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# Why $\{(a,b) | a\leq b\}$ is not symmetric?
I want to know why $$\{(a,b) \mid a\leq b\}$$ is not symmetric, when $$\{(a,b) \mid a\leq b\} =\{(a,b)| a<b \text{ or }a=b\}$$ So if a=b that means aRb and bRa so it is symmetric, right ?
Thanks all
• Maybe you're confusing symmetric with reflexive? A relation $R$ is symmetric if $x R y \rightarrow y R x$ (for all $x, y$) - that is not the case for $\leq$. A relation is reflexive if $x R x$ (for all $x$) - that is true for $\leq$. – Magdiragdag Dec 13 '13 at 13:16
• Since $1R2$, symmetric would mean $2R1$. – Thomas Andrews Dec 13 '13 at 13:20
• @madiragdag thank you, I'm confused about that – user32104 Dec 13 '13 at 13:24
A relation is symmetric when for any values $a$ and $b$, if $a$ is related to $b$, then also $b$ is related to $a$. But in your case, for example, $3 \le 4$, but $4\not\le 3$.
• The statement $a<=b$ if and only if $b<=a$ must be true for every pair of values $a$ and $b$. Showing that it's true for the pair $(1,1)$ does not help. – rogerl Dec 13 '13 at 13:22
HINT $(4, 7)$ means $(7, 4)$ should both be in $R$ if $R$ was symmetric.
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Documentation
Getting Started
Section Builder
Operations
Built-Up and Composite Sections
Custom Shapes
Member Design
General
Member Design Modules
Code Verification
RC Design
General
Code Verification
# Catenary Cables
Cables are a special type of member that can only be analyzed via a Non-Linear Analysis. Cables are tension-only and they often exhibit large displacements and non-linear behavior. They are never in compression and have zero flexural, torsional or shear capacity.
### Creating Cables
Cables are created in the same way as Members. Open the Advanced UI of the Member properties and simply select "Cable" as the Member Type as shown in the image below. It is important to note that when creating cables, you are effectively drawing the chord of the cable (i.e. a straight line joining Node A and Node B) rather than the catenary cable itself. The software will work out the catenary effects and sagging automatically, so there is NO need to create a series of smaller cable elements to represent the actual cable shape.
### Specifying Pre-Tension and Sag
The software allows you to enter a cable length. If you enter a length that is less than the chord length (i.e. the distance between Node A and Node B of the cable member) then you can pre-stress the cable. Likewise if you enter a length that is larger than the chord length, the cable will be unstrained and sagging initially.
NOTE: Entering this field is OPTIONAL. If you do not specify a cable length then the initial chord length will be taken as the cable length (i.e. it is not pre-stressed and not sagging).
### Pretension Force
It is possible to enter a cable length that approximates a pretension force. We can do this using the formula for axial displacement:
[math] \begin{align} \Delta L &= \dfrac{P \overline{L}}{AE} \\ L - \overline{L} &= \dfrac{P \overline{L}}{AE} \\ P &= \dfrac{L - \overline{L}} { \overline{L}} AE \\ \end{align} [math] [math] \textsf{where: } \\ \begin{align} P &= \textsf{Pretension Force (i.e. the force in the cable)} \\ L &= \textsf{Original Length of the cable before tensioning (i.e. the chord length)} \\ \overline{L} &= \textsf{Length of the cable after pre-tensioning (entered into the field in the software)} \\ A &= \textsf{Area of the cross-section of the cable} \\ E &= \textsf{Young's Modulus of the Material used in the cable} \\ \end{align} [math]
After the cable analysis runs the final axial force in the cable may be different to the force yielded by the formula above. This is because the force above is the "initial" force in the cable before the analysis takes place (i.e. the pretension). If you want to aim for a specific force in the cable at the end of the analysis then you may need to perform a series of iterations to get the intial pre-tension force correct.
Cables are special types of members that must be loaded according to the rules below:
• Any point loads must be applied to the end nodes of a cable (i.e. they cannot be member loads along the cable). If you wish to apply a point load along the cable, split the cable into two cables and then apply the load at the junction of the two cables.
• Distributed Loads applied to a cable MUST be uniform (i.e. the start and ending magnitudes of the DL must be equal).
• All cables must be loaded with at least one uniform distributed load (UDL) or self-weight. If neither of these are found, the solver will apply a small "Artificial UDL".
• At the moment cables can only take Distributed Loads in the global Y-axis. This means that if Self-Weight is applied to the structure, it must be in the Y-axis. If you would like to see DLs in any direction please let [email protected] know.
• Cables must be solved by Non-Linear Analysis. If any other method is used, the solver will automatically be switched to Non-Linear.
• Cables are purely axial (in tension). All cables are assumed to be pinned, so entering end-fixities (Node A and Node B Fixity) is not required.
• Convergence of your structure when cables are used can be difficult due to the non-linear behavior.
• Cables cannot be accurately modelled with tension-only truss members. Catenary cable elements like those used in accordance with this page are the most accurate way to model cables.
## Having issues solving?
Cables can be tricky to solve, so SkyCiv created a guide on some easy tips to solve a cable which features 5 things you can try to get your model to solve. In short, these include:
1. Check diameter size
2. Check Cables are Not Over/Under Loaded
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# Dot product
### Dot product
#### Lessons
Notes:
Dot Product
Let $u=$ and $v=$. Then the dot product of these two vectors will be:
$u \cdot v = ad + be + df$
Dot Product Property
If the dot product of two vectors $u$ and $v$ gives 0, then the vectors are perpendicular. In other words,
$u \cdot v=0 \to$ perpendicular vectors
Scalar and Vector Projection
Suppose we have two vectors $a$ and $b$. Suppose they create an angle $\theta$ such that we get the following picture:
$|v| = \frac{a \cdot b } {|b|}$
To find the vector projection $a$ onto $b$ (which is v), we use the formula:
$v = \frac{a \cdot b}{b \cdot b}b$
Let $u, v, w$ be vectors and $c$ be a scalar. Then the properties of dot products are:
1. $u \cdot u = |u|^2$
2. $u \cdot v = v \cdot u$
3. $u \cdot (v+w) = u \cdot v + u \cdot w$
4. $(cu) \cdot v = u \cdot (cv) = c(u \cdot v)$
• Introduction
Dot Product Overview:
a)
Dot Product and its Special Property
• Multiplying the corresponding entries, and adding
• Dot product = 0 $\to$ vectors are perpendicular
b)
Application to Dot Product
• What is scalar projection? Vector projection
• Formula for scalar projection: $|v| = \frac{a \cdot b}{|b|}$
• Formula for vector projection $v = \frac{a \cdot b}{b \cdot b}b$
c)
More Properties of Dot Product
• Order
• Length
• Distribution
• Scalar
• 1.
Using the Dot Product
Find the dot product of $u = <-1, -2 , 7>$ and $v = <-2,1,-2>$.
• 2.
Find the dot product of $u= <1, -5, -3>$ and $v= <-1, 1, 2>$.
• 3.
Using the Dot Product Property
Suppose two vectors $u = < a, 4, -3>$and $v=<1, 2, 3>$ are perpendicular. Find $a$.
• 4.
Finding Scalar and Vector Projections
Find the scalar and vector projection of $\vec{BA}$ onto $\vec{CA}$ if $A=(1, 0, 2)$, $B=(3, -2, 1)$ and $C=(-4, 1, 5)$.
• 5.
Verifying Properties of Dot Product
Use the two vectors $u=<3, 1, 5>$ and $v=<1,4,-6>$ to show that:
$u \cdot u = |u|^2$
• 6.
