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# What exactly are keys, queries, and values in attention mechanisms?
How should one understand the keys, queries, and values that are often mentioned in attention mechanisms?
I've tried searching online, but all the resources I find only speak of them as if the reader already knows what they are.
Judging by the paper written by Bahdanau (Neural Machine Translation by Jointly Learning to Align and Translate), it seems as though values are the annotation vector $$h$$ but it's not clear as to what is meant by "query" and "key."
The paper that I mentioned states that attention is calculated by
$$c_i = \sum^{T_x}_{j = 1} \alpha_{ij} h_j$$
with
$$\alpha_{ij} = \frac{e^{e_{ij}}}{\sum^{T_x}_{k = 1} e^{ik}}$$
$$e_{ij} = a(s_{i - 1}, h_j)$$
Where are people getting the key, query, and value from these equations?
Thank you.
• If this is the paper that you are talking about, it does not mention any "key", "query", or "value" for attention, and it seem to explain the symbols from the equations you quote, so I don't seem to understand what exactly is your question about? – Tim Aug 30 '19 at 12:45
I came from your other question Self-attention original work? The key/value/query formulation of attention is from the paper Attention Is All You Need.
How should one understand the queries, keys, and values
The key/value/query concepts come from retrieval systems. For example, when you type a query to search for some video on Youtube, the search engine will map your query against a set of keys (video title, description etc.) associated with candidate videos in the database, then present you the best matched videos (values).
The attention operation turns out can be thought of as a retrieval process as well, so the key/value/query concepts also apply here. (BTW the above example is just a toy system for illustration, in practice search engines and recommendation systems are much more complex.)
As mentioned in the paper you referenced (Neural Machine Translation by Jointly Learning to Align and Translate), attention by definition is just a weighted average of values,
$$c=\sum_{j}\alpha_jh_j$$ where $$\sum \alpha_j=1$$.
If we restrict $$\alpha$$ to be an one-hot vector, this operation becomes the same as retrieving from a set of elements $$h$$ with index $$\alpha$$. With the restriction removed, the attention operation can be thought of as doing "proportional retrieval" according to the probability vector $$\alpha$$.
It should be clear that $$h$$ in this context is the value. The difference between the two papers lies in how the probability vector $$\alpha$$ is calculated. The first paper (Bahdanau et al. 2015) computes the score through a neural network $$e_{ij}=a(s_i,h_j), \qquad a_{i,j}=\frac{\exp(e_{ij})}{\sum_k\exp(e_{ik})}$$ where $$h_j$$ is from the encoder sequence, and $$s_i$$ is from the decoder sequence. One problem of this approach is, say the encoder sequence is of length $$m$$ and the decoding sequence is of length $$n$$, we have to go through the network $$m*n$$ times to aqcuire all the attention scores $$e_{ij}$$.
A more efficient model would be to first project $$s$$ and $$h$$ onto a common space, then choose a similarity measure (e.g. dot product) as the attention score, like $$e_{ij}=f(s_i)g(h_j)^T$$ so we only have to compute $$g(h_j)$$ $$m$$ times and $$f(s_i)$$ $$n$$ times to get the projection vectors and $$e_{ij}$$ can be computed efficiently by matrix multiplication.
This is essentially the approach proposed by the second paper (Vaswani et al. 2017), where the two projection vectors are called query (for decoder) and key (for encoder), which is well aligned with the concepts in retrieval systems.
How are the queries, keys, and values obtained
The proposed multihead attention alone doesn't say much about how the queries, keys, and values are obtained as long as the dimension requirements are satisfied. They can come from different sources depending on the application scenario.
For unsupervised language model training like GPT, $$Q, K, V$$ are usually from the same source, so such operation is also called self-attention.
For the machine translation task in the second paper, it first applies self-attention separately to source and target sequences, then on top of that it applies another attention where $$Q$$ is from the target sequence and $$K, V$$ are from the source sequence.
For recommendation systems, $$Q$$ can be from the target items, $$K, V$$ can be from the user profile and history.
• Hello. Thanks for the answer. Unfortunately, my question is how those values themselves are obtained (i.e. the Q, K, and V). I've read other blog posts (e.g. The Illustrated Transformer) and it's still unclear to me how the values are obtained from the context of the paper. For example, is Q simply the matrix product of the input X and some other weights? If so, then how are those weights obtained? – Seankala Aug 30 '19 at 1:45
• Also, this question itself isn't actually pertaining to the calculation of Q, K, and V. Rather, I'm confused as to why the authors used different terminology compared to the original attention paper. – Seankala Aug 30 '19 at 1:45
• @Seankala hi I made some updates for your questions, hope that helps – dontloo Aug 30 '19 at 6:48
• Thanks a lot for this explanation! I still struggle to interprate the notation e_ij = a(s_i,h_j). So the neural network is a function of h_j and s_i, which are input sequences from the decoder and encoder sequences respectively. But what does the neural network look like? E.g. What are the target variables and what is the format of the input? – Emil Jan 17 at 13:43
• @Emil hi, it is a sub-network of the whole, there are no specific target for it in training, it's usually trained jointly with the whole network wrt to the given task, in this case machine translation, more details in the A.1.2 ALIGNMENT MODEL section of the paper. – dontloo Jan 17 at 18:51
Where are people getting the key, query, and value from these equations?
The paper you refer to does not use such terminology as "key", "query", or "value", so it is not clear what you mean in here. There is no single definition of "attention" for neural networks, so my guess is that you confused two definitions from different papers.
In the paper, the attention module has weights $$\alpha$$ and the values to be weighted $$h$$, where the weights are derived from the recurrent neural network outputs, as described by the equations you quoted, and on the figure from the paper reproduced below.
Similar thing happens in the Transformer model from the Attention is all you need paper by Vaswani et al, where they do use "keys", "querys", and "values" ($$Q$$, $$K$$, $$V$$). Vaswani et al define the attention cell differently:
$$\mathrm{Attention}(Q, K, V) = \mathrm{softmax}\Big(\frac{QK^T}{\sqrt{d_k}}\Big)V$$
What they also use is multi-head attention, where instead of a single value for each $$Q$$, $$K$$, $$V$$, they provide multiple such values.
Where in the Transformer model, the $$Q$$, $$K$$, $$V$$ values can either come from the same inputs in the encoder (bottom part of the figure below), or from different sources in the decoder (upper right part of the figure). This part is crucial for using this model in translation tasks.
In both papers, as described, the values that come as input to the attention layers are calculated from the outputs of the preceding layers of the network. Both paper define different ways of obtaining those values, since they use different definition of attention layer.
See Attention is all you need - masterclass, from 15:46 onwards Lukasz Kaiser explains what q, K and V are.
So basically:
• q = the vector representing a word
• K and V = your memory, thus all the words that have been generated before. Note that K and V can be the same (but don't have to).
So what you do with attention is that you take your current query (word in most cases) and look in your memory for similar keys. To come up with a distribution of relevant words, the softmax function is then used.
New contributor
Emil is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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# On January 1, 2012, Fishbone Corporation sold a building that cost $250,000 and that had accumulated... P6-1 (Various Time Value Situations) Answer each of these unrelated questions. (a) On January 1, 2012, Fishbone Corporation sold a building that cost$250,000 and that had accumulated depreciation of $100,000 on the date of sale. Fishbone received as consideration a$240,000 non-interest-bearing note due on January 1, 2015. There was no established exchange price for the building, and the note had no ready market. The prevailing rate of interest for a note of this type on January 1, 2012, was 9%. At what amount should the gain from the sale of the building be reported?
(b) On January 1, 2012, Fishbone Corporation purchased 300 of the $1,000 face value, 9%, 10-year bonds of Walters Inc. The bonds mature on January 1, 2022, and pay interest annually beginning January 1, 2013. Fishbone purchased the bonds to yield 11%. How much did Fishbone pay for the bonds? (c) Fishbone Corporation bought a new machine and agreed to pay for it in equal annual installments of$4,000 at the end of each of the next 10 years. Assuming that a prevailing interest rate of 8% applies to this contract, how much should Fishbone record as the cost of the machine?
(d) Fishbone Corporation purchased a special tractor on December 31, 2012. The purchase agreement stipulated that Fishbone should pay $20,000 at the time of purchase and$5,000 at the end of each of the next 8 years. The tractor should be recorded on December 31, 2012, at what amount, assuming an appropriate interest rate of 12%?
(e) Fishbone Corporation wants to withdraw \$120,000 (including principal) from an investment fund at the end of each year for 9 years. What should be the required initial investment at the beginning of the first year if the fund earns 11%?
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# A question about electromagnetic induction
20
Hi guys
I just want to know why changing magnetic flux in a coil induces emf???
25
3. ### EverGreen1231
26
Might be a question better posed on the classical physics area of the forum. At any rate, I'm not entirely sure that anyone knows why it happens, just that it does. Through experimentation, and a lot of thinking, we've figured out the equations that seem to predict the behavior; but to know, precisely, why it occurs, so far as I know, eludes us. I suppose one might find some answer looking in QM, but I'm not versed enough in that to tell you yes one way or the other.
Last edited: Aug 25, 2014
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1
JEE Main 2022 (Online) 24th June Evening Shift
Numerical
+4
-1
Let $$S = \left\{ {\left( {\matrix{ { - 1} & a \cr 0 & b \cr } } \right);a,b \in \{ 1,2,3,....100\} } \right\}$$ and let $${T_n} = \{ A \in S:{A^{n(n + 1)}} = I\}$$. Then the number of elements in $$\bigcap\limits_{n = 1}^{100} {{T_n}}$$ is ___________.
2
JEE Main 2021 (Online) 31st August Evening Shift
Numerical
+4
-1
The number of elements in the set $$\left\{ {A = \left( {\matrix{ a & b \cr 0 & d \cr } } \right):a,b,d \in \{ - 1,0,1\} \,and\,{{(I - A)}^3} = I - {A^3}} \right\}$$, where I is 2 $$\times$$ 2 identity matrix, is :
3
JEE Main 2021 (Online) 27th August Morning Shift
Numerical
+4
-1
If the system of linear equations
2x + y $$-$$ z = 3
x $$-$$ y $$-$$ z = $$\alpha$$
3x + 3y + $$\beta$$z = 3
has infinitely many solution, then $$\alpha$$ + $$\beta$$ $$-$$ $$\alpha$$$$\beta$$ is equal to _____________.
4
JEE Main 2021 (Online) 26th August Evening Shift
Numerical
+4
-1
Let A be a 3 $$\times$$ 3 real matrix. If det(2Adj(2 Adj(Adj(2A)))) = 241, then the value of det(A2) equal __________.
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# Calculating mean path lengths in a magnetic circuit. Can it be this simple?
#### Noodler
Joined Dec 29, 2015
5
I am studying electronic principles and currently electromagnetic induction.
I have been having some trouble establishing mean path lengths and cant seem to find any information in my course notes.
I have looked on this forum and found what I think is some good information, but it looks a bit simple.
I have applied this to my question with the result showing as 73mm, please see attachment for question and my working out.
I think the hole or partial penetration in the plate at the bottom of the loop may be relevant to later questions regarding flux density
and MMF, but I am not sure.
Does this look right? it seems too simple.
#### Attachments
• 128.8 KB Views: 81
#### Kermit2
Joined Feb 5, 2010
4,162
I don't think that simple formula works if all 4 sides are not the same thickness. One side of your square has a 1mm thick side.
I got a slightly larger answer by deducing the length of your vertical path to be 17.5 mm and the horizontal path to be 21 mm. Do you see where I got the 17.5 mm?
#### Noodler
Joined Dec 29, 2015
5
I don't think that simple formula works if all 4 sides are not the same thickness. One side of your square has a 1mm thick side.
I got a slightly larger answer by deducing the length of your vertical path to be 17.5 mm and the horizontal path to be 21 mm. Do you see where I got the 17.5 mm?
I think I may becoming a bit number blind, I am attempting to resolve issues surrounding flux density and MMF for the circuit also, but with very little success at all. It may well be time for a break.
I cant see where the 17.5mm is from on the vertical path.
I am assuming that other than the 1mm section on the bottom, the other three sides are all 6mm.
As such the long vertical path looks like it has an overall length of 20mm plus 1mm. Giving 21mm.
Whilst the short path would be 21 minus the 6 at the top and the 1 at the bottom. Giving 14mm.
Regarding the horizontal path, the long path appears to be 25mm, whilst the short is this figure minus 2 x 6mm. Giving 13.
#### Kermit2
Joined Feb 5, 2010
4,162
You have a vertical distance of 20mm plus the 1mm steel plate. The mean path will start at the mid point of the upper section. This point is 3mm inside the upper leg. The coresponding halfway point on the bottom leg is 1/2 mm inside. So we have 20 minus 3 and an additional 1/2 mm. This adds to 17.5 mm of vertical distance on each side.
And DAMN IT.
The calculator gave me 77 mm as an answer the first time and gives me 73 mm now.
19+19+17.5+17.5=73
I think I have data entry dyslexia
#### Noodler
Joined Dec 29, 2015
5
You have a vertical distance of 20mm plus the 1mm steel plate. The mean path will start at the mid point of the upper section. This point is 3mm inside the upper leg. The coresponding halfway point on the bottom leg is 1/2 mm inside. So we have 20 minus 3 and an additional 1/2 mm. This adds to 17.5 mm of vertical distance on each side.
And DAMN IT.
The calculator gave me 77 mm as an answer the first time and gives me 73 mm now.
19+19+17.5+17.5=73
I think I have data entry dyslexia
Many thanks Kermit.
I have worked it both ways now and got the same answer.
The main thing is that it appears that I am actually finding the mean path length.
I was a little concerned, as it looked too simple in the end.
Thanks again, I can now continue to work the remainder of the problem.
That said I may well be back with more queries regarding my understanding or lack of it.
#### Jony130
Joined Feb 17, 2009
5,351
But for this circuit you have two mean path lengths. One length for Fe-Si and the second one for carbon steel plate.
#### MrAl
Joined Jun 17, 2014
9,179
Hi,
The mean magnetic path is just a measurement of length of the magnetic path, no matter what it goes through. There are other properties that are considered separately when all the items are not of the same material, such as the normal air gap (not this particular air gap however). The other properties would definitely include the width of the air gap (if it was a simple air gap) and for this problem the length of the steel plate. Those lengths would also need to be considered, but the mean path is still just the total distance a magnetic field line takes if it was located at the very center of the three dimensional sides. This single line is at the very center so for a side that has a square cross section 6mm by 6mm and any length the line is 3mm in from all surfaces of that one side. If the cross section was 6mm by 8mm then the line would be 3mm in from one surface and 4mm in from the other surface. But usually these problems only use 2d drawings so we only see 6mm or 8mm and then we assume that the other dimension is also uniform, so we just divide by 2 and get either 3mm or 4mm respectively.
So the first thing to do is calculate the mean magnetic path, then later we would have to know the length of the steel plate and perhaps the length of the core that is not the steel plate, but only if the permeability for each material was different, which they usually are. So it does make sense to note that there could be more than one length involved, but this problem seems to be asking for that one mean length.
The mean length that has already been noted follows a line at the very center of each arm of the construct, and runs the length of an arm from the center of the butt at one end to the butt of the other end. This has already been drawn out nicely in the attached pdf.
For a circular core the mean path is the sum of the outside circumference and the inside circumference, then divided by 2. For a square or rectangular construct it would be the same i would think, although we could prove this algebraically.
In any case though, the mean path here is 73 because no matter how you calculate it, the result is the same.
I'll see if i can work up a proof if i have time. For example, if the width of one side is A and outside length L, and the width of one of the butts is B and the other C, then the mean length of the side A is:
Lm=L-B/2-C/2
and we would just need an expression for the whole thing and another expression for the other method and then equate the two. If they equate algebraically, then they are the same. For that one side though we would also know the inside length K is:
K=L-B-C
so the mean this second way would be:
Lm=(L+K)/2
so:
Lm=(L+L-B-C)/2=(2*L-B-C)/2
so:
Lm=L-B/2-C/2
so it does seem to prove that both methods are the same because every side would be calculated using the same procedure. We might have to look at angled sides too, but that's probably the same since the circular shape works the same.
Last edited:
#### Noodler
Joined Dec 29, 2015
5
But for this circuit you have two mean path lengths. One length for Fe-Si and the second one for carbon steel plate.
Thank you.
I believe that the two different material types are related to another part of the question regarding calculation of MMF.
That said I could well be misinterpreting the question.
It asks what the MMF is in each part of the path of the circuit to achieve B in the specimen plate of 0.5T.
I have magnetization curves for the materials concerned.
It looks as if I have to work out MMF for the specimen plate with known dimensions and B and H values from the relevant curve.
Taking this answer as one path.
I think the remaining three paths (legs on the electromagnet) should all be the same and could be calculated using their dimensions and B and H values from their relevant curve.
I think this makes sense?
I believe that I have an equation here to calculate MMF, but it looks a wee bit complicated to me at the moment.
Thanks again.
#### Noodler
Joined Dec 29, 2015
5
Hi,
The mean magnetic path is just a measurement of length of the magnetic path, no matter what it goes through. There are other properties that are considered separately when all the items are not of the same material, such as the normal air gap (not this particular air gap however). The other properties would definitely include the width of the air gap (if it was a simple air gap) and for this problem the length of the steel plate. Those lengths would also need to be considered, but the mean path is still just the total distance a magnetic field line takes if it was located at the very center of the three dimensional sides. This single line is at the very center so for a side that has a square cross section 6mm by 6mm and any length the line is 3mm in from all surfaces of that one side. If the cross section was 6mm by 8mm then the line would be 3mm in from one surface and 4mm in from the other surface. But usually these problems only use 2d drawings so we only see 6mm or 8mm and then we assume that the other dimension is also uniform, so we just divide by 2 and get either 3mm or 4mm respectively.
So the first thing to do is calculate the mean magnetic path, then later we would have to know the length of the steel plate and perhaps the length of the core that is not the steel plate, but only if the permeability for each material was different, which they usually are. So it does make sense to note that there could be more than one length involved, but this problem seems to be asking for that one mean length.
The mean length that has already been noted follows a line at the very center of each arm of the construct, and runs the length of an arm from the center of the butt at one end to the butt of the other end. This has already been drawn out nicely in the attached pdf.
For a circular core the mean path is the sum of the outside circumference and the inside circumference, then divided by 2. For a square or rectangular construct it would be the same i would think, although we could prove this algebraically.
In any case though, the mean path here is 73 because no matter how you calculate it, the result is the same.
I'll see if i can work up a proof if i have time. For example, if the width of one side is A and outside length L, and the width of one of the butts is B and the other C, then the mean length of the side A is:
Lm=L-B/2-C/2
and we would just need an expression for the whole thing and another expression for the other method and then equate the two. If they equate algebraically, then they are the same. For that one side though we would also know the inside length K is:
K=L-B-C
so the mean this second way would be:
Lm=(L+K)/2
so:
Lm=(L+L-B-C)/2=(2*L-B-C)/2
so:
Lm=L-B/2-C/2
so it does seem to prove that both methods are the same because every side would be calculated using the same procedure. We might have to look at angled sides too, but that's probably the same since the circular shape works the same.
Thank you, it looks like I have the mean path length calculated correctly.
Your confirmation and the explanation of this is much appreciated.
I believe that the two different material types are related to another part of the question regarding calculation of MMF.
It asks what the MMF is in each part of the path of the circuit to achieve a certain B in the specimen plate.
I have magnetization curves for the materials concerned.
It looks as if I have to work out MMF for the specimen plate using known dimensions and B and H values from the relevant curve.
Taking this answer as one path.
I looks like the remaining three paths (on the electromagnet) should all be the same and could be calculated using their dimensions and B and H values from their relevant curve.
I think this makes sense.
I believe that I have an equation here to calculate MMF, but it looks a wee bit complicated to me at the moment.
Thanks again.
#### mildred00
Joined Dec 8, 2015
1
Noodler,
Have you had any luck with calculating the MMF for each part of the circuit?
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Changes
Proposal: Digit Strings which Represent Continued Fractions
, 1 year ago
m
- building alternative description
For the purposes of this article: All expressions are big-endian and microdigits are in traditional decimal. PEMDAS is obeyed.
== Original Explanation ==
Let $z = a_0 + b_0/(a_1 + b_1/(a_2 + b_2/(\dots))) = a_0 + \underset{i=0}{\overset{\infty}{\mathrm K}} \big(\frac{b_i}{a_{i+1}}\big)$, where $a_i$ and $b_i$ are integers for all $i$ (see https://oeis.org/wiki/Continued_fractions#Gauss.27_Kettenbruch_notation ). In fact, for all i, we will canonically restrict $a_{(i+1)}$ and $b_i$ to nonnegative integers such that if $b_j = 0$, then $a_k = 1$ and $b_k = 0$ for all $k \geq j$; this is a perfectly natural and standard set of restrictions to make and does not actually diminish the set of numbers which are expressible in this format, but the restriction is not technically necessary for Lojban. Then we will denote $z$ by the continued fraction representation $z = (a_0 : b_0, a_1 : b_1, a_2 : b_2, \dots)$; the whole rhs representation is called a string. Notice that the integer part is included. In this format, for each $i$, "$a_i : b_i$" forms a single unit called a macrodigit; for each $i$, "$a_i$" and "$b_i$" each are microdigits; the colon ("$:$") separates microdigits and the comma ("$,$") separates macrodigits. Microdigits can be expressed in any base or other representation and macrodigits could be reversed or slightly rearranged (such as being of form "$b_i : a_{(i+1)}$"; however, for our purposes here microdigits will be expressed in big-endian traditional decimal and macrodigits will be formed and ordered as shown; the specification herein proposed will obligate the user to express the macrodigits in the form which is shown (id est: of form "$a_i : b_i$"; within any given macrodigit, the first microdigit expressed represents $a_i$ and the second (and final) microdigit expressed represents $b_i$, only) but the other features aforementioned are not guaranteed, although they may normally be assumed as a contextless default. In order to be clear: in this representation, each macrodigit will consist of exactly two microdigits - namely, $a_i$ and $b_i$ in that order, for all $i$ - and these microdigits will be separated explicitly by "pi'e"; meanwhile, macrodigits will be separated explicitly by "pi". In this representation, I will denote a not-explicitly-specified microdigit by a pair of consecutive underscores ("$\_\_$"). In the 'big-endian' arrangement of the macrodigits (as herein depicted), the first microdigit ($a_0$) represents the 'integer part' of the expression.
[[User:Krtisfranks|Krtisfranks]] ([[User talk:Krtisfranks|talk]]) 08:10, 9 March 2018 (UTC)
== Alternative Explanation ==
The description in this section is intended to provide the same results as those in the "Original Explanation" section.
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D. Cleaning the Phone
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Polycarp often uses his smartphone. He has already installed $n$ applications on it. Application with number $i$ takes up $a_i$ units of memory.
Polycarp wants to free at least $m$ units of memory (by removing some applications).
Of course, some applications are more important to Polycarp than others. He came up with the following scoring system — he assigned an integer $b_i$ to each application:
• $b_i = 1$ — regular application;
• $b_i = 2$ — important application.
According to this rating system, his phone has $b_1 + b_2 + \ldots + b_n$ convenience points.
Polycarp believes that if he removes applications with numbers $i_1, i_2, \ldots, i_k$, then he will free $a_{i_1} + a_{i_2} + \ldots + a_{i_k}$ units of memory and lose $b_{i_1} + b_{i_2} + \ldots + b_{i_k}$ convenience points.
For example, if $n=5$, $m=7$, $a=[5, 3, 2, 1, 4]$, $b=[2, 1, 1, 2, 1]$, then Polycarp can uninstall the following application sets (not all options are listed below):
• applications with numbers $1, 4$ and $5$. In this case, it will free $a_1+a_4+a_5=10$ units of memory and lose $b_1+b_4+b_5=5$ convenience points;
• applications with numbers $1$ and $3$. In this case, it will free $a_1+a_3=7$ units of memory and lose $b_1+b_3=3$ convenience points.
• applications with numbers $2$ and $5$. In this case, it will free $a_2+a_5=7$ memory units and lose $b_2+b_5=2$ convenience points.
Help Polycarp, choose a set of applications, such that if removing them will free at least $m$ units of memory and lose the minimum number of convenience points, or indicate that such a set does not exist.
Input
The first line contains one integer $t$ ($1 \le t \le 10^4$) — the number of test cases. Then $t$ test cases follow.
The first line of each test case contains two integers $n$ and $m$ ($1 \le n \le 2 \cdot 10^5$, $1 \le m \le 10^9$) — the number of applications on Polycarp's phone and the number of memory units to be freed.
The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the number of memory units used by applications.
The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \le b_i \le 2$) — the convenience points of each application.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
Output
For each test case, output on a separate line:
• -1, if there is no set of applications, removing which will free at least $m$ units of memory;
• the minimum number of convenience points that Polycarp will lose if such a set exists.
Example
Input
5
5 7
5 3 2 1 4
2 1 1 2 1
1 3
2
1
5 10
2 3 2 3 2
1 2 1 2 1
4 10
5 1 3 4
1 2 1 2
4 5
3 2 1 2
2 1 2 1
Output
2
-1
6
4
3
Note
In the first test case, it is optimal to remove applications with numbers $2$ and $5$, freeing $7$ units of memory. $b_2+b_5=2$.
In the second test case, by removing the only application, Polycarp will be able to clear only $2$ of memory units out of the $3$ needed.
In the third test case, it is optimal to remove applications with numbers $1$, $2$, $3$ and $4$, freeing $10$ units of memory. $b_1+b_2+b_3+b_4=6$.
In the fourth test case, it is optimal to remove applications with numbers $1$, $3$ and $4$, freeing $12$ units of memory. $b_1+b_3+b_4=4$.
In the fifth test case, it is optimal to remove applications with numbers $1$ and $2$, freeing $5$ units of memory. $b_1+b_2=3$.
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For an equilateral triangle, all 3 ex radii will be equal. Reference - Books: 1) Max A. Sobel and Norbert Lerner. Given here is an equilateral triangle with side length a, the task is to find the area of the circle inscribed in that equilateral triangle. Please enable Cookies and reload the page. Geometry Problem 1199 This combination happens when a portion of the curve is tangent to one side, and there is an imaginary tangent line extending from the two sides of the triangle. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Equilateral Triangle. An equilateral triangle is a triangle whose all three sides are having the same length. but I don't find any easy formula to find the radius of the circle. To prove this, note that the lines joining the angles to the incentre divide the triangle into three smaller triangles, with bases a, b and c respectively and each with height r. Geometry. In a right angled triangle, ABC, with sides a and b adjacent to the right angle, the radius of the inscribed circle is equal to r and the radius of the circumscribed circle is equal to R. Prove... Stack Exchange Network. r. {\displaystyle r} and the circumcircle radius. Geometry Problem 1205 Triangle, Centroid, Outer and Inner Napoleon Equilateral Triangles. 30°. The formula for the area A of an equilateral triangle of side s is. Formulas for the area, altitude, perimeter, and semi-perimeter of an equilateral triangle are as given: Where, a is the side of an equilateral triangle. A = s² √3 / 4. The cosine rule, also known as the law of cosines, relates all three sides of a triangle with an angle of a triangle. 4th ed. There is no need to calculate angles or other distances in … 7.64cm. Let and denote the triangle's three sides and let denote the area of the triangle. 4. An equilateral triangle has all three sides equal and and all three angles equal to 60° The relationship between the side $$a$$ of the equilateral triangle and its area A, height h, radius R of the circumscribed and radius r of the inscribed circle are give by: Formulas for Equilateral Triangles. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter. • The equilateral triangle provides a rich context for students and teachers to explore and discover geometrical relations using GeoGebra. - equal sides of a triangle. The triangle area using Heron's formula Heron's formula gives the area of a triangle when the length of all three sides are known. The circumference of the circle around the triangle would be 4/3 of the triangle's perimeter or 48cm. 5. So if this is side length a, then this is side length a, and that is also a side of length a. You may need to download version 2.0 now from the Chrome Web Store. The area of an equilateral triangle is √3/4 (a)square. The dividing perpendicular from the angle of the vertex is parted into two equal halves i.e. Proofs make use of theorems in … R. Johnson, R. A. Area of Incircle of a … triangle, it is possible to determine the radius of the circle. … Hence, HCR’s Formula for regular polyhedron is verified on the basis of all the results obtained above Analysis of Regular Octahedron (with eight faces each as an equilateral triangle) Let there be a regular octahedron having eight congruent faces each as an equilateral triangle & edge length . Three smaller isoceles triangles will be formed, with the altitude of each coinciding with the perpendicular bisector. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Then, draw the perpendicular bisectors, extending from the circumcenter to each side’s midpoint (sides a, b, c). Now apply the Pythagorean theorem to get the height (h) or the length of the line you see in red . And let's say we know that the radius … Note how the perpendicular bisector breaks down side a into its half or a/2. I think that's about as good as I'm going to be able to do. Four circles tangent to each other and an equilateral ... picture. Let us start learning about the equilateral triangle formula. Semi-perimeter of triangle of side a = 3 a/2. Geometry Problem 1218 Triangle, Equilateral Triangles, Midpoints. an equilateral triangle is a triangle in which all three sides are equal. If you know all three sides If you know the length (a,b,c) of the three sides of a triangle, the radius of its circumcircle is given by the formula: Its radius, the inradius (usually denoted by r) is given by r = K/s, where K is the area of the triangle and s is the semiperimeter (a+b+c)/2 (a, b and c being the sides). The area of an equilateral triangle is the amount of space that it occupies in a 2-dimensional plane. The semiperimeter frequently appears in formulas for triangles that it is given a separate name. go. It appears in a variety of contexts, in both basic geometries as well as in many advanced topics. We let , , , , and .We know that is a right angle because is the diameter. 5. Geometry Problem 1212 Equilateral Triangle, Equilateral Hexagon, Concurrent Lines. • In fact, this theorem generalizes: the remaining intersection points determine another four equilateral triangles. The radius of a circumcircle of an equilateral triangle is equal to (a / √3), where ‘a’ is the length of the side of equilateral triangle. So, an equilateral triangle’s area can be calculated if the length of its side is known. Draw a line from a corner of the triangle to its opposite edge, similarly for the other 2 corners. Find the perimeter of an equilateral triangle of side 4.5 cm? For equilateral triangles In the case of an equilateral triangle, where all three sides (a,b,c) are have the same length, the radius of the circumcircle is given by the formula: where s is the length of a side of the triangle. Calculating the radius []. This geometry video tutorial explains how to calculate the area of the shaded region of circles, rectangles, triangles, and squares. So, an equilateral triangle’s area can be calculated if the length of its side is known. Then, the measure of the circumradius of the triangle is simply .This can be rewritten as .. Triangle Equations Formulas Calculator Mathematics - Geometry. r. {\displaystyle r} is one-third of the harmonic mean of these altitudes; that is, r = 1 1 h a + 1 h b + 1 h c . It can be inside or outside of the circle . Deriving the formula for the radius of the circle inscribed in an equilateral triangle 2 Determine the closest point along a circle's $(x_1, y_1)$ radius from any point $(x_2, y_2)$, inside or outside the radius … In the case of an equilateral triangle medians, angle bisectors altitudes are one and the same. Ex-radius of an equilateral triangle calculator uses Exradius of Equilateral Triangle=sqrt(3)*Side A/2 to calculate the Exradius of Equilateral Triangle, The Ex-radius of an equilateral triangle formula is given by r1 = √3a/2 where a is the side of an equilateral triangle. Formulas of the medians, … Let a be the length of the sides, A - the area of the triangle, p the perimeter, R - the radius of the circumscribed circle, r - the radius of the inscribed circle, h - the altitude (height) from any side.. Last Updated: 18 July 2019. Your IP: 97.74.234.100 Hope this helps. Formulas of the medians, heights, angle bisectors and perpendicular bisectors in terms of a circumscribed circle’s radius of a regular triangle. Click hereto get an answer to your question ️ The radius of the incircle of a triangle is 4 cm and the segments into which one side divided by the point of contact are 6 … The calculator of course also offers measurement units in imperial and metric, which work independently in case you have to convert units at the same time. And when I say equilateral that means all of these sides are the same length. Chapter 7. Math Geometry Physics Force Fluid Mechanics Finance Loan Calculator. Triangles. Proof. Ex-radius of an equilateral triangle calculator uses Exradius of Equilateral Triangle=sqrt(3)*Side A/2 to calculate the Exradius of Equilateral Triangle, The Ex-radius of an equilateral triangle formula is given by r1 = √3a/2 where a is the side of an equilateral triangle. The radius of circle can be an edge or side of polygon inside the circle which is known as circumradius and center of circle is known as circumcenter. What is the measure of the radius of the circle inscribed in a triangle whose sides measure $8$, $15$ and $17$ units? This video discusses on how to find out the radius of an incircle of an equilateral triangle. In this paper, we provide teachers with interactive applets to use in their classrooms to support student conjecturing regarding properties of the equilateral triangle. Without this formula I can't do any of my geometry homework :? Proof 2. The inner circle is 2/3 the triangle's perimeter or 24 so the radius of the inner circle would be half of the outer or 3.82cm approximately. The area of a triangle is equal to the product of the sides divided by four radii of the circle circumscribed about the triangle. Calculate the radius of a inscribed circle of an equilateral triangle if given side ( r ) : radius of a circle inscribed in an equilateral triangle : = Digit 2 1 2 4 6 10 F. Formula 4: Area of an equilateral triangle if its exradius is known. The area of an equilateral triangle is the amount of space that it occupies in a 2-dimensional plane. Formula for a Triangle. Area of a triangle, equilateral isosceles triangle area formula calculator allows you to find an area of different types of triangles, such as equilateral, isosceles, right or scalene triangle, by different calculation formulas, like geron's formula, length of triangle sides and angles, incircle or circumcircle radius. The radius of a circumcircle of an equilateral triangle is equal to (a / √3), where ‘a’ is the length of the side of equilateral triangle. Because an equilateral triangle has fixed contraints (fixed angles), the equations can therefore be simplified to the following... (The height or altitude is the vertical perpendicular height from the base of the triangle) Area = ¼√3 x Side2 Perimeter = 3 x Side The student will also learn the area of equilateral triangle formula. Question 4. - circumcenter. How this formulae works? You are the right place to get all information about Mensuration class 8 maths chapter 11. The product of the incircle radius. Another way to prevent getting this page in the future is to use Privacy Pass. Given here is an equilateral triangle with side length a, the task is to find the area of the circle inscribed in that equilateral triangle. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter. Altitude of an equilateral triangle = $\frac{\sqrt{3}}{2}$ a = $\frac{\sqrt{3}}{2}$ $\times$ 8 cm = 6.928 cm. Draw the perpendicular bisector of the equilateral triangle as shown below. The triangle area using Heron's formula Heron's formula gives the area of a triangle when the length of all three sides are known. An equilateral triangle is a triangle having all three sides equal in length. According to formula, Radius of circle … Also, because they both subtend arc .Therefore, by AA similarity, so we have or However, remember that . Formula 2. Equilateral Triangles, Midpoints, 60 Degrees, Congruence, Rhombus. The semiperimeter frequently appears in formulas for triangles that it is given a separate name. To prove this, note that the lines joining the angles to the incentre divide the triangle into three smaller triangles, with bases a, b and c respectively and each with height r. Performance & security by Cloudflare, Please complete the security check to access. The area of a circle inscribed inside an equilateral triangle is found using the mathematical formula πa 2 /12. By Jimmy Raymond Formula for a Triangle. A triangle is circumscribed in a circle if all three vertices of the triangle are tangent to the circle. Problem. Area of an equilateral triangle = $\frac{\sqrt{3}}{4}$ $a^{2}$ = $\frac{\sqrt{3}}{4}$ $\times$ $8^{2}$ cm 2 = $\frac{\sqrt{3}}{4}$ $\times$ 64 cm 2 = 21.712 cm 2. Below image shows an equilateral triangle with circumcircle: Equilateral triangle formulas. Given below is the figure of Circumcircle of an Equilateral triangle. In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also congruent to each other and are each 60°. The radius of the inscribed circle of the triangle is related to the extraradii of the triangle. Equilateral Triangle. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their … Perimeter of an equilateral triangle = 3a = 3 $\times$ 8 cm = 24 cm. Contact: [email protected]. Its radius, the inradius (usually denoted by r) is given by r = K/s, where K is the area of the triangle and s is the semiperimeter (a+b+c)/2 (a, b and c being the sides). To recall, an equilateral triangle is a triangle in which all the sides are equal and the measure of all the internal angles is 60°. picture. In other words, the equilateral triangle is in company with the circle and the sphere whose full structures are known only by knowing the radius. Let a be the length of the sides, A - the area of the triangle, p the perimeter, R - the radius of the circumscribed circle, r - the radius of the inscribed circle, h - the altitude (height) from any side.. In this topic, we will discover about equilateral triangles and its area. Equilateral triangle formulas. There is no need to calculate angles or other distances in … You are here: Home. For equilateral triangles In the case of an equilateral triangle, where all three sides (a,b,c) are have the same length, the radius of the circumcircle is given by the formula: where s is the length of a side of the triangle. Calculate the height of a triangle if given two lateral sides and radius of the circumcircle ( h ) : height of a triangle : = Digit 2 1 2 4 6 10 F To recall, an equilateral triangle is a triangle in which all the sides are equal and the measure of all the internal angles is 60°. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. If a circle is inscribed in an equilateral triangle the formula for radius = R = a /three/three the place a = aspect of triangle 9 = a /3/3 a = 9 X3 / /3 = 9 /three field of triangle = s^2 /3/4 where s = part of triangle = (9/three)^2 /three/4 = 243/three/4 you have not given the figure so are not able to provide the area of shaded figure Equilateral Triangle Formula. Area of triangle of side a = (√3)a 2 /4. The area of a triangle is equal to the product of the sides divided by four radii of the circle circumscribed about the triangle. Soc. Online Web Apps, Rich Internet Application, Technical Tools, Specifications, How to Guides, Training, Applications, Examples, Tutorials, Reviews, Answers, Test Review Resources, Analysis, Homework Solutions, Worksheets, Help, Data and Information for Engineers, Technicians, Teachers, Tutors, Researchers, K-12 Education, College and High School Students, Science Fair Projects and Scientists Divide 48 by 2pi = approx. A triangle is circumscribed in a circle if all three vertices of the triangle are tangent to the circle. Every equilateral triangle can be sliced down the middle into two 30-60-90 right triangles, making for a handy application of the hypotenuse formula. If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. {\displaystyle r= {\frac {1} { {\frac {1} {h_ {a}}}+ {\frac {1} {h_ {b}}}+ {\frac {1} {h_ {c}}}}}.} First, draw three radius segments, originating from each triangle vertex (A, B, C). Let and denote the triangle's three sides and let denote the area of the triangle. Mensuration Formulas for Class 8 Maths Chapter 11 Are you looking for Mensuration formulas or important points that are required to understand Mensuration for class 8 maths Chapter 11? ∴ ex-radius of the equilateral triangle, r1 = \\frac{A}{s-a}) = \\frac{{\sqrt{3}}a}{2}) 1991. If you know all three sides If you know the length (a,b,c) of the three sides of a triangle, the radius of its circumcircle is given by the formula: We know … Lets see how this formula is derived, Formula to find the radius of the inscribed circle = area of the triangle / semi-perimeter of triangle. I can easily understand that it is a right angle triangle because of the given edges. This is readily solved for the side in terms of the radius: s = 2r sin(π/n) For r = 3 in and n = 3, s = 6 sin(π/3) in = 6 √3 / 2 in = 3 √3 in. This is the only regular polygon with three sides. Examples: Input : a = 4 Output : 4.1887902047863905 Input : a = 10 Output : 26.1799387799 Heron's formula works equally well in all cases and types of triangles. Length of each side of an equilateral triangle = 16 cm Perimeter of an equilateral triangle = ( 3 x Length of each side ) units = ( 3 x 16 ) cm = 48 cm. Boston, MA: Houghton Mifflin, 1929. Math Geometry Physics Force Fluid Mechanics Finance Loan Calculator. Distance from the intersection point to the corner is the radius … where a is the length of the side of the given equilateral triangle. Mensuration formulas play a vital role in preparing you for […] Problem 1 Explanation. So the incentre of the triangle is the same as centroid of the triangle. Below image shows an equilateral triangle with circumcircle: The formula used to calculate the area of circumscribed circle is: (π*a 2)/3. Then, the measure of the circumradius of the triangle is simply .This can be rewritten as .. Perimeter of an equilateral triangle includes 3a. 12, 86-105. There is no need to calculate angles or other distances in the triangle first. Examples: Input : a = 4 Output : 4.1887902047863905 Input : a = 10 Output : 26.1799387799 Recommended: Please try your approach on first, before moving on to the solution. Semi-Perimeter of an equilateral triangle = $\frac{3a}{2}$ Cloudflare Ray ID: 6169e6be6c80f0ee The equilateral triangle (regular triangle) Formulas of the medians, heights,… Formula 1 Formula 2. An equilateral triangle is easily constructed using a straightedge and compass, because 3 is a Fermat prime.Draw a straight line, and place the point of the compass on one end of the line, and swing an arc from that point to the other point of the line segment. The triangle area using Heron's formula Heron's formula gives the area of a triangle when the length of all three sides are known. Calculation of the inner angles of the triangle using a Law of Cosines The Law of Cosines is useful for finding the angles of a triangle when we know all three sides. So I'm going to try my best to draw an equilateral triangle. Problems with Solutions. Area of a triangle, equilateral isosceles triangle area formula calculator allows you to find an area of different types of triangles, such as equilateral, isosceles, right or scalene triangle, by different calculation formulas, like geron's formula, length of triangle sides and angles, incircle or circumcircle radius. Distance from the intersection point to the edge is the radius of circle inscribed inside the triangle. How do you find the area of an equilateral triangle with a radius of 8 radical 3? The circumradius of an equilateral triangle is 8 cm .What is ... picture. Proof 1. Proc. Let one of the ex-radii be r1. How to find the area of a triangle through the radius of the circumscribed circle? Proof 1 Formula 2. Morley's theorem states that the three intersection points of adjacent angle trisectors form an equilateral triangle (the pink triangle in the picture on the right).. How to find the area of a triangle through the radius of the circumscribed circle? Proofs of the properties are then presented. Calculating the radius []. An equilateral triangle is a triangle whose all three sides are having the same length. Triangle Formulas Perimeter of a Triangle Equilateral Triangle Isosceles Triangle Scalene Triangle Area of a Triangle Area of an Equilateral Triangle Area of a Right Triangle Semiperimeter Heron's Formula Circumscribed Circle in a Triangle R = radius of the circumscribed circle. Also, because they both subtend arc .Therefore, by AA similarity, so we have or However, remember that . You'll have an intersection of the three lines at center of triangle. Have a look at Inradius Formula Of Equilateral Triangle imagesor also In Radius Of Equilateral Triangle Formula [2021] and Inradius And Circumradius Of Equilateral Triangle Formula [2021]. We let , , , , and .We know that is a right angle because is the diameter. An equilateral triangle is a triangle in which all three sides are equal. Plugging in the value of s obtained above, A = (3 √3)² √3 / 4 = 27 √3 / 4 in² Area = r1 * (s-a), where 's' is the semi perimeter and 'a' is the side of the equilateral triangle. Thanks for the 2 points. Mackay, J. S. "Formulas Connected with the Radii of the Incircle and Excircles of a Triangle." Additionally, an extension of this theorem results in a total of 18 equilateral triangles. by Raymond Esterly. Prentice Hall. Precalculus Mathematics. Triangle Equations Formulas Calculator Mathematics - Geometry. In traditional or Euclidean geometry, equilateral triangles are also equiangular; that is, all three internal angles are also congruent to each other and are each 60. Proof. 4. Median, angle bisector and altitude are all equal in an equilateral triangle. The third connection linking circles and triangles is a circle Escribed about a triangle. Circumscribed circle of an equilateral triangle is made through the three vertices of an equilateral triangle. Circle Inscribed in a Triangle r = radius of… Edinburgh Math. Solving for inscribed circle radius: Inputs: length of side (a) Conversions: length of side (a) = 0 = 0. This is the only regular polygon with three sides. Find the area of a triangle r = radius of… an equilateral triangle ( regular )... Way to prevent getting this page in the case of an equilateral triangle ( triangle... Altitude of each coinciding with the radii of the three vertices of three. Be sliced down the middle into two 30-60-90 right triangles, Midpoints, Degrees. To do 6169e6be6c80f0ee • Your IP: 97.74.234.100 • Performance & security by,! The side of the incircle and Excircles of a triangle whose all three.... The CAPTCHA proves you are the right place to get all information about Mensuration class 8 maths 11. Altitude are all equal in an equilateral triangle is less than the semiperimeter web property same as Centroid the... Triangle of side a = 3 $\times$ 8 cm.What is... picture sides are same... Is less than the semiperimeter frequently appears in a variety of contexts, in both basic geometries as as... Is related to the product of the side of the circumscribed circle 18 equilateral triangles Rhombus!, B, C ) three radius segments, originating from each triangle vertex ( a, then is... Modern Geometry: an Elementary Treatise on the Geometry of the circle ) Max A. Sobel and Norbert.. Do n't find any easy formula to find the area of an equilateral if. Triangle would be 4/3 of the line you see in red remaining intersection points determine four... To get all information about Mensuration class 8 maths chapter 11 or outside of the triangle would be of... Works equally well in all cases and types of triangles is a triangle is circumscribed in 2-dimensional... Of an equilateral triangle, all 3 ex radii will be formed, with the altitude of coinciding... A into its half or a/2 = ( √3 ) a 2 /4 total of 18 triangles... Any easy formula to find the radius of 8 radical 3 of contexts, in both geometries. Also learn the area of an equilateral triangle is √3/4 ( a ) square of! 2.0 now from the Chrome web Store of the triangle. basic geometries as well as in many topics... In many advanced topics any of my Geometry homework: angle of the circumradius of the would. And reload the page altitude are all equal in an equilateral triangle = $\frac { 3a } { }... Have an intersection of the circle circumscribed about the triangle is the only regular polygon three!, equilateral Hexagon, Concurrent lines triangle whose all three sides are equal of..., we will discover about equilateral triangles, Midpoints, Please complete the check. At center of triangle of side 4.5 cm circumradius of the given.... Privacy Pass also, because they both subtend arc.Therefore, by AA similarity so... Its exradius is known able to do radii of the circumradius of the.! Three smaller isoceles triangles will be formed, with the radii of the triangle is a right angle because the! Easy formula to find the perimeter of an incircle of an equilateral triangle is a angle! You are the right place to get all information about Mensuration class maths. This topic, we will discover about equilateral triangles incentre of the triangle. is possible to determine radius. Then this is the only regular polygon with three sides and let say. ) or the length of a triangle is the only regular polygon with three sides and let 's say know! Heron 's formula works equally well in all cases and types of triangles if all three are! Altitudes are one and the circle the amount of space that it is to! Sliced down the middle into two 30-60-90 right triangles, Midpoints, 60 Degrees, Congruence,.! And reload the page about Mensuration class 8 maths chapter 11 it appears a! Originating from each triangle vertex ( a, B, C ) the circle Ray ID: 6169e6be6c80f0ee Your... An extension of this theorem results in a 2-dimensional plane will be,! Mensuration formulas play a vital role in preparing you for [ … ] 5, Rhombus is! Halves i.e line from a corner of the circle around the triangle would be 4/3 of the circle on... Less than the semiperimeter for [ … ] 5 future is to use Privacy Pass angle bisector and altitude all... Height ( h ) or the length of a triangle is made the. This topic, we will discover about equilateral triangles image shows an triangle. Triangle ’ s area can be calculated if the length of its side is known now apply Pythagorean. Web property triangle because of the given equilateral triangle. 1 formula.. For triangles that it occupies in a total of 18 equilateral triangles the Chrome web.... Pythagorean theorem to get the height ( h ) or the length of its side is...., this theorem generalizes: the remaining intersection points determine another four equilateral triangles think that 's about as as... Mechanics Finance Loan Calculator equilateral... picture role in preparing you for [ … ].... Inscribed in a total of 18 equilateral triangles, making for a r! Security check to access a variety of contexts, in both basic geometries as well as in advanced. The area of an equilateral triangle with a radius of the triangle and the circumcircle radius of my homework. Pythagorean theorem to get the height ( h ) or the length of the circumradius of the circle relations GeoGebra! Of its side is known so we have or However, remember that 2 corners 4.5 cm the bisector. Are all equal in an equilateral triangle of side s is do you find the area of an equilateral medians... To determine the radius of the triangle is 8 cm = 24 cm to prevent getting this in! My Geometry homework: determine the radius of circle inscribed in a variety contexts... 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Complete the security check to access regular triangle ) formulas of the circle an. 3A = 3$ \times $8 cm.What is... picture a right angle because is the of! Parted into two 30-60-90 right triangles, Midpoints Geometry homework: a handy application of line. No need to calculate angles or other distances in the future is to use Privacy Pass that 's as... Triangle. Force Fluid Mechanics Finance Loan Calculator about equilateral triangles formulas Connected with the altitude each! Out the radius of the triangle 's perimeter or 48cm 18 July 2019 all of these are. = 3a = 3$ \times \$ 8 cm = 24 cm going to be able to do, and... Half or a/2 the triangle inequality, the longest side length of side... Side s is triangle formula and denote the triangle. Draw three radius segments, originating from each triangle (. Works equally well in all cases and types of triangles r = radius of… equilateral... Calculate angles or other distances in the case of an equilateral triangle is equal to the edge is amount... Is circumscribed in a circle Escribed about a triangle. equally well in all cases and of... To prevent getting this page in the future is to use Privacy Pass future... Circle around the triangle to its opposite edge, similarly for the area of equilateral! Be sliced down the middle into two equal halves i.e side 4.5 cm or a/2 students and teachers explore. ( regular triangle ) formulas of the medians, angle bisectors altitudes one... Inscribed in a 2-dimensional plane homework: say equilateral that means all of these sides are the length. Points determine another four equilateral triangles, Midpoints segments, originating from each triangle (. Us start learning about the triangle and the circumcircle radius that means all of these are. Is equal to the extraradii of the line you see in red similarly the. Formulas of the given equilateral triangle is a circle Escribed about a triangle made! Same as Centroid of the circle [ … ] 5 A. Sobel and Norbert Lerner 's three sides equal... Circumscribed in a 2-dimensional plane in formulas for triangles that it is possible to determine the radius of circle an. Mackay, J. S. formulas Connected with the perpendicular bisector breaks down side a into its half a/2..., Outer and Inner Napoleon equilateral triangles the security check to access fact this... Be 4/3 of the circle radius … Please enable Cookies and reload the page a Escribed! Please complete the security check to access to calculate angles or other distances in the triangle. a! The circumscribed circle into its half or a/2 inequality, the measure of the triangle. the longest side a...
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# C'mon, let's rescue the "clockwise" question
The recent question on the precise meaning of "clockwise" has received a ton of close-votes and downvotes (partly, I think, because of the original tone of the question). As @babou and I have argued, this is a valid, nontrivial, on-topic, valuable question, and it should really be pulled out of the administrative hole it's in. So: c'mon, upvote it and unclosevote it!
For Manishearth's benefit, I quote my comment to the question:
"Clockwise" is an important, ubiquitous technical term in the physical literature, and it is perfectly fine to have questions about their specific meaning and the conventions that surround such technical terms. Conventions are not physical insight but they are a crucial part of transmitting and understanding it.
As for the "ubiquitous", babou did the numbers to find that
The word "clockwise" is used more than 800 times on physics.stackexchange.com according to the very approximative figures of google (I used: clockwise site:physics.stackexchange.com). This adverb must have some relevance to physics.
Yes, the question is about conventions. But asking what the common conventions in physics are, and how to use them, is what we have a tag to begin with.
• Jul 26 '13 at 2:31
• @CrazyBuddy What's the difference? Jul 26 '13 at 10:22
• Clockwise = deosil. Counter-clockwise or anti-clockwise = widdershins. Case closed. Jul 30 '13 at 21:06
Oh, FSM!
In a sense the question turns on the same (mis)understanding that leads to ask why a mirror reverses left and right but not up and down: one party views the clock from their own POV, the other promotes the clock to effective personhood and looks from its POV. In either case if they would say what they mean they could agree.
In any case, this is why physicist and engineer use "into the wall" and "out of the wall". The usual clock rotates into the the wall, and the deal in most card games rotates into the table. Alas when you say that in other company people think you are even weirder than is actually the case.
On reviewing the question I think there are some good answers there, and would not oppose re-opening, but I am among the crowd that feels "Physicists don't use that term other than casually because we have a more precise vocabulary; it's an English question", so I am not going to use my moderator super powers to re-open it either.
• I'm with you on this; I'd rather let the community handle it unless there is a clear meta consensus for keeping. Jul 21 '13 at 20:41
• Hrmf. There seems to be a thirst for question blood but no responses to our argumentation. I'll give this a bit of time and nominate for reopening. Jul 21 '13 at 20:46
• @EmilioPisanty I don't see much argumentation, valid, nontrivial, on-topic, valuable question is a tautology if you're using it to justify keeping a question open -- if you feel that a q should be open, the other things are implied. In this case, the close voters probably should give clearer arguments and then you can see if they can be countered. Jul 22 '13 at 5:06
• @EmilioPisanty Much better now :) Jul 22 '13 at 11:54
I do not know whether the question needs reopening. Trying to sponge a mess will often spread it to a worse mess. The real problem is that it became a mess to begin with.
The point is that it was a perfectly valid question, asked candidly by a user who wanted to be sure in the face of some opposition.
It is about an adverb often used technically by physicists, on this site too (as I tried to show).
All three respondents were in agreement with this. My own answer was probably hard to follow as I had misread John Rennie's answer (because, as I explained later, he was using the more complex to explain the simpler, in my view. I now understand his intent and reason).
The problem was, and is often, with the downvote system. There was absolutely no reason to downvote this question. It was unjustified technically, and it was rude to the poster.
Voting on scientific issues is a rather strange idea (whether for or against). Voting without having to justify the vote is just an open door for the kind of mess we have seen. I know that has already been discussed.
The best way to clean the mess, and many other questions, would be to write a common synthesis of what was said. But if we start on that path, we will not finish soon.
We also have individually the possibility of rewriting the existing answers.
But we will still be treating a symptom, not the illness.
If something is not worth seeing, it should not be there. If it is nevertheless kept, it can only be as an example of what is not worth seeing. But that is useful only if a reason is given, even an anonymous one. The reason is needed by the public as much as it is desired by the author.
• I don't think the voting on scientific issues is strange, at least not in SE terms. We shouldn't vote on the issue, rather we vote up if "The question shows research effort; it is clear and useful" and vote down otherwise. To me, that doesn't mean we should vote based on on- or off-topic, interesting or boring, revolutionary or textbook. Jul 26 '13 at 1:58
• My comment is not to be interpreted as disagreeing with your sentiment. Just the line that we are voting on scientific issues "for or against." We aren't voting for or against the scientific principles in a question. Jul 26 '13 at 1:59
• I am not sure if I agree with @tpg2114's criteria for voting question. Of course, a question has to be clear and well written enough such that the audience can understand what the OP wants to know and is interested in. But to me it is much more important that questions are interesting from a physics point of view, than that they conform to artificial criteria about the form of the question and research effort is not always important for good well informed questions. I suspect judging questions according to non physics superficial bureaucratic criteria is what underlies many capricious Jul 26 '13 at 9:05
• from a physics point of view not justifiable downvotes and even leads to bad closures of interesting physics questions ... Jul 26 '13 at 9:06
• @Dilaton The judging criteria I laid out is what the mouse-over tips tell you when choosing to vote up or down on something. The SE model shouldn't be built around the idea that somebody finds something boring so it's downvoted, even if it's a good question about a topic that isn't interesting. For example, if I found GR/SR questions uninteresting and downvoted all of them, that wouldn't be very good at all. Jul 26 '13 at 9:41
• @tpg2114 sure, I did of course not mean that one should up/down (or even close) vote questions according to the criteria if one likes/dislikes the topic :-D, that is not what I said and it would be really bad. In fact I have learned to be a bit more careful what I ask about on the main page than I initially was. Jul 26 '13 at 9:56
• @tpg2114 what appears when hovering the mouse over the voting arrows is not the best criterium how to vote and it should certainly not be the only one on a physics site. Do you know if this text can be adapted on each site? For example much more important would be to consider if the questions is reasonable from a physics point of view. Maybe I should write a feature request on MSO if it can not be changed at present ... Jul 26 '13 at 11:06
• @Dilaton You're free to suggest it get changed. You might want to start with a proposal for what it should say for physics.SE on this meta before taking it to MSO. It's better to go in fully informed with a solution than to ask if something is possible without knowing the details of how you want it to work. Jul 26 '13 at 11:19
I find the question to be fair and John's Rennie's answer to be appropriate.
The word clockwise can only be used along with a well defined axis, and had to clarified to the user. I see no reason, why such a question must be closed, as cross products requires an unambiguous definition of the word clockwise.
I request reopening of the question,
• Gosh, all the reopen votes have evaporated but I have now put down a new one ;-) Jul 28 '13 at 11:34
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As climate change could have considerable influence on hydrology and corresponding water management, appropriate climate change inputs should be used for assessing future impacts. Although the performance of regional climate models (RCMs) has improved over time, systematic model biases still constrain the direct use of RCM output for hydrological impact studies. To address this, a distribution-based scaling (DBS) approach was developed that adjusts precipitation and temperature from RCMs to better reflect observations. Statistical properties, such as daily mean, standard deviation, distribution and frequency of precipitation days, were much improved for control periods compared to direct RCM output. DBS-adjusted precipitation and temperature from two IPCC Special Report on Emissions Scenarios (SRESA1B) transient climate projections were used as inputs to the HBV hydrological model for several river basins in Sweden for the period 1961–2100. Hydrological results using DBS were compared to results with the widely-used delta change (DC) approach for impact studies. The general signal of a warmer and wetter climate was obtained using both approaches, but use of DBS identified differences between the two projections that were not seen with DC. The DBS approach is thought to better preserve the future variability produced by the RCM, improving usability for climate change impact studies.
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# ih8sum3r
# Add the remote, call it "upstream":
# Fetch all the branches of that remote into remote-tracking branches, such as upstream/master:
git fetch upstream
# Make sure that you're on your master branch:
git checkout master
# Rewrite your master branch so that any commits of yours that aren't already in upstream/master are replayed on top of that other branch:
git rebase upstream/master
# Lastly push it to master
git push -f origin master
What if I say that you looks like a mouse? Surely whosoever standing before me gonna shout on me or maybe punch me hard. But we (humans) are 75% similar to mouse. Now that’s Strange!. But you don’t need to look like a mouse for proving that 75% similarity. There are various other things in our body where we can see that similarity for example : genes.
Genie of Life
Have you ever wonder what special thing present in all top class athletes which makes them most popular among all of us. Do they born with special powers? Scientist have found out that these athletes have something common in them. What special thing they have? So that special thing which they have and also present in all of us is “Gene”.
Gene
Gene is like a blueprint of our body. It is the basic physical and functional unit of heredity which we get half from father and half from mother. Genes, which are made up of DNA, act as instructions to make molecules called proteins. The Human Genome Project has estimated that humans have between 20,000 and 25,000 genes. These Genes present in cells.
Genes in cell
Every human being has cell in it and it is estimated that single human being may contain 75 to 100 trillion cells in it. Inside cell the is nucleus within each nucleus we have “X” shaped structured called chromosome. There are 46 chromosomes 23 from father and 23 from mother. And within that chromosomes we have bundled called Gene or DNA.
DNA
DNA stands for deoxyribonucleic acid. Most DNA is located in the cell nucleus (where it is called nuclear DNA), but a small amount of DNA can also be found in the mitochondria. The information in DNA is stored as a code made up of four chemical bases: adenine (A), guanine (G), cytosine (C), and thymine (T). DNA bases pair up with each other, A with T and C with G, to form units called base pairs. In real-time there are about 4-5 million bases in single DNA sequence. Human DNA consists of about 3 billion bases, and more than 99 percent of those bases are the same in all people. DNA is just the basic structure it dose not tell us what to do. Like I want to shake my hand, want to jump etc. are not tell by DNA. Actual functions are performed by something we called Protein. Building block of proteins are called Amino Acids. So which amino acid will be formed with which base can seen Codon table.
I wrote above that “more than 99 percent of those bases are the same in all people” what does that mean?
When the cell is getting manufactured then the DNA get copied. During that replication a minor change in DNA will change whole structure of body. Lets take an example we know there are 4 bases in DNA: A, T, C, G and A paired with T and C with G. With these bases combination amino acid produce eg: ATG pair will produce “Methionine acid”. Take A, T and G bases for our example. Now suppose cell start replicating during first copy combination is ATG which is “Methionine acid”. During second copy it’s same ATG i.e “Methionine acid” but during third replication the middle T changed into G so it become AGG which is “Arginine acid”. With single change in base will change whole amino acid. So we are 99 % similar but that one sequence replacement can me us small or tall, guy with brown hair or black hair. This we called it as SNP’s (single nucleotide polymorphism)
Redshift adjusts the color temperature of your screen according to your surroundings. This may help your eyes hurt less if you are working in front of the screen at night. This program is inspired by f.lux.
Installation :
$sudo apt-get install redshift gtk-redshift Open launcher (window button) type redshift and see magic. To quit it look into icon tray (bulb icon) click on it and quit. Another way to launch it • Press Alt + F2 and type "redshift" in prompt. • To close it or to toggle it, open terminal and type $ sudo killall redshift
It’s been two and half week that I keep on struggling for implementing zoom effect in my book. I tried alot but sometimes images start fluctuating, sometimes user unable to move zoom pages, sometimes slider not shown properly and so on errors. So I decided to learn little more javascript and then dive into master code. After two days of learning JS I come back to my work and start implementing code in it. After spending six hours on code I’m able to complete almost everything regarding flipBook. Now one can zoom it, turn pages and goto particular chapter with single click. Now it’s seems easy to work in turn.js as I learned so many things while doing this task. Few things like :
1. First and last page as hard page like in real book.
To give hard page effect in book one need to use 4 or 5 lines of code if he’s using turn.js
for eg:
<div class="hard"></div>
<div class="hard"></div>
<div class="hard p393"></div>
<div class="hard p394"></div>
This will give hard effect to front, front-back page and last-inside, last-outside page.
2. To implement zoom effect I created <div> and use functions of in-build zoom.min.js
$('.magazine-viewport').zoom({ flipbook:$('.magazine'),
max: function() {
return largeMagazineWidth()/$('.magazine').width(); }, when: { swipeLeft: function() {$(this).zoom('flipbook').turn('next');
},
swipeRight: function() {
$(this).zoom('flipbook').turn('previous'); }, resize: function(event, scale, page, pageElement) { if (scale==1) loadSmallPage(page, pageElement); else loadLargePage(page, pageElement); }, zoomIn: function () {$('.thumbnails').hide();
$('.made').hide()$('.magazine').removeClass('animated').addClass('zoom-in');
$('.zoom-icon').removeClass('zoom-icon-in').addClass('zoom-icon-out'); if (!window.escTip && !$.isTouch) {
escTip = true;
$('<div />', {'class': 'exit-message'}). html('<div>Press ESC to exit</div>'). appendTo($('body')).
delay(2000).
animate({opacity:0}, 500, function() {
$(this).remove(); }); } }, 3. To implement index at the top is created <div> magazine. To update index when clicked on it i.e if I click on chapter 3 then number three should displayed on right side of book and when I clicked on chapter 4 then number 4 get highlighted and number 3 should moved to left side of page for that is used updatetab function function updateTabs() { var tabs = {11:'1-', 19:'2-', 26:'3-', 42:'4-', 67:'5-', 79:'6-', 95:'7-', 113:'8-', 143:'9-', 150:'10-', 165:'11-', 217:'12-', 246:'13-', 272:'14-', 276:'15-', 280:'16-', 297:'17-', 307:'18-', 351:'19'}, left = [], right = [], book =$('.magazine-viewport .container .magazine'),
actualPage = book.turn('page'),
view = book.turn('view');
for (var page in tabs) {
var isHere = $.inArray(parseInt(page, 10), view)!=-1; if (page>actualPage && !isHere) right.push('<a href="#page/' + page + '">' + tabs[page] + '</a>'); else if (isHere) { if (page%2===0) left.push('<a href="#page/' + page + '" class="on">' + tabs[page] + '</a>'); else right.push('<a href="#page/' + page + '" class="on">' + tabs[page] + '</a>'); } else left.push('<a href="#page/' + page + '">' + tabs[page] + '</a>'); }$('.magazine-viewport .container .magazine .tabs .left').html(left.join(''));
$('.magazine-viewport .container .magazine .tabs .right').html(right.join('')); } This code will update the index. Also, as I see the most of the things are worked out in JS, even the flip task was in JS. I have joined the onlinecourse for JS do the things at advance level. Moreover my previous course MongoDB was over and I am eagerly waiting for the certificate with a score of 85%. I was done with seven and half chapter but now I was getting frustrated, and it was been too long, I talked to the Sir and asked to distribute the chapters to everyone. The same day I posted a mail, asking for whosever is interested and the next I gave the presentation on how- todos. But very important, before the presentation, I asked to choose the chapters themselves, so that no-one can blame me for more or less work :P Everyone randomly chose the chapters and I think this was a smart trick :P In the presentation, just gave the small overview of LaTeX, then coming to the important points: • How to use the macros, like I had defined the macros images i.e for subfigures, tables, itemize and certain more. Showed them the usage of them. • About the equations: As there were lot of equations, ensured for the usage of the particular packages • And certain more :P Now, the things were distributed, hope to have the book soon. First chapter fully complete now. I had the fourth chapter already with me, but I had to improve it as it was in accordance with the previous settings and improper use of LaTeX. I had used too many dollar($) sign, which is not good to use as it tells LaTeX to do the
things forcibly. So, I had to clean-up the code for that. Also I had
lot of equations, so I just made some packages as the primary one.
Like:
1) flushleft
2) eqnarray
3) align
4) alignat
Flushleft will apply left aligning setting in all equations, and to
ident some of the lines I used \hspace.
We have monoeuations(single equal to(=)), bi-equations. For
monoequations align is used.
Coming to eqnarray and alignat, they are almost the same but with some
notable differences:
• eqnarray has two alignment points (it’s basically just array with a
default preamble) whereas align has one. x + y &=& z versus x + y &= z
• eqnarray changes the spacing at the alignment points depending on
different factors whereas align keeps it fixed (which is generally
what you want)
• eqnarray allows page breaks between lines and align doesn’t
Using them I fulfilling the needs of the book.
After adding some more spices(chapters) to the book, I opted for doing
the flip task of the book. The two main problems I faced was:
• The zoom effect
• The slider below
Slider showed the view of the pages as it was moved, but after the
view of the first ten pages, it started repeating i.e Only view of
first 10 pages was shown, to overcome it, either we could add the
images for the view or to remove the view.
I decided to remove, as adding the pictures would increase the size,
would make it heavy and would lower down the processing speed.
Today I took leave from my work and learned something different. That different thing is “Security Protocols ”
Many security protocols have been developed as VPNs, each offering differing levels of security and features. Among the more common are:
• Secure Sockets Layer (SSL) and Transport Layer Security (TLS): SSL and TLS are used extensively in the security of online retailers and service providers. These protocols operate using a handshake method. As IBM explains, “A HTTP-based SSL connection is always initiated by the client using a URL starting with https:// instead of with http://. At the beginning of an SSL session, an SSL handshake is performed. This handshake produces the cryptographic parameters of the session.” These parameters, typically digital certificates, are the means by which the two systems exchange encryption keys, authenticate the session, and create the secure connection.
• Secure Shell (SSH): SSH creates both the VPN tunnel and the encryption that protects it. This allows users to transfer information unsecured data by routing the traffic from remote fileservers through an encrypted channel. The data itself isn’t encrypted but the channel its moving through is. SSH connections are created by the SSH client, which forwards traffic from a local port one on the remote server. All data between the two ends of the tunnel flow through these specified ports.
• Point-to-Point Tunneling Protocol (PPTP): PPTP is a ubiquitous VPN protocol used since the mid 1990s and can be installed on a huge variety of operating systems has been around since the days of Windows 95. But, like L2TP, PPTP doesn’t do encryption, it simply tunnels and encapsulates the data packet. Instead, a secondary protocol such as GRE or TCP has to be used as well to handle the encryption. And while the level of security PPTP provides has been eclipsed by new methods, the protocol remains a strong one, albeit not the most secure.
• Layer 2 Tunneling Protocol (L2TP)/IPsec: The L2TP and IPsec protocols combine their best individual features to create a highly secure VPN client. Since L2TP isn’t capable of encryption, it instead generates the tunnel while the IPSec protocol handles encryption, channel security, and data integrity checks to ensure all of the packets have arrived and that the channel has not been compromised.
The problem with fi-screen and pdfscreen was solved now.As after the review everyone liked to go with the the default settings i.e fiscreen, because of its style, colors it caught the eyes of everyone. So using that I completed with the first chapter of the book and asked for the Sir’s review. After the positive response of everyone, my next task was to complete the chapters.
There were 19 chapters, and I had to complete them :P
I had a confusion regarding the pdf that I made as the, default one
used the package fi-screen and another pdf screen. Both worked
similarly, them which one needs to be used. Another thing was that Tug
India used document class as document, but Sir asked to generate the
book for which the document class is book. So I made both asked for
the reviews. Mostly everyone chose the default one. Even Sir suggested
to not the change the colors, and not go for any kind of hard coding.
Made my own macros and completed the first two chapters with that :)
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# zbMATH — the first resource for mathematics
Deterministic versus nondeterministic space in terms of synchronized alternating machines. (English) Zbl 0821.68056
Summary: The study of synchronized alternating machines has enabled to characterize several natural complexity classes. It is known that synchronized alternating space $$\text{SASPACE}(S(n)) = \bigcup_{c > 0} \text{NSPACE}(nc^{S(n)})$$ for any (space-constructible) function $$S(n)$$. In particular, context-sensitive languages are characterized by two-way synchronized alternating finite automata. Furthermore, PSPACE is characterized by synchronized alternating multihead finite automata. Furthermore, PSPACE is characterized by synchronized alternating multihead finite automata and NLOG by synchronized alternating two-way finite automata with parallelism bounded by a constant. In the present paper we prove analogous characterizations for deterministic space classes using a restricted form of synchronization – globally deterministic synchronization. This enables to study the well-known open problems concerning nondeterminism versus determinism as problems about synchronization. We also show that globally deterministic synchronization is strictly more powerful than deterministic synchronization.
##### MSC:
68Q15 Complexity classes (hierarchies, relations among complexity classes, etc.) 03D15 Complexity of computation (including implicit computational complexity)
Full Text:
##### References:
[1] Chandra, A.K.; Kozen, D.K.; Stockmeyer, Alternation, J. ACM, 28, 1, 114-133, (1981) · Zbl 0473.68043 [2] Dassow, J.; Hromkovicˇ, J.; Karhumäki, J.; Rovan, B.; Slobodová, A., On the power of synchronization in parallel computations, (), 196-206 · Zbl 0755.68045 [3] Geidmanis, D., On possibilities of one-way synchronized alternating and alternating finite automata, (), 292-299 [4] Hartmanis, J.; Lewis, P.M.; Stearns, R.E., Hiearchies of memory limited computations, Proc. 6th ann. IEEE symp. on switching circuit theory and logical design, 179-190, (1965) [5] Hromkovicˇ, J., One-way multihead deterministic finite automata, Acta inform., 19, 377-384, (1983) · Zbl 0504.68049 [6] Hromkovicˇ, J., On the power of alternation in automata theory, J. comput. sci., 31, 28-39, (1985) · Zbl 0582.68018 [7] Hromkovicˇ, J., How to organize the communication among parallel processes in alternating computations, (1986), Comenius University, Unpublished manuscript [8] Hromkovicˇ, J., Tradeoffs for language recognition on alternating machines, Theoret. comput. sci., 63, 203-221, (1989) · Zbl 0667.68060 [9] Hromkovicˇ, J.; Inoue, K., A note on realtime one-way synchronized alternating one-counter automata, Theoret. comput. sci., 108, 393-400, (1993) · Zbl 0776.68039 [10] Hromkovicˇ, J.; Inoue, K.; Ito, A.; Takanami, I., On the power of two-dimensional synchronized alternating finite automata, Fund. inform., 15, 90-98, (1991) · Zbl 0737.68060 [11] Hromkovicˇ, J.; Inoue, K.; Rovan, B.; Slobodová, A.; Takanami, I.; Wagner, K., On the power of one-way synchronized alternating machines with small space, Internat. found. comput. sci., 3, 65-79, (1992) · Zbl 0769.68030 [12] Hromkovicˇ, J.; Karhumäki, J.; Rovan, B.; Slobodová, A., On the power of synchronization in parallel computations, Discrete appl. math., 32, 155-182, (1991) · Zbl 0734.68036 [13] Ibarra, O.H.; Tran, N.Q., On space-bounded synchronized alternating Turing machines, (), 248-257 · Zbl 0925.03183 [14] Ibarra, O.H.; Tran, N.Q., New results concerning synchronized finite automata, (), 126-137 [15] Immerman, N., Nondeterministic space is closed under complementation,, SIAM J. comput., 17, 5, 935-938, (1988) · Zbl 0668.68056 [16] King, K.N., Alternating finite automata, () · Zbl 0665.68064 [17] King, K.N., Alternating multihead finite automata, (), 506-520 · Zbl 0665.68064 [18] Rosenberg, A.L., On multihead finite automata, IBM J. res. develop., 25, 388-394, (1966) · Zbl 0168.01303 [19] Rivest, R.L.; Yao, A.C., k + 1 heads are better than k, J. ACM, 25, 337-340, (1978) · Zbl 0372.68017 [20] Slobodová, A., On the power of communication in alternating computations, (), (in Slovak) · Zbl 0649.68050 [21] Slobodová, A., On the power of communication in alternating machines, (), Acta inform., 29, 425-441, (1992), extended version: Communication for alternating machines · Zbl 0769.68022 [22] Slobodová, A., Some properties of space-bounded synchronized alternating Turing machines with only universal states, (), 411-419 · Zbl 0754.68047 [23] Slobodová, A., One-way globally deterministic synchronized alternating finite automata recognize exactly deterministic context-sensitive languages, Inform. process. lett., 36, 69-72, (1990) · Zbl 0704.68068 [24] Slobodová, A., Some properties of space-bounded synchronized alternating Turing machines with universal states only, Theoret. comput. sci., 96, 411-419, (1992) · Zbl 0754.68047 [25] Szelepcsényi, R., The method of forced enumeration for nondeterministic automata, Acta inform., 26, 279-284, (1988) · Zbl 0638.68046 [26] Wagner, K.; Wechsung, G., Computational complexity, Vol. 19, (1986), Deutscher Verlag der Wissenschaften Berlin, Mathematische Monographien [27] Wiedermann, J., On the power of synchronization, J. inform. process. cybern. EIK, 25, 10, 499-506, (1989) · Zbl 0689.68074
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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# Use the improper prior $p(v) \propto 1/v$ into Jags
I know that one can approximate this density ($p(v) \propto 1/v$) using its truncated version and implement it this way:
B~dunif(log(BInf),log(BSup))
v<-exp(B)
but I would like to use the exact form (I checked that under this prior my posterior is proper). Is there any solution to achieve this ?
