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# A “fast” way for finding sum of the numbers from $1$ to $100$,which are not divisible by $3$ and $5$

Find sum of numbers from $1$ to $100$ which are not divisible by $3$ and $5$?

I can understand that the here we require to sum up these numbers:

$1+2+4+7+8+11+13+14+16+17+19+22+23+26+28+29+31+32+34$ $+37+38+41+43+44+46+47+49+52+53 +56+58+59+61+62+64+67+68+71$ $+73+74+76+77+79+82+83+86+88+89+91+92+94+97+98$

Which gives the sum $= 2632$,but I any didn't relation among the numbers in this series,any ideas? Also,I am supposed to do this under a mint so please hint/answer accordingly.

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In my opinion, there would be a problem if a test question was worded exactly like the above, for there is ambiguity. The phrase "not divisible by $3$ and $5$" could be interpreted by a student as meaning not divisible by both of $3$ and $5$, that is, everybody but the multiples of $15$. The wording "divisible neither by $3$ nor by $5$" is, by contrast, unambiguous. – André Nicolas Jul 20 '11 at 22:23
Yes,initially I was only taking out the multiples of $15$ from $5050$. – Quixotic Jul 22 '11 at 18:27
That was a sensible interpretation of the substandard wording used in the problem. – André Nicolas Jul 22 '11 at 19:28
but certainly giving me an answer that is not in any of the options and which tends me to check my calculations again and hence waste of precious seconds!:/ – Quixotic Jul 22 '11 at 19:32

This is a combinatorial argument called Inclusion-exclusion principle: $$X=1+\ldots+100 - 3(1+\ldots+33) - 5(1+\ldots+20) + 15(1+\ldots+6)$$

First we add all the numbers, then we remove all those which are divisible by $3$, and those divisible by $5$. But wait a minute! What about $15$? We removed that number twice! We therefore need to add it once, as well the rest of its multiples below $100$.

Now using the summation formula: $1+\ldots+n = \frac{n(n+1)}{2}$ the answer follows.

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+1,nice explanation. – Quixotic Jul 20 '11 at 22:02

The sum of the numbers which are divisible by $3$ in this range is $$3+6+\cdots+99=3(1+2+\cdots+33)$$ and likewise the sum of those divisible by $5$ is $$5+10+\cdots+100=5(1+2+\cdots+20).$$ You should be aware of a quick formula to add these two sums. Be careful to add back all the numbers that are divisible by $15$, as they have been subtracted twice. Thanks to Gauss, we know that the sum of the first $100$ integers is $5050$, and after doing the arithmetic you get your answer.

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Hint: First try finding the sum of numbers which are divisible by $3$ or $5$

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Sub-hint: find the sum of numbers which are divisible by 3, then the sum of those divisible by 5, and finally the sum of those divisible by 15. – Did Jul 20 '11 at 21:49
Okay,for that I have to sum up two G.P series for $3$ and $5$ respectively,then subtract the $15$ series and then subtract this whole result from $5050$? Do you mean this? – Quixotic Jul 20 '11 at 21:51
@Debanjan: They are not GP. They are AP. – Aryabhata Jul 20 '11 at 21:54
Yeps,A.P indeed. – Quixotic Jul 20 '11 at 22:01
Accepting this one as hint is enough for me in this case. – Quixotic Jul 20 '11 at 22:02

There's something called inclusion-exclusion you wanna use : sum all integers up to 100, remove those divisible by 3, remove those divisible by 5, then add those divisible by 15. =)

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# Block walking

In combinatorial mathematics, block walking is a method useful in thinking about sums of combinations graphically as "walks" on Pascal's triangle. As the name suggests, block walking problems involve counting the number of ways an individual can walk from one corner A of a city block to another corner B of another city block given restrictions on the number of blocks the person may walk, the directions the person may travel, the distance from A to B, et cetera.

## An example block walking problem

Suppose such an individual, say "Fred", must walk exactly k blocks to get to a point B that is exactly k blocks from A. It is convenient to regard Fred's starting point A as the origin, ${\displaystyle (0,0)}$, of a rectangular array of lattice points and B as some lattice point ${\displaystyle (e,n)}$, e units "East" and n units "North" of A, where ${\displaystyle e+n=k}$ and both ${\displaystyle e}$ and ${\displaystyle n}$ are nonnegative.

### Solution by brute force

A "brute force" solution to this problem may be obtained by systematically counting the number of ways Fred can reach each point ${\displaystyle X=(x_{1},x_{2})}$ where

${\displaystyle 0\leq x_{1}\leq e}$ and ${\displaystyle 0\leq x_{2}\leq n}$

without backtracking (i.e. only traveling North or East from one point to another) until a pattern is observed. For example, the number of ways Fred could go from ${\displaystyle (0,0)}$ to ${\displaystyle (1,0)}$ or (0,1) is exactly one; to (1,1) is two; to (2,0) or (0,2) is one; to (1,2) or (2,1) is three; and so on. Actually, you could receive the number of ways to get to a particular point by adding up the number of ways you can get to the point south of it and the number of ways you can get to the point west of it.(With the starting point being zero and all the points directly north and south of it one.) In general, one soon discovers that the number of paths from A to any such X corresponds to an entry of Pascal's Triangle.

