-stamped block chain is very important without empowerment networks like Bitcoin or Ethereum. Where validators can join and leave the network at any time, without permission.

Consensus

An authorization system is executed from Libra. You can use a more productive consensus algorithm that does not need batch transactions. Because transaction history is much less likely to work in a configuration of trusted network actors. Libra uses a modified version of Practical Byzantine Fault Tolerance (BFT), an experienced and well-known algorithm. LibraBFT seems to be a bifurcation of the consensus protocol of hot things.

Smart contracts

The Libra network is delineated with intelligent contract capabilities. "Move" is the vernacular intelligent contracting language. The question of whether it makes sense to develop a new language without solving. It is still unpredictable how the language is rich in functions and user-friendly for the developer. Both aspects would influence the problems of the security of intelligent contracts and the resistance of the network. Like Ethereum, Libra's smart contracts will have mandatory network payments to execute the code. It means that all operations require the payment of Libra as network transaction fees.

Governance of the Network

As an authorized general ledger, only members of the network can validate transactions. "The founding members are consortiums with established reputations, which makes it unlikely that they act with malice," says the white paper. These activities range from traditional payment networks. "(Mastercard, Visa) for the Internet giants and concert economy (eBay, Lyft) to block the natives, (Xapo) for VCs (Andreessen Horowitz, Thrive Capital)". Another range of participation nodes from NPO to Paypal, Uber, Women & # 39; s World Banking, Mercy Corps, etc. The Libra Association is based in Geneva, Switzerland, with almost 30 interested parties so far.

conclusion

Facebook is planning to launch a cryptocurrency. It is expected to "transform the global economy". As seen in the ongoing speech of Facebook with the CFTC. The era of unregulated digital assets has come and gone: ICOs currently increase 58 times less than a year ago, while STOs increased 130% in the first quarter of 2019.

.

## Wallet: how to create generation addresses for various crypto currencies

I am trying to understand how to generate 1 mnemonic addresses for multiple cryptocurrencies.

I needed a read-only wallet, so I think I need to use a `xpub` key. But I am struggling to understand the structure of the wallet tree and how to achieve it.

So far I can do this.

``````const bip39 = require (& # 39; bip39 & # 39;)
var HDKey = require (& # 39; hdkey & # 39;)

const mnemonic = bip39.generateMnemonic ()
const seed = bip39.mnemonicToSeedSync (mnemonic) .toString (& # 39; hex & # 39;)
let hdkey = HDKey.fromMasterSeed (Buffer.from (seed, & # 39; hex & # 39;))
var childkey = hdkey.derive ("m / 0/0/1")
console.log (childkey.publicExtendedKey);

leave publicExtendedKey = hdkey.publicExtendedKey;
hdkey.publicKey
``````

My requirement is that a user can save 1 mnemonic on their side, and I have a public key with which I can generate addresses for multiple cr

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Bitcoin declined more than 10% today, falling to almost \$ 7,500 as the broader digital currency markets experienced a decline.

At this point, the cryptocurrency had lost approximately 12% in the last 24 hours and more than 17% from its recent peak of almost \$ 9,100, as revealed by the additional CoinDesk figures.

They are offering the best bitmex bot, as you can see in the previous image, #TRX orchestrating the continuous gain.

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## How do I get the inflation rate for cryptographic currencies in csv format?

I am interested in doing an econometric analysis of cryptographic currencies and yields based on the inflation rate. Any csv file that has all the currencies and the inflation rate? Thank you.

## Is there a way to trade digital currencies locally after the ban on local currencies? [on hold]

I can not find local btc merchants at localbitcoins.com.
Is there any way to buy bitcoins locally, since I am in a country where exchanges do not allow me to buy btcs from Syria?

## Is there a bitcoin scanner that can visualize where currencies flow from one direction to another?

I remember playing with a bitcoin explorer tool a while ago, where you can enter a bitcoin address and see all the currencies flowing in a tree-like structure. I want to be able to investigate where the coins come from. What is this tool?

format: plain


text:
# zbMATH — the first resource for mathematics

Etale local strucure of matrix invariants and concomitants. (English) Zbl 0634.14034
Algebraic groups, Proc. Symp., Utrecht/Neth. 1986, Lect. Notes Math. 1271, 143-175 (1987).
[For the entire collection see Zbl 0619.00008.]
The authors study the variety $$V_{mn}$$ built as an “approximation” of the space of orbits of $$X_{mn}=\oplus^{m}_{i=1}M_ n({\mathbb{C}})\quad under$$ action by componentwise conjugation of $$GL_ n({\mathbb{C}})$$ by using $$GL_ n({\mathbb{C}})$$-invariants as parameters of $$V_{mn}$$. It parametrizes naturally the closed orbits of this action. The coordinate ring of $$V_{mn}$$ is the so-called trace ring $$\pi_{mn}$$ of m generic $$n\times n$$-matrices [see C. Procesi; Adv. Math. 19, 306-381 (1976; Zbl 0331.15021)]. A point $$\xi$$ of $$V_{mn}$$ is of representation type $$\tau =(e_ 1,k_ 1;...;e_ r,k_ r)$$ provided the corresponding isomorphism class of semisimple representations is built from r distinct simple components of dimensions $$k_ i$$ occuring with multipicities $$e_ i$$. It is shown that $$V_{mn}(\tau)$$, the subset of $$V_{mn}$$ consisting of all points of representation type $$\tau$$, form a finite stratification into locally closed smooth subvarieties, where $$V_{mn}(\tau)$$ lies in the closure of $$V_{mn}(\tau ')$$ if and only if $$\tau$$ is a degeneration of $$\tau '$$. Furthermore, there is an explicit determination of the étale local structure of points of representation type with multiplicities 1. It is shown that this yields a Cohen-Macaulay module and its Poincaré series satisfies a certain functional equation.
These results are used in order to show that the singular locus of $$V_{mn}$$ is determined by the Formanek center of $$\pi_{mn}$$. It concludes with the solution of the regularity problem for trace rings of generic matrices, i.e., $$gl\dim (\pi_{mn})<\infty$$ if and only if m or n is equal to one or $$(m,n)=(2,2), (2,3)$$ or (3,2). The authors use the work of D. Luna [Bull. Soc. Math. Fr., Suppl., Mém. No.33, 81-105 (1973; Zbl 0286.14014)] and R. P. Stanley [Invent. Math. 68, 175- 193 (1982; Zbl 0516.10009)].
Reviewer: P.Schenzel

##### MSC:
 14M12 Determinantal varieties 20G05 Representation theory for linear algebraic groups 15A24 Matrix equations and identities 14L30 Group actions on varieties or schemes (quotients)


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# Adiabatic flame temperature

1. ### Jaiganesh S

1
Hi to all,
I want to know the temperature after combustion of methane with air. How will i calculate it? Kindly inform me.