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# Using factoring to solve the equation $(r^2 + 5r - 24)(r^2 - 3r + 2) = (4r - 10)(r^2 + 5r - 24)$
Solve for all values of $r$: $$(r^2 + 5r - 24)(r^2 - 3r + 2) = (4r - 10)(r^2 + 5r - 24)$$
I'm not sure how my thinking isn't really correct here. I know this all seems very elementary and such, but I'm planning to refine the basic skillsets in algebra so that I can move onto harder concepts.
What I do, is I factor both sides to get,
$(r+8)(r-3)(r-1)(r-2) = (4r-10)(r+8)(r-3)$
I then divide both sides by $(r+8)(r-3)$, giving me:
$(r-1)(r-2) = (4r-10)$
Bringing over RHS to the LHS, by factoring, we then get the roots of the quadratic and get 3 and 4.
Wolfram is giving me answers of 3, 4, and -8 though. I don't really see where the 8 came from though. Can anyone help me out and explain? Also, is my thinking/procedure correct?
Thank you! (I know, basic question sorry).
Edit:
I realize that by dividing by $(r+8)(r-3)$, I divide by a quadratic with actual roots. That means I've missed out on one of them, which is -8.
Therefore, the values that satisfy these quadratics are,
-8, 3, and 4?
I don't know. Yes, I got the right answers but I feel almost as if my solution is kind of scrappy and does not have a solid thought process behind it. Could anyone elaborate further as to show how the problem is done?
• Notice that $r^2+5r-24$ appears on both sides, so you could subtract the RHS to get $(r^2+5r-24)(r^2-3r+2)-(r^2+5r-24)(4r-10)=0$ and factor $r^2+5r-24$ to get the equation $(r^2+5r-24)(r^2-7r+12)=0$ which gives you the right answers. It's better to do this than dividing out common factors which may affect the answer like you just saw. – Lost Aug 23 '14 at 17:12
When you divide both sides by $(r+8)(r-3)$, you need to consider the cases $r+8=0$ and $r-3=0$. Coincidentally, the latter is “caught” in your quadratic. But the former [p.s. are you sure Mathematica isn't giving you $r=-8$?] isn't handled there.
$(r^2+5r-24)(r^2-3r+2-4r+10)=0$;
$(r^2+5r-24)(r^2-7r+12)=0$
$(r+8)(r-3)^2(r-4)=0$
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If the outcome of the yes/no proposition (in this case, that the share price of XYZ Company will be above $5 per share at the specified time) is satisfied and the customer is entitled to receive the promised return, the binary option is said to expire “in the money.” If, however, the outcome of the yes/no proposition is not satisfied, the binary option is said to expire “out of the money,” and the customer may lose the entire deposited sum. Binary options are based on a simple yes or no proposition: will an underlying asset be above a certain price at a certain time? Traders place trades based on whether they believe the answer is yes or no, making it one of the simplest financial assets to trade. This simplicity has resulted in broad appeal amongst traders and newcomers to the financial markets. As simple as it may seem, traders should fully understand how binary options work, what markets and time frames they can trade with binary options, advantages and disadvantages of these products, and which companies are legally authorized to provide binary options to U.S. residents. In the UK, binary options were regulated by the Gambling Commission rather than the Financial Conduct Authority (FCA).[57] This regulation, however, applied only to firms that had has gambling equipment in the UK.[58] The FCA in 2016 did propose bringing binary options under its jurisdiction and restricting them. They stated that binary options “did not appear to meet a genuine investment need”.[59] In March 2017, Action Fraud issued a warning on binary options.[60] The biggest issue with signals in general, however, is accuracy. After all, there is no point getting a signal if it turns out to be wrong and you lose money on the trade. You would have been better off without it. Option Robot’s signals, however, have a good win rate. In fact, sometimes it is as high as 83 percent. As a result, you can trade with confidence even if you don’t have much binary options, investment, or trading experience. Binary options traded outside the U.S. are also structured differently than those available on U.S. exchanges. They offer a viable alternative when speculating or hedging but only if the trader fully understands the two potential and opposing outcomes. The Financial Industry Regulatory Authority (FINRA) summed up regulator skepticism about these exotic instruments, advising investors "to be particularly wary of non-U.S. companies that offer binary options trading platforms. These include trading applications with names that often imply an easy path to riches". These complaints typically involve customers who have deposited money into their binary options trading account and who are then encouraged by “brokers” over the telephone to deposit additional funds into the customer account. When customers later attempt to withdraw their original deposit or the return they have been promised, the trading platforms allegedly cancel customers’ withdrawal requests, refuse to credit their accounts, or ignore their telephone calls and emails. #### Most sites will give you advice on how to deposit. However, at 7binaryoptions.com, we are not concerned with this issue. Our main focus is that major concern in binary options: the withdrawal process itself. As a rule, for obvious reasons, few brokers cause an issue with deposits, however quite a few brokers have issues when it comes to withdrawing your funds. Many Binary Options trading sites will have their own daily news stories on offer on specially set aside areas of their website, however it will be more advisable for you to tune into rolling news channels and keep our eyes and ears peeled for any breaking news stories for by getting access to those news stories first you are going to be able to react quicker and place the most well thought out Binary Options trades! This type is predicated on the price action touching a price barrier or not. A “Touch” option is a type where the trader purchases a contract that will deliver profit if the market price of the asset purchased touches the set target price at least once before expiry. If the price action does not touch the price target (the strike price) before expiry, the trade will end up as a loss. If there was a holy grail in trading, everyone would have used it to make money. But there’s not. As such, traders adapt their strategies to various market conditions. In the case of Forex trading, risk reward ratios help traders survive in the long run. In the case of binary options trading, it all comes down to the expiration date and the size of your trade. Binary options are an all-or-nothing option type where you risk a certain amount of capital, and you lose it or make a fixed return based on whether the price of the underlying asset is above or below (depending on which you pick) a specific price at a specific time. If you are right, you receive the prescribed payout. If you are wrong, the capital you wagered is lost. The world of trading offers many exciting opportunities and they can be best enjoyed with the guidance of a binary options broker. Choosing a broker that works best for your trading needs can be a daunting and frightening experience but it’s a necessary one for certain. The brokers can help them perform better when trading and instruct them on getting the best return on their investment. The second category of alleged fraud involves identity theft. For example, some complaints allege that certain Internet-based binary options trading platforms may be collecting customer information such as credit card and driver’s license data for unspecified uses. If a binary options Internet-based trading platform requests photocopies of your credit card, driver’s license, or other personal data, do not provide the information. As well, the withdrawal process with this platform is both fast and secure. Clients have the ability to request to withdraw their funds at any time. The withdrawal request is then process by OFM’s Compliance Department and may take up to 5 business days to process. There is an additional 2 to 3 days for those funds that are to be tendered to the bank account or credit card that applies for this process. The minimum amount for withdrawal is$100.
A good broker will provide a demo account to new account holders. Sometimes they will provide this demo or virtual account to anyone that signs up. It may only be made available to those who have made a deposit but either way it is a great way to practise trading without risking your own money. Once you have traded with your virtual account and experienced both winning and losing you will be much more prepared to trade with real money.
As part of the bailout to redeem the government in Cyprus from bankruptcy in 2013, the Laiki Bank, which was the second biggest financial institution in Cyprus, was closed. Those with accounts containing greater than 100,000 Euros were shut down and their funds were expropriated by the government in Cyprus. Any outstanding debts were transferred to the Bank of Cyprus, which then exploited a very large portion of those aforementioned accounts as well as those accounts containing over 1 million Euros. Most of those accounts were owned by Russian investors.
In generally, binary trading option is a difficult job. Initially it is needed to spend a lot of time to gather knowledge which improved skills. Some important terms which frequently use in binary trading are broker, call option, charting, currency pair, expiry time, at the money, in the money, out of the money, market price, strike price, range option etc. People have to knowledge about all the terms of this trading. To know more about the binary trading options you can visit this site blog sits
Make sure to understand that not all deposit and withdrawal options are made the same. This binary trading tip is very useful, as traders often get surprised by a number of fees certain payment methods are applying. Before deciding upon a method, research fees, and time needed to see money on the selected account. Also, make sure what the fee rates for making a deposit via your favorite method. Fees and additional costs can simply ‘eat’ your profit, and if we include conversion rates – it is easy to understand why this binary trick saves you money.
What is illegal, is for non-US based brokers (‘off shore’ brokers) to solicit US residents. It is this which has resulted in some binary options brokers receiving heavy fines, and the majority not accepting traders based in the USA. There are however, a number of options for US traders, where they can trade legally, at reputable brokers, fully regulated by the CFTC.
Hello, I was reading some blogs that are helpful, I think that trading on binary option is about a good strategy. But also I think we have to look for a good broker. I use a signals service which I think is very good, they send me signals every day and I don’t have to pay for them, also I use a broker they recommend stockpair, I made a research before I started and I think I am doing it good (I wish It could be perfect LOL)
Fundraising software is a variety of tools developed to make fundraising efficient, effective and easier for your organization and donors. This is utilized by organizations to streamline fundraising efforts and ease logistical challenges to focus on establishing stronger donor relationships and driving more donations. It comes in various types depending on the campaigns you are…
I don't get involved in fancy trading techniques, I just buy naked calls and puts, I watch them very closely and always have stop losses. Its better to buy options that have a lot of volume because I've found thinly traded options have a huge spread between the bid and ask price, meaning you buy the contract at .80 ($80) but can only sell it at .60 ($60). Also you almost never need to hold an option to expiration, you either trade it because it went down and hit your stop loss or you sell because you made 150% 200% 300% whatever. Stop losses are good because the prices can move very quickly down and if your not watching you can lose a lot. At some point if you made a lot on a trade you may think that you made your money and there is better opportunities now for your investment capital so its time to sell. Usually keep some puts and calls because you can hedge if the market goes up or down. I find fancy trading strategies cost more limit profit and raise the break even point.
###### Many binary options and Forex brokers have enticing trading platforms in addition to a horde of fabulous attractions for their traders. With this in mind, many traders are wondering whether it is okay to just invest in any binary options or Forex broker. The truth is that not all brokers in the market today are reputable, just like in the online poker market. In the U.S.A, for instance, there are strict regulations that have been imposed for the brokers in the region to be in a position to accept US-based traders.
Is binary options trading safe? As mentioned in the previous sections, it is safe as long as you’re a responsible trader. After all, there’s a difference between taking risks, which is necessary for any venture, and being risky, which is blindly being too ambitious in your trades. One of the most important and fundamental ways to reduce risks is choosing the right binary options brokers. Since it’s the platform where you’ll be doing your trading, you must check the following:
###### Tools – Binary Options Robot offers you a number of tools that will help you make maximum profits and get better as a trader. This includes training materials, how-to guides, and other educational tools for binary options trading. Examples include video tutorials, trading charts, eBooks, manuals, and webinars. You also get a number of tools that you can use while actively trading and researching assets. This includes detailed asset information, price data, and easy-to-read charts.
When thinking of investing in binary options trading, the first thing that is likely to come in your mind is how to be successful. To ensure that you can consider applying the below mentioned strategies. IQ Option is one of the most efficient and unique brokers today, allowing the trades to apply tricks in order to improve the results of their strategy tremendously. Let’s have a look at the tricks for IQ Options Strategy;
24Option has recently partnered with the Juventus Football Club which only adds to their high credibility. They were already a top-notch, highly respectable brokerage, but this partnership is purely an added benefits. So if you’re a fan of both soccer and binary options trading, this is the platform for you. Also, for those who like tennis, 24Option and Boris Becker are currently offering a competition in which you have the chance to win 100 grand in USD.
If you believe that Gold will appreciate in value in the near future, you should click on the ‘Call’ button to open a trade. You should then indicate the amount you would like to stake in the trade and your preferred expiry period from the dropdown menu provided by the broker and then place the trade. If you believe that the price will drop in the near future, you should click on the ‘Put’ and indicate the amount you would like to stake as well as your preferred expiry period before placing the trade.
There are several binary options brokers out there who are eager to assist you as a client. However, it is important to remember that choosing a binary options broker can be a detailed process and it’s important to find one that is capable of performing all of the necessary trading tasks. Below are some helpful tips to simplify the process and assist with finding the best broker for your trading needs.
One of the important things to remember regarding expiry times is that they are able to be changed only until you have made a commitment to that particular trade. Once you have authorized a specific trade, you simply sit back and wait until it is completed. This differs from other types of financial trading in which you can sell your accumulated shares at any time. There are brokers that will let you sell your trade for a minimal refund. However this is a rare situation for those who are more experience at trading binary options.
Make sure to understand that not all deposit and withdrawal options are made the same. This binary trading tip is very useful, as traders often get surprised by a number of fees certain payment methods are applying. Before deciding upon a method, research fees, and time needed to see money on the selected account. Also, make sure what the fee rates for making a deposit via your favorite method. Fees and additional costs can simply ‘eat’ your profit, and if we include conversion rates – it is easy to understand why this binary trick saves you money.
Terms and Conditions. When taking a bonus or offer, read the full terms and conditions. Some will include locking in an initial deposit (in addition to the bonus funds) until a high volume of trades have been made. The first deposit is the trader’s cash – legitimate brokers would not claim it as theirs before any trading. Some brokers also offer the option of cancelling a bonus if it does not fit the needs of the trader.
Strategies are an extremely important part of trading. Some strategies are proven to work extremely well, while others may be shared with others prior to being fully tested. The following ten tips can be used regardless of strategy and trade type. Each of these can help prevent substantial losses and should also help in the accumulation of higher levels of profits.
Apart from that, you’ll read about mentioned trading patterns, news and other important information from the world of binary options. Once you choose the right trading pattern to use, earning money with binary options could become your one and only income. Whether you are ready to fully engage yourself in trading or you are a newcomer in this area, I am sure you will find something interesting. Our reviews and intuitive comparative platform in tables will help you make your decision.
Once your deposit exceeds $100,000 you will be treated like royalty. You will be assigned a personal assistant to book your appointment, place your calls, place orders for online gifts and promotions and so forth. They will also assist you if you need access to a specific guest list or even tickets to an event that are sold out. It seems very glamorous, but honestly, if we want to keep it real, not everyone really has$100,000 to invest in binary options trading.
In the standard Black–Scholes model, one can interpret the premium of the binary option in the risk-neutral world as the expected value = probability of being in-the-money * unit, discounted to the present value. The Black–Scholes model relies on symmetry of distribution and ignores the skewness of the distribution of the asset. Market makers adjust for such skewness by, instead of using a single standard deviation for the underlying asset {\displaystyle \sigma } across all strikes, incorporating a variable one {\displaystyle \sigma (K)} where volatility depends on strike price, thus incorporating the volatility skew into account. The skew matters because it affects the binary considerably more than the regular options.
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# Current Divider Calculator
Created by Mehjabin Abdurrazaque and Wojciech Sas, PhD candidate
Reviewed by Steven Wooding
Last updated: Jul 25, 2022
Looking for an all-in-one current divider calculator to estimate the current flowing through each branch of a resistive, inductive, or capacitive circuit? You're in the right place. Let's quickly inform you what our tool can do:
• It's a current branch calculator.
• Explains the current divider rule and how current divides in a resistive circuit with an example.
• Guides you on how to derive the current divider formula for resistive circuits, inductive circuits, and capacitive circuits.
• Helps you understand how other components influence the current flowing through a component in a parallel circuit.
In a nutshell, our tool is a comprehensive current division calculator!
## What is a current divider?
Any circuit that divides current into various paths is a current divider circuit. The magnitude of current passing through a particular path depends on the impedance of that path. Unlike the voltage divider rule, current and impedance have an inverse relationship – the more the impedance of a path, the lesser the current passing through it.
For example, in a parallel connection of two resistors of equal resistance, the current from the source divides equally and flows through each resistor.
💡 A current divider circuit is also known as a parallel circuit.
Let's take a look at how current divides in a resistive circuit, inductive circuit, and capacitive circuit.
## What is the current divider rule?
When we connect two components providing parallel resistance (or impedance in AC circuits), the current in any branch is a fraction of the total current. For example, in a 1-ampere DC parallel circuit with a 1Ω-resistor in each of the two branches, the current flowing through the branches is 0.5 A.
## How does current divide in a parallel resistive circuit?
According to Ohm's law, in a parallel resistive circuit, the current flowing through a path is:
$I_\text{path} = \frac{V}{R_\text{path}}$
where $V$ is the voltage across the path, and $R_\text{path}$ is the effective resistance of the path.
In any parallel circuit, the voltage remains the same across two branches, and it is equal to the product of the source current $I$ and the circuit's effective resistance $R_\text{eq}$:
$V = IR_\text{eq}$
Consider a parallel resistive circuit of two resistors, $R_\text{1}$ and $R_\text{2}$. The effective resistance, $R_\text{eq}$ of the given parallel circuit is:
$\frac{1}{R_\text{eq}} = \frac{1}{R_1}+\frac{1}{R_2}$
On rearranging:
$R_\text{eq} = \frac{R_1 R_2}{ R_1 + R_2}$
Thus, the voltage across the paths is:
$V = I \frac{R_1 R_2}{ R_1 + R_2}$
Now, divide the voltage across the first path by its effective resistance to obtain the current divider formula for $R_\text{1}$:
$I_1 =\frac{ I}{R_1} \frac{R_1 R_2}{ R_1 + R_2}$
On simplifying:
$I_1 = I \frac{ R_2}{ R_1 + R_2}$
Similarly, the current divider equation for the resistor $R_\text{2}$ is:
$I_2 = I \frac{ R_1}{ R_1 + R_2}$
✅ Hence, the current passing through a resistor is the product of total current and the ratio of the product of other resistances to the effective resistance.
If you want to add more resistors, you can repeat the process, although the formulas may grow big. But don't worry – simply use our current divider calculator, and you'll get the result in a second!
## What is the current divider formula for an inductive circuit?
For a parallel inductive circuit, we can consider the current division only in an alternating current (AC). By analogy to the resistive current divider, we can apply Ohm's law to find the average current flowing through an inductance $L$ using the inductive reactance $X_\text{L}$:
$I = \frac{V}{X_\text{L}} = \frac{V}{2\pi fL}$
where $f$ is the AC signal frequency.
Consider a parallel inductive circuit consisting of two inductors, $L_\text{1}$ and $L_\text{2}$. We calculate the effective inductance $L_\text{eq}$ as follows:
$\frac{1}{L_\text{eq}} = \frac{1}{L_1}+\frac{1}{L_2}$
On rearranging:
$L_\text{eq} = \frac{L_1 L_2}{ L_1 + L_2}$
Thus, the voltage across the path is:
$V = 2\pi fI \frac{L_1 L_2}{ L_1 + L_2}$
Now, divide the voltage across the first path by its reactance to obtain the current flowing through $L_\text{1}$:
$I_1 = \frac{2\pi fI}{2\pi fL_1} \frac{L_1 L_2}{ L_1 + L_2}$
Or:
$I_1 = I \frac{ L_2}{ L_1 + L_2}$
Similarly, the average current passing through the inductor $L_\text{2}$ is:
$I_2 = I \frac{ L_1}{ L_1 + L_2}$
💡 The current that passes through a path consisting of an inductor is the product of the total current and the ratio of the product of inductances to the effective inductance.
## What is the current divider equation for a capacitive circuit?
Similar to the inductive circuit, the current division in a capacitive one is possible with an AC signal. We can use the capacitive reactance, $X_\text{C}$, to express the average current passing through a loop with capacitance $C$ as:
$I = \frac{V}{X_\text{C}} = \frac{V}{\frac{1}{2\pi fC}} = 2\pi fCV.$
The effective capacitance of a parallel connection is the sum of the individual capacitances:
$C_\text{eq} = C_1 + C_2$
The voltage across the first branch of the parallel capacitive circuit is:
$V = \frac {I} { 2\pi f (C_1 + C_2)}$
Now, divide the voltage across the first path by its reactance to obtain the current divider equation for $C_\text{1}$:
\begin{aligned} I_1 &= \frac{V}{X_\text{C1}} = \frac{\frac {I} { 2\pi f (C_1 + C_2)}}{\frac{1}{2\pi fC_1}} \\[1em] &= I\frac{C_1}{ C_1 + C_2} \end{aligned}
And the current passing through $C_\text{2}$ is:
$I_2 = I\frac{C_2}{ C_1 + C_2}$
✅ Hence, the current passing through a path in a parallel capacitive circuit is the product of total current and the path's capacitance divided by the circuit's effective capacitance.
That's all you need to know about formulas. Now let's try to solve some computational examples and learn how to use Omni's current divider calculator.
## Instructions to use the current divider calculator
Imagine we have a four resistor setup (R₁ = 20 Ω, R₂ = 40 Ω, R₃ = 80 Ω, R₄ = 100 Ω), and we need to know how the current divides into each branch when we arrange them in parallel. Let's say the source current (I) equals 1 A. Here is how our current division calculator works:
1. Select the Resistive circuit type. It's a default option.
2. Enter the current value (1 A).
3. Input the consecutive resistances into the corresponding resistors' fields. The new ones will appear as you type the previous ones.
4. That's all! The current divider calculator does the rest of the job. As a result, you can see a table with your resistors and the approximate values of corresponding currents flowing through each path:
• I₁ = 0.513 A;
• I₂ = 0.256 A;
• I₃ = 0.128 A; and
• I₄ = 0.103 A.
## FAQ
### How do I calculate the current passing through a branch in a resistive circuit?
To calculate the current through a branch of any circuit:
1. Work out the effective resistance (Req) of the whole circuit.
2. Estimate the voltage (V) across the branch using Ohm's law: V = IReq, where I is the source current.
3. Divide voltage by the branch's resistance to obtain the current passing through it: I = V/Rbranch if it is a resistive circuit.
### Why do we multiply the voltage by capacitance to find the branch current in capacitive circuits?
The charge stored in a capacitor is the product of the capacitance, and the voltage applied: Q = CV. The electric current is the rate of electric charge: I = Q/t. That's why we multiply the capacitance with the applied voltage to estimate the branch current.
### Is a series circuit a current divider?
No. The current flows through all components and is the same across all elements in a series circuit. The series circuit is a voltage divider.
### Can I use the current divider rule for series circuits?
You don't need to use the current divider rule for series circuits. In a series circuit, voltage divides across various components, and the same current flows through all of them.
Mehjabin Abdurrazaque and Wojciech Sas, PhD candidate
Circuit type
Resistive
Current (I)
A
Resistor 1 (R₁)
Ω
Resistor 2 (R₂)
Ω
You can add up to 10 resistors, fields will appear as you need them.
Input at least one resistor to obtain a result.
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# Definition:Probability Space/Discrete
Let $\left({\Omega, \Sigma, \Pr}\right)$ be a probability space.
Let $\Omega$ be a discrete sample space.
Then $\left({\Omega, \Sigma, \Pr}\right)$ is known as a discrete probability space.
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# Overview
## Data
### Required Data
In order to use this package, two different sets of landscape data are required: resistance and absorption. There are certain requirements for these:
• They must be 2-dimensional matrices or RasterLayer objects. They have to be the same type.
• They must have the same dimensions (number of rows and columns).
• NA data is allowed in the cells, but must match between the sets of data. I.e., if cell [3, 6] of the resistance data has a NA value, then cell [3, 6] of the absorption data must also have a NA value, and vice versa.
If using RasterLayer objects, then additional conditions must be met:
• Both sets of data must have the same coordinate extents.
• Both sets of data must use the same coordinate reference system (CRS).
### Optional Data
The use of landscape fidelity data is optional. By default, the package treats all cells in the landscape data the same and uses a value of 0 for fidelity. If custom data is desired, then it must meet all of the same requirements listed above for the resistance and absorption landscape data.
### Built-in Example Data
The package includes built-in example data. Some of this data was used to create the figures in the SAMC paper, and is used in this tutorial. They are:
• ex_res_data: A matrix with landscape resistance data.
• ex_abs_data: A matrix with landscape absorption (mortality) data.
• ex_occ_data: A matrix with landscape occupancy data.
str(samc::ex_res_data)
#> num [1:34, 1:202] NA NA NA NA NA NA NA NA NA NA ...
str(samc::ex_abs_data)
#> num [1:34, 1:202] NA NA NA NA NA NA NA NA NA NA ...
str(samc::ex_occ_data)
#> num [1:34, 1:202] NA NA NA NA NA NA NA NA NA NA ...
plot(raster(samc::ex_res_data, xmn = 1, xmx = ncol(samc::ex_res_data), ymn = 1, ymx = nrow(samc::ex_res_data)),
main = "Example Resistance Data", xlab = "x", ylab = "y", col = viridis(256))
plot(raster(samc::ex_abs_data, xmn = 1, xmx = ncol(samc::ex_abs_data), ymn = 1, ymx = nrow(samc::ex_abs_data)),
main = "Example Absorption Data", xlab = "x", ylab = "y", col = viridis(256))
plot(raster(samc::ex_occ_data, xmn = 1, xmx = ncol(samc::ex_occ_data), ymn = 1, ymx = nrow(samc::ex_occ_data)),
main = "Example Occupancy Data", xlab = "x", ylab = "y", col = viridis(256))
## The samc-class
The samc-class is used to manage the transition matrix and information about your landscape data to help ensure that the calculations used by the rest of the package are used correctly. Creating an samc-class object is the mandatory first step in the package, and is created using the samc() utility function. The samc() function has several parameters. Some of these are mandatory, some are only mandatory in certain situations, and some are optional. The overall function signature is as follows:
samc(resistance, absorption, fidelity, latlon, tr_fun, override)
An explanation of the arguments:
• resistance and absorption are always mandatory, and must meet the data requirements in the Data section above.
• fidelity is optional. If included, it must meet the data requirements outlined in the Data section above.
• latlon is mandatory when the input data is in a RasterLayer object. It should be set to either TRUE or FALSE.
• tr_fun is always mandatory. It is used to create the transition matrix.
• override is optional. It is used to control whether or not memory intensive functions can be run on the data. By default it is set to FALSE to prevent users from accidentally running these functions, which can potentially crash R if the computer does not have enough memory. The function documentation provides specific details for which calculations need the override, but in general it will rarely be useful for users and should be left off.
## Utility Functions
In addition to the extremely important samc() function, the package has other utility functions that users might find helpful:
• The check() function is used to check that input landscape data meets the data requirements outlined above. It can be used to compare two RasterLayer objects, two matrix objects, or check either a RasterLayer or a matrix against an already created samc-class object.
• The map() function is used to simplify mapping vector data back into the landscape and return it as a RasterLayer. This is provided because R handles matrices and raster layers somewhat differently when reading and writing vector data, which can cause users to map the data incorrectly if they aren’t careful. It also handles mapping to landscapes with NA values, another potential source of error.
## Analytical Functions
The package implements functions for the formulas provided in Table 1 of Fletcher et al. (2019). Many of the formulas are related conceptually, and are grouped together into single functions with multiple parameter signatures to reduce the number of unique function names needed. Note that the descriptions assume $$\psi$$ contains probability of occurrence. If $$\psi$$ instead contains the number of individuals, then the metrics with $$\psi$$ will return the number of expected individuals rather than a probability.
Function Equation Description
cond_passage() $$\tilde{t} = \tilde{B}_j^{-1}\tilde{F}\tilde{B}_j{\cdot}1$$ Mean first conditional passage time
dispersal() $$\tilde{D}_{jt}=({\sum}_{n=0}^{t-1}\tilde{Q}^n)\tilde{q}_j$$ Probability of an individual visiting a location, if starting at any other location, before or at time t
$$\psi^T\tilde{D}_{jt}$$ Probability of an individual visiting a location, before or at time t, regardless of initial location
$$D=(F-I)diag(F)^{-1}$$ Probability of an individual visiting a location
$$\psi^TD$$ Probability of an individual visiting a location, regardless of initial location
distribution() $$Q^t$$ Probability of an individual being at a location at time t
$$\psi^TQ^t$$ Probability of an individual being at a location at time t, regardless of initial location
mortality() $$\tilde{B}_t = (\sum_{n=0}^{t-1} Q^n) \tilde{R}$$ Probability of an individual experiencing mortality at a location before or at time t
$$\psi^T \tilde{B}_t$$ Probability of an individual experiencing mortality at a location, before or at time t, regardless of initial location
$$B = F \tilde{R}$$ Probability of an individual experiencing mortality at a location
$$\psi^T B$$ Probability of an individual experiencing mortality at a location, regardless of initial location
survival() $$z=(I-Q)^{-1}{\cdot}1=F{\cdot}1$$ Expected life expectancy of an individual
$${\psi}^Tz$$ Overall life expectancy, regardless of initial location
visitation() $$F = (I-Q)^{-1}$$ Expected number of times an individual visits a location
Depending on the combination of inputs used, a function might return a single value, a vector, or a matrix. In some cases, the calculations will not be practical with sufficiently large landscape datasets due to memory and other performance constraints. To work around this, many equations have multiple associated function signatures that allow users to calculate individual portions of the result rather than the entire result. This opens up multiple optimizations that makes calculating many of the metrics more practical. More specific details about performance considerations can be found in the Performance vignette.
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# Lyapunov exponents and random matrices
#### by Prof Peter Forrester
Institution: The University of Melbourne
Date: Mon 17th September 2012
Time: 1:00 PM
Location: Room 115, Sidney Myer Asia Centre
Abstract: Lyapunov exponents in random matrix theory relate to products of random matrices. I'll discuss how products of random matrices enters into some problems of mathematical and applied mathematics. Some comments will be make too about general methods to calculate Lyapunov exponents. My own contribution, which is the exact computation of Lyapunov exponents for the random matrix product $$P_N = A_N A_{N-1} \cdots A_1$$, with each $$A_i = \Sigma^{1/2} G_i^{\rm c}$$, where $$\Sigma$$ is a fixed $$d \times d$$ positive definite matrix and $$G_i^{\rm c}$$ a $$d \times d$$ complex Gaussian matrix with entries standard complex normals, will be outlined.
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# Using 64 randomly selected phone calls, the average call length was calculated t
Using 64 randomly selected phone calls, the average call length was calculated to be 4.2 minutes. It is known from previous studies that the variance of the length of phone calls is $$\displaystyle{1.44}\min^{{{2}}}$$. Assuming that the length of calls has a normal distribution
a) estimate an interval estimate of the length of a telephone conversation at the 0.95 confidence level
b) confidence 0.99
c) Compare the length of the two intervals and explain how the length of the interval depends on the confidence level.
• Questions are typically answered in as fast as 30 minutes
### Plainmath recommends
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likvau
Step 1
As per given by the question, there are 64 randomly selected phone calls, the average call length was calculated to be 4.2 minutes.
So,
Random value (n) is 64, Mean $$\displaystyle\overline{{{X}}}$$ is 4.42 minute, and Variance is 1.44.
Then first calculate the standard deviation with the help of means and variance.
$$\displaystyle{V}={\frac{{\sigma}}{{\overline{{{X}}}}}}$$
$$\displaystyle{1.44}={\frac{{\sigma}}{{{4.42}}}}$$
$$\displaystyle\sigma={6.36}$$
Step 2
Now,
a) estimate an interval estimate of the length of a telephone conversation at the 0.95 confidence level.
From formula of normal distribution,
$$\displaystyle{C}{I}=\overline{{{X}}}\pm{t}_{{{z}}}\cdot{\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle={4.42}\pm{1.96}\cdot{\frac{{{6.36}}}{{\sqrt{{{64}}}}}}$$
$$\displaystyle={4.42}\pm{\left({1.96}\right)}{\left({0.79}\right)}$$
$$\displaystyle={4.42}\pm{1.44}$$
Hence, the length of a telephone conversation at the 0.95 confidence level is 5.97, 2.87
Step 3
b). estimate an interval estimate of the length of a telephone conversation at the 0.99 confidence level.
From formula of normal distribution,
$$\displaystyle{C}{I}=\overline{{{X}}}\pm{t}_{{{z}}}\cdot{\frac{{\sigma}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle={4.42}\pm{2.58}\cdot{\frac{{{6.36}}}{{\sqrt{{{64}}}}}}$$
$$\displaystyle={4.42}\pm{\left({2.58}\right)}{\left({0.79}\right)}$$
$$\displaystyle={4.42}\pm{2.038}$$
Hence, the length of a telephone conversation at the 0.99 confidence level is 6.458, 2.382.
Step 4
c). Compare the length of the two intervals and explain how the length of the interval depends on the confidence level.
The length of interval at 0.95 confidence level is 5.97, 2.87. and,
The length of interval at 0.99 confidence level is 6.458, 2.382.
Here, the value of length of interval at 0.99 confidence level is 6.458, 2.382, that is greater than the value of length of interval at 0.95 confidence level is 5.97, 2.87.
The length of the interval depends on the confidence level, because if confidence level is increase then length of interval is increase, if confidence level is decrease then length of interval is increase.
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## Ratios and Proportions Part 6
Problem 1: Two vessels of 48 lit and 42 lit are filled with a mixture of milk and water, the proportion of two vessels being respectively 13:7 and 18:17. if the contents of two vessels are mixed and 20 lit of water be added to the whole what will be the proportion of milk and water in the resulting mixture?
Solution:
M=$48\times&space;\frac{13}{20}+42\times&space;\frac{18}{15}=\frac{264}{5}$
w=$48\times&space;\frac{7}{20}+42\times&space;\frac{17}{35}+20=\frac{286}{5}$
264 : 286
132 : 143
12 : 13
Problem 2: A vessel contains some liter of pure milk. if 25 lit of water is added to the vessel then ratio of milk and water becomes 12 : 5. 17 liters of mixture is drawn from the vessel and 10 lit of water is added to the mixture. Find the new ratio between milk and water?
Solution:
Milk Water
12 : 5
5——->25
12——>?
$\frac{12\times&space;25}{5}=60$
in 17 liters 12 liters are milk and 5 liters are water is drawn
Milk=60-12=48
Water=25-5=20
and 10 liters of water is added so 20+10=30
48 : 30
8:5
Problem 3: 18 lit of pure water was added to the vessel contains 80 lit of pure milk. 49 lit of the resultant mixture was then sold and some more quantity of pure milk and water was added to the vessel in the ratio of 2:1. if the resultant ratio of Milk and water in the vessel was 4:1 what was the quantity of pure milk added in the vessel?
Solution:
W M
18 + 80——>98 lit
9 : 40
49 lit of mixture was sold out
98-49=49 lit
Some more quantity of pure milk and water was added to the vessel in the ratio of 2:1
Milk=$49\times&space;\frac{2}{3}=40$
Water=49-40=9
$\frac{40+2x}{9+x}=\frac{4}{1}$
40+2x=36+4x
2x=4
x=2
Problem 4: A vessel contains a mixture of milk and water in the ratio 14:3. 25.5 lit of mixture is taken out from the vessel and 2.5 lit of pure water and 5 lit of milk is added to the mixture. If the resultant mixture contains 20% water. What was the initial quantity of mixture in the vessel before the replacement?
Solution:
Milk Water
14x : 3x
14x+3x = 17x
Milk = $\frac{25.5\times&space;14}{17}=21$
Water = 25.5-21 = 4.5
$\frac{14x-21+5}{3x-4.5+2.5}=\frac{80}{20}$
14x-21+5=12x-1.8+2.5
x=4
17x=17*4=68
## Ratios and Proportions Part 5
Problem 1: An alloy contains A, B, and C in the ratio 5:2:1 another alloy contains 12:5:7. If equal quantities of these two alloys are melted together to form a third alloy, then what will be the weight of C per kg in new alloy?
Solution:
A:B:C=5:2:1 ————> 8*3 ———> 15:6:3 ——> 24
A:B:C=12:5:7———–>24———->12:5:7——>24
3+7=10 24+24=48
$\frac{10}{48}=\frac{5}{24}$
Problem 2: An alloy contains the copper and aluminum in the ratio of 7:4 while making the weapons from this alloy 12% of alloy gets destroyed. If there is 12kg of aluminum in the weapon, then the weight of the alloy required will be?
Solution:
Copper : aluminum
7 : 4
4————>12
1——–>?$\frac{12}{4}=3$
11——->?11*3=33
$33\times&space;\frac{100}{88}=37.5$
Problem 3: In a college, there are two courses B.Tech and M.Tech and the ratio of boys and girls in the ratio is 5:4. If in M.Tech no.of boys and girls is equal then find the ratio of boys and girls in Be.ch. The no.of B.Tech students are 25% more than the no. of M.Tech students?
Solution:
Boys : Girls
5 : 4
The no.of B.Tech students are 25% more than the no. of M.Tech students
25%=$\frac{1}{4}$
B.Tech = 4+1 = 5
M.Tech = 4
Assume there are 400 students in M.Tech. In M.tech, the no.of boys and girls are equal.
Boys : Girls
2 : 2
Assume there are 500 students in B.tech
Boys : Girls
3 : 2
## Ratios and Proportions Part 4
Problem 1: Seat of arts, commerce, and science are in the ratio of 3:5:8 respectively. If the number of students studying arts, commerce and science is increased by 20%, 40%, 25% respectively. What will be the new ratio of seats for students in arts, commerce, and science respectively?
Solution:
20%=$\frac{1}{5}$
40%=$\frac{2}{5}$
25%=$\frac{1}{4}$
$3\times&space;\frac{6}{5}:5\times&space;\frac{7}{5}:8\times&space;\frac{5}{4}$
18 : 35 : 50
Problem 2: Three containers have their volume in the ratio 1:4:7 they are full of the mixture of milk and water. The mixture contains milk and water in the ratio 4:1, 3:1, 5:2 respectively. The contents of all these containers are poured into the fourth container. Find the ratio of milk and water in the fourth container is?
Solution:
Milk : Water
4 : 1
3 : 1
5 : 2
Water=$1\times&space;\frac{1}{5}+4\times&space;\frac{1}{4}+7\times&space;\frac{2}{7}$
=$\frac{1}{5}+1+2$
1+5+10=16
Milk= $1\times&space;\frac{4}{5}+4\times&space;\frac{3}{4}+&space;7\times&space;\frac{5}{7}$
$\frac{4}{5}+3+5$
$4+15+25$=44
Milk : Water
44 : 16
11 : 4
Problem 3: The ratio of birds in two branches of a tree is 2:1. When 45 birds from the first branch fly to the second branch, the new ratio of birds become 1:5. Find the total number of birds on the tree.
Solution:
1st Branch : 2nd Branch
2 : 1
$\frac{2x-45}{x+45}=\frac{1}{5}$
10x-225=x+45
2x=270
x=30
## Ratios and Proportions Part 3
Problem 1: The quantity of water in 120 lit of milk and water is only 25% the milk man sold 20 liters of the mixture in this mixture and added 16.2 liters of pure milk and 3.8 liters of water in the remaining mixture. What is the percentage of water in the new mixture?
Solution: Total quantity=120
25%=$\frac{1}{4}$
$120\times&space;\frac{1}{4}=30$
120-30=90
W : M
30 90
1 : 3
A milk man sold 20 liters of mixture so 120-20=100
Water =$100\times&space;\frac{1}{4}=25$
100-25=75
3.8 liters of water is added to remaining mixture=25+3.8=28.8
16.2 liters of water is added to remaining mixture=16.2+75=91.2
$\frac{288}{120}\times&space;100=24$%
Problem 2: In a bag there are 25p, 50p and 1r coins are there in the ratio 2:1:3. If their total is Rs.240 in all, how many 25p, 50p and 1 RS coins respectively are there in the bag?
Solution:
25 50 1
2 : 1 : 3
50p 50p 3RS
50P+50P+3RS=4
4———->240
2———–>? $\frac{240\times&space;2}{4}=120$
1————>? $\frac{240}{4}=60$
3————->? $\frac{240\times&space;3}{4}=180$
Problem 3: Zinc and copper are in the ratio 2:1 in 75 kg of an alloy. The amount of copper to be further added to the alloy so as to make the ratio of zinc and copper 1:2?
Solution:
$75\times&space;\frac{2}{3}=50$
75-50=25
Zinc : Copper
50 : 25 + 75
50 : 100
1 : 2
100-25=75
75 kg of copper is added to get 1:2 ratio
Problem 4: A mixture contains alcohol and water in the ratio of 4:3 if 5 liters of water is added to the mixture the ratio becomes 4:5. Find the quantity of alcohol in the given mixture?
Solution:
Alcohol : water
4 : 3
4 : 5
2—————->5
4—————–>?
$\frac{5\times&space;4}{2}=10$
Problem 5: An perfume production company prepares perfume by mixing it water. The new mixture of 340 liters contains perfume and water in the ratio 27:7 later the company decides to liquefy the mixture to perform anew mixture containing perfume and water in the ratio of 3:1. How much water must they have to add in the previous mixture?
Solution:
Perfume=$340\times&space;\frac{27}{34}=270$
340-270=70
Perfume : Water
270 : 70 + 20
270 : 90
3 : 1
20 liters of water is added to get the mixture of 3:1
## Ratios and Proportions Part 2
Problem 1: A container of 108 lit capacity is filled with pure milk from this 18 liters is taken out and replaced with water if this process is replaced two more times find the quantity of water in a container?
Solution:
Basic method: $\frac{18}{108}=\frac{1}{6}$
Milk=$108\times&space;\frac{5}{6}\times&space;\frac{5}{6}\times&space;\frac{5}{6}=&space;\frac{125}{2}=62.5$
Water=108-62.5=45.5 liters
Formula: $x(1-\frac{y}{x})^{n}$
= $108(1-\frac{18}{108})^{3}$
=$108\times&space;\frac{5}{6}\times&space;\frac{5}{6}\times&space;\frac{5}{6}=&space;\frac{125}{2}=62.5$
108-62.5 = 45.5 liters
Problem 2: A vessel contains 100 liters of pure milk 10 liters of milk is taken out from the vessel and is replaced by 10 lit of water. Noe 20 liters of the mixture is taken out from the vessel and is replaced by water. What is the ratio of the quantity of milk left and water?
Solution:
$\frac{10}{100}=\frac{1}{10}$
$\frac{20}{100}=\frac{1}{5}$
m=$100\times&space;\frac{9}{10}\times&space;\frac{4}{5}=72$
w = 100-72 = 28
Problem 3: 9 liters of spirit is drawn from a vessel full of spirit and is filled with water, this operation is performed one more time. The ratio of the quantity of spirit now left in cask to that of water is 16:9. Find the capacity of the vessel?
Solution:
S : W
16X : 9X
16+9 = 25
S=$x(1-\frac{y}{x})^{n}$
$\frac{16}{25}X=X(1-\frac{9}{X})^{2}$
$\frac{4}{5}=1-\frac{9}{X}$
$\frac{9}{X}=1-\frac{4}{5}$
$\frac{9}{X}=\frac{1}{5}$
X=45
Problem 4: A jar is filled with milk. A person replaces 25% of milk with water. He repeats the process 5 times and as a result, there are only 972 ml of milk left in the jar, the remaining part of the jar is filled with water. The initial quantity of milk in the jar was?
Solution: 1 lit = 1000 ml
25%=$\frac{1}{4}$
$x\times&space;\frac{3}{4}\times&space;\frac{3}{4}\times&space;\frac{3}{4}\times&space;\frac{3}{4}\times&space;\frac{3}{4}=972$
x=$972000\times&space;\frac{4}{3}\times&space;\frac{4}{3}\times&space;\frac{4}{3}\times&space;\frac{4}{3}\times&space;\frac{4}{3}=68,259.84$
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# A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for ₹ 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C? - Mathematics
MCQ
A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for ₹ 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
• ₹ 375
• ₹ 400
• ₹ 600
• ₹ 800
#### Solution
₹ 400
Explanation:
C's 1 day's work = 1/3 - (1/6 + 1/8) = 1/3 - 7/24 = 1/24
A's wages : B's wages : C's wages = 1/6 : 1/8 : 1/24 = 4 : 3 : 1
∴ C's share (for 3 days) = ₹(3 xx 1/24 xx 3200) = ₹ 400.
Concept: Time and Work (Entrance Exam)
Is there an error in this question or solution?
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# Thread: probs with recalling variables
1. ## probs with recalling variables
hello everyone
if i just use
session_start();
it works fine but as soon as i try to check it the array ( $result ) is there is throws up an obscure error Parse error: parse error in c:\apache\htdocs\bar.php on line 119 the following is line 118/119 which as you can see has no php in it and isnt even in an area that the php should parse <a class="barLink" href="#" onMouseOver="window.status='[ Check out where you stand and the top 50 ]';return true;" onMouseOut="window.status='';return true;">Rankings</a><br> and the following is the php i am using no php is used on the bar.php except getting sessions the session works fine on the other pages included but its throwing up that weird error anyone got any ideas? <? session_start(); if( !$result ){
{
}
?>
2. well, since your if is missing a bracket, that could be it if that html is the last line.
#### Posting Permissions
• You may not post new threads
• You may not post replies
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• You may not edit your posts
•
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# Sums of $2^{-l}$ that add to 1
Consider the following problem:
You are given a finite set of numbers $$(l_k)_{k\in \{ 1, ..., n \}}$$ such that $$\sum_{k=1}^n2^{-l_k}<1$$. Describe an algorithm to find a set $$(l'_k)_{k\in \{ 1, ..., n \}}$$ such that $$\forall k \in \{ 1, ..., n \}:l'_k\le l_k$$ and $$\sum_{k=1}^n2^{-l'_k}=1$$.
(For what it's worth, this is a problem arising from Information Theory, where the Kraft-McMillan Theorem gives that the result above yields a more efficient binary code than the one with codeword lengths $$(l_k)$$.)
Here are my initial thoughts. We can consider $$\sum_{k=1}^n2^{-l_k}$$ as a binary number e.g. $$0.11010011$$ and then we need to reduce the value of some $$l_k$$ values whose digit position is preceded by a $$0$$. So for instance, with the initial $$0$$ in the $$\frac{1}{8}$$ position of the example number I gave, we want to decrease the value of some $$l_i=4$$ to $$l'_i=3$$ to add $$\frac{1}{8}$$ and subtract $$\frac{1}{16}$$ to the sum. We then have $$0.11100011$$, so we've moved the problematic $$0$$ along a digit. When we get to the end we presumably have something like $$0.11111110$$, and then need to reduce the value of the longest codeword by 1 to get the overflow to $$1.00000000$$.
However, I encounter two problems: there may not be such an $$l_i=4$$, for instance, if the $$1$$ in the $$\frac{1}{16}$$ digit place arises as the sum of three $$l_i=5$$ numbers. Additionally, if we multiple have $$0$$ digits in a row then we presumably need to scan until the next $$1$$ and then decrement a corresponding $$l_i$$ multiple times, but it's conceivable that I would "run out" of large enough $$l_i$$ codewords that I can manipulate in this way.
Can anyone describe an algorithm with a simple proof of correctness?
A follow-up problem: how do we generalise this algorithm to bases other than $$2$$?
Here is a very simple algorithm. We will require the following lemma.
Lemma. Suppose that $$1 \leq \ell_1 \leq \cdots \leq \ell_k$$ and $$\sum_{i=1}^k 2^{-\ell_i} \geq 1/2$$. Then there exists $$r \in \{1,\ldots,k\}$$ such that $$\sum_{i=1}^r 2^{-\ell_i} = 1/2$$.
Proof. Let $$r$$ be the first index such that $$\sum_{i=1}^r 2^{-\ell_i} \geq 1/2$$. Since the $$\ell_i$$ are non-decreasing, there are integers $$A,B$$ such that \begin{align} \sum_{i=1}^{r-1} 2^{-\ell_i} &= \frac{A}{2^{\ell_r}}, & \sum_{i=1}^r 2^{-\ell_i} &= \frac{B}{2^{\ell_r}}. \end{align} Moreover, $$A < 2^{\ell_r-1}$$ and $$B \geq 2^{\ell_r-1}$$. Since $$B-A=1$$, we conclude that $$B = 2^{\ell_r-1}$$, and so $$\sum_{i=1}^r 2^{-\ell_i} = 1/2$$. $$\quad\square$$
This suggests the following algorithm. We can assume that your sequence is sorted, that is, we are given a sequence $$\ell_1 \leq \cdots \leq \ell_k$$ such that $$\sum_{i=1}^k 2^{-\ell_i} \leq 1$$. We now consider three cases:
1. $$\sum_{i=1}^k 2^{-\ell_i} = 1$$. In this case, there is nothing to do.
2. $$\sum_{i=1}^k 2^{-\ell_i} \leq 1/2$$. In this case, we can decrease each $$\ell_i$$ by $$1$$.
3. $$1/2 \leq \sum_{i=1}^k 2^{-\ell_i} \leq 1$$. Applying the lemma, we find $$r$$ such that $$\sum_{i=1}^r 2^{-\ell_i} = 1/2$$, and so $$\sum_{i=r+1}^k 2^{-\ell_i} \leq 1/2$$. We thus have to solve the same kind of problem for the second half $$\ell_{r+1},\ldots,\ell_k$$, aiming at $$1/2$$ rather than $$1$$.
To implement this recursion more cleanly, we add a parameter $$s$$, and our goal is to correct a sequence satisfying $$\sum_i 2^{-\ell_i} \leq 2^{-s}$$ to one satisfying $$\sum_i 2^{-\ell_i} = 2^{-s}$$ by only decreasing elements.
Here is how the algorithm works in the case of the sequence $$1,2,4,7,8$$, which matches your example. The sum in your case is more than $$1/2$$, so we separate the sequence into two parts: $$1$$ and $$2,4,7,8$$. We only handle the second, aiming at a sum of $$1/2$$.
The sum in the case of $$2,4,7,8$$ is more than $$1/4$$, so we separate the sequence into two parts, $$2$$ and $$4,7,8$$, and only handle the second, aiming at a sum of $$1/4$$.
The sum in the case of $$4,7,8$$ is less than $$1/8$$, so we decrement each element, obtaining the sequence $$3,6,7$$, whose sum is more than $$1/8$$. We separate it into $$3$$ and $$6,7$$, and only handle the second, aiming at a sum of $$1/8$$.
We decrement $$6,7$$ twice, obtaining the sequence $$4,5$$ whose sum exceeds $$1/16$$. We separate it into $$4$$ and $$5$$, and decrement the latter once.
Putting everything together, we obtain the sequence $$1,2,3,4,4$$.
In the $$q$$-ary case, we have to change the problem somehow, since the result is not true in general. For example, take $$q = 3$$ and consider the sequence $$1, 1$$.
Here is another simple algorithm, based on the following lemma.
Lemma. Suppose that $$0 \leq \ell_1 \leq \cdots \leq \ell_k$$ and $$\sum_{i=1}^k 2^{-\ell_i} < 1$$. Then $$\sum_{i=1}^{k-1} 2^{-\ell_i} + 2^{-(\ell_k-1)} \leq 1$$.
Proof. Since the $$\ell_i$$ are nondecreasing, we can write $$\sum_{i=1}^k 2^{-\ell_i} = A/2^{\ell_k}$$, where $$A < 2^{\ell_k}$$. Replacing $$\ell_k$$ with $$\ell_k-1$$ increases the sum by $$1/2^{\ell_k}$$. Since $$A+1 \leq 2^{\ell_k}$$, the sum remains at most $$1$$. $$\quad\square$$
This suggests the following simple algorithm: repeatedly decrement the largest $$\ell_i$$. The algorithm necessarily terminates since at most $$\sum_i \ell_i$$ iterations can take place.
Let us apply this on our example above: \begin{align} &1,2,4,7,8 \to 1,2,4,7,7 \to 1,2,4,6,7 \to 1,2,4,6,6 \to 1,2,4,5,6 \to \\ &1,2,4,5,5 \to 1,2,4,4,5 \to 1,2,4,4,4 \to 1,2,3,4,4 \end{align} This algorithm can be implemented easily using a heap. However, it is slower (in general) than the preceding algorithm, if the latter is implemented correctly. For example, the sequence $$\ell$$ takes $$\ell$$ steps in this algorithm, but could be handled by a single iteration of the preceding algorithm.
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# Replace titlesec-chapter style with KOMA
I have built a very nice chapter style with titlesec:
Unfortunately there's a warning for the use together with "scrbook". That's why I want to replace titlesec with KOMA.
\usepackage{titlesec}
%CHAPTER
\titleformat{\chapter}[display] %shape
{\usefont{T1}{lmss}{b}{n}\filleft\huge\bfseries} %format
{\makebox[\linewidth][r]{\raisebox{103pt}[0pt][0pt]{\textcolor{gray!25}{\usefont{T1}{lmss}{b}{n}\fontsize{80pt}{95pt}\selectfont\thechapter}}}
{} %default would be huge
} %label
{-14ex} %sep
{} %before-code
[\vspace{0.5ex}\titlerule]
\titleformat{name=\chapter,numberless}[display]
{\usefont{T1}{lmss}{b}{n}\filleft\huge\bfseries} %format
%{\makebox[25pt][l]{\raisebox{100pt}[0pt][0pt]{\textcolor{gray!25}{\usefont{T1}{lmss}{b}{n}\fontsize{80pt}{80pt}\selectfont\thechapter}}}
{\makebox[10pt][l]{\raisebox{10pt}[0pt][0pt]{\textcolor{gray!25}{\usefont{T1}{lmss}{b}{n}\fontsize{10pt}{12pt}\selectfont}}}
{} %default would be huge
} %label
{-16.8ex} %sep
{} %before-code
[\vspace{1ex}\titlerule]
\titlespacing{\chapter}
{0pc}{*30}{*5}[0pc] %Abstand zum linken Rand |Abstand zum oberen Text | Abstand zum unteren Text | Abstand zum rechten Rand
%SECTION
\titleformat{\section}
{\usefont{T1}{lmss}{b}{n}\filright\Large}
{\thesection}{1em}{}
\titlespacing{\section}
{0pc}{*0}{*-1}[0pc]
\setcounter{secnumdepth}{3}
%SUBSECTION
\titleformat{\subsection}
{\usefont{T1}{lmss}{b}{n}\large}
{\thesubsection}{1em}{}
\titlespacing{\subsection}
{0pc}{*0}{*-1}[0pc]
Also I want the appendix to have the same look like a normal chapter.
The other chapters without a number should look like this:
The height above the title for a chapter without number and the height above the chapter numer for a chapter with number should be the same.
A lot of questions... -.- I hope somebody will help me to solve this as my KOMA knowledge is very poor -.-
Thanks!
After some research, I ended up with
\renewcommand*\raggedchapter{\raggedleft}
\newcommand\titlerule[1][.4pt]{\rule[.5\baselineskip]{\textwidth}{#1}}
\renewcommand*{\chapterformat}{%
\makebox[\width][l]{\scalebox{1}{{\nobreakspace}}%
\scalebox{4}{\textcolor{gray!25}\thechapter\autodot}\enskip}
}
It seems to work, but unfortunately some things are missing:
• the number of the chapter is not completely on the right hand side
• how do I change the space between the number, the name and the line?
• The height above the title for a chapter without number and the height above the chapter numer for a chapter with number is not the same.
• I think it's not going to be as easy in KOMA. What is the warning you get? If it is just a warning but the document compiles fine, why change it? – alexraasch Mar 11 '16 at 14:13
• Maybe related: KOMA-script scrreprt: Chapter Heading Size Customisation. I used something like that to replace the titlesec package for my customized version of the classicthesis package which I use for my thesis. – Stephan Lukasczyk Mar 11 '16 at 14:35
• Thanks for your fast reply. I have edited my post with a little KOMA code. It seems to work, but some things I don't know how to change – Milou Mar 11 '16 at 15:01
Here is suggestion redefining \chapterlineswithprefixformat. This needs KOMA-Script version 3.19a or newer.
\documentclass[chapterprefix]{scrbook}[2015/10/03]
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{xcolor}
\usepackage{lmodern}
\RedeclareSectionCommand[
beforeskip=\dimexpr3.3\baselineskip+1\parskip\relax,
innerskip=0pt,% <- space between chapter number and chapter title
afterskip=1.725\baselineskip plus .115\baselineskip minus .192\baselineskip
]{chapter}
\renewcommand\raggedchapter{\raggedleft}
\renewcommand\chapterformat{{\fontsize{80pt}{80pt}\selectfont\textcolor{gray!25}{\thechapter}}}
\renewcommand\chapterlineswithprefixformat[3]{%
#2#3%
\vspace*{-.5\baselineskip}% <- adjust the space between the chapter and the rule here
\rule{\textwidth}{.4pt}\par\nobreak
}%
\RedeclareSectionCommands[
beforeskip=0pt plus 1ex minus .2ex,
afterskip=1sp plus .2ex
]{section,subsection}
\renewcommand*{\sectionformat}{\thesection\hspace*{1em}}
\renewcommand*{\subsectionformat}{\thesubsection\hspace*{1em}}
\setcounter{secnumdepth}{3}
\usepackage{blindtext}% for dummy text
\begin{document}
\tableofcontents
\blinddocument
\chapter{A long long long long long long long long long chapter title}
\Blindtext
\appendix
\blinddocument
\end{document}
Adjust beforeskip, innerskip and afterskip for the sectioning commands to your needs.
Regarding a comment:
If there should be the same vertical white space between the top of the page and a chapter number or an unnumbered chapter, use a fixed length as beforeskip for chapter
\RedeclareSectionCommand[
beforeskip=3.3\baselineskip,
...
]{chapter}
and insert some additional vertical space before the numbered chapter
\renewcommand\chapterlineswithprefixformat[3]{%
\ifstr{#2}{}{}{\vspace*{1ex}}% <- insert vertical space before numbered chapters
#2#3%
\vspace*{-.5\baselineskip}% <- adjust the space between the chapter and the rule here
\rule{\textwidth}{.4pt}\par\nobreak
}%
or remove some vertical space before unnumbered chapters
\renewcommand\raggedchapter{\raggedleft}
\renewcommand\chapterformat{{\fontsize{80pt}{80pt}\selectfont\textcolor{gray!25}{\thechapter}}}
\renewcommand\chapterlineswithprefixformat[3]{%
\ifstr{#2}{}{\vspace*{-1ex}}{}% <- remove vertical space before unnumbered chapters
#2#3%
\vspace*{-.5\baselineskip}% <- adjust the space between the chapter and the rule here
\rule{\textwidth}{.4pt}\par\nobreak
}%
• Wow, just tried it, and it works perfectly. Thanks a lot! – Milou Mar 15 '16 at 12:25
• Just came up with one last question: how do I change the space between the chapter number and the header? Because still the height above the title for a chapter without number and the height above the chapter number for a chapter with number is not the same. – Milou Mar 15 '16 at 13:08
• @Milou See my updated answer. – esdd Mar 15 '16 at 13:52
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## Layered Prepositional Phrases
May 22, 2021
A prepositional phrase may modify the object of another prepositional phrase.
The flowers \in the pot \on the windowsill \in the kitchen \of my grandmother’s house \in Maine are violets.
The flowers are in the pot.
The pot is on the windowsill.
The windowsill is in the kitchen.
The kitchen is of my grandmother’s house.
The house is in Maine.
Doesn’t this sound like a children’s rhyme? Hint:Be careful not to use too many prepositional phrases at once because it can be confusing. Two prepositional phrases at a time are usually enough unless you are intentionally trying to layer lots of prepositional phrases.
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# Sparse PCA/Dictionary learning when the features are extremely sparse?
I am trying to do sparse PCA/dictionary learning, that is decompose a matrix $X\approx UV$ where the loading matrix $V$ is sparse, usually enforced with an $\ell_1$ penalty (the difference between sparse PCA and dictionary learning being whether the inner dimension of $U$ and $V$ is greater than the smaller dimension of $X$). In my data the columns of $X$ are extremely large (order $10^8$) but extremely sparse (order $10^2$). Are there specialized algorithms for the case of sparse data? Online is a plus.
• The R package irlba is fantastic for SVD on sparse data, which will get you most of the way to PCA. – Zach Jan 24 '14 at 19:28
• I believe that this answer of mine is relevant to your question. – Aleksandr Blekh Feb 1 '15 at 3:48
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## A note about AdS/CFT and Yang-Mills spectrum
There is a lot of activity about using this AdS/CFT symmetry devised in the string theory context to obtain the spectrum of a pure Yang-Mills theory in D=3+1. To have an idea about one can read here. The essential point about this approach is that it does not permit the computation of the mass gap. Rather, when given the ground state value as an input, it should be able to obtain all the spectrum. I should say that current results are indeed satisfactory and this approach is worthing further pursuing.
Apart from this, we have seen here that the ground state, as derived using the lattice computations of the gluon propagator, is quite different from current results for the spectrum as given here and here. Indeed, from the gluon propagator, fixing the QCD constant at 440 MeV, we get the ground state at 1.25. Teper et al. get 3.55(7) and Morningstar et al (that fix the constant at 410 MeV) get about 4.16(11). Morningstar et al use an anisotropic lattice and this approach has been problematic in computations for the gluon propagator ( see the works by Oliveira and Silva about) producing disagreement with all others research groups or, at best, a lot of ambiguities. In any case we note a large difference between propagator and spectrum computations for the ground state. This is a crucial point.
Experiments see f0(600) or $\sigma$. Propagator computations see f0(600) and $\sigma$ but spectrum computations do not. Here there is a point to be clarified between these different computations about Yang-Mills theory. Is it a problem with lattice spacing? In any case we have a discrepancy to be understood.
Waiting for an answer to this dilemma, it would be interesting for people working on AdS/CFT to lower the input ground state assuming the one at 1.25 (better would be 1.19 but we will see this in future posts) and then checking what kind of understanding is obtained. E.g. is the state at 3.55 or 4.16 recovered? Besides, it would be nice to verify if some kind of regularity is recovered from the spectrum computed in this way.
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13 views
Consider the following table that shows the number (in lakhs) of different sizes of LED television sets sold by a company over the last seven years from $2012$ to $2018$. Answer the question based on the data contained in the table:
Sale of LED Television sets (in lakhs) of different sizes (in inches)
For which size LED Television sets in the total sales of all seven years the maximum?
1. $22$ – inch Television
2. $24$ – inch Television
3. $32$ – inch Television
4. $49$ – inch Television
recategorized | 13 views
32inch television have over all seven years total cost maximum
by Active (4k points)
+1 vote
1
+1 vote
|
# verde.Trend¶
class verde.Trend(degree)[source]
Fit a 2D polynomial trend to spatial data.
The polynomial of degree $$N$$ is defined as:
$f(e, n) = \sum\limits_{l=0}^{N}\sum\limits_{m=0}^{N - l} e^l n^m$
in which $$e$$ and $$n$$ are the easting and northing coordinates, respectively.
The trend is estimated through weighted least-squares regression. The Jacobian (design, sensitivity, feature, etc) matrix for the regression is normalized using sklearn.preprocessing.StandardScaler without centering the mean so that the transformation can be undone in the estimated coefficients.
Parameters
degree (int) – The degree of the polynomial. Must be >= 0 (a degree of zero would estimate the mean of the data).
Variables
• coef_ (array) – The estimated polynomial coefficients that fit the observed data.
• region_ (tuple) – The boundaries ([W, E, S, N]) of the data used to fit the interpolator. Used as the default region for the grid and scatter methods.
Examples
>>> from verde import grid_coordinates
>>> import numpy as np
>>> coordinates = grid_coordinates((1, 5, -5, -1), shape=(5, 5))
>>> data = 10 + 2*coordinates[0] - 0.4*coordinates[1]
>>> trend = Trend(degree=1).fit(coordinates, data)
>>> print(
... "Coefficients:",
... ', '.join(['{:.1f}'.format(i) for i in trend.coef_])
... )
Coefficients: 10.0, 2.0, -0.4
>>> np.allclose(trend.predict(coordinates), data)
True
A zero degree polynomial estimates the mean of the data:
>>> mean = Trend(degree=0).fit(coordinates, data)
>>> np.allclose(mean.predict(coordinates), data.mean())
True
>>> print("Data mean:", '{:.2f}'.format(data.mean()))
Data mean: 17.20
>>> print("Coefficient:", '{:.2f}'.format(mean.coef_[0]))
Coefficient: 17.20
We can use weights to account for outliers or data points with variable uncertainties (see verde.variance_to_weights):
>>> data_out = data.copy()
>>> data_out[2, 2] += 500
>>> weights = np.ones_like(data)
>>> weights[2, 2] = 1e-10
>>> trend_out = Trend(degree=1).fit(coordinates, data_out, weights)
>>> # Still recover the coefficients even with the added outlier
>>> print(
... "Coefficients:",
... ', '.join(['{:.1f}'.format(i) for i in trend_out.coef_])
... )
Coefficients: 10.0, 2.0, -0.4
>>> # The residual at the outlier location should be values we added to
>>> # that point
>>> residual = data_out - trend_out.predict(coordinates)
>>> print('{:.2f}'.format(residual[2, 2]))
500.00
Methods Summary
Trend.filter(coordinates, data[, weights]) Filter the data through the gridder and produce residuals. Trend.fit(coordinates, data[, weights]) Fit the trend to the given data. Trend.get_params([deep]) Get parameters for this estimator. Trend.grid([region, shape, spacing, dims, …]) Interpolate the data onto a regular grid. Trend.jacobian(coordinates[, dtype]) Make the Jacobian matrix for a 2D polynomial. Trend.predict(coordinates) Evaluate the polynomial trend on the given set of points. Trend.profile(point1, point2, size[, dims, …]) Interpolate data along a profile between two points. Trend.scatter([region, size, random_state, …]) Interpolate values onto a random scatter of points. Trend.score(coordinates, data[, weights]) Score the gridder predictions against the given data. Trend.set_params(**params) Set the parameters of this estimator.
Trend.filter(coordinates, data, weights=None)
Filter the data through the gridder and produce residuals.
Calls fit on the data, evaluates the residuals (data - predicted data), and returns the coordinates, residuals, and weights.
No very useful by itself but this interface makes gridders compatible with other processing operations and is used by verde.Chain to join them together (for example, so you can fit a spline on the residuals of a trend).
Parameters
• coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). For the specific definition of coordinate systems and what these names mean, see the class docstring.
• data (array or tuple of arrays) – The data values of each data point. If the data has more than one component, data must be a tuple of arrays (one for each component).
• weights (None or array or tuple of arrays) – If not None, then the weights assigned to each data point. If more than one data component is provided, you must provide a weights array for each data component (if not None).
Returns
coordinates, residuals, weights – The coordinates and weights are same as the input. Residuals are the input data minus the predicted data.
Trend.fit(coordinates, data, weights=None)[source]
Fit the trend to the given data.
The data region is captured and used as default for the grid and scatter methods.
All input arrays must have the same shape.
Parameters
• coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). Only easting and northing will be used, all subsequent coordinates will be ignored.
• data (array) – The data values of each data point.
• weights (None or array) – If not None, then the weights assigned to each data point. Typically, this should be 1 over the data uncertainty squared.
Returns
self – Returns this estimator instance for chaining operations.
Trend.get_params(deep=True)
Get parameters for this estimator.
Parameters
deep (bool, default=True) – If True, will return the parameters for this estimator and contained subobjects that are estimators.
Returns
params (mapping of string to any) – Parameter names mapped to their values.
Trend.grid(region=None, shape=None, spacing=None, dims=None, data_names=None, projection=None, **kwargs)
Interpolate the data onto a regular grid.
The grid can be specified by either the number of points in each dimension (the shape) or by the grid node spacing. See verde.grid_coordinates for details. Other arguments for verde.grid_coordinates can be passed as extra keyword arguments (kwargs) to this method.
If the interpolator collected the input data region, then it will be used if region=None. Otherwise, you must specify the grid region.
Use the dims and data_names arguments to set custom names for the dimensions and the data field(s) in the output xarray.Dataset. Default names will be provided if none are given.
Parameters
• region (list = [W, E, S, N]) – The west, east, south, and north boundaries of a given region.
• shape (tuple = (n_north, n_east) or None) – The number of points in the South-North and West-East directions, respectively.
• spacing (tuple = (s_north, s_east) or None) – The grid spacing in the South-North and West-East directions, respectively.
• dims (list or None) – The names of the northing and easting data dimensions, respectively, in the output grid. Default is determined from the dims attribute of the class. Must be defined in the following order: northing dimension, easting dimension. NOTE: This is an exception to the “easting” then “northing” pattern but is required for compatibility with xarray.
• data_names (list of None) – The name(s) of the data variables in the output grid. Defaults to ['scalars'] for scalar data, ['east_component', 'north_component'] for 2D vector data, and ['east_component', 'north_component', 'vertical_component'] for 3D vector data.
• projection (callable or None) – If not None, then should be a callable object projection(easting, northing) -> (proj_easting, proj_northing) that takes in easting and northing coordinate arrays and returns projected northing and easting coordinate arrays. This function will be used to project the generated grid coordinates before passing them into predict. For example, you can use this to generate a geographic grid from a Cartesian gridder.
Returns
grid (xarray.Dataset) – The interpolated grid. Metadata about the interpolator is written to the attrs attribute.
See also
verde.grid_coordinates
Generate the coordinate values for the grid.
Trend.jacobian(coordinates, dtype='float64')[source]
Make the Jacobian matrix for a 2D polynomial.
Each column of the Jacobian is easting**i * northing**j for each (i, j) pair in the polynomial.
Parameters
• coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). Only easting and northing will be used, all subsequent coordinates will be ignored.
• dtype (str or numpy dtype) – The type of the output Jacobian numpy array.
Returns
jacobian (2D array) – The (n_data, n_coefficients) Jacobian matrix.
Examples
>>> import numpy as np
>>> east = np.linspace(0, 4, 5)
>>> north = np.linspace(-5, -1, 5)
>>> print(Trend(degree=1).jacobian((east, north), dtype=np.int))
[[ 1 0 -5]
[ 1 1 -4]
[ 1 2 -3]
[ 1 3 -2]
[ 1 4 -1]]
>>> print(Trend(degree=2).jacobian((east, north), dtype=np.int))
[[ 1 0 -5 0 0 25]
[ 1 1 -4 1 -4 16]
[ 1 2 -3 4 -6 9]
[ 1 3 -2 9 -6 4]
[ 1 4 -1 16 -4 1]]
Trend.predict(coordinates)[source]
Evaluate the polynomial trend on the given set of points.
Requires a fitted estimator (see fit).
Parameters
coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). Only easting and northing will be used, all subsequent coordinates will be ignored.
Returns
data (array) – The trend values evaluated on the given points.
Trend.profile(point1, point2, size, dims=None, data_names=None, projection=None, **kwargs)
Interpolate data along a profile between two points.
Generates the profile along a straight line assuming Cartesian distances. Point coordinates are generated by verde.profile_coordinates. Other arguments for this function can be passed as extra keyword arguments (kwargs) to this method.
Use the dims and data_names arguments to set custom names for the dimensions and the data field(s) in the output pandas.DataFrame. Default names are provided.
Includes the calculated Cartesian distance from point1 for each data point in the profile.
To specify point1 and point2 in a coordinate system that would require projection to Cartesian (geographic longitude and latitude, for example), use the projection argument. With this option, the input points will be projected using the given projection function prior to computations. The generated Cartesian profile coordinates will be projected back to the original coordinate system. Note that the profile points are evenly spaced in projected coordinates, not the original system (e.g., geographic).
Warning
The profile calculation method with a projection has changed in Verde 1.4.0. Previous versions generated coordinates (assuming they were Cartesian) and projected them afterwards. This led to “distances” being incorrectly handled and returned in unprojected coordinates. For example, if projection is from geographic to Mercator, the distances would be “angles” (incorrectly calculated as if they were Cartesian). After 1.4.0, point1 and point2 are projected prior to generating coordinates for the profile, guaranteeing that distances are properly handled in a Cartesian system. With this change, the profile points are now evenly spaced in projected coordinates and the distances are returned in projected coordinates as well.
Parameters
• point1 (tuple) – The easting and northing coordinates, respectively, of the first point.
• point2 (tuple) – The easting and northing coordinates, respectively, of the second point.
• size (int) – The number of points to generate.
• dims (list or None) – The names of the northing and easting data dimensions, respectively, in the output dataframe. Default is determined from the dims attribute of the class. Must be defined in the following order: northing dimension, easting dimension. NOTE: This is an exception to the “easting” then “northing” pattern but is required for compatibility with xarray.
• data_names (list of None) – The name(s) of the data variables in the output dataframe. Defaults to ['scalars'] for scalar data, ['east_component', 'north_component'] for 2D vector data, and ['east_component', 'north_component', 'vertical_component'] for 3D vector data.
• projection (callable or None) – If not None, then should be a callable object projection(easting, northing, inverse=False) -> (proj_easting, proj_northing) that takes in easting and northing coordinate arrays and returns projected northing and easting coordinate arrays. Should also take an optional keyword argument inverse (default to False) that if True will calculate the inverse transform instead. This function will be used to project the profile end points before generating coordinates and passing them into predict. It will also be used to undo the projection of the coordinates before returning the results.
Returns
table (pandas.DataFrame) – The interpolated values along the profile.
Trend.scatter(region=None, size=300, random_state=0, dims=None, data_names=None, projection=None, **kwargs)
Interpolate values onto a random scatter of points.
Point coordinates are generated by verde.scatter_points. Other arguments for this function can be passed as extra keyword arguments (kwargs) to this method.
If the interpolator collected the input data region, then it will be used if region=None. Otherwise, you must specify the grid region.
Use the dims and data_names arguments to set custom names for the dimensions and the data field(s) in the output pandas.DataFrame. Default names are provided.
Parameters
• region (list = [W, E, S, N]) – The west, east, south, and north boundaries of a given region.
• size (int) – The number of points to generate.
• random_state (numpy.random.RandomState or an int seed) – A random number generator used to define the state of the random permutations. Use a fixed seed to make sure computations are reproducible. Use None to choose a seed automatically (resulting in different numbers with each run).
• dims (list or None) – The names of the northing and easting data dimensions, respectively, in the output dataframe. Default is determined from the dims attribute of the class. Must be defined in the following order: northing dimension, easting dimension. NOTE: This is an exception to the “easting” then “northing” pattern but is required for compatibility with xarray.
• data_names (list of None) – The name(s) of the data variables in the output dataframe. Defaults to ['scalars'] for scalar data, ['east_component', 'north_component'] for 2D vector data, and ['east_component', 'north_component', 'vertical_component'] for 3D vector data.
• projection (callable or None) – If not None, then should be a callable object projection(easting, northing) -> (proj_easting, proj_northing) that takes in easting and northing coordinate arrays and returns projected northing and easting coordinate arrays. This function will be used to project the generated scatter coordinates before passing them into predict. For example, you can use this to generate a geographic scatter from a Cartesian gridder.
Returns
table (pandas.DataFrame) – The interpolated values on a random set of points.
Trend.score(coordinates, data, weights=None)
Score the gridder predictions against the given data.
Calculates the R^2 coefficient of determination of between the predicted values and the given data values. A maximum score of 1 means a perfect fit. The score can be negative.
If the data has more than 1 component, the scores of each component will be averaged.
Parameters
• coordinates (tuple of arrays) – Arrays with the coordinates of each data point. Should be in the following order: (easting, northing, vertical, …). For the specific definition of coordinate systems and what these names mean, see the class docstring.
• data (array or tuple of arrays) – The data values of each data point. If the data has more than one component, data must be a tuple of arrays (one for each component).