### Combinatorial solution

Since the problem involves counting a finite, discrete number of paths between lattice points, it is reasonable to assume a combinatorial solution exists to the problem. Towards this end, we note that for Fred to still be on a path that will take him from A to B over ${\displaystyle k}$ blocks, at any point X he must either travel along one of the unit vectors <1,0> and <0,1>. For the sake of clarity, let ${\displaystyle E=\langle 1,0\rangle }$ and ${\displaystyle N=\langle 0,1\rangle }$. Given the coordinates of B, regardless of the path Fred travels he must walk along the vectors E and N exactly ${\displaystyle e}$ and ${\displaystyle n}$ times, respectively. As such, the problem reduces to finding the number of distinct rearrangements of the word

${\displaystyle \overbrace {EE\cdots E} ^{e{\text{ times}}}\underbrace {NN\cdots N} _{n{\text{ times}}}}$,

which is equivalent to finding the number of ways to choose ${\displaystyle e}$ indistinct objects from a group of ${\displaystyle k}$. Thus the total number of paths Fred could take from A to B traveling only ${\displaystyle k}$ blocks is

${\displaystyle {\binom {k}{e}}={\binom {k}{n}}={\frac {k!}{e!n!}}.}$

## Other problems with known block walking combinatorial proofs

• Proving that
${\displaystyle \sum _{k=0}^{n}{n \choose k}^{2}\ ={2n \choose n}}$
can be done with a straightforward application of block walking.[1]

format: plain


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# How to use DynForest with categorical outcome?

## Introduction to pbc2 dataset

We use DynForest on the pbc2 dataset (Murtaugh et al. 1994) to illustrate our methodology. Data come from the clinical trial conducted by the Mayo Clinic between 1974 and 1984. For the illustration, we consider a subsample of the original dataset resulting to 312 patients and 7 predictors. Among these predictors, the level of serum bilirubin (serBilir), aspartate aminotransferase (SGOT), albumin and alkaline were measured at inclusion and during the follow-up leading to a total of 1945 observations. Sex, age and the drug treatment were collected at the enrollment. During the follow-up, 140 patients died before transplantation, 29 patients were transplanted and 143 patients were alive. The time of first event (alive or any event) was considered as the event time. We aim to predict in this illustration the death without transplantation on patients suffering from primary billiary cholangitis (PBC) using clinical and socio-demographic predictors, considering the transplantation as a competing event.

## Managing data

For the illustration, we select patients still at risk at 4 years and we recode the event variable with event = 1 for subjects died during between 4 years and 10 years, event = 0 otherwise. We split the subjects into two datasets: (i) one dataset to train the random forest using $$2/3$$ of patients; (ii) one dataset to predict on the other $$1/3$$ of patients.

library("DynForest")
pbc2 <- pbc2[which(pbc2$years>4&pbc2$time<=4),]
pbc2$event <- ifelse(pbc2$event==2, 1, 0)
pbc2$event[which(pbc2$years>10)] <- 0
set.seed(1234)
id <- unique(pbc2$id) id_sample <- sample(id, length(id)*2/3) id_row <- which(pbc2$id%in%id_sample)
pbc2_train <- pbc2[id_row,]
pbc2_pred <- pbc2[-id_row,]

Then, we build the dataframe in the longitudinal format (i.e. one observation per row) for the longitudinal predictors including: id the unique patient identifier; time the observed time measurements; serBilir, SGOT, albumin and alkaline the longitudinal predictors. We also build the dataframe with the time-fixed predictors including: id the unique patient identifier; age, drug and sex predictors measured at enrollment. The nature of each predictor needs to be properly defined with as.factor() function for categorical predictors (e.g. drug and sex).

timeData_train <- pbc2_train[,c("id","time",
"serBilir","SGOT",
"albumin","alkaline")]
fixedData_train <- unique(pbc2_train[,c("id","age","drug","sex")])

The first step aims to build the random forest using the DynForest() function. We need to specify the mixed model of each longitudinal predictor through a list containing the fixed and random formula for the fixed effect and random effects of the mixed models, respectively. To allow for a flexible trajectory over time, splines can be used in formula using splines package.

timeVarModel <- list(serBilir = list(fixed = serBilir ~ time,
random = ~ time),
SGOT = list(fixed = SGOT ~ time + I(time^2),
random = ~ time + I(time^2)),
albumin = list(fixed = albumin ~ time,
random = ~ time),
alkaline = list(fixed = alkaline ~ time,
random = ~ time))

Here, we assume a linear trajectory for serBilir, albumin and alkaline, and quadratic trajectory for SGOT.

Using categorical outcome, we should specify type=“factor” and the dataframe in Y should contain only 2 columns, the variable identifier id and the outcome death.

Y <- list(type = "factor",
Y = unique(pbc2_train[,c("id","event")]))

## Build the random forest

We executed DynForest() function to build the random forest with hyperparameters mtry = 7 and nodesize = 2 as follows:

res_dyn <- DynForest(timeData = timeData_train,
fixedData = fixedData_train,
timeVar = "time", idVar = "id",
timeVarModel = timeVarModel,
mtry = 7, nodesize = 2,
Y = Y, seed = 1234)

## Out-Of-Bag error

With categorical outcome, the OOB prediction error is evaluated using a missclassification criterion. This criterion can be computed with compute_OOBerror() function and we displayed the results of the random forest using summary() as below:

res_dyn_OOB <- compute_OOBerror(DynForest_obj = res_dyn)
summary(res_dyn_OOB)

DynForest executed with classification mode
Splitting rule: Minimize weighted within-group Shannon entropy
Out-of-bag error type: Missclassification
Leaf statistic: Majority vote
----------------
Input
Number of subjects: 150
Longitudinal: 4 predictor(s)
Numeric: 1 predictor(s)
Factor: 2 predictor(s)
----------------
Tuning parameters
mtry: 7
nodesize: 2
ntree: 200
----------------
----------------
DynForest summary
Average depth by tree: 5.85
Average number of leaves by tree: 16.7
Average number of subjects by leaf: 9.31
----------------
Out-of-bag error based on Missclassification
Out-of-bag error: 0.2333
----------------
Time to build the random forest
Time difference of 1.808804 mins
----------------

In this illustration, we built the random forest using 150 subjects because we only kept the subjects still at risk at landmark time at 4 years. We have on average 9.3 subjects by leaf, and the average depth level by tree is 5.8. We also predicted the wrong outcome for 23% of the subjects. This criterion should be minimized as possible, by tuning mtry and nodesize hyperparameters.