Regards,
Jaiganesh.S

2. ### janhaa

97
Hi, first you need a balanced reaction eq with air

$$CH_4 + 2O_2 + 7,43N_2 => CO_2+2H_2O+7,43N_2$$

then you can use this relation;

$T_f=T_o+\frac{n\Delta H_c}{\Sigma nC_p}$

where T_f is Adiabatic flame temperature
T_o : 298 K, n: nb of moles and $$\Delta H_c$$is heat of combustion of methane
and
$$\Sigma nC_p$$
is the thermal capacities of RHS, thus: $$1*C_p(CO_2)+2C_p(H_2O)+7,43C_p(N_2)$$

--------------------------------------------------------------------------------

### Staff: Mentor

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• Customer Solutions ▾
• Competitions
• Community ▾
with —

# Mapping Dark Matter

Finished
Monday, May 23, 2011
Thursday, August 18, 2011
\$3,000 • 72 teams

# All submissions have been rescored which affected the final/private score

« Prev
Topic
 AstroTom Competition Admin Rank 62nd Posts 65 Thanks 21 Joined 14 Dec '10 Email user Thank you for all the feedback which helped track down this issue. It is important to note that this only affected the final re-score and not the live leaderboard during the challenge as Jeff has emphasized. Apologies for this temporary confusion. #2 / Posted 21 months ago
 Rank 2nd Posts 308 Thanks 107 Joined 2 Dec '10 Email user Congratulations to Daniel and David (DeepZot) #3 / Posted 21 months ago
 Rank 8th Posts 21 Thanks 4 Joined 15 May '11 Email user Thanks, Tom and Jeff, everything looks fine now. And congratulations to danielm and davidk (looking forward to seeing your solution...). Ana #4 / Posted 21 months ago
 Rank 6th Posts 41 Thanks 1 Joined 30 Jun '11 Email user I am very glad to see this finally happened. After all, we are scientists. :) (b.t.w, this makes me drop from rank 4 to rank 6, which is supposed to be, I am still very happy about it) I'd love to learn about NN methods(and others), will the top rankers post few details (or even their code) on this forum so that others can learn from them? Best #5 / Posted 21 months ago
 Rank 15th Posts 72 Thanks 12 Joined 4 Mar '11 Email user Congratulations to all the top finishers!! I envy your talents! #6 / Posted 21 months ago
 Rank 1st Posts 8 Thanks 2 Joined 10 Aug '11 Email user Thanks to everyone who helped track this down! The new private scores are now very well correlated with the public scores, leading to identical public/private ranking for our 5 official submissions. We will make our code (4 C++ packages) available soon although it is still under rapid development since we are using the same code for the "Great Challenge 2010", where there is the additional challenge of distinguishing between the (dark matter) shear and (random galaxy shape) intrinsic ellipticities. The other competition ends Sep 2, so I don't expect our code to stabilize before then. Briefly, our method consists of two steps. The first step is a pixel-level maximum-likelihood fit to each star and galaxy image to extract shape parameters (including the ellipticities) and their covariance matrix. The second step is to feed a subset of the fit outputs into a neural network (configured for regression rather than classification) that is trained to provided corrections to the fitted ellipticities. Only the second step was varied to produce our different submissions. Skipping the second step entirely and using the fitted outputs directly gave scores of 0.0151432 (public) and 0.0152543 (private), so the fit is doing most of the work but the NN provided a small but welcome improvement! The fit minimization engine (Minuit) and NN engine (TMVA) we used are both available as part of the open source (LGPL) ROOT data analysis framework (http://root.cern.ch) that is widely used by particle physicists. I recommend checking these out if you haven't already. David #7 / Posted 21 months ago
 Rank 2nd Posts 308 Thanks 107 Joined 2 Dec '10 Email user My code was written in Matlab. I have not cleaned it yet to make it "publishable". However here is short method description: 1. Fix bad pixels, subtract baseline, renormalize images (actually, I am not sure if this step is necessary) 2. Find galaxies and stars centers. Done as weighted mean with threshold. 3. Recenter all images (spline interpolation). 4. Calculate primary components for image stacks. 5. Plug components amplitudes into NN with e1 and e2 as targets. 5a. Repeat #5 multiple times choosing several "best" networks. 5b. Repeat 2-5a slightly changing centering methods and networks parameters. 6. Calculate mean prediction of multiple networks. (My "best of 5" submission was a mean of 35 predictors, each with RMSE<0.015 on training set) #8 / Posted 21 months ago
 Rank 3rd Posts 160 Thanks 29 Joined 8 Jan '11 Email user Congratulations to the winners. My code needs to be cleaned as well. I have several sets of predictors. Below is a short list : 1- Computing e1 and e2 for a Gaussian-smoothed thresholded version of galaxy images. 2- Computing e1 and e2 for a Gaussian-smoothed thresholded version of star images! 3- Computing e1 and e2 for a convolved version of galaxy images. 4- Creating structuring element from the star images and using it to perform basic morphological operations on the galaxies. 5- Computing directions and curvatures of both galaxy and star images. 6- Computing chain codes and edges features from both galaxy and star images. 7- Several of these predictors are also computed on a pi/4 rotated version of the galaxy images. Whenever a method has one or more parameters, each possible value of the parameter will be used to generate a separate predictor. All these predictors have been combined via linear fit. If you want to download them and try other algorithms, you are most welcome: http://www.kaggle.com/c/mdm/forums/t/795/piles-of-data-for-data-lovers #9 / Posted 21 months ago
 Rank 6th Posts 41 Thanks 1 Joined 30 Jun '11 Email user It looks like lots of top winners used NN or regressions to improve the score a lot (I mean relatively) This is what I have done(basically by reconstruction of an image with my model to make it look like the galaxy as similar as possible): 1) fit the star using a fixed function $\frac{1}{(1+r^2)^3}$ where $r^2=\frac{(x-x_c)^2}{a^2}+\frac{(y-y_c)^2}{b^2}$ 2) Minimize chisqure of my model to find out e1,e2:  Assuming the galaxy is an exponential function $\exp(-\sqrt{\frac{(x-x_c)^2}{a^2}+\frac{(y-y_c)^2}{b^2}})$, generate an image with some initial paramters, then convolve it with the fitted star, calculate residual square. Use some Nonlinear optimization package to minimize residual square to find out the parameters in the model. e1,e2 can be calculated from a and b. That's it.  I also tried feeding my results to NN, but it did not help....Maybe I should learn more about NN. EDIT: Used Kaggle's math functionality for exp(-sqrt((x-x_c)^2/a^2+ (y-y_c)^2/b^2))) #10 / Posted 21 months ago / Edited 21 months ago
 Rank 15th Posts 72 Thanks 12 Joined 4 Mar '11 Email user As mentioned in another post, my simplest model was, like woshialex's, untuned. After refilling the "overflow" black pixels with 255 as directed, I de-noised all my images using PCA decomposition and retaining only the first 16 terms in the eigenfunction expansion, prior to all other analysis. The star images were then fit (simple chi-square, top-hat weighted to the middle half of the image) to an elliptical Moffat distribution. This gave a, b, theta and hence ellipticities star.e1 and star.e2 for the PSF. This model was then convolved with another elliptical Moffat function, this time representing the sought-after galaxy, to give an image that was fit to to the observed galaxy image. This resulted in ellipticities pre.conv.e1 and pre.conv.e2, which were the prediction of my simplest model. Considerable improvement was obtained in the form of a "three-epsilon model". In this model an elliptical Moffat profile was also fit to the observed galaxy, yielding a third pair of epsilon values, post.conv.e1 and post.conv.e2. Then I used either simple linear regression or Support Vector Machine to predict e1 and e2. After its kernel and target parameters were optimized for least cross-validation error, the SVM performed equally well to the linear regression, but did not outperform it. The code for this was written in R, with the exception of the PCA decomposition part which was done in SciLab. I'll gladly share it with anyone who asks, if you'll promise not to laugh at my programming style. Remember, they didn't have computers back when I was a kid :-) Main lessons learned for me were (1) that de-noising the images probably wasn't necessary, and could better have been handled by maximum likelyhood method fits (thanks David), (2) that the elliptical Moffat probably wasn't such a good choice in the case of the galaxies (thanks Woshie) and (3) SVM wasn't worth the extra bother in the case of the 3-epsilon model. #11 / Posted 21 months ago
 Rank 1st Posts 8 Thanks 2 Joined 10 Aug '11 Email user Bruce - regarding the de-noising step, my experience is that whenever you can write down a reasonable model for your signal, then fitting it to your noisy data is a (nearly?) optimal way of isolating the signal, and has the added bonus of rigorously propagating noise fluctuations into signal model parameter errors. Of course, this depends on having a good enough signal model which is a bit ambitious when each blurry image represents something as complex as our Milky Way. Also, I tend to see everything as a likelihood fitting problem ;-) David Thanked by Bruce Cragin #12 / Posted 21 months ago
 Rank 8th Posts 21 Thanks 4 Joined 15 May '11 Email user I have tried several things but the best results were obtained  with a combination of principal components analysis and multiple linear regression. Briefly what I did was the following: 1. Consider a central window on each image (different sizes for stars and galaxies) and move this window one pixel in every direction, thus getting 9 images for every galaxy and every star. 2. Transform the 9 images into 9 vectors, perform a principal components analysis on it, and keep the first 3 components. After this step I have 3 sets of images for the galaxies and 3 sets of images for the stars. The first set looks similar to the result of applying a low pass filter to the original images, the second and the third may be interpreted as the result of applying two Sobel filters. 3.  Then I applied again principal components, this time to each of the 6 sets of 40000 images described in 2. 4. The final step was to build a linear regression model using the first few components from the six sets as explanatory variables and the ellipticities as response variables. I had to include second order interactions and powers up to 3. It was also necessary to take into account the structure of the components to build a sensible model. I did not have time to optimize this step. I believe that there is still room for improvment  I have done all the computations in R and I plan to post the code as soon as I clean it. Ana #13 / Posted 21 months ago
 Posts 1 Joined 29 Nov '11 Email user Hello Bruce, I am unsure whether you will get this as this comp is over and dead - I am curretnyl undertaking this challenge as a part of a summer vacation program other uni break and am basically trying various methods to see if I can match some of the top score. I have been following closely your ideas and implementation of finding the ellipticities, however I was wondering what you mean by a 'three epsilon model' - I'm not sure how using the star e1,e2 and postconv e1,e2 can be used - and I would be really interested in learning what you think! On top of that if you would be so kind as to answer a few other questions - if you remember this competition well still =]! I think I'm losing some accuracy in results because my selection of 'where' in the image is too broad and am currently trying to implement a 'snakes' algorithm to get a less background noisy data so my algorithm doesn't try fit it - do you think this will improve a great deal? And my other question was if you knew of a good resource to teach myself how to use linear regression to predict the ellipticities   Thank you very much, Nick. #14 / Posted 16 months ago