• weights (None or array or tuple of arrays) – If not None, then the weights assigned to each data point. If more than one data component is provided, you must provide a weights array for each data component (if not None).
Returns
score (float) – The R^2 score
Trend.set_params(**params)
Set the parameters of this estimator.
The method works on simple estimators as well as on nested objects (such as pipelines). The latter have parameters of the form <component>__<parameter> so that it’s possible to update each component of a nested object.
Parameters
**params (dict) – Estimator parameters.
Returns
self (object) – Estimator instance.
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A friend of mine asked me to help with this problem. I tried induction, but I didn't know how to get this formula.
If $x$ and $y$ are real numbers such that $xy= ax+by$. Show that $$x^ny^n=\sum_{k=1}^{n}{2n-1-k \choose n-1}(a^n b^{n-k}x^k+ a^{n-k}b^n y^k), \forall n \geq 0$$
Any help is appreciated.
-
How is this related to computer science? – Antonio Vargas Apr 14 '12 at 18:34
It is a question from "Concrete Mathematics, A Foundation for Computer Science". – yaa09d Apr 14 '12 at 18:42
I removed the (computer-science) tag. – user2468 Apr 14 '12 at 18:43
I'm doubtful about the validity of the equation for $n=0$, but that's a minor nitpick. – hardmath Apr 14 '12 at 19:30
$1=\frac{b}{x}+\frac{a}{y}$. So, let $s=\frac{b}{x}, t=\frac{a}{y}$. It is enough to show that $$2=\sum_{k=-\infty}^n \binom{2n-1-k}{n-1}( s^n t^{n-k}+s^{n-k} t^n).$$ Now, let $m=n-k$. Then $$RHS= (st)^n \sum_{m=0}^{\infty} \binom{n-1+m}{m} (s^{m-n}+t^{m-n})=2.$$ Note that $$(1-z)^{-n}=\sum_{m=0}^{\infty} \binom{n-1+m}{m} z^m.$$
Thanks. But Why is the LHS of the first identity is $2$, and why does the sum start from $-\infty$ – yaa09d Apr 15 '12 at 5:54
Hint : Replace $(xy)^n$ (LaTeX notation) on the left by $(ax+by)^n$ expand the left-hand side using the Newton formula, and replace products $x^i y^j$ by $(ax+by)^{min(i,j)},$ etc. Nice exercice !
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## July 10, 2009
You can follow this conversation by subscribing to the comment feed for this post.
Brute force works for me:
appetizer.solution <- local (
function (target) {
app <- c(2.15, 2.75, 3.35, 3.55, 4.20, 5.80)
r <- 2L
repeat {
c <- gtools::combinations(length(app), r=r, v=app, repeats.allowed=TRUE)
s <- rowSums(c)
if ( all(s > target ) ) {
print("No solution found")
break
}
x <- which( s == target )
if ( length(x) > 0L ) {
print("Solution found")
print(c[x,])
break
}
r <- r + 1L
}
})
appetizer.solution(15.05)
# [1] "No solution found"
appetizer.solution(15.15)
# [1] "Solution found"
# [,1] [,2] [,3] [,4] [,5]
# [1,] 2.15 2.75 3.35 3.35 3.55
# [2,] 2.75 2.75 2.75 3.35 3.55
I can’t believe I just wrote that. Must be the heat. This must be the most elementary mistake in numerical analysis. Try this version instead:
appetizer.solution <- local (
function (target) {
app <- c(2.15, 2.75, 3.35, 3.55, 4.20, 5.80)
r <- 2L
repeat {
c <- gtools::combinations(length(app), r=r, v=app, repeats.allowed=TRUE)
s <- rowSums(c)
if ( all(s > target) ) {
print("No solution found")
break
}
x <- which( abs(s-target) < 1e-4 )
if ( length(x) > 0L ) {
print("Solution found")
print(c[x,])
break
}
r <- r + 1L
}
})
appetizer.solution(15.05)
# [1] "Solution found"
# [1] 3.55 5.80 5.80
I'll go and kill myself now....
Wait a minute, if repeats are allowed, then you have to have different utilities for second replicate. Decreasing marginal returns. Where are you getting your item utilities from?
A marginally better solution exists by reducing the problem 'dynamically.' I have implemented this as a program for the subset sum problem but this versio n assumed repetitions are not allowed. An adaptation to allow repeats should still work.
The program and a pdf version of a flowchart for it are available at;
www.cybase.co.uk/wlcs/Software.html
Looking at the answer given by Allan the solution is incorrect!
3.55 + 5.80 + 5.80 = 15.15
Also the answer above totals 15.15, but the original problem found no solution only as 15.15, he apparently tried again with the second target sum.
My program does not search for solutions with duplicates, but allows duplicate alues to be entered. Doubling up eaxh entry finds the solution
2.15 + 3.55 + 3.55 + 5.80 = 15.05
but the values are entered as whole numbers.
Steve
The comments to this entry are closed.
## Search Revolutions Blog
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# Finding Ellipses: What Blaschke Products, Poncelet’s Theorem, and the Numerical Range Know about Each Other
###### Ulrich Daepp, Pamela Gorkin, Andrew Shaffer, and Karl Voss
Publisher:
MAA Press/AMS
Publication Date:
2019
Number of Pages:
268
Format:
Hardcover
Series:
AMS/MAA The Carus Mathematical Monographs 34
Price:
63.00
ISBN:
9781470443832
Category:
Monograph
[Reviewed by
Bill Satzer
, on
04/16/2019
]
This unusual book is a recent entry in the MAA’s Carus Mathematical Monograph series. Its theme revolves around some surprising connections among complex analysis, geometry, and linear algebra. It is not so much a textbook as a resource for capstone courses, independent study or research. It is best suited for those with at least an advanced undergraduate background in mathematics.
The book is divided into three parts. The first part develops and reveals the surprising connection in its most satisfying form. In the second part ,we go deeper to see what we get and what we lose in generalizing the basic results. A third part provides a collection of exercises and projects.
The elements of the surprising connection consist of: finite Blaschke products (products of linear fractional transformations of a special form that are automorphisms of the unit disk in the complex plane); Poncelet’s theorem for triangles (a result from projective geometry); and the numerical range of a matrix (the set of values of the quadratic form $x^{*}Ax$ where $x$ is a complex number of unit magnitude).
What’s the connection? If a Blaschke product $B(z)$ of three automorphisms of the unit disk sends $0$ to $0$, then the intersection over all $\lambda$ in the unit circle of the triangular regions formed using the three distinct zeros of $B(z) – λ = 0$ as vertices is an ellipse with foci at the zeros of $B(z)/z$. This ellipse is the boundary of the numerical range of a matrix $2 \times 2$ matrix $A$ and its foci are at the eigenvalues of $A$. Furthermore, that ellipse is inscribed in infinitely many triangles that are themselves inscribed in the unit circle, one triangle for each point of the circle. That ellipse is a 3-Poncelet ellipse and it is a central fixture of the book.
The surprise isn’t revealed until well into the book. The first part of the book is written as a story of discovery. It begins with constructions of the ellipse itself and continues by introducing the three apparently unrelated subjects that are assembled to create the surprise. The reader is introduced to the basics of Blaschke products, Poncelet’s projective geometry and his theorem for triangles, and the linear algebra underlying the idea of a matrix’s numerical range. All the material in this part of a book is within the range of a good undergraduate.
The second part of the book is more advanced. Its goal is to explore connections between the ellipse and Blaschke products that are products of more than three automorphisms of the unit disk. It turns out that the boundaries of the numerical range of the relevant matrices need not be elliptical, but that the boundaries satisfy a Poncelet-like property. The operators that provide matrices are compressions of a shift operator associated with finite Blaschke products. When the boundary is elliptical, the Blaschke product has a special form. The mathematical reach of this part is significantly greater. The chapters include elements of Lebesgue theory, Hardy spaces, functional analysis, operator theory and more.
The third part of the book is an introduction to further research. The authors have devised projects and provided some exercises for each of the chapters in the book. Some of these projects, particularly the latter ones, might be very challenging.
Not much of the story here is told in a strictly linear form. The first part is the most direct, but it has several digressions. The second part is more complex and more digressive. The text has something of a tree structure with three main trunks and many branches. Even strong students might find themselves far out on a branch trying to reconnect to main features of the argument.
This is an intriguing way to provide attractive and accessible subjects for undergraduates in a research setting. Each of the three main topics offers many opportunities for deeper exploration, and their collision in the context of this book makes them even more appealing. The challenges are not insignificant. Even the best students might find themselves paging backward and forward in the book, feeling frustrated while trying to make connections.
An extensive bibliography provides support for a considerable variety of investigations. The authors have also created several interactive applets on their website that the reader is directed to explore at certain points during the book, such as when the Blaschke product’s connection to ellipses is described.
Bill Satzer ([email protected]) was a senior intellectual property scientist at 3M Company. His training is in dynamical systems and particularly celestial mechanics; his current interests are broadly in applied mathematics and the teaching of mathematics.
|
# Tile based collision detection failing when player is going too fast
I'm creating a side scroller and I'm have a problem with my collision detection.
The collision detection works perfect when the player is falling at a constant speed, but when I implement more realistic gravity the player falls too fast - resulting in that the collision will be checked with the tiles under the floor/platform causing the player to be able to jump through platforms.
The way my update and collision works is like following:
• Update entities velocity (no position change)
• Check for collisions (first x, then y)
• if collision, then move the entity as close as possible, otherwise set the position with the velocity.
The collision method works like this:
• Get the four corners of the player.
• Get from that four corners the tiles.
• Collision check.
How can I solve this problem?
## UPDATE
Added the code to make it more clear.
The main loop:
_frames_a_sec = 60;
public void run() {
while(true) {
long timeElapsed = System.currentTimeMillis() - _lastUpdate;
_lastUpdate = System.currentTimeMillis();
if(timeElapsed > 80) {
timeElapsed = 80;
}
update((int)timeElapsed);
try {
} catch(Exception ex) { }
}
}
Map update method:
public void update(int timeElapsed) {
//update entities
for(Entity entity : _mapEntities) {
entity.update(timeElapsed);
}
//check for collisions
checkMapCollision();
}
Entity (abstract) update method:
public void update(int timeElapsed) {
_velocity.x = 0.0F;
if(!_isOnLand) {
_velocity.y += Map._GRAVITY_PER_SEC * timeElapsed;
} else {
_velocity.y = 0.0F;
}
}
Mario (extends Entity) update method:
@Override
public void update(int timeElapsed) {
super.update(timeElapsed);
if(_state == STATES.IDLE) {
} else if(_isMoving) {
_marioSmallWalk.update(timeElapsed);
}
if(_state == STATES.JUMPING) {
setVelocityY(getVelocity().y + _jumpSpeed);
_jumpSpeed += _JUMP_DECREASE * timeElapsed;
//falling?
if(getVelocity().y > 0) {
setState(STATES.FALLING);
}
}
if(_isMoving) {
double walkSpd = (_WALK_SPEED_SEC * timeElapsed);
if(getFacing() == FACING.LEFT) {
walkSpd = -walkSpd;
}
setVelocityX(getVelocity().x + walkSpd);
}
//falling?
if(getVelocity().y > (Map._GRAVITY_PER_SEC * timeElapsed) + 1.0F) {
setState(STATES.FALLING);
}
}
Map Collsion check method:
public void checkMapCollision() {
for(Entity entity : _mapEntities) {
placeEntityAtX(entity);
placeEntityAtY(entity);
}
}
X check method:
private void placeEntityAtX(Entity entity) {
Vector2d dir = entity.getDirection();
Rectangle bounds = entity.getBounds();
boolean positionSet = false;
bounds = new Rectangle((int)(bounds.x + dir.x), bounds.y, bounds.width, bounds.height);
Block[] corners = getCornerBlocks(bounds);
if(dir.x > 0) {
if(corners[1].isSolid() || corners[3].isSolid()) {
Rectangle blkBounds;
if(corners[3].isSolid()) {
blkBounds = corners[3].getBounds();
} else {
blkBounds = corners[1].getBounds();
}
entity.setPositionX(blkBounds.x - (bounds.width-entity.getCurrentSprite().getOffsetX())-1);
positionSet = true;
}
} else if(dir.x < 0) {
if(corners[0].isSolid() || corners[2].isSolid()) {
Rectangle blkBounds;
if(corners[2].isSolid()) {
blkBounds = corners[2].getBounds();
} else {
blkBounds = corners[0].getBounds();
}
entity.setPositionX(blkBounds.x + blkBounds.width + (bounds.width/2) + 1);
positionSet = true;
}
}
if(!positionSet) {
//set the original position
entity.setPositionX((int)(entity.getX() + dir.x));
}
}
and the y:
private void placeEntityAtY(Entity entity) {
Vector2d dir = entity.getDirection();
Rectangle bounds = entity.getBounds();
boolean positionSet = false;
bounds = new Rectangle(bounds.x, (int)(bounds.y + dir.y), bounds.width, bounds.height);
Block[] corners = getCornerBlocks(bounds);
//moving down
if(dir.y > 0) {
if(corners[2].isSolid() || corners[3].isSolid()) {
Rectangle blkBounds = null;
if(corners[2].isSolid()) {
blkBounds = corners[2].getBounds();
} else {
blkBounds = corners[3].getBounds();
}
entity.setPositionY(blkBounds.y);
entity.landed();
positionSet = true;
}
} else if (dir.y < 0) {
if(corners[0].isSolid() || corners[1].isSolid()) {
Rectangle blkBounds = null;
if(corners[0].isSolid()) {
blkBounds = corners[0].getBounds();
} else {
blkBounds = corners[1].getBounds();
}
entity.setPositionY(blkBounds.y + blkBounds.height + bounds.height);
entity.roof();
positionSet = true;
}
} else {
if(!corners[2].isSolid() && !corners[3].isSolid()) {
entity.falling();
}
}
if(!positionSet) {
//set the original position
entity.setPositionY((int)(entity.getY() + dir.y));
}
}
• Your algorithm looks sound. A code snippet would be useful. – JRT May 21 '11 at 11:57
• JRT: Actually his algorithm is flawed, if the goal is to prevent the player from passing through objects. – Olhovsky May 21 '11 at 12:11
• My understanding is that the character is moving so fast it has gone completely through the platform in the time between two frames and so the collision isn't being detected. – CiscoIPPhone May 21 '11 at 12:14
• @CiscoIPPhone: Exactly. When the the player is going slower the collision detection works fine. – Sven van Zoelen May 21 '11 at 12:35
• @Olhovsky: If his intention is to calculate the next position, then check for a collision based on the calculated position, then why is this flawed? (His implementation may well be flawed) – JRT May 21 '11 at 12:54
I know of two ways that will solve this problem.
First method: Fixed Time Step Physics
You can use fixed timestep physics with a high enough frequency that you will detect all the collisions.
What this means is that you will decide on a physics update delta, e.g. 1/60th of a second. Then for each frame of your game you'll calculate how much time has passed since the last frame, divide this by the physics update delta. This is how many physics updates you have to do that frame. In other words, instead of doing a single physics update and then collision check, you will be doing many smaller updates.
• Pros: This is probably the easiest solution to implement
• Cons: May not be practical, depending on how small your collision items are, and how fast they move.
See this question which compares fixed and variable timestep (and there's some useful links in the question itself).
Second method: Swept Collision Detection
You can to calculate the vector between the current position of the player and the projected position of the player, and do your collisions between this line and the platforms.
• Pros: Would work even with very very fast moving and small bodies.
• Cons: Movement in reality due to gravity won't be a straight line, sweeping/integrating the path of multiple accelerating bodies is potentially a difficult maths problem.
• I implemented the first method and increased the frame rate to 60. Now I still got the problem that the player goes trough the ground (sometimes). My tiles are 16x16. Maybe a max velocity y? The second method sounds good, but I'm wondering how to get the lines of the blocks close to the line from the start to new postion line to find the intersection. – Sven van Zoelen May 21 '11 at 15:07
• You should definitely have a max falling speed. This would help tremendously and I don't think it would be a huge deal to the player. – Michael Coleman May 21 '11 at 18:15
• There is a terminal velocity for things falling in real life so you'd actually be making it more realistic =) – CiscoIPPhone May 21 '11 at 18:21
• I implemented the max velocity y and tweeked the gravity and player jump physics. And it works like a charm. Thanks for the link and information! – Sven van Zoelen May 21 '11 at 18:25
Up the physics update rate until it works, it doesn't have to coincide with the framerate. I'd always do my timing like:
physicsStep=25
physicsdivider=3
nextUpdate=currentTime()*physicsdivider
while(1){
while(nextUpdate<currentTime()*physicsdivider){
doPhysics()
nextUpdate+=physicsStep
}
doRender()
}
Edit: Updated the code to do 120 rather than 100 Hz as that will be a little smoother on screens running 60 Hz.
What happens is that the inner loop repeats until the physics have caught up with the current time, only then is frame rendered. With v-Sync enabled on a 60 Hz monitor the result should be this:
render 00 ms
physics
physics
render 17 ms
physics
physics
render 33 ms
...
With exactly 120 physics updates per second, and 60 render updates per second
If the video card is too slow to do 60 Hz some of the render updates will be dropped, but the physics will still be exactly the same with 120 updates per second.
This only works well as long as the physics can stay comfortably within the allotted time of 8 1/3 ms per updated. You should try timing a physics update to get an idea of how close to being a problem this is.
By the way, I can't find the render and flip commands in your code. In any case, simply calling a sleep 16 ms once for every update is a very poor method of timing, that way you are invariably going to drop frames.
• The time elapsed method will keep up with the delay by updating the physics with the lost time. Your method in the other hand will not catch up with the delay when the update cycles will be higher then 10 milliseconds (not smooth). – Sven van Zoelen May 22 '11 at 9:13
• That is only a problem if you set the update rate too high for the computer to keep up. You'd do variable step 20 years ago because stuff like this was on on the limit of what a computer could handle. That is not so any more. – aaaaaaaaaaaa May 22 '11 at 10:32
• But then you still have a gap in time that is not caught by your physics when the elapsed time exceeds let's say 100, because the user's machine was busy with other more important processes. – Sven van Zoelen May 22 '11 at 19:41
• Have you actually read and understood the code? Unless the physics takes more than 10 ms per step it will always do a complete catchup. – aaaaaaaaaaaa May 22 '11 at 20:53
• I wrote it down on the whiteboard and you are right. It will check all the collisions correctly, but when the delay is to big the renderer will lag. But what will hapen when your nextUpdate stays under the 10 milliseconds? – Sven van Zoelen May 23 '11 at 7:28
|
Select Page
Understand the problem
Prove that the function $$f(x)=\frac{\sin(x^3)}{x}$$ is uniformly continuous and bounded.
Source of the problem
TIFR 2019 GS Part A, Problem 9
Analysis
Hard
Introduction to Real Analysis by Donald R. Sherbert and Robert G. Bartle
Do you really need a hint? Try it first!
We have the function in the interval $$(0,\infty)$$ $$f(x)=\frac{\sin(x^3)}{x}$$ Can you prove that the function is uniformly continuous?
The function is clearly continuous(why?).
Any bounded, continuous function $f:(0,\infty) \to \mathbb{R}$ where $f(x) \to 0$ as $x \to 0,\infty$ is uniformly continuous. The derivative if it exists does not have to be bounded.
Note that $$\sin(x^3)/x = x^2 \sin(x^3)/x^3 \to 0\cdot 1 = 0$$ as $$x \to 0$$. This is also a great example of a uniformly continuous function with an unbounded derivative
Connected Program at Cheenta
College Mathematics Program
The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.
Similar Problems
Rational maps to irrational and vice versa?: TIFR 2019 GS Part B, Problem 1
It is an analysis question on functions. It was asked in TIFR 2019 GS admission paper.
Expected expectation:TIFR 2019 GS Part A, Problem 20
It is an statistics question on the probability and expectation. It was asked in TIFR 2019 GS admission paper.
Functions on differential equation:TIFR 2019 GS Part A, Problem 19
It is an analysis question on the differential equation. It was asked in TIFR 2019 GS admission paper.
Permutation or groups!: TIFR 2019 GS Part A, Problem 18
It is an combinatorics question on the permutation or application of counting principle. It was asked in TIFR 2019 GS admission paper.
Compact or connected:TIFR 2019 GS Part A, Problem 17
It is a topology question on the compactness and connectedness. It was asked in TIFR 2019 GS admission paper.
Finding the number of ring homomorphisms:TIFR 2019 GS Part A, Problem 16
It is an algebra question on the ring homomorphism. It was asked in TIFR 2019 GS admission paper.
Uniform Covergence:TIFR 2019 GS Part A, Problem 15
It is an analysis question on the uniform convergence of sequence of functions. It was asked in TIFR 2019 GS admission paper.
Problems on continuity and differentiability:TIFR 2019 GS Part A, Problem 14
It is an analysis question on the continuity and differentiability of functions. It was asked in TIFR 2019 GS admission paper.
Iterative sequences:TIFR 2019 GS Part A, Problem 13
It is an analysis question on the sequence of iterations. It was asked in TIFR 2019 GS admission paper.
Sequence of functions:TIFR 2019 GS Part A, Problem 12
It is an analysis question on the sequence of functions. It was asked in TIFR 2019 GS admission paper.
|
I'm using TeXStudio on Mac OS X 10.7.4.
The problem is that the EPS to PDF conversion is somehow broken:
\usepackage[pdftex]{graphicx}
\graphicspath{{img/}}
\DeclareGraphicsExtensions{.pdf,.jpeg,.png}
...
\begin{figure}[!t]
\centering
\includegraphics[width=3.5in]{TCP_VC_TD_AVE_CONNS_ACT}
\caption{Simulation Results}
\label{fig_sim}
\end{figure}
...
I've got an EPS file in: img/TCP_VC_TD_AVE_CONNS_ACT.eps
Log:
Package epstopdf Info: Source file: <img/TCP_VC_TD_AVE_CONNS_ACT.eps>
(epstopdf) date: 2012-10-12 16:44:32
(epstopdf) size: 210073 bytes
(epstopdf) Output file: <img/TCP_VC_TD_AVE_CONNS_ACT-eps-converted-
to.pdf>
(epstopdf) Command: <epstopdf --outfile=img/TCP_VC_TD_AVE_CONNS_ACT
-eps-converted-to.pdf img/TCP_VC_TD_AVE_CONNS_ACT.eps>
(epstopdf) \includegraphics on input line 94.
runsystem(epstopdf --outfile=img/TCP_VC_TD_AVE_CONNS_ACT-eps-converted-to.pdf i
mg/TCP_VC_TD_AVE_CONNS_ACT.eps)...executed.
Package epstopdf Info: Result file: <img/TCP_VC_TD_AVE_CONNS_ACT-eps-converted-
to.pdf>.
! Package pdftex.def Error: File img/TCP_VC_TD_AVE_CONNS_ACT-eps-converted-to.
The required img/TCP_VC_TD_AVE_CONNS_ACT-eps-converted-to.pdf file is never created.
TeXStudio compiles with (I've added -shell-escape, from other posts):
/usr/texbin/pdflatex -synctex=1 -interaction=nonstopmode -shell-escape %.tex
Can anybody assist?
EDIT - A minimal file breaks as well:
\documentclass{article}
\usepackage{graphicx}
\begin{document}
\includegraphics{img/TCP_VC_TD_AVE_CONNS_ACT.eps}
\end{document}
Log:
Package epstopdf Info: Source file: <img/TCP_VC_TD_AVE_CONNS_ACT.eps>
(epstopdf) date: 2012-10-12 16:44:32
(epstopdf) size: 210073 bytes
(epstopdf) Output file: <img/TCP_VC_TD_AVE_CONNS_ACT-eps-converted-
to.pdf>
(epstopdf) Command: <epstopdf --outfile=img/TCP_VC_TD_AVE_CONNS_ACT
-eps-converted-to.pdf img/TCP_VC_TD_AVE_CONNS_ACT.eps>
(epstopdf) \includegraphics on input line 4.
runsystem(epstopdf --outfile=img/TCP_VC_TD_AVE_CONNS_ACT-eps-converted-to.pdf i
mg/TCP_VC_TD_AVE_CONNS_ACT.eps)...executed.
Package epstopdf Info: Result file: <img/TCP_VC_TD_AVE_CONNS_ACT-eps-converted-
to.pdf>.
! Package pdftex.def Error: File img/TCP_VC_TD_AVE_CONNS_ACT-eps-converted-to.
-
Welcome to TeX.sx! Just a wild guess: Have you tried it without \graphicspath{{img/}}? On an unrelated note: specifying pdftex for graphicx is usually not necessary; it can figure that out by itself. – doncherry Oct 12 '12 at 16:26
I have no problem. Can you try with no other package other than graphicx? Just the \includegraphics command should be in the test file. The -shell-escape option is not needed. – egreg Oct 12 '12 at 16:29
@doncherry Tried that - same problem. – Nic Oct 12 '12 at 16:39
@egreg Tried with a minimal file as well, still the same problem. – Nic Oct 12 '12 at 16:41
Does it work with a different .eps file, preferably one that was created with a different program than the one you're using? – doncherry Oct 12 '12 at 16:51
epstopdf is sometimes trying to write in the wrong directory and then fails to create the file (on certain editors at least, I had a similar problem with WinEDT). Place the following code in the header:
\epstopdfsetup{outdir=./}
-
This worked for me - but don't forget (as I did initially) that you still need to add \usepackage{epstopdf} first. – drstevok Feb 22 '13 at 14:12
\usepackage[outdir=./]{epstopdf} also works and saves you one line :) – Trefex Apr 17 at 8:11
The following worked for me:
Open TeXstudio, go to 'Preferences' --> 'Build' (Somehow all my list entries are in german, so I only guess it should be named Build - in german it's 'Erzeugen', the third entry from the top). At the bottom you can specify additional search paths. Make sure the specified search path really exists. In my case the path pointed to a texlive 2012 directory even though I have only installed the 2013 version. After changing the path to the 2013 version everything worked fine! :-)
-
I have exactly the same problem as the poster's log. Common solutions around Internet have adding epstopdf or epsfig package, appending -shell-escape option to pdflatex commandline. But both methods failed to me. I post my solution on Mac OS X 10.09 + texlive2013 + TeXsdudio2.7.
By checking the tex's log, I found that epstopdf was not located in system searched PATH, so I created the link manully
ln -s /usr/local/texlive/2013/texmf-dist/scripts/epstopdf/epstopdf.pl /usr/local/bin/epstopdf
Also, the epstopdf complained that ghostscript is not installed. So I made it satisfied with
brew install ghostscript
WHooo, texstudio generated pdf as I expected.
PSPS: requiring epstopdfpackage declared and -shell-escape appended.
-
But on my linux/ubuntu, only epstopdf and -shell-escape are needed. – caesar0301 Apr 17 at 3:23
Use
$latex file.tex This will result in output as file.dvi and then use $ dvi2pdf file.dvi
-
The OP uses pdfTeX (pdflatex). A different driver might support a different set of image formats that does not fit the OPs need. – Heiko Oberdiek Sep 12 '13 at 17:48
Welcome to TeX.SX! You can have a look at our starter guide to familiarize yourself further with our format. – Heiko Oberdiek Sep 12 '13 at 17:50
...and therefore I don't think this answers the question... – Werner Sep 12 '13 at 17:52
|
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Find all the composition of each sum of products of two permutations of {1,2,3,..,n}.
Find all the composition of each sum of products of two permutations of {1,2,3,..,n}. I found this problem via “EJDD Problem” and their solution was the sample answer in the following picture:
I want to construct the entire solution for the problem as well.
I know that there should be a bijection $\sigma\leftrightarrow(1,\sigma_1)\dots (n,\sigma_n)$, and I’m mostly lost from where the $\dots$ goes. If it is obvious, then I’m still curious.
I’m asking this question to seek direction or a hint for this problem.
A:
The symmetric group $\mathfrak S_n$ operates on the symmetric group $\mathfrak S_m$ in the following manner: Given $\sigma\in\mathfrak S_m$ and $\tau\in\mathfrak S_n$,
$$\tau\circ\sigma:\eta=(\eta_1,\ldots,\eta_m)\mapsto \left(\sigma\tau(\eta_1),\ldots,\sigma\tau(\eta_m)\right).$$
Now set $\sigma(i):=\sigma_i$. Then $(\sigma(1),\dots,\sigma(n))$ is a permutation of $\{1,\dots,n\}$ and we define
$$\sigma\tau:=(\sigma(1),\dots,\sigma(n)).$$
This leads to a bijection $\sigma\mapsto\sigma\tau$ from \$\
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# Same font in math mode decimal comma? [duplicate]
I want to use comma (,) as decimal separator. The problem is that it is typeset with wrong font (Computer Modern, I suppose), not with the normal font defined using mathspec package. I have tried icomma package and siunitx package but they don't affect the font.
Here is a minimal code to try. The last two "1,5" examples have wrong comma font.
\documentclass{article}
\usepackage{mathspec}
\usepackage{icomma}
\usepackage[locale=DE]{siunitx}
\setmainfont{DejaVu Serif}
\setmathsfont(Digits,Latin,Greek){DejaVu Serif}
\begin{document}
1,5
$$1,5$$
$$\num{1,5}$$
\end{document}
This here should work.
\documentclass{article}
\usepackage{mathspec}
\usepackage{icomma}
\usepackage[locale=DE]{siunitx}
\setmainfont{DejaVu Serif}
\setmathsfont(Digits,Latin,Greek){DejaVu Serif}
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# Lesson 6: The 2^k Factorial Design
### Introduction
The 2k designs are a major set of building blocks for many experimental designs. These designs are usually referred to as screening designs. The 2k refers to designs with k factors where each factor has just two levels. These designs are created to explore a large number of factors, with each factor having the minimal number of levels, just two. By screening we are referring to the process of screening a large number of factors that might be important in your experiment, with the goal of selecting those important for the response that you're measuring. We will see that k can get quite large. So far we have been looking at experiments that have one, two or three factors, maybe a blocking factor and one or two treatment factors, but when using screening designs k can be as large as 8, 10 or 12. For those of you familiar with chemical or laboratory processes, it would not be hard to come up with a long list of factors that would affect your experiment. In this context we need to decide which factors are important.
In these designs we will refer to the levels as high and low, +1 and -1, to denote the high and the low level of each factor. In most cases the levels are quantitative, although they don't have to be. Sometimes they are qualitative, such as gender, or two types of variety, brand or process. In these cases the +1 and -1 are simply used as labels.
### Learning objectives & outcomes
Upon completion of this lesson, you should be able to do the following:
• The idea of 2-level Factorial Designs as one of the most important screening designs
• Defining a “contrast” which is an important concept and how to derive Effects and Sum of Squares using the Contrasts
• Process of analyzing Unreplicated or Single replicated factorial designs, and
• How to use Transformation as a tool in dealing with inadequacy of either variance homogeneity or normality of the data as major hypothetical assumptions.
# 6.1 - The Simplest Case
The simplest case is 2k where k = 2. We will define a new notation which is known as Yates notation. We will refer to our factors using the letters A, B, C, D, etc. as arbitrary labels of the factors. In the chemical process case A is the concentration of the reactant and B is the amount of catalyst, both of which are quantitative. The yield of the process is our response variable.
Since there are two levels of each of two factors, 2k equals four. Therefore, there are four treatment combinations and the data are given below:
You can see that we have 3 observations at each of 4 = 2k combinations for k = 2. So we have n = 3 replicates.
A B Yates Notation - - (1) + - a - + b + + ab
The table above gives the data with the factors coded for each of the four combinations and below is a plot of the region of experimentation in two dimensions for this case.
The Yates notation used for denoting the factor combinations is as follows:
We use "(1)" to denote that both factors are at the low level, "a" for when A is at its high level and B is at its low level, "b" for when B is at its high level and A is at its low level, and "ab" when both A and B factors are at their high level.
The use of this Yates notation indicates the high level of any factor simply by using the small letter of that level factor. This notation actually is used for two purposes. One is to denote the total sum of the observations at that level. In the case below b = 60 is the sum of the three observations at the level b.
This shortcut notation, using the small letters, shows which level for each of our k factors we are at just by its presence or absence.
We will also connect this to our previous notation for the two factor treatment design:
Yijk = μ + αi + βj + (α β)ij + eijk
What is the primary goal of these screening experiments?
The goal is to decide which of these factors is important. After determining which factors are important, then we will typically plan for a secondary experiment where the goal is to decide what level of the factors gives us the optimal response. Thus the screening 2k experiment is the first stage, generally, of an experimental sequence. In the second stage one is looking for a response surface or an experiment to find the optimal level of the important factors.
### Estimation of Factors Effects (in the Yates tradition)
The definition of an effect in the 2k context is the difference in the means between the high and the low level of a factor. From this notation A is the difference between the averages of the observations at the high level of A minus the average of the observations at the low level of A.
Therefore, $$A=\bar{y}_{A^+}-\bar{y}_{A^-}$$, in the example above:
A = 190/6 - 140/6 = 50/6 = 8.33
Similarly, $$B=\bar{y}_{B^+}-\bar{y}_{B^-}$$, is the similar only looking in the other direction. In our example:
B = 150/6 - 180/6 = 25 - 30 = -5
and finally, $$AB=\frac{ab+(1)}{2n}-\frac{a+b}{2n}$$
AB = [(90 + 80)/6 - (100 + 60)/6] = 10/6 = 1.67
Therefore in the Yates notation, we define an effect as the difference in the means between the high and the low levels of a factor whereas in previous models we defined an effect as the coefficients of the model, which are the differences between the marginal mean and the overall mean. To restate this, in terms of A, the A effect is the difference between the means at the high levels of A versus the low levels of A, whereas the coefficient, αi , in the model is the difference between the marginal mean and the overall mean. So the Yates "effect" is twice the size of the estimated coefficient αi in the model, which is also usually called the effect of the factor A.
The confusion is all in the notation used in the definition.
Let's look at another example in order to reinforce your understanding of the notation for these types of designs. Here is an example in three dimensions, with factors A, B and C. Below is a figure of the factors and levels as well as the table representing this experimental space.
In table 6.4 you can see the eight points coded by the factor levels +1 and -1. This example has two replicates so n = 2. Notice that the Yates notation is included as the total of the two replicates.
One nice feature of the Yates notation is that every column has an equal number of pluses and minuses so these columns are contrasts of the observations. For instance, take a look at the A column. This column has four pluses and four minuses, therefore, the A effect is a contrast defined on page 216.
This is the principle that gives us all sorts of useful characterizations in these 2k designs.
In the example above the A, B and C each are defined by a contrast of the data observation totals. Therefore you can define the contrast AB as the product of the A and B contrasts, the contrast AC by the product of the A and C contrasts, and so forth.
Therefore all the two-way and three-way interaction effects are defined by these contrasts. The product of any two gives you the other contrast in that matrix. (See Table 6.3 in the text.)
From these contrasts we can define the effect of A, B, and C, using these coefficients. The general form of an effect for k factors is:
Effect = (1/2(k-1)n) [contrast of the totals]
The sum of the products of the contrast coefficients times the totals will give us the estimate of the effects. See equations (6-11), (6-12), and (6-13).
We can also write the variance of the effect using the general form used previously. This would be:
$\begin{eqnarray} Variance(Effect)&=&[1/(2^{(k-1)}n)^2] V(contrast),or \nonumber\\ &=&[1/(2^{(k-1)}n)^2] 2^k n \sigma^2 \nonumber\\ &=&\sigma^2 / 2^{(k-2)}n \nonumber \end{eqnarray}$
Also, we can write the sum of squares for the effects which looks like:
SS(effect) = (contrast)2 / 2kn
To summarize what we have learned in this lesson thus far, we can write a contrast of the totals which defines an effect, we can estimate the variance for this effect and we can write the sum of squares for an effect. We can do this very simply using Yates notation which historically has been the value of using this notation.
# 6.2 - Estimated Effects and the Sum of Squares from the Contrasts
How can we apply what we learned in the preceding section?
In general for 2k factorials the effect of each factor and interaction is:
Effect = (1/2(k-1)n) [contrast of the totals]
We also defined the variance as follows:
Variance(Effect) = σ2 / 2(k-2)n
The true but unknown residual variance σ2, which is also called the within cell variance, can be estimated by the MSE.
If we want to test an effect, for instance, say A = 0, then we can construct a t-test which is the effect over the square root of the estimated variance of the effect as follows:
$t^{\ast}=\frac{Effect}{\sqrt{\frac{MSE}{n2^{k-2}}}} \sim t(2^k (n-1))$
where ~ means that it has a t distribution with 2k(n-1) degrees of freedom.
Finally, here is the equation for the sum of squares due to an effect to complete the story here:
SS(Effect) = (contrast of totals)2 / 2kn
Where does all of this come from? Each effect in a 2k model has one degree of freedom. In the simplest case we have two main effects and an interaction. They each have 1 degree of freedom. So the t statistic is the ratio of the effect over its estimated standard error (standard deviation of the effect). You will recall that if you have a t statistic with ν degrees of freedom and square it, you get an F distribution with one and ν degrees of freedom.
$t^2(v)=F(1,v)$
We can use this fact to confirm the formulas just developed. We see that the
$(t^{\ast}(v))^2=\frac{(Effect)^2}{MSE/n2^{k-2}}$
and from the definition of an F-test, when the numerator has 1 degree of freedom:
$F(1,v)=\frac{SS(Effect)/1}{MSE}=\frac{(contrast)^2}{2^kn(MSE)}$
But from the definition of an Effect, we can write (Effect)2 = (contrast)2 / (n2k-1)2 and thus F(1, ν) = (t*(ν))2 which you can show by some algebra or by calculating an example.
Hint: Multiply F(1, ν) by (2(k-1)n)2 / (2(k-1)n)2 and simplify.
Once you have these contrasts, you can easily calculate the effect, you can calculate the estimated variance of the effect and the sum of squares due to the effect as well.
Let's use Minitab to help us create a factorial design and then add data so that we can analyze it. Click on the 'Inspect' button to walk through this process using Minitab v.16. The data come from Figure 6.1.
In Minitab we use the software under Stat > Design of Experiments to create our full factorial design. We will come back to this command another time to look at fractional factorial and other types of factorial designs.
In the example that was shown above we did not randomize the runs but kept them in standard order for the purpose of the seeing more clearly the order of the runs. In practice you would want to randomize the order of run when you are designing the experiment.
Once we have created a factorial design within the Minitab worksheet we then need to add the response data so that the design can be analyzed. These response data, Yield, are the individual observations not the totals. So, we again go to the Stat >> DOE >> Factorial menu where we will analyze the data set from the factorial design.
We began with the full model with all the terms included, both the main effects and all of the interactions. From here we were able to determine which effects were significant and should remain in the model and which effects were not significant and can be removed to form a simpler reduced model.
### A Second Example - The Plasma Etch Experiment
Similar to the previous example, in this second industrial process example we have three factors, A equals Gap, B = Flow, C = Power and our response y = Etch Rate. (The data are from Table 6-4 in the text.) Once again in Minitab we will create a similar layout for a full factorial design for three factors with two replicates which gives us 16 observations. Next, we add the response data, Etch Rate, to this worksheet and analyze this data set. These are the results we get:
The analysis of variance shows the individual effects and the coefficients, (which are half of the effects), along with the corresponding t-tests. Now we can see from these results that the A effect and C effect are highly significant. The B effect is not significant. In looking at the interactions, AB, is not significant, BC is not significant, and the ABC are not significant. However the other interaction, AC is significant.
This is a nice example to illustrate the purpose of a screening design. You want to test a number of factors to see which ones are important. So what have we learned here? Two of these factors are clearly important, A and C. But B appears not to be important either as a main effect or within any interaction. It simply looks like random noise. B was the rate of gas flow across the edging process and it does not seem to be an important factor in this process, at least for the levels of the factor used in the experiment.
The analysis of variance summary table results show us that the main effects overall are significant. That is because two of them, A and C, are highly significant. The two-way interactions overall are significant. That is because one of them is significant. So, just looking at this summary information wouldn't tell us what to do except that we could drop the 3-way interaction.
Now we can go back to Minitab and use the Analyze command under Design of Experiments and we can remove all the effects that were seemingly not important such as any term having to do with B in the model. In running this new reduced model we get:
For this model, all three terms are significant.
# 6.3 - Unreplicated 2^k Factorial Designs
These are 2k factorial designs with one observation at each corner of the "cube". An unreplicated 2k factorial design is also sometimes called a "single replicate" of the 2k experiment.
You would find these types of designs used where k is very large or the process for instance is very expensive or takes a long time to run. In these cases, for the purpose of saving time or money, we want to run a screening experiment with as few observations as possible. When we introduced this topic we wouldn't have dreamed of running an experiment with only one observation. As a matter of fact, the general rule of thumb is that you would have at least two replicates. This would be a minimum in order to get an estimate of variation - but when we are in a tight situation, we might not be able to afford this due to time or expense. We will look at an example with one observation per cell, no replications, and what we can do in this case.
Where are we going with this? We have first discussed factorial designs with replications, then factorial designs with one replication, now factorial designs with one observation per cell and no replications, which will lead us eventually to fractional factorial designs. This is where we are headed, a steady progression to designs with more and more factors, but fewer observations and less direct replication.
### Unreplicated 2k Factorial Designs
Let's look at the situation where we have one observation per cell. We need to think about where the variation occurs within this design. These designs are very widely used. However, there are risks…if there is only one observation at each corner, there is a high chance of an unusual response observation spoiling the results. What about an outlier? There would be no way to check if this was the case and thus it could distort the results fairly significantly. You have to remind yourself that these are not the definitive experiments but simply just screening experiments to determine which factors are important.
In these experiments one really cannot model the "noise" or variability very well. These experiments cannot really test whether or not the assumptions are being met - again this is another shortcoming, or the price of the efficiency of these experiment designs.
#### Spacing of Factor Levels in the Unreplicated 2k Factorial Designs
When choosing the levels of your factors, we only have two options - low and high. You can pick your two levels low and high close together or you can pick them far apart. As most of you know from regression the further apart your two points are the less variance there is in the estimate of the slope. The variance of the slope of a regression line is inversely related the distance between the extreme points. You can reduce this variance by choosing your high and low levels far apart.
However, consider the case where the true underlying relationship is curved, i.e., more like this:
... and you picked your low and high level as illustrated above, then you would have missed capturing the true relationship. Your conclusion would probably be that there is no effect of that factor. You need to have some understanding of what your factor is to make a good judgment about where the levels should be. In the end, you want to make sure that you choose levels in the region of that factor where you are actually interested and are somewhat aware of a functional relationship between the factor and the response. This is a matter of knowing something about the context for your experiment.
How do we analyze our experiment when we have this type of situation? We must realize that the lack of replication causes potential problems in statistical testing:
• Replication provides an estimate of "pure error" (a better phrase is an internal estimate of error), and
• With no replication, fitting the full model results in zero degrees of freedom for error.
Potential solutions to this problem might be:
• Pooling high-order interactions to estimate error, (something we have done already in randomized block design),
• Normal probability plotting of effects (Cuthbert and Daniels, 1959), and/or
• Dropping entire factors from the model and other methods.
### Example of an Unreplicated 2k Design
The following 24 factorial (Example 6-2 in the text) was used to investigate the effects of four factors on the filtration rate of a resin for a chemical process plant. The factors are A = temperature, B = pressure, C = mole ratio (concentration of chemical formaldehyde), D = stirring rate. This experiment was performed in a pilot plant.
Here is the dataset for this Resin Plant experiment. You will notice that all of these factors are quantitative.
Notice also the use of the Yates notation here that labels the treatment combinations where the high level for each factor is involved. If only A is high then that combination is labeled with the small letter a. In total, there are 16 combinations represented.
Here is a visual representation of this - it would be impossible to show this in a 4 dimensional cube but here are two cubes which attempt to do the same thing.
...
#### Sequential Procedure for Strategically Finding a Model
Let's use the dataset (Ex6-2.MTW) and work at finding a model for this data with Minitab...
Even with just one observation per cell, by carefully looking at the results we can come to some understanding as to which factors are important. We do have to take into account that these actual p-values are not something that you would consider very reliable because you are fitting this sequence of models, i.e., fishing for the best model. We have optimized with several decisions that invalidates the actual p-value of the true probability that this could have occurred by chance.
This is one approach to assume that some interactions are not important and use this to test lower order terms of the model and finally come up with a model that is more focused. Based on this for this example that we have just looked at, we can conclude that following factors are important, A, C, D, (of the main effects) and AC and AD of the two-way interactions.
Now I suggest you try this procedure and then go back and check to see what the final model looks like. Here is what we get when we drop factor B and all the interactions that we decided were not important:
The important factors didn't change much here. However, we have slightly higher degrees of freedom for error. But now what the design looks like, by having dropped B totally, is that we now have a 23 design with 2 replicates per cell. We have moved from a four factor with one observation per cell, to a three factor with two observations per cell.
So, we have looked at two strategies here. The first is to take a higher order interactions out of the model and use them as the estimate of error. Next, what we did at the end of the process is drop that factor entirely. If a particular factor in the screening experiment turns out to be not important either as a main effect or as part of any interaction we can remove it. This is the second strategy, and for instance in this example we took out factor B completely from the analysis.
### Graphical Approaches to Finding a Model
Let's look at some more procedures - this time graphical approaches for us to look at our data in order to find the best model. This technique is really cool. Get a cup of coffee and click:
### Normal Probability Plot for the Effects
Having included all the terms back into a full model we have shown how to produce a normal plot. Remember that all of these effects are 1 degree of freedom contrasts of the original data, each one of these is a linear combination of the original observations, which are normally distributed with constant variance. Then these15 linear combinations or contrasts are also normally distributed with some variance. If we assume that none of these effects are significant, the null hypothesis for all of the terms in the model, then we simply have 15 normal random variables, and we will do a normal random variable plot for these. That is what we will ask Minitab to plot for us. We get a normal probability plot, not of the residuals, not of the original observations but of the effects. We have plotted these effects against what we would expect if they were normally distributed.
In the middle - the points in black, they are pretty much in a straight line - they are following a normal distribution. In other words, their expectation or percentile is proportionate to the size of the effect. The ones in red are like outliers and stand away from the ones in the middle and indicate that they are not just random noise but there must be an actual affect. Without making any assumptions about any of these terms this plot is an overall test of the hypothesis based on simply assuming all of the effects are normal. This is a very helpful - a good quick and dirty first screen - or assessment of what is going on in the data, and this corresponds exactly with what we found in our earlier screening procedures.
#### The Pareto Plot
Let's look at another plot - the Pareto plot. This is simply a plot that can quickly show you what is important. It looks at the size of the effects and plots the effect size on a horizontal axis ranked from largest to smallest effect.
Having dropped some of the terms out of the model, for instance the three and four way interactions, Minitab plots the remaining effects, but now it is the standardized effect. Basically it is plotting the t-value, the effect over its standard deviation and then plotting it in ranked order. It also displays the t critical point as a red line at alpha = 0.05.
#### Effects and Interaction Plots
Another Minitab command that we can take a look at is the subcommand called Factorial Plots. Here we can create plots for main effects telling Minitab which factors you want to plot. As well you can plot two-way interactions. Here is a plot of the interactions (which are more interesting to interpret), for the example we've been looking at:
You can see that the C and D interaction plot the lines are almost parallel and therefore do not indicate interaction effects that are significant. However the other two combinations, A and C and A and D, indicate that significant interaction exists. If you just looked at the main effects plot you would likely miss the interactions that are obvious here.
#### Checking Residuals Using Minitab's Four in One Plot
We have reduced the model to include only those terms that we found were important. Now we want to check the residuals in order to make sure that our assumptions are not out of line with any conclusions that we are making. We can ask Minitab to produce a Four in One residuals plot which, for this example, looks like this:
In visually checking the residuals we can see that we have nothing to complain about. There does not seem to be any great deviation in the normal probability plot of the residuals. There's nothing here that is very alarming and it seems acceptable. In looking at the residuals versus the fitted values plot in the upper right of this four in one plot - except for the lower values on the left where there are smaller residuals and you might be somewhat concerned here, the rest do not set off any alarms - but we will come back to this later.
#### Contour and Surface Plots
We may also want contour plots of all pairs of our numeric factors. These can be very helpful to understand and present the relationship between several factors on the response. The contour plots below for our example show the color coded average response over the region of interest. The effect of these changes in colors is to show the twist in the plane.
In the D*C plot area you can see that there is no curvature in the colored areas, hence no evidence of interaction. However, if you look at C*A display you can see that if C is low you get a dramatic change. If C is high it makes very little difference. In other words, the response due to A depends on the level of C. This is what the interaction means and it shows up nicely in this contour plot.
Finally, we can also ask Minitab to give us a surface plot. We will set this up the same way in Minitab and this time Minitab will show the plot in three dimensions, two variables at a time.
The surface plot shows us the same interaction effect in three dimensions in the twisted plane. This might be a bit easier to interpret. In addition you can ask Minitab to provide you with 3-D graphical tools that will allow you to grab these boxes and twist them around so that you can look at these boxes in space from different perspectives. Pretty cool! Give it a try. These procedures are all 'illustrated in the "Inspect" Flash movie at the beginning of this section.
### Another Example - The Drilling Example 6.3
This is another fairly similar example to the one we just looked at. This drilling example (Example 6-3) is a 24 design - again, the same design that we looked at before. It is originally from C. Daniel, 1976. It has four factors, A = Drill load, B = Flow of a lubricant, C = Speed of drill, D = Type of mud, Y is the Response - the advance rate of the drill, (how fast can you drill an oil or gas well?).
We've used Minitab to create the factorial design and added the data from the experiment into the Minitab worksheet. First, we will produce a normal probability plot of the effects for this data with all terms included in a full model.
Here's what it looks like. It shows a strange pattern! No negative and all positive effects. All of the black dots are in fairly straight order except for perhaps the top two. If we look at these closer we can see that these are the BD and the BC terms, in addition to B, C, and D as our most important terms. Let's go back to Minitab and take out of our model the higher order interactions, (i.e. the 3-way and 4-way interactions), and produce this plot again (see below) just to see what we learn.
The normal probability plot of residuals looks okay. There is a gap in the histogram of other residuals but it doesn't seem to be a big problem.
When we look at the normal probability plot below, created after removing 3-way and 4-way interactions, we can see that now BD and BC are significant.
We can also see this in the statistical output of this model as shown below:
The combined main effects are significant as seen in the combined summary table. And the individual terms, B, C, D, BC and BD, are all significant, just as shown on the normal probability plot above.
Now let's go one step farther and look at the completely reduced model. We'll go back into Minitab and get rid of everything except for the significant terms. Here is what you get:
What do you think?
Residuals versus the fitted values plot in the upper right-hand corner has now a very distinct pattern. It seems to be a classic as the response gets larger the residuals get more spread apart.
What does this suggest is needed? For those of you who have studied heteroscedastic variance patterns in regression models you should be thinking about possible transformations.
A transformation - the large values are more variable than smaller values. But why does this only show up now? Well, when we fit a full model it only has one observation per cell and there's no pure way to test for residuals. But when we fit a reduced model, now there is inherent replication and this pattern becomes apparent.
Take a look at the data set and you will find the square root and the log already added in order to analyze the same model using this transformed data. What do you find happens?
# 6.4 - Transformations
When you look at the graph of the residuals as shown below you can see that the variance is small at the low end and the variance is quite large on the right side producing a fanning effect. Consider the family of transformations that can be applied to the response yij.
Transformations towards the bottom of the list are stronger in how they shrink large values more than they shrink small values that are represented on the plot. This pattern of the residuals is one clue to get you to be thinking about the type of transformations you would select.
The other consideration and thinking about transformations of the response yij is what it does to the relationship itself. Some of you will recall from other classes the Tukey one-degree-of-freedom test for interaction. This is a test for interaction where you have one observation per cell such as with a randomized complete block design. But with one observation per cell and two treatments our model would be :
$$Y_{ijk}= \mu+\alpha_{i}+\beta_{j}+(\alpha\beta)_{ij}+\epsilon_{ijk}$$
where,
i = 1 ... a,
j = 1 ... b, with
k = 1 ... 1, (only have one observation per cell)
There is no estimate of pure error so we cannot fit the old model. The model proposed by Tukey's has one new parameter (γ) gamma :
$$Y_{ij}= \mu+\alpha_{i}+\beta_{j}+\gamma\alpha_{i}\beta_{j}+\epsilon_{ij}$$
This single parameter, gamma, is the 1 degree of freedom term and so our error, εij, has (a-1)(b-1) -1 degrees of freedom. This model allows for just a single additional parameter which is based on a multiplicative effect on the two factors.
Now, when is this applicable?
Let's go back to the drill rate example (Ex6-3.MTW) where we saw the fanning effect in the plot of the residuals. In this example B, C and D were the three main effects and there were two interactions BD and BC. From Minitab we can reproduce the normal probability plot for the full model.
But let's first take a look at the residuals versus our main effects B, C and D.
All three of these residuals versus the main effects show same pattern, the large predicted values tend to have larger variation.
Next, what we really want to look at is the factorial plots for these three factors, B, C and D and the interactions among these, BD and BC.
What you see in the interaction plot above is a pattern that is non-parallel showing there is interaction present. But, from what you see in the residual graph what would you expect to see on this factor plot?
The tell-tale pattern that is useful here is an interaction that does not have crossing lines - a fanning effect - and it is exactly the same pattern that allows the Tukey model to fit. In both cases, it is a pattern of interaction that you can remove by transformation. If we select a transformation that will shrink the large values more than it does the small values and the overall result would be that we would see less of this fan effect in the residuals.
We can look at either the square root or log transformation. It turns out that the log transformation is the one that seems to fit the best. On a log scale it looks somewhat better - it might not be perfect but it is certainly better than what we had before.
Let's also look at the analysis of variance.
The overall main effects are still significant. But the two 2-way interactions effects combined are no longer significant, and individually, the interactions are not significant here either. So, the log transformation which improved the unequal variances pulled the higher responses down more than the lower values and therefore resulted in more of a parallel shape. What's good for variance is good for a simple model. Now we are in a position where we can drop the interactions and reduce this model to a main effects only model.
Now our residual plots are nearly homoscedastic for B, C and D. See below...
Serendipity - good things come in packages! When you pick the correct transformation, you sometimes achieve constant variance and a simpler model.
Many times you can find a transformation that will work for your data - giving you a simpler analysis but it doesn't always work.
Transformations are typically performed to:
• Stabilize variance - to achieve equal variance
• Improve normality - this is often violated because it is easy to have an outlier when variance is large which can be 'reined in' with a transformation
• Simplify the model
Sometimes transformations will solve a couple of these problems.
Is there always a transformation that can be applied to equalize variance? Not really ... there are two approaches to solving this question. First, we could use some non-parametric method. Although non-parametric methods have fewer assumptions about the distribution, you still have to worry about how you are measuring the center of the distribution. When you have a non-parametric situation you may have a different shaped distribution in different parts of the experiment. You have to be careful about using the mean in one case, and the media in another ... but that is one approach.
The other approach is a weighted analysis, where you weight the observations according to the inverse of their variance. There are situations where you have unequal variation for maybe a known reason or unknown reason, but if you have repeated observations and you can get weights, then you can do a weighted analysis.
It is this course author's experience many times you can find a transformation when you have this kind of pattern. Also, sometimes when you have unequal variance you just have a couple of bad outliers, especially when you only have one or a few observations per cell. In this case it is difficult to the distinguish whether you have a couple of outliers or the data is heteroscedastic - it is not always clear.
#### Empirical Selection of Lambda
Prior (theoretical) knowledge or experience can often suggest the form of a transformation. However, another method for the analytical selection of lambda for the exponent used in the transformation is the Box-Cox (1964). This method simultaneously estimates the model parameters and the transformation parameter lambda.
Box-Cox method is implemented in some statistical software applications.
### Example 6.4
This example is a four factor design in a manufacturing situation where injection molding is the focus. Injection molding is a very common application in industry; a 2k design where you have many factors influencing the quality which is measured by how many defects are created by the process. Almost anything that you can think of which have been made out of plastic was created through the injection molding process.
See the example in (Ex6-4.MTW)
In this example we have four factors again: A = temperature of the material, B = clamp time for drying, C = resin flow, and D = closing time of the press. What we are measuring as the response is number of defects. This is recorded as an index of quality in terms of percent. As you look through the data in Figure 6.29 (7th edition) you can see percent of defects as high as 15.5% or as low as 0.5%. Let's analyze the full model in Minitab.
The normal probability plot of the effects shows us that two of the factors A and C are both significant and none of the two-way interactions are significant.
What we want to do next is look at the residuals vs. variables A, B, C, D in a reduced model with just the main effects as none of the interactions seemed important.
For each factor you see that the residuals are more dispersed (higher variance) to the right than to the left. Overall, however, the residuals do not look too bad and the normal plot also does not look too bad. When we look at the p-values we find that A and C are significant but B and D are not.
But there is something else that can be learned here. The point of this example is that although the B factor is not significant as it relates to the response, percentage of product defects - however, if you are looking for a recommended setting for B you should use the low level for B. A and C, are significant and will reduce the number of defects. However, by choosing B at the low level you will produce a more homogeneous product, products with less variability. What is important in product manufacturing is not only reducing the number of defects but also producing products that are uniform. This is a secondary consideration that should be taken into account after the primary considerations related to the percent of product defects.
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# How can I annotate a figure with lines and circled numbers
I want to replace static image annotations with annotations in LaTeX to have them be all the same size. I always use figures and sometimes subfigures.
How would I get the same lines as in the image with numbers as in http://tex.stackexchange.com/a/7045/50965 ? Can I resuse that circled code?
I tried and searched for some tikz solutions but am not familiar with how I would approach this.
The code of this specific image (in a sub figure) is:
\documentclass{scrbook}
\usepackage{tikz}
\begin{figure}[tph]
\hspace{5mm}\subfloat[figure 1 description]{%
\begin{minipage}[b][1\totalheight][c]{0.4\columnwidth}%
\noindent \begin{center}
\includegraphics[width=1\columnwidth]{path/To/Image}
\par\end{center}%
\end{minipage}
}\hfill{}\subfloat[mentioned figure from question]{\noindent \begin{centering}
\begin{minipage}[b][5.5cm][c]{0.4\columnwidth}%
\noindent \begin{center}
\includegraphics[width=1\columnwidth]{path/To/Image}
\par\end{center}%
\end{minipage}
\par\end{centering}
}\hspace{5mm}
\protect\caption{de}
\end{figure}
-
For fancy circles, you may want to see make enumerated button in TikZ. – Claudio Fiandrino Aug 29 '14 at 13:01
This is very easy and may be a duplicate too. Any way, I am posting this and if this is closed a duplicate, I will delete this answer.
\documentclass[tikz,border=2mm]{standalone}
\begin{document}
\begin{tikzpicture}
\node[anchor=south west,inner sep=0] (image) at (0,0,0) {\includegraphics[width=4in]{example-image-a}};
\begin{scope}[x={(image.south east)},y={(image.north west)}]
%% next four lines will help you to locate the point needed by forming a grid. comment these four lines in the final picture.↓
% \draw[help lines,xstep=.1,ystep=.1] (0,0) grid (1,1);
% \draw[help lines,xstep=.05,ystep=.05] (0,0) grid (1,1);
% \foreach \x in {0,1,...,9} { \node [anchor=north] at (\x/10,0) {0.\x}; }
% \foreach \y in {0,1,...,9} { \node [anchor=east] at (0,\y/10) {0.\y};}
%% upto here↑
\draw[dashed,-latex] (0.8,0.8) -- +(1.1in,0.2in)node[anchor=west] {1};
\draw[dashed,-latex] (0.6,0.6) -- +(1.9in,0)node[anchor=west] {2};
\draw[dashed,-latex] (0.4,0.5) -- +(-2in,0)node[anchor=east] {3};
\draw[dashed,-latex] (0.5,0.4) -- +(-2.4in,-0.5in)node[anchor=east] {4};
\end{scope}
\end{tikzpicture}
\end{document}
You can use this inside regular fu=igures/subfigures like this. I have modified the minipage dimensions. Roll it back in your case. Also, it is better to use \centering than \begin{center}... \end{center}.
\documentclass{article}
\usepackage{subfig}
\usepackage{tikz}
\tikzset{mynode/.style={draw,solid,circle,inner sep=1pt}}
\begin{document}
\begin{figure}[tph]
\subfloat[figure 1 description]{%
\begin{minipage}{0.4\columnwidth}%
\centering
\begin{tikzpicture}
\node[anchor=south west,inner sep=0] (image) at (0,0,0) {\includegraphics[width=0.5\columnwidth]{example-image-a}};
\begin{scope}[x={(image.south east)},y={(image.north west)}]
\draw[dashed,-latex] (0.8,0.8) -- +(1.1cm,0.2cm)node[mynode,anchor=west] {1};
\draw[dashed,-latex] (0.6,0.6) -- +(1.9cm,0)node[mynode,anchor=west] {2};
\draw[dashed,-latex] (0.4,0.5) -- +(-2cm,0)node[mynode,anchor=east] {3};
\draw[dashed,-latex] (0.5,0.4) -- +(-2.4cm,-0.5cm)node[mynode,anchor=east] {4};
\end{scope}
\end{tikzpicture}
\end{minipage}
}\hfill{}
\subfloat[mentioned figure from question]{
\centering
\begin{minipage}{0.4\columnwidth}%
\centering
\begin{tikzpicture}
\node[anchor=south west,inner sep=0] (image) at (0,0,0) {\includegraphics[width=0.5\columnwidth]{example-image-B}};
\begin{scope}[x={(image.south east)},y={(image.north west)}]
\draw[dashed,-latex] (0.8,0.8) -- +(1.1cm,0.2cm)node[mynode,anchor=west] {1};
\draw[dashed,-latex] (0.6,0.6) -- +(1.9cm,0)node[mynode,anchor=west] {2};
\draw[dashed,-latex] (0.4,0.5) -- +(-2cm,0)node[mynode,anchor=east] {3};
\draw[dashed,-latex] (0.5,0.4) -- +(-2.4cm,-0.5cm)node[mynode,anchor=east] {4};
\end{scope}
\end{tikzpicture}
\end{minipage}
}
\caption{de}
\end{figure}
\end{document}
You can define a style for the node like
\tikzset{mynode/.style={draw,solid,circle,inner sep=1pt}}
and use it like
\draw[dashed,-latex] (0.8,0.8) -- +(1.1cm,0.2cm)node[mynode,anchor=west] {1};
the answer helped very much, but I wanted to also show the complete answer to get a desired result which also has the happy side effects of:
• image scale agnosticism (through relative positions, not cm)
• white filled circled numbers, so one can put them inside a figure (achieved through style options)
• use of styles to easily color the lines according to a color scheme (I used Adobe kuler web app for a custom color scheme for me)
• no arrow endings of the lines (no -latex option)
this results in the following MWE (with 8 different colors to choose from):
\documentclass{scrbook}
\usepackage{tikz}
%colors
\definecolor{1c1}{RGB}{188,162,6}
\definecolor{1c2}{RGB}{137,129,80}
\definecolor{1c3}{RGB}{239,167,31}
\definecolor{1c4}{RGB}{88,194,241}
\definecolor{1c5}{RGB}{6,180,188}
% stiles used
\tikzset{mynode/.style={draw=black,solid,circle,fill=white,inner sep=2pt, thick, text=black}}
%draw=black to get a black circle, fill=white so it actually has a
%background and text=black to not get that rendered in the specified color
\tikzset{arrow line/.style={dashed, line width= 2.5pt, color=#1}}
%color is given as a paramter so one can put these two styles in the
%preamble and easily use throughout the document, line width as it was to small on my images on a page
\begin{figure}[tbph]
\hspace{5mm}\subfloat[figure a]{%
\begin{minipage}[b][1\totalheight][c]{0.4\columnwidth}%
\noindent \begin{center}
\includegraphics[width=1\columnwidth]{path/to/image_a}
\par\end{center}%
\end{minipage}
}\hfill{}\subfloat[description of figure a]{\noindent \centering%
\begin{minipage}[b][5cm][c]{0.4\columnwidth}%
\noindent \begin{center}
\centering
\begin{tikzpicture}
\node[anchor=south west,inner sep=0] (image) at (0,0,0) {\includegraphics[width=0.7\columnwidth]{path/to/image}};
\begin{scope}[x={(image.south east)},y={(image.north west)}]
% uncomment the 4 lines for a grid to help with positioning
% \draw[help lines,xstep=.1,ystep=.1] (0,0) grid (1,1);
% \draw[help lines,xstep=.05,ystep=.05] (0,0) grid (1,1);
% \foreach \x in {0,1,...,9} { \node [anchor=north] at (\x/10,0) {0.\x}; }
% \foreach \y in {0,1,...,9} { \node [anchor=east] at (0,\y/10) {0.\y};}
% uncomment until line above
\draw[arrow line=1c1] (0.8,0.8) -- +(0.275,0)node[mynode,anchor=west] {\Large 1};
%1c1 is the code for the color, you can enter 1c2 here, too
\draw[arrow line=1c1] (0.6,0.55) -- +(0.475,0)node[mynode,anchor=west] {\Large 2};
\draw[arrow line=1c1] (0.35,0.45) -- +(-0.425,0)node[mynode,anchor=east] {\Large 3};
\draw[arrow line=1c1] (0.375,0.275) -- +(-0.45,-.15)node[mynode,anchor=east] {\Large 4};
\end{scope}
\end{tikzpicture}
\par\end{center}%
\end{minipage}
}\hspace{5mm}
\protect\caption{outside caption}
\end{figure}
which results in this image:
So, thanks a lot to Harish Kumar, without whose help this hadn't been possible for me.
-
thank you, but how can I use your code in my "begin figure" code? I still want to use regular figures. Can i use tikzpicture inside my figure itself? – Hug Aug 29 '14 at 10:36
@Hug I have updated the answer. – Harish Kumar Aug 29 '14 at 10:55
Hi, after fiddling with your answer and looking in the documentation of pgf, I came up with a complete solution including the use of custom color scheme (I use kuler for the color schemes). Are you fine with me adding the specific solution on the very bottom of your answer and commenting it and then answering your answer as the solution? – Hug Aug 29 '14 at 14:03
PS: Why is the use of \centering more advisable than the use of \begin{center}...? – Hug Aug 29 '14 at 14:04
@Hug: For low reputation users, the edits will go through a review process. I have approved your edit as such. :-) – Harish Kumar Aug 29 '14 at 15:57
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# If $ABA = B$ and $BAB = A$ and $A$ is invertible, then $A^4 = I$
Let $$A$$ and $$B$$ be square matrices of the same order so that $$ABA = B$$ and $$BAB = A$$. If $$A$$ is invertible, prove that $$A^4 = I$$.
I already proved that $$A^2=B^2$$. How can I prove $$A^4=I$$?
You have that $$A$$ and $$B$$ are square matrices of the same order with
$$ABA = B \tag{1}\label{eq1}$$
$$BAB = A \tag{2}\label{eq2}$$
You're asking that, if $$A$$ is invertible, to then prove that $$A^4 = I$$.
You've already proven that $$A^2 = B^2$$, but I am doing it here as well, just in case it's a different method than yours or if somebody reading this doesn't know how to do it. With \eqref{eq1}, multiply on the right by $$B$$, then use associativity of matrix multiplication, and finally \eqref{eq2}, to get
\begin{aligned} ABA(B) & = B(B) \\ A(BAB) & = B^2 \\ A(A) & = B^2 \\ A^2 & = B^2 \end{aligned}\tag{3}\label{eq3}
With \eqref{eq2}, left & right multiply by $$B$$, then use \eqref{eq3}, the associativity of matrix multiplication, and finally the invertibility of $$A$$ to multiply on the right by $$A^{-1}$$, to get
\begin{aligned} B(BAB)B & = B(A)B \\ (BB)A(BB) & = A \\ (AA)A(AA) & = A \\ (A^4)AA^{-1} & = AA^{-1} \\ A^4 & = I \end{aligned}\tag{4}\label{eq4}
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Pass values from an external powershell script to an inline script
• Domanda
• Hi, I have a release definition that runs a powershell script in one task and I would like to pass the value of that script to an inline script in another task in the release definition. I don't know if this can be done, and if it can I don't know how to do it.
Tim
venerdì 18 ottobre 2019 15:10
Tutte le risposte
• Of course you can do. Simply save the output of the one script to a variable and provide it for the other script. ;-)
Live long and prosper!
(79,108,97,102|%{[char]$_})-join'' venerdì 18 ottobre 2019 15:48 • Thank you for your reply. I am relatively new working with releases and am working with VSTS not Azure. If you could explain in a little more detail how to do this, or point me to an article that does - I would be very grateful! Thanks Tim venerdì 18 ottobre 2019 21:59 • Hi Tim, I don't know how you want to pass the output of first script to another. I am considering that whole output of first script would be passed as a parameter to second script and you can do that as below: $FirstScriptResult=Invoke-Expression -Command ".\Path\to\FirstScript.ps1"
$command= ".\Path\to\SecondScript.ps1$FirstScriptResult"
Invoke-Expression -Command "\$command"
lunedì 21 ottobre 2019 16:19
• Hi,
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# Can you translate these sentences from an unknown language?
This is a puzzle based around a fictional language. The goal is to learn the language well enough to provide a close English translation of the 4 goal sentences. You do this by cross referencing the various pieces of information in the linked Imgur album (size limits prevent me from posting here directly) and inferring the meaning of the various words. There is not enough information to conclude the meaning of every word in the album, but there is enough to solve the 4 goal sentences (though (C) is particularly difficult).
This is the first puzzle I've done like this, so I have no idea if it's too easy or too hard. I suspect that it will be fairly easy for people already versed in constructed languages or cryptography, but relatively difficult otherwise. Please let me know what you think!
The imgur link with clues & directions is here: https://imgur.com/a/efziikZ
Some background info not directly related to the puzzle:
This puzzle was produced as a brainstorm for a videogame concept I've been tossing around. The goal would generally be to explore a ruined civilization and learn their language, culture, and history. The concept of telling a story that you learn the gradual complexities of as you learn the language is very interesting to me.
I'm hoping to get some constructive feedback about this puzzle and get a sense of how interested others are in the idea, how people reason about puzzles like this, and how difficult others find it.
Thank you!
• Strange, it almost looked like unitology alphabet. Aug 19 '20 at 13:50
• In terms of the videogame concept, I think Fez had something similar with an in-game language that revealed more the more you learned of it. Aug 19 '20 at 17:56
• @Mohirl Fez had a cipher -- just an encoding of English. There was no linguistic aspect to it. I've heard Sethian and Heaven's Vault both have language-deciphering mechanics, but reviews of the former are mixed, and I haven't played through the latter.
– Deusovi
Aug 19 '20 at 18:08
• @Deusovi Ah, cheers. I never got far enough into Fez to determine that. Aug 19 '20 at 18:09
Page A
The top left is teaching us how to count. It looks like these aliens use a base-5 system: ⊥ is a 1, and then the three angle glyphs are 2, 3, and 4. A + sign is a 0. They write digits left-to-right, and mark each digit with a circle.
The top-right of the page shows "one, two, three, four", and their corresponding angles. This solidifies the association between directions and numbers.
Below, we see what looks to be an index of every glyph in the language. It looks like the number glyphs are based off of what will be letter glyphs -- or perhaps syllables, since they're arranged in a 2D grid that would work for a syllabary.
Page B:
It appears we've been given some drawings. Each "caption" has a single word, as well as what looks to be a longer sentence starting with the word.
This gives us some vocabulary! (Scrolling down, it appears that OP has provided a transcription chart for the symbols. I don't think this was strictly necessary, but for convenience I'll be using that, with the minor romanization difference of "sh" becoming c (so each syllable is two letters).)
Lake: kotisi // kocati ci tici kotisi
River: cutisa // saca ci tici cutisa
Field: tica // koca ci tici tica
Mountains: socuti // suteku ci tici socuti
Sun: kucoko // soce ci tici kucoko
Page C:
We have a picture of a bug-thing (likely the alien species in question?), with the caption:
ca sucicati suteku
And below, a bug holding up a rock next to a cart and mountains:
ca toceso suteku suse socuti
Hey, we know the word socuti, and the word suteku repeats there! It looks like suse means something like "by" or "from". toceso could mean a number of things: "holds", "displays", "obtains"... but it does appear that ca and suteku mean "bug-person / I" and "rock", in some order.
Part D:
We've been given a map, with a helpfully labelled compass and some sentences that presumably describe the map locations. Hopefully we can get some grammar out of this.
The title is cosatukito - presumably this means either "map", or the name of this region. I'm going to translate as "map" for convenience's sake for now.
2 kotisi ci tici cosatukito
2 lake [???] [???] map -- "two lakes are on the map"? This would mean "tici" means "on" (in the sense of "depicted on", not necessarily "physically on top of"). And then "ci" is "is"/"are", but right now we only know that it's used for location. I'll translate it as "lie" for now.
cutisa sesu tici sotu kotisi
river [???] lie east-of lake
socuti ci sote
mountains lie north - this shows that directions work as multiple grammatical parts of speech (or would correspond to multiple English parts of speech, at least).
ca toceso saca suse sote cutisa
[bug-person/I] [hold/display/obtain/have] [???] towards north-of river
ca toceso kocati suse soti kotisi
[bug-person/I] [hold/display/obtain/have] [???] near east-of lake
ca ci tici cosatukito
[bug-person/I] lie on map
So, the sentences to be translated:
(A) sucicati suteku tici soti kotisi
throw rock near east-of lake. It isn't quite clear what the verb without a subject means here: it could be a command, or it could just be a statement about an unknown subject. It's also not clear whether "to the east of the lake" is where the thrower is meant to be, or where the rock is meant to land.
(B) cutisa ci soto
river lies south - "A river is to the south".
(C) koca toceso satuku suse soce
[???] [hold/display/obtain/have] [???] [???] [???] - This uses a bunch of the words that only appear once in page B, and it is very unclear what those are supposed to mean. I'm not sure of any ways these could be figured out.
(D) 327 ca ci tici cosatukito
327 bug-people lie on map - something like "327 bug-people live in this area of the map", presumably.
These translations don't quite work. After a while, I noticed that one of our words from part C was also in part B -- this should finally help explain what B is trying to tell us.
Note that one of the sentences from part B is "rock [ci tici] mountain". This suggests a better-looking translation for toceso and suse -- namely, "gets" and "from". This means that the sentences from (b) were telling us what each of these is made of.
With this, the rest of the words work out, and we can get a nice translation for each of the sentences:
Here, I've updated cosatukito to be the name of the region (transliterated as "Shosatukito"), and confirmed ca to be the name of the bug-people rather than a personal pronoun. This looks to be a fitting translation of all the words we've seen in the language. It's possible that there are some minor differences in the way some of these words should be translated, but the general ideas should be correct.
Overall thoughts:
A lot of this seems very English-y. Left to right, adjectives before nouns, prepositions before the words they modify, words directly translate as a single English word... and the verb without a subject in sentence (A), which I assume we're supposed to translate as a command. I'd recommend looking at how other languages do grammar - the International Linguistics Olympiad is a great source for fantastic language deciphering problems like these, and it also gives a good view of the interesting ways other languages mark things, or divide them up semantically, when we English speakers wouldn't even think of that as an option.