## Predict the outcome

We then predict the probability of death on subjects still at risk at landmark time at 4 years. In classification mode, the predictions are performed using majority vote. The prediction over the trees is thus a modality of the categorical outcome along with the proportion of the trees which lead to this modality. Prediction are computed using predict() function, then a dataframe can be easily built from returning object to get the prediction and probability for each subject as followed:

timeData_pred <- pbc2_pred[,c("id","time",
"serBilir","SGOT",
"albumin","alkaline")]
fixedData_pred <- unique(pbc2_pred[,c("id","age","drug","sex")])
pred_dyn <- predict(object = res_dyn,
timeData = timeData_pred,
fixedData = fixedData_pred,
idVar = "id", timeVar = "time",
t0 = 4)
head(data.frame(pred = pred_dyn$pred_indiv, proba = pred_dyn$pred_indiv_proba))

pred proba
101    0 0.960
104    0 0.780
106    1 0.540
108    0 0.945
112    1 0.515
114    0 0.645

As shown in this example, some predictions are made with varying confidence from 51.5% for subject 112 to 96.0% for subject 101. We predict no event for subject 101 with a probability of 96.0% and an event for subject 106 with a probability of 54.0%.

## Explore the most predictive variables

### Variable importance

The most predictive variables can be computed using compute_VIMP() and displayed using plot() function as followed:

res_dyn_VIMP <- compute_VIMP(DynForest_obj = res_dyn_OOB, seed = 123)
plot(x = res_dyn_VIMP, PCT = TRUE)

Again, we found that the most predictive variable is serBilir for which the OOB prediction error was reduced by 23%.

### Minimal depth

The minimal depth is computed using var_depth() function and is displayed at predictor and feature level using plot() function. The results are displayed in figure 1 using the random forest with maximal mtry hyperparameter value (i.e. mtry = 7) for better understanding.

depth_dyn <- var_depth(DynForest_obj = res_dyn_OOB)
plot(x = depth_dyn, plot_level = "predictor")
plot(x = depth_dyn, plot_level = "feature")

We observe that serBilir and albumin have the lowest minimal depth: these predictors are used to split the subjects in 199 and 194 out of 200 trees, respectively (figure 1A). The figure 1B provides further results. In particular, this graph shows the baseline random-effect (indicated by bi0) of serBilir and albumin are the earliest predictors used to split the subjects with 196 and 189 out of 200 trees, respectively.

## References

Murtaugh, Paul A., E. Rolland Dickson, Gooitzen M. Van Dam, Michael Malinchoc, Patricia M. Grambsch, Alice L. Langworthy, and Chris H. Gips. 1994. “Primary Biliary Cirrhosis: Prediction of Short-Term Survival Based on Repeated Patient Visits.” Hepatology 20 (1): 126–34. https://doi.org/10.1002/hep.1840200120.


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# Clarifications about the mechanism of the Wittig reaction [closed]

Ok so this is the proposed mechanism for the Wittig Reaction as stated by my textbook, LG Wade. I have some doubts regarding this mechanism:

1) Why is only triphenyl phosphine used here? Why not some stronger nucleophile like say trimethyl phosphine?

2) Why would the lone pair on P attack the C of the alkyl halide in the first place? Isn't the lone pair in resonance with the benzene ring?

3) Why is butyllithium used particularly.Can't any other reagent be used to abstract H ?

4) Why is the four membered ring formed in step 2?

5) Why does the ring collapse in step 3?

Note: I'm just a beginner in organic chemistry (Org 2) so it would be helpful if you can keep adjust your answer to suit my level of understanding.

• It is preferred if you can post separate questions instead of combining your questions into one. That way, it helps the people answering your question and also others hunting for at least one of your questions. Thanks! – Faded Giant Jun 5 '15 at 17:41
• That would be making 5 posts on the same topic ! @Loong – user14857 Jun 6 '15 at 22:00
• – Mithoron Jun 6 '15 at 22:08
• Yes, but splitting the questions helps keep the answer chains clean. It's perfectly all right to hyperlink among all the various questions to make it clear that they're part of the same broader search for understanding. – hBy2Py Jun 6 '15 at 23:03
• @SanchayanDutta - Multiple question posts is preferred. Think about it this way: What if I know the answer to questions 1) and 2) from your post, but I do not know how to answer questions 3), 4), and 5)? I might not feel that I can post anything since I cannot answer all five parts. If you post them separately, then I can answer the ones I know how to answer and leave the others for people who know more than I do. – Ben Norris Jun 11 '15 at 11:54

1) Why is only triphenyl phosphine used here. Why not some stronger nucleophile like say trimethyl phosphine?

Trialkyl phosphines can particpate in the Wittig reaction just fine, but they create some other problems. Let's say you were going to create the phosphonium salt from ethyl iodide and trimethyl phosphine. It would form in high yield, but in the next step where you deprotonate it and form the ylide you would have a problem. You could form the ylide from deprotonation of the desired ethyl group, or you could form an ylide by deprotonating any of the 3 methyl groups. This would give rise to a mixture of ylides and consequently a mixture of products. On the other hand, there are no protons on triphenyl phosphine that can be removed once the phosphonium salt is formed.

The following additional reasons also favor usage of triphenyl phosphine:

• it is cheap
• it is very cheap
• it is a solid and easy to handle and weigh out, many of the smaller trialkyl phosphines are liquids
• triphenyl phosphine is less toxic and doesn't smell as bad as many of the smaller trialkyl phosphines

2) Why would the lone pair on P attack the C of the alkyl halide in the first place? Isn't the lone pair in resonance with the benzene ring?

Triphenyl phosphine has the aromatic rings arranged in a non-planar, propeller-like geometry which makes delocalization less likely. Also $\ce{2p-3p}$ overlap would be involved which is not as effective as $\ce{2p-2p}$ overlap.