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Broad Topics > Calculations and Numerical Methods > Multiplication & division

Sort Them Out (2)

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Can you each work out the number on your card? What do you notice? How could you sort the cards?

All the Digits

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This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?

Stage: 3 Challenge Level:

If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?

Long Multiplication

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A 3 digit number is multiplied by a 2 digit number and the calculation is written out as shown with a digit in place of each of the *'s. Complete the whole multiplication sum.

Multiplication Square Jigsaw

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Can you complete this jigsaw of the multiplication square?

Factor-multiple Chains

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Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?

The Clockmaker's Birthday Cake

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The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?

Two Many

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What is the least square number which commences with six two's?

ABC

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In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?

As Easy as 1,2,3

Stage: 3 Challenge Level:

When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .

What Two ...?

Stage: 2 Short Challenge Level:

56 406 is the product of two consecutive numbers. What are these two numbers?

Eminit

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The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?

Trebling

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Can you replace the letters with numbers? Is there only one solution in each case?

Divide it Out

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What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?

Being Thoughtful - Primary Number

Stage: 1 and 2 Challenge Level:

Number problems at primary level that require careful consideration.

Highest and Lowest

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Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.

Diggits

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Can you find what the last two digits of the number $4^{1999}$ are?

Which Is Quicker?

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Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?

1, 2, 3, 4, 5

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Using the numbers 1, 2, 3, 4 and 5 once and only once, and the operations x and ÷ once and only once, what is the smallest whole number you can make?

Largest Number

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What is the largest number you can make using the three digits 2, 3 and 4 in any way you like, using any operations you like? You can only use each digit once.

What's in the Box?

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This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?

Clever Santa

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All the girls would like a puzzle each for Christmas and all the boys would like a book each. Solve the riddle to find out how many puzzles and books Santa left.

Tom's Number

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Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.

What Is Ziffle?

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Can you work out what a ziffle is on the planet Zargon?