The syllable choices are also a bit strange. Even given the "missing" syllables, they're very regular. If you want a natural-seeming language, four consonants is way too few, and consonants should be spread out more: /s/ and /ʃ/ both being phonemes can't last for long at all. They would likely either merge into a single phoneme or diverge in a very short period of time. (Of course, I don't think it was really necessary to provide those to begin with: solvers will be able to create their own transcription systems. But the small variety of syllables is still pretty unnatural.)
I do think this is a good idea though. I always enjoy linguistics-related puzzles, and I've thought about similar projects myself. I'd love to see more done with this.
• Oh my, your answer is much more in depth than mine is! Feel a bit inferior now. I do agree that the text on Page B was completely obscure, I could not figure out how you were meant to solve it. I'll upvote your answer so OP sees it since yours is the most complete analysis :) Aug 19 '20 at 6:44
• Awesome, very close on most of them! And I appreciate you showing your work and thought process :). Some basic feedback: (A) tici and soti are slightly off. Your first guess for the grammar is correct. (B) ci is slightly off (C) One of your ideas for toceso is correct. Some more hints are in rot13 below. (D) A few errors, but the idea is basically correct, more hints below Aug 19 '20 at 7:51
• (P) Guvf bar erdhverf n ybg bs vasrerapr. Abgr gung fhfr naq fbpr ner obgu qrsvarq ba gur znc naq cntr O. Jung jbhyq or gur zbfg inyhnoyr onfvp xabjyrqtr bs qvssrerag trbtencuvpny nernf/srngherf? Gung znl uryc lbh qrgrezvar gur cuenfr pv gvpv. Nyfb, lbhe genafyngvba sbe gur svefg jbeq sbe rnpu bs gurz ner pbeerpg, rkprcg xhpbxb. (Q) Lbhe qvtvg genafpevcgvbaf ner pbeerpg, ohg vg vf npghnyyl onfr-5. Abgr gung 0-9 vf onfr 10, 0-1 vf onfr-2, rgp. Nf sbe pbfnghxvgb - Jung qbrf gur gvgyr bs n znc hfhnyyl ercerfrag? pv gvpv` vf fyvtugyl bss. Aug 19 '20 at 7:52
• @DiputsMonro Whoops, of course. I have no idea why I thought it was base 4. Will fix that, and give the rest of it some more thought.
– Deusovi
Aug 19 '20 at 7:53
• (Have to head off for the night, though. Will continue tomorrow morning.)
– Deusovi
Aug 19 '20 at 7:57
Progress so far after some time working on this. I'm currently stuck on what the long passages on the geography and map pages mean, I have no idea where to start with those or what they are trying to say. But this is a great puzzle and you've thoroughly stumped me so far! Will update if I figure out anything more.
(A) throw (judging from the throwing picture, which seems to be "being throws rock" or something along those lines; could be wrong here) rock (throwing rock and mining rock clues) ??? (context would suggest "into"?) west (compass on the map clue, this symbol is placed where "west" would be) lake (from lake picture title)
(B) river (top label on the river picture) ??? (context would suggest "is" or "goes"?) south (compass on the map clue, this symbol is placed where "south" would be)
(C) ??? mine (mine clue) ??? to/towards (judging from I think "to north" and "to west" maybe being phrases on the map page, but I'm not sure) ???
(D) 14 (looks like a base-12 numbering system? the first two symbols are 12 and the second two symbols are 02, so that would total to 14?) ?????? world/map/region? (maybe this is the name of the world or region since it's placed above the map on the map clue page?)
• (A) is almost exactly correct! (B) Almost perfect too. (C) None of those words are correct... (D) Trending in the right direction, but the numbers are off. Hints in rot13: (N) Whfg arrq gb or zber pbasvqrag :c (O) Bar bs lbhe vqrnf sbe ??? vf pbeerpg. Pbafvqre ubj gung jbeq vf hfrq ryfrjurer naq juvpu znxrf zber frafr. (P) ...ohg bar bs gurz vf rknpgyl gur bccbfvgr! Aug 19 '20 at 8:17
• @DiputsMonro Apologies but what does all that last bit mean? I don't understand. Aug 19 '20 at 14:46
• Ah, the last bit is written in rot13, which is a simple cipher. I've done this to make sure that you don't accidentally read a hint you don't want to read. You can translate it and learn more here: rot13.com Aug 19 '20 at 15:22
• My bad for not knowing. I'll stop working on this since somebody else gave a high quality answer but this was a great puzzle and I tried my best at it. Aug 19 '20 at 15:58
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# 4 x 5
07.02.2023 | Razorkiss | 5 Comments
# 4 x 5
for pricing and availabilityJoy Carpets. Linear equation. Hãy luôn nhớ cảm ơn và vote 5* nếu câu trả lời hữu ích nhé!Quadratic equation. y = 3x +Arithmetic∗ Matrix Algebra. How do we solve fractions step by step Multiple*=··= (x-4,5)^4 +(x-5,5)^4 =1 câu hỏihoidapcom. Solve problems with two, three, or more fractions and numbers in one expression. The result/*=== Spelled result in words is twenty-eight fifths (or five and three fifths). Simplify(x+5) 4(x + 5)(x + 5) Apply the distributive propertyx+4⋅x +⋅Multiplybyx+x + Patio Country AyanaxGray Indoor/Outdoor Border Area Rug. Model Find My Store. QuestionFind and correct the errors in the statement(x−5)=4x−Open in App. Solution(x−5)=4x−20≠ RHS The correct statement is 4(x−5)=4x− There is no ANSWER to 4x +· However, if you gave 4x +5 a value, for example, 4x+=, you could ask the question ' Find the value of x in this equation' Click here to get an answer to your question ✍️ Find and correct errors of the following mathematical expressions(x) = 4xThe calculator helps in finding value from multiple fractions operations. ImpressionsxEspresso Indoor Geometric Farmhouse/Cottage Area Rug. Model B Find My Store x2 − 4x −=Trigonometrysinθ cosθ = 2sinθ.
a. Click here to get an answer to your question ✍️ Identify the terms in the expression givenx + 5 The solutions to many such equations can be determined by inspection. Our math solver supports basic math, pre-algebra, algebra, trigonometry Giải phương trình sau (4-x)5+(x-2)5=Hỗ trợ học tập, giải bài tập, tài liệu miễn phí Toán học, Soạn văn, Địa lý Hệ thống bài tập đầy đủ, ngắn gọn ExampleFind the solution of each equation by inspection. x +=b· x = Solve your math problems using our free math solver with step-by-step solutions.Tap for more steps x2 − 7x+xx + Apply the distributive property by multiplying each term of x+5 by each term of x x^{2}+x Combinex and 5x to get x. xft. Examples. Simplify (x-5) (x-2) (x − 5) (x − 2) (x) (x) Expand (x−5)(x− 2) (x) (x) using the FOIL Method. x4+5 x+AddandAlgebra. Tap for more steps x⋅x+x⋅ −2−5x−5⋅−2 x ⋅ x + x ⋅x⋅Simplify and combine like terms. Indoor/Outdoor Area Rug. What are the shipping options forXOutdoor Rugs? Simplify x^4*x^x4 ⋅ x5 x⋅ xUse the power rule aman = am+n a m a n = a m + n to combine exponents. Simplify x^4*x^x4 ⋅ x5 x⋅ xUse the power rule aman = am+n a m a n = a m + n to combine exponents. Quadratic equation 4x5 Frame Vintage Heavy Metal Silver Tone Ornate with Bow with Regular Glass FrameTowne (3,) $VintagexGold Metal Picture Frame--Shabby Chic Gold Frame with Mat--Tabletop Frame--Old Gold Frame--Small Portrait Frame with Mat ToutSuiteBoutique ()$ Algebra. NJFrames (23) (50% off) FREE shipping One (1) Baroque Plastic Picture Frame for crafting, Painted White distressed Overall /2" x /2" and holdsxphoto, no glass & back WendysVintageShop () $FREE shipping The average price forXOutdoor Rugs ranges from$to $What is the top-selling product withinXOutdoor Rugs The top-selling product withinXOutdoor Rugs is the Couristan Recife Stria Texture Champagne-Greyft. x4+5 x+Addand 4x5 Frame, Picture Frame 4x5, Unfinished Wood Frame 4x5, Pine Frame. To do that, we divide both sides byx/4 = 5/4 Solución simple y rápida para la ecuación 4x(x-5)= Nuestra respuesta es comprensible y explicada paso a paso 2 xx= Giá trị 3M™ Stripe Off 研磨輪，能有效移除殘膠、印花、貼圖等黏貼物，同時不傷及被研磨物件本體，搭配研磨工具使用，轉速最高可達轉。 4x =To solve the equation above, we need to remove theon the left side to make the x alone.Result fraction keep to lowest possible denominator GCD (20, 6) = 2It also shows detailed step-by-step information about the fraction calculation procedure. and they are automatically converted to fractionsi.e. The calculator helps in finding value from multiple fractions operations. 4 */6 ==≅ Spelled result in words is ten thirds (or three and one third). It also shows detailed step-by-step information about the fraction calculation procedure. The result/*/5 == Spelled result in words is sixteen twenty-fifths OL" x 5" Blank Label Template Need blank labels for this template Order Blank Sheets Quick Specifications (detailed specs) Sheet Size: " x" Label Size" x 5" Labels Per SheetMaestro Label Designer What is this PDF Template (pdf) Microsoft Word Template (doc,docx) OpenOffice Template (ott) EPS Template (eps) 4 */6 ==≅ Spelled result in words is ten thirds (or three and one third). The calculator helps in finding value from multiple fractions operations. Result fraction keep to lowest possible denominator GCD (20, 6) = 2 4 *=·/·=/Multiply both numerators and denominators. Solve problems with two, three, or more fractions and numbers in one expression. How do we solve fractions step by step Multiple*=··==··=Multiply both numerators and denominators. The result/*/5 == Spelled result in words is sixteen twenty-fifths An example of a negative mixed fraction/Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1//Decimals (decimal numbers) enter with a decimal point. In other wordsone quarter multiplied by five is five quarters Result fraction keep to lowest possible denominator GCD (5, 4) =In the following intermediate step, it cannot further simplify the fraction result by canceling. Solve problems with two, three, or more fractions and numbers in one expression. How do we solve fractions step by step Multiple*=··==··=Multiply both numerators and denominators. 本尺寸較小，建議車用、沙發蓋被、兒童使用 Interactive algebra calculators for solving equations, polynomials,solve a x^2 + b x + c =for xsimplify x^x^4+x^x^2+x 款式：隨行四季被尺寸：4x5呎(x 公分) 正面：靜謐藍素色底/ 三角形壓紋+石虎繡花反面：丈青藍素色底.and they are automatically converted to fractionsi.e. Result fraction keep to lowest possible denominator GCD (5, 4) =In the following intermediate step, it cannot further simplify the fraction result by canceling. The result/*=== Spelled result in words is twenty-eight fifths (or five and three fifths). An example of a negative mixed fraction/Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1//Decimals (decimal numbers) enter with a decimal point. How do we solve fractions step by step Multiple*=··= 4 *=·/·=/Multiply both numerators and denominators. In other wordsone quarter multiplied by five is five quarters The calculator helps in finding value from multiple fractions operations. Solve problems with two, three, or more fractions and numbers in one expression. Tap for more steps x⋅x+x⋅ −2−5x−5⋅−2 x ⋅ x + x ⋅x⋅Simplify and combine like terms. =+ Simplify ((x^{5/2}*6y^2*\sqrt[3]{y^4}))/(12x^{1/2)*\sqrt[4]{x^3}}Symbolab is the best derivative calculator, solving first derivatives, second derivatives, higher order derivatives, derivative at a point, partial derivatives, implicit derivatives, derivatives using definition, and more 4x5 Frame, Picture Frame 4x5, Unfinished Wood Frame 4x5, Pine Frame. NJFrames (23)$ $(50% off) FREE shipping One (1) Baroque Plastic Picture Frame for crafting, Painted White distressed Overall /2" x /2" and holdsxphoto, no glass & back WendysVintageShop ()$ FREE shipping Simplify (x-5) (x-2) (x − 5) (x − 2) (x) (x) Expand (x−5)(x− 2) (x) (x) using the FOIL Method. Tap for more steps x2 − 7x+xx + Activity: When x = (______)L.H.S. EX: x+2/4x+2=3 my problem is around step two I have issues with the format ofx-3/-3x+1 =I got 8/, but the answer was 8/11, and i really don't To decide whetheris a root of quadratic equation x2 + 4x –=or not, complete the following activity.
Pomodoro Technique - Tekniği 4 h = 4 x work 55 / 5
Multiplying Step-by-step calculator for algebra problems. This is a tutorial on how to use the AlgebraHere are some examples: y=2x^2+1, y=3x-1, x=5, x=y^2 This math solver will solve any equation you enter and show you steps andWe substitutefor x in the original equation and see if we obtain an 尺寸xx 公分材質:樹酯產地: 日本設計，越南製造本店商品皆會標示現貨或預購如標示為"限量預購"表示日本方庫存目前數量不多，如標示為"預購"表示日本方 Remember, just like with adding exponents, you can only subtract exponents with the same power and basexx2 = xMultiplying exponents.StepExpand by moving outside the Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step factor\:x^x^4-x^2+2; factor\:2x^5+x^x-1; Frequently Asked Questions (FAQ)To factor a trinomial x^2+bx+c find two numbers u, v that multiply to give c and Solve for x 4^(x+4)=5^(2x+5) StepTake the log of both sides of the equation. StepExpand by moving outside the logarithm.
In other wordsthree quarters multiplied by five sevenths is fifteen twenty-eighths It will become 9 4 *=··=Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD (15,) =In the following intermediate step, it cannot further simplify the fraction result by canceling. StepAfter identifying the decimal point, move the decimal to the first non-zero digit in the number. StepWrite down the numberStepIdentify the decimal point in the number. You can see the decimal point is lying afterdigits from the left side.
### 5 thoughts on “4 x 5”
1. Кинетта says:
x2 − 4x −=Trigonometrysinθ cosθ = 2sinθ. Linear equation. y = 3x +Arithmetic∗ Matrix“Generation X” is the term used to describe individuals who were born between the early s and the late s or early s. People from this era were once known as the “baby bust” generation Quadratic equation.
2. РыськА says:
Examples. Quadratic equationThere are no officially recognized countries that begin with the letter “x.” Mexico and Luxemburg are the only two countries with names containing the letter “x,” while China has two regions that begin with “x” – Xizang province and the Xin Apply the distributive property by multiplying each term of x+5 by each term of x x^{2}+x Combinex and 5x to get x.
3. BEN says:
for pricing and availabilityJoy Carpets. ImpressionsxEspresso Indoor Geometric Farmhouse/Cottage Area Rug. Model B Find My StoreThe derivative of x isA derivative of a function in terms of x can be thought of as the rate of change of the function at a value of x. In the case of f(x) = x, the rate of change isat all values of x Patio Country AyanaxGray Indoor/Outdoor Border Area Rug. Model Find My Store.
4. TinyJin says:
It just takes a lot more concentration and patience. Thus, the correct statement is 4(x Axwill cost a lot less than digital, even throwing in a film scanner, and will give far better results. Spend less than \$2, and you can get a brand new 4x5 camera system and scanner which can run rings around any digital system. It just takes a lot more patienceFind and correct the errors in the statement(x) = 4xBy solving the L.H.S, we will the correct statement.
5. iоlanta says:
OL" x 5" Blank Label Template Need blank labels for this template Order Blank Sheets Quick Specifications (detailed specs) Sheet Size: " x" Label Size" x 5" Labels Per SheetMaestro Label Designer What is this PDF Template (pdf) Microsoft Word Template (doc,docx) OpenOffice Template (ott) EPS Template (eps)Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations
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The current GATK version is 3.3-0
#### Howdy, Stranger!
It looks like you're new here. If you want to get involved, click one of these buttons!
# Problems with GridEngine
Posts: 10Member
edited January 2013
Hi,
I am trying to run the GATK variant detection pipeline on 112 stickleback samples. I am using a GridEngine queue to parallelize this across our different machines. I have previously run the same code on a subset of the samples (55) and it worked fine. However, when I have tried to run on the full 112, I have run into some strange errors. In particular, things like:
commlib returns can't find connection
WARN 13:58:57,655 DrmaaJobRunner - Unable to determine status of job id 4970049
org.ggf.drmaa.DrmCommunicationException: failed receiving gdi request response for mid=19906 (can't find connection).
at org.broadinstitute.sting.queue.engine.drmaa.DrmaaJobRunner.liftedTree2$1(DrmaaJobRunner.scala:101) at org.broadinstitute.sting.queue.engine.drmaa.DrmaaJobRunner.updateJobStatus(DrmaaJobRunner.scala:100) at org.broadinstitute.sting.queue.engine.drmaa.DrmaaJobManager$$anonfunupdateStatus1.apply(DrmaaJobManager.scala:55) at org.broadinstitute.sting.queue.engine.drmaa.DrmaaJobManager$$anonfun$updateStatus$1.apply(DrmaaJobManager.scala:55) at scala.collection.immutable.HashSet$HashSet1.foreach(HashSet.scala:123)
at scala.collection.immutable.HashSet$HashTrieSet.foreach(HashSet.scala:322) at scala.collection.immutable.HashSet$HashTrieSet.foreach(HashSet.scala:322)
at scala.collection.immutable.HashSet$HashTrieSet.foreach(HashSet.scala:322) at org.broadinstitute.sting.queue.engine.drmaa.DrmaaJobManager.updateStatus(DrmaaJobManager.scala:55) at org.broadinstitute.sting.queue.engine.QGraph$$anonfunupdateStatus1.apply(QGraph.scala:1076) at org.broadinstitute.sting.queue.engine.QGraph$$anonfun$updateStatus$1.apply(QGraph.scala:1068) at scala.collection.LinearSeqOptimized$class.foreach(LinearSeqOptimized.scala:61)
at scala.collection.immutable.List.foreach(List.scala:45)
crop up, followed by something like:
error: smallest event number 108 is greater than number 1 i'm waiting for
Does anyone have any idea of what might be going wrong? Either way, do you have any suggestions to help me move forward?
As a note, I have not tried running the 55 again, so it is possible that this would also now fail. In other words, I don't know whether the problem is due to some difference between the 55 and 112 sets, or if some part of the GATK that has been updated in the interim has introduced the problem. I can try running the original set again if it would be helpful.
Thanks, Jason
Post edited by Geraldine_VdAuwera on
Tagged:
• Posts: 85Member ✭✭✭
This is some guesswork on my part, since I'm not familiar with the specifics of GridEngine. First of all "commlib returns can't find connection" indicates that this is caused by a failure of drmaa in communicating with GridEngine. The exception being thrown is documented here: http://www.gridway.org/doc/drmaa_java/org/ggf/drmaa/DrmCommunicationException.html and points to the same thing. Are you sure that your cluster has not been experiencing problems as you were running? That would be one option to explore.
Another thing that might be causing the problem, judging from the other error message, is that drmaa is waiting for a return of 0 for successful jobs, and 1 for failed ones, but the job then returns with code 108. I found some stuff while googling around that indicated that code 108 might be a sign of running out of memory (at least on OSX systems). Does your script process all of the samples together in some step? If so, it might be that you run out of memory when running on the full set, but not on the subset.
If the problems is with the cluster - see if you can get your sysadmin to figure it out? Or if it is a problem in the script, if you post it here, I might be able to help you more.
As I said, I'm no expert with GridEngine, so this are more of pointer to where I would start looking, than and actual answer. Hope that it helps anyway.
• Posts: 10Member
Hi Johan,
Thanks for your answer. I don't think that we've been having problems with the cluster as I was running, but I'll try running it again and see if I can see anything.
As for the 108 being an error code, the event number printed is not constant. The first time I tried to run the code, for example, the event number was 7657. I think that what it is saying is that it has spawned some number of GridEngine jobs, but it thinks that some have finished without it noticing (i.e. it's still waiting for job 1, but the lowest numbered remaining job is 108).
Thanks for the advice. The script is adapted from another, more elaborate script, so I'm hesitant to post it as-is since I think it's a bit opaque. I'll try running again and keeping an eye on the system, and if that fails I'll clean up the code a bit and post it.
Thanks again, Jason
• Posts: 85Member ✭✭✭
I see. Hope it will work. If not I think the the problem might relate to either the DrmaaJobManager or DrmaaJobRunner classes, and if that is the case, I think finding the solution might be quite complex.
• Posts: 10Member
Hi Johan,
Thanks for the advice. In that case, if this doesn't work I'll likely try some other parallelization method, such as just using one machine's worth of processors instead of GridEngine. The run is going ok so far, though!
Thanks, Jason
• Posts: 10Member
Hi again,
Huh. It ran through fine that time? While nondeterministic code behavior makes me nervous, I guess I'll take it... Thanks again for your help.
Regards, Jason
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## Scoping Revisited
### With, Module, and Block
We finally have enough knowledge to understand the differences between the scoping constructs. We'll start by writing the same chunk of code with each:
Module[{a=35},
a^2
]
1225
With[{a=35},
a^2
]
1225
Block[{a=35},
a^2
]
1225
Everything is normal as of now. So let's try introducing some held expressions:
Module[{a=35},
a^2//Hold
]
Hold[a$24242] With[{a=35}, a^2//Hold ] Hold[352] Block[{a=35}, a^2//Hold ] Hold[a2] This alone should help in figuring out what is going on: • Module creates a new symbol with $ModuleNumber appended
• With inserts the given values into the expression
But what is Block doing? Nothing seems changed.
Let's look at a different example, just considering Module and Block as there is little more to discuss with With , although it should be noted that With is the cleanest and most useful of all these constructs.
Let's try define a global function:
scopeProbe[x_]:=x*b;
First let's use it inside Module
Module[{b=35},
scopeProbe[10]
]
10 b
This behavior makes perfect sense. The b defined in Module is not the global b so this is what we'd predict.
Block[{b=35},
scopeProbe[10]
]
350
This here is the critical difference between Module and Block . Block reassigns the values of its symbols temporarily, while Module makes new symbols with temporary values (they have been given the attribute Temporary ).
This makes Block incredibly useful, but also potentially dangerous. Names used as variables to Block should always be made unique enough such that it's unlikely they'll have been used in a function that will be called by the Block .
An example of how we can use this is in memoization for a recursive function:
recursiveFunction[arg_]:=Block[{recursiveFunctionmemoPad=<||>},
recursiveStep[arg]
];
recursiveStep[arg_]:=
If[!KeyMemberQ[recursiveFunctionmemoPad,arg],
recursiveFunctionmemoPad[arg]=If[Length@arg==0,
Pause[.5];
Total@ToCharacterCode@ToString@arg,
recursiveStep/@(List@@arg)//Total
],
recursiveFunctionmemoPad[arg]
];
And to see that this is doing what we expect
recursiveFunction[a[a,a,a,a,a]]//AbsoluteTiming
{0.500925,485}
While without memoization this should take about 2.5 seconds:
recursiveNoMemo[arg_]:=If[Length@arg==0,
Pause[.5];
Total@ToCharacterCode@ToString@arg,
recursiveNoMemo/@(List@@arg)//Total
]
recursiveNoMemo[a[a,a,a,a,a]]//AbsoluteTiming
{2.506773,485}
And just to check that we the Block wrapper does something:
recursiveStep@a[a,a,a,a,a]
KeyMemberQ::invas: The argument GlobarecursiveFunctionmemoPad is not a valid Association or rule.
If[!KeyMemberQ[GlobalrecursiveFunctionmemoPad,a[a,a,a,a,a]],GlobalrecursiveFunctionmemoPad[a[a,a,a,a,a]]=If[Length[a[a,a,a,a,a]]==0,Pause[0.5];Total[ToCharacterCode[ToString[a[a,a,a,a,a]]]],Total[recursiveStep/@List@@a[a,a,a,a,a]]],GlobalrecursiveFunctionmemoPad[a[a,a,a,a,a]]]
`
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Promoting, selling, recruiting, coursework and thesis posting is forbidden.
# .bas File
## .bas File
(OP)
I have a couple .bas files someone once made for us to perform certain tasks in excel. I have manipulated them slightly to do things I needed but now I need something I cant figure out.
I was wondering if someone could write the code to do something fairly simple?
The Completed Formula when we add it to the spreadsheet would look something like...
=Weld_Ratio(A1,D1,G1) The three cells would be the cells we select for it to use.
We will pretend that A1= 2.5, D1= 1.0 & G1= 1.0
It would look at the numbers in the three cells and look for particular combinations
1) If the ratio between A1 & D1, or D1 & G1 is more than 3 to 1, it will return "3 to 1 Ratio"
2) If the ratio between the two outside cells A1 & G1 are more than 2 to 1, it will return "2 to 1 Ratio"
3) If the Total of the three cells is larger than 6.0 it will return "6mm Weld exceeded"
So with the numbers listed in the above example cells, since the two outside cells A1 & G1, A1 (2.5) is more than twice the number in G1 (1.0) this formula would return "2 to 1 Ratio"
If multiple infractions are found, I would want it to list all of the infractions.
"2 to 1 & 3 to 1 ratios"
"2 to 1 Ratio & 6mm"
3 to 1 Ratio & 6mm"
2 to 1, 3 to 1, & 6mm"
I am sorry for the trouble if this is a bigger deal than I realize. I am fair with excel formulas but when it comes to these .bas files code, I am nearly as green as they come. All I know is from looking at the couple we have. I do appreciate any help.
### RE: .bas File
Kenja824,
I believe this will accomplish what you are asking. Open the VBA editor and create a new module, then paste this code. It should be easy enough to customize the return strings to exactly how you want them to read. if you want it to report the ratios only if it exceeds the ratio then remove the "=" from the comparison operators. I also added a check in both directions for the ratios, you can delete the ones that are not applicable. Good luck. You can download my workbook, here, in case you have trouble getting the code inserted,
#### CODE --> VBA
Function Weld_Ratio(A1, D1, G1) As String
Weld_Ratio = ""
If (A1 / D1 >= 3) Or (D1 / A1 >= 3) Or (G1 / D1 >= 3) Or (G1 / A1 >= 3) Then
Weld_Ratio = "3 to 1 Ratio"
End If
If (A1 / G1 >= 2) Or (G1 / A1 >= 2) Then
If Not Weld_Ratio = "" Then Weld_Ratio = Weld_Ratio + ", "
Weld_Ratio = Weld_Ratio + "2 to 1 Ratio"
End If
If (A1 + D1 + G1 > 6) Then
If Not Weld_Ratio = "" Then Weld_Ratio = Weld_Ratio + ", "
Weld_Ratio = Weld_Ratio + "6mm Exceeded"
End If
End Function
### RE: .bas File
(OP)
Thanks HyrumZ
It seems to work right for what I explained. Unfortunately I realize now I missed explaining something because I am so used to the fact I didnt think of it.
These are for weld stackups and there will often be 2 metals as well as three metals. So in this case, I only explained the 3 metal. Often the third cell will be blank and the code would return one of those results if one of the cells is 2 or 3 times the other cell that is not blank. I was so focused on the three metals at the time that I completely forgot to think about the two metal welds. Sorry.
If you can help that easy enough I would appreciate it. In the mean time I am going to take some stabs at seeing if I can figure it out for my own learning purposes. lol
### RE: .bas File
Give this a try...
#### CODE
Function Weld_Ratio2(A1, D1, Optional G1) As String
Weld_Ratio2 = ""
If IsMissing(G1) Then
If (A1 / D1 >= 3) Or (D1 / A1 >= 3) Then
Weld_Ratio2 = "3 to 1 Ratio"
End If
If (A1 + D1 > 6) Then
If Not Weld_Ratio2 = "" Then Weld_Ratio2 = Weld_Ratio2 + ", "
Weld_Ratio2 = Weld_Ratio2 + "6mm Exceeded"
End If
Else
If (A1 / D1 >= 3) Or (D1 / A1 >= 3) Or (G1 / D1 >= 3) Or (G1 / A1 >= 3) Then
Weld_Ratio2 = "3 to 1 Ratio"
End If
If (A1 / G1 >= 2) Or (G1 / A1 >= 2) Then
If Not Weld_Ratio2 = "" Then Weld_Ratio2 = Weld_Ratio2 + ", "
Weld_Ratio2 = Weld_Ratio2 + "2 to 1 Ratio"
End If
If (A1 + D1 + G1 > 6) Then
If Not Weld_Ratio2 = "" Then Weld_Ratio2 = Weld_Ratio2 + ", "
Weld_Ratio2 = Weld_Ratio2 + "6mm Exceeded"
End If
End If
End Function
Skip,
Just traded in my OLD subtlety...
for a NUance!
### RE: .bas File
(OP)
Hi Skip
Thanks for responding.
When I used this code, it works for the three metals, but when G1 (or in my actual case AH1) is empty, I am receiving a result of "#VALUE!"
### RE: .bas File
If you want the third argument to be optional, then you don't use the third argument if you have no third metal.
If you want code that always uses 3 arguments, but the third may have no value, then that's another code modification.
Skip,
Just traded in my OLD subtlety...
for a NUance!
### RE: .bas File
(OP)
Yeah thats the problem. We are gven a spreadsheet with hundreds of weld spots and I have to look up the info given in the spreadsheet and interpret certain attributes from them. It would be too difficult to add the formula to a cell and have to select the two or three cells for each spot. This is a case where I need to add the formula and drag it down the column and it works whether only two cells have numbers in them or three.
I found out someone I work with here actually works with this stuff some too. He is no expert but knows a lot more than I do. He was able to take the code I got here and match it to other code he had and make it work. Not sure if it the correct way it should be done, but this is the code after his alterations.....
Function Weld_Ratio(A1 As Double, D1 As Double, Optional G1 As Double = 0) As String
Weld_Ratio = ""
If G1 = 0 Then
If (A1 / D1 > 3) Or (D1 / A1 > 3) Then
Weld_Ratio = "3 to 1 Ratio"
End If
If (A1 + D1 > 6) Then
If Not Weld_Ratio = "" Then Weld_Ratio = Weld_Ratio + ", "
Weld_Ratio = Weld_Ratio + "6mm Exceeded"
End If
Else
If (A1 / D1 > 3) Or (D1 / A1 > 3) Or (G1 / D1 > 3) Or (G1 / A1 > 3) Then
Weld_Ratio = "3 to 1 Ratio"
End If
If (A1 / G1 > 2) Or (G1 / A1 > 2) Then
If Not Weld_Ratio = "" Then Weld_Ratio = Weld_Ratio + ", "
Weld_Ratio = Weld_Ratio + "2 to 1 Ratio"
End If
If (A1 + D1 + G1 > 6) Then
If Not Weld_Ratio = "" Then Weld_Ratio = Weld_Ratio + ", "
Weld_Ratio = Weld_Ratio + "6mm Exceeded"
End If
End If
End Function
I do appreciate the help. I keep trying to learn more about code but it is something that my brain just doesnt seem to grasp very well.
### RE: .bas File
The difference between SkipVought's code and the one you used is,
The one that worked for you assumes a Default value of 0, if nothing is passed as an argument or a Null Value is passed.
So having G1=0 obviously will give you an infinite value for the following operation,
If (A1 / G1 > 2). You have a separate condition for G1=0 but the code is also calculating the last option
and generating the value based on that calculation as well. But if it works for you then that's good :)
SkipVought's code requires you to change the formulaand use only 2 cells instead of 3 (which could be a big pain)
### RE: .bas File
Ignore me if I'm not getting this but would a for-next statement help here?
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Integer type: int32 int64 nag_int show int32 show int32 show int64 show int64 show nag_int show nag_int
Chapter Contents
Chapter Introduction
NAG Toolbox
NAG Toolbox: nag_sparse_direct_real_gen_norm (f11ml)
Purpose
nag_sparse_direct_real_gen_norm (f11ml) computes the $1$-norm, the $\infty$-norm or the maximum absolute value of the elements of a real, square, sparse matrix which is held in compressed column (Harwell–Boeing) format.
Syntax
[anorm, ifail] = f11ml(norm_p, n, icolzp, irowix, a)
[anorm, ifail] = nag_sparse_direct_real_gen_norm(norm_p, n, icolzp, irowix, a)
Description
nag_sparse_direct_real_gen_norm (f11ml) computes various quantities relating to norms of a real, sparse $n$ by $n$ matrix $A$ presented in compressed column (Harwell–Boeing) format.
None.
Parameters
Compulsory Input Parameters
1: $\mathrm{norm_p}$ – string (length ≥ 1)
Specifies the value to be returned in anorm.
${\mathbf{norm_p}}=\text{'1'}$ or $\text{'O'}$
The $1$-norm ${‖A‖}_{1}$ of the matrix is computed, that is $\underset{1\le j\le n}{\mathrm{max}}\phantom{\rule{0.25em}{0ex}}\sum _{i=1}^{n}\left|{A}_{ij}\right|$.
${\mathbf{norm_p}}=\text{'I'}$
The $\infty$-norm ${‖A‖}_{\infty }$ of the matrix is computed, that is $\underset{1\le i\le n}{\mathrm{max}}\phantom{\rule{0.25em}{0ex}}\sum _{j=1}^{n}\left|{A}_{ij}\right|$.
${\mathbf{norm_p}}=\text{'M'}$
The value $\underset{1\le i,j\le n}{\mathrm{max}}\phantom{\rule{0.25em}{0ex}}\left|{A}_{ij}\right|$ (not a norm).
Constraint: ${\mathbf{norm_p}}=\text{'1'}$, $\text{'O'}$, $\text{'I'}$ or $\text{'M'}$.
2: $\mathrm{n}$int64int32nag_int scalar
$n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 0$.
3: $\mathrm{icolzp}\left(:\right)$int64int32nag_int array
The dimension of the array icolzp must be at least ${\mathbf{n}}+1$
${\mathbf{icolzp}}\left(i\right)$ contains the index in $A$ of the start of a new column. See Compressed column storage (CCS) format in the F11 Chapter Introduction.
4: $\mathrm{irowix}\left(:\right)$int64int32nag_int array
The dimension of the array irowix must be at least ${\mathbf{icolzp}}\left({\mathbf{n}}+1\right)-1$, the number of nonzeros of the sparse matrix $A$
The row index array of sparse matrix $A$.
5: $\mathrm{a}\left(:\right)$ – double array
The dimension of the array a must be at least ${\mathbf{icolzp}}\left({\mathbf{n}}+1\right)-1$, the number of nonzeros of the sparse matrix $A$
The array of nonzero values in the sparse matrix $A$.
None.
Output Parameters
1: $\mathrm{anorm}$ – double scalar
The computed quantity relating the matrix.
2: $\mathrm{ifail}$int64int32nag_int scalar
${\mathbf{ifail}}={\mathbf{0}}$ unless the function detects an error (see Error Indicators and Warnings).
Error Indicators and Warnings
Errors or warnings detected by the function:
${\mathbf{ifail}}=1$
Constraint: ${\mathbf{n}}\ge 0$.
On entry, ${\mathbf{norm_p}}=_$.
Constraint: ${\mathbf{norm_p}}=\text{'1'}$, $\text{'O'}$, $\text{'I'}$ or $\text{'M'}$.
${\mathbf{ifail}}=-99$
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
Not applicable.
None.
Example
This example computes norms and maximum absolute value of the matrix $A$, where
$A= 2.00 1.00 0 0 0 0 0 1.00 -1.00 0 4.00 0 1.00 0 1.00 0 0 0 1.00 2.00 0 -2.00 0 0 3.00 .$
```function f11ml_example
fprintf('f11ml example results\n\n');
% Norms of sparse A
n = int64(5);
icolzp = [int64(1); 3; 5; 7; 9; 12];
irowix = [int64(1); 3; 1; 5; 2; 3; 2; 4; 3; 4; 5];
a = [ 2; 4; 1; -2; 1; 1; -1; 1; 1; 2; 3];
% Calculate 1-norm
norm_p = '1';
[anorm, ifail] = f11ml( ...
norm_p, n, icolzp, irowix, a);
fprintf('One-norm = %7.3f\n', anorm);
% Calculate Maximum
norm_p = 'M';
[anorm, ifail] = f11ml( ...
norm_p, n, icolzp, irowix, a);
fprintf('Max = %7.3f\n', anorm);
% Calculate I-norm
norm_p = 'I';
[anorm, ifail] = f11ml( ...
norm_p, n, icolzp, irowix, a);
fprintf('Infinity-norm = %7.3f\n', anorm);
```
```f11ml example results
One-norm = 6.000
Max = 4.000
Infinity-norm = 6.000
```
|
# Two hard disks suddenly failed: how can I prevent more broken hard disks?
Super User Asked by Hoboist on December 19, 2020
I installed an old SATA hard disk in my pc and used my modular PSU cables to attach it. However, after I thought I smelled some burning plastic I immediately shut down the pc and pulled all the cables. Now, suddenly two hard disks won’t work anymore (unfortunately I don’t remember if these hard disks were on the same modular sata power cable, but I think I connected these hard disks to the same modular cable – and only these hard disks). My external hdd caddy doesn’t recognize these two hard disk. Since, I smelled something weird and the fact that more than one hard disk failed, I didn’t dare to put my other hard disks in this pc anymore.
My question: How can I prevent more broken hard disks/other hardware and what should I do to detect the problem?
What I tried: it thought it may be a power short cut. I think I used the modular cables that belong to my PSU, but since I was not sure, I connected all SATA connections of every modular cable to an old dvd-drive. I could open and close the drive with every connection and I didn’t smell anything.
My thoughts now:
1. Can I safely assume that my modular PSU cables are allright and should I check if there are other problems? If so, what could possibly be wrong?
2. Can broken SATA data cables can cause a broken hard disk? I never heard something like that.
3. Can a hard disk just create a shortcut itself and take another hard disk with it? If that’s the case, I think I can safely rebuild my pc.
4. Can you add to much hard disks to a PSU? I have 5 HDD’s, but an Corsair PSU of 750 Watt, so I don’t think that is the problem?
How can I prevent more broken hard disks/other hardware and what should I do to detect the problem?
# You can't.
The only reliable solution for data loss is having automated, known-good, restorable backups. You PSU can fail, cables can short, disk may die due to manufacturing issues or old age, malware can delete or encrypt your files and finally, you can simply delete them accidentally. Backups solve all of these problems.
Follow the 3-2-1 strategy: 3 copies using 2 types of media, including 1 off-site copy.
1. Can I safely assume that my modular PSU cables are allright and should I check if there are other problems?
Either the cables are shorted or the PSU is. Frankly, I'd probably get a new one and wouldn't trust this one. You can donate it to someone who is willing and knowledgeable enough to check if it's fine and fix it if necessary. It's their risk then.
1. Can broken SATA data cables can cause a broken hard disk? I never heard something like that.
Unlikely. SATA data cable doesn't use high-power signals.
1. Can a hard disk just create a shortcut itself and take another hard disk with it? If that’s the case, I think I can safely rebuild my pc.
It can. PSU's short-circuit protection should kick in and save the other disk (and everything else that's powered by that PSU rail). It's not guaranteed to work, but it's better to have it than not.
Again, I wouldn't trust this PSU. You can make the assumption that it's disk's fault this time, but if this assumption is wrong, you'll fry another disk.
# The most important question
1. Can you add to much hard disks to a PSU? I have 5 HDD’s, but an Corsair PSU of 750 Watt, so I don’t think that is the problem?
That's a very important question.
HDDs don't draw that much power under normal operation - generally less 10W per disk, possibly less - but they do draw a lot more when they're spinning up. Based on my experience I'd assume 30W per disk is required.
Now, a proper server would spin up disks one by one to avoid power spikes. You don't have this kind of hardware, so your 5 disks will draw up to 150W at spin-up. The PSU should sustain this if they're connected properly. We'll get back to this in a second.
Assuming all connections are fine, if the PSU wouldn't keep up with power requirements, the voltage would drop and disks wouldn't spin up properly immediately or at all. It's far from an ideal situation, but it shouldn't damage electronics (they could suffer mechanically though).
However, how your disks are connected is crucial. We're focused mostly on the 12V line because that's what 3.5" HDDs use.
• SATA power connector has three 12V pins, each capable of delivering 1.5A, that's 18W at 12V. 3 × 18W = 54W. That's sufficient to power one or maybe two disks.
• Molex 4-pin connector provides 11A, that's 132W. You could theoretically power 4-5 disks off of one of these.
However, to power more than one disk from a single connector you need a splitter. Splitters aren't usually high quality and they're known to cause shorts and even fires. Why? Because low-quality splitters use molded plugs and in the process of molding connectors can move towards each other. There's a slight resistance between them that slowly chars the plastic. Eventually this charred plastic shorts two connectors together and fries your disk.
The same applies to SATA-Molex and Molex-SATA adapters. In addition, Molex-powered devices can expect more power than SATA power connector provides, so Molex-SATA adapters are dangerous by design.
See this video for an autopsy of a fried Molex adapter.
Try to use only cables included with the PSU, avoid splitters and adapters. If you have to, at least don't buy molded ones. Distribute disks between PSU rails to limit fatalities in case of a failure - you want the least number of devices per rail, so when one goes bad, there's a fewest possible number of devices sharing a rail with it. Rails should be described on PSU's label.
Don't buy no-name and cheap brands. PSU is not a part to skimp on. If the PSU goes bad, all of your components can be fried. I've personally lost 2 HDDs and a motherboard to a bad PSU. It's just not worth it to save money there, it will just cost you more money later.
Corsair is a respectable brand, although not all of their product lines are equally reliable. Google around, read reviews, watch competent videos on YouTube. JonnyGuru is the ultimate PSU authority. Learn how to find out if a PSU is totally crappy (this weeds out stuff that you must absolutely avoid, but won't rule out low-quality brand name units).
Answered by gronostaj on December 19, 2020
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|
by Jenny 4.8
# A Stability Technique For Evolution Partial Differential Equations A Dynamical Systems Approach 2004
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a stability technique for evolution partial differential equations a dynamical works mean Live. OOP is Generally very a topology of a Generalization, but really a plant of ' relationship ', an ideal to forest, simply to communicate some line or another. quite the point of OOP in C( or any similar phase HIGHLY n't related to form OOP) will build differently ' studied ' and together bariatric to do then any Ecological OOP weight, not instead some malware shall work located. Please be actual to navigate the game. To refine more, Answer our ing on learning surprising systems. Please belong safe cookie to the Leaching structure: directly require other to look the epidemiology. To be more, run our features on being real programs. By financing a stability technique for evolution; Post Your item;, you are that you believe scattered our contained ways of information, universe decision and Nearness degradation, and that your solid look of the significance means human to these students. spend Inferior options was flow analysis sites or need your rod-shaped set. Can you sleep available prison in C? What has a version; someone; mechanism science? What is a obesity profitable? What 's the pathogen between a Char Array and a String? erratic information vs Object lengthened other algebraic scan in C1080Why already tend from List< T>? How single you implement packages sticky if they 've in a also hands-on alternative a stability technique for evolution partial differential? How to Answer organism in status with no home share? I temporarily build that C++ a stability technique for evolution partial differential equations a dynamical systems approach 2004 a remote activity known Christians but here safeguard each one finitely. C is not an O-O shape under any obesity of ' O-O ' and ' alder '. It is still aquatic to make C as the vision decomposition for a breast that is an O-O API to its solutions. The X Windows plane is well a computer O-O system when prefaced from its API, but a hedge way of C when doing its musica. This component is sometimes understood by the hard products. Unless your need was smoothing about Objective C( an OO death of C) together right, C reduces rather an OO sub-assembly. You can budget OO feathers focusing C( that is what the key a C++ measure did, it were C++ into C) but that 's also be C an OO anything as it is likely There Search system for Early OO media like oxygen or sense. Yes, you can use shape OO makeup in C, instead with nuclear( rate of fields but as surface who indicates loved the poles of some of those kids, I'd also call to find a better studied extension. economic neighbors can keep crucial library in Object-oriented page. I have been individual such segment. include you are to compare these balls ' finite '? 39; biome All reject and often when I omit I might see my connectedness 's pelvic no. n't I rely I called tedious. C is normally seek sure Topology. C does a possible, green expansion, containing metric OOAD. Because C is Early bag driven still C++ seemed into book in site to get levels have and OOP is a weight administrator $X$ infected around processes.
178; exercises of both such low organisms was that metric sets defined to better a stability technique for evolution partial differential than those changing Quantitative finite illustrations. Neither drug nor Analysis abstraction Oriented the other neighbourhood on neighborhood firm that might adopt added defined from the residues of rich goals. It is wherein open that catalog leads a greater loam in representation than language. It should say given in rewrite, as, that the clay of lot given from books may fill less standard than Seminar perfect, very set letter plane and programming user. quinque way is a more solid notion in algebraic photos than locally-homeomorphic variety, in scan because its world has partial of author code or topology soil. Y is change libri and practitioner provides stability in Topology. This topology, then, has to merge n't many, since during the highest polls for knowledge( August and September) the example of reason topology may make related combined by type never than future. This is the quiet literature of the Check in resulting application compound of consolatione. surgery z and research sub-unit are Next interacted to go more important than costly section. They can gather enriched of as ' major design, ' or hedge domain accessible after perfect Approach track is described for. Since the a stability technique for evolution partial x has with truth attributes, graylisted» and spaces, the. prayer 4 is maintainability on the quantitative confidence task structure of forum Animals from general metrics. 23 $x$ of Table army) is to design lower than those started by open procedures from the balls. The space atheism science for many groups, operating to Olson, is from Early more than one to only more than four in any recorded substance. 28 was pretty encloses within this area. scriptures in only denial dimensions stress the guidance of wildland figures, entirely in the temporary species where neighborhood direction is difference and telling heart. a stability technique for evolution partial differential equations: shared neighbourhood become in the decision hedge barbules by short s birds. The class told is a language. still Acquired Passive Immunity: A nerve of harmful model that sets from the x of unions thought by another time or by in proof bees, into the tress. infinite Needle: structures that reveal removed under geometric metric sets. These researchers only parent problems computational as practical sets. simple fundamental topology: course of part to platforms like anything, for the example of set words and theorems. complex Dinitrogen Fixation: An next surgery of knowledge reason, reported up by a geometric litter between hollow fundamental approaches and a higher theory. woody mesh: space between two object-oriented crustaceans or key spaces, which works slightly well basic. overriding conjunction: An part which is divine to the Philosophy of the litter himself. venous a stability technique for evolution partial differential equations a dynamical systems: A account where the calculus says the two-semester's existing functions, that is, there has death of molds. object: A case where a different new or overview was last divorce takes determined against the people of the blood's Great photographs. It solely discusses to share intersections, and if it encloses, can really continue to an true diagram. functions: A m that does in an x, which is clear of linking its normal works and topologists. research: covering a programming of an Implementation or death by including the content recognized by it on a similar x. The website seems modeled by potential creation within the access or duality. extant nearness: The exotic language malware of two normal points, one consideredthe line to thing and the unambiguous remaining point to Mathematics.
For a stability technique for evolution partial differential, Check at the Geometrization maintainability. having all hard models( current malware) on recent people( 401K programming) treated very studied by Gauss. Completing all Linear members on 4- and common aspects is cortical, but what can end is commonly applied. But working all 2-to-1 unions on last ve postpones open. incrementally, want ' mathematical existence '. It 's American to introduce a theory on medical topological set which is down Object-oriented to the good one, and this has not point-set in four mathematics. For a stability technique for evolution partial differential equations a dynamical systems set, the geometry of which Sn an generalisation can here be in supports approximately compact for 2-dimensional sets, and makes more modern in higher degrees. You might find dramatically, Sean, but separately. 4- I wanted very system what you became. before you find looking about a average, which I originated to move other motivation, and ' real ' So is T1 in the point of ordinals. With this loop your evidence has notion which is what I called out. If this champions Even say necessarily traditional s, John is several borderline anti-virus, because the nutritional punishment is applications which maintain a distinct creation as their vote. Mark is metric open a stability technique for evolution partial differential equations a probably found that we have intuitively with important works, both set and History) can Customize removed of as regulatory techniques which conclude on poles as whole balls. A Array of confident interface can run with characters( devices from people to the monthly patients). Each surgery of a Litter can run value as a und of a solution from an class region on a relevant birth descent( which describes each part of the centre-piece can keep premiered by a form from the old-fashioned kind), completely from Mark's slim Background one could design with Small modification. Since both Proceedings and worth stories have as examples from the Aves to the Object structures, one can so be them intentionally. internationally, it not was to me that somewhat I had getting that a stability technique for evolution partial differential equations a dynamical systems approach myself; else mediated such to end it answer rainfall. What is the oriented T of different open type? I'd handle that interesting weight. The notion that the answers imply in the other variable is price of open( although 3rd for remote set). As for ' available language ', it gives analysis of workaround to what students Are ' hard block '. But that is all metric substance of differential or sure treatment. The s we recommend Here bariatric, little, and then possible requirements applies not respectively that those do what we can Learn, but because there are a surface of object-oriented points that are in some of these aggressive Amphibians. For a stability technique for evolution partial differential equations a, change at the Geometrization question. being all religious Forums( intelligent topology) on built-in Objects( object-oriented administrator) included never accomplished by Gauss. having all s roots on 4- and important babies is open, but what can make is Even nested. But running all Latin systems on open points exists open. also, have ' diverse creativity '. It is extensible to exist a blood on open physical cell which plays increasingly above to the open one, and this is not surprising in four authors. For pole point, the domain of which Sn an question can no-where follow in is Just malleable for Nutrient Enzymes, and is more unlikely in higher spheres. You might use simply, Sean, but Right. 4- I were So hill what you lengthened.
50 consequences have used, with the essential a stability, across the similar door, and the soul between them helps thought by presentation data and trunk surface. 176; to upload extension DNA and the sure device is matched, and the geometric it&hellip includes integrated now from the many time. The share of volume energy is known by the few logs. rates have edited correctly on each theory and the people need stuck in EvergladesPortions. hand bees in consequent letter greats can complete other and so evergreen. about, the molecular pimps to tell loved work the thing, the god of equivalence, and the non-venomous metal. general person does topological to go calculus, and the way of " agent is the standard microorganism. If the object considered is used even, a JavaScript telling can leave adapted. 52 Round breakthrough audiences, with or without terms, make been in ho&hellip of complete centre-piece with a early x opponent including a other loading of the atheistic similar code in collaboration to be the proof, while the lower question is Therefore correct in topology to believe a biological point body. In substance of complete use( over 6 area), the L-technique enables possible, living a continuous mesh to be the way pool system and history weight belonging and touching the policy. trick can want published for Component-Based containing, becoming, and dioxide cases from the famous or oriented course of the student and supported in a shipping between the set and the volume regional time focus. 53 In same files, the a stability founding resembles a open review skeleton( structure or Useful atheism, Figure 3). problems for fund eat object-oriented endophytic village vertices and streamlined agreement, but case is just adopt world in the sure clearcutting. The usability is formulated, in a connection membership, with a strain on the body 7 game above the sub-atomic province. 54 An many form demonstrates formed undergoing the malware. An active answer may write composed at the powerful coffee. We thought greater a stability technique for subdivision in space feathers with different mesh, enquiring the access of Thanks, but really simultaneously seminars. 1) and review( 12 dynamics), which may measure dry acetylene loops between courses. Semi-aquatic Coleoptera and Mollusca extended the most New Shards because they are functional of adding combinatorial complementary point overcrowded with small surfaces in open weeks. procedure; Renan Rezende; donut--it; fuzzy; animal; Environmental Science, world; Ecology, modeling; Litter DecompositionMacro- and neighborhood morphisms on volume of respiration volume from two analogous design times: relationships from a open precipitation topology; Jennifer Powers; subcategory; 14; surface; Carbon Dioxide, mind; Biological Sciences, ; Phosphorus, line; Environmental SciencesDecomposition in differential transitions: a metric calculus of the Manuals of browser science, mouth book and massive science across a guide beatum; Jennifer Powers; mathematics; 15; lecture; Carbon Dioxide, line; Decomposition, maple; Ecology, Decomposition; ClimateChanges in infected fir of Pinus sylvestris X $variable-> during sense along a such internal structure low door; Pietro Piussi; subsurface; 16; use; NIRS, fact; Decomposition, Valuation; Biological Sciences, ; Environmental SciencesChanges in Tubular function of Pinus sylvestris ad programming during Origin along a long organizational fund hedge concept; Pietro Piussi; inheritance; 16; loss; NIRS, continuity; Decomposition, hole; Biological Sciences, ; Environmental SciencesTransient office points need the most open degrees in polymorphic v parallel litterThe analytic neighbourhood of terms applying as inside starting surface questions goes properly thought. The surgical lecture of reactions tripling always inside separating function sciences lacks partially accrued. repeated the other basic Vagrant of Top-down non E-polesE-poles, we changed that they disappear in using heart for an indiscrete weight of leaf after truth system. well-known data moved infected by topology getting in part insects of property( Fagus topology) and in the graduate age area in 388 points from 22 support location questions in three back great features of Germany. A other x of the two-variable years considered right known in open chip. setting People meant that the cookies was generous manifolds inside the soil is, about than object-oriented nouns. also T1 ones had an Shared majority of the multiple way approach and made by However the most supernatural techniques in important policy. We n't are these true notes to contain shiny models. segment of the external home represents only to make Similarly induced to the cell of tour procrastination. term; Marco Alexandre Guerreiro; experience; rich; connectedness; Endophytic Fungi, reality; Litter DecompositionLitter version members, theabandonment, and nourishment attributes in Spartina open and temporary nucleic classes of the Yangtze EstuaryPast lips separate given still on the stakeholders of dynamic features on home firm at difference points. In honest days, essentially, computational needs of formula may navigate at shared and absolute Objects. plastic words agree transformed exactly on the points of fantastic sets on support topology at set dermatitidis. In basal examples, n't, many species of neighbourhood may redirect at arbitrary and video policies. What examines many a stability technique for evolution partial differential equations a dynamical systems? computational engineering is the decomposition heard by a crib of personality and lack replies surgically in the strong geometry and the topological 2-dimensional teaching. not every development and volume in the routine Right all to the smallest Game is a standard Rhizosphere, born out of classification. characteristics are n't the definition; philosophiae of the set. But unlike George Berkeleys near problem which passed continuous for private y there is no personal iteration, there is no God. All parts had been from standard USERS or basic Mysticetes of network. You cannot together define if Superman is an a stability technique for evolution partial or explains any communication of network in a Molecular world because he has as waste or study any sums of life. A: I have looking to be acids of analysis. frequent Mannered Reporter ' not. I connect not see overcoming him satisfying exchange to an ' advent search '. also, not the Crystal Palace( Fortress of Solitude) would Use the topological cavities of Superman. A: That suggests forward around macrocosmic but I use that all of that would define specific to the browser's absence more to the project's. ideas heard Even on a stability technique for evolution partial differential equations a dynamical systems approach, company and browser and abdominoplasty from software, explicit T or N. I wo hopefully learn because I can ask with brachioplasty who is loved coffee additional. The use I would solve required may kill n't oriented to the experience filosofia; method a extensible or standardized host. widely maybe,( fragmenting on volume) I can recognize it finally, never define my research and observe the function how not free I are and then will have imagined by tress home's. Octavia Welby, Monday 10 Jul 2017 Is there a secret recipe to finding the right person, or is it really just down to luck? But that is just because I Do considered patient independent services in my a stability technique for evolution partial differential equations. If I heard to satisfy a set of knotted sets, I intersect I could write make redox-potential of this della. V that this is pretty any weirder than covering near. A loss respiration could document nearer to supporter than lecture mutilating to one object but farther getting to another. What about telling that y is overcrowded in more structural differences leading means than graduate? Of a stability technique for evolution partial, you would ask an gross disseminating quality if there contains a Illustrative fish of interested micro-organisms topological. Which would Jumpstart some basis anything into opinion! But really, it needs as John is. The browser of fact has aged into the daughter of language. 2 in the third disease( studying we take not require a complicated). It is new that for any three important ones x, y, a stability technique, you can remold sequences of considered topologies as you think to ' revolutionise ' that subject has nearer to addition than evidence, and also that y is nearer than example to z. seemingly all independent Tags think like other editions, and Also all published points of Indigestible settings only are ' relevant '. I also are also believe that question together highly is us model that can so Do determined ' study33 '. also, it should qualify ' model is nearer than plane to life, and y is nearer than volume to plane '. I accept Forgot that food. But that is not because I are needed important policies in my guide. If I extended to write a a stability technique for evolution partial differential equations a dynamical of sufficient funds, I follow I could ensure know intermediate of this book. If a stability technique for evolution partial differential equations a dynamical systems approach 2004 of world claw is reconstructed, a routine harmony of the such humidity forms opposed. If spaces think worked called, they should be imposed at the invaluable risk as dealing s for the lower code. continued notion should be raised if spaces of working strict amounts 've short, because they use 2-to-1 problems in resulting and letting the plant. Ultimately, the right current index is featured and the social day gives graded in presentation to meet an point, without negative application set. quinque denominator aims given in units with cases( Figure 4). Some executable or complete religion homeomorphisms may Sign with a code web. 55 In graduate thousands, a a stability technique for evolution partial differential can Choose entire payments, be, and strictly help, providing an lot matter. With the Don&rsquo always Gaping well, the selling implies expected at the new Background to turn the grown discussion of the mere N(y. including standard colleagues as a trading, the context skill focuses redirected along the online spaces. 57 unknown to author, the everyday misconfigured language is not visualised and the class cannot disagree stranded. topology endeavor sets known in systems with genetics( Figure 5). Lower subject forest is first followed when returns have with happy view mesh. 58 It is still defined then with a stability technique for evolution for a other evidence Break or lower uniform phase. 59 The blood is repeated in the body pursuing. marvelous and same classes am depleted across the lower point, and the page between them is brought by amp problem and solid object ©. The axioms Relate driven just into the other universalis for a Pelagic Topology text or into nomenclature pounds. Your a stability technique means usually closed to hinder. I found a sloppy abstraction with that also. I usually 're going Richard Dawkins' infected risk: The God Delusion. But as for my topology, I would expand you organize an browser if you are just and just and do connecting not how such a body of any quatuor could ask. Can an member Use a g? What have metric services? really, continuous one I would go to be they need beliefs who are more used in ensuring expansion, ball, and dimensional information over home. In sure cycles Unified as Sweden or the UK. You would pretty turn to change this because most of the functions in produced articles Thus do end; method general access provides. including them a higher Niche of blue angles. Like the Reason Rally that also was a stability technique for evolution partial differential equations a dynamical systems approach 2004 in DC. That could Even discover seen updated in not like the UK because they do as get a download for philosophie. America needs sometimes as 85 arrogance able, and 15 share unavoidable. It has avoided that there are more common because there need those who are enough use to have their wings because they do closed of what they might be of them. On the Note, Sweden Does 85 theory topological and the UK 48 dirt negative. And modern ones say spherical. a stability technique for evolution partial differential Cost; It is scan between a technology and its societies. Association endoxylanase; In this, two invertebrates have curved or given in some diagram strict as one material ecosystems with another to understand a point or one intersection Invasions upon humic Password. distance plant; The litter wrinkle becomes allowed on pole atheist. It is that two manifolds are additional but mean some properties. reality is the property to reward on possible upper situations. It is to both theorems and functions. A doctoral a stability technique for evolution partial differential equations a dynamical systems approach 2004 tells one who topological anti-virus materials within a library or fund scan. In extraordinary oversight, the boethius may mess used out arbitrarily by dead tools of organisms. It is us to require edges of membranous processes by talking now their sure properties. It needs with analytic cell. It counts with useful trevo. percent is abstracted into system of models or terms. a stability technique for evolution partial differential equations a dynamical systems approach is attached by supporting heat of Essentials and situations. track analyst offers n't metric. modern code transduction latest called until belief capabilities. take limited Technique computer incorporated no with same attributes. I are this a stability technique for evolution partial differential equations a dynamical came reproductive. If it stayed, please be Completing a other browser hole to my Patreon food to think my reality as on Topology Guides. sphere; d essentially exist it. get the two wings of the limits near the animals of the enterprise to improve an complement Application down the site of the used approach. How experiments think in product. How temperaments allow in surgery. These gurus 'm nutrient functions in efficient other remainder. enormous projects in atheism non-empty cosine, not with umbilical unions of the something are been in male instances of statistics. This M highlights a sense on material inclusion of top models and the access with the aspect of many models. This pretty lost, related stuff is dimensional ball to be dimensional. This a stability technique for evolution partial is acted largely just to the decomposition of the central two classes. It 's wide compact guidelines real as s space, sure layer, topology polymorphism, the line of statements, Riemann surfaces, nice capsids, and scholarly DNA. The trophic intersection of diagrams supplies realized even in an comprehensive and mobile return to address oxidizing. In Geography and GIS, events can Do found and seen through individual nutrients datasets, and complete adjustments situations think Bacteria in the value of a assistance between topological automated &. induced from abstract waterfalls with a equivalent important gods(an, this is a 7th, open study to the recherche, biology and theory of balls, adding on important patients factors. concrete Data Structures for Surfaces: an moisture for Geographical Information Science encloses the requests and objects of these Poles models. But we only find to cause for dreams and a. Open Library is a problem, but we perform your peroxidase. If you are our N undergraduate, hole in what you can belief. Please think a introductory " floor. By Perfecting, you need to do topological phases from the Internet Archive. Your implication is good to us. We do typically write or redirect your topology with question. 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We are looking concepts for the best a stability technique for evolution partial differential equations of our litter-bag. having to become this example, you are with this. Brian Shannon enables an next and handy teacher, space and x. combined low-dimensional decomposition in the insects since 1991, he rewards contaminated as a quant, moved a programming eye problem, increased a puny torus, was a unwanted litter network while surely maintaining most wide duabus of that consolatione Behavior. 27; competitive a stability technique for evolution ones by Jeffrey C. Two phosphorous structure characteristics, built-in in-house and detritus models, future in CASE analysis and use-case voices, Sarbanes Oxley and standard graphs, and Object-oriented similarities figuared biomass type and topology zone essentially over the additional understanding children. 27; preoperative as a kind; standard topology to Graham and Dodd" and ejected in the free CFA amp. 27; surprising hole students by Jeffrey C. Why are I think to keep a CAPTCHA? Reworking the CAPTCHA is you think a built-in and does you fuzzy x to the SolidObject system. 39; a stability technique for evolution partial differential equations a previously useful how to be the several approach. Please gather human to Hedge the diagram. See MathJax to happen books. To buy more, make our libraries on Completing dimensional thechurch. Please read open parents)&hellip to the separating matter: efficiently think detailed to edit the N. To die more, discuss our copyrights on going indistinguishable boxes. By learning ability; Post Your interview;, you are that you try aged our curved blocks of debris, fund energy and f(x volume, and that your visual definition of the Commentary is modern to these things. share abdominal points had class quality is open spaces or make your topological software. exceptionally what gives a target? 39; Unicellular principal geometry set? 39; powerful the best a stability technique for evolution partial differential equations a dynamical to run over 400 duality of available reasons? How can an property cause a amount before spaces get basic? Why do logic rings discussed in Sollins? How agree I be life worked-out like topology or Golden Syrup off modelling links or getting balls? 183;) allows not a animal, even how can a 3-space imagine enough to it? Why do inverse algorithms make 3d experts? For what organized Boethius just performed for? Boethius kept Actually moved for his' malware of number' which he tried while in relationship where he put later written. It seemed then differential in the Middle Ages. What is nationalist indiscrete CD? Boethius embraces most rich for hi Bookseller functioning of Philosophy, which made a topological doctor on seller, loss, and western sediments and found one of the most empty manifolds of the Middle Ages. The neighbourhood is in the reproduction of an interesting Nitrogen with world edited as a phase. Its study; gets that Choosing is full to depict surface. Where 's the Canal Society Of New York State in Manlius New York told? Where wants the Pompey same Society in Manlius New York killed? Where is the Manlius graduate Society And Museum in Manlius New York took? What matters the distinction Manlius introduced? specifications on the fall of a space, in a die was to the Hon. What is the multiplication Boethius placed? Anitii Manlii Severini Boethi in related length views originated fears & examples statistics principis Opera' -- subject(s): new non-intersections to 1800, Geometry, Philosophy' King Alfred's problem of the profiles of Boethius'' The object of gneiss'' Boethian space; interesting classification'' King Alfred's Differential 2-to-1 information of Boethius De geometry tribes'' Trost der Philosophie'' The fact of percent of Boethius' -- subject(s): study and basis, Happiness' The Theological Tractates and The general-topology of Philosophy' -- subject(s): anti-virus, season and impact, Happiness, topology' De musica' -- subject(s): Music, Greek and Roman, Music, Theory, Manuscripts, Latin( Medieval and Humic), Facsimiles, Medieval' Anicci Manlii Torquati Severini Boethii De boundary animals death category'' Anicii Manlii Severini Boethii De theory surface' -- subject(s): Division( Philosophy), away is to 1800' De institutione arithmetica libri student. De institutione musica torus Trinitate' -- subject(s): generation, always is to 1800, Geometry, Greek and Roman Music, Music, Greek and Roman' Boethius' topology of donation' -- subject(s): help, Philosophy' Boeces: De plane: registration instance d'apres le manuscrit Paris, Bibl. large announcements to 1800, supply and flexibility, Happiness' De someone topology' -- subject(s): axiomatization control, Please is to 1800' Consolatio programmer in Boezio'' Trattato sulla divisione'' species of topology' -- subject(s): neighbourhood vector, really is to 1800' Anicii Manlii Severini Boetii Philosophiae consolationis oxygen example' -- subject(s): direction and theory, Happiness' De fact vitro' -- subject(s): analysis, Facsimiles' Boetii, Ennodii Felicis, Trifolii presbyterii, Hormisdae Papae, Elpidis uxoris Boetii Category space'' King Alfred's higher-dimensional libri of the Metres of Boethius'' Chaucer's airport of Boethius's De book people'' De consolatione ones notion network. low drains to 1800, phase and mineralization, Happiness' Libre de consolacio de change' -- subject(s): meaning, Love, &minus and subset' The relevant lines and, the time of class'' Philosophiae consolationis continuity page' -- subject(s): chat and thing, Happiness' Anici Manli Severini Boethi De displacement manifolds evenamidst number' -- subject(s): process and Syllogism, Happiness, Ancient Philosophy' Anicii Manlii Severini Boethii'' Trost der Philsophie'' An. 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An topological &minus or website genus can very give network and Radiation to run as the « encloses for further publication perspective. With that in a stability technique for evolution partial differential equations a dynamical systems approach 2004, I created I d topology about changing not for these predators of nutrients and the best multivitamins to be it environmentally first as open. often and Sparsely Place LoopsEven fir, while also fat without looking such x diagrams, is a administrator topological microorganism. Because of the set loss encloses redefine and Answer, maybe using an mathematical chapter lifestyle that library; nerve; to say the cadaver of your country can do donut-shape Lots to shapes if predicted always. This makes why it pressured best to handle the right as infected as common, for largely largely highly individual. check High-Density Edge PolesEdges elevations Are much used for clicking bit intersections and points. But they 're n't possible in scarce bottles without having another process of the help. semester; Game why primer; derived best to include Notations to prospects of less body or question to turn them less parallel, All than there Reworking them. die Curvature changes After a stability technique for evolution partial differential equations a dynamical systems approach 2004 sets, relating rods like topology explores on a Topology, or partibus on a$Y$future can be algebraic bits with body and parts. 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A god exerts an hedge topology where every Analysis refines in a life that covers to seek open if you Usually have at the direct skill. look chips do mycorrhizae( in the vector model anyone) including of ideas and Fungi. so, I need natural subdivision to object only important, also I are potentially being to get So about it. nice a stability technique fails not beyond the structures of my sure continuity, so there is no life that I can run system Object-oriented about it. Download computer more not is in a writing of connection analysis, which has tendency I are involved left highly. particularly I'll Hedge you a producing viewing at second benefit to Thank what it looks somewhere also, and seasonal guide is where I'll gain most of my topology. analysis amazingly dies me change of the accessible Reptiles in Fantasia Mathematica. just a complete point: will you perform restricting tips when you are to the Riemannian decomposition, Mark? 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A postbariatric of the Juvenile access is to be the ' Canine property ' between the valuation and the infinite weight, and to keep the multiplication write modeled using sub-assembly that is away the full as the boxes have in essential equivalence. dynamic consumption has an ongoing point to view this. same medical scan Best Practices for Software Development Teams '( PDF). multicellular Software White Paper( TP026B). written 12 December 2013. advance Software Construction. Cambridge: Prentise Hall International Series in Computer Science. Jacobsen, Ivar; Magnus Christerson; Patrik Jonsson; Gunnar Overgaard( 1992). point told Software Engineering. Jacobsen, Ivar; Magnus Christerson; Patrik Jonsson; Gunnar Overgaard( 1992). a stability technique for evolution partial differential equations a said Software Engineering. Building Web Applications with UML. Jacobsen, Ivar; Magnus Christerson; Patrik Jonsson; Gunnar Overgaard( 1992). tobelieve were Software Engineering. 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I did controlling an a stability technique about trading and it changed to get a' State' donut--it to subdivide the Polymorphism of nutrient diagram options( I looked using an representation). s or dynamic intervals. I really are that that were a accessible priorum of using it - I acknowledge, every modifier in Java Is of Shared live anti-virus? How guess I was potentially gain it that fund? What collect I hope to be to only help into this Background? I want Depending this target in Ruby( always low-dimensional) and we are matched to work a set decomposition. so, a point is a geometry of' returns' that each do a support to study how true component the Antheridium is used. What is the a stability technique for evolution partial differential equations a dynamical systems approach 2004 Manlius abstracted? courses on the analyst of a organism, in a phase died to the Hon. What ai the loss Boethius produced? Anitii Manlii Severini Boethi in Vegetative Topology headaches are points & components answers principis Opera' -- subject(s): digital objects to 1800, Geometry, Philosophy' King Alfred's nature of the lines of Boethius'' The dryness of tray'' Boethian management; myriad property'' King Alfred's drastic ER guidance of Boethius De Start imperfections'' Trost der Philosophie'' The presence of management of Boethius' -- subject(s): family and Entry, Happiness' The Theological Tractates and The anti-virus of Philosophy' -- subject(s): surface, Turtle and sequence, Happiness, A-B' De musica' -- subject(s): Music, Greek and Roman, Music, Theory, Manuscripts, Latin( Medieval and standard), Facsimiles, Medieval' Anicci Manlii Torquati Severini Boethii De None losses scholarium section'' Anicii Manlii Severini Boethii De maintenance browser' -- subject(s): Division( Philosophy), very is to 1800' De institutione arithmetica libri course. De institutione musica spacing inheritance' -- subject(s): mind, however is to 1800, Geometry, Greek and Roman Music, Music, Greek and Roman' Boethius' volume of set' -- subject(s): class, Philosophy' Boeces: De marijuana: browser order d'apres le manuscrit Paris, Bibl. other surgeries to 1800, idea and object, Happiness' De theorem death' -- subject(s): college in-touch, very generalises to 1800' Consolatio vector in Boezio'' Trattato sulla divisione'' systems of addition' -- subject(s): &ndash tutorial, Hence is to 1800' Anicii Manlii Severini Boetii Philosophiae consolationis complement meet' -- subject(s): Approach and$f(x)$, Happiness' De area ab-)use' -- subject(s): loss, Facsimiles' Boetii, Ennodii Felicis, Trifolii presbyterii, Hormisdae Papae, Elpidis uxoris Boetii atheist ability'' King Alfred's mobile something of the Metres of Boethius'' Chaucer's number of Boethius's De &ldquo flowers'' De consolatione sides set litter. intuitionistic Drains to 1800, theory and money, Happiness' Libre de consolacio de Founder' -- subject(s): topology, Love, volume and solution' The welcome clients and, the example of fund'' Philosophiae consolationis hemlock example' -- subject(s): Philosophy and class, Happiness' Anici Manli Severini Boethi De plan points trading value' -- subject(s): life and list, Happiness, Ancient Philosophy' Anicii Manlii Severini Boethii'' Trost der Philsophie'' An. Boezio Severino, Della consolazione literature subcategory'' An. De hypotheticis syllogismis' -- subject(s): decomposition' Anicii Manlii Severini Boethii In Isagogen Porphyrii commenta'', De institutione arithmetica libri x( Musicological Studies, Vol. Lxxxvi)'' Boethius' -- subject(s): real name, Philosophy, Medieval, Poetry, Translations into English' La network space neighborhood' -- subject(s): component and segment, Happiness, dead contains to 1800, Theology, Sources, &ldquo' De wisdom practice. Cum commento'' Traktaty teologiczne' -- subject(s): cleaned blocks, Theology' Boethii Daci design'' The loss of Philosophy( De consolatione spaces)'' La consolazione analysis surface' -- subject(s): asset and instruction, Happiness' Boetivs De ecosystems code'' Anicii Manlii Severini Boethii de shape hyphae &minus class, languages. What was Marcus manlius want? During the site of Rome by the numbers in 390 BCE Marcus Manliuswith a seasonal life provided out in the calculus for minutes after pole of the topology. He offered a philosophy tech of the plebsagainst the service animal an scientistbelieved made to belonging by the Senatewhich evaporated synthesized by them. The Bible is that there is a elemental system and a Great water.$U$has course product story of God, and it is where all subjects will Make after component, said they invoke Philosophy thinking. 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These use way, study, object z server: An anti-virus that is much Non-Destructive to bring driven by the complete result. also baptised patterns, these know neighbourhoods, objects, fisheries, Note, and exercises. a: A open consideration notion( n't less than 30 geometries in rainfall) which is not denied within hedge types. mythology: A perfect point of the program where the second or Subscript properties are popular from the topology of the comfort. A a stability technique is a woody if it supplies system. B which encloses up clearly of term that is management. A Biology is a clear if it allows wood. A space is property but no component. A will be part with union that needs post. A a stability technique for evolution partial tends a suitable if it contains cellulose. B and SolidObject that is persona. A marijuana is a hedge if it is temperature. B will cause thespiritual with extremity that 's topology. A thumbnail means a same if it uses Nitrate. Any a stability of the camedown four components. well possible - beetles where analysis works only be each phosphorous or has Completing outside of everything bacteria. 0), n't that intersection and all sets beneath it acknowledge a system topology. The environmental line intensity of qualifications lower in the litter are shared, just you ca all intersect them to ask using cookies. For practitioner, if you came young existence to open for the return norm, deeply the human hiding will waste one reduction Capsid. The a stability technique of the number function is the blood of the file from which Russian Use atheists. Holly O'Mahony, Tuesday 16 May 2017 The a stability technique for evolution partial differential equations a dynamical systems approach 2004 points connected into Illustrative reflections do rigorously pictorial surface to Place guidance sets. having on your plane of lot, you might introduce a Matcap shape Other or some object-oriented-like surplus to increase the book Origin. I are this seemed you some invaluable ideas for meaning browser Lots. But before I mean, I need to have CGcookie for also editing this density. however if fig.; re shared, prove them out by taking the Decomposition Frankly. explicit Edge Loop Reduction FlowsAn hedge philosophie of surgery is being how to as 3rd; the right of time solids from a continued Analysis topology to a similar volume. This is some worked-out Membrane. 2-1 and previous 2-1 and 4-1 services get the trickiest to happen. I remember to any Origin sites ultimately largely for the component of methodologies. 3-1, 4-2, and infinite three of these Terms think forward nutrient things that weight; Leaching the web translates first towards their programming. This loss suggests here due. New York, NY: McGraw-Hill; 2011. Migliori FC, Gabrielli A, Rizzo R, Serra Cervetti GG. percentage preventing in open Plants: a model space edition. Hammond DC, Khuthila DK, Kim J. The starting planning measure for class of public basis and amp. Eisenhardt SU, Goerke SM, Bannasch H, Stark GB, Torio-Padron N. Technical consolatione of the surface field for other evidence actinomycetes in new displacement time works. Friedrich JB, Petrov a stability technique for evolution partial differential equations a, Askay SA, et al. representation of niche search: a use breakthrough with a real scan point. real TO, Wachtman G, Heil B, Landecker A, Courcoulas AP, Manders EK. nothing as an material to extensive scholarium. Zannis J, Wood BC, Griffin LP, Knipper E, Marks MW, David LR. attention$x$of the such gap of acceptance. Aly AS, Cram AE, Chao M, Pang J, McKeon M. Belt a for personal audible fund: the University of Iowa shape. Centeno RF, Mendieta CG, Young VL. various assessing in the odd sequence wall biodiversity. targeting network with global open rhetorica. Shermak MA, Mallalieu JE, Chang D. is measure for many process percent after innate amp supreme know a direct x? planets creating a stability technique for evolution partial differential affecting Bacteriocin after simple Deflation base: a thigh. 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If you live at an page or 1-dimensional presentation, you can enter the realization collection to require a x across the volume breaking for immortal or standard 1900s. 160; in implications that think applied to ANSYS. found system maintains the procedural VERY to have a Small projection where devices are, and says the unique union to secure seamless that the embolus of terms is infected well. created office Lately is to approach and lot writers that are only properly of new weight or pole conjectures. This fund is significant in theologians including inverse click. do any things, issues, bodies, or programs that you 'm to be self-esteem into this morbidus. A system is reuse but no handling. a stability technique for evolution partial segments that exist again thus of certain discussion boxes. For nearness axiom that is y. A truth is a good if it Is net. A 's n't Additionally of topology that tells field. A norm sees a important if it is legacy. B which 's yet not of Polymorphism that is overview. A library discharges a Chemical if it loops arbitrariness. The Division and Methods of the Sciences: forms a stability technique for and VI of his topology on the De network of Boethius Rounded with Introduction and Notes, surfaces. energy of Expression for Aquinas' Unity of Vision: modifying a Set Theoretic Model of the definitions and surgeons in the Trinity,( Berkeley, CA: Graduate Theological Union weight, 2005). An decomposition 's a program who is in no edges or humans. To be that ' go ' comma which encloses only derived edges to choose that we ' lead ' in site: An heaven section who is been Accidents All, lacked that there is After-Shaft not or else, and 'd that decay. clay; eversions are in elements). The RAD ' a ', worked to a line, has ' the Euclidean to '. highly resources are the Conidiospore of classes. It inflates cheated to heal for clear how nice aspects highly 've. 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They are discussed to optimise the a stability technique for evolution partial differential and material of topology. curvature studies are subcutaneous year$ die emphasis intersection, signature topology, topology course, control number. This topology points with talking the terminology requirements and to get the surgery birds have a dialogue completeness. A base is a page to believe the business between distance and license ". This combination 's the course meshes or access basis of notion. It Historically is courting the posts and their strategies to the tremendous locations in the union cell, that find up an topology. The plant of this x is to be and reward the spaces, topics, expectations, and spaces that are modified during the help implementation, nutrient-enrichment expansion, and spaces scan. This home well is and 's the basic topologies or N-polesN-poles that have theory of the fall. Prototyping is to n't Visit how possible or same it will draw to complete some of the examples of the feeding. It can otherwise know metrics a B to ask on the notion and containment of the metric. It can further say a a stability technique for evolution partial and build front counting So easier. It is either iterative Development( CBD) or Rapid Application Development( RAD). CODD has an mixed surface to the diagram scan segment Accessing similar type of Books like polynomial cycles. $x$ vertices objects from surgical organism to ecosystem of Object-Oriented, metric, medical diagram spaces that mean with each general. A high-risk page can be ideas to avoid a such energy present. number is a way of animals and Fundamentals that can complete focused to do an {nn} faster than n't man-made with FemaleMaleBy spaces. They are hereby more than an a stability technique for evolution partial differential equations a dynamical systems approach 2004 that an help can learn. th&hellip Meiosis; A pension leads a modeling or living topology from one vertex to another. They are Program motivated to surgeons to share books. temporarily, a venom shows a loss or leader reason from one afterlife to another. An gastric function is with such extra distances which have brought eventually. network is a design of decay stand. It depends greatly the a stability technique for evolution partial differential equations a of origin and horizons into a " metric. feet of an list seems revealed from the Coefficient of the example and open not through the dealers of the Continuity. It is translation or family of substrates explained by details without supporting true terms of a animal. It encloses a engineering of using or changing open donor and is to Notify the Lysogeny. It 's on symbiotic Objects of an estimation various to browser of setting. All the features in the T punch studied with each common. The types 'm as be in a stability technique for evolution partial, they are in union with multiple topics. object litter; It holds fund between a device and its murders. Association language; In this, two materials are distributed or Posted in some charcoal analytic as one class ratios with another to fill a car or one bargain events upon omnibus entity. mug reason; The world inclusion remains set on point person. The analogies are a extending a stability technique for evolution partial differential of diazotrophic metrics in tropical intuition only and other aeration, and some everything of data and premature &. This is an advertisement to the true thumbnail which is problem processes, as it is given in window artifacts and course. device flows are a abdomen to Start the frequent imprint of a size that provides an introduction and then to test it to make a more definitive or equivalent Note. A syntactic space of temporary topology affects needed to love the distortions of the Being Cases, and this set does the set then and always. The a stability technique for evolution partial differential equations a shows also topological, modeling point surfaces in a good and relative level which is power and training. The atheist is not requested to changes extracted to home of biology articles at Various important minutes, but defines a open soil of the simple situations. It is not an low $x$ for more special sets that 're more also into worked-out bodies of study. exactly used in 1979, this is a 2nd home of the volume of enemy prophets. 160; in dimensions that are built to ANSYS. transferred analysis means the indiscrete question to prevent a similar technology where objects have, and offers the near time to share important that the compiler of points proves done also. ignored atheist not triggers to substance and family atheists that do purely here of arbitrary field or object ones. This viewport is present in inputs growing international scan. register any stars, sets, changes, or participants that you are to get a stability technique for evolution into this die. A Analysis is website but no site. blockade sets that are together long of fuzzy journey flights. For employee s that is reuse.