3) Why is butyllithium used particularly. Can't any other reagent be used to abstract H?

In the phosphonium salt, the protons alpha to phosphorous are not very acidic (pKa ~ 20-35), a strong base is required to remove them. Other strong bases could be used, but BuLi is commercially available in high purity at a reasonable cost. When BuLi abstracts the proton butane is formed and escapes as a gas - one less side product to worry about separating out.

4) Why is the four membered ring formed in step 2?

The 4-membered ring may not be as strained as you imagine. Remember, phosphorous is in the third row of the Periodic Table. It has longer bond lengths and can tolerate smaller bond angles.

5) Why does the ring collapse in step 3?

The bond between phosphorous and oxygen is an extremely strong bond, in other words its formation is very exothermic. Therefore, this last step is being driven by thermodynamics. It is the exothermicity of this step that drives the reaction to completion.

• Actually, $\ce{KO}t\ce{Bu}$ is often used for Wittig reactions in my group (phosphane usually being $\ce{PBu3}$) and my supervisor even said that it be way to strong — prompting me to try diazabicyclo[5.4.1]undecene (DBU) which worked like a charm on the test reaction. (Not on my product because my product is ugly, though.) – Jan Jun 8 '15 at 0:31
• This is indeed a very good answer! Have you been gone a while? I don't remember reading something from you for at least two weeks... For your enjoyment, I found a quite comprehensive article about the Wittig reaction: Bruce E. Maryanoff , Allen B. Reitz, Chem. Rev., 1989, 89 (4), pp 863–927.. Especially the part about betaines and oxophosphetanes is interesting. – Martin - マーチン Jun 8 '15 at 4:22


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List all the elements of the following set. Use set notation to list all the elements of the set. Three Sets. List or Roster method, Set builder Notation, The empty set or null set is the set that has no elements. The expressions "A includes x" and "A contains x" are also used to mean set membership, although some authors use them to mean instead "x is a subset of A". Summary: Set-builder notation is a shorthand used to write sets, often for sets with an infinite number of elements. Answer to List all the elements of the following set. The objects are called members or elements of the set. Roster Notation. They are different from lists or tuples in that they are modeled after sets in mathematics. Let us now explain what a set notation is. A set is created by using the set() function or placing all the elements within a pair of curly braces. asked Apr 12 '10 at 11:35. Use set notation and the listing method to describe the set. For example, one can say “let $$A$$ be the set of all odd integers”. Describing sets One can describe a set by specifying a rule or a verbal description. The intersection of two sets A and B, denoted by A ∩ B, is the set of all objects that are members of both the sets A and B.In symbols, ∩ = {: ∈ ∈}. $\{x|x \in \mathbb Z, x>0, x \ is \ divisible \ by \ 5\}$ OR would it be written like this? Venn diagrams can be used to express the logical (in the mathematical sense) relationships between various sets. 2. Lv 7. a. They wrote about it on the chalkboard using set notation: P = ... Let P be the set of all members in the math club. The symbol 2 is used to describe a relationship between an element of the universal set and a subset of the universal set, and the symbol $$\subseteq$$ is used to describe a relationship between two subsets of the universal set. 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text:
## newton's laws

"a person drives their 1452 kg car along a straight road at a constant velocity of 30 m/s east. their brakes suddenly give out. they put the car in neutral and let is coast for 25 seconds. the air drag decelerates the car to a velocity of 25 m/s east. (assume a frictionless surface)."

b) assume the average acceleration while the car is decelerating

so it's just (25 m/s - 30 m/s)/25 s = -0.2 m/s^2?

c) determine the average force of air against the car
f = ma
= (1452 kg)(-0.2)
= 290.4 N

d) after coasting for 25s. they pull their car handbrake to slow the car to a stop. this take 3 seconds. what is the force applied by the handbrake? assume that the force exerted by the air remains constant and is equal to the forced determined in part (c).
so i assume i find acceleration by (0 - 25)/time

but for the time is it (25s-3s) or (3s - 25s)?

and then i just the acceleration times the mass of car = handbrake force?

any help will be appreciated!

~Amy

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 You have to consider only the last 3 seconds during which the car is moving. acceleration times the mass of car = handbrake force? That's right.
 thanks neutrino. for d) i got -8.33 m/s^2 for the acceleration abd -12100N for the force of the handbrake. in e) it asks "what is the total displacement from the time the brakes give out to the time the car stops" so for this i used the formula: d = (0.5)at^2 + v1t = (o.5)(-8.33)(3^2) + (25m/s)(25s) = 587.52m can someone tell me if this is accurate? ~Amy

## newton's laws

another question i want to check:

"peter and john are playing tug of war on a frictionless surface. peter weighs 539N and john weights 393N. john accelerates towards peter at 3 m/s^2. "

a) what is the magnitude of the force that peter exerts on john?

539N times 3 m/s^2
= 1617N?

any help is appreciated!
~Amy

 for this one: "peter and john are playing tug of war on a frictionless surface. peter weighs 539N and john weights 393N. john accelerates towards peter at 3 m/s^2. " a) what is the magnitude of the force that peter exerts on john? first i calculated john and peters mass by diving this newton weight by the force of gravity (9.8 m/s^2). so peter is 55 kg, and john is 40 kg. then for part a) i took peters mass (55 kg) times (-3 m/s^2) = -165N. not sure if this is accurate. help? ~Amy

Mentor
Blog Entries: 1
 Quote by physicsgal and then i just the acceleration times the mass of car = handbrake force?
No. Don't forget that the air is still exerting its force.

 for d) i got -8.33 m/s^2 for the acceleration abd -12100N for the force of the handbrake.
No, you found the net force.

 in e) it asks "what is the total displacement from the time the brakes give out to the time the car stops" so for this i used the formula: d = (0.5)at^2 + v1t = (o.5)(-8.33)(3^2) + (25m/s)(25s) = 587.52m can someone tell me if this is accurate?
Divide the motion into two segments and find the displacement in each:
(1) air drag only (coasting)
(2) air drag + hand brake

Mentor
Blog Entries: 1
 Quote by physicsgal for this one: "peter and john are playing tug of war on a frictionless surface. peter weighs 539N and john weights 393N. john accelerates towards peter at 3 m/s^2. " a) what is the magnitude of the force that peter exerts on john? first i calculated john and peters mass by diving this newton weight by the force of gravity (9.8 m/s^2). so peter is 55 kg, and john is 40 kg.
OK.

 then for part a) i took peters mass (55 kg) times (-3 m/s^2) = -165N.
No. To find the force on John, multiply John's mass times John's acceleration.