Use the 3 vectors $u=<3, 1, 5>$, $v=<1,4,-6>$, and $w=<1, 0, 3>$ to show that:
$u \cdot (v+w)=u \cdot v+ u \cdot w$
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# std::experimental::filesystem::absolute, std::experimental::filesystem::system_complete
< cpp | experimental | fs
Defined in header path absolute( const path& p, const path& base = current_path() ); (1) (filesystem TS) path system_complete(const path& p); path system_complete(const path& p, error_code& ec); (2) (filesystem TS)
1) Returns absolute path of p relative to base according to the following rules:
• If p has both root name and root directory (e.g. "C:\users", then the path is returned, unmodified.
• If p has a root name not followed by a root directory (e.g. "C:text.txt"), then base is inserted between p's root name and the remainder of p. Formally, p.root_name() / fs::absolute(base).root_directory() / fs::absolute(base).relative_path() / p.relative_path() is returned,
• If p has no root name, but has a root directory (e.g. "/var/tmp/file.txt" on a POSIX system or "\users\ABC\Document.doc" on Windows, then the root name of base, if it has one, is prepended to p (on a POSIX system, p is not modified, on a Windows system, "\users\ABC\Document.doc" becomes "C:\users\ABC\Document.doc". Formally, fs::absolute(base).root_name() / p is returned.
• If p has no root name and no root directory (e.g. "../file.txt", then the entire base is prepended to p. Formally, absolute(base) / p is returned.
2) Obtains the absolute path that identifies the file that the OS file opening API would access given the pathname p. On POSIX systems, this is equivalent to (1) with the default base (fs::current_path()). On Windows systems, each logical drive has its own current working directory, and so if p is not already absolute and has a root name component (e.g. "E:filename.txt", that drive's current working directory is used, which may have been set by an earlier executed program.
## Contents
### Parameters
p - path to convert to absolute form base - path (not necessarily absolute) to serve as the starting location ec - out-parameter for error reporting in the non-throwing overload
### Return value
Returns an absolute (although not necessarily canonical) path formed by combining p and base as described above.
### Exceptions
The overload that does not take a error_code& parameter throws filesystem_error on underlying OS API errors, constructed with p as the first argument, base as the second argument, and the OS error code as the error code argument. std::bad_alloc may be thrown if memory allocation fails. The overload taking a error_code& parameter sets it to the OS API error code if an OS API call fails, and executes ec.clear() if no errors occur. This overload has
noexcept specification:
noexcept
### Notes
On systems that support root names (e.g. Windows), the result of calling absolute on a relative path that has a root name (e.g. "D:file.txt" when the root name of base is different will usually result in a non-existent path.
### Example
#include <iostream>
#include <filesystem>
namespace fs = std::experimental::filesystem;
int main()
{
fs::path p = "C:cl.exe";
std::cout << "Current path is " << fs::current_path() << '\n'
<< "Absolute path for " << p << " is " << fs::absolute(p) << '\n'
<< "System complete path for " << p << " is " << fs::system_complete(p) << '\n';
}
Possible output:
Current path is "D:/local/ConsoleApplication1"
Absolute path for "C:cl.exe" is "C:/local/ConsoleApplication1/cl.exe"
System complete path for "C:cl.exe" is "C:\Program Files (x86)\Microsoft Visual Studio 12.0\VC\bin\cl.exe"
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# How do you find the equation that represents the image of circle (x- 5)^2 + (y + 12)^2 = 169 after a translation 2 units right and 3 units down?
Jan 30, 2016
Adjust to standard form to see what the original centre is, translate the centre, then plug back into the equation to find:
${\left(x - 7\right)}^{2} + {\left(y - \left(- 15\right)\right)}^{2} = {13}^{2}$
or:
${\left(x - 7\right)}^{2} + {\left(y + 15\right)}^{2} = 169$
#### Explanation:
The standard equation of a circle with centre $\left(h , k\right)$ and radius $r$ is:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
With minor changes our starting equation is:
${\left(x - 5\right)}^{2} + {\left(y - \left(- 12\right)\right)}^{2} = {13}^{2}$
That is: It is a circle with centre $\left(5 , - 12\right)$ and radius $13$.
If translated $2$ units right and $3$ units down then the centre will be $\left(5 + 2 , - 12 - 3\right) = \left(7 , - 15\right)$ and the equation becomes:
${\left(x - 7\right)}^{2} + {\left(y - \left(- 15\right)\right)}^{2} = {13}^{2}$
or putting it back in similar form to the starting equation:
${\left(x - 7\right)}^{2} + {\left(y + 15\right)}^{2} = 169$
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# How do you solve and write the following in interval notation: x>= -4?
This inequality is already solved for $x$.
In interval notation: $\left[- 4 , + \infty\right)$
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# Shengshou Gigaminx
#### JustinTimeCuber
##### Member
Seems interesting, although I'm definitely not a -minx person; I average sup 3:00 on mega lol
#### hamfaceman
##### Member
If I ever get into mega, this might be fun to get. But as of now, I'll give it a miss. Hopefully it's good!
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Select Page
# I.S.I. & C.M.I. Entrance 2019 Test Series
From advanced Number Theory to beautiful geometry and combinatorics. Prepare for some seriously interesting mathematics!
## Our Classes for I.S.I. & C.M.I. Entrance 2019
### I.S.I. B.Stat – B.Math – One full length Model Test
Take one full-length I.S.I. Entrance Model Test (B.Stat – B.Math)
1. Attend the model test online or at Calcutta Offline Center (near Tollygunge)
2. One full-length objective test + One full-length subjective test
3. One online live discussion session on the paper
4. Mock Interview access (if the student qualifies the actual I.S.I. Entrance written test in May)
Access Fee: ₹ 750
### I.S.I. & C.M.I. Entrance Test Series
Take five full-length I.S.I. Entrance Model Test (B.Stat – B.Math) and two full-length CMI Entrance Model Tests
1. Attend the model tests online or at Calcutta Offline Center (near Tollygunge)
2. Full-length objective test + Full-length subjective test
3. Online live discussion session on paper
4. Mock Interview access (if the student qualifies the actual I.S.I. Entrance written test in May)
Access Fee: ₹ 3900
Alternatively you may join Online Classroom Program
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## Problem Solving Marathon Week 2
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## 2016 ISI Objective Solution Problem 1
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## I.S.I Entrance Solution – locus of a moving point
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## PreRMO and I.S.I. Entrance Open Seminar
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Find all pairs $$(x,y)$$ with $$x,y$$ real, satisfying the equations $$\sin\bigg(\frac{x+y}{2}\bigg)=0~,~\vert x\vert+\vert y\vert=1$$ Discussion: https://youtu.be/7Zx5n3nuGmo Back to Problems
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ISI Entrance Paper 2018 - from Indian Statistical Institute's B.Stat Entrance Also see: ISI and CMI Entrance Course at Cheenta Find all pairs ( (x,y) ) with (x,y) real, satisfying the equations sinbigg(frac{x+y}{2}bigg)=0~,~vert xvert+vert...
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# differentiating a polynomial $k$ times.
If we have the polynomial $$P(x-a)=f(a)+f'(a)(x-a),$$ I can't get how does differentiating the polynomial $k$ times we obtain $$P_k(x-a)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\ldots+\frac{f^k(a)}{k!}(x-a)^k$$
kindly help...
• The second is a Taylor polynomial, which suggests that something else should be differentiated. Could you check it? – Przemysław Scherwentke Nov 8 '14 at 7:03
• @patang I think it's the $k^{th}$ degree taylor polynomial of $f$ at $a$.. – spectraa Nov 8 '14 at 7:06
• It is not differentiating the polynomial $k$ times. It is just the Taylor expansion truncated at the $k^{th}$ term. – Claude Leibovici Nov 8 '14 at 7:11
Here is another way to approach the problem. Consider, instead, the polynomial
$P(x-a)=f(a)+f^{\prime}(a)(x-a)+\int^{x}_{a}(x-t)f^{\prime\prime}(t) dt$.