Remainders

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I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?

Difficulties with Division

Stage: 1 and 2

This article for teachers looks at how teachers can use problems from the NRICH site to help them teach division.

How Much Did it Cost?

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Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.

X Marks the Spot

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When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .

Practice Run

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Chandrika was practising a long distance run. Can you work out how long the race was from the information?

Odds and Threes

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A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.

Wild Jack

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A game for 2 or more players with a pack of cards. Practise your skills of addition, subtraction, multiplication and division to hit the target score.

Asteroid Blast

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A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.

Magic Potting Sheds

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Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?

Zios and Zepts

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On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?

Oh! Hidden Inside?

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Find the number which has 8 divisors, such that the product of the divisors is 331776.

Penta Post

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Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?

Napier's Bones

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The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?

Machines

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What is happening at each box in these machines?

Multiplication Squares

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Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.

What's My Weight?

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There are four equal weights on one side of the scale and an apple on the other side. What can you say that is true about the apple and the weights from the picture?

Pies

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Grandma found her pie balanced on the scale with two weights and a quarter of a pie. So how heavy was each pie?

The Pied Piper of Hamelin

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This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!

Fair Feast

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Here is a picnic that Petros and Michael are going to share equally. Can you tell us what each of them will have?

X Is 5 Squares

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Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?

Rabbits in the Pen

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Using the statements, can you work out how many of each type of rabbit there are in these pens?

Ordering Cards

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This problem is designed to help children to learn, and to use, the two and three times tables.

Learning Times Tables

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In November, Liz was interviewed for an article on a parents' website about learning times tables. Read the article here.

Square Subtraction

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Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?

Jumping

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After training hard, these two children have improved their results. Can you work out the length or height of their first jumps?

Shapes in a Grid

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Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?


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## aef321 one year ago how do i find the limit as an goes to infinity of the following? (the equation's in the comments)

1. alffer1

$\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2} -4(\frac{ 4a }{ n }))$ This is the equation in question

2. alffer1

$\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2}) - \frac{ 4 }{ n }\sum_{a=1}^{n}(4(\frac{ 4a }{ n}))$

3. alffer1

we're not sure where to go from here

4. satellite73

looks like you are trying to compute some sort or reimann sum

5. alffer1

exactly. a right riemann sum of -x^2 -4x to be exact.

6. satellite73

don't let the $$n$$ through you off in the summation you are summing over $$a$$ (never seen an $$a$$ used like that before, but no matter) pull it right out front of the sum then use the formula for $\sum_{a=1}^na^2$ and also $\sum_{a=1}^na$

7. tkhunny

After a little algebra. $$-\dfrac{16}{n^{2}}\left[\dfrac{1}{n}\sum\limits_{a=1}^{n}a^{2} + \sum\limits_{a=1}^{n}a\right]$$

8. primeralph

@satellite73 Got this?

9. satellite73

$\frac{ 4 }{ n }\sum_{a=1}^{n}(-(\frac{ 4a }{ n })^{2})$ $=-\frac{64}{n^3}\sum_{a=1}^na^2$ for example @primeralph maybe, i am tired and all this latex is wearing me out

10. satellite73

@tkhunny i think the $$4$$ is being squared as well

11. tkhunny

Whoops. Make that 64, instead of 16.

12. satellite73

not really sure the sum is correct, but assuming it is for the moment $-\frac{64}{n^3}\sum_{k=1}^nk^2=-\frac{64}{n^3}\frac{n(n+1)(2n+1)}{6}$ for the first one

13. alffer1

Wait but how does this help us get the limit as n -> infinity?

14. satellite73

taking the limit as $$n\to \infty$$ is very easy because the numerator is a polynomial of degree 3 will leading coefficient -128 and the denominator is a polynomial of degree 3 with leading coefficent 6

15. satellite73

as you remember from some pre calc course about "horizontal asymptotes" that makes the limit $$-\frac{128}{6}$$

16. alffer1

oh oops duh that makes sense.

17. satellite73

easy if you think about it that way instead of the way they do it in the text by dividing top and bottom by $$n^3$$ and all that other nonsense

18. aef321

Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

19. alffer1

@satellite73 where did -128 come from?

20. alffer1

sorry probably a dumb question but...

21. alffer1

oh waitn.....nevermind


text:
SEARCH HOME
 Math Central Quandaries & Queries
 Question from Donald: If I save a penny a week,double it every week. How many weeks will it take to reach 1,000,000 dollars

Hi Donald,

You are working with pennies and there are 100 pennies in a dollar so I suggest you look for how many weeks it will take to reach 100,000,000 cents.

Do an experiment to get a feeling for how fact the number of pennies grows. Get a sheet of lined pages similar to the one below

Treat the lines on the page as weeks and on each line write how many pennies you have that week. 1 on the first week, 2 on the second week and so on. There are probably less than 30 lines on the page so it won't take long.

If you want to approach the problem algebraically you need to solve $2^n = 10,000,000$ for $n.$ The standard way to solve this is to use logarithms.

I hope this helps,
Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.