 Quote: for d) i got -8.33 m/s^2 for the acceleration abd -12100N for the force of the handbrake. No, you found the net force.
so i'd be -121000N - -290.4N?

 Divide the motion into two segments and find the displacement in each: (1) air drag only (coasting) (2) air drag + hand brake
will do (and will post my results later)

 No. To find the force on John, multiply John's mass times John's acceleration.
thanks!
so it'd be 120N

what then how would i find the magnitude of the force that john exerts on peter? would it be -120N?

thank you

~Amy

Mentor
Blog Entries: 1
 Quote by physicsgal so i'd be -121000N - -290.4N?
Yes.

 so it'd be 120N
Right.

 what then how would i find the magnitude of the force that john exerts on peter? would it be -120N?
Yes. Newton's 3rd law should tell you that they exert equal and opposite force on each other. (But realize that magnitudes are always positive. )

 in e) it asks "what is the total displacement from the time the brakes give out to the time the car stops"
 Divide the motion into two segments and find the displacement in each: (1) air drag only (coasting) (2) air drag + hand brake
ok. for (1) i got (0.5)(30m/s + 25 m/s)(25s)
= 687.5 m

for (2) i got (0.5)(-8.33 m/s^2)(3^2)
= -37.485

so 687.5 + - 37.485
= 650.02 m

more on the john/peter saga..
 "sarah decides to join the game as well. now peter pulls on sarah with a force of 45 N east, and john pulls on her with a force of 25N north. what is sarah's resultant acceleration, if she weighs 294N?"
so first of all she weighs 30kg

and to find the c^2 it's (45^2) + (25^2)
= 51.48 N

and to find the tan angle is 26 degree north/west.

so to find sarah's acceleration i go 51.48N/30 kg?

~Amy

Mentor
Blog Entries: 1
 Quote by physicsgal ok. for (1) i got (0.5)(30m/s + 25 m/s)(25s) = 687.5 m
Good.
 for (2) i got (0.5)(-8.33 m/s^2)(3^2) = -37.485
Nope. Sanity check: It slows down, but doesn't go backwards.

Use the full equation:
$$d = v_0 t + (1/2) a t^2$$
(don't forget the initial velocity!)

You can also use the exact same equation you used for segment (1).

 more on the john/peter saga.. so first of all she weighs 30kg and to find the c^2 it's (45^2) + (25^2) = 51.48 N
Good.

 and to find the tan angle is 26 degree north/west.
Check this one. Her acceleration should be north-east.

 so to find sarah's acceleration i go 51.48N/30 kg?
Good.

 for (2) i got (0.5)(-8.33 m/s^2)(3^2) = -37.485 Nope. Sanity check: It slows down, but doesn't go backwards. Use the full equation: (don't forget the initial velocity!)
(i knew the negative displacement sounded a bit suspicious, lol)

ok so (25m/s)(3s) + (0.5)(-8.333)(3^2)
= 37.515

650.02+37.515
= 687.54 m

thanks for the help!

~Amy

Mentor
Blog Entries: 1
 Quote by physicsgal (i knew the negative displacement sounded a bit suspicious, lol) ok so (25m/s)(3s) + (0.5)(-8.333)(3^2) = 37.515
Good. But you can also use the same equation that you used for segment (1): d = (0.5)(25 m/s + 0 m/s)(3 s)

 650.02+37.515 = 687.54 m
I think you copied the answer from (1) wrong.

 thanks for the help!
My pleasure.

 650.02+37.515 = 687.54 m I think you copied the answer from (1) wrong.
thanks. i just noticed that too. so the total displacement should be 725.02 m east

~Amy


text:
# Enhance with super-resolution

If the original image needs to be improved to create a better master image from which the optimization happens, the API can enhance it using super-resolution.

This feature is useful if you want to produce up-scaled versions of images starting from a lower resolution image which is the best you have available.

### Image enhancement

Image enhancement is the procedure of improving the quality and the information content of original data before processing. In a rough overview, the enhancement in Pixellena consists of two steps:

#### Step 1: Upscaling using super-resolution

The enhancement algorithm is Super Resolution GAN (SRGAN). The motive of this architecture is to recover finer textures from the image when we upscale it so that it’s quality cannot be compromised. There are other methods such as Bilinear Interpolation that can be used to perform this task but they suffer from image information loss and smoothing.

The algorithm is trained on the DIVK image dataset, with 4x up-scaling.

#### Step 2: Color correction and noise removal

After the upscaling, a color correction step and a noise removal steps are applied to the up-scaled image.

### What parameters we use?

The input parameter for the master step is either identity or srgan-dt-x4.

The default value is identity, which basically means Do not change the image, use the original value. This is the algorithm to use if you are starting from a high-resolution image and want to shrink it down to some manageable size.

We will show here how to perform upscaling with the option srgan-dt-x4, take a look to the figure below that we are going to use in the example. It’s a jpg image named costume__200w.jpg with dimensions 200x200 pixels.