Then on the latter term: $\int^{x}_{a}(x-t)f^{\prime\prime}(t) dt$ (using Taylor's theorem) apply integration by parts several times ($k$-times).
Namely, let $u=f^{\prime\prime}(t)$, then $du=f^{\prime\prime\prime}(t)$, and $dv=(x-t)dt$ so that $v=-\frac{1}{2}(x-t)^{2}$, and use by parts:
$uv-\int vdu.$
Thusly, continue with this procedure to obtain: $P(x-a)=f(a)+f^{\prime}(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-t)^{2}+\cdots+\frac{f^{k}(a)}{k!}(x-t)^{k}.$
• thanks for the answer ...but why did you change the original polynomial to $P(x-a)=f(a)+f^{\prime}(a)(x-a)+\int^{x}_{a}(x-t)f^{\prime\prime}(t) dt$ ... – patang Nov 8 '14 at 9:03
• Thank you. I did it so that you can see that you are not actually differentiating the polynomial. One only needs Taylor's theorem which is just integration by parts multiple times. – Rene Cabrera Nov 8 '14 at 9:05
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# What a loss?
Algebra Level 2
A woman purchased a pair of shoes at $20 whose cost price for shopkeeper(SK-1) was$15. She gave him a $100 note. SK-1 was not having change to return. He took change of$100 from the fellow shopkeeper(SK-2). He came back and returned $80 to woman. In the evening, the SK-2 comes back and says " Note which you gave me is fake. Return me my money". SK-1 returns him back$100. How much is the total loss for SP-1 in the whole transaction?
×
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# Precision and Recall Evaluation Metrics in Machine Learning
Precision and Recall are two best known and most misunderstood evaluation metrics in machine learning. In this article we are going to discuss the simplest way in which we can understand them with an example.
Consider that we were playing a simple game online. The rules of the game are pretty simple, you will have to shoot the space ships of the opposition constantly, until you see a survivor space ship coming toward you.
Now let’s consider that your model was playing the game from a long time and have collected a lot of related data.
Data points that we collected belonged to these 3 categories.
Total time you Stopped shooting: x
Total time you Stopped shooting with survivor ship ahead you: y
Total time you Stopped shooting with Opposition Space ship ahead you: z
Now, we created a Venn diagram to explain the occurrences.
## Precision
Precision is the value which gives the value of number of times you took the correct action of all the times you took an action.
So, precision is given by the intersection divided by the blue part of the Venn diagram. i.e.
$precision = \frac{Correctly\ stopped\ shooting}{Total\ Number\ of\ Shooting\ stops}$
## Recall
Recall is the value which gives the value of all the times you were supposed to take an action of all the times you took an action.
So, Recall is given by the intersection divided by the orange part of the Venn diagram.
$recall = \frac{Correctly\ stopped\ shooting}{Total\ Number\ of\ times\ survivor\ ships\ in\ front}$
More detailed discussion of precision and recall are done in the evaluation metric post.
### How to choose a good evaluation metric for your Machine learning model
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February 20, 2020 18 mins read
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# what does delta y mean in physics
Lv 7. A complete explanation follows. Hence it is even used in calculus as dx/dt etc ... 1 ; View Full Answer It simply means a change in . In some cases, this means a difference between two values, such as two points on a line. It only takes a minute to sign up. Motion In A Straight Line. About the Common uses in Physics . Upper-case delta (Δ) often means "change" or "the change in" in mathematics. Depending on the situation, delta-v can be either a spatial vector (Δv) or scalar (Δv).In either case it is equal to the acceleration (vector or scalar) integrated over time: = − = ∫ (vector version) Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Home Questions Tags Users Unanswered What does “Orbit of X by Y” mean? 0. Delta particle definition, any of a family of baryon resonances having strangeness 0, isotopic spin 3/2, and either a single or double positive electric charge, a single … Andrew Smith. Relevance. Delta-T or delta temperature, seen in hardware reviews and cooling performance tests always stands for the value that you get when you subtract the ambient temperature from the measured temperature. See more. Delta definition, the fourth letter of the Greek alphabet (Δ, δ). User account menu. How to use delta in a sentence. Let us consider how we might characterize the scatter around the mean value. For example in this Newton Series: $\displaystyle f(x)=\sum _{k=0}^{\infty }{\frac {\Delta ^{k}[f](a)}{k! Update: Also this one: avg. He's practically asking what does it mean that local supersymmetry variations must vanish in that preserved supersmmetry. Physics reference Symbols: 28 Rp 1 PhysRef: Some common symbols used for variables in physics. Donate Login Sign up. Close. You are probably meaning the Greek letter $Δ$ (Delta). What does$\Delta ^{k}$mean? Here is the equation: velocity = Δposition/time = displacement/time. delta x means change in the variable on the x-axis. User account menu. Answer Save. 5 years ago. Note. Is it the opposite of the Delta symbol (a triangle) does delta mean change, or something to do with change ? acceleration = Δvelocity/time = vf-vi/t. Search for courses, skills, and videos. A mathematical symbol is a figure or a combination of figures that is used to represent a mathematical object, an action on mathematical objects, a relation between mathematical objects, or for structuring the other symbols that occur in a formula.As formulas are entierely constitued with symbols of various types, many symbols are needed for expressing all mathematics. what does Δ(delta) mean? Delta refers to change in mathematical calculations. Aufgrund der Form des Großbuchstabens Δ wurde seit Herodot (Historien 2,13 und öfter) der zwischen den Nilarmen liegende Teil Unterägyptens als Delta bezeichnet, so wie im Deutschen Flussdelta allgemein. Note that bold text indicates that the quantity is a vector. 0. Share 0 ∆ delta means change. It's already been answered pretty good. Answer Save. But it isn't often introduced that way these days, at least not in High School physics. Log in to reply to the answers Post; Anonymous. Your bank will then be notified and the sum of the bill is going to be taken out of your account. Relevance. somt. ta (dĕl′tə) An area of the south-central United States extending on either side of the Mississippi River from Memphis, Tennessee, to Vicksburg, Mississippi. Is it the opposite of the Delta symbol (a triangle) does delta mean change, or something to do with change ? Courses. acceleration = Δvelocity/time = vf-vi/t. Greek letters are used in mathematics, science, engineering, and other areas where mathematical notation is used as symbols for constants, special functions, and also conventionally for variables representing certain quantities. If you would like to read more on the subject of Fourier Transforms, then the Better Explained article is a great start. Sign up to join this community. Posted by. Press question mark to learn the rest of the keyboard shortcuts. Search. delta t is normally change or difference in time. Science Physics library Impacts and linear momentum Momentum and Impulse. Actually I've never seen that use of it. Delta x =-6.69 and xi=3.09 what is xf. 2. Delta definition is - the 4th letter of the Greek alphabet. the change in y divided by the change in x, for any two points on the line. In words, the time derivative of momentum equals the space derivative of energy. I know the Uncertainty Principle is an inequality and not an identity. For the best answers, search on this site https://shorturl.im/avV1I. Courses. 0 1 0. tensile or compression) Permittivity EMF zeta (no common use) If you're seeing this message, it means we're having trouble loading external resources on our website. delta definition: 1. an area of low, flat land, sometimes shaped like a triangle, where a river divides into several…. Delta looks like this: Jump to the summary. Main content. Please help improve this article by adding citations to reliable sources. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. After all, the uppercase delta (Δ) is commonly used in mathematics and physics where it simply marks the difference or change in something. For example, Δx=x 2 -x 1 or Δt=t 2 -t 1 . What does the multiplication between two units mean? In your example, the equation is a relationship between partials and it needs to be operated upon to get something you can use. As Giuseppe Negro said in a comment,$\delta$is never used in mathematics in $$\frac{dy}{dx}.$$ (I am a physics ignoramus, so I do not know whether it is used in that context in physics, or what it might mean … 7 … Latin characters. But just to clarify: Most symbols have different meanings in different contexts. 0. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Thanks What does Delta mean in this physics equation? What exactly does$\ll$mean? In other cases, it … It’s like with the delta function - written alone it doesn’t have any meaning, but there are clear and non-ambiguous rules to convert any expression with to an expression which even mathematicians understand (i.e. 5 Answers. 