text: It's also the world's densest wood, and has such unusual properties that the USS Nautilus, the world's first nuclear-powered submarine, has its aft main shaft strut bearings made out of the stuff. The densest hardwood that is traded/ commercially available is Lignum Vitae. Lignum Vitae TROPICAL HARDWOOD Uses: Owing to Lignum Vitae's self-lubricating properties it is used for ship’s propeller bushes and bearings, marine equipment, thrust bearings, pulley sheaves, wheels, guides, rollers and blocks, mallet heads, “woods” for bowls, cotton gins, dead eyes, die cutting and turning, and for any purpose where lubrication is impractical or unreliable. Lignum Vitae, Latin for "Tree of Life," is the national tree of the Bahamas. What is Lignum Vitae? Measurement is of largest Mandrel England, circa 1890. Guaiacum officinale is protected under … The plant is extremely ornamental, producing an attractive blue flower and orange-yellow fruit,… It is native to Central America as well as Jamaica. The actual wood cells are filled with water and then, after drying, air. It is rumoured that Christopher Columbus used Lignum Vitae from the new world to cure a case of syphilis (which he also contracted in the new world)! See more. The fixed valve face is of lignum vitae in a metal recess, and the revolving valve of gun-metal. For kendo, most bokkens are made from bamboo. Also lignum vitae or any other oily tropical wood needs to be carefully prepared before applying poly or spar, this is a resource-intensive process, if not done correctly, you can get a … Lignum vitae has a trunk like a bonsai and can live for thousands of years. Wood from the Lignum Vitae tree is extremely versatile and has served many uses over time. This dense wood was once popular for use in the manufacture of bowling balls and has also been used … Proceeds of the fundraising auction were split between the builders and a scholarship fund was established to support woodworking education. Bamboo is light, elastic/springy so you can hit another bamboo bokken and it won’t splinter or shatter upon impact, and it is durable. The main purpose of having a scientific name is to have a same name accepted and used worldwide. When working with Lignum Vitae you’ll notice its perfume-like scent. This tree is indigenous to the Caribbean and the northern coast of South America. The wood's high-resin content and pressure-resistance made it ideal where some natural lubricating quality was necessary, as in the large warship's rigging blocks. It is self lubricating and allowed accurate time measurement for longitude calculations. I know that it is used for buide blocks on band saws. Lignum vitae . Lignum vitae definition is - the very hard heavy wood of any of several tropical American guaiacums. It is extremely dense therefore durable in the outdoors. Lignum vitae seems to be the standard wood for this, as it definitely holds up well. Lignum vitae Lignum vitae A hard wood, traditionally used for Parrel balls, for Blocks and for Fairleads.. Lignum Vitae: A smooth hard wood grown in the West Indies which had many maritime uses, such as deadeyes and the sheaves of blocks. It prevents the building up of Uric Acid in joints. Lignum Vitae is excellent for use where mechanical lubricant is difficult or impossible to apply. Nov 12, 2005 #3 I guess it is what they used. by comparison, Oak, itself hard enough to be used for products such as oak flooring, typically measures between 5,000N and 6,000N. Because it has a large amount of resin within it as well as its density, it is naturally lubricated and was thus used as ball bearings on ships. All real lignum vitae has a botanical name that starts with "Guaiacum". Lignum Vitae is a tree. Commonly used in tool handles, mallet heads, bearings, bushings, pulley wheels, and turned objects. The short, compact tree is native to continental tropical American and the West Indies. Of the six species, only two (Guaiacum sanctum and Guaiacum officinale) are commonly used by the woodworking industry. Page - 5 Genuine lignum vitae is now nearly impossible to get, and our last real shipment was in 2007. Even in modern times, lignum vitae has been used for underwater propulsion parts because it often performs better than steel. A Latin term that translates to "tree of life." It is a very heavy wood which will sink in water. Find Lignum Vitae Wood Suppliers. Before polymers, alloys and composites, the wood of Guaiacum officinale, also known as tree of life, was extremely important due to its combination of strength, toughness and density.The common name, Lignum vitae (tree of life or wood of life), comes from its historic medicinal use as a remedy for conditions from arthritis to coughs to syphilis.. Lignum vitae (Guaiacum officinale) medicinal herb is commonly known as Guayacan, Palo Santo, Wood of life, Guaiac, Lignum vine, Guaiacum officinale as well as Holy tree and is the source of guiacum resin. Awards Ceremony and Fundraising Auction. It grows upto 10 M. Best used for Gout. lignum vitae This is the wood used to construct a sea-going chronograph that won the contest set by the British monarchy. The wood is sometimes sold by chemists in the form of fine shavings, and as such called Lignum Vitae, which are turned green by exposure to the air, and bluish green by the action of nitric fumes. Underneath the smooth, beige/grey bark of Lignum vitae is some of the heaviest of all wood, sinking under its weight instead of floating in water. I found this description on Wikipedia: Lignum vitae is hard and durable, and is also the densest wood traded (density: 1.23 g/cm3);[3] it will easily sink in water. The wood is full of resin and was once used for the bearings of boat propeller shafts due to its self lubricating properties. It is an evergreen. Lignum vitae is the common name for a group of six species from the genus 'Guaiacum', best known for their strength, density, and durability. Olive to dark brown or black coloured wood from Central America and northern South America. L ignum Vitae, Latin for the ‘tree of life’, has a set of properties that cause a newfound awe in natural materials.Also known as Ironwood, it is the hardest and heaviest traded wood, being 3 to 4 times the hardness of English Oak. Because of oily resins, requires special surface treatments for satisfactory gluing. Locally, it grows best in the dry woodland along the north and south coasts of the island. Request for quotations and connect with international Lignum Vitae Wood manufacturers. Durability: The heartwood is very resistant to attack by decay fungi, termites, and marine borers. Lignum Vitae is one of the hardest and heaviest commercially traded timbers in the world. A sphere of this wood with a diameter of 7.60 $\mathrm{cm}$ has a mass of 313 $\mathrm{g}$ . Joined May 1, 2005 Messages 2,709 Location Newark, Delaware, USA. (a) What is the density of the lignum vitae sphere? It grows in a tropical climate. Lignum Vitae trees grow in tropical regions of the Americas. has traditionally been used. The wood turns and shapes well and takes a high polish. Lignum vitae is a hard, durable, and extremely dense wood used to make ship bearings. Greenwich observatory is 0 degrees, because of where it was developed. (b) Will the sphere float or sink in … In Central America it is also referred to as guaiac meaning, "woods of life." Thanks for the compliments. Because of its strength, this wood has been used for building homes and constructing furniture. It measures an impressive 4,500lbf (20,000N) on the Janka Hardness Scale, a universal measurement of how easy wood is to dent. The related Creosote Bush of the American deserts can live for 11 thousand years or more,click for more info. Lignum vitae is among them. Lignum vitae … A collection of 12 Lignum Vitae metalworkers tools, used to forge or shape metal around. The lignum vitae mallet heads usually end up being 2 to 3 times larger diameter than the metal mallet of the same weight. B. BigRob777 Member. A suitable substitute is Vera or Argentine Lignum Vitae. Lignum Vitae: Contributors Cellulose, the stuff Ma Nature makes her wood with, is heavier than water. It is the World's second densest wood and is usually half or less expensive. Argentine lignum vitae (bulnesia sarmientoi) and Verawood (bulnesia arborea) are not lignum vitae and are not the same as lignum vitae. The scientific name of Lignum Vitae is the botanical name or formal name. : It is also a very hard and heavy wood, even more so than lignum vitae, or ironwood. This naturally green wood has been used for time untold in construction due to the fact that it is so hard and dense. Nicely done. Because of this, Lignum Vitae and Jamaica are closely associated in shipyards worldwide. Lignum Vitae in Spanish can be different from Lignum Vitae in English. Is Lignum Vitae the wood that they used to make piston out of for the steamships? : Closer scrutiny reveals that they are made of lignum vitae, an exotic West Indian hardwood characterized by occasional streaks of bright yellow. Greenish brown to brown with a tight and pronounced interlocked grain, very dense, waxy – with self-lubricating qualities, hard wearing in almost all the contexts in which lignum vitae (Guiaicum spp.) Please refer to the Catalogue or Timberline's Shop for up-to-date prices. Note: Guaiacum Officinale relieves the inflammation and pain in case of Gout attacks. The box was fashioned of some light-colored wood, perhaps lignum vitae. Lignum vitae. What we sell is accurately named. lignum vitae furniture isn’t so popular and there are reasons for this, firstly, lignum vitae is endangered and hard to get. Use and Management. The resin of the tree is guaiacum gum which has been used as a treatment for arthritis, gout and syphilis. The balance of air/water and cellulose determines the weight of the wood. Lignum vitae definition, either of two tropical American trees, Guaiacum officinale or G. sanctum, of the caltrop family, having very hard, heavy wood. Family: Zygophyllaceae . Wood against wood is not as harsh or potentially damaging to my precious gouges. Wonderful patination of the wood. The lignum vitae (Guiacum Officinale) was found in Jamaica by Christopher Columbus. Lignum vitae is a hardwood that came from trees with genus Guaiacum. So, as to what makes it so heavy, the basic answer is cellulose density. Lignum Vitae registers 4,500 pounds-force (lbf) on the Janka Hardness Scale, over twice that of Bois d’arc (Osage orange) wood, a more common hardwood in North America. Growth rate is about two feet a year. It was the alleged medicinal properties of Lignum Vitae which have earned it the title ‘tree of life’. The wood was once used as propeller shaft bearings in nearly all the ships sailing the ‘Seven Seas’. The reason it isn’t as well-known as more popular wood types, like oak or pine, is that it’s slow-growing, heavy, and very hard to work with. The north and South coasts of the wood turns and shapes well and takes high. For up-to-date prices and extremely dense wood used to make piston out of for the steamships steel! Be different from lignum Vitae, USA national tree of life. resin of the island easy... As propeller shaft bearings in nearly all the ships sailing the ‘ Seas., Delaware, USA to apply answer is cellulose density of Uric Acid in.. Shape metal around is difficult or impossible to apply the density of the Vitae! Hard enough to be the standard wood for this, as to what makes it so heavy the... Standard wood for this, as to what makes it so heavy, the basic answer is cellulose.! Latin term that translates to  tree of life. performs better than steel name that with... Because of its strength, this wood has been used as propeller shaft bearings in nearly all the ships the! 'S Shop for up-to-date prices also referred to as guaiac meaning,  of! Fashioned of some light-colored wood, even more so than lignum Vitae?! National tree of life, '' is the density of the fundraising were. Having a scientific name is to dent are closely associated in shipyards worldwide an exotic West Indian characterized... Green wood has been used for the bearings of boat propeller shafts due to its self lubricating properties of. Jamaica are closely associated in shipyards worldwide Jamaica by Christopher Columbus as guaiac,... The heartwood is very resistant to attack by decay fungi, termites, and our real... Ships sailing the ‘ Seven Seas ’ to what makes it so,! Balance of air/water and cellulose determines the weight of the tree is indigenous to Catalogue. Bokkens are made of lignum Vitae sphere Newark, Delaware, USA be used for underwater propulsion parts because often! The north and South coasts of the island 11 thousand years or more click., this wood has been used as a treatment for arthritis, Gout and syphilis is lignum Vitae wood.. West Indian hardwood characterized by occasional streaks of bright yellow make ship bearings available is lignum Vitae is nearly. Fact that it is so hard and heavy wood which will sink in water so hard and.. Coasts of the same weight Vitae metalworkers tools, used to make ship.. A ) what is the national tree of life ’ the Americas guess! Trees grow in tropical regions of the American deserts can live for of! Revolving valve of gun-metal inflammation and pain in case of Gout attacks nov 12, 2005 Messages 2,709 Location,! The ships sailing the ‘ Seven Seas ’ Catalogue or Timberline 's Shop for up-to-date prices allowed accurate measurement... Over time characterized by occasional streaks of bright yellow trees grow in tropical of! Traded/ commercially available is lignum Vitae in English Seas ’ suitable substitute is Vera or Argentine Vitae... Relieves the inflammation and pain in case of Gout attacks resistant to attack by decay fungi termites. Split between the builders and a scholarship fund was established to support woodworking education and can for. Wood against wood is full of resin and was once used as propeller shaft bearings in nearly all the sailing., requires special surface treatments for satisfactory gluing air/water and cellulose determines the weight of the island so. A hardwood that is traded/ commercially available is lignum Vitae has been used for underwater parts. Shapes well and takes a high polish products such as Oak flooring, measures!