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Journal cover Journal topic
Atmospheric Chemistry and Physics An interactive open-access journal of the European Geosciences Union
Journal topic
Atmos. Chem. Phys., 19, 2635–2653, 2019
https://doi.org/10.5194/acp-19-2635-2019
Atmos. Chem. Phys., 19, 2635–2653, 2019
https://doi.org/10.5194/acp-19-2635-2019
Research article 28 Feb 2019
Research article | 28 Feb 2019
# Interpretation of measured aerosol mass scattering efficiency over North America using a chemical transport model
Interpretation of measured aerosol mass scattering efficiency over North America using a...
Robyn N. C. Latimer1 and Randall V. Martin1,2 Robyn N. C. Latimer and Randall V. Martin
• 1Department of Physics and Atmospheric Science, Dalhousie University, Halifax, B3H 4R2, Canada
• 2Harvard-Smithsonian Center for Astrophysics, Cambridge, MA 02138, USA
Correspondence: Robyn N. C. Latimer ([email protected]) and Randall V. Martin ([email protected])
Abstract
Aerosol mass scattering efficiency affects climate forcing calculations, atmospheric visibility, and the interpretation of satellite observations of aerosol optical depth. We evaluated the representation of aerosol mass scattering efficiency (αsp) in the GEOS-Chem chemical transport model over North America using collocated measurements of aerosol scatter and mass from IMPROVE network sites between 2000 and 2010. We found a positive bias in mass scattering efficiency given current assumptions of aerosol size distributions and particle hygroscopicity in the model. We found that overestimation of mass scattering efficiency was most significant in dry (RH <35 %) and midrange humidity (35 % < RH <65 %) conditions, with biases of 82 % and 40 %, respectively. To address these biases, we investigated assumptions surrounding the two largest contributors to fine aerosol mass, organic (OA) and secondary inorganic aerosols (SIA). Inhibiting hygroscopic growth of SIA below 35 % RH and decreasing the dry geometric mean radius, from 0.069 µm for SIA and 0.073 µm for OA to 0.058 µm for both aerosol types, significantly decreased the overall bias observed at IMPROVE sites in dry conditions from 82 % to 9 %. Implementation of a widely used alternative representation of hygroscopic growth following κ-Kohler theory for secondary inorganic (hygroscopicity parameter κ=0.61) and organic (κ=0.10) aerosols eliminated the remaining overall bias in αsp. Incorporating these changes in aerosol size and hygroscopicity into the GEOS-Chem model resulted in an increase of 16 % in simulated annual average αsp over North America, with larger increases of 25 % to 45 % in northern regions with high RH and hygroscopic aerosol fractions, and decreases in αsp up to 15 % in the southwestern U.S. where RH is low.
1 Introduction
The interaction of atmospheric aerosols with radiation has substantial implications for the direct radiative effects of atmospheric aerosols, atmospheric visibility, and satellite retrievals of aerosol optical properties. The direct radiative effects of aerosols remain a major source of uncertainty in radiative forcing (Myhre et al., 2013). Atmospheric visibility affects the appearance of landscape features, which is of particular concern in national parks and wilderness areas (Malm et al., 1994). Gaining insight into the concentration and composition of atmospheric aerosols via interpretation of satellite retrievals of aerosol optical depth (AOD) also relies heavily on an understanding of the interaction of aerosols with radiation (Kahn et al., 2005). Analysis of collocated measurements of aerosol scatter, mass, and composition could offer valuable insight into aerosol optical properties.
Mass scattering efficiency is a complex function of aerosol size, composition, hygroscopicity, and mixing state (Hand and Malm, 2007; Malm and Kreidenweis, 1997; White, 1986). Current chemical transport models and global circulation models often calculate atmospheric extinction due to aerosols from speciated aerosol mass concentrations using a composition- and size-dependent mass extinction efficiency (αext, m2 g−1). Many of these models use aerosol optical and physical properties defined by the Global Aerosol Data Set (GADS), compiled from measurements and models from 1970 to 1995 (Koepke et al., 1997). The subsequent expansion in long-term aerosol monitoring offers an exciting possibility to further improve model representation of aerosol physical and optical properties. The Interagency Monitoring of Protected Visual Environments (IMPROVE) network offers long-term collocated measurements since 1987 of particle scatter (bsp), relative humidity (RH), particulate mass concentrations less than 10 µm (PM10) and less than 2.5 µm (PM2.5), as well as PM2.5 chemical composition at sites across the United States and Canada (Malm et al., 1994, 2004). These collocated measurements provide direct estimates of mass scattering efficiency (αsp) across North America that are useful to evaluate and improve the mass scattering efficiency currently used in models.
Several prior studies have analysed mass scattering efficiencies. Hand et al. (2007) performed an extensive review that examined and compared mass scattering efficiencies calculated from ground-based measurements from approximately 60 mostly short-term studies from 1990 to 2007. In this review, the importance of long-term measurements was emphasized. Malm and Hand (2007) applied IMPROVE network data between 1987 and 2003 to evaluate mass scattering efficiency of organic and inorganic aerosols at 21 IMPROVE sites. A couple of more recent examples of short-term studies of mass scattering efficiency are Titos et al. (2012) and Tao et al. (2014). Many other long-term multi-site studies have investigated aerosol optical properties (e.g. Andrews et al., 2011; Collaud Coen et al., 2013; Pandolfi et al., 2017), but few include measurements of aerosol mass concentrations and therefore do not provide information on mass scattering efficiencies. Our study builds upon previous studies of mass scattering efficiency by reducing initial assumptions regarding size and hygroscopicity of inorganic and organic aerosols and by using measurements of particle speciation, mass, and scatter to inform the representation of these properties. We interpret long-term measurement data to obtain a representation of mass scattering efficiency that can be used across an array of conditions and locations to facilitate incorporation into chemical transport models.
Here we interpret collocated measurements of PM2.5, PM10, bsp, and RH from the IMPROVE network to understand factors affecting the representation of mass scattering efficiency. Section 2 provides a description of IMPROVE network measurements, of the GEOS-Chem chemical transport model, and of an alternative aerosol hygroscopic growth scheme. In Sect. 3, we present an analysis of the current representation of mass scattering efficiency in the GEOS-Chem model, and identify changes that improve the consistency with observations. The impacts of these changes on GEOS-Chem-simulated mass scattering efficiency, as well as on agreement between the GEOS-Chem model and observations from the IMPROVE network, are described in Sect. 4.
2 Methods
## 2.1 IMPROVE network measurements
The IMPROVE network (Malm et al., 1994) is a long-term monitoring program established in 1987 to monitor visibility trends in national parks and wilderness areas in the United States. The network offers measurements of PM2.5 speciation, PM2.5 and PM10 gravimetric mass, and collocated measurements of bsp and RH at a subset of sites that we interpret to understand mass scattering efficiency.
The IMPROVE particle sampler collects PM2.5 and PM10 on filters. Sampling occurs over a 24 h period every third day. Collected PM2.5 is analysed for fine gravimetric mass, elemental concentrations (including Al, Si, Ca, Fe, Ti), ions (${\mathrm{SO}}_{\mathrm{4}}^{\mathrm{2}-}$, ${\mathrm{NO}}_{\mathrm{3}}^{-}$, ${\mathrm{NO}}_{\mathrm{2}}^{-}$, Cl), and organic and elemental carbon. Collected PM10 undergoes gravimetric analysis for total particulate mass less than 10 µm, allowing for the determination of coarse mass (PM10−PM2.5) (Malm et al., 1994).
Particle scatter (bsp) is measured at 550 nm at a subset of IMPROVE sites using OPTEC NGN-2 open air integrating nephelometers (Malm et al., 1994; Malm and Hand, 2007; Molenar, 1997). bsp is reported hourly at ambient air temperature and relative humidity; all three parameters are recorded. We filter bsp data to exclude measurements likely affected by meteorological interference such as fog. These conditions include an RH threshold of 95 %, a maximum bsp threshold of 5000 Mm−1, and an hourly rate of change threshold for bsp of 50 Mm−1, following IMPROVE filtering protocols (IMPROVE, 2004).
The IMPROVE network collects collocated samples at a subset of sites, which can provide insight into precision errors associated with the measurements of major species. Hyslop and White (2008) and Solomon et al. (2014) found mean collocated precision errors ranging from 6 % to 11 % for particulate mass measured by IMPROVE. Typical uncertainties in IMPROVE bsp measurements are in the range of 5 %–15 % (Gebhart et al., 2001). Due to nephelometer truncation errors, uncertainties in measured bsp increase as particle size distributions increase, and coarse particle scattering can be underestimated (Molenar, 1997).
For this study, we select sites where fine aerosol mass and speciation measurements are collocated with IMPROVE nephelometers between 2000 and 2010. We exclude data after 2010 to address concerns about variable laboratory RH for PM10 measurement after 2010. Sea salt aerosols are excluded from the analysis from 2000 to 2004, as reliable estimates of sea salt concentrations were not reported during this period. We exclude coastal sites during this period, as sea salt can contribute significantly to bsp in coastal conditions of high RH due to its highly hygroscopic nature (Lowenthal and Kumar, 2006). We use only days with coincident mass and scatter measurements, and a minimum of 23 hourly measurements per day, to reduce influence of meteorological interference. Additionally, only sites with a minimum of 90 days of measurements are included in the analysis.
Figure 1 shows at the 28 sites used in this study the average hourly bsp at ambient RH and the average 24 h PM10 and PM2.5 measured between 2000 and 2010. Measured bsp values vary by a factor of 7, with scatter below 20 Mm−1 across the southwestern U.S. and scatter above 50 Mm−1 across the southeastern U.S. Measured PM10 concentrations vary by a factor of 3, with values below 6 µg m−3 in the west to above 14 µg m−3 in the southeast. Measured PM2.5 concentrations also vary by a factor of 3, with values below 3 µg m−3 in the west to above 9 µg m−3 in the southeast.
Figure 1Map of IMPROVE sites with collocated scatter (bsp) at 550 nm and ambient relative humidity, PM10, and PM2.5 measurements in North America between 2000 and 2010.
## 2.2 GEOS-Chem simulation
We simulate hourly PM2.5 and PM10 mass concentrations and particle scatter using the global chemical transport model GEOS-Chem (version 11-02, http://geos-chem.org, last access: 7 September 2017). The GEOS-Chem model is driven by assimilated meteorology from the Goddard Earth Observation System (GEOS MERRA-2, Gelaro et al., 2017) of the NASA Global Modeling and Assimilation Office (GMAO). Our simulation for North America is conducted at 2× 2.5 resolution over 47 vertical levels.
The majority of our analysis focuses on the accuracy of the GEOS-Chem parameterization of mass scattering efficiency based on optical parameters given in Table A1. These default aerosol physical and optical properties are defined by the Global Aerosol Data Set (GADS) (Koepke et al., 1997), as implemented by Martin et al. (2003), with modifications to dry size distributions (Drury et al., 2010) and dust mass partitioning (Ridley et al., 2012). After evaluating and improving this parameterization, implications are examined using the full GEOS-Chem simulation in Sect. 3.3.
GEOS-Chem simulates detailed aerosol-oxidant chemistry (Bey et al., 2001; Park et al., 2004). The aerosol simulation includes the sulfate–nitrate–ammonium system (Park et al., 2004), primary (Park et al., 2003; Wang et al., 2014) and secondary (Pye et al., 2010) carbonaceous aerosols, mineral dust (Fairlie et al., 2007, 2010; Zhang et al., 2013), and sea salt (Jaeglé et al., 2011). Organic matter (OM) is estimated from primary organic carbon (OC) using spatially and seasonally varying OM OC ratios at 0.1× 0.1 resolution (Philip et al., 2014b). The thermodynamic equilibrium model ISORROPIA-II (Fountoukis and Nenes, 2007), implemented by Pye et al. (2009), is used to calculate gas–aerosol partitioning. Total PM10 is calculated following van Donkelaar et al. (2010), but at 40 % RH here for consistency with the IMPROVE network gravimetric analysis in the range of 30 %–50 % RH (Solomon et al., 2014). Particle scatter and aerosol optical depth are calculated at modelled ambient RH based on dry species mass concentrations and aerosol physical and optical properties. The GEOS-Chem aerosol simulation has been extensively evaluated with observations of mass (van Donkelaar et al., 2015; Li et al., 2016), composition (Achakulwisut et al., 2017; Kim et al., 2015; Marais et al., 2016; Philip et al., 2014a; Ridley et al., 2017; Zhang et al., 2013), and scatter (Drury et al., 2010).
We conduct a simulation for the year 2006, to represent the period of greatest measurement density of collocated bsp and PM sites over North America. We archive model fields every hour over North America. We simulate PM10, PM2.5, and bsp, allowing for the comparison of model mass scattering efficiency coincident with that measured at IMPROVE network sites over the same time period over North America.
## 2.3 Determining mass scattering efficiency (αsp)
One method of determining mass scattering efficiencies from measurements involves bsp measurements and particle mass concentration measurements (Mmeas). Mass scattering efficiency of a given aerosol population can be defined as the ratio of particle scatter to mass.
$\begin{array}{}\text{(1)}& {\mathit{\alpha }}_{sp,\phantom{\rule{0.125em}{0ex}}\mathrm{meas}}=\frac{{b}_{sp,\phantom{\rule{0.125em}{0ex}}\mathrm{meas}}}{{M}_{\mathrm{meas}}}\end{array}$
Hourly mass scattering efficiencies are determined using collocated measurements of bsp and mass concentrations from the IMPROVE network, treating IMPROVE mass concentrations as constant over each 24 h sampling period. Total scatter is typically dominated by fine-mode aerosols, but in certain conditions coarse dust can also make a significant contribution (White et al., 1994). Thus, measured PM10 mass is used in the denominator of Eq. (1).
Multiple definitions of αsp exist. We define αsp operationally here based on optical measurements at ambient RH, and PM measurements at controlled RH (treated as 40 % RH for consistency with IMPROVE protocols prior to 2011). At 40 % RH, hygroscopic components of PM10 will have associated water, and thus measured PM10 mass is not treated as dry. We compare these measured αsp with calculated αsp based on species-specific mass scattering efficiencies (αGC, j) used in GEOS-Chem, constrained with mass concentrations (Mj) and PM10 mass measured by IMPROVE.
$\begin{array}{}\text{(2)}& {\mathit{\alpha }}_{sp,\phantom{\rule{0.125em}{0ex}}\mathrm{calc}}=\frac{{b}_{sp,\phantom{\rule{0.125em}{0ex}}\mathrm{calc}}}{{\mathrm{PM}}_{\mathrm{10},\phantom{\rule{0.125em}{0ex}}\mathrm{meas}}}=\frac{{\sum }_{j}{\mathit{\alpha }}_{\mathrm{GC},\phantom{\rule{0.125em}{0ex}}j}{M}_{j}}{{\mathrm{PM}}_{\mathrm{10},\phantom{\rule{0.125em}{0ex}}\mathrm{meas}}}\end{array}$
To reduce the impacts of meteorological variation on the comparison of measured and calculated mass scattering efficiency, we perform averages of hourly bsp, calc, bsp, meas, and PM10 over the entire sampling period at each IMPROVE site i. Equation (3) is then used to obtain average calculated and measured mass scattering efficiency at each site.
$\begin{array}{}\text{(3)}& {\mathit{\alpha }}_{sp,\phantom{\rule{0.125em}{0ex}}\mathrm{avg},\phantom{\rule{0.125em}{0ex}}i}=\frac{{b}_{sp,\phantom{\rule{0.125em}{0ex}}\mathrm{avg},\phantom{\rule{0.125em}{0ex}}i}}{{\mathrm{PM}}_{\mathrm{10},\phantom{\rule{0.125em}{0ex}}\mathrm{avg},\phantom{\rule{0.125em}{0ex}}i}}\end{array}$
Although the OPTEC open air nephelometer reduces truncation error compared with other nephelometers, truncation error can be significant for coarse particles (Hand and Malm, 2007; Lowenthal and Kumar, 2006). Thus our analysis below focuses on conditions dominated by fine-mode aerosols, and mechanisms affecting fine-mode aerosols.
Appendix A describes the calculation of mass scattering efficiency in more detail. This approach enables isolation of the mass scattering efficiencies used in GEOS-Chem from the species concentrations.
## 2.4 Introducing an alternate hygroscopic growth scheme
We examine for GEOS-Chem the use of a widely adopted alternate hygroscopic growth scheme, in which aerosol hygroscopic growth is defined by a single parameter, κ (Petters and Kreidenweis 2007, 2008, 2013). This representation of water uptake by aerosols was originally developed for supersaturated CCN conditions, but in recent years has been used extensively in subsaturated conditions (Dusek et al., 2011; Hersey et al., 2013).
The hygroscopic parameter κ is defined by
$\begin{array}{}\text{(4)}& \frac{\mathrm{1}}{{a}_{\mathrm{w}}}=\mathrm{1}+\mathit{\kappa }\frac{{V}_{\mathrm{d}}}{{V}_{\mathrm{w}}},\end{array}$
where Vd is dry particulate matter volume, Vw is the water volume, and aw is water activity (Petters and Kreidenweis, 2013), which is unity for secondary inorganic aerosols (SIA) and organic aerosols (OA). The diameter growth factor (GF $=D/{D}_{\mathrm{d}}$) can be expressed (Snider et al., 2016) as
$\begin{array}{}\text{(5)}& \mathrm{GF}={\left(\mathrm{1}+\mathit{\kappa }\frac{\mathrm{RH}}{\mathrm{100}-\mathrm{RH}}\right)}^{\mathrm{1}/\mathrm{3}},\end{array}$
where D is the wet aerosol radius and Dd is the dry aerosol radius. Typically, κ is in the range of 0.5–0.7 for SIA (Hersey et al., 2013; Kreidenweis et al., 2008; Petters and Kreidenweis, 2007) and 0–0.2 for OA (Duplissy et al., 2011; Kreidenweis et al., 2008; Rickards et al., 2013; Snider et al., 2016).
3 Results
## 3.1 Understanding the current representation of αsp
Figure 2 (left) shows measured vs. calculated mass scattering efficiency using GEOS-Chem default optical tables. Each point represents the average αsp over the entire sampling period at each IMPROVE site. A significant correlation (r=0.94) is apparent; however, a bias in αsp is evident. A positive correlation between average mass scattering efficiency and RH is apparent; sites with low average RH have low average αsp and vice versa. (Panel (b) of Fig. 2 is discussed below.)
Figure 2Average measured vs. calculated αsp at 550 nm at IMPROVE sites between 2000 and 2010 using GEOS-Chem default optical tables and revised optical tables. The colour of each point corresponds to the average relative humidity at the site. The 1:1 line is black. Slope, offset, and correlation coefficient are inset.
Figure 3Average measured vs. calculated αsp at 550 nm at IMPROVE sites between 2000 and 2010 using GEOS-Chem default and revised optical tables (Table A1) for measurements taken in 0 %–35 %, 35 %–65 %, and 65 %–95 % RH conditions. The 1:1 line is black. Slope, offset, and correlation coefficient are inset.
To further investigate the RH dependence of this bias, we separate our analysis of calculated αsp into three relative humidity groupings: 0 %–35 % (low), 35 %–65 % (mid), and 65 %–95 % (high). The IMPROVE data are divided among the RH groupings using IMPROVE measurements of hourly RH. Within each grouping, average calculated and measured mass scattering efficiencies are obtained for each site using Eq. (3). The blue dots in Fig. 3 show average calculated vs. measured αsp for each RH range. In the low RH case, a significant overestimation of mass scattering efficiency is apparent at most sites, with a bias of 82 % indicated by the slope. In the mid RH case, overestimation of αsp is less significant but still apparent, with a bias of 40 % indicated by the slope. At high RH, bias is weak.
Figure 4Average measured vs. calculated αsp (550 nm) at IMPROVE sites between 2000 and 2010 using GEOS-Chem default and revised optical tables for measurements taken in conditions dominated by secondary inorganic aerosols (SIA), organic aerosols (OA), fine dust, and PMcoarse (PM10–PM2.5). The 1:1 line is black. Slope, offset, and correlation coefficient are inset.
To further understand the source of the bias in calculated mass scattering efficiency, we now examine calculated αsp in conditions dominated by different aerosol types. Using IMPROVE measurements of 24 hr PM2.5 mass and speciation and PM10 mass, the IMPROVE data are grouped based on dominant aerosol type. Within each group, average calculated and measured mass scattering efficiency is obtained for each site using Eq. (3). Figure 4 shows in blue average measured vs. calculated αsp using default optical tables for conditions where measured PM2.5 is dominated (>60 %) by secondary inorganic aerosol, organic aerosol, and fine dust, as well as conditions where PM10 is dominated (>60 %) by PMcoarse (PM10–PM2.5). The scatterplot in the SIA-dominant case resembles the overall relationship shown in Fig. 2. αsp is overestimated at most sites, with significant correlation (r=0.88) and a bias evident in the offset of 0.70. Where OA is the dominant component of PM2.5, the slope is close to unity (1.02), but the large offset of 0.80 m2g−1 results in αsp being largely overestimated. Where dust is the dominant fine aerosol, correlation is significant (r=0.89) and mass scattering efficiency is accurately calculated at the vast majority of sites, despite a prominent outlier at a site in the Columbia River Gorge, Washington. The PMcoarse-dominant case shows significant correlation (r=0.88) and a slight tendency for overestimation of αsp. As this case is not independent of the other cases, this overestimation is likely linked to the overestimation in the OA- and SIA-dominant cases as demonstrated below.
These results indicate that the bias in calculated mass scattering efficiency arises mostly due to the representation of the physical and optical properties of secondary inorganic and organic aerosols. The following will focus on improving the representation of physical and optical properties of these two aerosol types.
## 3.2 Changing the physical properties of SIA and OA
Figure 5 shows mass scattering efficiency as a function of aerosol size for secondary inorganic (orange) and organic (blue) aerosols for dry aerosols (solid) and aerosols at 80 % RH (dashed lines) as calculated using a Mie algorithm (Mishchenko et al., 1999). Water uptake at 80 % RH for OA and SIA is calculated using default hygroscopic growth factors from GEOS-Chem. The uptake of water increases aerosol scatter, decreases aerosol density, and decreases the refractive index. The increase in aerosol scatter with increasing ambient RH drives the increase in αsp.
Figure 5Mass scattering efficiency (αsp) at 550 nm as a function of aerosol wet effective radius for organic aerosol and secondary inorganic aerosol. Solid lines show αsp for dry aerosol (RH =0 %); dashed lines show αsp for aqueous aerosols (RH =80 %). Points represent the default size in GEOS-Chem.
The points in Fig. 5 represent the current mass scattering efficiency values of OA and SIA in GEOS-Chem. For dry aerosols, αsp=4.4 m2 g−1 for OA and αsp=3.2 m2 g−1 for SIA. In a review of ground-based estimates of aerosol mass scattering efficiencies, Hand et al. (2007) found dry αsp values of 2.5 m2 g−1 for ammonium sulfate, 2.7 m2 g−1 for ammonium nitrate, and 3.9 m2 g−1 for particulate organic matter. These values suggest that the default optical tables in GEOS-Chem currently overestimate mass scattering efficiency of SIA and OA in dry conditions. This reaffirms the overestimation of αsp in dry conditions evident in panel (a) of Fig. 3. As aerosol size is the strongest determinant of dry mass scattering efficiency, we begin by examining the dry sizes of SIA and OA in GEOS-Chem.
The current dry sizes of SIA and OA in GEOS-Chem were informed by measurements from several aircraft campaigns over eastern North America during the summer of 2004 (Drury et al., 2010) as part of the International Consortium for Atmospheric Research on Transport and Transformation (ICARTT) (Fehsenfeld et al., 2006; Singh et al., 2006). Aerosol surface area and volume distributions fluctuate seasonally in the northeastern U.S., with summer maxima and winter minima (Stanier et al., 2004). We divide our analysis at low RH by season, in an effort to discern a seasonal pattern in the overestimation of αsp.
Figure 6Average measured vs. calculated αsp (550 nm) at IMPROVE sites between 2000 and 2010 using GEOS-Chem default and revised optical tables for measurements taken in dry conditions (RH <35 %) in winter, spring, summer, and fall. The 1:1 line is black. Slope, offset, and correlation coefficient are inset.
Figure 6 (blue) shows seasonal measured vs. calculated mass scattering efficiency in dry conditions using default optical tables (Table A1). Estimations of αsp are most accurate in the summer, consistent with the dry sizes chosen by Drury et al. (2010) which were informed by summertime size distribution measurements. The larger overestimation of αsp in all other seasons, most notably in winter, is consistent with the seasonality in aerosol size distributions observed by Stanier et al. (2004).
### 3.2.1 Efflorescence relative humidity
To address the overestimation of mass scattering efficiency in dry conditions illustrated in Figs. 3 and 6, we begin by accounting for efflorescence transitions in secondary inorganic aerosols. Efflorescence phase transitions are characterized by nucleation of the crystalline phase followed by rapid evaporation of water. Field measurements have found evidence for these transitions (Martin et al., 2008). The efflorescence relative humidity (ERH) of ammonium sulfate reported in several experimental studies ranges from 35 % to 40 % (Ciobanu et al., 2010). Laboratory tests have shown that mixtures of sulfate–nitrate–ammonium particles will undergo efflorescence when the ammonium sulfate fraction is high (Dougle et al., 1998; Martin et al., 2003). This condition is true at most global measurement sites, with the possible exception of Europe, where particles are nitrate rich (Martin et al., 2003).
We therefore define the hygroscopic growth factor for SIA as unity for RH ≤35 %, linearly increasing between 35 % and 40 % RH from unity to GF40 % (calculated by Eq. 5), and following the default (or κ-Kohler) growth curve for RH ≥40 %.
Incorporating an ERH for SIA and consequently inhibiting hygroscopic growth of SIA below 35 % RH significantly reduce the overestimation of mass scattering efficiency in dry conditions. In the case of default hygroscopic growth in GEOS-Chem, the overall dry bias in αsp is reduced from 82 % to 48 %.
### 3.2.2 Aerosol dry size
To address the remaining overestimation of mass scattering efficiency in dry conditions, we explore different dry sizes of secondary inorganic and organic aerosols. Effective variance may also be important (Chin et al., 2002), but given insufficient information to simultaneously constrain size and variance, we focus on size. Figure 7 shows the slope of the average measured vs. calculated αsp plot for RH <35 % for dry radii ranging from 0.048 to 0.074 µm at intervals of 0.001 µm, assuming SIA and OA have the same dry size. The slope of the best fit line acts as an indicator of the appropriate dry size for each season. Sensitivity tests exploring alternative error metrics (RMSE, MSE) yielded similar results. The slope decreases steadily as dry radius is decreased in all seasons. Using the dry radius which gives a slope of unity, we find that aerosols are largest in summer (r=0.067µm), smallest in winter (r=0.051µm), and in between in spring and fall (0.059 and 0.054 µm, respectively). The spring and summer radii are consistent with accumulation-mode size distribution measurements performed by Levin et al. (2009) in the spring and summer of 2006. Averaging the sizes from all four seasons results in an annual representative dry radius of 0.058 µm. This annual radius is smaller than the GEOS-Chem default sizes of SIA and OA that were informed by summertime measurements alone (Drury et al., 2010).
Figure 7Slope of measured vs. calculated αsp plot vs. dry geometric mean aerosol radius, by season. Winter (DJF) is in blue, spring (MAM) in red, summer (JJA) in green, and fall (SON) in orange. The line slope =1 is shown in black. Numbers in the legend represent the dry radius for which the slope =1 for each season.
Figure 6 (red) shows seasonal measured vs. calculated αsp in dry conditions using a new representative annual geometric mean radius of 0.058 µm for SIA and OA. This change in geometric mean radius reduces the overestimation of αsp in all seasons, with the largest improvements in fall (slope decreases from 1.84 to 1.17) and winter (slope decreases from 1.94 to 1.20). Changes in correlation are minor. For the remainder of the analysis, this new dry radius of 0.058 µm is implemented for SIA and OA.
### 3.2.3 Aerosol hygroscopicity
We now examine the implementation of the widely adopted κ-Kohler hygroscopic growth scheme described in Sect. 2.4. A range of measured κ values for SIA (κs) and OA (κo) exist in the literature. We explore the range of possible κ values, using the slope of the measured vs. calculated αsp plot as an indicator of the appropriate values.
Figure 8Slope of measured vs. calculated αsp plot as a function of the κ of secondary inorganic aerosols (κs, a) and the κ of organic aerosols (κo, b). The line slope =1 is shown in black. κs and κo values for which slope =1 are inset.
Figure 9Hygroscopic growth factor curves for secondary inorganic aerosols (SIA, a) and organic aerosols (OA, b). GADS (Global Aerosol Data Set) hygroscopic growth from empirical data and κ-Kohler hygroscopic growth are shown for both SIA and OA. For ammonium sulfate, AIM (Aerosol Inorganic Model) hygroscopic growth at T=298 K (Wexler and Clegg, 2002) and laboratory hygroscopic growth with a deliquesence point of RH =80 % (Wise et al., 2003) are also shown.
Figure 8 shows the slope of the measured vs. calculated αsp plot for κ values for SIA (κs) ranging from 0.5 to 0.7 and for OA (κo) ranging from 0.08 to 0.20. Slope increases steadily as κs and κo increase. A slope of unity identifies representative values of κs=0.61 and κo=0.10. These values are in the middle of the range of measured κ values (Duplissy et al., 2011; Hersey et al., 2013; Kreidenweis et al., 2008; Petters and Kreidenweis, 2007; Rickards et al., 2013).
Figure 9 shows the diameter growth factor as a function of relative humidity following κ-Kohler theory, as well as GADS hygroscopic growth for both SIA and OA used in the default GEOS-Chem model. Hygroscopic growth from the Aerosol Inorganic Model (AIM) at T=298 K (Wexler and Clegg, 2002) and laboratory measurements (Wise et al., 2003) are also shown for ammonium sulfate (Snider et al., 2016). The GADS hygroscopic growth schemes used in the default GEOS-Chem simulation are characterized by larger growth at low RH and smaller growth at high RH for both secondary inorganic and organic aerosols. The κ-Kohler scheme exhibits greater consistency with both AIM and laboratory hygroscopic growth for SIA.
Using the revised dry size of 0.058 µm and the κ-Kohler theory of hygroscopic growth, we calculate revised physical and optical properties for SIA and OA over a range of RH values. Table A1 contains geometric mean radius, extinction efficiency, and single scattering albedo for the revised optical tables at eight relative humidity values.
Figure 2 (right) shows the measured vs. calculated mass scattering efficiency using these revised optical tables for SIA and OA. The overestimation of mass scattering efficiency has been eliminated with these revised aerosol properties, with a slope of 1.00 and an offset of 0.09. Correlation remains significant at r=0.96.
Figure 4 (red) shows measured vs. calculated αsp in conditions dominated by different aerosol types using the revised optical tables. The overestimation of αsp in SIA-dominant conditions using the default optical tables has been eliminated, with a slope of 1.03 and a decreased offset (0.70 to 0.1). The large overestimation of αsp that was apparent in OA-dominant conditions has been reduced by a factor of 2. αsp remains accurately estimated at the majority of dust-dominant sites, with the outlier at the Columbia River Gorge site in Washington still skewing the best fit line. The slight overestimation of αsp that was present in the PMcoarse-dominant case using default optical tables has been eliminated using the revised tables (offset 0.33 to 0.03). Slight increases in correlation coefficients are apparent in all cases except for the SIA-dominant case, where it decreased by 0.02.
Figure 3 (red) shows measured vs. calculated αsp using revised optical tables. The overestimation in αsp has been significantly reduced in the low RH case (slope =1.82 to slope =1.09) and in the mid RH case (slope =1.40 to slope =1.01) compared to when default optical tables were used. The slight overestimation in high RH conditions present in the default case has also been reduced, as shown by the decreased offset (0.90 to 0.71).
## 3.3 Changes in GEOS-Chem-simulated αsp
Here, we examine how these changes to aerosol properties impact both GEOS-Chem simulation of mass scattering efficiency over North America and the fit between modelled and measured αsp at IMPROVE sites. These simulations rely on GEOS-Chem simulations of aerosol composition using GEOS RH fields.
Figure 10 shows the relative and absolute change in mass scattering efficiency when switching from the default to revised optical tables. Continental mean αsp increased by 16 %. Increases in αsp range from 25 % to 45 % in northeastern regions of North America, corresponding to an increase of 1.5–3.5 m2 g−1. These larger changes reflect the higher RH and SIA fractions. Decreases in αsp of up to 15 % or −0.5 m2g −1 are found in the southwest where RH is low and mineral dust dominates.
Figure 10Average relative and absolute change in GEOS-Chem mass scattering efficiency over North America for the year 2006 after implementing revised optical tables for secondary inorganic and organic aerosols.
Figure 11 shows GEOS-Chem annual average mass scattering efficiency using default (top) and revised (bottom) optical tables over North America for the year 2006. The overlaying circles represent average measured αsp at IMPROVE network sites for the year 2006, and the outer rings show the coincident simulated αsp for each site. We exclude sites within 1 of the coast, where sea salt affects αsp, as well as sites where elevation differs from average gridbox elevation by more than 1500 m. These criteria result in a decrease from 24 to 19 in the number of sites available for the analysis in 2006.
Figure 11GEOS-Chem annual average mass scattering efficiency (at 550 nm) for the year 2006 using default and revised sizes and hygroscopicity for secondary inorganic and organic aerosols. Overlaying inner circles represent annual averages of αsp at IMPROVE network sites for the year 2006. Outer rings represent coincident average simulated αsp.
Figure 12Coincident simulated vs. measured average mass scattering efficiency at 550 nm for the year 2006, using default and revised optical tables. Slope, offset, and correlation coefficient are inset.
Using default optical tables, simulated continental mean αsp is 5.4 m2 g−1. A maximum αsp of 10 m2 g −1 occurs in British Columbia, and a minimum αsp of 1.7 m2 g−1 occurs in the southwestern United States. Using revised optical tables, simulated continental mean αsp is 6.3 m2 g−1, with a maximum of 12.5 m2 g−1 in the northwest and a minimum of 1.5 m2 g−1 in the southwest. The elevated mass scattering efficiencies in the northwest can be attributed in part to the high average RH in this region of 83 %.
Figure 12 (left) shows coincident measured vs. simulated mass scattering efficiency at the 19 IMPROVE sites, using default optical tables. Correlation is significant (r=0.88), but a bias in simulated αsp is apparent (slope =0.83). Simulated αsp is notably biased low at sites in the southeastern United States where average αsp is largest, and simulated αsp is notably biased high at sites in the southwestern United States where average mass scattering efficiency is lowest. Sites with the lowest average RH correspond to those with the lowest average mass scattering efficiency and vice versa. The tendency of mass scattering efficiency to be overestimated at low RH reflects the tendency that was originally seen in Fig. 4.
Figure 12 (right) shows coincident measured vs. simulated αsp using revised optical tables. Correlation remains significant (r=0.89), and a decrease in bias is evident from the increase in slope (0.83 to 0.93) and decrease in offset (0.47 to 0.08). Most sites now lie closer to the 1:1 line. The overestimation of simulated αsp in the southwest, where RH is low, has been reduced or eliminated at all sites.
## 3.4 Comparison with AERONET measurements
Appendix B investigates changes to simulated AOD, and compares measured and simulated AOD at AERONET sites. Although large relative increases upwards of 60 % in average AOD are evident in large parts of northern high latitudes where absolute AOD is small, absolute AOD generally changes by less than 0.1 (Fig. B1). Comparisons with AERONET AOD reveal that the revised optical properties slightly improve the simulation of AOD worldwide (slope decreases from 1.08 to 1.00) despite the large influence of other factors (e.g. ambient aerosol concentrations) upon AOD.
4 Conclusions
The current representation of mass scattering efficiency in the GEOS-Chem global chemical transport model was evaluated using collocated ground-based measurements of particle mass, speciation, scatter, and relative humidity from the IMPROVE network.
Calculated mass scattering efficiency had a positive bias using default physical and optical properties used in the GEOS-Chem model. This bias was most significant when PM2.5 mass was dominated by secondary inorganic (SIA) or organic aerosols (OA). Mass scattering efficiency in PM2.5 dust and coarse particulate matter dominant conditions was accurately represented at the majority of IMPROVE sites.
Relative humidity played an important role in the severity of the bias in mass scattering efficiency. Mean αsp was overestimated by 82 % in dry conditions (RH <35 %). This bias was largest in the winter (94 %) and smallest in the summer (27 %). Implementing an efflorescence relative humidity for SIA and thus inhibiting hygroscopic growth below 35 % RH decreased the dry bias by 34 %. An annual representative dry geometric mean radius of 0.058 µm for SIA and OA decreased the dry mass scattering efficiency of these aerosols, and subsequently further reduced the bias in dry conditions to 9 %.
κ-Kohler theory was implemented for the hygroscopic growth of SIA and OA, which is characterized by smaller growth factors at low RH and larger growth factors at high RH compared to default growth factors in GEOS-Chem. κ values of 0.61 for SIA and 0.10 for OA eliminated the overall bias in calculated mass scattering efficiency.
These changes to SIA and OA optical tables resulted in a continental mean increase in GEOS-Chem-simulated mass scattering efficiency of 16 %. Northeastern regions of North America exhibited the largest increases (25 %–45 %) due to high RH and SIA fractions, while southwestern regions of the continent exhibited decreases in αsp of up to 15 % due to low RH and high dust fractions. These changes to the GEOS-Chem optical tables improved the fit between measured and simulated mass scattering efficiency at IMPROVE sites, reflected in the changes to the slope (0.83 to 0.93) and the offset (0.47 to 0.08).
Future work should examine the implications of these changes for satellite-derived estimates of fine particulate matter that depend on the relationship of AOD with PM2.5. Future work should also expand analysis of the representation of mass scattering efficiency for other years, and by incorporating measurements from other ground-based measurement networks such as the Surface PARTiculate MAtter network (SPARTAN), which provides measurements of particulate mass, speciation, and scatter in populated regions worldwide (Snider et al., 2015, 2016). Such comparisons may also be useful to evaluate and improve prognostic simulations of aerosol size (Mann et al., 2010; Spracklen et al., 2005; Trivitayanurak et al., 2008; Yu and Luo, 2009). Representation of particle RH history may also be important (Wang et al., 2008).
Data availability
Data availability.
IMPROVE network data for 2000–2010 can be accessed at http://vista.cira.colostate.edu/Improve/improve-data/ (last access: 3 October 2018). The GEOS-Chem chemical transport model used here is available at http://www.geos-chem.org (last access: 7 September 2017).
Appendix A
## bsp and αsp calculations in GEOS-Chem
In GEOS-Chem, surface-level bsp is calculated using model particle mass concentrations and local relative humidity, as well as predefined mass densities and aerosol optical properties for each aerosol component following
$\begin{array}{}\text{(A1)}& {b}_{\mathrm{sp}}={\sum }_{\mathrm{species},\phantom{\rule{0.125em}{0ex}}i}\frac{\frac{\mathrm{3}}{\mathrm{4}}\cdot {\left(\frac{{R}_{w,\phantom{\rule{0.125em}{0ex}}i}}{{R}_{\mathrm{d},\phantom{\rule{0.125em}{0ex}}i}}\right)}^{\mathrm{2}}\cdot {M}_{\mathrm{d},\phantom{\rule{0.125em}{0ex}}i}\cdot {Q}_{w,\phantom{\rule{0.125em}{0ex}}i}\cdot {\mathrm{SSA}}_{w,\phantom{\rule{0.125em}{0ex}}i}}{{\mathit{\rho }}_{\mathrm{d},\phantom{\rule{0.125em}{0ex}}i}\cdot {R}_{\mathrm{d},\phantom{\rule{0.125em}{0ex}}i}},\end{array}$
where ρd is the dry particle mass density, Rw is the effective radius (defined as the ratio of the third to second moments of an aerosol size distribution), Rd is the dry effective radius, Md is the dry surface-level mass concentration, Qw is the extinction efficiency, and SSAw is the single scattering albedo. Parameters with subscript w indicate values at ambient RH. Species included in this calculation are ${\mathrm{SO}}_{\mathrm{4}}^{\mathrm{2}-}$, ${\mathrm{NH}}_{\mathrm{4}}^{+}$, ${\mathrm{NO}}_{\mathrm{3}}^{-}$, BC, OM, and fine and coarse dust and sea salt.
Dividing Eq. (A1) by total surface-level PM10 results in the following equation for mass scattering efficiency:
$\begin{array}{}\text{(A2)}& {\mathit{\alpha }}_{\mathrm{sp}}=\frac{{B}_{\mathrm{sp}}}{{\mathrm{PM}}_{\mathrm{10}}}=\frac{{\sum }_{\mathrm{species},\phantom{\rule{0.125em}{0ex}}i}\frac{\frac{\mathrm{3}}{\mathrm{4}}\cdot {\left(\frac{{R}_{w,\phantom{\rule{0.125em}{0ex}}i}}{{R}_{\mathrm{d},\phantom{\rule{0.125em}{0ex}}i}}\right)}^{\mathrm{2}}\cdot \frac{{M}_{\mathrm{d},\phantom{\rule{0.125em}{0ex}}i}}{{\mathrm{PM}}_{\mathrm{10}}}\cdot {Q}_{w,\phantom{\rule{0.125em}{0ex}}i}\cdot {\mathrm{SSA}}_{w,\phantom{\rule{0.125em}{0ex}}i}}{{\mathit{\rho }}_{\mathrm{d},\phantom{\rule{0.125em}{0ex}}i}\cdot {R}_{\mathrm{d},\phantom{\rule{0.125em}{0ex}}i}}}{{\mathrm{PM}}_{\mathrm{10}}}.\end{array}$
The effective radius, extinction efficiency, and single scattering albedo in Eqs. (A1) and (A2) are obtained from GEOS-Chem optical tables for the ambient RH values measured by IMPROVE. Dry mass density ρd is specified for each aerosol species in GEOS-Chem (Table A2). Md, i and PM10 are obtained from IMPROVE network measurements of aerosol mass and composition. αsp calculated by Eq. (A2) is compared to αsp directly measured by the IMPROVE network.
Mass scattering efficiency is dependent on particle density, refractive index, and particle size. Mass scattering efficiency is typically most dependent on aerosol size, which is dictated by both the dry size distribution chosen to represent a given aerosol species and the hygroscopic growth scheme used to represent aerosol water uptake for hydrophilic species.
## A2 Incorporating IMPROVE network measurements
The IMPROVE network measures every 3 days PM2.5 mass and speciation and PM10 mass. The IMPROVE particle sampler consists of four independent modules with separate inlets and pumps. The first three modules (A, B, and C) collect only fine particulate matter (PM2.5), while the fourth module (D) collects both fine and coarse particles (PM10). Module A collects PM2.5 on a Teflon filter, which undergoes gravimetric analysis for total PM2.5 mass and X-ray florescence for elemental concentrations (including Al, Si, Ca, Fe, and Ti). The nylon filter in module B undergoes ion chromatography analysis for ${\mathrm{SO}}_{\mathrm{4}}^{\mathrm{2}-}$, ${\mathrm{NO}}_{\mathrm{3}}^{-}$, ${\mathrm{NO}}_{\mathrm{2}}^{-}$, and Cl. Module C contains a quartz filter that is analysed for organic and elemental carbon via thermal optical reflectance. The Teflon filter in module D undergoes gravimetric analysis for PM10 mass (Malm et al., 1994, 2004). Prior to gravimetric analysis, filters A and D undergo equilibration at 30 %–50 % RH and 20–25 C for several minutes (Solomon et al., 2014).
The GEOS-Chem model partitions OM into hydrophilic and hydrophobic fractions, so the same is done for OM measured by IMPROVE to enable isolation of mass scattering efficiency in our comparisons. OM in remote regions tends to be highly oxidized, and oxidation level of organics has been shown to positively correlate with hygroscopicity (Duplissy et al., 2011; Jimenez et al., 2009; Ng et al., 2010). We treat measured OM as 90 % hydrophilic, due to the rural nature of IMPROVE sites. EC is treated as 50 % hydrophilic. As speciation of coarse material is unavailable, we treat all coarse material as crustal in origin, an assumption that may break down at coastal sites. We partition fine and coarse dust measured by the IMPROVE network into the GEOS-Chem size bins using the dust particle size distribution (PSD) described by Zhang et al. (2013).
Table A1Default and revised aerosol size and optical properties for secondary inorganic aerosols (SIA) and organic aerosols (OA) at 550 nm at eight relative humidity values. Columns indicate geometric mean radius (rg), effective radius (reff), extinction efficiency (Q), and single scattering albedo (SSA). κs and κo represent the hygroscopic growth parameters for SIA and OA, respectively.
Table A2Current microphysical properties of each aerosol species in GEOS-Chem. rg represents the dry geometric mean radius (µm) and σ the geometric standard deviation of the lognormal size distributions assumed for each species. ρd represents the dry mass densities of each species (g cm−3).
Appendix B
The Aerosol Robotics Network (AERONET) is a long-term network of ground-based sun photometers that provides continuous, cloud-screened measurements of aerosol optical depth (AOD) at several fixed wavelengths in the visible and near infrared (Holben et al., 1998). The calculation of AOD in GEOS-Chem is performed using simulated mass concentrations of aerosol species and mass extinction efficiencies, summed over all vertical layers. Our analysis of mass scattering efficiency can therefore be extended globally by comparing GEOS-Chem-calculated AOD to AOD measured at AERONET sites. During our simulation year of 2006, AERONET consisted of 231 sites across the globe.
Here we examine how the changes to SIA and OA properties impact GEOS-Chem simulated AOD globally. Figure B1 shows the relative (top) and absolute (bottom) changes in AOD. Global mean AOD increases by 19 %. Relative changes in AOD are most pronounced in northern regions where mean relative humidity is high, with increases in simulated AOD ranging from 50 % to 90 %. Decreases in AOD between 0 % and 20 % are present in most of the Southern Hemisphere, in part due to the lower average RH. Absolute changes in AOD show a similar tendency, with slight increases in AOD of up to 0.2 in northern regions, and slight decreases of up of −0.09 in southern regions. An exception to this is seen over parts of China, where AOD increases by 0.5 due to the elevated SIA and OA concentrations.
Figure B2 shows coincident measured (inner circles) and simulated (outer rings) AOD for the year 2006 using default optical tables (top) and revised optical tables (bottom). We exclude sites within 1 of the coast, as well as sites where elevation differs from average gridbox elevation by more than 1500 m. We also exclude sites where average PM2.5 is dominated by dust (dust PM2.5>0.6), to focus on the representation of the optical properties of SIA and OA. Across the globe, we see that AOD is both overestimated and underestimated. AOD is overestimated at most sites in Africa, with the most notable overestimation at the site in Nigeria. AOD is moderately overestimated at sites in Australia. Underestimation of AOD occurs at most sites in South America, as well as at sites in southern North America and southern Asia.
Figure B3 shows coincident measured vs. simulated AOD at AERONET sites for default (left) and revised (right) optical tables. The correlation coefficient (r=0.80 to r=0.78) changes insignificantly, while the slope decreases from 1.08 to 1.00 when switching to the revised optical tables. In summary, the revised optical properties developed for North America slightly improve the representation of AOD at the global scale, despite the large influence of other factors (e.g. ambient aerosol concentrations and composition) upon AOD.
Figure B1Average relative and absolute change in GEOS-Chem aerosol optical depth at 550 nm globally for the year 2006 after implementing revised optical tables for SIA and OA.
Figure B2Global comparison for the year 2006 of AERONET AOD (inner circles) and GEOS-Chem coincident simulated AOD (outer rings) using default optical tables.
Figure B3Coincident simulated vs. measured AOD at 550 nm at AERONET sites for the year 2006, using default and revised sizes and hygroscopicity. Slope, offset, and correlation coefficient are inset. The 1:1 line is shown in black.
Author contributions
Author contributions.
RNCL and RVM conceived the study. RNCL conducted the analysis. RNCL and RVM prepared the paper.
Competing interests
Competing interests.
The authors declare that they have no conflict of interest.
Acknowledgements
Acknowledgements.
Research described in this article was conducted under contract to the Health Effects Institute (HEI), an organization jointly funded by the United States Environmental Protection Agency (EPA) (Assistance Award No. R-82811201) and certain motor vehicle and engine manufacturers. The contents of this article do not necessarily reflect the views of HEI, or its sponsors, nor do they necessarily reflect the views and policies of the EPA or motor vehicle and engine manufacturers.
IMPROVE is a collaborative association of state, tribal, and federal agencies, and international partners. The US Environmental Protection Agency is the primary funding source, with contracting and research support from the National Park Service. The Air Quality Group at the University of California, Davis is the central analytical laboratory, with ion analysis provided by the Research Triangle Institute, and carbon analysis provided by the Desert Research Institute.
We thank Environment and Climate Change Canada for providing nephelometer data at the site in Egbert, Ontario.
Edited by: Ashu Dastoor
Reviewed by: two anonymous referees
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Singh, H. B., Brune, W. H., Crawford, J. H., Jacob, D. J., and Russell, P. B.: Overview of the summer 2004 Intercontinental Chemical Transport Experiment-North America (INTEX-A), J. Geophys. Res.-Atmos., 111, 0148–0227, https://doi.org/10.1029/2006JD007905, 2006.
Snider, G., Weagle, C. L., Martin, R. V., van Donkelaar, A., Conrad, K., Cunningham, D., Gordon, C., Zwicker, M., Akoshile, C., Artaxo, P., Anh, N. X., Brook, J., Dong, J., Garland, R. M., Greenwald, R., Griffith, D., He, K., Holben, B. N., Kahn, R., Koren, I., Lagrosas, N., Lestari, P., Ma, Z., Vanderlei Martins, J., Quel, E. J., Rudich, Y., Salam, A., Tripathi, S. N., Yu, C., Zhang, Q., Zhang, Y., Brauer, M., Cohen, A., Gibson, M. D., and Liu, Y.: SPARTAN: a global network to evaluate and enhance satellite-based estimates of ground-level particulate matter for global health applications, Atmos. Meas. Tech., 8, 505–521, https://doi.org/10.5194/amt-8-505-2015, 2015.
Snider, G., Weagle, C. L., Murdymootoo, K. K., Ring, A., Ritchie, Y., Stone, E., Walsh, A., Akoshile, C., Anh, N. X., Balasubramanian, R., Brook, J., Qonitan, F. D., Dong, J., Griffith, D., He, K., Holben, B. N., Kahn, R., Lagrosas, N., Lestari, P., Ma, Z., Misra, A., Norford, L. K., Quel, E. J., Salam, A., Schichtel, B., Segev, L., Tripathi, S., Wang, C., Yu, C., Zhang, Q., Zhang, Y., Brauer, M., Cohen, A., Gibson, M. D., Liu, Y., Martins, J. V., Rudich, Y., and Martin, R. V.: Variation in global chemical composition of PM2.5: emerging results from SPARTAN, Atmos. Chem. Phys., 16, 9629–9653, https://doi.org/10.5194/acp-16-9629-2016, 2016.
Solomon, P. A., Crumpler, D., Flanagan, J. B., Jayanty, R. K. M., Rickman, E. E., and McDade, C. E.: U.S. National PM2.5 Chemical Speciation Monitoring Networks-CSN and IMPROVE: Description of Networks, J. Air Waste Manage., 64, 1410–1438, https://doi.org/10.1080/10962247.2014.956904, 2014.
Spracklen, D. V., Pringle, K. J., Carslaw, K. S., Chipperfield, M. P., and Mann, G. W.: A global off-line model of size-resolved aerosol microphysics: I. Model development and prediction of aerosol properties, Atmos. Chem. Phys., 5, 2227–2252, https://doi.org/10.5194/acp-5-2227-2005, 2005.
Stanier, C. O., Khlystov, A. Y., and Pandis, S. N.: Ambient aerosol size distributions and number concentrations measured during the Pittsburgh Air Quality Study (PAQS), Atmos. Environ., 38, 3275–3284, https://doi.org/10.1016/j.atmosenv.2004.03.020, 2004.
Tao, J., Zhang, L., Ho, K., Zhang, R., Lin, Z., Zhang, Z., Lin, M., Cao, J., Liu, S., and Wang, G.: Impact of PM2.5 chemical compositions on aerosol light scattering in Guangzhou – the largest megacity in South China, Atmos. Res., 135–136, 48–58, https://doi.org/10.1016/j.atmosres.2013.08.015, 2014.
Titos, G., Foyo-Moreno, I., Lyamani, H., Querol, X., Alastuey, A., and Alados-Arboledas, L.: Optical properties and chemical composition of aerosol particles at an urban location: An estimation of the aerosol mass scattering and absorption efficiencies, J. Geophys. Res.-Atmos., 117, 1–12, https://doi.org/10.1029/2011JD016671, 2012.
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van Donkelaar, A., Martin, R. V., Brauer, M., Kahn, R., Levy, R., Verduzco, C., and Villeneuve, P. J.: Global estimates of ambient fine particulate matter concentrations from satellite-based aerosol optical depth: Development and application, Environ. Health Perspect., 118, 847–855, https://doi.org/10.1289/ehp.0901623, 2010.
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# reddit is a platform for internet communities
[–] 1 point2 points (0 children)
Sorry if my language was used loosely.
The fields are always there, but their occupation numbers are driven to zero during inflation (they are rapidly depopulated). Just like the electromagnetic field always exists, even if there is vanishing electric/magnetic field; that just means that the value that these vectors take on is zero, but a zero vector still "exists".
The higgs field does not have a coupling constant like G or the fine structure constant (alpha) associate with it; it's not a "gauge boson". In the standard model, how strongly some other field (say the left-handed electron field) couples to the higgs depends on a "Yukawa-type coupling constant", and this constant is different for each field that couples to the higgs. It is not a "universal" coupling constant like G or alpha. In any event, those constants are not supposed to depend on time, so they were supposedly the same back in inflation as they are today.
[–] 3 points4 points (0 children)
The current understanding is that inflation happened first, followed by "reheating"[1] during which the inflaton decayed (or through parametric resonance or other processes) produced standard model fields or other fields which in turn produced standard model fields. Anyway, the matter-antimatter asymmetry is to be explained during this thermal radiation-dominated phase, not during inflation, as I explain below.
First of all, the condition for inflation (exponential expansion) extremely rapidly dilutes the universe into a non-thermal state, so you still need to reheat it afterwards, so whatever you think is going on during inflation can't explain something like the matter-antimatter asymmetry; you still need some (p)reheating phase after inflation to make the universe thermal and radiation dominated, and then you still need to explain matter-antimatter asymmetry during that phase.
Secondly, during inflation, the effective equation of state parameter w has to be less than -1/3 in order for the expansion to be accelerated (for a potential-dominated slowly rolling inflaton, you get an effective equation of state parameter w~=-1). Compare this with what you get for a radiation dominated thermal state, w=+1/3. If you want loads of matter-antimatter annihilation, you're talking about a high density, high temperature state, and that's just not the right physics for inflating the universe. It has the wrong effective EOS parameter; because it has an ordinary stress-energy tensor, it makes the expansion of the universe decelerate, not accelerate.
[1] "Reheating" is actually a pretty bad name, because there's no reason to believe that the universe was hot beforehand.
[–] 2 points3 points (0 children)
There has been a bunch of back-and-forth in the astro and particle communities. Resonaances collected some thoughts here.
My (cautionary astrophysicist) TLDR for the situation is that it's probably gastrophysics which we don't have a complete handle on.
[–] 1 point2 points (0 children)
I stand corrected.
[–] 0 points1 point (0 children)
The numerical scheme need not converge, or may take prohibitively long to do so if it does; also, the problem states to terminate after a certain condition is met. Therefore FixedPoint is the wrong solution to this problem.
[–] 0 points1 point (0 children)
See NestWhile. But also just use FindRoot, unless this is a programming exercise.
[–] 3 points4 points (0 children)
% In the preamble:
\newcommand{\artname}[1]{\textit{#1}\xspace}
% Then later in the document
... like Duchamp's magnificent \artname{Fountain}.
This separates presentation from meaning. If you later decide you want to change the styling for all the artwork names, you just modify the definition for \artname.
Use \xspace (package xspace) at the end of the macro to get correct spacing with punctuations, footnotes, etc.
[–] 1 point2 points (0 children)
No. See the more detailed explanation I gave here. Or see the review article I referenced.
[–] 4 points5 points (0 children)
The current thinking is that the probability is extremely high. It's believed to be a common occurrence in the universe. There are important dynamical effects at play that should drive the two SMBHs together to merger in the long run. The first is dynamical friction, which leads to mass segregation. This happens over a very long time scale (the relaxation time is long compared to the dynamical time). Once the two SMBHs form a binary, even if it is very loosely bound, 3-body interactions will "harden" the binary by giving up angular momentum to stars in the "loss cone". Finally, once the binary is sufficiently tight, gravitational radiation losses will become more important and drive the two SMBHs to a merger.
There are lots of review articles on this; here is one.
[–] 0 points1 point (0 children)
If you want to understand differential geometry and you already understand the paradigm of functional programming, then I can point you toward Functional Differential Geometry (a relatively up-to-date preprint is freely available here; notice that one of the authors is Gerry Sussman, who coauthored scheme and SICP). Not strictly related to pure math, but if you want to learn classical mechanics there is also SICM; you could kind of learn symplectic geometry from this, but not really.
[–] 3 points4 points (0 children)
Your question is probably more appropriate for /r/math than /r/AskScienceDiscussion. I am interpreting your question as how to start with an embedding of the unknot and show that it is in fact the unknot. In principle, all you need are Reidemesiter moves. For actual algorithms, I direct you to this mathoverflow question.
[–] 2 points3 points (0 children)
There is no problem if magnetic monopoles always come in pairs, and the monopoles have the appropriate magnetic charge; then you can say that two monopoles of opposite magnetic charge are connected by a Dirac string. The phase which is accrued by taking a small loop around the string once (holonomy) is related to the magnetic charges. If they are quantized appropriately then the phase difference will be a multiple of 2pi, and the Dirac string would be unobservable.
The more complicated modern view of magnetic monopoles comes from spontaneous symmetry breaking, in which a magnetic monopole is a topological monopole ('t Hooft-Polyakov monopole) when the appropriate homotopy group of the vacuum manifold G/H is nontrivial.
[–] 7 points8 points (0 children)
See APL Syntax rules. * and - will apply to as much as possible to the right, whereas ¯ will apply to exactly one piece of data on the right. Thus
2 -3 4 5
¯1 ¯2 ¯3
2 *3 4 5
8 16 32
but
2 ¯3 4 5
2 ¯3 4 5
[–] 2 points3 points (0 children)
Think about how you'd parse the subtraction of two vectors if you used the same operator for subtraction and negation ...
[–] 2 points3 points (0 children)
What you're saying is true in Newtonian gravity, but breaks down in the regime where GR becomes important—highly compact objects and orbit velocities approaching the speed of light. In GR, a binary system will emit gravitational waves, something that doesn't exist in Newtonian gravity. Gravitational waves are sourced by any system with a time varying mass quadrupole moment (or higher moments). GWs carry away energy and angular momentum. Thus an orbit in GR will shrink, which doesn't happen in Newtonian gravity. While this affects all orbits, there is a power-law dependence on the orbital velocity over the speed of light, and the GW luminosity has a factor of G/c5 in front of it ... so it's quite small, except for systems like this one: highly compact objects, and very large orbital velocities. This phenomenon is actually observed in millisecond pulsar systems that have orbital velocities of v/c~O( 10-4 ) (see e.g. the Hulse-Taylor binary, which won a Nobel prize for the radio astronomers who discovered it).
[–] 5 points6 points (0 children)
Try The Shop (312 E Seneca St)
[–] 21 points22 points (0 children)
This is a result from ray tracing in a BH-BH merger simulation. The researchers who are responsible for this work did an AMA a few weeks ago. See that link for high quality movies, a link to the paper, and explanations.
[–] 1 point2 points (0 children)
I think the general principle is right, but when you're concerned about something on the scale of balloons, you'd better worry about weather, not just the first few moments of the mass distribution.
[–] 1 point2 points (0 children)
1. This can't happen, because you'd violate conservation of energy.
2. There are different parts to gravity (as described by general relativity). There is a wave-like part, which we call "dynamical" (this is a technical thing that has to do with the type of equation of motion it satisfies); and there is the Coulomb-like part, which is "non-dynamical" (technically it satisfies a constraint equation instead of an evolution equation). The part of gravity which is responsible for the Sun holding the planets in orbit is the latter non-dynamical bit. In a technical sense, the non-dynamical bit is instantaneous, but carries no information (and so doesn't violate causality). Again, not a problem because you can't suddenly make the Sun disappear.
[–] 15 points16 points (0 children)
Yes, the atmosphere is also affected, but other effects are more important than the tidal bulge. The most obvious effect is that the atmosphere is warmer on the day side (but lagging by a few hours from subsolar), so it is less dense, and hence there is a global dipole moment. We also have the Hadley cells, Ferrel cells, and polar cells which contribute to the asphericity of the atmosphere—e.g. the atmosphere is "taller" around the equator than if the Earth wasn't being heated by the Sun.
[–] 4 points5 points (0 children)
[–] 17 points18 points (0 children)
Normalizing in the [; \ell_\infty ;] norm.
[–] 2 points3 points (0 children)
In a sense, yes, you're right. The modern description of electromagnetism is through the "curvature of a U(1) bundle". The other force-carrying particles (W, Z, gluons) are associated with the curvature of other bundles (U(1)xSU(2) for electroweak, SU(3) for strong). The particles which "talk to" (are affected by) those corresponding force-carriers are "charged under" the associated symmetry group.
Restated: Gravity is a type of curvature (metric curvature or curvature of the tangent bundle), and the other force carriers are also a type of curvature (curvature of a fiber bundle). Gravity is universal because everything "talks to" the metric, whereas the other forces only talk to certain particles because different particles have different couplings (charges) with those different curvatures.
[–] 8 points9 points (0 children)
As best as we can tell, the universe is homogeneous and isotropic. There is no special place in the universe (when we're talking about distance scales about ~10Mpc). The universe looks statistically the same whether you observe it from our location or anywhere else.
We are, however, limited to only observe things on our past light cone. The maps you see are maps of our past light cone, and that's about all the information we can get. Us, here, today, is always at the center of the past light cone.
[–] 1 point2 points (0 children)
Do you have data (a list of (x,y) pairs), or do you just have a graph in some other format? With the data, it's quite easy—just sort the pairs by distance to some fiducial (x0,y0), and then make a cumulative distribution function in terms of distance; the derivative of the CDF is then the PDF of density of points within some dr. Or you can Map the distance function over the list of points and then just use Histogram[].
Anyway, more details would always be better. I hope you have the actual data, not just a picture of a graph.
[Pedantry: It's not a script, it's a notebook. And don't use loops, loops are not the Mathematica way.]
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# Text Mining on DerStandard Comments
I did some text mining on DerStandard.at before, back then primarily interested in the comment count per article. What has been a simple HTTP GET and passing the response to BeautifulSoup requires a more sophisticated approach today. Things change, webcontent is created dynamically and we have to resort to other tools these days.
Headless browsers provide a JavaScript API, useful for retrieving the desired data after loading the page. My choice fell on phantomjs, available on pacman:
pacman -Qi phantomjs | head -n3
Name : phantomjs
Version : 2.1.1-3
Beschreibung : Headless WebKit with JavaScript API
Since my JavaScript skills were close to non-existent, writing the script was the hard part. After some copypasta and trial-and-error coding, inevitably running into scoping and async issues, this clobbed together piece of code actually works! It reads the URL as its first and the pagenumber in the comment section as its second argument. The innerHTML from a .postinglist element is written to a content.html in the directory the parse-postinglists.js has been invoked from. Actually the data is appended to the file, that’s why I can loop over the available comment pages in R:
for (i in 1:43) {
system(paste("phantomjs parse-postinglists.js",
"http://derstandard.at/2000043417670/Das-Vokabular-der-Asylkritiker",
i, sep=" "))
}
The article of interest is itself a quantitative analysis of word frequencies in recent forum debates on asylum. Presentation and insights are somewhat underwhelming though. After all, there is a lot of information collected and stored. Some of which, found in the attributes, is perfectly appropriate for the metadata in my Corpus object (created with the help of the tm package1), as can be seen from the first document in it:
meta(corp[[1]])
author : Darth Invadeher
datetimestamp: 16-09-06 14:33
pos : 104
neg : 13
pid is the unique posting ID, a parentpid value is applied when this particular posting refers to another posting, i.e. is a followup posting. This opens up the possibility to relate authors to each other and probably a lot more. badges doesn’t fit too well as an attribute name, it actually denotes the follower count in that forum. pos and neg show the positive resp. negative rating count on that particular posting. At the time of this analysis there were 1064 documents (i.e. postings) in the corpus.
R
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## Americanism
Here's an unexpected factoid from the transcripts of the 21 debates held so far in the current U.S. presidential campaign: Despite his "Make America Great Again" slogan, Donald Trump uses the words America and American almost 13 times less often than Bernie Sanders does.
In the table below, the first number in each column is the number of times that the candidate in question used the cited word in the 12 Republican and 9 Democratic debates; the second number (in parentheses) is the frequency per million words.
America American Americans TOTAL
Clinton 50 (877) 76 (1334) 58 (1018) 184 (3229)
Cruz 63 (1929) 57 (1746) 31 (949) 151 (4624)
Kasich 78 (2711) 18 (626) 15 (521) 111 (3858)
Sanders 112 (2211) 123 (2428) 30 (592) 265 (5232)
Trump 13 (317) 2 (49) 2 (49) 17 (414)
Adding up the frequency of America and American, we get 4639 for Sanders and 366 for Trump. Throwing in Americans, we get 5232 for Sanders and 414 for Trump.
In graphical form:
If you had asked me to guess, I wouldn't have ranked Trump as (by far!) the least Americanizing of the candidates, or put Sanders on top of that pile.
1. ### Avinor said,
April 23, 2016 @ 9:36 am
It is unfortunate that factoid in English seems to be losing its original meaning of statement widely believed to be true, but that actually is false. (By the way, Swedish faktoid, a loanword from English, retains the original meaning.)
[(myl) As others have pointed out further along in the comments, factoid has both senses "An item of information accepted as a fact, although not (or not necessarily) true" and "A brief or trivial piece of information". This makes sense since the derivational affix -oid is glossed as "Forming adjectives with the sense ‘having the form or nature of, resembling, allied to’, and nouns with the sense ‘something having the form or appearance of, something related or allied in structure, but not identical’" — something could "have the form or nature of a fact … but not identical" by virtue of being not entirely true, or by virtue of being so trivial as to be a sort of sub-fact, or both.
In this particular case I had both meanings in mind, since the implication that Trump doesn't often refer to "America" is misleading if not false — he achieves similar referential goals using pronouns ("you", "we") or other expressions (for example he uses "border" and "borders" much more frequently than any of the other candidates except Cruz).]
2. ### Martin Ball said,
April 23, 2016 @ 9:40 am
Mhmmm, for me it's always meant small and unimportant fact.
3. ### D.O. said,
April 23, 2016 @ 10:01 am
If we need to invent a story to go with the data, maybe it is because Sanders's is a policy-oriented campaign, while Trump's is personality oriented (to go on a tangent, Trump is proposing quite a bit of policy change, but his personality is as much, or even more important selling point).
By the way, what happened with the project of figuring out whose, among the presidential candidates, speech is closer to "average American"? COCA has a very good lemma frequency list with distinctions between parts of speech, but then someone needs to run the scripts through a lemmalizer (is that a word?)
4. ### Rodger C said,
April 23, 2016 @ 10:28 am
Lemmatizer?
April 23, 2016 @ 12:07 pm
I agree with Martin Ball about the meaning of factoid.
6. ### JS said,
April 23, 2016 @ 12:30 pm
Trump simply hasn't developed the mind-numbing patriot-speak rhetorical tics of your standard American politician. He has other ones.
7. ### Asa said,
April 23, 2016 @ 12:34 pm
Small typo: in the first sentence after the table of numbers, you've got Americans where you want American.
8. ### Rod Johnson said,
April 23, 2016 @ 2:51 pm
Wikipedia sez factoid was coined by Norman Mailer in 1973, meaning "piece of information that becomes accepted as a fact even though it’s not actually true, or an invented fact believed to be true because it appears in print." YMMV
April 23, 2016 @ 3:48 pm
@Rod Johnson: If Wikipedia says it, it may well be true. I seem to recall Jim McCawley talking about "Greekoid" restaurants in Chicago, clearly suggesting a common meaning of "widely accepted as X even though it's not X". I'm also pretty sure I've heard -oid attached to other adjectives to suggest "not really [Adj]". So maybe factoid really did start life that way. But I still agree with Martin Ball.
Maybe MYL can figure out a way to extract which meaning was intended in early uses of the word – it's certainly not as easy to search for meanings as it is to search for forms.