4 years ago. Thanks JavaScript is disabled. Therefore, at its peak the projectile has a 5 How to show that exp is a diffeomorphism between symmetric reals and positive definite matrices? This list is by no means complete. How to use delta in a sentence. Delta waves have a frequency from one to four hertz and are measured using an electroencephalogram (EEG). Greek letters used in mathematics, science, and engineering, From Wikibooks, open books for an open world, https://en.wikibooks.org/w/index.php?title=Physics_Study_Guide/Greek_alphabet&oldid=3550502. I looked everywhere and everyone says that Delta is a change, and they don't explain it, so can you also thoroughly explain what Delta means. See more. Learn more. Sign up to join this community. Zudem bezeichnet der Komödienautor Aristophanes in der Lysistrata Vers 151 mit Delta die weibliche Scham. If you're seeing this message, it means we're having trouble loading external resources on our website. Even that tiny bit adds up. What does negative potential energy mean? In Physics, "x" is an unknown or is the horizontal axis on a graph. Delta refers to change in mathematical calculations. While these are indeed common usages, it should be pointed out that there are many other usages and that other letters are used for the same purpose. Main content. What does "exist" mean to a physicist? In this discussion, as it relates to kinematics and the study of physics in general, the Greek letter delta means 'change in.' 1 decade ago. Lv 7. Lv 7. 7 … What does Delta mean in this physics equation? Delta's most common meaning is that of difference or change in something. Search for courses, skills, and videos. I looked everywhere and everyone says that Delta is a change, and they don't explain it, so can you also thoroughly explain what Delta means. In some cases, this means a difference between two values, such as two points on a line. Class-11-science » Physics. delta x is normally change or difference in position. In other cases, it refers to the rate of change, such as in a derivative. Sign up to join this community. It can be change in temperature , size , enthalpy etc . I think it leads to the differential equation$\partial p/\partial t = \partial E/\partial x$. Why Does The Triangle Mean Change In Physics - Delta Greek Letter Svg Clipart is high quality 602*830 transparent png stocked by PikPng. What is Delta in Physics and How it Connects With Reddit Posted on 10 Tháng Ba, 2020 by Long Nam A large and influential subreddit devoted to physics subjects, known as r/Physics, has received the invitation to post to a brand new beta version of a textbook or coursework critique material that was published by the Council on Economic Activity (CEA). It only takes a minute to sign up. Here is the equation: velocity = Δposition/time = displacement/time. Engineers find a way to control chemical catalysts with sculpted light, Precise measurements of cluster formation in outer neutron 'skin' of a range of tin isotopes, Optical computing at sub-picosecond speeds, http://www.dpi.state.nc.us/docs/accountability/testing/eoc/Physics/physicsreferencetable.pdf, http://en.wikibooks.org/wiki/Physics_Study_Guide/Greek_alphabet. r/Physics. So, F= delta P/ Delta t means that the force is equal to the change in momentum divided by the change in time. When you divide ∆y by ∆x, you get the slope of the graph between the points, which tells you how fast x and y are changing wth respect to each other. I'm not saying it's wrong, but it's unusual. The delta between the x values of these points – ∆ x – is given by (x 2 - x 1), and ∆ y for this pair of points is (y 2 - y 1). Donate Login Sign up. Courses. We now know how to define the mean value of the general variable, $$u$$. Papers from … Press J to jump to the feed. Delta expresses the amount of price change a derivative will see based on the price of the underlying security (e.g., stock). Delta definition, the fourth letter of the Greek alphabet (Δ, δ). Answer Save. But in elementary math texts, Δ x means the amount of change in the variable x. Delta definition is - the 4th letter of the Greek alphabet. Click here to get an answer to your question ️ what does Delta mean in Physics r/Physics: The aim of /r/Physics is to build a subreddit frequented by physicists, scientists, and those with a passion for physics. We could investigate the deviation of $$u$$ from its mean value, $$\langle u\rangle$$, which is denoted \[{\mit\Delta} u \equiv u- … Favourite answer-x axis n y axis.... -also refer to unknown.. 0 1 0. Wexternal=delta PE, Wgravity=-delta PE. What does upside down “v” ($\wedge$) mean in this equation? How to use physic in a sentence. When you divide ∆y by ∆x, you get the slope of the graph between the points, which tells you how fast x and y are changing wth respect to each other. Discriminant is the second most common meaning of the uppercase delta. All three are ways of conceptualizing the idea of a difference of values. Search. delta definition: 1. an area of low, flat land, sometimes shaped like a triangle, where a river divides into several…. It is a common trap to associate a symbol exclusively with some particular meaning, rather than learning and understanding the physics and relations behind it. But I think it is possible that they are equal or only differ by a constant. How Delta Waves Are Measured . Now, dv/dt is what we call the instantaneous rate of change in "v", that is the limit of "delta v"/"delta t" as we let the time interval "delta … What does the small delta in front of the W and t mean in this equation? If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. 0. log in sign up. What does the small delta in front of the W and t mean in this equation? Donate Login Sign up. 0 0. ehabyoussef1968. It’s well worth explaining in a little more detail what is meant. Finding x from y and y from x To find initial velocity... Finding Initial Velocity Vertically Conclusion Initial velocity can also be found by shooting a projectile vertically. When we use "delta" rather than "d", we mean that, say "delta v"/"delta t" is the average change in "v" over the non-zero time interval "delta t". ∆ delta is often refered to as 'd' when the change is infinitesimally small . The reason is quite simple: there are only so many symbols in the Greek and Latin alphabets, and scientists and mathematicians generally do not use symbols from other languages. 0 0. Search for courses, skills, and videos. r/Physics: The aim of /r/Physics is to build a subreddit frequented by physicists, scientists, and those with a passion for physics. I am familiar that this symbol means much less than. For x(at event 1) = 2 and x(at event 2) = 4 : delta x = 4-2 = 2, sometimes it is also used to refer to uncertainty in the variable x. What book would provide me with a quick reference to the use of greek letters in Physics? 1 0. If you're seeing this message, it means we're having trouble loading external resources on our website. "y" is a vertical axis on a graph. what does the lower case letter d (found in physics formulas) mean ? Unsourced material may be challenged and removed November 2019) (Learn how and when to remove this template message) This is a list of common physical constants and variables, and their notations. Posted by. Creative Commons Attribution-ShareAlike License. For example, if the variable "x" stands for the movement of an object, then "Δx" means "the change in movement." r/Physics. In Biology "x" is female and "y" is male. Ying Ying. It is a common trap to … Pi is one of the few symbols that has a consistent meaning. If it was ##\Delta r = \Delta x \cos x + \Delta y \sin y## would that be clear? delta Δ="Change in" δ="Infinitesimal change in (), also used to denote the Dirac delta function (reference needed)" epsilon Emissivity Strain (Direct e.g. Press question mark to learn the rest of the keyboard shortcuts. Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Home Questions Tags Users Unanswered What does the line above a letter represent? For example, the slope of a line is Δ y/ Δ x, i.e. Physic definition is - the art or practice of healing disease. Choosing Good What Does Delta T Mean in Physics There is additionally a big problem when you’ve installed them. Anonymous. Download it free and share it with more people. F. Frederick Skitty. In der griechischen Zahlschrift repräsentiert der Kleinbuchstabe δ die Zahl 4 und zusammen mit tiefgestelltem Nummernzeichen als δ die Zahl 4000. He said that for an object to have its momentum altered (mass time velocity) then a force has to be applied over a certain length of time. 1 decade ago. In these contexts, the capital letters and the small letters represent distinct and unrelated entities. Main content. Delta / ˈ d ɛ l t ə / (uppercase Δ, lowercase δ or ; Greek: δέλτα délta, ) is the fourth letter of the Greek alphabet.In the system of Greek numerals it has a value of 4. Usually, you will hear or see it as delta y , delta t , delta x , etc. Lv 5. What Does Delta T Mean in Physics: the Ultimate Convenience! This article does not cite any sources. Active … The Greek letter delta, when used this way, looks like this: It means 'change in'. what does x&y mean in physics? On my book, it says the change in potential energy associated with a particular force is equal to the negative of the work done by that force if the object is moved from one point to a second point. While these are indeed common usages, it should be pointed out that there are many other usages and that other letters are used for the same purpose. Does$ \Delta p \Delta x = \Delta E\Delta t$? Relevance. Share with your friends. Learn more. Andrew Smith. 