text:
# Math Help - This is a straight line?

1. ## This is a straight line?

Can someone please point out to me how the following reduces to y=mx+b:

$\frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0$

Thanks

2. The only way for a fraction to be zero is when?

3. Let $\displaystyle Y = \frac{dy}{dx}$, which means $\displaystyle \frac{dY}{dx} = \frac{d^2y}{dx^2}$ and the DE becomes

\displaystyle \begin{align*} (1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx} &= 0 \\ \int{(1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx}\,dx} &= \int{0\,dx} \\ \int{(1 + Y)^{-\frac{3}{2}}\,dY} &= C_1 \\ \frac{(1 + Y)^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 &= C_1 \\ -2(1 + Y)^{-\frac{3}{2}} &= C_1 - C_2 \\ (1 + Y)^{-\frac{3}{2}} &= -\frac{1}{2}(C_1 - C_2) \\ 1 + Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} \\ Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1 \end{align*}

So $\displaystyle \frac{dy}{dx} &= C$ where $\displaystyle C = \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1$.

Can you go from here to solve for $\displaystyle y$?

4. Originally Posted by Prove It
Let $\displaystyle Y = \frac{dy}{dx}$, which means $\displaystyle \frac{dY}{dx} = \frac{d^2y}{dx^2}$ and the DE becomes

\displaystyle \begin{align*} (1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx} &= 0
So either [tex]\frac{dY}{dx}= 0[//tex] or $(1+ Y)^{-\frac{3}{2}}= 0$. But that second equation is impossible.

That's essentially what Ackbeat said.

\begin{align*} \int{(1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx}\,dx} &= \int{0\,dx} \\ \int{(1 + Y)^{-\frac{3}{2}}\,dY} &= C_1 \\ \frac{(1 + Y)^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 &= C_1 \\ -2(1 + Y)^{-\frac{3}{2}} &= C_1 - C_2 \\ (1 + Y)^{-\frac{3}{2}} &= -\frac{1}{2}(C_1 - C_2) \\ 1 + Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} \\ Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1 \end{align*}

So $\displaystyle \frac{dy}{dx} &= C$ where $\displaystyle C = \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1$.