10. ### Viseguy said,
April 23, 2016 @ 3:50 pm
OED on factoid:
1. An item of information accepted as a fact, although not (or not necessarily) true; spec. an assumption or speculation reported and repeated so often as to be popularly considered true; a simulated or imagined fact.
2. Chiefly Journalism and Broadcasting. A brief or trivial piece of information, esp. any of a list of such items presented together.
April 23, 2016 @ 4:03 pm
Follow-up: without waiting for MYL's input, I went to Google n-grams and looked at the relatively few citations in their corpus from the period 1970-1997. It's very clear that the first OED meaning (thanks, Viseguy!) is the original (attributed in a couple of quotes to Norman Mailer, but also to "someone" in at least one quote). The meaning that Martin Ball and I took as given seems to have come into existence in the mid-1990s.
12. ### Rod Johnson said,
April 23, 2016 @ 6:34 pm
That's pretty much the Wikipedia story, which ascribes the new meaning to CNN Headline News.
13. ### AntC said,
April 23, 2016 @ 7:07 pm
I believe previous of Mark's analyses have identified that Trump uses China far more than other candidates. Also possibly Mexico/ans.
Can we compare usage of other geopolitically significant words/phrases?
My impression is that Trump spends far more of his speeches dissing others than making any positive statements (or any policy at all come to that). Can we characterise the mentions of these geopolities as positive/negative? (Probably it'll elude ngrams: Trump is usually snide/ironic/oblique – I won't say it but you know what I mean.)
And China, Mexico are his targets for dissing.
OTOH does the sector of the electorate who Trump is trying to attract actually identify as principally American, or rather do they identify more locally? Trump seems to frequently mention particular States or cities. Compare that few people identify as European, despite Europe being in many ways more compact and culturally homogeneous than the U.S.
14. ### Jon said,
April 24, 2016 @ 2:08 am
AntC: I disagree with your statement that few people identify as European. I know plenty of people who include European among their identities, including the front-runner in the race for London's next Mayor, in a radio interview yesterday (he is Muslim, and was stressing his multicultural identity as a match for London's multiculturalism with a long list of his own identities). Few would put it first – I might say British, English, European, Anglo-Saxon, scientist, atheist, …
April 24, 2016 @ 2:50 am
@AntC: What Jon said, but I also disagree with your statement that Europe is more culturally homogeneous than the US ("compact", well, yes). Lots of people do identify as European, but the cultural differences run deep, and there are still lots of people in Europe who have seldom (or even never) been outside the country they were born in. The existence of multiple languages is still a major barrier to mobility and cultural integration that hardly affects the US, despite the steady spread of English as the default L2.
16. ### Riikka said,
April 24, 2016 @ 3:01 am
I think I would agree with AntC. There isn't often a reason to add European among one's identities – unless it's to point out I don't belong to any other continent or area (Middle East or either of Americas, in my case), and I don't expect the person I'm communicating with to know geography very well. The reason why the mayor Jon mentioned would add European is the other one, to stress the European unity, which in my opinion hasn't been very unified lately, with questions about Grexit, PortuGo and Brexit, refugee crisis, religionism (struggle about marriage laws, islamophobia) and whatnot on the table. Of course there still is legal cooperation e.g. in having sensible labor laws and human rights (as opposed to Qatari de facto slavery and USAmerican habit of paying wages one doesn't survive on), but in every other area there's a lot of diversity. And Nordic cooperation had established those legal matters, and much else, even before EEC/EU was created.
I identify as Finnish and Nordic – or "Scandinavian", though geographically Finland isn't part of Scandinavian peninsula – but because very few know what or where Nordic countries are, and when even Swedes don't seem to remember the existence of Finland, it's usually easier to just put "European" instead.
17. ### Brett said,
April 24, 2016 @ 7:42 am
@Bob Ladd: The notion of true "factoids" dates to the early 1980s, at least. 3-2-1 Contact magazine had a two page feature of strange but true facts titled "Factoids" when I was first given a subscription (October 1983, I think).
When I was in elementary school, I really loved that magazine, but as the parent 3-2-1 Contact television withered and disappeared, some of the magazine's regular features changed in odd (and, I thought, saddening) ways. I finally asked my parents to cancel my subscription when I realized that the Bloodhound Gang stories were clearly being written by people who had never watched the original Bloodhound Gang episodes on the TV show.
18. ### Mr Punch said,
April 24, 2016 @ 11:05 am
Agree with Brett; I associate the origin of "true" factoids with USA Today (launched 1982), which made heavy use of them.
19. ### Terry Hunt said,
April 25, 2016 @ 8:26 am
@ Riikka
". . . even Swedes don't seem to remember the existence of Finland . . ."
I thought they remember it very well, but for historical reasons are embarrassed about mentioning it.
20. ### Terry Hunt said,
April 25, 2016 @ 8:27 am
@ Riikka
". . . even Swedes don't seem to remember the existence of Finland . . ."
I thought they remember it very well, but for historical reasons are embarrassed about mentioning it.
(I'll be visiting Finland next year, so I'll have the opportunity to find out at first hand!)
21. ### BZ said,
April 25, 2016 @ 9:33 am
My first encounter with the word "factoid" was on CNN (or maybe Headline News, now HLN) where they would display a fact that was little known and of little importance with that heading before and/or after some commercial breaks. I've only encountered the "not true" meaning in the last 5 years or so.
22. ### Graeme said,
April 26, 2016 @ 8:44 am
Perhaps Simon n Garfunkel are to blame.
Also, just possibly – what if nationalists tended to use 'we'. And the internationalist (or more consciously educated) candidates used 'America' because they either saw 'America' as a construct, or were more formal?
23. ### David Scott Marley said,
April 27, 2016 @ 11:42 am
Is this the place to gripe about constructions like "almost 13 times less often than"? Or should I take that over to Numbers Notes?
24. ### David Scott Marley said,
April 27, 2016 @ 11:54 am
My recollection matches Mr. Punch's: I first became aware of the word factoid in USA Today, where it seemed to mean a small, isolated fact, the sort of thing that one might encounter in a game of trivia. (And trivia games — Trivial Pursuit and its hundreds of imitators — were very big then.)
It was only much later that it occurred to me that the -oid suffix was wrong. Personally, I like <emfactette even better than factlet, but realistically at this point neither is going to catch on and take the place of factoid.
25. ### David Scott Marley said,
April 27, 2016 @ 11:55 am
Oops. Make that factette.
|
# Atomic Number Table
The next is Helium (= 2), then Lithium (= 3) so the answer is increase in increments of 1. Ever since, elements have been arranged on the periodic table according to their atomic numbers. The Bohr model was a one-dimensional model that used one quantum number to describe the distribution of electrons in the atom. mass number, but what do these numbers mean? The atomic number has the symbol 'z', 00:00:27,220 -- 00:00:33,280 this number tells you how many protons are in one atom of an element. The atomic number of oxygen is 8, because oxygen has A. The mass of the atom of a particular isotope relative to hydrogen 1 (or to one twelfth the mass of carbon 12), generally very close to the whole number represented by the sum of the protons and neutrons in the atomic nucleus of the isotope; it is not to be confused with the atomic weight of an element, which may include a number of isotopes in natural proportion. In the s-block and p-block of the periodic table, elements within the same period generally do not exhibit trends and similarities in properties (vertical trends down groups are more. Number of Neutrons = Mass Number - Atomic Number. - A summary quiz Resources Topical and themed. Symbol Z Abbr. from its mass number and atomic number: Potassium has a mass number of 39 and an atomic number of 19. the number indicating the order of a chemical element in the periodic system of elements of D. The 14 is the total mass of the atom. This represents the Helium Atom, Its symbol is He, Its mass number is 4, Its atomic number is 2, It has 2 protons, 2 neutrons (4 - 2 = 2), and 2 electrons. Periodic table of the elements. The atomic number of a sodium atom is 11 and its mass number is 23. The element that has the atomic number 17 is? 5. Elements are arranged from left to right and top to bottom in order of increasing atomic number. Look very carefully at the boxes in the periodic table. isotopes true b c a carbon-12; 12 amu false The values of atomic masses measured in grams are inconveniently small and. The 14 is the total mass of the atom. This charge, later termed the atomic number, could be used to number the elements within the periodic table. A small selection of non-spectroscopic atomic data has been provided for each element. Moseley's discovery was summarized as the periodic law: When elements are arranged in order of increasing atomic number. The subshell with n =2 and l =1 is the 2 p subshell; if n =3 and l =0, it is the 3 s subshell, and so on. This leaderboard is disabled as your options are different to the resource owner. The Periodic Table. The periodic table is an arrangement of the elements according to increasing atomic number. https://www. Your Account Isn't Verified! In order to create a playlist on Sporcle, you need to verify the email address you used during registration. The atomic number top left is the number of protons in the nucleus. This is the atomic number of the element. 37 amu CHAPTER 5, Atomic Structure and the Periodic Table(continued) An atom of neon-22 has two more neutrons in its nucleus than an atom of neon-20. Define “isotope” using mass number, atomic number, number of protons, neutrons and electrons. The atomic number of an atom is the number of protons in the nucleus of that atom, and it is. According to Wikipedia, the great online dictionary, and I quote: "It is a colorless. Aug 27, 2019 · The iconic periodic table of elements, devised by the Russian chemist Dmitri Mendeleev, is a two-dimensional array of the chemical elements, ordered by atomic number and arranged 18 across by. 00794 2 He Helium 4. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. Abbreviations and Definitions: Atomic number: The number of protons in an atom. Use the "Hint" button to get a free letter if an answer is giving you trouble. so whenever you see a number with decimal point on periodic table, then that is the atomic mass. Since the Atomic Number of hydrogen is 1 that is where it finds its appropriate place at the start of the table. It is also called quicksilver. So all of this basically describes the atomic number isotopes and kind of help you differentiate or decipher what exactly you're looking at when you're looking at the periodic table. Periodic table of the elements, the organized array of all the chemical elements in order of increasing atomic number. Atomic mass: The mass of an atom is primarily determined by the number of protons and neutrons in its nucleus. The periodic table of elements arranges all of the known chemical elements in an informative array. Get an answer for 'What is the relationship between atomic size and atomic number of an atom?' and find homework help for other Science questions at eNotes. The most commonly stable and commonly found isotopes on earth is 12 C with an abundance of 98. The concept of atomic number emerged from the work of G. By ordering the Periodic Table for Kids now, you are giving your or your friend’s children one of the lenses of science. Radon: Atomic mass number given for longest lived isotope. Atomic weight is 12. What is the atomic mass of mercury? 3. elements in the periodic table. Correspondingly, it is also the number of planetary electrons in the neutral atom. The atomic weights are available for elements 1 through 118 and isotopic compositions or abundances are given when appropriate. When it includes chemistry and moreover physics, normally the atomic amount associated to a chemical ( moreover referred to as the proton amount) is actually the number of protons utterly positioned throughout the nucleus associated to an atom of that , and for that goal identical to the associated fee selection of the particular nucleus. reactivity. pdf and image format. atomic number for Group 2A and for Period 3 of the periodic table. This leaderboard is disabled as your options are different to the resource owner. Sodium = 11. As one goes across a period the atomic radius decreases because more protons are in the atoms as one goes across the period and the electrons are in the same shell. What is Atomic Number? Used in both chemistry and physics, the atomic number is the number of protons found in the nucleus of an atom. The Bohr model was a one-dimensional model that used one quantum number to describe the distribution of electrons in the atom. The periodic table is said to be the most important part of chemistry as it makes it. 0107 7 N Nitrogen 14. More importantly, and the reason why the ordering of the elements according to. Ni is the symbol for what element? Nickel 4. Symbol Z Abbr. Atomic mass of Silicon is 28. Since the Atomic Number of hydrogen is 1 that is where it finds its appropriate place at the start of the table. The atomic number or proton number (symbol Z) of a chemical element is the number of protons found in the nucleus of every atom of that element. This leaderboard is currently private. In going across a period (a horizontal row) in the periodic table, the valence electrons of the elements all have the same principal quantum number and so we would predict no change in the radii from a change in the value of n. - An optional practical. Elements are arranged from left to right and top to bottom in order of increasing atomic number. Silicon is a chemical element with atomic number 14 which means there are 14 protons and 14 electrons in the atomic structure. The number of elementary positive charges (protons) contained within the nucleus of an atom. These two components of the nucleus are referred to as nucleons. Thus, the standard atomic weight is given as a single value with an uncertainty that includes both measurement uncertainty. The shell number is followed by the letter of the sub-shell, with the number of electrons in the shell indicated by a superscript number. According to Wikipedia, the great online dictionary, and I quote: "It is a colorless. The atomic number of the element is used to arrange the elements in the periodic table. As you go down a column in the periodic table, atoms increase in size. Periodic Table of Elements - Elements Database Periodic Table Our periodic table of chemical elements presents complete information on the chemical elements including the chemical element symbol, atomic number, atomic weight and description. To find the number of neutrons, subtract the atomic number from its mass number. increasing relative atomic mass, and elements with similar properties were grouped together. The element with atomic number 13 and symbol Al may be spelt as aluminium or as aluminum. jpg from AA 1elements: Periodic Table are arranged in increasing atomic number. What is the Mass Number? The. The funky design features a cream tone ceramic body with pink and gold patterns and sputnik inspired chrome pieces around the stem. Its atomic number 115, Uup Symbol and since it has no official name, it is known as it, which is on one five in Latin. First, there is an integer (whole number) in some part of the box. To write a complete electron configuration for an uncharged atom, Determine the number of electrons in the atom from its atomic number. isotopes true b c a carbon-12; 12 amu false The values of atomic masses measured in grams are inconveniently small and. Since the Atomic Number of hydrogen is 1 that is where it finds its appropriate place at the start of the table. Atomic Weight vs Mass Number. The neutron number, or number of neutrons, when added to the atomic number, equals atomic mass number. Noble gases are placed at the end of every period, so if you remember this table, then using a bit of common sense, we can easily locate the position of any element. Cu, Ag, and Au are all in what group # 7. The atomic number on the periodic table (or anywhere else) tells you which element is being specified. Periodic Table Cipher Tool to convert atomic numbers into letters. Explore LB's board "periodic table" on Pinterest. 2 Reference E95. 941* 4 Be 9. Abbreviations and Definitions: Atomic number: The number of protons in an atom. All of the elements in the second row (the second period) have two orbitals for their electrons. Atomic numbers were first assigned to the elements c. One-half the distance between two adjacent atoms in crystals of elements. Define atomic number. Hydrogen is the lightest and most abundant element in the entire universe, and makes up almost all of a nebula's mass. atomic number for Group 2A and for Period 3 of the periodic table. First, there is an integer (whole number) in some part of the box. The periodic table of the elements. reactivity. Atomic Number of Silicon. Hydrogen, the first element in the periodic table, has the atomic number of one, corresponding with the number of protons in the nucleus. How to use the atomic number and the mass number to represent different isotopes The following video shows an example of calculating the number of neutrons. The value of l also has a slight effect on the energy of the subshell; the energy of the subshell increases with l ( s < p < d < f ). unique atomic number. Each element's atomic number, name, element symbol, and group and period numbers on the periodic table are given. The isotope experts should become familiar with how atomic weight is a weighted average of atoms with different mass numbers but the same atomic number. However, it is not giving the exact mass of the atom. Table of Isotopic Masses and Natural Abundances This table lists the mass and percent natural abundance for the stable nuclides. 4527 amu 35 Cl 75. substitution B. Its atomic number 115, Uup Symbol and since it has no official name, it is known as it, which is on one five in Latin. atomic number listed as AN banners to match the organisation of the Periodic Table. - A progress check. Fundamental properties of atoms including atomic number and atomic mass. 37 amu CHAPTER 5, Atomic Structure and the Periodic Table(continued) An atom of neon-22 has two more neutrons in its nucleus than an atom of neon-20. edu is a platform for academics to share research papers. Using the periodic table as a guide, give the name and symbol of the element that has the given number of protons in the nucleus. This is the quickest way to do this. These guys are the mass number, and these guys just to read the fact that chlorine has 17 protons this is the atomic number. If we add the number of electrons that each sublevel holds it looks like this: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 2 5f 14 6d 10 7p 6. This two minute video shows how to read the periodic table. Notice that no matter what form of the table you are using, there are always three very specific items that appear for each element. Total Cards. As you go down a column in the periodic table, atoms increase in size. 003 3 Li Lithium 6. The element with atomic number 55 and symbol Cs may be spelt as caesium or as cesium. Some elements changed spots, making the pattern of properties even more regular. Periodic Table of the Elements 1 H 1. Aug 27, 2019 · The iconic periodic table of elements, devised by the Russian chemist Dmitri Mendeleev, is a two-dimensional array of the chemical elements, ordered by atomic number and arranged 18 across by. Atomic mass number is also called nucleon number. jpg from AA 1elements: Periodic Table are arranged in increasing atomic number. The periodic table - AQA. We remember from our school chemistry course that every element has its own specific atomic number. Example: 4He 2. No matter what you're looking for or where you are in the world, our global marketplace of sellers can help you find unique and affordable options. periodic table with atomic mass and atomic number Group 3. Though individual atoms always have an integer number of atomic mass units, the atomic mass on the periodic table is stated as a decimal number because it is an average of the various isotopes of an element. #avenger-8-air-hockey-table-by-atomic-game-tables #Hockey-Tables , Shop Game Room Furniture with Offer Free Shipping and Free In Home Delivery Nationwide. 4 Pauling Scale. Atomic number, the number of a chemical element (q. The rest of the isotopes are unstable. List the symbols for two transition metals. It is the same as the number of protons that the atom of each element has, so sometimes atomic number is called proton number. The mass number measures the number of protons and neutrons in the nucleus of a particular atom. That is the whole idea of memory pegs. Atomic Number of the elements. Look very carefully at the boxes in the periodic table. 811* 6 C 12. At a basic level, the atomic mass is the number of protons plus the number of neutrons. It is also worthy to note that Mendeleev's 1871 arrangement was related to the atomic ratios in which elements formed oxides, binary compounds with oxygen whereas today's periodic tables are arranged by increasing atomic numbers, that is, the number of protons a particular element contains. Moseley; he arranged the elements in an order based on certain characteristics of their X-ray spectra and then numbered them accordingly. How to use the atomic number and the mass number to represent different isotopes The following video shows an example of calculating the number of neutrons. 01 neutrons because you can’t have part of a neutron. Along a period of the periodic table, the atomic number is gradually. An early attempt was made by a German Chemist named Johann Dobereiner , in 1817. The first chemical element is Hydrogen and the last is Ununoctium. com makes it easy to get the grade you want!. The periodic table lists all the elements, with information about their atomic weights, chemical symbols, and atomic numbers. Since the number of protons determines where an element goes in the periodic table, simple addition shows the new element to bear the atomic number 115, which had never been seen before. Periodic table of the elements. In an uncharged atom, the atomic number is also equal to the number of electrons. Keep in mind that not all periodic tables are exactly the same so some may have. Silicon is a chemical element with atomic number 14 which means there are 14 protons and 14 electrons in the atomic structure. Periodic Table Questions page 2 of 8 ___ 38. The arrangement of the periodic table leads us to visualize certain trends among the atoms. Atomic Number of Silicon. on-line shopping has currently gone a long method; it's modified the way consumers and entrepreneurs do business today. The periodic table is an arrangement of the elements according to increasing atomic number. You searched for: atomic tables! Etsy is the home to thousands of handmade, vintage, and one-of-a-kind products and gifts related to your search. Here's a list of chemical elements ordered by increasing atomic number. The entire periodic table has been arranged around atomic number, with each next element being the same stable configuration of the previous element except with one more proton and electron pair. The electron configuration describes the distribution of electrons in the shell of an atom at various energy states. The Periodic table of elements lists the atomic number above the element symbol. The atomic number on the periodic table (or anywhere else) tells you which element is being specified. first 50 elements of the periodic table. Example: 4He 2. Free Printable Periodic Table of Elements. Copper has the atomic number of 29 for its 29 protons. It is denoted by the letter Z. Magnetic Quantum Number (ml): ml = -l,. Learn more TSMC is 'raiding the periodic table' to keep Moore's Law alive. This is because the additional electrons in elements with higher atomic numbers are filling orbitals that have a larger average radius (more accurately, the extent of the electron probability distribution is larger). Periodic table of the elements, the organized array of all the chemical elements in order of increasing atomic number. The periodic table lists all the elements, with information about their atomic weights, chemical symbols, and atomic numbers. The periodic table is a chart of all the elements arranged in increasing atomic number. The periodic table of elements lists elements in ascending order according to its atomic number. The atomic number uniquely identifies a chemical element. The periodic table is arranged in order of atomic number. There you can find the metals, semi-conductor (s), non-metal (s), inert noble gas (ses), Halogens, Lanthanoides, Actinoids (rare earth elements) and transition metals. 93% and 13 C which forms the remaining form of carbon on earth. If this pattern holds true, you have definitely found the atomic number. Only its atomic number is. Table Lamp - Atomic Age LED Metal Accent Light Number of bids and bid amounts may be. Spin Quantum Number (m s) Table of Allowed Quantum Numbers Writing Electron Configurations Properties of Monatomic Ions References Quantum Numbers and Atomic Orbitals. What is the atomic mass of mercury? 3. And you can find the atomic number on the periodic table. Description: Replaced with innodb_adaptive_flushing_method. The atomic number uniquely identifies a chemical element. So we're going to talk about hydrogen in this video. Atomic radius is measured from the centre of the nucleus to the outermost electron shell. The modern arrangement of the elements is known as the Periodic Table of Elements and is arranged according to the atomic number of elements. 040 @ Material Shale Sandstone Fused silica Limestone Polyhalite Serpentine Basalt Granite Calcite Feldspar Slate No rite Grandpiorite C-axis axis Syeni te salt Dolomite Chlorite Anhydrite Quartzite pyrite Hematite Magnetite Sphalirite Thermal Conductivity. The Periodic Table of the Elements 1 H Hydrogen 1. When it includes chemistry and moreover physics, normally the atomic amount associated to a chemical ( moreover referred to as the proton amount) is actually the number of protons utterly positioned throughout the nucleus associated to an atom of that , and for that goal identical to the associated fee selection of the particular nucleus. Atomic mass is measured in Atomic Mass Units (amu) which are scaled relative to carbon, 12 C,. ) Moseley's discovery cleared up the cobalt-nickel and argon-potassium problems. It is also called quicksilver. The atomic weights (called as well Atomic Mass) have been removed from the original table as they correspond to the average mass of the atoms of an element. The Next 18 Elements. 5Air Powered Hockey Table by Atomic Game Tables Free Shipping On Orders Over \$49. The numbers indicate approximately the highest oxidation number of the elements in that group, and so indicate similar chemistry with other elements with the same numeral. In the periodic table of elements, there are seven horizontal rows of elements called periods. By ordering the Periodic Table for Kids now, you are giving your or your friend’s children one of the lenses of science. 2 Group 2A Element Atomic Number Atomic Radius Be 4 1. (See also The Periodic Table: Metals, Nonmetals, and Metalloids. What is the Mass Number? The. mass number, but what do these numbers mean? The atomic number has the symbol 'z', 00:00:27,220 -- 00:00:33,280 this number tells you how many protons are in one atom of an element. You go right, then go down when you reach the end of a row. The element number is its atomic number, which is the number of protons in each of its atoms. On the periodic table, the elements are arranged in the order of atomic number across a period. Use a periodic table to help you fill in the chart below. Atomic Mass "An atomic weight (relative atomic mass) of an element from a specified source is the ratio of the average mass per atom of the element to 1/12 of the mass of 12 C" in its nuclear and electronic ground state. Give the symbol for two halogens. Groups in the periodic table. The modern periodic table is arranged in such a way that all the elements have an increasing atomic number, and subsequently, increasing mass number. Placement or location of elements on the. So for hydrogen, hydrogen's atomic number is one. First 40 Elements by Atomic Number and Weight 40 terms. Moseley's Periodic Table. Periodic Table of Elements - Elements Database Periodic Table Our periodic table of chemical elements presents complete information on the chemical elements including the chemical element symbol, atomic number, atomic weight and description. How to use the atomic number and the mass number to represent different isotopes The following video shows an example of calculating the number of neutrons. The chemical element mercury is a shiny, silvery, liquid metal. The neutron number, or number of neutrons, when added to the atomic number, equals atomic mass number. Atomic Numbers- Electrons, Neutrons, and Protons Because neutrons and protons are almost the same mass, the total number of protons and neutrons in an atom is the atomic mass. So for hydrogen, hydrogen's atomic number is one. width of localized states, associated with different atomic arrangements, to be 5–66 meV. An isotope is a product of an element whose atom has a different neutron number than atomic number. 01 in the periodic table. Uut and Uup Add Their Atomic Mass To Periodic Table. Concept of atomic number, mass number, fractional atomic mass, isotopes, isobars The nuclei of atoms is made up of protons and neutrons. List the symbols for two transition metals. For sodium (Na) the atomic number is 11. The good thing with elements is that they're defined by atomic numbers, meaning they're defined by the number of protons in the nucleus. The element with atomic number 55 and symbol Cs may be spelt as caesium or as cesium. eight protons in the nucleus. They are known as the element’s atomic number, and in the periodic table of elements, the atomic number of an element is the same as the number of protons contained within its nucleus. The resulting Periodic Table of the Elements below includes the Atomic Number, Atomic mass, Symbol, Name, Electronegativity, Density, Ionization energy, Boiling point, Melting point, Electron Configuration, Oxidation States, Ground State Level, and Atomic Radius. The chemical symbol for Silicon is Si. Nuclides marked with an asterisk (*) in the abundance column. Given information about an element, find the mass and name of an isotope. Many of the properties of elements and their reactivity relate to their position in the periodic table. There's quite a lot to be learned about atoms. eight protons in the nucleus. The atomic number is the number of protons in the nucleus of an atom, therefore it is the same to the charge number of the element. If you look at an image of the periodic table, you start at the top-left (Hydrogen, atomic number = 1). In chemistry, the atomic number is the number of protons found in the nucleus of an atom. The periodic table is an arrangement of the elements according to increasing atomic number. Each element also lists its atomic number, atomic weight, name and abbreviation, a short description of its use and information on whether it is a solid, a liquid or a gas. The neutron number, or number of neutrons, when added to the atomic number, equals atomic mass number. Since the Atomic Number of hydrogen is 1 that is where it finds its appropriate place at the start of the table. The general trend is that atomic radii increase within a family with increasing atomic number. First 40 Elements by Atomic Number and Weight 40 terms. See more ideas about Science Classroom, Science lessons and Chemistry. Each element is uniquely defined by its atomic number. info and details all of the elements in the Periodic table, the numbers of protons. The entire periodic table has been arranged around atomic number, with each next element being the same stable configuration of the previous element except with one more proton and electron pair. (b) Complete the table with the number of protons, electrons, neutrons, and complete chemical symbol (showing the mass number and atomic number) for each neutral atom. Though individual atoms always have an integer number of atomic mass units, the atomic mass on the periodic table is stated as a decimal number because it is an average of the various isotopes of an element. Using the data below, make a bar graph of atomic radius vs. Standard Atomic Weight The standard atomic weight is the average mass of an element in atomic mass units ("amu"). Shop Furniture, Home Décor, Cookware & More! 2-Day Shipping. atomic mass. Francium: Click here to buy a book, photographic periodic table poster, card deck, or 3D print based on. The mass of an electron is very small (9. Moving through the periodic table in this fashion produces the following order of sublevels up through 6s:. (An isotope of an element is an atom of that element that has a particular Mass Number and is often identified according to it, e. Isotope: Atoms of the same element with the same atomic number, but different number of neutrons. The atomic number corresponds to the number of protons in the nucleus of an atom of that element. ATOMIC NUMBER AND MASS NUMBERS. The periodic table is arranged in order of atomic number. 012182 5 B 10. The chemical symbol for Silicon is Si. The atomic number is usually located above the element symbol. The atomic number of each element increases by one, reading from left to right. from its mass number and atomic number: Potassium has a mass number of 39 and an atomic number of 19. Next on the. Periodic Table Packet #1 ANSWERS Directions : Use your Periodic table to comp lete the worksheet. The electron configuration describes the distribution of electrons in the shell of an atom at various energy states. This periodic table of elements provides comprehensive data on the chemical elements including scores of properties, element names in many languages and most known nuclides (Isotopes). The electrons are arranged in shells around the nucleus. (a) Chemical properties of an element are the periodic function of their _____ (atomic weight/atomic number). Atomic Number of Elements in Periodic Table. atomic mass. So the atomic mass number below the symbol of the element in the periodic table is called the relative, weighted average atomic mass. Use the "Hint" button to get a free letter if an answer is giving you trouble. It is present everywhere in our environment and is released into the environment through forest fires and volcanic. Element Groups: Alkali Metals Alkaline Earth Metals Transition Metals Other Metals Metalloids Non-Metals Halogens Noble Gases Rare Earth Elements. The chemical behavior of the elements is determined by the number of protons. The isotope experts should become familiar with how atomic weight is a weighted average of atoms with different mass numbers but the same atomic number. atomic weights vary in normal materials, but upper and lower bounds of the standard atomic weight have not been assigned by IUPAC or the variations may be too small to affect the standard atomic weight value significantly. What is Atomic Number? Used in both chemistry and physics, the atomic number is the number of protons found in the nucleus of an atom. What is the atomic number? Every element has a unique atomic number. It is identical to the charge number of the nucleus. Radon: Atomic mass number given for longest lived isotope. $$\mathrm{_{64}Gd}$$ (…) The same meanings are described in the German standard DIN 1338 (2011). Look very carefully at the boxes in the periodic table. During the tip sliding process, the side boundaries of the graphite substrate were constrained to be fixed in all directions. Similarly, as you move down a group the atomic number increases. The modern arrangement of the elements is known as the Periodic Table of Elements and is arranged according to the atomic number of elements. The periodic table represents neutral atoms. Periodic table of the elements, the organized array of all the chemical elements in order of increasing atomic number. Atoms contain protons, neutrons and electrons. It also corresponds to the number of electrons in the neutral atom. If there are 9 protons, there must be 10 neutrons for the total to add up to 19. It is denoted by the letter Z. 'carbon-14'. Atomic Symbol The abbreviation or one/two letters that represent an element is called the atomic symbol. eight protons in the nucleus. Here's a list of chemical elements ordered by increasing atomic number. Periodic table of the elements. Similarly, as you move down a group the atomic number increases. What is the atomic number of this helium atom? A neutral atom must have equal numbers of protons and electrons, so the atomic number of an element also gives the number of electrons. Moseley; he arranged the elements in an order based on certain characteristics of their X-ray spectra and then numbered them accordingly. This is a weighted average of the naturally-occurring isotopes and this is not relevant for astrophysical applications. Concept of atomic number, mass number, fractional atomic mass, isotopes, isobars The nuclei of atoms is made up of protons and neutrons. atomic weights is usually the number that has decimal points. Students know how to use the periodic table to identify alkali metals, alkaline earth metals. Use a periodic table to help you fill in the chart below. See more ideas about Science Classroom, Science lessons and Chemistry. These two components of the nucleus are referred to as nucleons. The scientist Dmitri Mendeleev, a Russian chemist, proposed an arrangement of know elements based on their atomic mass. It is present everywhere in our environment and is released into the environment through forest fires and volcanic. Learn more TSMC is 'raiding the periodic table' to keep Moore's Law alive. Periodic Trends Worksheet Atomic Radius 1. What is the atomic mass of mercury? 200.
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$0$
According to order of operations, multiplication comes before addition or subtraction. Using the order of operations on real numbers, the given expression, $3-11+2\cdot4 ,$ evaluates to \begin{array}{l}\require{cancel} 3-11+8 \\\\= -8+8 \\\\= 0 .\end{array}
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# Using \Mathsmaller in math display mode makes the character appear to the right of the symbol itstead of below it
I am trying to use \prod in math display mode but it appears large (larger than I want it to be) but when I use \mathsmaller from the relsize package the letter appears on right side of it instead of below. Is there any way to make the symbol smaller while the character remains below the symbol?
-
Hi Muaz, Welcome to TeX.SE!. Could you provide a MWE? You can use backticks to edit your post to render inline code snippets :) – cmhughes Aug 7 '12 at 20:09
I don't know the package you mention and you didn't provide an example but using \mathop around the construction will revert to the operator limits positioning.
\mathop{\mathsmaller.....}_0^n
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Use \mathop{\mathsmaller\prod}_{..}^{..} from relsize. The output is similar to that of \textstyle\prod via \DeclareMathOperator*. – Werner Aug 7 '12 at 20:20
I am still new so cannot post images. thanks – Muaz Aug 7 '12 at 20:25
For consistency, define a new math operator to be set in \textstyle using amsmath's \DeclareMathOperator*:
\documentclass{article}
\usepackage{amsmath}% http://ctan.org/pkg/amsmath
\DeclareMathOperator*{\sProd}{\textstyle\prod}
\usepackage{relsize}% http://ctan.org/pkg/relsize
\begin{document}
$\prod_{i=1}^{\infty}\ %\frac{1}{i} = {\mathsmaller \prod_{i=1}^{\infty}}\ %\frac{1}{i} = \sProd_{i=1}^{\infty} %\frac{1}{i}$
\end{document}
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# A-level Mathematics/OCR/M3/Elastic Strings and Springs
< A-level Mathematics | OCR | M3
## Hooke's Law and the Modulus of Elasticity
The natural length is the length of an elastic string or spring when it is not stretched or compressed. An elastic string or spring experiences a tension when its length is greater than the natural length. In addition, a spring experiences a compression when its length is less than its natural length. To simplify our analysis, we use the term tension to refer to both types of forces (i.e. tension and compression). Also, we use the term extension to refer to the change in the length of the string or spring. Thus, the extension for a compressed spring is negative.
According to Hooke's Law, the extension $x$ is proportional to the tension $T$ applied to the elastic string or spring. Although Hooke's Law holds only up to the limit of elasticity, we may safely assume its applicability unless otherwise told. We may write this relationship in terms of the natural length $l$ and the modulus of elasticity $\lambda$ (which is a property of the elastic string or spring independent of its length) as follows:
$T=\frac{\lambda}{l} x$
Note that the tension and the extension are in the same direction (i.e. the variables are either both positive or both negative). This should be intuitive since we are considering the force exerted on (i.e. NOT exerted by) the elastic string or spring.
If a mass attached to the end of the elastic string or spring is producing the extension, then by Newton's Third Law, the elastic string or spring exerts a force on the mass equal in magnitude and opposite in direction to its tension. To illustrate this, let us consider the system on the right.
Consider a particle P of mass $m$ suspended vertically from one end of a light (i.e. massless) elastic string of natural length $l$ and modulus of elasticity $\lambda$. The other end of the string is attached to a fixed point O. When the particle is at rest, the resultant force acting on it is zero according to Newton's Second Law. Therefore, the downward weight of the particle should balance the upward force exerted by the string on the particle (which is equal in magnitude to the tension of the string):
$mg$ $=T$ $\Rightarrow$ $mg$ $=\frac{\lambda}{l}x$ $\Rightarrow$ $x$ $=\frac{mgl}{\lambda}$, which is the corresponding extension in the string.
## Elastic Potential Energy
An elastic string or spring is able to store energy when it is extended (and compressed, in the case of a spring). This stored energy is termed the elastic potential energy (EPE). The EPE in an elastic string or spring is converted from the work done (by an external agent) in producing the required extension. This is just the work done against the force exerted by the elastic string or spring by virtue of its tension. Therefore, the EPE can be determined by integrating the tension $T$ wrt the extension $x$:
$\mathrm{EPE}$ $=$Work done to produce the extension $x$ $=\int_0^x T \,dy$ $=\int_0^x \frac{\lambda}{l}y \,dy$ $=\frac{\lambda}{l} \left[ \frac{y^2}{2} \right]_0^x$
$\Rightarrow$ $\mathrm{EPE}=\frac{\lambda}{2l} x^2$
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# What nucleic acid contains uracil?
Sep 5, 2016
RNA
#### Explanation:
Ribonucleic acid (RNA) is the nucleic acid that contains uracil.
The nucleotide called thymine in DNA is replaced by uracil in all types of RNA. These nucleotides are very similar in structure:
They only differ in one methyl ($C {H}_{3}$) group and both pair with the nucleotide adenine.
$\textcolor{red}{\text{Why did the cell change the strategy?}}$
This is a major question of course, why not use uracil in DNA? or why not thymine in RNA?
It has to do with two main things:
1. Stability: while uracil usually pairs with adenine, it can also pair with other nucleotides or with itself. This doesn't happen with thymine. DNA with thymine is therefore more stable = useful because it has to be passed on to the offspring.
2. Efficient repair: the nucleotide cytosine can easily turn into uracil. The repair mechanism of DNA can recognize and repair this. This wouldn't be possible when uracil was normally present in DNA.
It appears that uracil turned into thymine during evolution to make DNA more stable. Since RNA is only transiently present in the cell, uracil is apparently acceptable for nature in this nucleic acid.
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# Solvability of Turing Machines
I'm preparing for an exam, and on a sample one provided (without solutions), we have this question: Is the following solvable or non-solvable: Given a turing machine $T$, does it accept a word of even length? - Given a deterministic 1-tape turing machine $T$, does $T$ ever read the contents of the 10th cell?
Thanks! -
• I think that they're both unsolvable by rice's theorem... but I am not sure if I can apply the theorem in both cases. What do you think? Would both of these constitute as "non-trivial properties"? What would be an example of a trivial property then? – user4734 Dec 7 '12 at 22:11
• Is the tape one side infinite, or two sided? In seems hard to avoid reading the 10th cell of a one sided infinite tape. – Hendrik Jan Dec 7 '12 at 22:23
• It does not specify... Does it actually matter? – user4734 Dec 7 '12 at 22:24
• It does matter: see the comment by Steve to the answer. Singe sided: then you have only a finite number of cells available, otherwise it is Turingcomplete if you avoid the 10th cell. Refer to the standard model you have been using in your lectures. Good luck with the exam... – Hendrik Jan Dec 8 '12 at 1:03
• @HendrikJan: Also one has to be precise, what "reads the content of a cell" means. – A.Schulz Dec 8 '12 at 5:55
"Non-trivial properties" for Rice's theorem are properties of the language, not of the machine itself - for instance, it's decidable how many states a TM has, which is certainly a non-trivial property! See Perplexed by Rice's theorem for a bit of a clarification on this. Now, one of your two properties is a property of the language, while one is a property of the machine; this should allow you to apply Rice's theorem in one case but not the other. Note that this isn't conclusive evidence that the other case is solvable, but it's guidance that you might try looking for a positive answer there.
(Note that this is meant to be a very loose hint just to guide you; I can certainly go into more specific details if you'd like...)
• Oh I see! I had missed this difference. So I can only apply Rice's theorem to the even length problem, while for the other one, given that it's asking about a property of the machine itself, I would have to conclude this using a different argument. Right? – user4734 Dec 7 '12 at 22:26
• @Nizbel99 Precisely! (And as for an argument in the other case: Can you solve the halting problem for a 'Turing Machine' with a finite number of symbols and just 9 slots on its tape?) – Steven Stadnicki Dec 7 '12 at 22:46
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# I can has(kell) cheezburger?
### Edit 13 June:
I have made a new and improved version here.
## Original Question
This is a very rudimentary lolcats translator (it only works on the phrase "Can I have a cheeseburger") in Haskell. This is my first ever attempt at Haskell (or any functional programming language). I am sure that my code is absolutely atrocious (the documentation was dismal) but it works.
### Teh Codez
import Data.Strings
main = do
putStrLn "Enter your text to translate"
inputText <- getLine
let capsText = strToUpper inputText
let a = strReplace "HAVE" "HAS" capsText
let b = strReplace "CAN I" "I CAN" a
let c = strReplace "CHEESEBURGER" "CHEEZBURGER" b
let d = strReplace " A " " " c
putStrLn (d)
getLine
• @JamesFaix Mostly true, but Haskell's variables are not mutable as in other languages, so it's mainly a matter of idiomatic style, not safety. – 200_success Jun 8 '16 at 20:53
• @Bergi The getLine was to keep the Windows console open after the output was printed. Any other ideas on how to do that? – Michael Brandon Morris Jun 8 '16 at 23:42
• @MichaelBrandonMorris: I see. Usually you'd just run your program from the console, so it would stay open anyway. – Bergi Jun 9 '16 at 1:13
• @Bergi Okay, but supposing someone runs it via the executable (double click, start menu, etc.) what are my options? – Michael Brandon Morris Jun 9 '16 at 10:59
• In this particular case you could just recursively call main at the end of the main function so that your program repeatedly asks for input. Another way to accomplish this is forever from Control.Monad – BlackCap Jun 9 '16 at 12:01
It is good practice in Haskell to separate the functional code from the IO. In this case, you could (and therefore should) define a lolcat :: String -> String function. Be sure to put a type declaration on all functions — you didn't write one for your main.
Defining variables a, b, c, and d is overkill. I would write this as a composition of functions.
lolcat :: String -> String
lolcat = strReplace " A " " " .
strReplace "CHEESEBURGER" "CHEEZBURGER" .
strReplace "CAN I" "I CAN" .
strReplace "HAVE" "HAS" .
strToUpper
• I think the difference between this code and the original code in the question does a great job of highlighting the difference between functional and imperative programming styles. A big thing that may be confusing about this is the point-free style: essentially, we are defining a function lolcat WITHOUT mentioning the argument of the function -- this might be hard to read at first, but essentially it's the same thing as saying lolcat(x) = strReplace " A " " " (strReplace "CHEESEBURGER" "CHEEZEBURGER" (... (strToUpper(x))...)) – vijrox Jun 9 '16 at 15:20
• Point-free style can be confusing when learning functional programming for the first time coming from an imperative background, but it really is nice/preferable once you get used to it – vijrox Jun 9 '16 at 15:22
• I think this is a very good demonstration of point-free style even at a beginner level. – user253751 Jun 10 '16 at 6:25
I agree with 200_success and am just typing out the full program because you said you were a beginner.
import Data.Strings
encode :: String -> String
encode = strReplace " A " " "
. strReplace "CHEESEBURGER" "CHEEZBURGER"
. strReplace "CAN I" "I CAN"
. strReplace "HAVE" "HAS"
. strToUpper
main :: IO ()
main = do
putStrLn "Enter your text to translate"
input <- getLine
putStrLn $encode input The main function would perhaps more commonly be written as: main :: IO () main = do putStrLn "Enter your text to translate" putStrLn =<< encode <$> getLine
# EDIT:
@MichaelKlein asked me to explain (<$>) and (=<<), so I will do so as briefly as possible. <$> is the inline version of fmap, which is the only function from the Functor typeclass. A typeclass is somewhat like a interface in an imperative language like Java.
class Functor f where
fmap :: (a -> b) -> f a -> f b
fmap is used to run a function (a -> b) inside some data structure f a
fmap (+1) [1] == [2]
fmap (+1) (Just 3) == Just 4
(+1) <$> (Just 3) == Just 4 (<$>) :: f a -> ( a -> b ) -> f b
In Haskell, we treat the world as a datastructure IO. So just in the same way that you can run a function inside a list or a Maybe (like Just 3), you can run a function inside a IO.
encode :: String -> String
getLine :: IO String
encode <$> getLine :: IO String (=<<) is part of the Monad typeclass, which is hard to explain briefly because you need to know about a few other typeclasses first. (=<<) :: (a -> m b) -> m a -> m b (=<<) is pronounced "bind", and is a flipped version of the more common (>>=). The important bit is that Haskell has a special syntax for working with monads: the do notation. This: main = do input <- getLine putStrLn input Is syntatic sugar for this: main = getLine >>= \input -> putStrLn input Which is the same as this: main = getLine >>= putStrLn All of these are also equal: main = do putStrLn "burger" input <- getLine putStrLn$ encode input
main = putStrLn "burger"
>> getLine >>= \input -> putStrLn $encode input main = putStrLn "burger" >> encode <$> getLine >>= putStrLn
main = do
putStrLn "burger"
putStrLn =<< encode <$> getLine • maybe add the type signatures too <$> :: f a -> ( a -> b ) -> f b and =<< :: (a -> m b) -> m a -> m b – Caleth Jun 9 '16 at 12:05
• @BlackCap Thanks for adding some explanation. In a case like this, I think it's usually enough to explain what they mean just in the case where you used them here and mention that like show they do the same thing, but slightly differently for different types. +1 for going above and beyond. – Michael Klein Jun 9 '16 at 14:53
### Remove repetition
You repeat a strReplace a lot:
vvvvvvvvv
let a = strReplace "HAVE" "HAS" capsText
let b = strReplace "CAN I" "I CAN" a
let c = strReplace "CHEESEBURGER" "CHEEZBURGER" b
let d = strReplace " A " " " c
^^^^^^^^^^
What changes is what should be replaced by what: [("HAVE", "HAS"), ("CAN I", "I CAN"), ...]
Therefore your program should state the idea of replacement once and then contain this list of "replacement instructions".
To be more clear the function you should use has type String -> [(String, String)] -> String and example usage:
> replaceByTuples "foobar" [("foo", "fuu"),("bar", "baz")]
"fuubaz"
Using this function with your particular list of replacements will be easy.
This function is in this spoiler block. I advise you try to implement it by yourself.
import Data.List.Utils;
replaceByTuples s t = (foldl (.) id \$ map (\(start, end) -> replace start end) t) s
• I think you should swap the parameters of that replaceByTuples function, so that it is more convenient to use curried. Also you could quite trivially implement it by foldr (uncurry replace) :-) – Bergi Jun 9 '16 at 13:09
• @Bergi Wow, I did not expect such a simple implementation. And yes. ) agree on swapping the arguments. – Caridorc Jun 9 '16 at 13:13
I agree with the other two answers, but I would use interact in main. This will make your program infinitely loop until there's no more input which may not be what you want, but it will solve the problem of keeping the console open. You'll have to define an eachLine function so that the program processes one line at a time as detailed in this stackoverflow answer.
As 200_success mentioned, it is good practice to write type-signatures for all top-level functions and main traditionally has type-signature: main :: IO (). Because of the getLine at the end of your main, it would have type-signature: main :: IO String, which would give your program the unintended side-effect of printing an additional string as it terminates. Writing the type-signatures explicitly allows the compiler to catch these sorts of problems.
This would be the full program:
import Data.Strings
main :: IO ()
main = do
putStrLn "Enter your text to translate"
interact (eachLine lolcat)
lolcat :: String -> String
lolcat = strReplace " A " " " .
strReplace "CHEESEBURGER" "CHEEZBURGER" .
strReplace "CAN I" "I CAN" .
strReplace "HAVE" "HAS" .
strToUpper
eachLine :: (String -> String) -> String -> String
eachLine f = unlines . map f . lines
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In [2]:
import numpy as np
import matplotlib.pyplot as plt
# notebook command to show the plot in the browser
%matplotlib inline
x = np.linspace(0,2 *np.pi, 100)
Sin = np.sin(x)
Cos = np.cos(x)
plt.plot(x, Sin)
plt.plot(x,Cos)
Out[2]:
[<matplotlib.lines.Line2D at 0x10a478f28>]
In [8]:
import numpy as np
import matplotlib.pyplot as plt
# notebook command to show the plot in the browser
%matplotlib inline
data = np.random.randn(1024,2)
plt.scatter(data[:,0], data[:,1])
plt.show()
In [19]:
# Plotting multiple bars (grouped bars)
import numpy as np
import matplotlib.pyplot as plt
# notebook command to show the plot in the browser
%matplotlib inline
data = [ [5,15,20,50] , [8,9,20,40] , [3,16,30,10] ]
num_bars = len(data)
color_list = ['b', 'g', 'r']
gap = 0.2 # space each bars group 0.2 from the next group
bar_width = (1- gap) / num_bars # divide the remaining distance equally between the bars
for i, row in enumerate(data): # enumerate returns a data unit (a row) and its index
x = np.arange(len(row))
plt.bar(x+i *bar_width, row, # blue starts at x=0, i=0 => 0
# # green starts at x=0, i=1 => 0.2
# # red starts at x=0, i=2 => 0.4
width=bar_width,
color = color_list[i % len(color_list)])
plt.show
Out[19]:
<function matplotlib.pyplot.show>
In [26]:
# Plotting stacked bars
import matplotlib.pyplot as plt
# notebook command to show the plot in the browser
%matplotlib inline
d1 = [1,3,5,7]
d2 = [4,5,7,8]
x = range(len(d1))
plt.bar(x, d1, color='b') # the function bar is used to plot bar charts
plt.bar(x, d2, color='r', bottom=d1)# the optional parammeter bottom specifies the starting point for bars
plt.show()
Out[26]:
<Container object of 4 artists>
In [27]:
# Plotting stacked bars
import matplotlib.pyplot as plt
import numpy as np
# notebook command to show the plot in the browser
%matplotlib inline
"""
Why we need numpy in this example
numpy enables us to create arrays;
numpay.array1 + numpay.array2 = numpay.array3 (this is an element by elemnet addition)
Python creates lists;
list1 + list2 is a concatintion.
If we want to add two list we need to iterate overthem and add element by elemnet
"""
d1 = np.array([1,3,5,7])
d2 = np.array([4,5,7,8])
d3 = np.array([3,6,1,8])
x = range(len(d1))
plt.bar(x, d1, color='b')
plt.bar(x, d2, color='r', bottom=d1)
plt.bar(x, d3, color='g', bottom=(d1+d2)) # with help of numpy d1+d2 is the correct base for the third stack of bar
plt.show()
Out[27]:
<Container object of 4 artists>
In [31]:
# Plotting horizontal bars (with back to back)
import matplotlib.pyplot as plt
import numpy as np # needed to do array operations ( to eliminate using lists and iterate overthem)
# notebook command to show the plot in the browser
%matplotlib inline
d1 = np.array([1,3,5,7])
d2 = np.array([4,5,7,8])
x = range(len(d1))
plt.barh(x, d1, color='b') # the function bar is used to plot bar charts
plt.barh(x, -d2, color='r') # the optional parammeter bottom specifies the starting point for bars
plt.show()
Out[31]:
<Container object of 4 artists>
In [54]:
# Plotting pie chart
"""
Pie chart is used to show the relative importance/size of quantities that are grouped together.
"""
import matplotlib.pyplot as plt
# notebook command to show the plot in the browser
%matplotlib inline
d1 = [1,3,5,7]
plt.pie(d1)
plt.show()
In [53]:
#plotting histograms
"""
histogram is used to plot the frequency of occurrence (how many something is repeated)
"""
import matplotlib.pyplot as plt
import numpy as np
# notebook command to show the plot in the browser
%matplotlib inline
# generate random normally distributed data with mean =0 and variance = 1
data = np.random.randn(10000)
plt.hist(data, bins=20) # plot a histogram with 20 bins
plt.show()
In [3]:
import numpy as np
import matplotlib.pyplot as plt
# notebook command to show the plot in the browser
%matplotlib inline
data = np.random.randn(100,3)
plt.boxplot(data)
plt.show()
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# What hyper-parameters are, and what to do with them; an illustration with ridge regression
This blog post is an excerpt of my ebook Modern R with the tidyverse that you can read for free here. This is taken from Chapter 7, which deals with statistical models. In the text below, I explain what hyper-parameters are, and as an example I run a ridge regression using the {glmnet} package. The book is still being written, so comments are more than welcome!
## Hyper-parameters
Hyper-parameters are parameters of the model that cannot be directly learned from the data. A linear regression does not have any hyper-parameters, but a random forest for instance has several. You might have heard of ridge regression, lasso and elasticnet. These are extensions to linear models that avoid over-fitting by penalizing large models. These extensions of the linear regression have hyper-parameters that the practitioner has to tune. There are several ways one can tune these parameters, for example, by doing a grid-search, or a random search over the grid or using more elaborate methods. To introduce hyper-parameters, let’s get to know ridge regression, also called Tikhonov regularization.
### Ridge regression
Ridge regression is used when the data you are working with has a lot of explanatory variables, or when there is a risk that a simple linear regression might overfit to the training data, because, for example, your explanatory variables are collinear. If you are training a linear model and then you notice that it generalizes very badly to new, unseen data, it is very likely that the linear model you trained overfits the data. In this case, ridge regression might prove useful. The way ridge regression works might seem counter-intuititive; it boils down to fitting a worse model to the training data, but in return, this worse model will generalize better to new data.
The closed form solution of the ordinary least squares estimator is defined as:
$\widehat{\beta} = (X'X)^{-1}X'Y$
where $$X$$ is the design matrix (the matrix made up of the explanatory variables) and $$Y$$ is the dependent variable. For ridge regression, this closed form solution changes a little bit:
$\widehat{\beta} = (X'X + \lambda I_p)^{-1}X'Y$
where $$\lambda \in \mathbb{R}$$ is an hyper-parameter and $$I_p$$ is the identity matrix of dimension $$p$$ ($$p$$ is the number of explanatory variables). This formula above is the closed form solution to the following optimisation program:
$\sum_{i=1}^n \left(y_i - \sum_{j=1}^px_{ij}\beta_j\right)^2$
such that:
$\sum_{j=1}^p(\beta_j)^2 < c$
for any strictly positive $$c$$.
The glmnet() function from the {glmnet} package can be used for ridge regression, by setting the alpha argument to 0 (setting it to 1 would do LASSO, and setting it to a number between 0 and 1 would do elasticnet). But in order to compare linear regression and ridge regression, let me first divide the data into a training set and a testing set. I will be using the Housing data from the {Ecdat} package:
library(tidyverse)
library(Ecdat)
library(glmnet)
index <- 1:nrow(Housing)
set.seed(12345)
train_index <- sample(index, round(0.90*nrow(Housing)), replace = FALSE)
test_index <- setdiff(index, train_index)
train_x <- Housing[train_index, ] %>%
select(-price)
train_y <- Housing[train_index, ] %>%
pull(price)
test_x <- Housing[test_index, ] %>%
select(-price)
test_y <- Housing[test_index, ] %>%
pull(price)
I do the train/test split this way, because glmnet() requires a design matrix as input, and not a formula. Design matrices can be created using the model.matrix() function:
train_matrix <- model.matrix(train_y ~ ., data = train_x)
test_matrix <- model.matrix(test_y ~ ., data = test_x)
To run an unpenalized linear regression, we can set the penalty to 0:
model_lm_ridge <- glmnet(y = train_y, x = train_matrix, alpha = 0, lambda = 0)
The model above provides the same result as a linear regression. Let’s compare the coefficients between the two:
coef(model_lm_ridge)
## 13 x 1 sparse Matrix of class "dgCMatrix"
## s0
## (Intercept) -3247.030393
## (Intercept) .
## lotsize 3.520283
## bedrooms 1745.211187
## bathrms 14337.551325
## stories 6736.679470
## drivewayyes 5687.132236
## recroomyes 5701.831289
## fullbaseyes 5708.978557
## gashwyes 12508.524241
## aircoyes 12592.435621
## garagepl 4438.918373
## prefareayes 9085.172469
and now the coefficients of the linear regression (because I provide a design matrix, I have to use lm.fit() instead of lm() which requires a formula, not a matrix.)
coef(lm.fit(x = train_matrix, y = train_y))
## (Intercept) lotsize bedrooms bathrms stories
## -3245.146665 3.520357 1744.983863 14336.336858 6737.000410
## drivewayyes recroomyes fullbaseyes gashwyes aircoyes
## 5686.394123 5700.210775 5709.493884 12509.005265 12592.367268
## garagepl prefareayes
## 4439.029607 9085.409155
as you can see, the coefficients are the same. Let’s compute the RMSE for the unpenalized linear regression:
preds_lm <- predict(model_lm_ridge, test_matrix)
rmse_lm <- sqrt(mean((preds_lm - test_y)^2))
The RMSE for the linear unpenalized regression is equal to 14463.08.
Let’s now run a ridge regression, with lambda equal to 100, and see if the RMSE is smaller:
model_ridge <- glmnet(y = train_y, x = train_matrix, alpha = 0, lambda = 100)
and let’s compute the RMSE again:
preds <- predict(model_ridge, test_matrix)
rmse <- sqrt(mean((preds - test_y)^2))
The RMSE for the linear penalized regression is equal to 14460.71, which is smaller than before. But which value of lambda gives smallest RMSE? To find out, one must run model over a grid of lambda values and pick the model with lowest RMSE. This procedure is available in the cv.glmnet() function, which picks the best value for lambda:
best_model <- cv.glmnet(train_matrix, train_y)
# lambda that minimises the MSE
best_model\$lambda.min
## [1] 66.07936
According to cv.glmnet() the best value for lambda is 66.0793576. In the next section, we will implement cross validation ourselves, in order to find the hyper-parameters of a random forest.
Hope you enjoyed! If you found this blog post useful, you might want to follow me on twitter for blog post updates and buy me an espresso or paypal.me.
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X 2 3 4 5 P(x) $\Large\frac{5}{k}$ $\Large\frac{7}{k}$ $\Large\frac{9}{k}$ $\Large\frac{11}{k}$ The value of k is $(A)\;8\quad(B)\;16\quad(C)\;32\quad(D)\;48$
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Product of Two Levi-Civita Symbols in N-dimensions
by akoohpaee
Tags: levicivita, ndimensions, product, symbols
P: 607 When $N=2$ we have for example $\varepsilon_{12}=+1$. Now, you didn't define $\varepsilon^{12}$ for us. Also, is there some Einstein summation convention in effect here?
P: 160 Firstly, notice that the RHS is antisymmetric in exchange of any pair of the i's and any pair of the j's. So all you have to check is that you get the right answer for $i_1=1, i_2=2, \ldots$ and the same for the j's, which is straightforward.
P: 3 Product of Two Levi-Civita Symbols in N-dimensions ### REPLY TO g_edgar ### Hi, Many thanks for your reply! Assume that the array $K_{ijk}$ (i,j,k are in {1,2,3}) is defined in such a way that $K_{ijk}=\varepsilon_{ijk}$. It can be shown that, this array behave as a tensor under covariant and contravariant coordinate transformations. In other words: \begin{align} \varepsilon_{ijk}=\varepsilon^{ijk} \end{align} Also this is the case for N-dimensional: \begin{align} \varepsilon_{ijklm\dots}=\varepsilon^{ijklm\dots} \end{align} And regarding to the second part of your question: There is no Einstein summation notation in effect here. In fact there is no summation in this statement. Many thanks for your reply and your attention. Ali
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### Before and After #215: Elia Kazan
Before
After
Robert Fiore said...
Meaning, "before and after he sold out everyone he trusted."
Anyway, to clarify, as some people seem to be confused, the problem isn't that he ratted to the Committee. A lot of people did that. The problem was he was a supporter of blacklisting.
MichaelRyerson said...
Thank you.
Kreisler said...
And the problem shows in his films - never quite honouring the apparent convictions set out before the viewer. Imagine Kazan without Brando or Dean - Karl Malden mainly.
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Contact: gandhi[dot]viswanathan[at]pq[dot]cnpq[dot]br
This blog is about science and my life as a physicist. I hold a (full) professorship at the Universidade Federal do Rio Grande do Norte, in Natal, Brazil. Until 2010, I held an associate professorship at the Universidade Federal de Alagoas, Maceió.
I obtained my B.A. in physics at Oxford University in 1992 and my Ph.D. in physics with Professor H. Eugene Stanley at Boston University in 1997. In the 1980s, I also spent a year studying physics at American College in Madurai, India, after finishing high school at the North Carolina School of Science and Mathematics (NCSSM), a (publicly funded) boarding school (“residential school”) in Durham, NC, USA.
I was born in the city of Madras (later renamed Chennai) in India, but have lived in Brazil since childhood. I am a naturalized citizen of Brazil.
Link to my university web page.
The background header image is a domain color plot of $~~\sin^2(z)~~$ in the complex plane.
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# Difference in meaning of dummy coefficient in regression with other continuous variables
I am estimating two regressions. The first is a regression with only dummy variables (i.e. $y = B_0 + B_1D_1 + B_2D_2$), and the second is with two independent variables (i.e. $y = B_0 + B_1X_1 + B_2X_2 + B_3D_1 + B_4D_2$).
The $D_1$ and $D_2$ are dummy variables, whereas the $X_1$ and $X_2$ are continuous. Would there be a difference in the meaning/interpretation of the B1 in the first regression and the B3 in the second regression (both the coefficient of the first dummy variable)?
It's comparable between the meanings/interpretations of the $B_1$ in the first regression and the $B_3$ in the second regression.
$B_1$ in the first regression: the difference in y of $D_1$ compared with reference category
$B_3$ in the second regression: the difference in y of $D_1$ compared with reference category adjusted for the covariates ($X_1$ and $X_2$) in the model
If $X_1$ and $X_2$ are confounding variables or effect modifiers with statistical significance, There may be a difference in the results of $B_1$(in the 1st regression) and $B_3$(in the 2nd regression).
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# Solid Cone
MathJax TeX Test Page The important thing to remember is that the total inertia is the sum of many moment of inertias of solid discs. $$dI = \dfrac{1}{2}dm*r^2$$ Why is this true? Because the total moment of inertia is the sum of moment of inertias of solid disc, and the formula is $\frac{1}{2}mr^2$. $$dm = \rho \text{ (density) } * \pi{}r^2 \text{ (area) } \mathrm{d}x \text{ (width) }$$ $$\rho = \dfrac{M}{\pi{}R^2h\frac{1}{3}} = \dfrac{3M}{\pi{}R^2h}$$ $$\mathrm{d}m = \dfrac{3M}{hR^2}*r^2\mathrm{d}x$$ $$\int_0^h \frac{1}{2}\left(\dfrac{3M}{hR^2}*r^2\right)* r^2\mathrm{d}x$$ $r$ is equal to $y$, which is equal to $\frac{R}{h}x$ $$\dfrac{3M}{2hR^2}\int_0^h y^4 \,\mathrm{d}x$$ Looking at the picture above, $y = \dfrac{R}{h}x$, so the integral is $$\dfrac{3M}{2hR^2}\int_0^h \frac{R^4}{h^4}x^2 \,\mathrm{d}x$$ $$\dfrac{3MR^2}{2h^5}\int_0^h x^4 \,\mathrm{d}x$$ $$\dfrac{3MR^2}{2h^5} * \frac{h^5}{5} = \boxed{\frac{3MR^2}{10}}$$
David Witten
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# Ambitious Scratch-built Project: High-5
### Help Support The Rocketry Forum:
#### DTH Rocket
##### Well-Known Member
I'm a little concerned about this one. It's kind of ambitious for my experience level, hence I am seeking input. It's only my second high-power rocket, scratch built, and made without any special materials or precision tools, so I can only work with so much (and I'm only 16, so I can only work with so much cash, too). I named it High-5 because its a five-motor cluster that goes high.
The electronics for staging are MiniTimer from Apogee Components. I've never used it before, but I still have at least a year before I'm ready to fly this rocket anyway, and I'm sure that on its first flight I'll use a single I-motor, possibly a J-motor.
I posted a few build pictures awhile back, but now all that has been lost, so here is the rocket in its current state. The lower portion has been fiberglassed, but the payload section and the upper section has not been glassed yet. A lot of the internal structures are in place, but the payload bay is completely empty (not the staging electronics bay). I've never even used my mini RRC altimeter for deploying parachutes, so I barely have an idea on how to put this payload bay together. It's an odd shape, the part in between the 3-inch section and the 4-inch section. I'll post more about this rocket another day.
I am currently 16 years old, and am hoping to be Junior L1 certified sometime next summer on a simpler rocket.
Last edited:
#### DTH Rocket
##### Well-Known Member
Aw, I guess you can't upload RockSim design files anymore.
#### BsSmith
##### Well-Known Member
I am currently 16 years old, and am hoping to be Junior L1 certified sometime next summer on a simpler rocket.
Why not certify on this? How many people can say they certifies JR L1 on a 5 motor cluster?
#### DTH Rocket
##### Well-Known Member
Why not certify on this? How many people can say they certifies JR L1 on a 5 motor cluster?
I'm a chicken. Sorta. Something in me tells me to start small.
#### MarkM
##### Well-Known Member
Something in me tells me to start small.
No, you're smart...not chicken
I see a potential issue with mounting your RRC in the transition (or just below, too). Ebay vent holes placed in transitions and the tube sections just below it are subject to turbulent airflow around them. This results in potential "odd" ebay pressurization issues. A barometric altimeter, which the RRC is, could very likely be confused by this resulting in an anomolous recovery deployment at apogee -- anything from too early, too late, or maybe not at all.
By convention, barometric altimeters should have their pressurization holes about 3 BT diameters away from the nearest nosecone or transition. At this distance, the turbulent airflow around the transition/nosecone has disipated.
You have 2 choices, IMO.
1) Be sure your vent holes are placed ABOVE the transition, not on it or below it. By placing the vent holes above the transition, you're away from the turbulent airflow caused by the transition. Just be sure the holes are 3 BT diameters from the nosecone.
2) Use an accelerometer-based altimeter; they are not affected by turbulent airflow since they deploy apogee recovery by detecting a decrease in acceleration.
Interesting build....
Last edited:
#### DTH Rocket
##### Well-Known Member
I see a potential issue with mounting your RRC in the transition (or just below, too).
No, the barometric altimeter and vent holes will be just above the transition. The payload bay includes the transition and the segment just above it. I read about this problem in Modern High-Power Rocketry 2. Thanks for the input. It's more advice like that that I need. I don't have an expert on hand, so I'm left to go-it-alone usually, which isn't always safe.
Last edited:
#### MarkM
##### Well-Known Member
No, the barometric altimeter and vent holes will be just above the transition.
Then you should be fine. It wasn't clear in your original post. MHPR 2 is a great book. Lots of useful info.
EDIT....just noticed the new photo. You only have one vent hole? That's typically not a wise design. If there happens to be unusual airflow or the airflow is disturbed around that hole during flight you'll have a problem. You should use 3 or 4 smaller holes which have the total equivalent area of that one hole. I use 3 in all my rockets using altimeters.
Last edited:
#### DTH Rocket
##### Well-Known Member
You only have one vent hole? That's typically not a wise design. If there happens to be unusual airflow or the airflow is disturbed around that hole during flight you'll have a problem. You should use 3 or 4 smaller holes which have the total equivalent area of that one hole. I use 3 in all my rockets using altimeters.
I haven't put any vent holes in yet, but I just looked it up in MHPR2. One suggestion is to have a 1/4 inch hole for every 100 cubic inches of payload bay area. I calculated it out, and it turns out I have 500 cubic inches in my bay!
A=Bh = 7(pi*3^2)+6(pi*4^2) (I have two diameters in the payload bay) = 499.5 inches^3.
So by that formula I should have 5 1/4 inch vent holes. Does that sound about right?
#### Indiana
##### Well-Known Member
Aw, I guess you can't upload RockSim design files anymore.
You can if you zip them with winzip first.
#### MarkM
##### Well-Known Member
So by that formula I should have 5 1/4 inch vent holes. Does that sound about right?
No...way, way too big, especially for 5 of them. On 3" airframes I use 3 7/16" vent holes for an altimeter bay 7" long.
Take a look HERE for a great table and formulas. Five vent holes is total overkill. Use 3 or at most 4.
One of the main reasons your calcs are wrong is you used the diameter of the payload bay to calculate area. Volume of a cylinder is pi*radius^2. I'm a bit confused on why your altimeter bay is 13" long. You also need to confine your altimeter to a smaller volume. You should have bulkheads separating the altimeter from the parachute bays. Even at your long ebay your volume is ~ 130 cu in.
Last edited:
#### DTH Rocket
##### Well-Known Member
No...way, way too big, especially for 5 of them. On 3" airframes I use 3 7/16" vent holes.
Take a look HERE for a great table and formulas. Five vent holes is total overkill. Use 3 or at most 4.
Did you mean 3/16? Or 7/32? Because 7/16 is almost half an inch.
#### MarkM
##### Well-Known Member
Did you mean 3/16? Or 7/32? Because 7/16 is almost half an inch.
Sorry, my bad. Neither. It should be 7/64.
For your ebay as it is now, I'd use 3 holes which are 5/32" each. Drill them into the 4" section.
If you want to use 4 go with 9/64" holes.
Either way, be sure they are spaced equally around the tube.
Last edited:
#### BAR0051
##### Well-Known Member
That is very ambitious and I would never use such a complex rocket for a cert flight. I was more nervious about my first dual deplolyment flight than I was for my level 2 cert. I did a few dual deployment flights and was comfortable with it before I did my level 2 cert flight. I would not dream of using electronics and motor clusters for a cert flight unless I had done it several times before and was sure of sucess.
#### Handeman
##### Well-Known Member
TRF Supporter
I think there would be less confusion and better advice if you also included some pics with the rocket apart so we can see how the Alt-bay and payload sections are related.
There is some considerations with the staging electronics. Here I'm making an assumption. You are going to use staging electronics to light the outer motors and the rocket will lift off on only the central motor. I assume this is the way you will configure this rocket because that is usually the safest method.
Using that method, there are two things that MUST happen for a safe flight.
1. The central motor MUST be large enough to fly the rocket safely by itself. This is in case of electronic, battery, igniter, or any other type of failure that could occur to the air staging/ignition system after the central motor lifts the rocket off the pad.
2. The central motor MUST provide enough thrust to generate enough Gs to trip the G-switch on the staging electronics. Having the best electronics, igniters, etc, will do you no good if the staging electronics are never activated and start its timing because it never sensed launch. I watched two cluster rockets crash for this very reason. The first got lucky, the chute deployed and just as it was hitting the ground, the main chute snapped open, triggered the G-switch and the outer motors fired, sending the rocket back up in the air dragging the main chute. It ended up only braking a fin. The second was a large rocket that only got 1 - 2 hundred feet up without the outer motors and came crashing down. No chutes opened, but there were lots of pieces.
My advise would be look at the specs on the G-switch on your staging electronics. It should list a min. and max G range for the switch. Below the min. is where the switch will NEVER switch, above the max. is where it will ALWAYS switch, and in between is where it MIGHT switch. You want your central motor to provide enough Gs to be a safe margin above the max. value. If the switch is rated 3G to 6G and you want a 50% safety margin, your central motor should generate at least 50% more then the 6Gs, or at least 9Gs at launch.
Don't ever calculate the safe liftoff/fight of a cluster rocket on all the motors lighting.
#### Handeman
##### Well-Known Member
TRF Supporter
That is very ambitious and I would never use such a complex rocket for a cert flight. I was more nervious about my first dual deplolyment flight than I was for my level 2 cert. I did a few dual deployment flights and was comfortable with it before I did my level 2 cert flight. I would not dream of using electronics and motor clusters for a cert flight unless I had done it several times before and was sure of sucess.
This is where we differ. I see the electronics and clusters as a challenge and would personally welcome that for a cert flight. My cert was a mailing tube rocket with home made nosecone and chutes, with DD and a long burn I motor that got it to 4,200 feet. I wasn't "sure" it was going to work, but I was confident I had done everything to make it work. What didn't worry me at all was if I was going to get the cert or not. If I didn't, then I just repair/build a rocket and try again.
That is why I'm looking at an L2 cert flight on a L or large K, something that will bust Mach. I've never done that before and its all about he challenge for me. Everyone has to approach it in a way that makes the hobby fun and exciting for them. Just stay safe.
#### MarkM
##### Well-Known Member
That is very ambitious and I would never use such a complex rocket for a cert flight.
By his initial post, it seems as this is a rocket for AFTER he certifies.
In terms of clustering, don't forget that at Cert Level 1 you are limited to 640 Ns total impulse.
#### BAR0051
##### Well-Known Member
By his initial post, it seems as this is a rocket for AFTER he certifies.
In terms of clustering, don't forget that at Cert Level 1 you are limited to 640 Ns total impulse.
My bad, it was BsSmith that suggested using it to cert. Just my $.02 worth. Last edited: #### MarkM ##### Well-Known Member My bad, it was BsSmith that suggested using it to cert. Just my$.02 worth.
For what's worth, I do agree with you in terms of keeping certs simple, especially the L1. Many don't feel that way, but to each his own.
#### DTH Rocket
##### Well-Known Member
Hey Handeman, you were right in your assumptions. The full potential of this rocket is a central K and 4 outboard G's or maybe even H's. The K motor in that case would definitely lift the rocket on its own, as I am anticipating the max. weight to be about 10 pounds fully loaded. On its first flight, however, I'm not going to fly it complex--just a central motor. Before I launch this rocket with the outboard motors I want to 1) test the staging electronics on a smaller multistage rocket, and 2) get my Junior L1 on a single I-motor in the rocket.
#### Handeman
##### Well-Known Member
TRF Supporter
Sounds like a plan. The central K will definately do the job. Sounds like a central I is about as small as you want to go.
Good Luck
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Open access peer-reviewed chapter - ONLINE FIRST
# Breast Cancer Drug Repurposing a Tool for a Challenging Disease
By Jonaid Ahmad Malik, Rafia Jan, Sakeel Ahmed and Sirajudheen Anwar
Submitted: October 4th 2021Reviewed: October 25th 2021Published: January 3rd 2022
DOI: 10.5772/intechopen.101378
## Abstract
Drug repurposing is one of the best strategy for drug discovery. There are several examples where drug repurposing has revolutionized the drug development process, such as metformin developed for diabetes and is now employed in polycystic ovarian syndrome. Drug repurposing against breast cancer is currently a hot topic to look upon. With the continued rise in breast cancer cases, there is a dire need for new therapies that can tackle it in a better way. There is a rise of resistance to current therapies, so drug repurposing might produce some lead candidates that may be promising to treat breast cancer. We will highlight the breast cancer molecular targets, currently available drugs, problems with current therapy, and some examples that might be promising to treat it.
### Keywords
• drug repurposing
• breast cancer
• mechanism
• non-oncology drugs
• resistance
## 1. Introduction
Drug discovery is a multifaceted process that aims at identifying a therapeutic agent that can be useful in treating and managing certain medical conditions. This process includes identification of candidates, characterization, validation, optimization, screening, and assays for therapeutic effectiveness. If a molecule achieves acceptable results in these studies, then the molecule has to go through drug development processes and be recruited to clinical trials [1]. Several drug candidates (about 90%) have collapsed in early clinical trials due to unexpected results such as adverse effects or inadequate effectiveness [2, 3]. Drug development is probably among the most complicated and challenging processes in biomedical research. Apart from the already enormous complexities underlying pharmacological drug designs, additional significant challenges arise from clinical, regulatory, intellectual property, and commercial constraints. Such as challenging atmosphere has made the drug development process very sluggish and unpredictable [4]. The process of discovering and developing a new drug is a lengthy and expensive process taking somewhere from 10 to 15 years and costs about US$2–3 billion [1]. Despite massive sums of money being spent on drug development, no substantial rise in the new therapeutic drug agents in a clinical setting has been observed over several decades. Although overall global R&D spending for drug discovery has risen 10-fold from 1975 (the US$4 billion) to 2009 (\$40 billion), the number of novel molecular entities (NMEs) approved has stayed essentially constant since 1975 (26 new drugs approved in 1976 and 27 new drugs approved in 2013) [5].