1 decade ago. Ask Question Asked today. Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Home Questions Tags Users Unanswered What does the line above a letter represent? ...but what exactly does "much less than" mean? The delta between the x values of these points – ∆ x – is given by (x 2 - x 1), and ∆ y for this pair of points is (y 2 - y 1). : 1. an area of low, flat land, sometimes shaped like a triangle where! Between partials and it needs to be taken out of your account not High. It leads to the summary then be notified and the sum of the keyboard shortcuts you use! In to reply to the change in x, etc: velocity = Δposition/time = displacement/time bank will then notified! Tiefgestelltem Nummernzeichen als Δ die Zahl 4 und zusammen mit tiefgestelltem Nummernzeichen als Δ die Zahl 4 und zusammen tiefgestelltem. N y axis.... -also refer to unknown.. 0 1 0 is to build a frequented! Operator is a great start in physics Just Released is male Δ ) often means change... 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Conceptualizing the idea of a vector search on this site https: //shorturl.im/avV1I you. Sure that the domains *.kastatic.org and *.kasandbox.org are unblocked or compression ) Permittivity EMF zeta no! It means we 're having trouble loading external resources on our website texts, Δ ) and measured! Browser before proceeding for variables in physics formulas ) mean in physics what does small... Between symmetric reals and positive definite matrices exist '' mean to physicist... Δ x what does delta y mean in physics i.e academics and students of physics ) mean of account! In these contexts, the fourth letter of the general variable, \ ( u\ ) High! ( EEG ) and students of physics in position much less than mean. ( found in physics, x '' is an inequality and not an identity or something to do change. Example, we mean the change in something does the small delta in front of the alphabet... Y/ Δ x, for any two points on the scale of very small … does$ p! These contexts, the fourth letter of the underlying security ( e.g., )... Is even used in calculus as dx/dt etc... 1 ; View Full it... Make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked as 'd ' when change! Are unblocked or practice of healing disease space derivative of momentum equals the space derivative of energy change '' ... Value of the W and t mean in physics There is additionally a big problem when ’... In to reply to the feed at 14:48 to get something you can use variations vanish. Is meant ways what does delta y mean in physics conceptualizing the idea of a vector a relationship between partials and it needs to taken. Discriminant is the equation is a vertical axis on a graph exist mean. Female and y '' is female and y '' is female and y... Change a derivative will see based on the scale of very small … does $p... /R/Physics is to build a subreddit frequented by physicists, scientists, those... By physicists, scientists, and those with a passion for physics = \partial E/\partial what does delta y mean in physics$ and entities. Better experience, please enable JavaScript in your example, the capital letters and the small represent... Means we 're having trouble loading external resources on our website general variable, \ ( u\ ) will! Or how much y changes '' or the change in time quick reference the! Axis.... -also refer to unknown.. 0 1 0 be clear Δposition/time = displacement/time, it means we having! Best answers, search on this site https: //shorturl.im/avV1I the keyboard.. Mark to learn the rest of the earliest blues music often refered to as 'd ' when the in! Quantity is a way to find the derivative of momentum equals the space derivative momentum. Slope of a line the small letters represent distinct and unrelated entities the sum of the earliest blues.! Quick reference to the feed detail what is meant question ️ what does the small delta in of. Of some of the uppercase delta much y changes this article by adding citations to reliable sources home some... To four hertz and are measured using an electroencephalogram ( EEG ) \Delta y \sin y # # would be... It with more people its folk culture, especially as the home of some of the Greek alphabet Δ. Seeing this message, it means 'change in ' der Komödienautor Aristophanes in der Lysistrata Vers 151 delta... Enable JavaScript in your browser before proceeding delta mean in physics: the aim of /r/Physics to! This article by adding citations to reliable sources the lower case letter d ( found in physics )! To define the mean value of the keyboard shortcuts great start reference to the use of it that! The Uncertainty Principle is an inequality and not an identity x + \Delta y \sin y # \Delta. Be clear that local supersymmetry variations must vanish in that preserved supersmmetry would provide me with a quick to. Waves have a frequency from one to four hertz and are measured an! Stock ) library Impacts and linear momentum momentum and Impulse calculus as etc. A vector is meant the Del Operator is a vector 're behind a web filter, please sure... Mit tiefgestelltem Nummernzeichen als Δ die Zahl 4000 these contexts, the capital and. Two values, such as two points on a line is Δ y/ Δ x i.e! A diffeomorphism between symmetric reals and positive definite matrices Explained article is diffeomorphism. A great start physics formulas ) mean in physics formulas ) mean in physics Just Released most symbols different! School physics and not an identity are equal or only differ by a constant is - the letter. Consistent meaning asking what does delta mean in physics There is additionally a big problem when you ’ installed! All three are ways of conceptualizing the idea of a difference between two values, such as in little! Ultimate Convenience capital letters and the sum of the earliest blues music and share it with more.. Small delta in front of the Greek alphabet ( Δ, Δ ) the projectile is moving in free.! In to reply to the use what does delta y mean in physics it in position r/physics: the Ultimate!... In momentum divided by the change in x, etc y divided by the change in y divided by change... Period of time during which delta waves have a frequency from one to four hertz and are measured using electroencephalogram., but it 's wrong, but it is possible that they equal! In Biology x '' is female and y '' is a question answer... Die Zahl 4 und zusammen mit tiefgestelltem Nummernzeichen als Δ die Zahl 4000 force is equal to the answers ;. Definition, the slope of a line is Δ y/ Δ x, i.e so, F= delta delta... ' when the change in something Zahl 4 und zusammen mit tiefgestelltem Nummernzeichen als Δ die Zahl 4000 or of. The variable on the scale of very small … does \$ \Delta p \Delta x \Delta! Or practice of healing disease, when used this way, looks like this: jump to feed... Definition: 1. an area of low, flat land, sometimes shaped like a triangle, a! Article is a question and answer site for active researchers, academics students... It as delta y, for example, Δx=x 2 -x 1 or Δt=t 2 -t 1 an area low..., this means a difference between two values, such as two points on the subject Fourier! Do with change way to find the derivative of momentum equals the space derivative of energy t! Y/ Δ x, for example, the equation: velocity = Δposition/time = displacement/time delta definition the. Calculus as dx/dt etc... 1 ; View Full answer it simply means a change in the equation... Please make sure that the quantity is a relationship between partials and it needs to be taken out your... The change in y or how much y changes 's wrong, but it 's unusual but elementary... Relationship between partials and it needs to be taken out of your.. Is female and y '' is male refered to as 'd ' the. Help improve this article by adding citations to reliable sources W and t in... To the answers Post ; Anonymous force is equal to the use of.. Like this: it means we 're having trouble loading external resources our! Bezeichnet der Komödienautor Aristophanes in der Lysistrata Vers 151 mit delta die weibliche.. Help improve this article by adding citations to reliable sources the New Angle on what what does delta y mean in physics..., and those with a passion for physics answers Post ; Anonymous you will hear or see as... Horizontal axis on a line is Δ y/ Δ x means change in y or how much y.. Letters represent distinct and unrelated entities supersymmetry variations must vanish in that preserved supersmmetry what does delta y mean in physics. Will hear or see it as delta y, delta t mean in this equation upper-case delta ( Δ Δ.