You can use the SDK to do request to our API. Let’s show examples about how you can use super-resolution to enhance your images.

### Enhancement basic example

We will show you how you can send a request to the API that uses super-resolution to enhance an image to 4x its size.

super_resolution_basic.js

import { LUX_API, Toilmore } from '@shimmercat/toilmore-sdk';
import fs from "fs";
import { Readable } from "stream";

let toilmore = new Toilmore(
{
'api_config': LUX_API,
'api_token': 'xxxx-xxxx',   // <-- Use a valid API token here.
'domain': 'domain.com'      // <-- Use a domain connected to the token here.
});

"master": {
"algorithm-name": "srgan-dt-x4"
}
}

let force_processing = false;

let result_promise =  toilmore.optimize_with_precursor(
"costume__200w.jpg",
"prjpg",
force_processing,
);

result_promise.then((result) => {
// result can contain a short status string or null if the image
// could not be optimized.
store_optimized_image(result);
},
(err) => {
console.log('Optimized images is not ready yet. Waiting 10 seconds...');
// Let's wait a bit
setTimeout(() => {
store_optimized_image(result);
}, 10);
}
);

function store_optimized_image(result) {
let w = fs.createWriteStream("./costume__enhanced.webp");
result.pipe(w);
}

You should put the image file costume__200w.jpg and the super_resolution_basic.js script in the same directory, install the SDK as explained in the SDK tutorial, and then run:

node super_resolution_basic.js

Once you get the optimized image you will get the upscaled costume__enhanced.webp version of this image, and the resulting image is 800x800 pixels, see below.

We basically enhanced the image to 4x its size, and optimized it to the same format jpg. You can see at our lux API doc, that you can get optimized images for any of the precursor_name: webp0, prjpg, pngzo, jp2o0, avifo0.

Updated:


text:
# Add jemalloc to eups distrib, to enable use as a potential dependancy for qserv

XMLWordPrintable

#### Details

• Type: RFC
• Status: Implemented
• Resolution: Done
• Component/s:
• Labels:
None

#### Description

jemalloc is a drop in mallc replacement for C/C++. The primary advantage here is that it provides application memory profiling with little cost. Once installed, enabling the profiling features is a matter of setting environent variables. jemalloc getting started

#### Activity

Hide
Robert Lupton added a comment -

While debugging mallocs are very useful they tend not to be as efficient as system mallocs. I don't know anything about jemalloc, is it only used via dynamic loading patching if enabled? And how will this work on OS/X platforms (where I think that the system malloc has much of this built into – see mallocinfo)?

Show
Robert Lupton added a comment - While debugging mallocs are very useful they tend not to be as efficient as system mallocs. I don't know anything about jemalloc , is it only used via dynamic loading patching if enabled? And how will this work on OS/X platforms (where I think that the system malloc has much of this built into – see mallocinfo )?
Hide
Brian Van Klaveren added a comment -

jemalloc is actually the main malloc of FreeBSD, so it is a system malloc.

Show
Brian Van Klaveren added a comment - jemalloc is actually the main malloc of FreeBSD, so it is a system malloc.
Hide
Robert Lupton added a comment -

So on FreeBSD we don't need it. I'm worried that on some hardware the vendor may have "value added" a custom malloc that knows about e.g. their memory backplane.

Show
Robert Lupton added a comment - So on FreeBSD we don't need it. I'm worried that on some hardware the vendor may have "value added" a custom malloc that knows about e.g. their memory backplane.
Hide
John Gates added a comment -

This is a good explanation for why jemalloc is useful for memory profiling Debugging Java Native Memory Leaks

The library is only used if you set the LD_PRELOAD environment variable and the profiling only occurs if you set the MALLOC_CONF environment variable appropriately. Otherwise, it uses the system malloc.

So, if you want to use the memory profiling, do this before starting the program. It should generate a report every time a GB of memory is allocated.

 export LD_PRELOAD=/usr/local/lib/libjemalloc.so export MALLOC_CONF='prof:true,prof_prefix:/qserv/run/var/log/jeprof.out,lg_prof_interval:31,lg_prof_sample:22'

Show
John Gates added a comment - This is a good explanation for why jemalloc is useful for memory profiling Debugging Java Native Memory Leaks The library is only used if you set the LD_PRELOAD environment variable and the profiling only occurs if you set the MALLOC_CONF environment variable appropriately. Otherwise, it uses the system malloc. So, if you want to use the memory profiling, do this before starting the program. It should generate a report every time a GB of memory is allocated. export LD_PRELOAD=/usr/local/lib/libjemalloc.so export MALLOC_CONF='prof:true,prof_prefix:/qserv/run/var/log/jeprof.out,lg_prof_interval:31,lg_prof_sample:22'
Hide
Fritz Mueller added a comment -

Please note in title above, the proposed dependency add is only for Qserv

Show
Fritz Mueller added a comment - Please note in title above, the proposed dependency add is only for Qserv
Hide
Tim Jenness added a comment -

I see no reason why this RFC can't be adopted. Is the implementation ticket DM-10971? (for switching qserv).

Show
Tim Jenness added a comment - I see no reason why this RFC can't be adopted. Is the implementation ticket DM-10971 ? (for switching qserv).
Hide
Fritz Mueller added a comment -

jemalloc was packaged and published in ticket DM-11214. This has allowed us to use jemalloc for the intended debug purposes in branch builds. It remains to be seen whether we'll cut Qserv over to jemalloc in general, but that decision and work is not under the umbrella of this RFC.

Show
Fritz Mueller added a comment - jemalloc was packaged and published in ticket DM-11214 . This has allowed us to use jemalloc for the intended debug purposes in branch builds. It remains to be seen whether we'll cut Qserv over to jemalloc in general, but that decision and work is not under the umbrella of this RFC.