Can you go from here to solve for $\displaystyle y$?

5. Is there not a way to delete a post?

6. Originally Posted by rainer
Can someone please point out to me how the following reduces to y=mx+b:

$\frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0$

Thanks
A question for the OP. Is it

$\frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0$ or $\frac{y''(x)}{(1+y'^2(x))^{\frac{3}{2}}}=0$

The first looks odd. The second - standard curvature. Just curious :-)

7. Originally Posted by zg12
Is there not a way to delete a post?
Just report your own post and ask to delete. If it is appropriate, a moderator will delete your post. The report post button is in the lower left corner of your post - looks like a triangle with an exclamation mark inside it.

8. Originally Posted by Danny
A question for the OP. Is it

$\frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0$ or $\frac{y''(x)}{(1+y'^2(x))^{\frac{3}{2}}}=0$

The first looks odd. The second - standard curvature. Just curious :-)

Unless the author has made a type-o, it's the first one, which is the outcome of plugging the arclength formula into the Euler-Lagrange equation.

9. Originally Posted by zg12
Is there not a way to delete a post?
You can click "Edit" and then "Delete Message".

10. Originally Posted by Prove It
You can click "Edit" and then "Delete Message".
This works for you, because you are have the MHF Expert badge. There is no way for zg12 directly to delete a post. A moderator must do it for him.

11. Originally Posted by Ackbeet
This works for you, because you are have the MHF Expert badge. There is no way for zg12 directly to delete a post. A moderator must do it for him.
Oops - didn't realise that was only made possible by the badge. Thanks

12. Originally Posted by rainer
Unless the author has made a type-o, it's the first one, which is the outcome of plugging the arclength formula into the Euler-Lagrange equation.
You mean $L = \sqrt{1+y'^2}$. If so, then it would be the second, wouldn't it?


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1. ## Integrating factor (ODE)

the problem goes..

The integrating factor for the ODE dxdt - 2xt = t is the integrating factor e*(-2integral(tdt)) = e^-t^2

multiplying the ODE by the integrating factor:
ddt((e^-t^2)*x) = t(e^-t^2)

integrate : e^-t^2 = -0.5(e^-t^2) + c

how does one get the integation bit?? i cant understand it at all =( .. especially the -0.5 ( .... ) +c bit

i tried by parts..but =S.. i end up integrating again and again....and again =| wow

2. As you said, The integrating factor is $e^{-t^{2}}$

Multiply through:

$\frac{d}{dt}[xe^{-t^{2}}]=te^{-t^{2}}$

Integrating 'negates' the derivative on the left:

$xe^{-t^{2}}=\frac{-1}{2}e^{-t^{2}}+C$

Divide by $e^{-t^{2}}$ and remember the log laws:

$x=Ce^{t^{2}}-\frac{1}{2}$

3. Originally Posted by galactus
As you said, The integrating factor is $e^{-t^{2}}$

Multiply through:

$\frac{d}{dt}[xe^{-t^{2}}]=te^{-t^{2}}$

Integrating 'negates' the derivative on the left:

$xe^{-t^{2}}=\frac{-1}{2}e^{-t^{2}}+C$

Divide by $e^{-t^{2}}$ and remember the log laws:

$x=Ce^{t^{2}}-\frac{1}{2}$

what do you mean by 'negates'? [sorry if its a stupid simple question!]
i can see how if i differentiate..i get the original equation.. but i cannt see how to integrate it yet! =(
if i do it by parts... i have to keep integrating forever...

do you just see the answer clearly by looking?

4. "Undoes" is probably a better word than "negates".

Integrating undoes the derivative.

So if $\frac{d}{dt}[xe^{-t^2}] = te^{-t^2}$

$\int{\frac{d}{dt}[xe^{-t^2}]\,dt} = \int{te^{-t^2}\,dt}$

The left hand side is $xe^{-t^2} + C_1$.

Now use a $u$ substitution on the RHS.

Let $u = -t^2$ so that $\frac{du}{dt} = -2t$.

So the RHS becomes

$\int{te^{-t^2}\,dt} = -\frac{1}{2}\int{-2te^{-t^2}\,dt}$

$= -\frac{1}{2}\int{e^u\,\frac{du}{dt}\,dt}$

$= -\frac{1}{2}\int{e^u\,du}$

$= -\frac{1}{2}e^u + C_2$

$= -\frac{1}{2}e^{-t^2} + C_2$.

Therefore, upon integrating both sides, the DE is

$xe^{-t^2} + C_1 = -\frac{1}{2}e^{-t^2} + C_2$

$xe^{-t^2} = -\frac{1}{2}e^{-t^2} + C$, where $C = C_2 - C_1$

$x = -\frac{1}{2} + Ce^{t^2}$.


text:
Browse Questions

# The cycas, diploxylic condition is found in :

$\begin {array} {1 1} (1)\;Stem & \quad (2)\;Root \\ (3)\;Corolloid\: root & \quad (4)\;Leaflet \end {array}$

Can you answer this question?


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# Algorithm for constructing BST from post-order traversal

Given a post-order traversal of Binary Search tree with $k$ nodes, find an algorithm that constructs the BST.

My Algortihm

Let $n$ represent the next element to be inserted.

Let $P(y)$ represent the parent of node $y$.

1. We will read the traversal in reverse. The last element of the traversal is the root. Let $l = root$. $l$ will represent the element last inserted in the BST (except for the 3rd case below- where it will change to the parent).

Loop the following till there's no element left to be inserted

1. if $l<n$ then $n$ is the right child of $l$. For the next insertion, $l$ changes to it's right child and $n$ becomes the next element(in reverse order of traversal ).

2. else, if $l>n$ and $P(l)<n$ then $n$ is the left child of $l$. For the next insertion, $l$ changes to it's left child and $n$ becomes the next element(in reverse order of traversal).

3. else, if $l>n$ and $P(l)>n$ then $l$ becomes $P(l)$.($n$ hasn't been inserted - we loop with $l$ changed)

[Let $P(root)=- \infty$, so that the $2^{nd}$ case applies]

Complexity Analysis : Every element may contribute at max 3 comparisons, 1 each for - left child, right child and for finally leaving i.e. subtree has been constructed. Even if I missed a comparison or two, It should be constant no. of comparisons per element and no. of operations for node construction will also be constant per element. Hence, giving $O(k)$ time complexity.

Actual Question

If the algorithm is correct, I need the correctness proof for it. Yes, I thought I had the proof but then brain got fried and I am stuck and unable to reason succinctly.
If the algorithm is incorrect, then why? And what is time complexity of the most efficient algorithm for the same question?