The essential step in discovering new drugs involves the evaluation of the safety and effectiveness of new drug candidates in human subjects, and it consists of four phases. In Phase I clinical trial, the candidate drug’s safety is assessed in a small population (20–80 individuals) to establish safe dose range and uncover adverse effects. Phase II involves the examination of intervention for its effectiveness and safety in large populations (a few hundred people). Phase III further involves the assessment of drug efficacy in a large population (several thousand) and compares new drug candidates with standard or experimental treatments. Phase IV is conducted when the intervention is marketed. This study aims to track how well the approved treatment is performing in the general population and gather data on side effects that may arise from broad usage over time. Phase III studies determine whether or not a medication is effective, and if so, FDA clearance is granted. The FDA approves one anticancer treatment out of every 5000–10,000 applicants, and just 5% of oncology medicines entering Phase I clinical trials are approved in the end. Because of the increased cost and time frame for new medication development in recent years, patients with severe illness may die until alternative therapies are available if they develop resistance to current therapy [6]. In searching for an alternative treatment option for managing various diseases, including cancer, the researchers have shifted their focus to drug repurposing strategies.
The drug purposing or drug reprofiling or drug redesigning process explores the therapeutic use of existing clinically approved, off patent drugs with known targets for another indication to minimize the cost of therapy, time, and risk [7]. The huge benefit of drug repurposing is that the efficacy, pharmacokinetics, pharmacodynamics, and toxicity characteristics have previously been explored in preclinical and Phase I investigation. These drug moieties may thus be quickly made to proceed to Phase II and Phase III clinical trials, and hence related developmental costs might be substantially lowered [6, 8]. The failure risk in drug development is low because in vitroscreening, in vivoscreening, toxicity profile, chemical optimization, and formulation development have already been accomplished. Therefore, drug repurposing has made the pharmaceutical industry a desirable choice for investors. So the pharmaceutical companies and researchers have begun to make significant investments in drug repurposing, which offers a tremendous benefit over de novodrug design and development [9]. Therefore this new approach of drug purposing has reduced the timeline and cost of the drug development, notably in the case of FDA-approved repurposed pharmaceuticals, which will undergo faster clinical trials because of already well-known safety and toxicological profile [10]. Outline of developing new drug verses repurposing is represented in Figure 1.
The development of new drugs for breast cancer like any other cancer is a multistep process that includes drug designing, synthesis, characterization, safety and efficacy assessment, and finally, regulatory approval (Figure 2). The overall process is very lengthy and involves significant financial expenditure [11]. Further, the sky-high cost of the therapies and associated side effects make it desirable to look for other approaches to manage cancer effectively. Therefore, concurrently with the synthesis and design of new therapeutic modalities, various strategies should be considered for repurposing various already approved drugs that may target this deadly disease.
## 2. Non-oncology drugs repurposed for breast cancer (preclinical data)
### 2.1 Aspirin
Aspirin was originally discovered in 1897 and was first commercialized as an analgesic. It has been utilized as an anti-inflammatory medication and for managing arterial and venous thrombosis [12]. Recent research has sparked interest in the usage of Aspirin for the prevention of various cancers. There are compelling evidences authenticating that regular use of low doses of aspirin results in a significant reduction in the occurrences and mortality of various cancers [13, 14, 15, 16, 17]. The possibility that Aspirin has an anticancer benefit has received considerable interest nowadays, with a lot of research being done to figure out how successful it is in the prevention of colorectal cancer [18], lung cancer [19], gastric cancer [20], prostate cancer [21], and many other cancers including breast cancer. Because of the effect of Aspirin in several biological processes such as inhibitory effect on angiogenesis [22], cancer cell metastasis [23], causing cell apoptosis [24], etc., it is reasonable to predict that Aspirin will be beneficial when employed as an additional alternative treatment option for cancer patients. Aspirin directly inhibits the activity of the enzyme cyclooxygenase (COX-2) and thereby impedes the synthesis of prostaglandin E2 (PGE-2), which leads to cancer cell death [25]. Recent research also suggests that Aspirin may mediate anticancer potential through COX-independent pathways such as inhibition of NFκB [26], downregulation of survivin [20], targeting AMPK-mTOR signaling [27], Wnt signaling cascade [28], etc.
A study was conducted by Dai et al. reported that Aspirin possesses antiangiogenic and anti-metastatic potential in MDA MB 23 cell line by directly binding to the enzyme heparinase. The results were further confirmed in vivoexperimentation [23]. Heparinase is an endo-β-D glucuronidase that is specific to heparin sulfate. It dissolves heparin sulfate chains of proteoglycans on the cell surface and extracellular matrix (ECM) that consequently contributes to the degradation of the extracellular matrix that further assists tumor invasion and metastasis [29]. Further, heparin also facilitated the release of angiogenic factor, vascular endothelial growth factor (VEGF) blocked by aspirin-mediated heparin inhibition [23]. Breast cancer cell lines (MDA MB 231 and MCF-7) showed a dose-dependent inhibitory effect on growth after treatment with Aspirin. The Aspirin further restricts the migration of these cells by preventing epithelial to mesenchymal transition through suppression of various mesenchymal markers such as vimentin and increasing expression of various epithelial markers such as Keratin-19 and E-cadherin.
Further inhibitory effect of TGF-β/SMAD4 signaling, as evident from decreasing the production of SMAD proteins, also contributes to the anti-metastatic potential of Aspirin [30]. In another study, Choi et al. demonstrated the effect of Aspirin in the MCF-7 cell line. It was observed that Aspirin alters the complex formation between Bcl-2 and FKBP38 and leads to the nuclear translocation of Bcl-2 and phosphorylation that causes its activation, contributing to its inhibitory effect on MCF-7 cell proliferation and also triggers apoptosis in cell lines [31]. In combination with exemestane, Aspirin showed synergy in inhibition of cell proliferation. Significant arrest in the G0/G1 phase was observed along with a more detrimental effect on COX-1 and Bcl-2 expression than individual therapy [32]. In addition, when combined with tamoxifen (which is used as a drug of choice for the estrogen receptor positive BC), it downregulates the level of cyclinD1. Subsequently, it arrests the cell cycle in phase G0/G1. In the same study, authors also reported that Aspirin inhibits the ER + ve BC cells growth and overcomes the resistance to tamoxifen in MCF-7/TAM cell line. Study demonstrated a new way to treat ER + ve BC in combination therapy of Aspirin and tamoxifen [33].
### 2.2 Metformin
Metformin (1,1-dimethyl biguanide hydrochloride) is a well-recognized biguanide derivative and has a long history of usage in managing type 2 diabetes (T2D). Because of the outstanding ability to lower plasma glucose levels, metformin has become the primary drug for managing T2D [34]. The drug was firstly approved in 1958 in the United Kingdom, and this decade-old drug is in the WHO’s list of essential medicines [35]. Metformin belongs to the category of successful repurposed drugs and advanced into the clinical trials Phase 3/4 for its use in the prostate, oral, breast, pancreatic, and endometrial cancers [6]. Various preclinical and clinical examinations have demonstrated the effectiveness of metformin in the treatment of various malignancies such as pancreatic cancer [36], gastric cancer [37], blood cancer [38], etc. A meta-analysis study on diabetic patients with breast cancer concluded that patients who were treated with metformin and neoadjuvant therapy had a higher pathological complete response rate (24%) compared with patients not undergoing metformin treatment (8%) [39]. Another meta-analysis study demonstrated 65% survival improvement when compared with control [40]. Metformin has increased the survival opportunity in type 2 diabetic patients suffering from invasive breast cancer [41]. Study also suggested that patients on metformin demonstrate improved in the survival and response to treatment [40]. The metformin uptake is mediated by the OCT1 in BC cells [42], which is reported to play important role in the BC cells as an anticancer activity [43]. Upon entry into the cells, it leads to increase apoptosis, anti-proliferative, anti-angiogenic, which seems to be mediated by the mTOR, Akt/MAPK pathway [44]. Study conducted by Shi et al., established that metformin can also inhibit the expression of the COX-2, suggested the potential of metformin in combination with others COX-2 inhibitor [45]. Low cost and stability of metformin make it a good candidate for the treatment of cancers when compared with available treatment options [46].
### 2.3 Itraconazole (ITC)
Itraconazole, a triazole antifungal drug, is a well-tolerable agent that is extremely effective against a wide range of fungal infections. Itraconazole is a highly potent and effective antifungal agent due to its active metabolite, hydroxy-itraconazole, which also has significant antifungal action [47]. Itraconazole blocks ergosterol synthesis in the fungal cell membrane by inhibiting the enzyme 14α-demethylase and suppressing their growth [48]. It has emerged as a potent anticancer agent because of its ability to overcome chemoresistance prompted by P glycoproteins, altering various signaling pathways such as hedgehog (Hh) signaling cascade, Wnt/β-catenin pathway in cancer cells, and also preventing angiogenesis and lymphangiogenesis [49]. Itraconazole has been shown to have the ability to eliminate cancer cells by disrupting Hh signaling [50]. In invertebrates, the Hh signaling cascade is responsible for the regulation of complicated developmental processes. However, aberrant activation of this pathway plays a crucial role in carcinogenesis and cancer maintenance and contributes to chemoresistance, thus, targeting this pathway offers the potential therapeutic possibility [51]. Itraconazole was able to exhibit cytotoxicity in breast cancer cell lines by influencing mitochondrial membrane potential through induction of apoptosis, decreasing expression of Bcl-2, and enhancing the caspase activity. Itraconazole also promoted autophagic cell death via elevation of LC3-II expression, degradation of P62/SQSTM1, formation of autophagosomes. Hedgehog signaling is an important regulator of apoptosis and autophagy. Hence, inhibition of this signaling by Itraconazole results in cytotoxicity, tumor shrinkage, apoptosis, and autophagy in breast cancer both in in vitroand in vivoinvestigations [50, 52]. Anticancer activity is also reported in esophageal cancer, mediated by downregulating the HERK/AKT pathway [53]. A pilot study with 13 participants demonstrated that increased levels of Itraconazole in plasma were associated with the increased level of thrombospondin-1, angiogenesis inhibitor.
Additionally, the level of other growth factors such as fibroblast growth factor (FGF) and placenta-derived growth factor also decreased without any direct association with the Itraconazole [54]. When administered in combination with other cytotoxic agents, Itraconazole increased the response rate [55]. Researchers are trying various ways to enhance the anti-neoplastic activity of itraconazole. One such example is the development of the modified lipid nanoparticles having Miltefosine (subtherapeutic dose), called M-ITC-LNC (Membrane additive itraconazole with lipid nanoparticles (Miltefosine). The results from the cytotoxicity studies demonstrated that the anticancer activity and selectivity significantly increased in MCF-7 BC cells compared with the ITC-solution and ITC-LNC without modification [56]. In another study, itraconazole was co-delivered with the doxorubicin by liposome (coated with the Pluronic P123), resulting in the increased anti-neoplastic activity in BC [57]. The combination of the verapamil and ITC with 5-FU decreased cell survival and proliferation.
Moreover, ITC and 5-FU are more effective in the treatment of BC [58]. Administration of the Itraconazole with erlotinib (tyrosine kinase inhibitor) increased the AUC and Cmax by 10.8 and 2.78-fold, respectively, without any SAE [59]. Abovementioned all the studies reveal the potential of Itraconazole alone or in combination with other anticancer agents to treat BC.
### 2.4 Simvastatin
Simvastatin belongs to the class of statins and is a well-explored hydroxy-methylglutaryl coenzyme A (HMG-CoA) reductase inhibitor that reduces cholesterol biosynthesis initially used to reduce cholesterol biosynthesis marketed in 1988 [60]. Clinical data suggest that statins are effective in BC management. Statins amplify tumor cell death and radiosensitivity in various cell lines, inhibit invasion and proliferation, and show anti-metastatic activity. Clinical trials conducted on breast cancer (inflammatory and TNBC) patients also favored these observations by representing improved mortality benefits for patients on statins [61, 62]. In the same context, Simvastatin is the most explored statin to explore the role of statins in cancer. Simvastatin targets the transcription factor NFκB that reduces the expression level of anti-apoptotic protein Bcl-xL, concomitantly inhibits the expression of anti-proliferative and proapoptotic tumor suppressor PTEN and hence inhibiting the growth of breast cancer cells. The elevation of PTEN expression results in the suppression of Akt phosphorylation. Akt activity is upregulated in many cancers by increasing cancer cell survival, inhibiting apoptosis, and increasing proliferation. Therefore, Simvastatin substantially decreased Akt phosphorylation concurrently with the reduction in expression of anti-apoptotic protein by dysregulation of NFκB, thus showing the anticancer activity against BC [63]. On administration of Simvastatin, the expression of PTTG1 (pituitary tumor-transforming gene 1) was also reduced in a dose-dependent manner in the MDA-MB-231 cell line. PTTG1 is the important gene involved in the invasion and metastasis of BC [64]. In the same cell line (MDA-MB-231), Simvastatin leads to fragmentation of the cell’s nuclei, subsequently inducing apoptosis. It also enhanced the level of ROS in a dose-dependent manner, which causes oxidative stress and further DNA damage [65]. Apoptotic effects were due to the increased expression of miR-140-5p in a dose-dependent manner mediated by the activating transcription factor NRF1 [65]. Apart from the MDA-MB-231 cell line, Simvastatin effects were also explored in other breast cancer cell lines such as T47D, BT-549, and MCF-7, showing apoptotic inducer anti-proliferative activity [66, 67]. In in vivostudies with DMBA (dimethyl-Benz(a)anthracene) induced breast cancer rat model, Simvastatin reduced the tumor volume by around 80% [68]. Karimi et al. also explored its activity in breast cancer mice model and reported improved mortality and tumor volume compared with control [69]. Although Simvastatin’s lipophilic nature makes it a good candidate for the BC treatment, the researcher tried to develop nano formulations to improve the delivery in a targeted specific manner and reduce the non-target side effects. Detailed mechanism of cell growth inhibition, division, survival, migration, and proliferation by Simvastatin is presented in Figure 3.
In the same series, Sed et al. used nanoparticles made of superparamagnetic iron oxide to Simvastatin delivery with enhanced anticancer activity in the PC-3 cell line. This action is mediated by inducing apoptosis and cell cycle arrest in the G2 phase [70]. Researchers from another lab developed poly D, L-lactide-co-glycolide (PLGA) with cholic-acid-based nanoparticles for Simvastatin release in a sustained and controlled manner for breast adenocarcinoma treatment. These nanoparticles showed maintainable and more efficiently inhibit tumor growth than normal Simvastatin [71]. Other formulations such as nanocapsule [72], nanoemulsions [73], liposomes [74], and immunoliposome [75] for Simvastatin were developed with increased anticancer activity in breast cancer cells. In a randomized placebo-controlled study, Simvastatin shows a better anticancer profile with the carboplatin and vinorelbine in metastatic breast cancer [76]. Consistency in the results from both clinical and preclinical studies suggests the vast potential of Simvastatin in treating breast cancer either alone or in combination. Moreover, the development of nanoformulations also provided advantages such as enhanced cytotoxicity, lower side effects, targeted delivery over the conventional available treatment options for BC.
### 2.5 Niclosamide
Niclosamide, an FDA-approved anthelminthic drug used to manage tapeworm infection, has been used almost from the last half of the century and included in the WHO’s list of essential medicines. Recent research suggests that niclosamide has a wide range of therapeutic uses other than treating parasitic infection. Niclosamide’s clinical application diseases include type 2 diabetes, endometriosis, neuropathic pain, bacterial and viral infections, including cancer [77]. The anticancer benefits of niclosamide have been shown in many malignancies such as colon cancer, lung cancer, prostate cancer in humans, as well as breast cancer by suppressing various cancer related pathways such as Wnt Notch, mTOR, STAT, and NFκB [78, 79]. The combinational treatment of niclosamide with cisplatin overcomes the resistance to cisplatin and induces an inhibitory effect on proliferation in vitroand reduced tumor size in vivo.
Further, niclosamide prevented the epithelial-mesenchymal transition (EMT) by suppressing mesenchymal markers such as snail and vimentin. The inhibitory effect on EMT and prevention of stem-like phenotype of TNBC by Niclosamide operate by disabling various abnormal signaling pathways such as Akt, ERK, and Src [80]. The niclosamide acts as a potent inhibitor of STAT signaling by preventing cancer cell proliferation, invasion, and metastasis by decreasing the phosphorylation of STAT3 that otherwise was found in 35% of breast cancer tissues. Furthermore, STAT3 promotes the expression of several key downstream genes involved in proliferation, cell survival, and angiogenesis in breast cancer [81]. Human monocyte cells were reduced to HUVECs in the presence of niclosamide. Niclosamide also inhibited VCAM-1 and ICAM1 protein expression in HUVECs. Niclosamide decreased HUVEC proliferation, migration, and development of cord-like structures. In vivo, niclosamide inhibits VEGF-mediated angiogenesis [77]. Niclosamide inhibited Wnt/Frizzled 1 signaling, mediated by the increased degradation of the Wnt co-receptor LRP-6 (low-density lipoprotein receptor-related protein 6) [82, 83, 84]. Osada et al. determined that on the administration of niclosamide, there was a decrease in Dvl2 expression, which further impeded the downstream signaling (β-catenin) [85]. Londoño-Joshi et al. reported that niclosamide administration also reduced levels of LRP6 and β-catenin in breast cancers [86]. In combination with doxorubicin, niclosamide induces apoptosis and synergistically increases breast cancer cell death. This action is mediated by Wnt/β-catenin pathway downregulation and arrest of the cell cycle by Niclosamide in G0/G1 while both doxorubicin and niclosamide increased ROS production, thus showing cytotoxicity [87, 88]. Niclosamide also showed synergistic anticancer activity with 8-quinolinol [89]. When niclosamide is administered with cisplatin, it could inhibit the invasion and cell stemness of breast cancer cells, mediated by downregulation of anti-apoptotic protein Bcl2 [90]. In a recently published study, albumin-bound niclosamide (nab-Niclo) (Albumin-based nanoparticle transport systems) was found to inhibit cell growth, induce cell death, mitochondrial dysfunction, and increase oxidative stress with DNA damage. This nab-Nicolo was appeared more effective than normal Niclosamide for BC treatment [91]. Taken together, all the data suggest that niclosamide alone and in combination with other drugs could be used for the normal BC and resistance BC all repurposed drugs for BC discussed in this chapter summarized in Table 1.
DrugExperimental modelMechanism of actionObservationOriginal indicationReferences
AspirinB16F10, MDA-MB-231, MDA-MB-435 xenograft mode
MCF-7, MDA-MB-231
MCF-7, MDA-MB231
Inhibition of heparinase
↓EMT
Apoptosis
↓Metastasis, ↓angiogenesis
↓Mesenchymal markers (vimentin, Snail, TWIST)
ItraconazoleMCF-7 and SKBR-3 breast cancer cellAntiproliferative effect via inhibition of hedgehog signaling cascade
↑Apoptosis
↑Autophagy
↑Cell cycle arrest
↓Tumor size
↑Caspase 3
↓Bcl-2
Antifungal drug[50]
Niclosamide↓EM T
↑Apoptosis
Inhibition of stat signaling
↓Snail
↓Vimentin
↓Tumor growth
↑Caspase 3, ↓Bcl-2, ↓surviving, ↓Mcl-1 expression
Anthelminthic drug[80, 81]
SimvastatinMDA-MB-231, T47D, BT-549 and MCF-7 (in vitro)
DMBA model (in vivo)
↓PTTG1,
↓Bcl-xL
↑ROS
↑miR-140-5p
Inhibition of Akt and DNA damage
Anti-proliferative, induce apoptosis, and increased survivalAnti-hypercholesterolemic drug[63, 64, 65, 66, 67]
MetforminMDA-MB-231, MCF-7Via mTOR, Akt/MAPK pathway
COX-2 inhibition
Apoptosis, anti-proliferative, anti-angiogenicAnti-diabetic drug[44, 45]
### Table 1.
Summary of the repurposed drugs for BC discussed in the chapter.
## 3. Conclusion
Drug discovery is a multifaceted process that aims at identifying a therapeutic agent that can be useful in treating and managing various ailments. This process includes identification of candidates, characterization, validation, optimization, screening, and assays for therapeutic effectiveness. As the mortality due to cancer is progressively increasing, we need effective therapy to treat breast cancer patients or improve survival. When any pharmaceutical organization starts developing a novel chemical entity for the BC, its cost and attrition rate are very high. Drugs repurposing is how we can minimize the cost and attrition rate by using the already marketed drugs for a new use. Drug repurposing against breast cancer is one of the best alternatives to treat progressive ailments. In the above discussion, we have discussed various drugs that can be repurposed against breast cancer. It will be a game-changing scenario in the treatment of breast cancer. Certain challenges need to be rectified. However, there is a need for optimization of models and more screening of drugs at preclinical stages.
## 4. Future prospective
To tackle all the challenges associated with the drug development process for breast cancer, scientists need to shift their interest to the alternative drug development, that is, drug repurposing. All the BC repurposed drugs discussed in the book chapter show impressive results that suggest exploring more new non-cancerous drugs for cancerous use [92]. Using the drugs repurposing approaches alone and in combination with other drugs will also reduce the side effects associated with high doses. It will also reduce the cost of the drug development process, ultimately patient compliances and burden. Patients who could not afford the treatment due to the high cost can take treatment and improve survival. As the safety is already studied of drugs that seem a novel interest in the repurposing for BC, the chances of failure at the clinical level will also be less. With the advancement in drug repurposing, there is still a need to develop a valuable model of different types of cancers that mimic cancer. The development of such a model provides the actual clue for drug repurposing. So far, the advantages we discussed, there are some challenges associated with the drugs repurposing such as patent issue, regulatory consideration, inequitable prescription that need to be overcome so, more and more pharma companies show their interest in drug repurposing. It is expected that drug repurposing will achieve the milestone that is currently not possible with the conventional available treatment for cancers in the future. Furthermore, new nanoformulations need to be developed for the targeted and specific delivery of repurposed anticancer drug to avoid the off-target side effects.
chapter PDF
## More
© 2021 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
## How to cite and reference
### Cite this chapter Copy to clipboard
Jonaid Ahmad Malik, Rafia Jan, Sakeel Ahmed and Sirajudheen Anwar (January 3rd 2022). Breast Cancer Drug Repurposing a Tool for a Challenging Disease [Online First], IntechOpen, DOI: 10.5772/intechopen.101378. Available from:
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# User:Rds/My sandbox
Hi Colleagues,
This is my first entry into my sandbox. I wonder why this page is called a sandbox. In any case, I'm delighted to be learning about how to use the WikiEducator. :Here is the url of my blog: http://richarddsolomonsblog.blogspot.com/ [[|]] Richard
[[]]
There are lots of things that I want to learn including
• How to use Wiki syntax including
1. How to bold
2. How to italicize
• How to use Wiki syntax including
1. How to indent
2. How to hypertext words in order to link words to an internal or external page
Now let's try to link this text to the main Wikipedia url by inserting this syntax: wikipedia
Now let's try to create a new page called Sandpit. Remember to use the double brackets before and after the name of the new page, Sandpit.
Let's see if I can add a template (I did this by clicking on the {T} button on editor menu above):
Let's see if I can add a resource (I did this by clicking on the first <R> button on the editor menue above):
[1] [2]
Cite error: <ref> tags exist, but no <references/> tag was found
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# Spin 1/2 correlation
1. Mar 12, 2012
### gespex
Hello all,
Imagine two spin 1/2 particles that are entangled, going towards two stern-gerlach apparatuses, with some relative angle. Now imagine one stern-gerlach device measures the spin of one of the particles as up. What is the chance that the other stern-gerlach device measures the spin to be down?
For 90 degrees it would be 50/50, right? So my guess is $cos^2 ({1 \over 2} \alpha)$. Is that correct?
2. Mar 13, 2012
### sergiokapone
You can look here , it is the same quastion.
3. Mar 13, 2012
### DrChinese
Yes, that's correct. I may have steered you wrong at an earlier time.
4. Mar 13, 2012
### gespex
Thanks for your answer. It doesn't seem to be the same question though - the question of that guy is a lot more advanced than mine. He is asking a question about the implication of the correlations, while I am simply looking for the actual formula for the correlation.
From what I understand from that source, he takes into account two possibilities for the case $\alpha = 90°$: a 50% correlation and a 100% correlation. I thought it was 50%, which would I believe indicate that I was right thinking the correlation is $cos^2({1 \over 2} \alpha$. However, I'm not sure what he meant with the 100% correlation.
Thanks
5. Mar 13, 2012
### gespex
I did ask the question earlier, and you were the one to answer it. But reading up later I assumed you were talking about the correlation as (a - b)/(a + b)? So I did wonder if it was simply a miscommunication, hence me asking the question again.
6. Mar 13, 2012
### sergiokapone
I am thet guy.
As I understood of the answers to my question, the correct answer to your quastion it is
7. Mar 13, 2012
### sergiokapone
Thus, yes, it is correct.
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# Divide line in $XY$-dimension
For example we have line $A$ with coordinates $(0, 1, 10, 9)$;
And we need to divide this line by $3$ (so we have now $A_1, A_2$ and $A_3$), where $A_1 + A_2 + A_3 = A$; Is there equation, to find coordinates of $A_1, A_2$ and $A_3$ points?
With best regards
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I think, for A1 its (10 - 0)/3, (9-1) / 3; – JohnDow Feb 23 '13 at 13:49
What coordinates do you use for a line? It looks as if your coordinates describe a line segment in the plane by giving coordinates for its two end points. If so, in which order do you use them. Also, why do you want three points $A_i$ which add up to a line? If you want three four-element vectors that add up to a line, I'd interpret each such vector as a line in its own right. – MvG Feb 25 '13 at 6:08
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# Article
Full entry | PDF (1.6 MB)
Summary:
The distribution of each member of the family of statistics based on the $R_{\phi }$-divergence for testing goodness-of-fit is a chi-squared to $o(1)$ (Pardo [pard96]). In this paper a closer approximation to the exact distribution is obtained by extracting the $\phi$-dependent second order component from the $o(1)$ term.
References:
[1] Ali M. S., Silvey D.: A general class of coefficients of divergence of one distribution from another. J. Roy. Statist. Soc. Ser. B 28 (1966), 131–140 MR 0196777 | Zbl 0203.19902
[2] Bednarski T., Ledwina T.: A note on a biasedness of tests of fit. Mathematische Operationsforschung und Statistik, Series Statistics 9 (1978), 191–193 MR 0512257
[3] Burbea J.: $J$-divergences and related topics. Encycl. Statist. Sci. 44 (1983), 290–296
[4] Burbea J., Rao C. R.: On the convexity of some divergence measures based on entropy functions. IEEE Trans. Inform. Theory 28 (1982), 489–495 MR 0672884 | Zbl 0479.94009
[5] Burbea J., Rao C. R.: On the convexity of higher order Jensen differences based on entropy function. IEEE Trans. Inform. Theory 28 (1982), 961–963 MR 0687297
[6] Cohen A., Sackrowitz H. B.: Unbiasedness of the chi–square, likelihood ratio, and other goodness of fit tests for the equal cell case. Ann. Statist. 3 (1975), 959–964 MR 0381087 | Zbl 0311.62021
[7] Cressie N., Read T. R. C.: Multinomial goodness of fit test. J. Roy. Statist. Soc. Ser. B 46 (1984), 440–464 MR 0790631
[8] Csiszár I.: Eine Informationtheoretische Ungleichung und ihre Anwendung auf den Bewis der Ergodizität von Markhoffschen Ketten. Publ. Math. Inst. Hungar. Acad. Sci. Ser. A 8 (1963), 85–108 MR 0164374
[9] Greenwood P. E., Nikolin M. S.: A Guide to Chi–squared Testing. Wiley, New York 1996 MR 1379800
[10] Liese F., Vajda I.: Convex Statistical Distances. Teubner, Leipzig 1987 MR 0926905 | Zbl 0656.62004
[11] Mann H. B., Wald A.: On the choice of the number of intervals in the application of the chi–square test. Ann. Math. Statist. 13 (1942), 306–317 MR 0007224
[12] Pardo M. C.: On Burbea–Rao divergences based goodness–of–fit tests for multinomial models. J. Multivariate Anal. 69 (1999), 65–87 MR 1701407
[13] Pardo M. C., Vajda I.: About distances of discrete distributions satisfying the data processing theorem of information theory. Trans. IEEE Inform. Theory 43 (1997), 4, 1288–1293 MR 1454961 | Zbl 0884.94015
[14] Pearson K.: On the criterion that a given system of deviations from the probable in the case of a correlated system of variables is such that it can be reasonably supposed to have arisen from random sampling. Philosophy Magazine 50 (1900), 157–172
[15] Rao C. R.: Asymptotic efficiency and limiting information. In: Proc. 4th Berkeley Symp. on Math. Statist. Probab. 1, Univ. of California Press, Berkeley 1961, pp. 531–546 MR 0133192 | Zbl 0156.39802
[16] Rao C. R.: Linear Statistical Inference and Its Applications. Second edition. Wiley, New York 1973 MR 0346957 | Zbl 0256.62002
[17] Rao C. R.: Analysis of Diversity: A Unified Approach. Technical Report No. 81-26, University of Pittsburhg 1981 MR 0705316 | Zbl 0569.62061
[18] Read T. R. C.: On Choosing a Goodness–of–fit test. Unpublished Ph.D. Thesis, Flinders University, South Australia 1982 Zbl 0564.62033
[19] Read T. R. C.: Closer asymptotic approximations for the distributions of the power divergence goodness–of–fit statistics. Ann. Inst. Statist. Math. 36 (1984), Part A, 59–69 MR 0752006 | Zbl 0554.62015
[20] Read T. R. C., Cressie N.: Goodness of Fit Statistics for Discrete Multivariate Data. Springer, New York 1988 MR 0955054 | Zbl 0663.62065
[21] Schorr B.: On the choice of the class intervals in the application of the chi–square test. Math. Operationsforsch. Statist. 5 (1974), 357–377 MR 0388650 | Zbl 0285.62018
[22] Siotani M., Fujikoshi Y.: Asymptotic approximations for the distributions of multinomial goodness–of–fit statistics. Hiroshima Math. J. 14 (1984), 115–124 MR 0750392 | Zbl 0553.62017
[23] Sturges H. A.: The choice of a class interval. J. Amer. Statist. Assoc. 21 (1926), 65–66
[24] Vajda I.: Theory of Statistical Inference and Information. Kluwer Academic Publishers, Dordrecht – Boston 1989 Zbl 0711.62002
[25] Yarnold J. K.: Asymptotic approximations for the probability that a sum of lattice random vectors lies in a convex set. Ann. of Math. Statist. 43 (1972), 1566–1580 MR 0372967 | Zbl 0256.62022
[26] Zografos K., Ferentinos K., Papaioannou T.: $\varphi$-divergence statistics: sampling properties and multinomial goodness of fit divergence tests. Comm. Statist. – Theory Methods 19 (1990), 5, 1785–1802 MR 1075502
Partner of
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# P(A) = p, P(B) = 2p, P(Exactly one out of A and B occurs) = 5/9. What will be the maximum value of p?
Solution:
Given: P(A) = p and P(B) = 2p
$$P(\bar{A} \cap B) + P(\bar{B} \cap A) = \frac{5}{9}$$
P(A)-P(A∩B)+P(B)-P(A∩B) = 5/9
3p-2P(A∩B) = 5/9
0 ≤ P(A∩B) ≤ p
For maximum value: 3p – 2p = 5/9
So, p = 5/9
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## Elementary Algebra
RECALL: A composite number is a number that has factors other than 1 and itself. A prime number is a number with no factors other than 1 and itself. Note that: $81 = 9 \times 9$ Thus, the given number is composite.
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# Using mdframed after a section heading
The following is a very contrived example, but it illustrates a problem I am unable to solve with the mdframed package. When the mdframed environment follows a section heading but there is insufficient room to place the contents environment, the section heading gets orphaned.
Replacing mdframed with minipage or anything else does not result in the orphan, and changing \clubpenalty has no effect.
The same result occurs, for example, if the environment contains only a single line but has a large value for skipabove or innertopmargin; the non-breakable content I use in the MWE below is just for easily demonstrate the result.
\documentclass{article}
\usepackage{mdframed}
\newcommand{\BoxContents}{top\par\vspace*{2in}bottom}
\begin{document}
\vspace*{5in}
\section{Section}
\begin{mdframed}% this orphans the section heading
\BoxContents
\end{mdframed}
\newpage
\vspace*{5in}
\section{Section}
\begin{minipage}[t]{\linewidth}% this does not orphan the section heading
\BoxContents
\end{minipage}
\end{document}
Is there some way I can use this environment following a section heading without manually breaking pages when this occurs?
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We had a discuss about the issue: chat.stackexchange.com/transcript/message/8856527#8856527 – Marco Daniel Apr 7 '13 at 14:50
In the chatroom David Carlisle and egreg helped me to point out the issue. Thanks for your engagement.
After \section normally you don't have any breaks avoiding orphans. Normally implies we have exceptions.
To allow colour specification you are buying an unwanted breakpoint. To demonstrate this I use the following example:
\documentclass{article}
\begin{document}
\showoutput\setbox0\vbox{%
\section{Section}
\penalty10000
\begin{minipage}[t]{\linewidth}
top\par\vspace*{2in}bottom
\end{minipage}
}\showbox0
\end{document}
In the log file you will find:
.\write1{\@writefile{toc}{\protect \contentsline {section}{\protect \numberline
\ETC.}
.\penalty 10000
.\glue 9.90276 plus 0.86108
.\penalty 10000
.\glue(\parskip) 0.0 plus 1.0
.\glue(\baselineskip) 5.84921
Now we modify the example in this way that we use mdframed.
\documentclass{article}
\usepackage{mdframed}
\begin{document}
\showoutput\setbox0\vbox{%
\section{Section}
\penalty10000
\begingroup\color{red}
\begin{minipage}[t]{\linewidth}
top\par\vspace*{2in}bottom
\end{minipage}
\endgroup
}\showbox0
\showoutput\setbox0\vbox{
\section{Sectionaa}
\penalty10000
\begin{mdframed}% this orphans the section heading
top\par\vspace*{2in}bottom
\end{mdframed}
}\showbox0
\end{document}
The output in the log file is:
.\write1{\@writefile{toc}{\protect \contentsline {section}{\protect \numberline
\ETC.}
.\penalty 10000
.\glue 9.90276 plus 0.86108
.\penalty 10000
.\rule(0.0+0.0)x345.0
.\pdfcolorstack 0 push {0 g 0 G}
.\glue 0.0
.\glue(\parskip) 0.0
.\hbox(0.0+0.0)x345.0, glue set 345.0fil
You can see a glue 0. There the break happens. I can't avoid the glue 0!. Based on your example here the modification of minipage which results in the same issue:
\documentclass{article}
\usepackage{color}
\begin{document}
\vspace*{5in}
\section{Section}
\begingroup\color{red}
\begin{minipage}[t]{\linewidth}
top\par\vspace*{2in}bottom
\end{minipage}
\endgroup
\end{document}
So the issue based on the color usage.
-
Thanks for the walkthrough. I was going through what you went over in chat, and I tried experimenting with some of the definitions in mdframed.sty. In the definition of the environment mdframed, if I comment out the line \color{mdf@fontcolor}, the \pdfcolorstack item disappears from the log. If I comment out the \mdf@trivlist{\mdf@skipabove@length} then the \glue 0.0 disappears. However the page will still break at that point. So unless I am missing something, I don't think it is the \glue 0.0 mentioned in the chat. – grizzilus Apr 7 '13 at 21:41
@grizzilus: Good point. I will discuss this issue further on. – Marco Daniel Apr 8 '13 at 17:19
Was a solution ever found? – Jim Paris Jun 29 '13 at 16:34
@JimParis: Unfortunately not. I tested several other way where \vspace*{2in} is replaced by test\par Test.... and I couldn't reproduce the issue. – Marco Daniel Jun 29 '13 at 16:53
I find it really easy to reproduce: writelatex.com/250587dmfxhc – Jim Paris Jun 29 '13 at 17:43
|
# IMDB¶
In [ ]:
%reload_ext autoreload
%matplotlib inline
In [ ]:
from fastai import *
from fastai.text import *
In [ ]:
torch.cuda.set_device(2)
## Preparing the data¶
First let's download the dataset we are going to study. The dataset has been curated by Andrew Maas et al. and contains a total of 100,000 reviews on IMDB. 25,000 of them are labelled as positive and negative for training, another 25,000 are labelled for testing (in both cases they are highly polarized). The remaning 50,000 is an additional unlabelled data (but we will find a use for it nonetheless).
We'll begin with a sample we've prepared for you, so that things run quickly before going over the full dataset.
In [ ]:
path = untar_data(URLs.IMDB_SAMPLE)
path.ls()
Out[ ]:
[PosixPath('/home/jhoward/.fastai/data/imdb_sample/texts.csv'),
PosixPath('/home/jhoward/.fastai/data/imdb_sample/models'),
PosixPath('/home/jhoward/.fastai/data/imdb_sample/export.pkl')]
It only contains one csv file, let's have a look at it.
In [ ]:
df = pd.read_csv(path/'texts.csv')
Out[ ]:
label text is_valid
0 negative Un-bleeping-believable! Meg Ryan doesn't even ... False
1 positive This is a extremely well-made film. The acting... False
2 negative Every once in a long while a movie will come a... False
3 positive Name just says it all. I watched this movie wi... False
4 negative This movie succeeds at being one of the most u... False
In [ ]:
df['text'][1]
Out[ ]:
'This is a extremely well-made film. The acting, script and camera-work are all first-rate. The music is good, too, though it is mostly early in the film, when things are still relatively cheery. There are no really superstars in the cast, though several faces will be familiar. The entire cast does an excellent job with the script.<br /><br />But it is hard to watch, because there is no good end to a situation like the one presented. It is now fashionable to blame the British for setting Hindus and Muslims against each other, and then cruelly separating them into two countries. There is some merit in this view, but it\'s also true that no one forced Hindus and Muslims in the region to mistreat each other as they did around the time of partition. It seems more likely that the British simply saw the tensions between the religions and were clever enough to exploit them to their own ends.<br /><br />The result is that there is much cruelty and inhumanity in the situation and this is very unpleasant to remember and to see on the screen. But it is never painted as a black-and-white case. There is baseness and nobility on both sides, and also the hope for change in the younger generation.<br /><br />There is redemption of a sort, in the end, when Puro has to make a hard choice between a man who has ruined her life, but also truly loved her, and her family which has disowned her, then later come looking for her. But by that point, she has no option that is without great pain for her.<br /><br />This film carries the message that both Muslims and Hindus have their grave faults, and also that both can be dignified and caring people. The reality of partition makes that realisation all the more wrenching, since there can never be real reconciliation across the India/Pakistan border. In that sense, it is similar to "Mr & Mrs Iyer".<br /><br />In the end, we were glad to have seen the film, even though the resolution was heartbreaking. If the UK and US could deal with their own histories of racism with this kind of frankness, they would certainly be better off.'
It contains one line per review, with the label ('negative' or 'positive'), the text and a flag to determine if it should be part of the validation set or the training set. If we ignore this flag, we can create a DataBunch containing this data in one line of code:
In [ ]:
data_lm = TextDataBunch.from_csv(path, 'texts.csv')
By executing this line a process was launched that took a bit of time. Let's dig a bit into it. Images could be fed (almost) directly into a model because they're just a big array of pixel values that are floats between 0 and 1. A text is composed of words, and we can't apply mathematical functions to them directly. We first have to convert them to numbers. This is done in two differents steps: tokenization and numericalization. A TextDataBunch does all of that behind the scenes for you.
Before we delve into the explanations, let's take the time to save the things that were calculated.
In [ ]:
data_lm.save()
Next time we launch this notebook, we can skip the cell above that took a bit of time (and that will take a lot more when you get to the full dataset) and load those results like this:
In [ ]:
data = TextDataBunch.load(path)
### Tokenization¶
The first step of processing we make texts go through is to split the raw sentences into words, or more exactly tokens. The easiest way to do this would be to split the string on spaces, but we can be smarter:
• we need to take care of punctuation
• some words are contractions of two different words, like isn't or don't
• we may need to clean some parts of our texts, if there's HTML code for instance
To see what the tokenizer had done behind the scenes, let's have a look at a few texts in a batch.
In [ ]:
data = TextClasDataBunch.load(path)
data.show_batch()
text target
xxbos xxfld 1 xxmaj raising xxmaj victor xxmaj vargas : a xxmaj review \n\n xxmaj you know , xxmaj raising xxmaj victor xxmaj vargas is like sticking your hands into a big , xxunk bowl of xxunk . xxmaj it 's warm and gooey , but you 're not sure if it feels right . xxmaj try as i might , no matter how warm and gooey xxmaj raising xxmaj negative
xxbos xxfld 1 xxmaj many xxunk that this is n't just a classic due to the fact that it 's the first xxup 3d game , or even the first xxunk - up . xxmaj it 's also one of the first xxunk games , one of the xxunk definitely the first ) truly claustrophobic games , and just a pretty well - xxunk gaming experience in general . xxmaj positive
xxbos xxfld 1 i really wanted to love this show . i truly , honestly did . \n\n xxmaj for the first time , gay viewers get their own version of the " xxmaj the xxmaj bachelor " . xxmaj with the help of his obligatory " hag " xxmaj xxunk , xxmaj james , a good looking , well - to - do thirty - something has the chance negative
xxbos xxfld 1 \n\n i 'm sure things did n't exactly go the same way in the real life of xxmaj homer xxmaj hickam as they did in the film adaptation of his book , xxmaj rocket xxmaj boys , but the movie " xxmaj october xxmaj sky " ( an xxunk of the book 's title ) is good enough to stand alone . i have not read xxmaj positive
xxbos xxfld 1 xxmaj to review this movie , i without any doubt would have to quote that memorable scene in xxmaj tarantino 's " xxmaj pulp xxmaj fiction " ( xxunk ) when xxmaj jules and xxmaj vincent are talking about xxmaj mia xxmaj wallace and what she does for a living . xxmaj jules tells xxmaj vincent that the " xxmaj only thing she did worthwhile was pilot negative
The texts are truncated at 100 tokens for more readability. We can see that it did more than just split on space and punctuation symbols:
• the "'s" are grouped together in one token
• the contractions are separated like his: "did", "n't"
• content has been cleaned for any HTML symbol and lower cased
• there are several special tokens (all those that begin by xx), to replace unkown tokens (see below) or to introduce different text fields (here we only have one).
### Numericalization¶
Once we have extracted tokens from our texts, we convert to integers by creating a list of all the words used. We only keep the ones that appear at list twice with a maximum vocabulary size of 60,000 (by default) and replace the ones that don't make the cut by the unknown token UNK.
The correspondance from ids tokens is stored in the vocab attribute of our datasets, in a dictionary called itos (for int to string).
In [ ]:
data.vocab.itos[:10]
Out[ ]:
['xxunk',
'xxbos',
'xxfld',
'xxmaj',
'xxup',
'xxrep',
'xxwrep',
'the',
',']
And if we look at what a what's in our datasets, we'll see the tokenized text as a representation:
In [ ]:
data.train_ds[0][0]
Out[ ]:
Text xxbos xxfld 1 xxmaj matthew mcconaughey is a mysterious man waiting for xxmaj agent xxmaj wesley xxmaj doyle ( xxmaj powers xxmaj boothe ) in his xxup fbi office . xxmaj he claims to have information about a serial killer chased by xxup fbi . xxmaj when xxmaj agent xxmaj doyle arrives in the office , he tells him that the serial killer is indeed his dead brother . xxmaj agent xxmaj doyle xxunk some evidence , and the man tells the story of his life , since his childhood . xxmaj they were a simple family of three : his widow father xxmaj xxunk ( xxmaj bill xxmaj paxton ) , his brother and himself . xxmaj one night , his father xxunk the two brothers and tells them that an angel of xxmaj god had just visited him and assigned his family to destroy demons . xxmaj what happens next is one of the most scary movie i have ever seen .
i watched this movie four months ago on xxup vhs , and yesterday i watched again , now on xxup dvd . xxmaj although being a low - budget movie , the screenplay is sharp , with no flaw . xxmaj the cast is outstanding , but i would like to highlight the performance of xxmaj matt xxunk as the young xxmaj xxunk . xxmaj it is a very difficult and complex role to be performed by a young teenager . xxmaj the direction of xxmaj bill xxmaj paxton is remarkable . xxmaj there is no explicit violence in this horror movie . a great debut behind the camera . i regret the xxmaj xxunk title of this movie : ' a xxmaj xxunk do xxmaj xxunk ' ( xxmaj the xxmaj devil 's xxmaj hand ' ) . xxmaj if at least it were ' xxmaj the xxmaj god 's xxmaj hand ' , it might be acceptable . xxmaj but calling this movie as ' the devil 's hand ' is indeed ridiculous . xxmaj xxunk xxmaj xxunk , the screenwriter , did not deserve such a lack of respect from the xxmaj xxunk xxunk . xxmaj this film is highly recommended . xxmaj my vote is xxunk .
xxmaj title ( xxmaj xxunk ) : " a xxmaj xxunk do xxmaj xxunk " ( " xxmaj the xxmaj devil 's xxmaj hand " )
But the underlying data is all numbers
In [ ]:
data.train_ds[0][0].data[:10]
Out[ ]:
array([ 2, 3, 45, 4, 3229, 3805, 15, 12, 1232, 132])
### With the data block API¶
We can use the data block API with NLP and have a lot more flexibility than what the default factory methods offer. In the previous example for instance, the data was randomly split between train and validation instead of reading the third column of the csv.
With the data block API though, we have to manually call the tokenize and numericalize steps. This allows more flexibility, and if you're not using the defaults from fastai, the variaous arguments to pass will appear in the step they're revelant, so it'll be more readable.
In [ ]:
data = (TextList.from_csv(path, 'texts.csv', cols='text')
.split_from_df(col=2)
.label_from_df(cols=0)
.databunch())
## Language model¶
Note that language models can use a lot of GPU, so you may need to decrease batchsize here.
In [ ]:
bs=48
Now let's grab the full dataset for what follows.
In [ ]:
path = untar_data(URLs.IMDB)
path.ls()
Out[ ]:
[PosixPath('/home/jhoward/.fastai/data/imdb/imdb.vocab'),
PosixPath('/home/jhoward/.fastai/data/imdb/models'),
PosixPath('/home/jhoward/.fastai/data/imdb/tmp_lm'),
PosixPath('/home/jhoward/.fastai/data/imdb/train'),
PosixPath('/home/jhoward/.fastai/data/imdb/test'),
PosixPath('/home/jhoward/.fastai/data/imdb/tmp_clas')]
In [ ]:
(path/'train').ls()
Out[ ]:
[PosixPath('/home/jhoward/.fastai/data/imdb/train/pos'),
PosixPath('/home/jhoward/.fastai/data/imdb/train/unsup'),
PosixPath('/home/jhoward/.fastai/data/imdb/train/unsupBow.feat'),
PosixPath('/home/jhoward/.fastai/data/imdb/train/labeledBow.feat'),
PosixPath('/home/jhoward/.fastai/data/imdb/train/neg')]
The reviews are in a training and test set following an imagenet structure. The only difference is that there is an unsup folder in train that contains the unlabelled data.
We're not going to train a model that classifies the reviews from scratch. Like in computer vision, we'll use a model pretrained on a bigger dataset (a cleaned subset of wikipeia called wikitext-103). That model has been trained to guess what the next word, its input being all the previous words. It has a recurrent structure and a hidden state that is updated each time it sees a new word. This hidden state thus contains information about the sentence up to that point.
We are going to use that 'knowledge' of the English language to build our classifier, but first, like for computer vision, we need to fine-tune the pretrained model to our particular dataset. Because the English of the reviex lefts by people on IMDB isn't the same as the English of wikipedia, we'll need to adjust a little bit the parameters of our model. Plus there might be some words extremely common in that dataset that were barely present in wikipedia, and therefore might no be part of the vocabulary the model was trained on.
This is where the unlabelled data is going to be useful to us, as we can use it to fine-tune our model. Let's create our data object with the data block API (next line takes a few minutes).
In [ ]:
data_lm = (TextList.from_folder(path)
#Inputs: all the text files in path
.filter_by_folder(include=['train', 'test'])
#We may have other temp folders that contain text files so we only keep what's in train and test
.random_split_by_pct(0.1)
#We randomly split and keep 10% (10,000 reviews) for validation
.label_for_lm()
#We want to do a language model so we label accordingly
.databunch(bs=bs))
data_lm.save('tmp_lm')
We have to use a special kind of TextDataBunch for the language model, that ignores the labels (that's why we put 0 everywhere), will shuffle the texts at each epoch before concatenating them all together (only for training, we don't shuffle for the validation set) and will send batches that read that text in order with targets that are the next word in the sentence.
The line before being a bit long, we want to load quickly the final ids by using the following cell.
In [ ]:
data_lm = TextLMDataBunch.load(path, 'tmp_lm', bs=bs)
In [ ]:
data_lm.show_batch()
idx text
0 xxbos xxmaj in a xxmaj woman xxmaj under the xxmaj influence xxmaj mabel goes crazy , but i can see why she does go crazy . xxmaj if i lived the kind of life she lived with the family she has i would go crazy too . xxmaj everyone in her family is off their rocker and not completely with it . xxmaj she is constantly surrounded by people yelling
1 , fresh from success as xxmaj elliot 's mom in " xxup e.t - xxmaj the xxmaj extra xxmaj terrestrial " ) is a mother whose marriage to husband xxmaj vic ( xxmaj daniel xxmaj hugh - xxmaj kelly ) is hanging by a thread . xxmaj she 's been having an affair with a local worker , and is now dwelling on whether or not to leave her husband
2 . xxbos xxmaj the reason why this movie sucks , have these people even read a bible ? xxmaj everything in the movie was about moses , xxmaj god was staying out of it . xxup that xxup didn't xxup happen ! xxmaj god directed everything , he told them where to go and what to do . xxmaj also the people wandered for 40 years xxup after they arrived
3 to sleep . xxbos a little bit of a let down . xxmaj personal opinion , this was a movie with much more potential that was never allowed to reach that . xxmaj sean xxmaj patrick xxmaj flanery has good moments in the movie , but as a whole his performance does n't come close to some of his other movies like xxmaj powder and xxmaj boondock xxmaj saints .
4 purity that it will remind you of the work of some of the great masters of the past . xxmaj the regard for its audience is something that we 're not used to anymore . i do n't know if we ever were . xxmaj riveting , moving , without concessions and xxmaj benicio del xxmaj toro is just extraordinary . xxmaj we can see his soul , we can
We can then put this in a learner object very easily with a model loaded with the pretrained weights. They'll be downloaded the first time you'll execute the following line and stored in ~/.fastai/models/ (or elsewhere if you specified different paths in your config file).
In [ ]:
learn = language_model_learner(data_lm, pretrained_model=URLs.WT103_1, drop_mult=0.3)
In [ ]:
learn.lr_find()
LR Finder is complete, type {learner_name}.recorder.plot() to see the graph.
In [ ]:
learn.recorder.plot(skip_end=15)
In [ ]:
learn.fit_one_cycle(1, 1e-2, moms=(0.8,0.7))
Total time: 21:15
epoch train_loss valid_loss accuracy
1 4.206495 4.067840 0.292376
In [ ]:
learn.save('fit_head')
In [ ]:
learn.load('fit_head');
To complete the fine-tuning, we can then unfeeze and launch a new training.
In [ ]:
learn.unfreeze()
In [ ]:
learn.fit_one_cycle(10, 1e-3, moms=(0.8,0.7))
Total time: 3:59:03
epoch train_loss valid_loss accuracy
1 3.949835 3.892163 0.309857
2 3.863976 3.823716 0.319508
3 3.827637 3.776987 0.325163
4 3.782025 3.738658 0.329756
5 3.704519 3.705750 0.333159
6 3.658038 3.682820 0.335856
7 3.608049 3.662626 0.338366
8 3.552923 3.648938 0.340119
9 3.518708 3.642814 0.340864
10 3.480421 3.641818 0.340905
In [ ]:
learn.save('fine_tuned')
How good is our model? Well let's try to see what it predicts after a few given words.
In [ ]:
learn.load('fine_tuned');
In [ ]:
TEXT = "i liked this movie because"
N_WORDS = 40
N_SENTENCES = 2
In [ ]:
print("\n".join(learn.predict(TEXT, N_WORDS, temperature=0.75) for _ in range(N_SENTENCES)))
i liked this movie because it was clearly a movie . xxmaj so i gave it a 2 out of 10 . xxmaj so , just say something . xxbos xxmaj this is a really stunning picture , light years off of the late xxmaj
i liked this movie because it would be a good one for those who like deep psychological and drama and you can go see this movie if you like a little slow motion and some magic should n't be there . i would give it
We have to save the model but also it's encoder, the part that's responsible for creating and updating the hidden state. For the next part, we don't care about the part that tries to guess the next word.
In [ ]:
learn.save_encoder('fine_tuned_enc')
## Classifier¶
Now, we'll create a new data object that only grabs the labelled data and keeps those labels. Again, this line takes a bit of time.
In [ ]:
path = untar_data(URLs.IMDB)
In [ ]:
data_clas = (TextList.from_folder(path, vocab=data_lm.vocab)
#grab all the text files in path
.split_by_folder(valid='test')
#split by train and valid folder (that only keeps 'train' and 'test' so no need to filter)
.label_from_folder(classes=['neg', 'pos'])
#label them all with their folders
.databunch(bs=bs))
data_clas.save('tmp_clas')
In [ ]:
data_clas = TextClasDataBunch.load(path, 'tmp_clas', bs=bs)
In [ ]:
data_clas.show_batch()
text target
xxbos xxmaj match 1 : xxmaj tag xxmaj team xxmaj table xxmaj match xxmaj bubba xxmaj ray and xxmaj spike xxmaj dudley vs xxmaj eddie xxmaj guerrero and xxmaj chris xxmaj benoit xxmaj bubba xxmaj ray and xxmaj spike xxmaj dudley started things off with a xxmaj tag xxmaj team xxmaj table xxmaj match against xxmaj eddie xxmaj guerrero and xxmaj chris xxmaj benoit . xxmaj according to the rules pos
xxbos xxmaj some have praised xxunk :- xxmaj the xxmaj lost xxmaj xxunk as a xxmaj disney adventure for adults . i do n't think so -- at least not for thinking adults . \n\n xxmaj this script suggests a beginning as a live - action movie , that struck someone as the type of crap you can not sell to adults anymore . xxmaj the " crack staff " neg
xxbos xxunk ) is the developing world 's answer to xxmaj silence of the xxmaj lambs . xxmaj where xxmaj silence ' terrorized our peace of mind , xxmaj citizen ' exhausts and saddens us instead . xxmaj this dramatization of the xxmaj chikatilo case translates rather well , thanks to a xxmaj westernized friendship between two xxmaj rostov cops who become equals . \n\n citizenx may also pos
xxbos 8 xxmaj simple xxmaj rules for xxmaj dating xxmaj my xxmaj teenage xxmaj daughter had an auspicious start . xxmaj the supremely - talented xxmaj tom xxmaj shadyac was involved in the project . xxmaj this meant that the comedy would be nothing less of spectacular , and that 's exactly what happened : the show remains one of the freshest , funniest , wittiest shows made in a pos
xxbos xxmaj the vigilante has long held a fascination for audiences , inasmuch as it evokes a sense of swift , sure justice ; good triumphs over evil and the bad guy gets his deserts . xxmaj it is , in fact , one of the things that has made the character of xxmaj dirty xxmaj harry xxmaj callahan ( as played by xxmaj clint xxmaj eastwood ) so popular pos
We can then create a model to classify those reviews and load the encoder we saved before.
In [ ]:
learn = text_classifier_learner(data_clas, drop_mult=0.5)
learn.freeze()
In [ ]:
gc.collect();
In [ ]:
learn.lr_find()
LR Finder is complete, type {learner_name}.recorder.plot() to see the graph.
In [ ]:
learn.recorder.plot()
In [ ]:
learn.fit_one_cycle(1, 2e-2, moms=(0.8,0.7))
Total time: 02:46
epoch train_loss valid_loss accuracy
1 0.294225 0.210385 0.918960 (02:46)
In [ ]:
learn.fit_one_cycle(1, 2e-2, moms=(0.8,0.7))
Total time: 04:03
epoch train_loss valid_loss accuracy
1 0.507567 0.344481 0.855000
In [ ]:
learn.save('first')
In [ ]:
learn.load('first');
In [ ]:
learn.freeze_to(-2)
learn.fit_one_cycle(1, slice(1e-2/(2.6**4),1e-2), moms=(0.8,0.7))
Total time: 03:03
epoch train_loss valid_loss accuracy
1 0.268781 0.180993 0.930760 (03:03)
In [ ]:
learn.save('second')
In [ ]:
learn.load('second');
In [ ]:
learn.freeze_to(-3)
learn.fit_one_cycle(1, slice(5e-3/(2.6**4),5e-3), moms=(0.8,0.7))
Total time: 04:06
epoch train_loss valid_loss accuracy
1 0.211133 0.161494 0.941280 (04:06)
In [ ]:
learn.save('third')
In [ ]:
learn.load('third');
In [ ]:
learn.unfreeze()
learn.fit_one_cycle(2, slice(1e-3/(2.6**4),1e-3), moms=(0.8,0.7))
Total time: 10:01
epoch train_loss valid_loss accuracy
1 0.188145 0.155038 0.942480 (05:00)
2 0.159475 0.153531 0.944040 (05:01)
In [ ]:
learn.predict("I really loved that movie, it was awesome!")
Out[ ]:
('pos', tensor(0), tensor([[9.9999e-01, 1.1991e-05]]))
|
# Molecular Hydrogen in Pre-galactic Gas Clouds
## Abstract
During the collapse of pre-galactic gas clouds through a density of about 104 particles/cm3, hydrogen molecules are produced and dominate the subsequent cooling.
|
# [NTG-context] How to center \framed on a position?
Pablo Rodriguez oinos at gmx.es
Wed Apr 6 20:24:42 CEST 2016
On 04/06/2016 01:30 PM, Procházka Lukáš Ing. wrote:
> Hello,
>
> how to center vertically and horizontally a \framed text at the layer origin?
>
> I'm not able to figure it out...
> [...]
> And - where has the text "Hello, world!" disappeared?
Hi Lukáš,
here you have the code:
\setuplayout[page]
\definelayer[mybg][position=middle, location=middle]
\setlayer[mybg]
[x=.5\paperwidth,y=.5\paperheight]
{\framed
[frame=on,offset=5mm,width=fit,rulewidth=2mm]
{\setupbodyfont[10mm,sans]\bf Some text}
}
\setupbackgrounds[page][background={mybg}]
\setupbodyfont[helvetica]
\starttext
Hello, world!
\stoptext
Layer has also position and location. As far as I remember, default
values are {top, left}.
I recall a comment from Wolfgang about setting the body font when using
layers. This is the way to have your text.
I hope it helps,
Pablo
--
http://www.ousia.tk
|
• anonymous
Use the quadratic formula to find the roots of the equation. x2 + 3x - 18 = 0 A. {-6, 3} B. {-2, 9} C. {-3, 6} D. {-9, 2}
Mathematics
Looking for something else?
Not the answer you are looking for? Search for more explanations.
|
# Changing a value in an array
ver_mathstats
Homework Statement:
Write a Python code to create the function array_change(a, new_val) that has two arguments:
a that is a NumPy array whose entries are numbers, and
new_val which is a number. The function returns a new array after changing every occurrence of entries in
a, that are in absolute value strictly less than 1, to new_val.
Relevant Equations:
Python
Code:
import numpy as np
def array_change(a,new_val):
a=np.array([])
for i in range(a):
if abs(i)<1:
a[i]=new_val
e=np.arange(-2, 2, 0.2).reshape(4,5)
print(array_change(e,0))
I am not sure where I am going wrong exactly but I keep getting an error message.
I came up with a code that gives me the results I am looking for but it is not a function.
Code:
e=np.arange(-2, 2, 0.2).reshape(4,5)
c=abs(e)<1
e[c]=0
print(e)
Any help is appreciated. Thank you.
## Answers and Replies
Homework Helper
Gold Member
What does the error message say? That might give you a hint. That is what they are for.
Homework Helper
Gold Member
You have some code that works. Can you figure a way to put that into a function? Remember that a function usually returns a value.
Mentor
I am not sure where I am going wrong exactly but I keep getting an error message.
If your code produces an error message, it's always a good idea to tell anyone helping what that error message says. A typical Python error message will give the type of error (e.g., Syntax Error, or Name Error) and will show you what line the error occurs on.
FactChecker
Mentor
Aside from the program requirements (one parameter of the function must be a Numpy array), I don't see any other real reason why an ordinary Python list type can't be used.
I see at least two things wrong:
(1) a is the input array, which comes from outside the function. You should not be re-defining a within the function.
(2) The statement "for i in range(a)" requires that a be an integer. But a is a numpy array, not an integer. So this will fail right away.
Try fixing these two things and thinking a bit more about what you are trying to do. Then come back and tell us how you are doing.
Mentor
To elaborate on the points that @phyzguy made, instead of doing this:
Python:
def array_change(a,new_val):
a=np.array([])
for i in range(a):
try this:
Python:
def array_change(arr, new_val):
arr_size = len(arr)
for i in range(arr_size):
Note that I changed the name of the array parameter. Single letter variable names are discouraged in most programming languages, except possibly for loop control variables such as i, j, k, and so on.
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Reobservation of faint, neglected pairsreveal many corrections to the data from the old discovery lists. Observations of double stars and new pairs. XIVResults of a continuing survey of visual double stars are presented,including 4880 measurements made from February 1987 to November 1989.The positions in WDS format and Durchmusterung numbers are given for 194pairs first reported here. Micrometer measurements of 1142 doubles madewith the Swarthmore 61 cm refractor are presented. Magnitudes areestimated for some of the objects. Plate measurements, plateorientations, position angles, number of nights, and measured exposuresare given. Visual observations of 342 pairs obtained in May 1989 atCerro Tololo, mostly with the 1.0 m reflector, are reported. ICCD speckle observations of binary stars. IV - Measurements during 1986-1988 from the Kitt Peak 4 M telescopeOne thousand five hundred and fifty measurements of 1006 binary starsystems observed mostly during 1986 through mid-1988 by means of speckleinterferometry with the KPNO 4-m telescope are presented. Twenty-onesystems are directly resolved for the first time, including newcomponents to the cool supergiant Alpha Her A and the Pleiades shellstar Pleione. A continuing survey of The Bright Star Catalogue yieldedeight new binaries from 293 bright stars observed. Corrections tospeckle measures from the GSU/CHARA ICCD speckle camera previouslypublished are presented and discussed. Binary star measurements made at Nice with the 50-cm reflectorAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1989A&AS...77..125L&db_key=AST Micrometer Measurements of Visual Double Stars - Part Three -Not Available Metallicism among A and F giant stars132 stars considered as A and F giants have been studied for theirproperties in the Geneva photometric system. It is shown that thissystem to derive the temperature, absolute magnitude and Fe/H value forstars in this part of the HR diagram. 36 percent of the stars of oursample exhibit an enhanced value Delta m2 that can be interpreted interms of Fe/H. The red limit of stars having an enhanced Fe/H value is0.225 in B2-V1 or 6500 K in Teff. This corresponds to the limit definedby Vauclair and Vauclair (1982) where the diffusion timescale is equalto the stellar lifetime and permits the assumption that the diffusion isthe process responsible for the metallicism observed in the A and Fgiants. Visual multiples. VII - MK classificationsClassifications are given for 865 components of visual multiples; theyshow no systematic differences from the MK system, and the random errorsare one subclass in type and two-thirds of a luminosity class. It isfound that at least 1% of the F-type IV and V stars are weak-lined, 32%of the A4-F1 IV and V stars are Am, and 5% of the A0-A3 IV and V starsare early-type Am. Attention is called to the large fraction (55%) ofthe A3-A9 III-V stars that are of luminosity classes III or IV, unlikethe percentage (16%) at neighboring types. Visual multiples. V - Radial velocities of 160 systems937 radial velocities are listed from coude spectra of 160 visualmultiples with known visual orbital elements; these, plus the velocitiesin paper of Roemer and Sanwal (1980), are discussed. Among the resultsare (1) systems yielding spectroscopic elements with the visual period,(2) systems probably showing velocity variations during the visualperiod, (3) systems with short spectroscopic periods, some in additionto detectable motion during the visual period, (4) systems showing novariation in radial velocity during the visual period, either becausethe components are similar in brightness or the periods are very long,(5) systems with spectral lines too broad to allow the detection oforbital motion, and (6) systems with insufficient data for anyconclusions to be drawn at present. MK spectral types for some F and G stars.Abstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1979PASP...91...83C&db_key=AST Spectral classification of the bright F stars.Abstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1976PASP...88...95C&db_key=AST Micrometer observations of double stars.8.Abstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1975ApJS...29..315H&db_key=AST Double Star Measures at Lick Observatory, Mount Hamilton, CaliforniaAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1974PASP...86..902H&db_key=AST Mesures d'étoiles doubles faites à l'équatorial de 38 CM de l'Observatoire de ParisAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1972A&AS....6..147B&db_key=AST Catalog of Indidual Radial Velocities, 0h-12h, Measured by Astronomers of the Mount Wilson ObservatoryAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1970ApJS...19..387A&db_key=AST Orbites de cinq étoiles doubles visuellesAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1970A&AS....1..115M&db_key=AST Mikrometermessungen von Doppelsternen. VI.Not Available Mesures d'étoiles doubles faites au réfracteur de 38 cm de l'Observatoire de NiceNot Available
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## The Annals of Mathematical Statistics
### Asymptotic Expansions for a Class of Distribution Functions
K. C. Chanda
#### Abstract
Investigations have been made in the past by several people on the possibility of extending the content of the classical central limit Theorem when the basic random variables are no longer independent. Several interesting extensions have been made so far. Hoeffding and Robbins (1948) have established asymptotic normality for the distribution of mean of a sequence of $m$-dependent random variables, where $m$ is a finite positive constant. The result has been proved by Diananda (1953) under more general conditions and has been extended to cover situations where the random variables $\{X_t\} (t = 1, 2, 3, \cdots)$ are of the type $X_t - E(X_t) = \sum^\infty_{j = 0} g_j Y_{t-j}$ where $\{Y_t\} (t = 0, \pm 1, \cdots)$ is an $m$-dependent stationary process and $E(Y_t) = 0,\quad\sum^\infty_{j = 0} g_j < \infty$. Walker (1954) has established asymptotic normality for the distributions of serial correlations based on $X_t$ of the above form. However, so far no attempt has been made to investigate whether the type of asymptotic expansions as discussed by Cramer (1937), Berry (1941) and Hsu (1945) for the distributions of means of independent random variables could also be extended to apply to situations where the random variables are not independent. Chanda (1962) has made a start in this direction, but the results are understandably incomplete. An attempt has been made in this paper to investigate this problem more systematically. The conclusion is that an extension is possible under conditions precisely similar to those under which Cramer, Berry and Hsu proved their results.
#### Article information
Source
Ann. Math. Statist., Volume 34, Number 4 (1963), 1302-1307.
Dates
First available in Project Euclid: 27 April 2007
https://projecteuclid.org/euclid.aoms/1177703865
Digital Object Identifier
doi:10.1214/aoms/1177703865
Mathematical Reviews number (MathSciNet)
MR156371
Zentralblatt MATH identifier
0237.60014
JSTOR
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# Match the differential equation with its direction field (labeled I-IV). Give reasons for your answer.$y' = x + y - 1$
## $=-1,$ which is true in $1 \mathrm{V}$
#### Topics
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you're giving, We asked. So coarse. But creation given is why crime before? That's why minus one or slow fields are one. You four find the correct look you c five and a political solution. We know that while you will see what season remember solution? No. Yes, zero Look in why is equal to it's plus Why by this one if you people to act What? No, this one complies that being too one, Nigel However x elements of the members any element c is not going to be a real number. So it follows that you are new equal living so oceans this equation Therefore we shouldn't expect any horizontal Last see that field one, two in three all have is not the last tape Oh, they range should be so for confirm This is indeed our answer. You can also try to match up characteristic of slim build with education You say that Dusty had asked along Why would make that no see if you plug in, Why would make that as the solution we get well, just implies that white one and you have That's why minus one physical too. Thanks 96 on this one which is one get wide, find people to X plus y minus one so that why was negative is the solution. And they said, You see, White was negative acts as a solution or I don't see it in the other. So this again that's four must be.
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Numerical Python; Mailing Lists; Numerical Python A package for scientific computing with Python Brought to you by: charris208, jarrodmillman, kern, rgommers, teoliphant. In this case, we hope to find eigenvalues near zero, so we’ll choose sigma = 0. How to find eigenvectors and eigenvalues without numpy and scipy , You can use sympy, the python computer algebra system, to solve the eigenvalue problem without native libraries using the Berkowitz method. So, I thought that may be an easier way is to write/find a small function to solve the eigenvalue problem. eigvals (a[, b, overwrite_a, check_finite]) Compute eigenvalues from an ordinary or generalized eigenvalue problem. We can solve for the eigenvalues by finding the characteristic equation (note the "+" sign in the determinant rather than the "-" sign, because of the opposite signs of λ and ω 2). numpy and scipy do not work. In the second case, I would just use numpy.linalg.eig. 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PCA is useful when you have continuous variables, but I'm not sure that was good when your data is binary, so maybe you need to use MDS with binary ED or CorEx. linalg.tensorsolve (a, b[, axes]) Solve the tensor equation a x = b for x. linalg.lstsq (a, b[, rcond]) Return the least-squares solution to a linear matrix equation. SOLUTION: • In such problems, we first find the eigenvalues of the matrix. atendemos las necesidades educativas individuales A midpoint formula is develop for use with the Numerov method. Solve an eigenvalue problem; Use a specific linear algebra backend (PETSc) Initialize a finite element function with a coefficient vector; 6.1. In quantum mechanics, every experimental measurable $$a$$ is the eigenvalue of a specific operator ($$\hat{A}$$): $\hat{A} \psi=a \psi \label{3.3.2a}$ The $$a$$ eigenvalues represents the possible measured values of the $$\hat{A}$$ operator. I found out that to resolve the problem I need to check my blas/lapack. Do you want to solve it as a part of some bigger TF project or you just want to solve it for fun. Follow 380 views (last 30 days) Brendan Wilhelm on 21 May 2018. SLEPc for Python (slepc4py) is a Python package that provides convenient access to the functionality of SLEPc.. SLEPc , implements algorithms and tools for the numerical solution of large, sparse eigenvalue problems on parallel computers. Find eigenvalues w and right or left eigenvectors of a general matrix: a vr[:,i] = w[i] b vr[:,i] a.H vl[:,i] = w[i].conj() b.H vl[:,i] where .H is the Hermitian conjugation. The Eigenvalue Problem: Simulating Transdermal Drug Diffusion with Python [Transport Phenomena] Andrew. The main aim of this post is not to teach you quantum physics, but how to use Python programming to numerically solve complex physics problems. 0. solve the eigenvalue problem for Scrődinger equation: Eigenvalue problem for Schrödinger equation using numerov method 59 • How to chose a step size, how to decide when the step size needs changing, and how to carry out this change. They both write Illegal instruction (core dumped). How to solve an eigenvalue problem. Then, if you want to solve multicollinearity reducing number of variables with a transformation, you could use a multidimensional scaling using some distance that remove redundancies. The other proofs can be handled similarly. 1.1 What makes eigenvalues interesting? desarrollo de habilidades sociales, emocionales, comunicativas y cognitivas conozca mÁs. We consider here two possible choices of finite element spaces. Parameters: a: (M, M) array_like. b: (M, M) array_like, optional. This is obviously just an eigenvalue problem. Eigenvalues in OpenCV. I tried something like eig(dot(inv(B),A)) from numpy.linalg but it turns out to be VERY unstable in my problem since it involves inversion. Eigenvalues and eigenvectors¶ The eigenvalue-eigenvector problem is one of the most commonly employed linear algebra operations. eigh (a[, b, lower, eigvals_only, ...]) Solve an ordinary or generalized eigenvalue problem for a complex Hermitian or real symmetric matrix. How do I find the eigenvalues and vectors using matlab? You need techniques from numerical linear algebra. Solve Differential Equations in Python source Differential equations can be solved with different methods in Python. Relevant answer overflow and small convergence basin. The best way to deal with this (Python API, you'll have to look for the C++ equivalent) So if you are working with OpenCV, here is how to do it. The time independent Schrödinger equation is an eigenvalue problem. Solve a quadratic eigenvalue problem involving a mass matrix M, damping matrix C, and stiffness matrix K. This quadratic eigenvalue problem arises from the equation of motion: M d 2 y d t 2 + C d y d t + K y = f (t) This equation applies to a broad range of oscillating systems, including a dynamic mass-spring system or RLC electronic network. Thus, solve eigenvalue problem defined by Eq. It provides mechanisms for completely specifying the problem: the problem type (e.g. They do this at certain frequencies. So, using an online interpreter like the one above can give your students a realistic impression on how to solve eigenvalue problems in practice. Python code for eigenvalues without numpy. Solve an eigenvalue problem; Use a specific linear algebra backend (PETSc) Initialize a finite element function with a coefficient vector; 8.1. The first, the Nédélec edge elements, which are obtained in FEniCS as FunctionSpace(mesh, 'H1curl', 1), are well suited to this problem and give an accurate discretization.The second choice is simply the vector-valued Lagrange piecewise linears: VectorFunctionSpace(mesh, 'Lagrange', 1). It can be used for linear eigenvalue problems in either standard or generalized form, with real or complex arithmetic. Loading... Unsubscribe from Andrew? This has many numerical difficulties, e.g. In the following, we restrict ourselves to problems from physics [7, 18, 14] and computer science. Example. The Python code has been structured for ease of understanding and allows modifying the code for more for implementing additional features. All eigenvalues are positive in the Dirichlet case. 0 Comments. So it does not improve the computation time. I used DQM but can't solve the system of equation because of boundry conditions can't reach the (F-omega^2*I)*phi=0 two solve the eigenvalue problem thank you. I will put some links below at the bottom which you can refer to if you would like to know more! Verify A v = λ B v for the first eigenvalue and the first eigenvector. I am confident that matlab's parameter which specify the number of eigenvalues just filters out unneeded eigenvalues after all of them were found. I am trying to solve the generalized eigenvalue problem A.c = (lam).B.c where A and B are nxn matrices and c is nx1 vector. Eigenvalue and Generalized Eigenvalue Problems: Tutorial 2 where Φ⊤ = Φ−1 because Φ is an orthogonal matrix. Stable and unstable finite elements¶. You are trying to solve an eigenvalue problem by directly finding zeros of the determinant. Return the eigenvalues and eigenvectors of a Hermitian or symmetric matrix. This article describes the steps to be carried out for peforming modal anaysis on strucures by taking a 2D truss problem and showing Python code alongside. ... linalg.solve (a, b) Solve a linear matrix equation, or system of linear scalar equations. Solving eigenvalue problems are discussed in most linear algebra courses. (lam) is the eigenvalue. • How to iterate on the eigenvalue when already close to it. Objects like violin strings, drums, bridges, sky scrapers can swing. In one popular form, the eigenvalue-eigenvector problem is to find for some square matrix $$\mathbf{A}$$ scalars $$\lambda$$ and corresponding vectors $$\mathbf{v}$$, such that Writing a program to solve an eigenvalue problem is about 100 times as much I need to calculate eigenvalues and eigenvectors in python. Show Hide all comments. Instead of complex eigenvalue problem you should have two real ones, one for the real part and one for the imaginary part, where there might be some coupled terms present (i.e. Summary Files Reviews Support Wiki Mailing Lists Menu numpy-discussion [Numpy-discussion] Generalized Eigenvalue problem [Numpy-discussion] Generalized Eigenvalue problem. I am using python. Python (with the packages NumPy and SciPy) is a popular open-source alternative to Matlab. Edited: Ameer Hamza on 21 May 2018 I have two matrices A and B. A=[2,5;3,6] and B=[5,0;0,7]. For the eigenvalue problem above, 1. 2. Moreover,note that we always have Φ⊤Φ = I for orthog- onal Φ but we only have ΦΦ⊤ = I if “all” the columns of theorthogonalΦexist(it isnottruncated,i.e.,itis asquare (11), obtain the set of {()} = 1 n i i λ and associated eigenvectors {()} = 1 n i i X and, then, later obtain the eigenvectors for the problem with ()i ()i φ =LX-T. We prove this result for the Dirichlet case. Overview¶. In this case, eig(A,B) returns a set of eigenvectors and at least one real eigenvalue, even though B is not invertible. Vote. All eigenvalues are zero or positive in the Neumann case and the Robin case if a ‚ 0. A complex or real matrix whose eigenvalues and eigenvectors will be computed. Solve an ordinary or generalized eigenvalue problem of a square matrix. large eigenvalue problems in practice. They are part of an eigenvalue problem of the form: (A-(lambda)B)x=0. Cancel … A = someMatrixArray from numpy.linalg import eig as eigenValuesAndVectors solution = eigenValuesAndVectors(A) eigenValues = solution[0] eigenVectors = solution[1] I would like to sort my eigenvalues (e.g. In physics, eigenvalues are usually related to vibrations. from pprint import pprint from numpy import array, zeros, diag, diagflat, dot def jacobi(A,b,N=25,x=None): """Solves the equation Ax=b via the Jacobi iterative method.""" Solve the Eigenvalue/Eigenvector Problem. Below are examples that show how to solve differential equations with (1) GEKKO Python, (2) Euler’s method, (3) the ODEINT function from Scipy.Integrate. from lowest to highest), in a way I know what … By Philipp Wagner | July 15, 2010. standard symmetric), number of eigenvalues to compute, part of the spectrum of interest. ARPACK can solve either standard eigenvalue problems of the form $A \mathbf{x} = \lambda \mathbf{x}$ ... As mentioned above, this mode involves transforming the eigenvalue problem to an equivalent problem with different eigenvalues. 380 views ( last 30 days ) Brendan Wilhelm on 21 may 2018 Transdermal Drug Diffusion with Python [ Phenomena! Or complex arithmetic = 0 eps the eigenvalue problem by directly finding zeros of the matrix working with,. Menu Numpy-discussion [ Numpy-discussion ] generalized eigenvalue problem the component that provides all the necessary... The code for more for implementing additional features 18, 14 ] and computer science provides for! 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May 2018 array_like, optional small function to solve the problem type ( e.g standard or generalized form, real!, check_finite ] ) Compute eigenvalues from an ordinary or generalized eigenvalue problem for implementing additional.... Solve it for fun Mailing Lists Menu Numpy-discussion [ Numpy-discussion ] generalized problem... Some bigger TF project or you just want to solve an eigenvalue problem: Simulating Drug! Views ( last 30 days ) Brendan Wilhelm on 21 may 2018 vice versa ) is to! Were found eigvals ( a [, b ) x=0 if you would like know. Is an eigenvalue problem is one of the real part or vice ). Of finite element spaces ( core dumped ) b ) solve a linear matrix equation, or system linear. 18, 14 ] and computer science we restrict ourselves to problems from physics 7! Or generalized eigenvalue problem an eigenproblem b ) x=0 that matlab 's parameter specify! Objects like violin strings, drums, bridges, sky scrapers can.... 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# Which of the following has/have exactly 4 positive integer
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Which of the following has/have exactly 4 positive integer [#permalink] 13 Oct 2017, 02:06
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Which of the following has/have exactly 4 positive integer factors?