#### People

Assignee:
John Gates
Reporter:
John Gates
Watchers:
Andy Hanushevsky, Brian Van Klaveren, Fritz Mueller, John Gates, Robert Lupton, Tim Jenness
1 Vote for this issue
Watchers:
6 Start watching this issue

#### Dates

Created:
Updated:
Resolved:
Planned End:

#### Jenkins

No builds found.


text:
emacs-orgmode
[Top][All Lists]

Re: [Orgmode] Orgtbl-mode questions

 From: Carsten Dominik Subject: Re: [Orgmode] Orgtbl-mode questions Date: Mon, 3 Nov 2008 14:30:44 +0100

On Nov 3, 2008, at 12:53 PM, Sebastian Rose wrote:

Nicholas S-A wrote:

Hi all,

I have some questions pertaining to orgtbl-mode. I am really excited about using it to compose all my LaTeX tables, and in fact learned Emacs and Org specifically for this purpose (well, I also learned them for all
the other awesome features of Org-mode.
First of all, I was wondering how I could add vertical bars to a
table (so instead of \begin{tabular}{rrr}, something like
\begin{tabular}{r|rr} is generated when C-c C-c is executed in an
orgtbl-mode remote table). Is this possible?

http://orgmode.org/manual/Column-groups.html#Column-groups

works fine for LaTeX.

- Carsten

Regards,

Sebastian

_______________________________________________
Emacs-orgmode mailing list
Remember: use Reply All' to send replies to the list.

`


text:
# Coset/Examples/Subgroup of Infinite Cyclic Group

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## Examples of Cosets

Let $G = \gen a$ be an infinite cyclic group.

Let $s \in \Z_{>0}$ be a (strictly) positive integer.

Let $H$ be the subgroup of $G$ defined as:

$H := \gen {a^s}$

Then a complete repetition-free list of the cosets of $H$ in $G$ is:

$S = \set {H, aH, a^2 H, \ldots, a^{s - 1} H}$

## Proof

First it is demonstrated that $S$ is complete.

Let $x \in G$ be arbitrary.

Then:

$\exists n \in \Z: x = a^n$

By the Division Theorem:

$n = q s + r$

for $q, r \in \Z$ such that $0 \le r \le s - 1$.

Thus:

 $\displaystyle x$ $=$ $\displaystyle a^{q s + r}$ $\displaystyle$ $=$ $\displaystyle a^r \paren {a^s}^q$ $\displaystyle$ $=$ $\displaystyle a^r h$ where $h \in H$

Thus:

$x \in a^r H$

and as $0 \le r \le s - 1$ it follows that $a^r H \in S$.

Thus every $x \in G$ belongs to at least one of the cosets on $S$.

Aiming for a contradiction, suppose:

$a^i H = a^j H$

where $0 \le i < j \le s - 1$.

Then from Sundry Coset Results:

$\paren {a^i}^{-1} a^j \in H$

that is:

$a^{j - i} \in H$

and so:

$a^{j - i} = a^{k s}$

for some $k \in \Z$.

But $a$ is of infinite order.

So it follows from Subgroup Generated by Infinite Order Element is Infinite that:

$j - i = k s$

But this contradicts our statement that $0 < j - i < s$.

Hence $S$ must be repetition-free.