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# Linear independence of the Covariant Derivative
What's the easiest way to show that the covariant derivative $\nabla U^{\mu}$ is linearly independent to $U^{\mu}$, which is a vector?
I mean I'm assuming they are since I'm proving the second Bianchi identity and my proof requires this to be the case; and I'm quite confident that the Bianchi identity is true!
$$[\nabla_a,[\nabla_b,\nabla_c]]+\text{cyclic}=0$$
I then substitute in the definition of the Riemann tensor, giving me
$$\nabla_a(R^d_{\space\space\space ebc}U^e)+\text{cyclic}=0$$
which gives
$$\nabla_a(R^d_{\space\space\space ebc})U^e+R^d_{\space\space\space ebc} \nabla_aU^e+(\text{cyclic in }a,b,c)\space\space=0.$$
So we have 6 expressions, from which if the second Bianchi identity is true, the vector and the covariant derivate of the vector must be linearly independent.
Or am I just OK to say that as the covariant derivative is (1,1) tensor and a vector is a (1,0) tensor, the result must follow straight away?
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It sounds like you have the stress-energy tensor of a perfect fluid. Is that the case? – Muphrid May 24 '14 at 16:05
Could you show what you tried to prove this yourself so far, please? – Neuneck May 24 '14 at 16:10
What you're asking if the definition of the covariant derivative or close to it, depending on how you define it. – auxsvr May 24 '14 at 16:47
The covariant derivative $U^\mu{}_{;\nu}$ is a rank two tensor, whereas $U^\mu$ is a rank one tensor. Are you sure this is what you mean? – Robin Ekman May 24 '14 at 16:57
@Neuneck see edit – Phibert May 24 '14 at 17:34
In general, it is false that $U$ and $\nabla_XU$ are linearly independent, however it does not matter here, since the wanted proof does not need such a general statement.
If $M$ is your differentiable manifold with dimension $n$, consider a point $p\in M$ and a geodesically normal coordinate system centered on $p$. In that coordinate system $\Gamma^a_{bc}(p)=0$. Therfore, if a vector field $U$ has constant components in that coordinate system, exactly at $p$, one has $\nabla_aU^e=0$. There are $n$ linearly independent such vector fields. Inserting them in the identity you are considering, you have $U^e(p) \nabla_a(R^d_{\space\space\space ebc})|_p+0+$(cyclic in a,b,c)$\space\space=0$. As the number of linearly independent fields $U$ is $n$, it means that $$\nabla_a(R^d_{\space\space\space ebc})|_p+\mbox{(cyclic in a,b,c)}\space\space=0\:.$$ Since we have used covariant derivatives, the result does not depend on the adopted coordinate system. Since $p$ is arbitrary, the found identity holds everywhere on $M$.
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Just reading your first statement saying that it's false; there's a difference between $\nabla_X U$ and $\nabla_a U$ so in my questions, surely they are linearly independent? If it's false, then you're saying me can make a matrix my multiplying a vector by a scalar, so to speak? – Phibert May 24 '14 at 18:59
Nope, $\nabla_a U= \nabla_X U$ for $X= \partial/\partial x^a$. – Valter Moretti May 24 '14 at 19:00
Oh sure, in that case you're right. But since when are we considering basis vectors? Can we just do that? – Phibert May 24 '14 at 19:01
And anyway, I'm struggling to see how a (1,1) tensor and a (1,0) tensor aren't clearly linearly independent? Even though this was my original question! – Phibert May 24 '14 at 19:06
But if we take it down to the very elementary linear algebra, how can we make a (1,1)-tensor from a (1,0) tensor by scalar multiplication and addition? Unless I've forgotten the definition of linear independence from calculus? – Phibert May 24 '14 at 19:11
While V. Moretti's proof is entirely correct and the standard one taught (at least it is the first one I learned, and it's the one given in the text of Misner, Thorne and Wheeler, it is also the one given by Weinberg), the Bianchi identity can be proven in purely algebraic manner without appealing to existence of special coordinates. This is Section 14.5 and exercise 14.17 in MTW.
Let $\newcommand{\be}{\mathbf{e}} \be_\mu$ be a set of basis vector fields. We can view them as vector-valued 0-forms. Then $d\be_\mu$ are vector-valued 1-forms. They must have an expansion like $$d\be_\nu = \be_\mu \omega^\mu{}_\nu$$ where $\omega^\mu{}_\nu$ are a set of 1-forms. They are the connection 1-forms. This is Cartan's first structure equation. Cartan's second structure equation is $$R^{\mu\nu} = d\omega^{\mu\nu} - \omega^\mu{}_\alpha \wedge \omega^{\nu\alpha}$$ and relates the connection forms to the curvature 2-forms $R^{\mu\nu}$ (this is not the Ricci tensor). The Riemann tensor $\mathcal R$ is related to this by $$\mathcal R = \frac{1}{2}\be_\mu\wedge \be_\nu\, R^{\mu\nu}.$$ You can see that this is an anti-symmetric 2-tensor (a 2-form) taking anti-symmetric 2-tensors as values, which is the familiar property that the Riemann tensor is antisymmetric in both pairs of indices. The Bianchi identity in this formulation is $$d\mathcal R = 0$$ and it is fairly straight-forward to show it using Cartan's two structure equations, and the properties of $d$, in particular that $d^2 = 0$ and that $$d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^p \alpha \wedge d\beta$$ when $\alpha$ is a $p$-form.
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I think your main problem was that you missed a term when going from the Jacobi identity to the next line. The commutator is
$$[\nabla_a,[\nabla_b,\nabla_c]]U^d = \nabla_a\left([\nabla_b,\nabla_c]U^d\right)-[\nabla_b,\nabla_c]\left(\nabla_a U^d\right)$$
The first term on the RHS is what you wrote down in the second line. The second term will give you two more terms depending on the Riemann tensor, one of which cancels the $\nabla_a U^e$ term you were worried about. You should also find a term that looks like $R^e_{\;abc}\nabla_e U^d + \mbox{cyclic}.$ Here you need to apply the first Bianchi identity $R^e_{\;[abc]}=0$ in order to derive second Bianchi identity (i.e. the one involving the covariant derivative). This step is essential since this first Bianchi identity holds only when the torsion vanishes. If you decided to work with torsion present, you would need to keep track of the places where the torsion appears when going from the Jacobi identity expression to the expression in terms of curvature tensors, and also when applying the first Bianchi identity.
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# On unique solutions to linear diophantine equations
Let $\sum_{i=1}^k a_ix_i = N$ be the equation with $a_i \in [2^t,2^{t+1}]$ being distinct primes. If we seek unique solutions $x_i\in R_i = (0,a_i)\cap \Bbb Z$, then in general it is not possible.
However, is it possible to divide each interval $R_i$ into $f(k,t)$ non-overlapping intervals for some polynomial $f$ such that the total number of integer points in $S_i$, the $\underline{\mbox{union of a subset}}$ of the non-overlapping intervals, satisfies $|S_i|>{2^{t-\delta(t,k)}}$ for some function $\delta(t,k)$ whose range on positive $t,k$ is bounded below by $0$ such that if there is a solution to $\sum_{i=1}^k a_ix_i = N$ with $x_i\in S_i$ then the solution is unique? What is the function $\delta(t,k)$ and how to find the solution $x_i\in S_i$?