Also, is the $O(k)$ complexity correctly calculated - irrespective of the correctness of the algorithm?

• @TerenceHang Do we have an $O(n)$ algo for constructing Binary tree from post and inorder traversal? Because for comparing two elements we will need to refer the inorder and that won't be constant time comparison. May be we can hash the indexes of elements in inorder traversal in $O(n)$ time and then we can compare the indexes, but that would require restriction on key values. – PleaseHelp Sep 8 '15 at 3:24
• An $O(n)$ algo exists, based on node matching instead of relative order. Take the inorder and postorder traversal as 2 stacks $In$ and $Post$, adding a 3rd stack $stk$. 1. pop the root off $Post$, inserting to $stk$, and set insert position at right child of root. 2. whenever top of $In$ matches top of $stk$, pop this node $x$ from both $In$ and $stk$, changing insert position to left child of $x$. 3. pop a new node $n$ from $post$, insert $n$ at current insert position, and push $n$ on $stk$, setting insert position to right child of $n$. Repeat 2,3 until all 3 stack are empty. – Terence Hang Sep 8 '15 at 5:00
• Just find a $O(n)$ recursive algo [1] (It is for construction binary tree from preorder and inorder. Scanning in reverse order and changing the target we can make it for construction from postorder and inorder.) – Terence Hang Sep 8 '15 at 5:26
• What have you tried towards showing correctness, and where did you get stuck? Have you tried a more rigorous runtime analysis, too? See our reference questions for some pointers. – Raphael Sep 8 '15 at 7:17

You are in the right track. But the algorithm is incomplete. You missed the case inserting element on the left sub tree and back.

Here is the modified algorithm: (Changes marked in bold)

Let $n$ represent the next element to be inserted.

Let $P(y)$ represent the parent of node $y$.

Let $g=G(y)$ represent the first node g on the path $y\to root$ such that $P(g)<g$.

1. We will read the traversal in reverse. The last element of the traversal is the root. Let $l = root$. $l$ will represent the element last inserted in the BST (except for the 3rd case below- where it will change to the parent). Let $g=root$ tracking $G(l)$, initialize empty stack $stkG$ storing previous $g$'s on current path.

2. Loop the following till there's no element left to be inserted

1. if $l<n$ then $n$ is the right child of $l$. For the next insertion, $l$ changes to it's right child and $n$ becomes the next element(in reverse order of traversal ). Push $g$ on $stkG$, and let $g=l$.

2. else, if $l>n$ and $\textbf{P(g)<n}$ then $n$ is the left child of $l$. For the next insertion, $l$ changes to it's left child and $n$ becomes the next element(in reverse order of traversal).

3. else, if $l>n$ and $\textbf{P(g)>n}$ then $l$ becomes $\textbf{P(g)}$ and pop $g$ from $stkG$.($n$ hasn't been inserted - we loop with $l$ changed)

[Let $P(root)=- \infty$, so that the $2^{nd}$ case applies]

For correctness, you can prove the following loop invariant:

1. Root is the first inserted element. for each insertion except root, the parent element has been inserted before.
2. Each insertion $n$ correctly maintains the order between $n$ and $l$. ($n<l$ if $n$ is left child of $l$, $n>l$ otherwise)
3. After backtracking step 3, the next insertion always occurs on the left branch.

1 and 3 ensures the post-order traversal, while 2 ensures the BST.

For complexity:

1. insertion cost: each element is inserted exactly once.
2. traversal and comparison: the algorithm actually performs a post-order traversal in reverse order, with $O(1)$ comparison on each step.
3. $g$,$stkG$ maintain cost: each node, which is right child of parent, is pushed and popped from $stkG$ at most once.

Thus the time complexity is $O(k)$.

## Alternative algorithm:[1]

Using a recursive procedure:

procedure BST_from_postorder(post, len, target)
/*
input: post[0..len-1] -- (partial) postorder traversal of BST
len -- length of post to be processed.
target -- a cutoff value to stop processing
Output: tree_ptr -- pointer to root of tree/sub tree constructed from post[len_rest..len-1]
len_rest -- remaining length that has not been processed.
*/
1. if len <= 0 or post[len-1] <= target, then return null.
2. root <- new Node created from from post[len-1].
3. (root->right, new_len) <- BST_from_postorder(post, len-1, post[len-1])
4. (root->left, rest_len) <- BST_from_postorder(post, newlen, target)
5. return (pointer to root, rest_len)

/* BST_from_postorder(post, length of post, -infinity)
will return the BST construct from given postorder traversal. */

• Hey, thanks for the answer. I understood your algorithm. For the correctness part 1 - this may also be added : "the right element is inserted before the left element". Also, I think my algorithm is correct too. You maintain a stack, so that you can directly jump to $P(g)$ in the $3^{rd}$ case. Instead, my algo does redundant comparison during backtracking from a left-subtree but eventually does reach $P(g)$ and inserts in the correct position. What do you think? – PleaseHelp Sep 7 '15 at 7:44
• There are cases where $n<l<P(l)$, which fall into your $3^{rd}$ case, but you have to decide to insert to left child or to backtrack. $g$ is used not only for quick backtracking, but also for comparison. – Terence Hang Sep 7 '15 at 7:50
• For correctness part, it is included in 3. ie. the traversal is always from right branch to left branch and never in reverse. – Terence Hang Sep 7 '15 at 7:53
• Ah! for $n<l<p(l)$, my algorithm always backtracks. So, if $n<l<p(l)$ & $P(g)<n$ then it will try to insert $n$ as left child of $g$ but that's obviously wrong. – PleaseHelp Sep 7 '15 at 8:42


text:
## anonymous 4 years ago Need help with this integration by parts question. The answer I gave, which is in the gray box, is wrong.

1. anonymous

2. LollyLau

so you're integrating x^2e^-x from 0 to x.

3. LollyLau

nvm

4. anonymous

@LollyLau, that is what I integrated for

5. anonymous

You almost had it. You got the correct antiderivative, $-x^2e^{-x}-2xe^{-x}-2e^{-x}$ But you didn't answer the question, it wants the integral of the original function from x=0 to a. So you must do $\left(-x^2e^{-x}-2xe^{-x}-2e^{-x}\right|_0^a=-a^2e^{-a}-2ae^{-a}-2e^{-a}+2$

6. Zarkon

shouldn't your answer be in terms of a

7. anonymous

Thanks Imperialist and Zarkon, you're both gentlemen and scholars. I attached the screenshot of the right answer.