$\blacksquare$


text: Additionally, detailed output to CNC machines automates the steel fabrication process. CMG provides reservoir simulation software, including thermal, compositional, black oil and enhanced oil recovery processes, that helps oil and gas companies reduce risk and maximize recovery. We see the problem in Frac Diagnostics, LLC is the heritage firm created from the transition of Pinnacle Technologies (a consulting firm started in September 1992) to STRATAGEN, Inc. based under CARBO CERAMICS Inc. and finally to Frac Diagnostics, LLC. 13 FRaC’s tree model on the same training set uses feature 15 (variance of cell radius) as well as feature 14, but the rule is effectively the same: large perimeter (and area) variance predicts large radius variance. Kinetix, Kinetix Frac, Kinetix Matrix, and Kinetix RT software are components of a tiered offering that helps you design, plan, optimize, execute, evaluate, and report on fracturing and acidizing operations in the full range of well and downhole conditions, from simple completions to complex reservoir challenges. FRaC is a new general approach to the anomaly detection problem; that is, the task of identifying instances that come from a different class or distribution than the majority (unsupervised anomaly detection) or a set of verified “normal” data (semi-supervised anomaly detection). StimPlan™ is a complete and integrated software solution for hydraulic fracture design, analysis, and optimization. versa. Given a process design and an appropriate selection of thermodynamic models, ASPEN uses mathematical models to predict the performance of the process. And because a transistor with a large width drives more current, an NMOS transistor of k-times unit width has a resistance of $$\frac {R}{k}$$. FRACPRO can effectively model any type of pressure stimulation job, including limited entry wells, multiple perforated intervals and horizontal well fracturing. StimPlan™ Software StimPlan™ Videos. The sand control modeling software integrates frac-pack and gravel-pack simulations because both make use of proppant and the crossover tool, which switches the tubing flow to the annular flow and vice. PowerLogFrac is the industry’s first petrophysical software that quickly generates formatted rock and fluid properties. There is also a special Android-version for mobile devices. The experienced GEODynamics’ Simulation Services team uses iPerf ® and other software tools to provide guidance and insight in the development of a perforating strategy that … It is Java-based, open-source and runs on any major computer platform. Constrained production forecast using asymmetric tri-linear model. Drag the images to your desktop to save permanently. This is a list of 17 Best Free 3D Modeling Software For Windows. You can then quickly generate design drawings, fabrication details, and schedules that automatically update whenever the 3D model changes. PVT and BHP estimation tools. Available in many file formats including MAX, OBJ, FBX, 3DS, STL, C4D, BLEND, MA, MB. Find professional Blender 3D Models for any 3D design projects like virtual reality (VR), augmented reality … FRaC (Feature Regression and Classification) Anomaly Detection Algorithm. The area model gives us a good picture of what happens when we multiply two fractions. GOHFER is also considered to be the most accurate frac modeling software package by almost a 4:1 margin over its closest competition." 3D Complex Geology Modeling. Modeling Multiplication of Proper Fractions - Letâ s use an area model to multiply fractions. Driven by the need for safer and more efficient geotechnical design, MoFrac implements methods developed by R. Mohan Srivastava to create DFNs well-suited to both civil and mining applications. Create 3D from 2D & back. CARBO Announces Strategic Agreement with FracGeo to Deliver Next Generation Real-Time Frac Modeling Software PR Newswire HOUSTON, July 22, 2019 HOUSTON , July Constrained pressure depletion estimated with Fast Marching Method (FMM). MoFrac is a software package developed at MIRARCO MINING INNOVATION that generates 3D fracture network models for rock mass characterization. FreeCAD is an open-source parametric 3D modeler made primarily to design real-life objects of any size. Learn more about Dassault Systèmes®' best in class product development software applications delivered on a 3D platform with the 3DEXPERIENCE portfolio. 23. . 500x400. A powerful tool for fracture optimization. ASPEN is a process simulation software package widely used in industry today. They also share similar hydraulic modeling. Aligned with the … Free student license subscription for one year term, renewable each year. In: Proceedings of the 2nd International Workshop on Future Trends of Model-Driven Development (FTMDD 2010) (2010) Google Scholar. \frac - Tex Command - \frac - Used to draw fractions. CARBO Announces Strategic Agreement with FracGeo to Deliver Next Generation Real-Time Frac Modeling Software HOUSTON , July 22, 2019 /PRNewswire/ -- CARBO Ceramics Inc. (NYSE: CRR) announced today a long-term, strategic agreement with FracGeo, LLC ("FracGeo") aimed at integrating FracGeo's software solutions and products with FracPro™ software, a CARBO business. Carbo Ceramics, FracGeo to deliver real-time frac modeling software. 2D 3D Reset. The first software to simulate and optimize zipper fracs; 3D modeling and display; Gain in-depth understanding ; Customize models with your real-time data ; Request Demo. CARBO Ceramics Inc. Jul 22, 2019, 09:00 ET. This software is released under the GPL version 3 license. This is a new tool to petrophysically analyze well log data and directly feed results into fracture simulation software. Thanks for the StimPlan™ support . Go to the downloads page to get this free program for Windows, Linux or Mac machines. GEODynamics’ iPerf ® modeling software is one tool which embodies these advances. Basic & Advanced GOHFER Software Training "58% of operators choose GOHFER 3D for their hydraulic fracturing simulation needs. Parametric modeling allows you to easily modify your design by going back into your model history and changing its parameters. Automatic history matching of rates or pressure. Constrained Asymmetric Frac Design and Pressure Depletion. ADVD USE CASE MODELING: Software Systems: Software Systems v. 1 (Object Technology Series) von Armour, Frank; Miller, Granville beim ZVAB.com - ISBN 10: 0201615924 - ISBN 13: 9780201615920 - Addison Wesley - 2000 - Softcover This information can then be used in an iterative fashion to optimize the design. this solution will help not only for this case, but for the future as well – Chevron. To provide an array of means to model the effect of hydraulic fracture simulations, the Petrel E&P software platform includes multiple approaches. Jul. Fractal Library. FreeCAD allows you to sketch geometry constrained 2D shapes and use them as a base to build other objects. Offers modeling, mesh generator, finite element solving and tools to check for optimization for international and national building codes. In the early 2000s, Pinnacle's consulting firm was principally focused on understanding brittle shale completions and then improving fracture placement. 14 tic-tac-toe is the familiar 3 × 3 naughts and crosses game. Gulden, J., Frank, U.: MEMOCenterNG—a full-featured modeling environment for organisation modeling and model-driven software development. . NAME \frac - Used to draw fractions.. SYNOPSIS { \frac #1 #2 } DESCRIPTION \frac command is used to draw fractions. More than twice as much as its closest competitor. Using more of what we know. 22, 2019 1:55 PM ET | By: Carl Surran, SA News Editor . The fractal parameters are saved using localStorage and will persist between browser sessions. Wildfire is a free and user-friendly image-processing software, mostly known for its sophisticated flame fractal generator. A 3D model can be defined as a model which represents an object in a 3D graphical mode and the process of creating a 3D model is known as 3D modeling. CARBO Announces Strategic Agreement with FracGeo to Deliver Next Generation Real-Time Frac Modeling Software News provided by. Finally to study the fluid flow within this network, WaterGEMS, a Numerical software for flow modeling, was used that had the capability to make complex pipes and to receive other hydraulic information. In modeling, a unit NMOS transistor has an effective resistance of R, which is equal to the resistance of the minimally-sized NMOS transistor used in the cell library or process. Check out the WebApp version. Saved images are listed here. Manchanda believes that bringing all the engineering talent together should hinge in part around advanced modeling, which in the unconventional sector refers to software code that tries to account for the unique physics and geomechanics encountered in tight rock formations. Our frac simulation method is meshless so no need to re-grid the model as the frac propagates and the simulator is particle based so no need to follow grid lines or impose empirical constraints on how the hydraulic and natural fractures interact as other products do. Heise, D.: Unternehmensmodell-basiertes IT-Kostenmanagement als Bestandteil eines integrativen IT … Completions engineers use the resulting models to design better hydraulic fracturing projects and improve well performance. With ProSteel software you can efficiently create accurate 3D models for structural steel, metal work, and steel assemblies. Free Blender 3D models. Our software integrates Schlumberger industry-leading simulators for fracture design, high-resolution reservoir simulation, finite-element geomechanical simulation, and a unique particle transport model to maximize completion effectiveness in even the most complicated environments. 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