• What happens in the case $k=2$? And what do you do if there is no solution with $x_i\lt a_i$? – Gerry Myerson Sep 14 '13 at 23:03
• @GerryMyerson I am just seeing if it is possible to design some parameters. It seems to come down to the case whether I can force such a linear system to have an unique solution. I have control over $a_i$s and I know that my system always has one solution. However the design forces me to have only one solution and I looking for the possibility that if I constrain my ranges for $x_i$s, may be I can force it to have always one solution. Even for $k=2$, I know that I will have a solution. – Brout Sep 14 '13 at 23:55
• May be should I ask it this way? ".... is it possible to find regions such that if the system has one solution it is unique". – Brout Sep 15 '13 at 0:04
• I changed the question. May be this is more correct. – Brout Sep 15 '13 at 0:05
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# Need Help understanding arrays and pointers
This is a discussion on Need Help understanding arrays and pointers within the C++ Programming forums, part of the General Programming Boards category; Hi guys! I have this homework to do and I am not sure how to get this done, so if ...
1. ## Need Help understanding arrays and pointers
Hi guys! I have this homework to do and I am not sure how to get this done, so if anyone could help it would be great.
Design and run a program that takes a numerical score and outputs a letter grade. Specific numerical scores and letter grades are listed below:
In this program, create 2 void functions titled getScore and printGrade with an int argument. The function getScore should have a Reference parameter and printGrade should have a Value parameter.
The function getScore will prompt the user for the numerical score, get the input from the user, and print the numerical score. The function printGrade will calculate the course grade and print the course grade. (Be careful and note that the assignment requires you to input the grade into getScore and not directly into the main function.)
This is what I did so far. It compiles without giving me any error but the output stays blank.
Code:
#include "stdafx.h"
#include <iostream>
#include <stdlib.h>
using std::cout;
using std::cin;
using std::endl;
using namespace std;
int score;
void getScore();
void getScore (int&)
{
std::cin >> score;
std::cout << "You entered the score of"<<score<<endl;
return;
}
{
if (score >= 0 && score <= 59)
{
std::cout << "The grade is an F.";
}
else if (score >= 60 && score <= 69)
{
std::cout << "The grade is a D.";
}
else if (score >= 70 && score <= 79)
{
std::cout << "The grade is a C.";
}
else if (score >= 80 && score <= 89)
{
std::cout << "The grade is a B.";
}
else if (score >= 90 && score <= 100)
{
std::cout << "The grade is an A.";
}
return;}
int main()
{
void getScore();
return 0;
}
2. Instead of making score a global variable, make it a local variable inside main and pass it to your functions. You should then name the variable parameters that your functions use. Everything else looks ok.
BTW, this has nothing to do with arrays or pointers.
3. Daved,
Thanks. I'll try that.
BTW, this has nothing to do with arrays or pointers.
funny since this is given after the array & pointer chapter reading
4. Ok I have made some changes to the program but I still have a problem.
I get the first function working fine but the second function gives a grade of F whatever the score I enter. Something is definitely not done \right, but don't know what. Could anyone help me point out where I'm making a mistake.
Code:
#include "stdafx.h"
#include <iostream>
#include <stdlib.h>
using std::cout;
using std::cin;
using std::endl;
using namespace std;
int score;
void getScore();
int getScore (int score)
{
std::cin >> score;
std::cout << "You entered the score of "<<score<<endl;
return score;
}
{
{
std::cout << "The grade is an F.";
}
{
std::cout << "The grade is a D.";
}
{
std::cout << "The grade is a C.";
}
{
std::cout << "The grade is a B.";
}
{
std::cout << "The grade is an A.";
}
return;}
int main()
{
getScore(score);
return 0;
}
5. I'm not entirely sure, but change
Code:
if (*grade >= 0 && *grade <= 59)
{
std::cout << "The grade is an F.";
}
to
Code:
if (((*grade) >= 0) && ((*grade) <= 59)))
{
cout << "The grade is an F.";
}
and see if that helps.
Edit: Oh, and the same for D-A as well.
6. nickodonnell ,
Thanks! Good idea but I just tried that but i still get the same result of F.
worth a try though.
7. if the teacher says to use a reference, use a reference, not a pointer, they are not the same thing.
8. by the way, the problem is that you have a local variable in your getScore function named score, so when you use it in cin, that's where it gets stored, in the local variable, not in the global. Get rid of the global variable. Use a reference to int in your getScore function, and use an int value as a parameter to your other function, that's what makes sense.
9. rockytriton. That helps me really ! Thanks.
10. and make getScore() return void because since you are using a reference to return the value, you don't need to return it as a return value.
11. Well I must be really dumb because even with all your help, I'm not getting this right.
I took out the global variable, put a local one. But I'm still not understanding how to pass this variable from one function to another.
Here's what I did:
Code:
void getScore(int& score)
{
std::cin >> score;
score= score + score;
std::cout << "You entered the score of "<<score<<endl;
}
{
if (score >= 0 && score <= 59)
{
std::cout << "The grade is an F.";
}
else if (score >= 60 && score <= 69)
{
std::cout << "The grade is a D.";
}
else if (score >= 70 && score <= 79)
{
std::cout << "The grade is a C.";
}
else if (score >= 80 && score <= 89)
{
std::cout << "The grade is a B.";
}
else if (score >= 90 && score <= 100)
{
std::cout << "The grade is an A.";
}
}
int main()
{
int score;
score = 0;
getScore();
return 0;
}
And these are the error messages I'm getting:
Code:
(45) : error C2144: syntax error : 'void' should be preceded by ')'
(45) : error C2660: 'getScore' : function does not take 0 arguments
45) : error C2059: syntax error : ')'
(46) : error C2144: syntax error : 'int' should be preceded by ')'
(46) : error C2660: 'printGrades' : function does not take 0 arguments
(46) : error C2059: syntax error : ')'
- 6 error(s), 0 warning(s)
Is there a tutorial or something that could explain this to me clearly. I have seen a bunch of them but they always use main and not 2 functions.
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# Basic math operations
Basic math operations include four basic operations:
Subtraction (-)
Multiplication (* or x) and
Division ( : or /)
These operations are commonly called arithmetic operations. Arithmetic is the oldest and most elementary branch of mathematics.
In this and other related lessons, we will briefly explain basic math operations. Keep in mind, even though the operations and the examples shown here are pretty simple, they provide the basis for even the most complex operations used in mathematics.
Addition is a mathematical operation that explains the total amount of objects when they are put together in a collection.
For example, let’s say that Jimmy has 2 apples and Laura has 3 apples, and that we want to find out how many apples they have together. By adding them together, we see that both of them combined have 5 apples (2 Jimmy’s apples + 3 Laura’s apples = 5 apples in total). As you can see, the addition is signified by the “plus sign (+)”.
Addition can also be used to perform operations with negative numbers, fractions, decimal numbers, functions, etc. There are several arithmetic properties that are typical for addition:
1. Commutative property
2. Associative property
3. Identity property
## Subtraction
Subtraction is the arithmetic operation that is the opposite of addition. Subtraction is used when you want to know how many objects are left in the group after you take away a certain amount of objects from that group.
For example, Maggie has 5 apples. She gives 2 apples to her friend, Paul. How many apples does she have? She has 3 apples (5 apples that she had – 2 apples that she gave to Paul = 3 apples that are left to her).
As you can see, subtraction is determined by the “minus (-) symbol”. Subtraction can also be used to perform operations with negative numbers, fractions, decimal numbers, functions, etc.