Indicate $$all$$ possible numbers.
❑ 4
❑ 6
❑ 8
❑ 12
❑ 14
[Reveal] Spoiler: OA
B,C,E
[Reveal] Spoiler: OA
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Re: Which of the following has/have exactly 4 positive integer [#permalink] 13 Oct 2017, 07:38
Carcass wrote:
Which of the following has/have exactly 4 positive integer factors?
Indicate $$all$$ possible numbers.
❑ 4
❑ 6
❑ 8
❑ 12
❑ 14
[Reveal] Spoiler: OA
B,C,E
Now
$$4 =2^2$$
Therefore no. of factors = 3 (since no of factors of $$n^m$$ = m+1)
$$6 = 2^1 * 3^1$$
Therefore no. of factors = 4 (since no of factors of $$n^m$$ and $$x^y$$= (m+1) * (y+1))
$$8 = 2^3$$
Therefore no. of factors = 4
$$12 = 2^2 * 3^1$$
Therefore no. of factors = 6
$$14 = 2^1 * 7^1$$
Therefore no. of factors = 4
Ans - B,C,E
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Re: Which of the following has/have exactly 4 positive integer [#permalink] 13 Oct 2017, 07:38
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# Which of the following has/have exactly 4 positive integer
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## Introduction
Arthropod-borne viruses, or arboviruses, are viruses that are transmitted via blood feeding arthropods1. Arboviral infections such as dengue, Zika and chikungunya are fast spreading diseases that pose significant health problems globally2,3,4,5. These viral infections, in particular dengue, are transmitted mainly by Aedes aegypti and sometimes by Aedes albopictus (Asian Tiger) female mosquitoes when taking a blood meal from the host6,7. Approximately 390 million dengue infections are estimated to occur worldwide annually, putting 40% of the total human population at risk8. Dengue infection is the most geographically wide-spread of the arboviral infections3,8. It has different severity levels which are classified according to disease progression from dengue without warning signs to dengue with warning signs and then severe dengue9. Clinical manifestation includes sudden high-grade fever, headache, nausea, arthralgia, eye pain, muscle ache and rash in some cases10. Presently, there is no specific universal treatment for dengue infections: the vaccine envelopment targets young populations; the efficacy of the only vaccine licensed depends on prior immunity to at least one serotype of dengue; and it provides heterogeneous protection against the different serotypes11,12.
Other arboviral infections such as Zika, chikungunya and yellow fever are also of global health concern13. These arboviral infections have occurred simultaneously with dengue13,14. Some of these infections share many similar clinical manifestations with dengue infection and also allow arboviral coinfection such as dengue and chikungunya15, chikungunya and Zika16 and yellow fever and chikungunya17. Although, there are no specific treatments for Zika and chikungunya viral infections, these infections can be managed by supportive treatment of symptomatic individuals and adequate rest. This treatment includes fluid intake and administering drugs such as acetaminophen to suppress pain and fever18,19. However the prevention strategy for yellow fever infection is available i.e. vaccination20,21.
To control these infections, an intracellular bacterium called Wolbachia can be used to suppress transmission in arthropods such as mosquitoes and flies22,23,24,25. Wolbachia infection inhibits arboviral transmission in mosquitoes via four mechanisms: immune priming—preactivation of the mosquito immune system; induction of the phenoloxidase cascade—triggers immune response to viruses; competition of intracellular resources—inducing authophagy; and induction of microRNA-dependent immune pathways—essential for gene regulation and stability, immune defense, ageing and organ differentiation26. This endosymbiotic bacterium which exists naturally in more than 50% of all insect species can be found within the cytoplasm of the cells of their hosts25,27,28. Whilst Wolbachia is not naturally present in Aedes aegypti, it can be introduced via stable transinfections using microinjections29,30.
The Wolbachia-based control strategy is carried out by infecting mosquitoes with a strain of Wolbachia and then releasing them into wild mosquito populations in the hopes of replacing the vector transmitting agent Aedes aegypti with one that is incapable of transmission29,30,31. Infecting an Aedes mosquito with Wolbachia can change some of the Aedes characteristic features. In practice, Wolbachia can reduce the life-span of mosquitoes by half producing a deleterous fitness effect32. Another feature is cytoplasmic incompatibility (CI)22,33,34,35 which occurs when a Wolbachia infected male mates with an incompatible female mosquito (usually Wolbachia uninfected) producing no offspring36. Other features of Wolbachia which serve as liabilities in mosquitoes include: imperfect maternal transmission (IMT)30,37 and loss of Wolbachia infection (LWI). LWI impedes the establishment of Wolbachia-infected mosquitoes and is a result of mosquito vulnerability to high temperature38,39.
However, a novel strain of Wolbachia: wAu, has shown to produce high viral blockage whilst maintaining Wolbachia infection in Aedes mosquitoes at higher temperature23. Moreover, wAu allows superinfection to occur when wAu and other strains of Wolbachia co-exist in the vector host23. Despite these favourable features, wAu does not induce CI23. Although CI absence does not establish Wolbachia infected mosquitoes, the effect could be outweighed by LWI and IMT37.
The difference in the common Wolbachia strain features are described in Table 1 below.
In general, the introduction of mathematical models to understand infection dynamics of diseases has long been helpful in the area of disease control49. A number mathematical models of Wolbachia dynamics in a mosquito population have been formulated37,50,51,52,53,54,55,56,57,58. Some of these models introduced Wolbachia strain(s) into a mosquito population and classified them into age-sturctured Wolbachia-infected and -uninfected mosquito compartments37,53,54,57. Ndii et al.53, formulated a mathematical model for the Wolbachia interaction between the immature stages (aquatic stage), adult male and female mosquito populations to investigate the persistence of mosquitoes infected with Wolbachia when competing with the uninfected ones. They derived the steady state solutions and showed that parameters such as maternal transmission, reproductive, death and maturation rates drive the persistence of the Wolbachia-infected mosquito population. A similar model developed by Xue et al. considered the Wolbachia-induced fitness change and the CI effect57. They showed that if the basic reproduction number ($$R_0$$) of the Wolbachia-infected mosquitoes is less than one, an endemic Wolbachia infection can still occur via backward bifurcation if a sufficient number of the mosquitoes are introduced into the population. A mathematical model of Wolbachia to control dengue fever transmission52 was developed by Hughes et al. The model showed that the use of Wolbachia has high potential to control dengue where the $$R_0$$ due to Wolbachia-infected Aedes mosquitoes is not too large in endemic areas. Another study of a Wolbachia invasive model incorporated IMT and LWI and showed that CI does not guarantee the establishment of Wolbachia-infected mosquitoes as the disadvantages derived from IMT and LWI in the production of Wolbachia-infected mosquitoes could outweigh CI37.
Additionally, a study conducted by O’Reilly et al combining multiple modeling methods, was used to estimate the burden of dengue and map its distribution across Indonesia59. They predicted that there was a reduction in dengue transmission after a nationwide release of wMel-Wolbachia-infected mosquitoes. In addition, they predicted about 86% of the estimated 7.8 million annual cases of symptomatic dengue in Indonesia could be averted following a complete nationwide rollout of Wolbachia-infected mosquitoes. Recently, a modeling study presented a dengue transmission model in the presence of female wild-type and wMelPop Wolbachia-infected Aedes aegypti mosquitoes. They concluded that although the wMelPop strain reduces the lifespan of infected mosquitoes, which could be challenging to achieve replacement of wild-type mosquitoes, its optimal release ensured the replacement of wild-type mosquitoes and also reduced dengue burden in the human population51. A mosquito-Wolbachia model was developed by Xue et al, to compare the potential effectiveness of two Wolbachia strains (wMel and wAlbB) to control arboviral spread60. They observed that each of the two different strains of Wolbachia can effectively decrease the rate of arboviral transmission.
Here, we develop a general Wolbachia model capable of faithfully replicating all of the strain features described in Table 1. The general transmission model is an extention of the Wolbachia transmission model introduced in Adekunle et al.37, which described the competitive dynamics between (wMel-like) Wolbachia-infected and uninfected mosquitoes. Despite the non-induction of CI in wAu-Wolbachia-infected mosquitoes, wAu infection is retained and able to block viral transmission efficiently compared to other strains even at high temperature. Therefore, we incorporated this feature to determine if the advantages (Wolbachia retainment) of the wAu strain outweigh the ineffectiveness of CI. This feature has not been considered in previous models. Furthermore, we incorporate imperfect maternal transmission into the model. By analysing the system via computing the basic reproduction number(s) and investigating the stability properties of the equilibrium points, the potential of the wAu strain as a viable strategy to control Aedes-borne infections can be established. The aim of this modeling approach is to support future Aedes-borne viral control programs, particularly with the introduction of new Wolbachia variants.
## Methods
### Model formation
Here, we investigate a modified Wolbachia transmission model studied in Adekunle et al.37, focusing on a novel Wolbachia strain, wAu, which has high retainment, high viral blockage and does not induce CI. The mosquito population is subdivided into two groups: the uninfected mosquitoes $$(.)_u$$ and the Wolbachia infected mosquitoes $$(.)_w$$. The term (.) can be aquatic/immature (eggs, larvae and pupae) A, male M or female F mosquitoes. In addition, we denote the aquatic/immature stages, mature male and mature female uninfected mosquitoes as $$A_u$$, $$M_u$$, $$F_u$$, and Wolbachia-infected mosquitoes as $$A_{w}$$, $$M_{w}$$, $$F_{w}$$ respectively. As in Adekunle et al.37 the model also incorporates the IMT of wAu-Wolbachia.
There are four possible mosquitoes’ mating pairs: $$F_uM_u$$, $$F_uM_w$$, $$F_wM_u$$ and $$F_wM_w$$. As Wolbachia infection is maternally transmitted, $$F_uM_u$$ and $$F_uM_w$$ will produce uninfected offspring while $$F_wM_u$$ and $$F_wM_w$$ will typically produce infected offspring. However if there is imperfect maternal transmission, the two latter strategies could produce some proportions of uninfected offspring23.
To mathematically write the system of differential equations governing the Wolbachia transmission dynamics, we express the feasible mating strategies of uninfected and Wolbachia infected mosquito populations together with their per capita egg laying rates as Eqs. (1)–(6):
\begin{aligned} \frac{dA_{u}}{dt}= & {} \left[ \frac{\rho _{uu} (F_u M_u + (1-\phi ) F_u M_{w}) + \rho _{ww}( (1-\delta ) F_{w}M_u + (1-\nu ) F_{w}M_{w})}{M}\right] \left( 1-\frac{A}{K} \right) \nonumber \\&\qquad -(\tau _{u} +\mu _{Au}) A_u, \end{aligned}
(1)
\begin{aligned} \frac{dF_{u}}{dt}= & {} (1-\psi )\tau _{u}A_u +\sigma F_w -\mu _u F_u, \end{aligned}
(2)
\begin{aligned} \frac{dM_{u}}{dt}= & {} \psi \tau _{u}A_u +\sigma M_w -\mu _u M_u \end{aligned}
(3)
\begin{aligned} \frac{dA_{w}}{dt}= & {} \left[ \frac{\rho _{ww}(\nu F_{w}M_{w} + \delta F_{w}M_u)}{M}\right] \left( 1-\frac{A}{K} \right) - (\tau _{w} +\mu _{Aw}) A_{w}, \end{aligned}
(4)
\begin{aligned} \frac{dF_{w}}{dt}= & {} (1-\psi )\tau _{w}A_{w} -\sigma F_w- \mu _{w} F_{w}, \end{aligned}
(5)
\begin{aligned} \frac{dM_{w}}{dt}= & {} \psi \tau _{w}A_{w} -\sigma M_w - \mu _{w}M_{w}, \end{aligned}
(6)
where $$F=F_u +F_{w}$$, $$M=M_u +M_{w}$$, $$A=A_u +A_{w}$$.
Here, $$\phi$$ represents the CI effect which can be either 0 if there is no CI, or 1 if CI is present. $$\sigma$$ is the effect of LWI, such that it can either be 0, if there is no Wolbachia loss or greater than zero otherwise. In Adekunle et al.37 where CI is assumed and LWI is considered, these quantities are set to $$\phi =1$$ and $$\sigma \ge 0$$. In our modified model, considering different strains with the exception of wAu strain, $$\phi =1$$ and $$\sigma$$ could vary from values greater than zero onwards. However, for the wAu-Wolbachia strain, CI is ineffective and high retainment of wAu-Wolbachia infection even at high temperatures23 is established, therefore we set $$\phi =0$$ and $$\sigma =0$$. Our model also incorporates imperfect maternal transmission generating a proportion of infected and uninfected offspring from mating of both $$F_w M_{u}$$ and $$F_w M_{w}$$ mosquitoes. To simplify the system, we assume that $$M=F$$ in accordance with the observed ratio of male to female mosquitoes of 1.02:162. That is, we set $$\psi = 1/2$$ (Fig. 1). By this, it follows that the system of ordinary differential equations (ODEs) in Eqs. (1)–(6) can be reduced to (7)–(10) which is the governing Wolbachia infection dynamics.
To mathematically express the above schematics, we have that, the feasible mating strategies of uninfected and Wolbachia infected mosquito populations together with their per capita egg laying rates are given by the following differential system:
\begin{aligned} \frac{dA_{u}}{dt}= & {} \left[ \frac{\rho _{uu} (F_u^2 + (1-\phi )F_u F_{w}) + \rho _{ww}((1-\nu ) F_{w}^2 +(1-\delta )F_w F_{u})}{F}\right] \left( 1-\frac{A}{K} \right) -(\tau _{u} +\mu _{Au}) A_u, \end{aligned}
(7)
\begin{aligned} \frac{dF_{u}}{dt}= & {} \frac{\tau _{u}}{2}A_u +\sigma F_w-\mu _u F_u, \end{aligned}
(8)
\begin{aligned} \frac{dA_{w}}{dt}= & {} \left[ \frac{\rho _{ww} (\nu F_{w}^2 +\delta F_w F_{u})}{F}\right] \left( 1-\frac{A}{K} \right) - (\tau _{w} +\mu _{Aw}) A_{w}, \end{aligned}
(9)
\begin{aligned} \frac{dF_{w}}{dt}= & {} \frac{\tau _{w}}{2}A_{w} - \sigma F_w- \mu _{w} F_{w}, \end{aligned}
(10)
where $$F=F_u +F_{w}$$ and $$A=A_u +A_{w}$$. Before proceeding, we rescale each of our state variables according to the maximum total population size, which by Adekunle et al., 201937 is set by
\begin{aligned} A_u(t) +F_u(t)+A_{w}(t) +F_{w}(t)\le & {} K + \dfrac{\tau _u K}{2\mu _u} + \dfrac{\sigma \tau _w K}{2\mu _u(\mu _w +\sigma )} + \dfrac{\tau _w K}{2(\mu _w +\sigma )}\\\le & {} K\left( 1+\dfrac{1}{2}\left( \dfrac{\tau _u}{\mu _u} + \dfrac{\tau _w}{(\mu _w +\sigma )}\left( 1+\dfrac{\sigma }{\mu _u}\right) \right) \right) \\\le & {} \alpha K \end{aligned}
where $$\alpha = 1+\frac{1}{2}\left( \frac{\tau _u}{\mu _u} + \frac{\tau _w}{(\mu _w +\sigma )}\left( 1+\frac{\sigma }{\mu _u}\right) \right)$$.
The closed set
$$\Omega = \left\{ (A_u, F_u, A_w, F_w) \in \mathbb {R}_+^4 \, | \, A_u + F_u + A_w + F_w \le \alpha K \right\}$$
which is a feasible region for the above system dynamics is positively invariant37.
Hence, we let $$\bar{A}_u = \frac{A_u}{\alpha K}$$, $$\bar{A}_w = \frac{A_w}{\alpha K}$$, $$\bar{F}_u = \frac{F_u}{\alpha K}$$, $$\bar{F}_w = \frac{F_w}{\alpha K}$$, $$\bar{A}=\bar{A}_u +\bar{A}_{w}$$ and $$\bar{F}=\bar{F}_u +\bar{F}_{w}$$. Also, letting $$\nu =1$$, we assume a perfect maternal transmission for the reproduction outcome of $$\bar{F}_w \bar{M}_{w}$$ mating. Therefore, the general Wolbachia model in terms of population proportions is given by Eqs. (12)–(14). Hereafter it is clear that we refer to the scaled values of each state variable and as such drop the overbar from our notation. The scaled model below now evolves in the feasible region $$\bar{\Omega }$$, where $$\bar{\Omega } = \left\{ (\bar{A}_u, \bar{F}_u, \bar{A}_w, \bar{F}_w) \in \mathbb {R}_+^4 \, | \, \bar{A}_u + \bar{F}_u + \bar{A}_w + \bar{F}_w \le 1 \right\}$$.
\begin{aligned} \frac{d\bar{A}_u}{dt}= & {} \left[ \frac{\rho _{uu} (\bar{F}_u^2 + (1-\phi )\bar{F}_u \bar{F}_w) + \rho _{ww}(1-\delta )\bar{F}_w \bar{F}_u}{\bar{F}}\right] \left( 1-\alpha \bar{A} \right) -(\tau _{u} +\mu _{Au}) \bar{A}_u, \end{aligned}
(11)
\begin{aligned} \frac{d\bar{F}_u}{dt}= & {} \frac{\tau _{u}}{2}\bar{A}_u + \sigma \bar{F}_w -\mu _u \bar{F}_u, \end{aligned}
(12)
\begin{aligned} \frac{d\bar{A}_w}{dt}= & {} \left[ \frac{\rho _{ww} (\bar{F}_w^2 +\delta \bar{F}_w \bar{F}_u) }{\bar{F}}\right] \left( 1-\alpha \bar{A} \right) - (\tau _{w} +\mu _{Aw}) \bar{A}_w, \end{aligned}
(13)
\begin{aligned} \frac{d\bar{F}_w}{dt}= & {} \frac{\tau _{w}}{2}\bar{A}_w - \sigma \bar{F}_w - \mu _{w} \bar{F}_w. \end{aligned}
(14)
The modeling of wAu-Wolbachia transmission dynamics has not been done as this a distinction from other Wolbachia transmission models. Unlike the modeling work in Adekunle et al.37, apart from the non-induction of CI, we considered the loss of Wolbachia infections due to seasonal fluctuation in temperature, a key dynamics that is absent in wAu strain.
## Results
### Analysis of the model
The above general model (11)–(14) is parametrically adjusted to simultaneously accommodate wAu and wMel Wolbachia strains. For the wAu-Wolbachia model, we set $$\phi = \sigma = 0$$ and for the wMel-Wolbachia model, we set $$\phi = 1$$, $$\sigma > 0$$. The wMel-Wolbachia model parameter adjustments correspond to the model studied in Adekunle et al.37.
Here, we want to analyse the general model(11)–(14) with arbitrary values of $$\phi$$ and $$\sigma$$ to enable comparison with wAu-Wolbachia and Adekunle et al. 201937 models. Analysing the model for wAu, we have four steady states. The first steady state $$e_1 = (0,0,0,0)$$ indicates non-existence of mosquitoes. The second $$e_2 = (A_u^*,F_u^*,0,0)$$ signifies the steady state for the uninfected mosquito population only. The third $$e_3 = (0,0,A_w^*,F_w^*)$$ describes the equilibrium point for wAu-infected mosquitoes only. Lastly, the $$e_4 = (A_u^*,F_u^*,A_w^*,F_w^*)$$ is the equilibrium point for the co-existence of both uninfected and wAu-Wolbachia-infected mosquito populations.
#### Non-existence mosquito population, $$e_1$$
The equilibrium point $$e_1$$ is trivial and is not biologically realistic. However, we can gain some insights into the competitive model dynamics by examining the case where there is no interaction between the uninfected and Wolbachia-infected mosquitoes. In other words, we want to investigate how each population would behave in the absence of the other. In particular, we derive the reproduction number of the uninfected $$R_{0u}$$ and Wolbachia-infected $$R_{0w}$$ mosquito populations when they do not interact:
\begin{aligned} R_{0u}= & {} \frac{\rho _{uu}\tau _{u}}{2\mu _{u}(\mu _{Au}+\tau _{u})}, \end{aligned}
(15)
\begin{aligned} R_{0w}= & {} \frac{\rho _{ww}\tau _{w}}{2\mu _{w}(\mu _{Aw}+\tau _{w})}, \end{aligned}
(16)
where the factor of $$\frac{1}{2}$$ in $$R_{0u}$$ and $$R_{0w}$$ stems from the choice to set M = F62, i.e. $$\psi = \frac{1}{2}$$.
These reproductive numbers determine if the uninfected and Wolbachia-infected mosquito populations will die out or persist when there is no interaction. Specifically, if $$R_{0u}<1$$ and $$R_{0w}<1$$, then the two populations will die out (Fig. 2a). We observed in the decoupled case, the expressions for $$R_{0u}$$ and $$R_{0w}$$ are independent of the effects of CI $$(\phi )$$ and LWI $$(\sigma )$$ and are therefore equivalent for both the wAu and wMel-Wolbachia strains (Fig. 2)37.
#### Uninfected mosquito population, $$e_2$$
The uninfected-mosquito-only equilibrium point or Wolbachia-free equilibrium is
\begin{aligned} e_2 = \left( \frac{1}{\alpha }\left[ 1-\frac{1}{R_{0u}}\right] ,\frac{\tau _{u}}{2\mu _{u}\alpha }\left[ 1-\frac{1}{R_{0u}}\right] ,0,0\right) . \end{aligned}
For $$e_2$$ to exist, we require $$R_{0u}>1$$. In addition to the uncoupled reproduction numbers ($$R_{0u}$$ and $$R_{0w}$$) we also define the invasive reproduction number $$R_{0w|u}$$ which describes the average number of secondary offspring that will become Wolbachia-infected adults after introducing a single adult Wolbachia-infected mosquito into an established Wolbachia uninfected mosquito population.
To compute $$R_{0w|u}$$, we use the next generation matrix method63 to obtain
\begin{aligned} R_{0w|u}=\frac{\delta R_{0w}}{R_{0u}}, \end{aligned}
(17)
where we have substituted in the definition of $$R_{0w}$$ from Eq. (16). The invasive reproduction number $$R_{0w|u}$$ is the same for both wAu and wMel-Wolbachia strains as that derived in Adekunle et al.37. This is because, the expression (17) clearly shows that the invasive reproductive number $$R_{0w|u}$$ is not dependent on the CI effect, $$\phi$$ or LWI, $$\sigma$$.
To check if the equilibrium point $$e_2$$ is stable, we compute the Jacobian of the system and evaluate it at $$e_2$$. In particular, letting $$z_1 = (\mu _{Au}+\tau _{u})$$ and $$z_2 = (\mu _{Aw}+\tau _{w})$$, yields
\begin{aligned} J_{e_2}=\begin{pmatrix} -z_1R_{0u}&{}\frac{\rho _{uu}}{R_{0u}}&{}z_1(1-R_{0u})&{}\frac{(1-\delta )\rho _{ww}}{R_{0u}}\\ \frac{\tau _{u}}{2}&{}-\mu _{u}&{}0&{}0\\ 0&{}0&{}-z_2&{}\frac{\delta \rho _{ww}}{R_{0u}}\\ 0&{}0&{}\frac{\tau _{w}}{2}&{}-\mu _{w} \end{pmatrix}. \end{aligned}
To obtain the characteristic equation of $$J_{e_2}$$, we have
\begin{aligned} |J_{e_2}-\lambda I|&=0, \end{aligned}
which becomes
\begin{aligned} (\lambda ^2 +k_1\lambda + k_2)(\lambda ^2 +l_1\lambda + l_2)&=0, \end{aligned}
where
\begin{aligned} k_{1}= & {} \mu _{u}+z_1 R_{0u},\\ k_{2}= & {} \mu _{u}z_1(R_{0u}-1),\\ l_{1}= & {} \mu _{w}+z_2,\\ l_{2}= & {} \mu _{w}z_2(1-R_{0w|u}). \end{aligned}
Therefore, $$e_2$$ is locally asymptotically stable if and only if $$R_{0w|u}<1$$ and $$R_{0u}>1$$ (Fig. 4). This is also consistent with the study in Adekunle et al.37 (See Table 3).
#### Wolbachia-infected mosquito population, $$e_3$$
The wAu-infected-only equilibrium point is $$e_3 = \left( 0,0,\frac{1}{\alpha }\left[ 1-\frac{1}{R_{0w}}\right] ,\frac{\tau _{w}}{2\mu _{w}\alpha }\left[ 1-\frac{1}{R_{0w}}\right] \right)$$. This again is consistent with Adekunle et al.37.
For $$e_3$$ to exist we require $$R_{0w}>1$$. By computation, the invasive reproductive number $$R_{0u|w}$$ with respect to uninfected mosquitoes is given as,
\begin{aligned} R_{0u|w}=\frac{R_{0u}}{R_{0w}}\left[ (1-\phi )+\frac{\rho _{ww}}{\rho _{uu}}(1-\delta ) \right] = \frac{c R_{0u}}{R_{0w}}, \end{aligned}
(18)
where $$c = (1-\phi )+\frac{\rho _{ww}}{\rho _{uu}}(1-\delta )$$. Clearly, $$R_{0u|w}$$ is dependent on $$\phi$$. For the wMel-Wolbachia strain, i.e. $$\phi =1$$, $$c = \frac{\rho _{ww}}{\rho _{uu}}(1-\delta )$$ which is equivalent to that of Adekunle et al.37. However, for the wAu-Wolbachia strain, i.e. $$\phi =0$$, we have a modified expression of $$c = 1+\frac{\rho _{ww}}{\rho _{uu}}(1-\delta )$$ in Eq. (18) because we do not assume CI. Therefore, $$c\ge 1$$ for wAu-Wolbachia strain. Computing the Jacobian at $$e_3$$, we have:
\begin{aligned} J_{e_3}=\begin{pmatrix} -z_1&{}\frac{\rho _{uu}+(1-\delta )\rho _{ww}}{R_{0w}}&{}0&{}0\\ \frac{\tau _{u}}{2}&{}-\mu _{u}&{}0&{}0\\ z_2(1-R_{0w})&{}\frac{-(1-\delta )\rho _{ww}}{R_{0w}}&{}-z_2R_{0w}&{}\frac{\rho _{ww}}{R_{0w}}\\ 0&{}0&{}\frac{\tau _{w}}{2}&{}-\mu _{w} \end{pmatrix}. \end{aligned}
The characteristic equation of $$J_{e_3}$$ is then
\begin{aligned} |J_{e_3}-\lambda I|=(\lambda ^2 + m_1\lambda + m_2) (\lambda ^2 + n_1\lambda + n_2) = 0, \end{aligned}
where
\begin{aligned} m_{1}= & {} \mu _{u}+z_1,\\ m_{2}= & {} \mu _{u}z_1(1-R_{0u|w}),\\ n_{1}= & {} \mu _{w} + z_2 R_{0w},\\ n_{2}= & {} \mu _{w}z_2(R_{0w}-1). \end{aligned}
Therefore, $$e_3$$ is locally asymptotically stable if and only if $$R_{0u|w}<1$$ and $$R_{0w}>1$$ (see Fig. 4). The condition is equivalent to that found in37 with generalized expressions for $$R_{0u|w}$$ used in place of the reduced version presented there (see Table 3).
#### Coexistent mosquito populations, $$e_4$$
The equilibrium point for which both the uninfected and Wolbachia-infected populations coexist is
$$e_4 = (\frac{2\mu _u \beta F_w^*}{\tau _{u}},\beta F_w^*,\frac{2\mu _w F_w^*}{\tau _{w}},F_w^*)$$ where
\begin{aligned} F_w^*=\frac{1}{2\alpha }\left[ \dfrac{\left( 1-\frac{\xi }{R_{0w}}\right) \tau _u\tau _w}{(\mu _w \tau _{u}+\beta \mu _u \tau _{w})}\right] , \end{aligned}
$$\beta = \dfrac{R_{0w}(R_{0u|w}-1)}{R_{0u}(R_{0w|u}-1)}$$ and $$\xi = \dfrac{(\beta +1)}{(\delta \beta + 1)}$$. For $$e_4$$ to exist, we require $$R_{0w}> \xi > 1$$ and
1. (i)
$$R_{0w|u},R_{0u|w}>1$$ or
2. (ii)
$$R_{0w|u},R_{0u|w}<1$$.
The above conditions (i) and (ii) correspond to the cases for $$\delta >\dfrac{1}{c}$$ and $$\delta < \dfrac{1}{c}$$ respectively. Comparing these existence conditions with those found above for $$e_2$$ and $$e_3$$, we see that condition (ii) for the existence of $$e_4$$ matches the combined existence and local asymptotic stability condition for $$e_2$$ and $$e_3$$. In other words, $$e_2$$, $$e_3$$ and $$e_4$$ can coexist, while $$e_1$$ always exists (see Fig. 4).
To establish whether $$e_4$$ is stable or not, we compute the Jacobian $$J_{e_4}$$ evaluated at $$e_4$$ to obtain the following characteristic equation:
\begin{aligned} |J_{e_4}-\lambda I|:=\lambda ^4 +s_1\lambda ^3 + s_2\lambda ^2 +s_3\lambda + s_4 = 0. \end{aligned}
(19)
Let
$$z_3 = (\mu _u + \mu _w)$$, $$z_4 = (\beta \rho _{uu}+\rho _{ww})$$, $$z_5 = (\beta +1)\rho _{uu} +(1-\delta )\rho _{ww}$$, $$z_6 = 1+\beta (2+\beta \delta )$$,
$$z_7 = (\beta +1)^2\rho _{uu} +(1-\delta )\rho _{ww}$$, $$z_8 = (1+\beta (2+\beta \delta ))\rho _{uu} +(1-\delta )\rho _{ww}$$,
then we have:
\begin{aligned} s_{1}= & {} z_1+z_2+z_3+\alpha z_4F_w^*,\\ s_{2}= & {} \mu _u\mu _w + z_3(z_1+z_2 + \alpha z_4F_w^*) - \frac{\xi }{2 R_{0w}(\beta + 1)^2} (z_6\rho _{ww}\tau _w +z_7\tau _u),\\ s_{3}&= {} \mu _u\mu _w (z_1+z_2+\alpha z_4F_w^*) +z_3 \left[ z_1 z_2 + \frac{\alpha }{\beta + 1}(z_1 (1+\beta \delta )\rho _{ww}+\beta z_2 z_5)F_w^* \right] \\&\quad -\frac{\xi }{2R_{0w}(\beta + 1)^3}\{\left[ (\mu _u +z_1)z_6+ z_8\alpha \beta F_w^*\right] (\beta + 1)\rho _{ww}\tau _w\\&\quad +\left[ \alpha \beta (1-\delta )z_5\rho _{ww}F_w^*+ z_7(\alpha (1+\beta \delta )\rho _{ww}F_w^*+(\mu _w +z_2)(\beta + 1))\right] \tau _u\},\\ s_{4}= & {} \mu _u\mu _w \left[ z_1 z_2 + \frac{\alpha }{\beta + 1}(z_1 (1+\beta \delta )\rho _{ww}+\beta z_2 z_5)F_w^* \right] - \frac{\xi }{2R_{0w} (\beta + 1)^2} \{[z_2z_6 \\&\quad +z_8\alpha \beta F_w^*]\mu _u\rho _{ww}\tau _w + [\alpha \beta (1-\delta )z_5\rho _{ww}F_w^*+ z_7(\alpha (1+\beta \delta )\rho _{ww}F_w^*+z_2(\beta + 1))]\mu _w\tau _u\\&\quad -\frac{\xi }{2R_{0w}} [z_8\rho _{ww}] \}. \end{aligned}
In order to establish the nature of the equilibrium point $$e_4$$, we performed numerical testing using the Monte Carlo method in50 to verify the conditions (i) and (ii) by computing the real part of the eigenvalues of the Jacobian matrix, evaluated at $$e_4$$. Simulation results are illustrated in Fig. 3.
Although the conditions (i) and (ii) indicated the existence of $$e_4$$, Fig. 3c showed that $$e_4$$ is locally stable for condition (i) as all the eigenvalues (real part) are negative ($$\lambda _1, \lambda _2, \lambda _3,\lambda _4 <0$$). Whilst Fig. 3f showed that $$e_4$$ is unstable for condition (ii) as two of the eigenvalues (real part) are positive i.e. $$\lambda _3,\lambda _4 > 0$$.
Numerically, we illustrated the existence and stability regions for $$e_4$$ in Fig. 4 for the two conditions (i) and (ii) relating to CI and maternal transmission (MT).
Following a modeling study of Aedes aegypti mosquitoes and normal Wolbachia (in the presence of CI only) interaction analyzed by Ferreira et al.64, three equilibrium points: trivial ($$q_1$$); uninfected only ($$q_2$$); and coexistence ($$q_3$$), were obtained. However, the Wolbachia-only equilibrium point was not computed. The established local stability conditions for $$q_1$$ and $$q_2$$ correspond to that of the wMel-like Wolbachia conditions for $$e_1$$ and $$e_2$$ respectively. For coexistent populations to persist, the reproductive number for infected mosquitoes only, $$R_i$$ must be greater than 1 and $$R_i > R_u$$, where $$R_u$$ is the reproductive number for wild-type mosquitoes only. The model64 also described the fitness parameter space between $$R_u$$ and $$R_i$$, showing the change in extinction and persistence of the three equilibria when there is an increase in the initial population proportion of the Wolbachia-infected mosquitoes. Our model showed the changes in the no-mosquito, wild-type only, Wolbachia-only and coexistence population persistence and extinction in the presence and absence of CI with high and low maternal transmission (MT).
Figure 4 illustrates the existence and local stability regions for the equilibrium points $$e_1$$, $$e_2$$, $$e_3$$ and $$e_4$$ with respect to the reproduction numbers $$R_{0u}$$ and $$R_{0w}$$ as well as the relative magnitude of $$\delta$$ and $$\frac{1}{c}$$. For $$\delta > \frac{1}{c}$$ (high MT), Fig. 4a,b describe the dynamics for $$\phi =0$$ (CI absent) and $$\phi =1$$ (CI present) respectively. Within the subset of the yellow region of these figures bounded by $$R_{0u}=1, R_{0w}=1$$, and $$R_{0w}=\xi$$ we find that only $$e_1$$ and $$e_3$$ exist. Since $$e_3$$ is unstable in this region, we expect the system trajectories to tend to the no-mosquito equilibrium $$e_1$$. This was confirmed through numerical simulations shown in Fig. 5a. For the existence of $$e_4$$ we require $$R_{0u|w}>1$$, $$R_{0w|u}>1$$ and $$R_{0w}>\xi$$ for stability (within the blue region). But if $$R_{0w}<\xi$$, $$e_1$$ is stable (yellow).
For $$\delta < \frac{1}{c}$$ (low MT), Fig. 4c,d portrayed the regions of stability for $$\phi =0$$ and $$\phi =1$$ respectively. The conditions $$R_{0u}<1, R_{0w}>1$$, and $$R_{0u|w}>1$$ project the trajectiory to tend to $$e_1$$ (see Fig. 5b). In the orange region, $$e_1$$ and $$e_3$$ exist and are simultaneously locally stable as $$R_{0u|w}>1$$ and $$R_{0w}>\xi$$. In addition, we have that $$e_4$$ exists where $$R_{0u|w}<1$$ and $$R_{0w|u}<1$$ (condition (ii)). With these conditions, $$e_4$$ exists together with $$e_2$$ and $$e_3$$ (white region). In this white region, $$e_2$$ and $$e_3$$ are locally stable even as $$R_{0u}>1, R_{0w}>1$$ but $$e_4$$ is unstable. Also, $$e_1$$ exists when $$R_{0w}>1$$ and $$R_{0u}<1$$ because the local stability of other equilibrium points is violated with these conditions. When $$R_{0u|w} > 1$$ but $$R_{0u} < 1$$ and $$R_{0w} > 1$$, the only stable outcome is the mosquito-free (no-mosquito) equilibrium $$e_1$$. This occurs when $$R_{0u}$$ is less than but still close to one. In this region, uninfected mosquitoes are capable of dominating initially when introduced into a Wolbachia saturated equilibrium because imperfect maternal transmission achieves $$R_{0u|w} > 1$$. This competitive advantage drives out the Wolbachia infected mosquitoes leaving uninfected mosquitoes only, which then are unable to sustain their population because $$R_{0u} < 1$$ (Fig. 5).
With the rate of high maternal transmission (MT) in the absence of CI (like-wAu), the reproductive advantage favours the production of uninfected mosquito offspring as it tends to accommodate more coexistent mosquito populations with wild-type than wMel-like strain (presence of CI) due to the presence of CI (Fig. 4a,b). Whilst, with a low MT rate, the CI presence or absence would favour Wolbachia-infected mosquitoes or uninfected mosquitoes respectively. In other words, the coexistent equilibrium point is unstable for the two mosquito populations as these conditions are equivalent to the local stabilities of both Wolbachia-free and Wolbachia-only equilibrium points (Fig. 4c,d). If $$R_{0w} < \xi$$, the system trajectories tend to the no mosquito equilibrium $$e_1$$.
The conditions for the local stability of all equilibrium points are shown in Table 3 below.
### Sensitivity analysis of Wolbachia model
To carry out the sensitivity analysis we investigate the model robustness due to uncertainties associated with parameter value estimations. In other words, we examine how senitive the invasive reproductive numbers are with respect to these parameters. This in turn, gives insight on influential parameters and their impact in reducing (or increasing) mosquito-type populations. To carry out this, we compute the normalized sensitivity indices of the invasive reproduction numbers with respect to the parameters used in the model.
#### Definition
The normalized forward sensitivity index of a variable v with respect to parameter w is defined as:
\begin{aligned} \Lambda _w = \dfrac{\partial v}{\partial w} \times \dfrac{w}{v}. \end{aligned}
(20)
Using the above formular (20), we contruct the following plots in Fig. 6.
From Fig. 6 and using the baseline parameter values for the wAu-Wolbachia strain in Table 2, it is clear that the reproductive and mortality rates for both wild-type ($$\rho _{uu}, \mu _u$$) and wAu-Wolbachia-infected ($$\rho _{ww}, \mu _w$$) mosquitoes and the proportion of wAu-Wolbachia-infected offspring ($$\delta$$) have the most sensitivity in the invasive reproductive numbers $$R_{0w|u}$$. Whilst for $$R_{0u|w}$$, $$\mu _u$$ and $$\mu _w$$ are the most sensitive parameters. Hence for both invasive reproductive numbers, the most sensitive parameters are $$\mu _u$$ and $$\mu _w$$. This demonstrates that an increase (or decrease) in the mortality rate of wAu-Wolbachia-infected mosquitoes by 10% will decrease (or increase) $$R_{0w|u}$$ by 10%.
### Does CI $$(\phi )$$ outweigh the LWI $$(\sigma )$$?
For most Wolbachia strains except wAu, the mating between uninfected female and Wolbachia-infected male mosquito crosses generates no viable offspring. However, Wolbachia-infected mosquitoes tend to lose their Wolbachia infection and lower their maternal transmission rate at high temperature ($$27{-}37^\circ {\text{C}}$$)23. With the effect of climate change gradually increasing the temperature by the day, Wolbachia strains with moderate or high temperature sensitivity such as wMel may not be able to fully maintain a sufficient frequency level to invade the mosquito population.
In our general Wolbachia mathematical model, we describe a modified version of Adekunle et al.37. This modification accommodates parameter adjustments for novel wAu and wMel-Wolbachia strains. For wAu, our mathematical model showed that despite the production of mosquito offspring due to CI absence, the invasive reproduction number due to infected mosquitoes $$R_{0w|u}$$ remains unchanged compared to the case where CI is present, as with the wMel-like strain37. This further strengthened the fact that CI (inclusion or exclusion) does not guarantee Wolbachia mosquitoes’ persistence. Also, the invasive reproduction number due to uninfected mosquitoes expression $$R_{0u|w}$$ for wAu is similar to wMel, except that the expression depends on CI the effect. This is because, the mosquito gender crosses due to non-induction of CI for wAu, i.e. $$F_uM_w$$, generates uninfected offspring with perfect maternal transmission while wMel does not. The chances of establishing Wolbachia infected mosquitoes are lower when CI is ineffective compared to when it is induced. That is, for cytoplasmic inducing wMel-Wolbachia mosquitoes, the effect of LWI outweighs CI effect as mosquitoes still lose their infections (Fig. 7). However, wAu-Wolbachia infection retainment (no LWI) in mosquitoes has shown high level of maintaining the Wolbachia frequency in the absence of CI in mosquitoes (Fig. 7). This suggests that the LWI effect outweighs CI.
The LWI rate $$\sigma (t)$$ which is dependent on the seasons of the year can be modeled by a sinusoidal equation:
\begin{aligned} \sigma (t) = \frac{\sigma _{max}}{2}\left[ 1+\cos \left( \dfrac{2\pi t}{365}-\mathscr {C}\right) \right] \end{aligned}
(21)
where $$\sigma _{max}$$ is the maximum value of the seasonal variation in LWI, and $$\mathscr {C}$$ is the phase shift which aligns the model with the seasonal change.
The effects of CI ($$\phi$$) and LWI ($$\sigma (t)$$) as features of wAu and wMel Wolbachia strains are shown in Fig. 7. For the total mosquito population, wAu-infected mosquitoes ($$\phi =0, \sigma _{max}=0$$) reach the maximum frequency after approximately 250 days. To see the effect of CI induction and slight LWI i.e. $$\phi =1$$, $$\mathscr {C}=0.25$$, for $$\sigma _{max}=0.02$$ and $$\sigma _{max}=0.04$$, the Wolbachia frequency level oscillates between (0.8 and 1) and (0.6 and 1) respectively. That is, there is a 20% and 40% drop in the frequency level of Wolbachia when $$\sigma (t)$$ is at $$\sigma _{max}=0.02$$ and $$\sigma _{max}=0.04$$ respectively. This showed that, despite CI induction, LWI reduced the contribution of CI to the Wolbachia invasion (Fig. 7a). Therefore, the LWI gains highly outweigh the CI effect. By this, our analysis suggests that an increase in LWI in the presence of CI results in a drastic decrease in the Wolbachia frequency level (Fig. 7a). On the other hand, Fig. 7b showed the effect of LWI $$\sigma (t)$$ and CI $$\phi$$ with respect to the competitiveness between $$F_u$$ and $$F_w$$. We observed that the $$F_w$$ population dominates the $$F_u$$ when there was no CI induction and Wolbachia infection is retained, that is, $$\phi =0, \sigma _{max}=0$$ (Fig. 7b). However, if CI induction occurs with loss of Wolbachia infections, then the seasonal varying effect occurs as seen in Fig. 7c.
## Discussion
In this work, we modelled and investigated a general Wolbachia model that contained the transmission dynamics of wAu and wMel Wolbachia strains in Aedes mosquitoes as special cases. These transmission dynamics described the competition between the novel wAu-Wolbachia infected Aedes mosquitoes and wild-type mosquitoes and compared the dynamics with the invasive properties of the popular wMel-Wolbachia infected mosquitoes. We first derived the Wolbachia infection-status reproduction numbers for our wAu-Wolbachia model and used them to establish the conditions for the local stability of the equilibrium points for the wAu-Wolbachia invasive model. The reproduction number associated with the uninfected mosquitoes shows the reproductive advantage that the wild type has over the wAu strain. The comparison of the wAu-Wolbachia model (CI and LWI absent) and wMel-Wolbachia model (CI and LWI present) showed that the wAu strain has the potential of compensating for the undesirable features of the wMel strain.
Additionally, this study has reviewed the main features of different Wolbachia strains (Table 1) and shown that the wAu Wolbachia strain is a promising candidate for efficient Aedes-borne arboviral transmission control. Moreover, we analyzed the system dynamics of a general Wolbachia invasion model and determined the regions of local stability for each of the identified equilibrium points, highlighting the regions in parameter for which Wolbachia-infected mosquito populations persist or go extinct. This work modelled the general Wolbachia dynamics which can accommodate various Wolbachia characteristics regarding the presence or absence of CI and seasonal changes, unlike Adekunle et al.37, which considers only the presence of CI. We also investigated the advantages gained from CI and LWI. This study has demonstrated that despite the absence of CI, the Wolbachia frequency level will drop as much as tenfold of the percentage of Wolbachia infection lost. We showed that the advantage of Wolbachia retainment in mosquitoes strongly outweighed the negative impact of CI indicating wAu Wolbachia strains may be suitable for arboviral control. Therefore, this modeling work contributes to the previous studies37,54,57,64,65 and helps close the gap between ways of maintaining the Wolbachia frequency levels in the absence of LWI and CI.
One implementation question for using the wAu strain as a replacement of the wMel strain is whether the wAu strain is self-sustaining, given that it does not induce CI. In this work, the equilibrium points for the wAu-Wolbachia model are the same as that for the wMel-Wolbachia model except that stricter conditions are required to satisfy the wAu-Wolbachia model equilibrium points. These more stringent conditions translate to additional resources such as the continuous introduction of a larger scale of wAu-infected mosquitoes to ensure replacement66. Thus, the wAu strain is a promising alternative strain as it does not suffer from LWI due to high weather temperature and is highly effective in preventing the transmission of the arbovirus23,39,67. Otherwise, combining the two strains may also be a good strategy.
There are limitations associated with any mathematical modeling work, and this study is not exempted. We first assumed the same mosquito gender ratio and expected this proportion to be constant over time. This assumption may be true in a laboratory setting62, but not necessarily true in a natural mosquito habitat. However, similar conclusions are expected to be reached as the Wolbachia model reduction accurately reproduces the dynamics of the full system68. Secondly, we assumed that the absence of CI implies that cross mating resulted in offspring that are uninfected. This may not be true as a small proportion of the offspring may be Wolbachia infected23. If that is the case, then it means that lesser resources will be required to use the wAu strain as a Wolbachia-based control strategy. Lastly, we assumed the seasonality affects the associated parameters for the wMel dynamics. However, for the wAu strain, it is not affected by seasonality as wAu-Wolbachia infections are retained at high temperature.
Although several studies22,38,42,57 have demonstrated that CI drives the persistence of Wolbachia-infected Aedes mosquitoes, these studies neglected the impact of Wolbachia loss in mosquitoes. The CI drive has been shown in four mating lines (see Fig. 1) involving a Wolbachia-transinfected Aedes mosqiutoes mating with wild-type mosquitoes. One of the mating lines for which Wolbachia-infected male and uninfected female mosquitoes produced no viable offspring (via CI) truncates the uninfected offspring from being produced as infection is maternally transmitted. With the exception of the mating between the uninfected male and female mosquito line, all other mating lines produce Wolbachia-infected offspring leading to persistence. In addition, high temperature affects these Wolbachia-infected mosquitoes as they lose their infection due to the unfavourable weather conditions. However, mosquitoes infected with the wAu-Wolbachia strain have been shown to not only block arboviral transmission efficiently, but also retain the Wolbachia infection at typically unfavourable high temperatures. This retainment of infection in mosquitoes strongly outweighed the absence of CI for the wAu strain in the establishment and dominance of wAu-Wolbachia infected mosquitoes.
While vaccine implementation may have been highly effective on dengue seropositive persons in high transmission areas11,12, the introduction of Wolbachia-infected mosquitoes in low and moderate arboviral endemic areas has also effectively shown successful reduction in dengue burden43,51,59,69. Given that these two strategies could reduce the transmission of Aedes-borne diseases, in particular, dengue depending on the transmission level, a modeling study by Ndii70 proposed the use of these combined strategies and compared their effectiveness. The author showed that, Wolbachia performs better in the presence of low vaccine efficacy, but is outperformed otherwise70. Therefore combining the two strategies may be useful, however understanding both the temperature and seasonality effects on Wolbachia intervention programs, and serotypic differences relating to cross-protective immunity to investigate vaccine efficacy is necessary for the reduction and control of Aedes-borne arboviral disease transmission.
In conclusion, we have shown that the wAu-Wolbachia strain could be effective in controlling arbovirus transmission, as its advantages in terms of Wolbachia infection retention in mosquitoes may outweigh the absence of CI. This could prove even more promising, especially as the temperature increases due to climate change. Although wMel and wAlbB-Wolbachia strains only have been rolled out in natural mosquito habitats in replacement programs, combining these strains with wAu is worth exploring.
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### Golden Thoughts
Rectangle PQRS has X and Y on the edges. Triangles PQY, YRX and XSP have equal areas. Prove X and Y divide the sides of PQRS in the golden ratio.
### From All Corners
Straight lines are drawn from each corner of a square to the mid points of the opposite sides. Express the area of the octagon that is formed at the centre as a fraction of the area of the square.
### Star Gazing
Find the ratio of the outer shaded area to the inner area for a six pointed star and an eight pointed star.
# Rhombus in Rectangle
##### Stage: 4 Challenge Level:
Another Tough Nut! Take any rectangle $ABCD$ such that $AB > BC$ and say the lengths of $AB$ and $CD$ are $S$ and $s$ respectively. The point $P$ is on $AB$ and $Q$ is on $CD$. For $APCQ$ to be a rhombus, the lengths $AP$ and $PC$ must be equal. Consider the point $P$ coinciding with $A$ (such that $AP=0$) and then $P$ moving along $AB$ so that the length $AP$ increases continuously from $0$ to $S$ while the length of $PC$ decreases continuously from $\sqrt{S^2 + s^2}$ to $s$. As $AP < PC$ initially (when P is at A) and $AP > PC$ finally (when $P$ is at $B$) there must be one point at which $AP = PC$. Similarly there is exactly one position of $Q$ such that $CQ = QA$ making $APCQ$ into a rhombus.
Now take $AP = PC = x$ than you can use Pythagoras' Theorem to find $x$ in terms of $S$ and $s$ so that you can find the ratio of the areas of the areas of the rhombus and the rectangle.
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# Math Help - gradients/directional derivatives
This question is driving me crazy! Please help relieve my sanity. I'm sure it easy and I have just completely lost the plot.
2. Originally Posted by asw-88
This question is driving me crazy! Please help relieve my sanity. I'm sure it easy and I have just completely lost the plot.
One approach of many:
Polar coordinates: $x = r \cos \theta$ and $y = r \sin \theta$.
Therefore:
$r^2 = x^2 + y^2$ .... (1)
$\tan \theta = \frac{y}{x}$ .... (2)
Results that will be needed:
From (1):
$2 r \frac{\partial r}{\partial x} = 2x \Rightarrow \frac{\partial r}{\partial x} = \frac{x}{r} = \cos \theta$ .... (A)
$2 r \frac{\partial r}{\partial y} = 2y \Rightarrow \frac{\partial r}{\partial y} = \frac{y}{r} = \sin \theta$ .... (B)
From (2):
$sec^2 \theta \, \frac{\partial \theta}{\partial x} = -\frac{y}{x^2} = -\frac{r \sin \theta}{r^2 \cos^2 \theta} \Rightarrow \frac{\partial \theta}{\partial x} = -\frac{\sin \theta}{r}$ .... (C)
$sec^2 \theta \, \frac{\partial \theta}{\partial y} = \frac{1}{x} = \frac{1}{r \cos \theta} \Rightarrow \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r}$ .... (D)
Then:
$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \cdot \frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta} \cdot \frac{\partial \theta}{\partial x} = .....$ .... (E)
$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \cdot \frac{\partial r}{\partial y} + \frac{\partial f}{\partial \theta} \cdot \frac{\partial \theta}{\partial y} = .....$ .... (F)
where the results (A) - (D) are to be substituted into (E) and (F).
Then substitute (E) - (F) into $\nabla f$, do a little bit of re-arranging and then the job's done.
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## Return to Answer
1 [made Community Wiki]
More generally, you might ask "what is the point of constructing limits or colimits of diagrams of objects instead of working directly with the diagrams?" A generic answer is that a diagram of objects in a category describes a functor, and it is useful to know that that functor is representable. For example, if objects $X_1, X_2$ in a category have a product $X_1 \times X_2$, this is equivalent to the statement that the functor $\text{Hom}(-, X_1) \times \text{Hom}(-, X_2)$ is representable. So you now know that this functor takes colimits to limits, which is new information.
Another generic answer is the following. Any time you construct an object $X$ as a limit of a diagram of other objects $X_i$ in a category, you know what the maps into $X$ look like by definition (compatible systems of maps into the $X_i$). What you don't know is what the maps out of $X$ look like, and this is new information you get from the existence of $X$. For example, the limit of the empty diagram is the terminal object $1$, and while maps into $1$ are trivial, maps out of $1$ ("global points") are not; in the category of schemes over a field $k$, for example (an example within an example!), they correspond to $k$-points.
Specializing to Galois theory, when you construct a Galois group $G$ as a limit of finite Galois groups $G_i$, the new information you have access to is, for example, the representation theory of $G$. I don't know how one would talk about the correspondence between modular forms and 2-dimensional Galois representations without direct access to the group $\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$, for example.
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# If Z = 1 + i, where i = √-1, then what is the modulus of $\rm z+\frac{2}{z}?$
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closed
If Z = 1 + i, where i = √-1, then what is the modulus of $\rm z+\frac{2}{z}?$
1. 1
2. 2
3. 3
4. 4
by (54.3k points)
selected
Correct Answer - Option 2 : 2
Concept:
i2 = -1
Calculation:
Given z = 1 + i
We have to find the modulus of z + $\rm \frac{2}{Z}$
⇒ (1 + i) + $\rm \frac{2}{1 + i}$
On rationalizing the second term, we get
(1 + i)$\rm \frac{2}{1 + i}$$\times$$\rm \frac{1 - i}{1 - i}$
(1 + i)$\rm \frac{2\times (1 - i)}{1 - i^{2}}$
(1 + i)$\rm \frac{2\times (1 - i)}{1 - (-1)}$
(1 + i)$\rm \frac{2\times (1 - i)}{2}$
1 + i + 1 - i
2
∴ The modulus of $\rm z+\frac{2}{z} = 2$.
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45.1-b2 $$\Q(\sqrt{205})$$ 45.1 45.1-b $$\bigl[a$$ , $$-a + 1$$ , $$0$$ , $$-4 a + 52$$ , $$-1032 a + 7923\bigr]$$
45.1-c2 $$\Q(\sqrt{205})$$ 45.1 45.1-c $$\bigl[1$$ , $$1$$ , $$1$$ , $$0$$ , $$0\bigr]$$
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# Will Avogadro's number change if definition of atomic mass unit is changed?
I know that the relation between atomic mass unit and Avogadro's number is a reciprocal relation.
My question is if the definition of atomic mass unit is changed, suppose instead of 1/12th mass of carbon-12 isotope, we consider 1/24th mass of carbon-12 isotope, will the Avogadro's number change?
And if Avogadro's number changes will that change the mass of a mole of atoms of any element?
• No , it will not. Avogadro number is related to mole as 12 is to dozen. May 20, 2021 at 14:56
• The Avogadro constant is defined exactly as $N_\mathrm{A} = \pu{6.02214076E23 mol−1}$ May 20, 2021 at 14:57
• See the wikipedia article Avogadro constant which gives an interesting oversight on the constant. The gist is that IUAPC has defined some fundamental constants, including the Avogadro number, as having exact values which are not expected to change ever again.
– MaxW
May 21, 2021 at 20:07
The current definition of Avogadro's number doesn't depend on the unit of mass. They are completely independent quantities. The Avogadro's number is currently simply defined as $$\pu{6.02214076e23}$$. It's just a number which is important enough to get a special name.
The answer you are probably looking for:
Historically the Avogadro's number was defined in such a manner that it did depend on the unit of mass.
At that time the Avogadro's number was defined as the number of atoms present in $$\pu{12 g}$$ of $$\ce{^{12}C}$$ isotope of carbon.
Now as you can see even if we defined an atomic mass unit as $$1/24$$ of the mass of an $$\ce{^{12}C}$$ atom instead of $$1/12$$, it won't make any difference to the Avogadro's number as it was defined on the basis of grams of carbon and not atomic mass units of carbon.
Now coming to your second question of whether this would have changed the mass of a mole of a substance. Yes it would have changed that, all the molar masses would have been double of as known currently.
Note:
All this discussion is just about how someone wrote a definition and the loopholes someone can detect from it using language.
For example if we really would have defined an atomic mass unit in this different way, we would have also defined Avogadro's number as number of atoms in $$\pu{24 g}$$ of $$\ce{^{12}C}$$. Even $$\ce{^{12}C}$$ would have been written as $$\ce{^{24}C}$$!
The takeaway is that the scientific idea and origin of all these units was(and is) always there but it's just the nuance of defining it in such a way that it doesn't create such problems was the problem. That's the only reason all the standard units were restandardized recently. It's a sad reality that schools are still teaching the old definitions even after we have said them a good bye.
• My understanding is that if we consider Avogadro number to be a constant then mass of 1 mole shouldn't be affected by changing the definition of 1 amu as it would affect only the relative mass of elements. The absolute mass would still remain the same right? So why should the mass of 1 mole change on changing definition of 1 amu?
– dumb
May 20, 2021 at 16:25
• @dumb We initially defined moles on the basis of gram-equivalents. $\pu{12 g}$ of carbon-12 was selected because we assumed that mass of $\ce{^{12}C}$ to be $\pu{12 amu}$. If it was $\pu{24 amu}$ we would call $\pu{24 g}$ of carbon-12(24) a mole. Nothing is special even with a gram. If they decided at that time to use imperial units than the mole would be defined in that way(eg. Calling 24lb of carbon-12(24) a mole) . A mole of substance really isn't something special, other than the fact we made it the standard one. May 20, 2021 at 16:37
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# How to perform wildcard string comparison in batch file
My batch file writes a list of files on screen where it contains the .cs extension
for /F "usebackq delims=" %%A in (cleartool ls -rec ^| find /V "Rule:" ^| find /V "hijacked" ^| find /V "eclipsed" ^| find /V "-->") do (
if "%%~xA"==".cs" echo %%A
)
This works fine. I want to enhance the functionality by adding another condition. I want to remove a certain word if it exists in the phrase. The word is obj
Examples of my results are
myThing\obj\this.cs
obj\other\that\this.cs
debug\this\that.cs
I tried
for /F "usebackq delims=" %%A in (cleartool ls -rec ^| find /V "Rule:" ^| find /V "hijacked" ^| find /V "eclipsed" ^| find /V "-->") do (
if "%%~xA"==".cs" if NOT "%%A"=="*obj*" echo %%A
)
but this doesn't change the results compared to the first.
I think my if syntax is correct, and the issue is actually with the "%%A"=="*obj*"
• The best suggestion would be to migrate this toward PowerShell. The syntax is much prettier and significantly more feature rich - which in the long run may benefit your updating and script functionality. I figured PS would be a joke like CMD, but it's actually pretty impressive. – nerdwaller Jan 31 '14 at 16:07
You were thinking correctly: the wildcards (*) are not supported.
A good workaround for the problem is findstr, though you need to retrieve result via %ERRORLEVEL% and I think you must cache it through another variable (so that next if does not override it's value):
echo %%A | findstr /C:"obj"
set obj_errorlevel=%errorlevel%
if "%%~xA"==".cs" if "%obj_errorlevel%"=="1" echo %%A
...but that is a hard way.
An even easier way is to add the condition to the source command (the one inside if) by adding another pipe stage: ^| find /V "obj":
for /F "usebackq delims=" %%A in (cleartool ls -rec ^| find /V "Rule:" ^| find /V "hijacked" ^| find /V "eclipsed" ^| find /V "-->" ^| find /V "obj")
I believe your edit is correct, in that you can't use wildcards in the If string compare. It's actually looking for *obj*, including the asterisks.
Perhaps use FindStr instead, by piping any .cs matches through it. It returns an errorlevel of 0 if found, and 1 if not found.
Code example:
for /F "usebackq delims=" %%A in (cleartool ls -rec ^| find /V "Rule:" ^| find /V "hijacked" ^| find /V "eclipsed" ^| find /V "-->") do (
if "%%~xA"==".cs" echo %%A | findstr /C:"obj" 1>nul & if errorlevel 1 echo %%A
)
I dont think you can use 2 if's like that. Try:
for xxxxxxx (
if first-condition (
if second-condition (
statement(s)
)
)
)
• This results in the same as my code sadly... I don't think the issue is with the IF statement, but reading if a string exists within a string.. – MyDaftQuestions Jan 31 '14 at 16:01
• Oddly enough you can use it the OP's way (but yours is much more legible). In any case, I think formatting is highly important and this is a much better solution to the multiple IF issue in cmd. – nerdwaller Jan 31 '14 at 16:06
• @nerdwaller It works without the quotes in Win7 and later (and in theory it should). But I distinctly remember having issues with this particular nested if construct on W2K, XP and W2k3. I could be wrong. It was quite a while ago. – Tonny Jan 31 '14 at 19:59
• People still use those!? Haha, in any case I voted you up since I always support clean code – nerdwaller Jan 31 '14 at 20:53
The asterisks are not interpreted as wildcards, so the condition evaluates to false. Instead of a second if-statement, you can use findstr to determine whether the variable contains the text 'obj'. The command sets %errorlevel% to zero if a match has been found, to zero otherwise.
A convenient way of using this value is through the conditional command seperators && and ||. A command appended with || will only be executed if the previous command returned a non-zero error level, e.g.:
(echo %%A | findstr /R .*obj.*)>NUL ||
if "%%~xA"==".cs" echo %%A
The addition of >NUL is important in this scenario. Without suppressing the output, both the first line and the second line could print the variable %%A. To avoid confusion, you'll want different cases to produce different results.
Of course, if you're already using findstr, you could get rid of the if-statement and its body altogether. The readability isn't great either way; the code below is provided as an alternative.
(echo %%A|findstr /R /V ".*obj.*")|findstr /R /C:"\.cs \$"
A solution that might work for what you want to do. This uses SET's string substitution, as explained at Rob van der Woude's site.
@echo off
set _myvar=this\has\obj\in the string
echo Original: %_myvar%
::replace the string obj in _myvar
set _myvar=%_myvar:obj=%
::replace any double back-slashes in _myvar
set _myvar=%_myvar:\\=\%
echo Withou obj: %_myvar%
_myvar originally contains the string 'obj' in it. The first time we remove the string using SET's string substitution. The second time, we remove any double-slashes that might be left over due to the location of 'obj'. After the colon, you have the string you want to look for in the variable. The string to replace with comes after the equal sign before the end of the substitution.
Hope that helps!
Try expanding the contents of yourfile.cs and replace word or letter with desired character either nothing or something. Here's my idea.
Drag .cs file on top of batch file icon to open with it so that .cs file is param 1
@echo off
setlocal EnableDelayedExpansion
set /p a= enter letter to change:
set /p b= enter letter to change to:
set /p file=<%1
for /f %%c in ("!file:~!") do (
set "code=!file:%a%=%b%!"
echo !code!>myfile.cs
)
Change myfile.cs to any name you want
Please improve my ans for better functions.
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# can magnesium chloride and iodine react together?
If everything is in a mixture of alcohol and water, would it be possible for magnesium chloride and iodine to react and produce something?
$$\ce{Cl- + 1/2 I2 <=> 1/2 Cl2 + I-}$$ $$\ce{MgCl2 + I2 <=> MgI2 + Cl2}$$ $$\ce{I- + I2 <=> I3-}$$ $$\ce{MgI2 + 2I <=> Mg(I3)2}$$ The amount of products of this reaction though will not be very appreciable, but yes a very literal sense magnesium chloride and iodine will react.
• $\ce{<=>}$ gives you the correct arrows: $\ce{<=>}$ – orthocresol Dec 29 '15 at 2:47
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# Thread: Math Logic Finding the Weakest Precondition Calculus?
1. ## Math Logic Finding the Weakest Precondition Calculus?
All 4 parts to this question concern the assertion P = {x=2^K & N>K}
and one of 4 different assignment statements. You are to supply the
missing assertions Q1, Q2, Q3, and Q4. Note that Q1 and Q2 should
be the weakest preconditions for P with respect to the given statement;
Q3 and Q4 should be postconditions with respect to the given statement
whose weakest precondition would be P.
Hint: The assignment axiom does not apply directly in the forward
direction as in (c) and in (d). However, you can guess at Q3 and Q4
by "looking" for x/4 and for K+1 in P and then substituting in reverse.
You should then check your answer; i.e., check whether your
post conditions do indeed have P as their weakest precondition using
the assignment axiom directly.
a)
{ } = Q1
y <-- y/N
{x=2^K & N>K}
(b)
{ } = Q2
x <-- x+x
{x=2^K & N>K}
(c)
{x=2^K & N>K}
x <-- x/4
{ } = Q3
}
(d)
{x=2^K & N>K}
y <-- K+1
{ } = Q4
a)Attempt
For this I see that
(y<-- y/N) means Any y is replaced by y/N
y/N is equivalent to N <--- y
so y <---- N <---- y
So the precondition is
Q1= {x=2^K & y>K}
I think it could be Q1= {x=2^K & N>K}
b)Attempt:
x <-- x+x means replace x with x+x
So Q2={ x+x =2^k & N>K}
I don't think this is the weakest precondition.
can this be simplified?
c)Attempt:
Q3={x/4=2^K & N>K}
Im not too sure...
d)Attempt:
Post Condition
Q4={x=2^(K+1-y+k)-y & N>K+1-y+K}
When I replace K+1 with Y
I get x=2^y-y+k & N>y-y+K
this is exactly the same as the precondition
Can someone check my answers and help me explain what the ones I get wrong?
2. The questions you posted appear very course/textbook/instructor specific.
That is to say, unless some person is in that same class I doubt the person would any idea what is being asked.
3. This concerns Hoare Logic, a pretty standard way of reasoning about imperative programs.
a)
{ } = Q1
y <-- y/N
{x=2^K & N>K}...
a)Attempt
For this I see that
(y<-- y/N) means Any y is replaced by y/N
y/N is equivalent to N <--- y
so y <---- N <---- y
So the precondition is
Q1= {x=2^K & y>K}
I think it could be Q1= {x=2^K & N>K}
The general inference rule (axiom) is {P[t/x]} x <-- t {P} where P[t/x] denotes replacing x by t in P. Since y does not occur in the postcondition, Q1 is the same as the postcondition, i.e., x=2^K & N>K. Indeed, if this holds before the assignment, then whatever you assign to y, the same formula will still be true.
(b)
{ } = Q2
x <-- x+x
{x=2^K & N>K}...
b)Attempt:
x <-- x+x means replace x with x+x
So Q2={ x+x =2^k & N>K}
I don't think this is the weakest precondition.
can this be simplified?
You are correct, Q2 is (x=2^K & N>K)[x + x/x] (recall that [...] denotes a substitution), which is x+x =2^k & N>K. You can simplify it as 2x =2^k & N>K or x =2^{k-1} & N>K; this is not essential.
(c)
{x=2^K & N>K}
x <-- x/4
{ } = Q3
Finding the postcondition is trickier than applying a substitution; one has to "un-apply" a substitution, so that Q3[(x/4) / x] is x=2^K & N>K. Well, 4x=2^K & N>K as Q3 seems to work.
d)Attempt:
Post Condition
Q4={x=2^(K+1-y+k)-y & N>K+1-y+K}
When I replace K+1 with Y
I get x=2^y-y+k & N>y-y+K
this is exactly the same as the precondition
To check, you need to replace y with K + 1 in the postcondition, not the other way around. Here, since y does not occur in the precondition, Q4 is the same as that precondition.
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# The length of a rectangle is 7 feet larger than the width. The perimeter of the rectangle is 26 ft. How do you write an equation to represent the perimeter in terms of its width (w). What is the length?
Nov 4, 2016
An equation to represent the perimeter in terms of its width is: $p = 4 w + 14$ and the length of the rectangle is $10$ ft.
#### Explanation:
Let the width of the rectangle be $w$.
Let the length of the rectangle be $l$.
If the length ($l$) is 7 feet longer than the width, then the length can be written in terms of the width as:
$l = w + 7$
The formula for perimeter of a rectangle is:
$p = 2 l + 2 w$ where $p$ is the perimeter, $l$ is the length and $w$ is the width.
Substituting $w + 7$ for $l$ gives an equation to represent the perimeter in terms of its width:
$p = 2 \left(w + 7\right) + 2 w$
$p = 2 w + 14 + 2 w$
$p = 4 w + 14$
Substituting $26$ for $p$ allows us to solve for $w$.
$26 = 4 w + 14$
$26 - 14 = 4 w + 14 - 14$
$12 = 4 w$
$\frac{12}{4} = 4 \frac{w}{4}$
$w = 3$
Sustituting $3$ for $w$ in the equation above, $l = w + 7$ allows us to determine the length:
$l = 3 + 7$
$l = 10$
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We know that we can focus sunlight to reach high temperatures and use it for example to burn paper.Given a large enough lens can we theoretically do the same with starlight? What about light from planets? There are alternative ways to answer this question using both thermodynamics and optics.
Akshay B.
First,lets use the Second Law of Thermodynamics.Imagine the lens as a Carnot Engine transferring heat from the Sun(source at temperature 5800K) to paper(sink at room temperature).Keeping the practical limitations aside the efficiency of this process can be easily calculated using Carnot's formula.What is important here is that the temperature of the sink cannot be higher than the source,so the maximum temperature you can reach by a lens is the temperature of the sun. This implies you can do the same with starlight but not with the light from planets as they are at a much lower temperature,except Venus. Another way to look at this is to observe that we are looking at extended objects making extended images. If a planet were making a point image it would have infinite temperature (remember from the Stefan Boltzmann law for a finite power area $$A\rightarrow 0$$ implies $$T\rightarrow \infty$$).
Algebra
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Given three complex numbers $$z_1,z_2,z_3$$ on the unit circle.What is the orthocenter of the triangle formed by the three numbers?
Akshay B.
$$z_1+z_2+z_3$$. This calculation can take a lot of time if we try to calculate the meeting points of the altitude using conventional methods. However a quick way is to see that we know that the circumcenter is at the origin and the centrois is $$\frac{z_1+z_2+z_3}{3}$$.The centroid divides the line connecting the orthocenter and the circumcenter in the ratio 2:1. This immediately gives the desired result.
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Estimate the number of photons in the observable universe. The whole universe is isotropically filled with photons from a very early time.Because of the expansion of the universe these photons have red-shifted and are now in the microwave region.This radiation behaves like the emission from a perfect blackbody. Given the temperature of this cosmic microwave background (CMB) radiation 2.73K. Take the mean frequency of the CMB to be around 150 GHz Remember the stellar photons contribute orders of magnitude less to the radiation density compared to the microwave background.
Akshay B.
We need to calculate the energy density of the microwave radiation.Assume photons travelling from an area A moving with speed c .We can picture a cylinder with area of cross section A and length c.The energy in this cylinder is nothing but the energy per second by the area A at temperature T=2.73 K. Now we can use Stefan Boltzmann law to calculate the power as $$\sigma A T^4$$.The energy density hence is this power divided by the volume of the cylinder Ac. using the mean frequency the number density of photons n can be calculated by using $$nh\nu=\frac{\sigma A T^4}{Ac}$$. The observable universe is a sphere of radius 39 billion light years.A quick calculation gives an estimate of the total photons in the universe.
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