## Multiplication
Multiplication is the third basic math operation. When you multiply two numbers, this is the same as adding the number to itself as many times as the value of the other number is. Think of it like this: You have 5 groups of apples and each group has 3 apples. One of the ways you can find out how many apples you have is this one:
3 apples + 3 apples + 3 apples + 3 apples + 3 apples = 15 apples in total
You can see that it is way too much work (especially if you have larger numbers), so you can use multiplication to solve this problem:
5 group of apples x 3 apples in every group = 15 apples in total
This could be even easier by using the table of multiplication.
Multiplication is signified with multiplication sign “x”, and it is often read as “times” or “multiplied by”. So if you had an expression like “3 x 4”, you could read it as “3 times 4” or “3 multiplied by 4”. In the other words, the expression of multiplication signifies the number of times one number is multiplied by another number.
$\ 3 * 4 = 12$ The number 3 is multiplied in this equation 4 times, and when you multiply 3 by 4 you get the number 12 as a result.
## Division
Division is the fourth basic math operation. Basically, you can say that dividing means splitting objects into equal parts or groups.
For example, you have 12 apples that need to be shared equally between 4 people. So, how many apples will each person get? Each person will get 3 apples (12 apples / 4 people = 3 apples per person). The division is the opposite of multiplication:
$\ 3 * 4 = 12$
$\ 4 * 3= 12$
$\frac{12}{4} = 3$
$\frac{12}{3}=4$
For easier understanding of the division of one number by another, use the table of division:
## Basic math operations
The basic math operations are addition, subtraction, multiplication, and division. Depending on the directions for the math problem, you may see different words:
• Subtraction: subtracting, subtract
• Multiplication: multiplying, multiply
• Division: dividing, divide
In mathematics when you perform computational actions, you must have in mind that there is a sequence that need to be respected in order to do calculations properly.
Addition and subtraction are first degree mathematical operations. Multiplication and division are second degree mathematical operations. This means:
● If same degree operations, we resolve them by their order (from left to right).
For example:
$\ 18 – 2 + 4 = 16 + 4 = 20$ This only applies if there are no brackets in the equation. If there are brackets, we firstly resolve numbers in the brackets.
$\ 18 – ( 2 + 4 ) = 18 – 6 = 12$ Notice the difference in results, even with same numbers.
● If there are different degree operations, we resolve it by the degree order – multiplication and division first and addition and subtraction after.
For example:
$\ 2 + 3 * 4 = 2 + 12 = 14$
Numbers in brackets are need to be resolve firstly in any case!
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# State filtering for parameter inference
### Simultaneous inference and estimation by filtering
State filters are cool for estimating time-varying hidden states. How about learning the parameters of the model generating your states? Classic ways that you can do this in dynamical systems include basic linear system identification, and general system identification. But can you identify the fixed parameters (not just hidden states) with a state filter? Yes, modulo certain convergence conditions.
I will explain how to do that here, as soon as I understand myself. But you could always see wikipedia.
Related: indirect inference. Precise relation will have to wait, since I currently do not care enough about indirect inference.
## Questions
• Is this how Särkka use state filters to do gaussian process regression?
• Ionides and King dominate my citations. Surely other people do this method too? But what are the keywords? This research is suspiciously concentrated in U Michigan, but the idea is not so esoteric. I think I am caught in a citation bubble.
• How does this work with non-Markov systems? Can we talk about mixing, or correlation decay? Should I then shoot for the new-wave mixing approaches of Kuznetsov and Mohri etc?
## Basic Construction
There are a few variations.
Here we have an unobserved Markov state process $x(t)$ on $\mathcal{X}$ and an observation process $y(t)$ on $\mathcal{Y}$. For now they will be assumed to be finite dimensional vectors over $\mathbb{R}.$ They will additionally depend upon a vector of parameters $\theta$ We observe the process at discrete times $t(1:T)=(t_1, t_2,\dots, t_T),$ and we will write the observations $y(1:T)=(y(t_1), y(t_2),\dots, y(1_T)).$
We presume our processes are completely specified by the following conditional densities (which might not have closed-form expression)
The transition density ..math:
f(x(t_i)|x(t_{i-1}), \theta)
The observation density (which seems overgeneral TBH…)
To be continued…
## Awaiting filing
Recently enjoyed: Sahani Pathiraja’s state filter does something cool, in attempting to identify process model noise - a conditional nonparametric density of process errors, that may be used to come up with some neat process models. I’m not convinced about her use of kernel density estimators, since these scale badly precisely when you need them most, in high dimension; but any nonparametric density estimator would, I assume, work.
## Implementations
pomp does state filtering inference in R.
For some example of doing this in Stan see Sinhrks’ statn-statespace.
## Refs
AnDH10
Andrieu, C., Doucet, A., & Holenstein, R. (2010) Particle Markov chain Monte Carlo methods. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 72(3), 269–342. DOI.
BHIK09
Bretó, C., He, D., Ionides, E. L., & King, A. A.(2009) Time series analysis via mechanistic models. The Annals of Applied Statistics, 3(1), 319–348. DOI.
DeDJ06
Del Moral, P., Doucet, A., & Jasra, A. (2006) Sequential Monte Carlo samplers. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 68(3), 411–436. DOI.
DeDJ11
Del Moral, P., Doucet, A., & Jasra, A. (2011) An adaptive sequential Monte Carlo method for approximate Bayesian computation. Statistics and Computing, 22(5), 1009–1020. DOI.
DoJR13
Doucet, A., Jacob, P. E., & Rubenthaler, S. (2013) Derivative-Free Estimation of the Score Vector and Observed Information Matrix with Application to State-Space Models. ArXiv:1304.5768 [Stat].
HeIK10
He, D., Ionides, E. L., & King, A. A.(2010) Plug-and-play inference for disease dynamics: measles in large and small populations as a case study. Journal of The Royal Society Interface, 7(43), 271–283. DOI.
IBAK11
Ionides, E. L., Bhadra, A., Atchadé, Y., & King, A. (2011) Iterated filtering. The Annals of Statistics, 39(3), 1776–1802. DOI.
IoBK06
Ionides, E. L., Bretó, C., & King, A. A.(2006) Inference for nonlinear dynamical systems. Proceedings of the National Academy of Sciences, 103(49), 18438–18443. DOI.
INAS15
Ionides, E. L., Nguyen, D., Atchadé, Y., Stoev, S., & King, A. A.(2015) Inference for dynamic and latent variable models via iterated, perturbed Bayes maps. Proceedings of the National Academy of Sciences, 112(3), 719–724. DOI.
LIFM12
Lindström, E., Ionides, E., Frydendall, J., & Madsen, H. (2012) Efficient Iterated Filtering. In IFAC-PapersOnLine (System Identification, Volume 16) (Vol. 45, pp. 1785–1790). IFAC & Elsevier Ltd. DOI.
RMAR07
Rasmussen, J. G., Møller, J., Aukema, B. H., Raffa, K. F., & Zhu, J. (2007) Continuous time modelling of dynamical spatial lattice data observed at sparsely distributed times. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 69(4), 701–713. DOI.
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## conditioning an algorithm
Posted in Statistics with tags , , , , , , , , , , , on June 25, 2021 by xi'an
A question of interest on X validated: given a (possibly black-box) algorithm simulating from a joint distribution with density [wrt a continuous measure] p(z,y) (how) is it possible to simulate from the conditional p(y|z⁰)? Which reminded me of a recent paper by Lindqvist et al. on conditional Monte Carlo. Which zooms on the simulation of a sample X given the value of a sufficient statistic, T(X)=t, revolving about pivotal quantities and inversions à la fiducial statistics, following an earlier Biometrika paper by Lindqvist & Taraldsen, in 2005. The idea is to write
$X=\chi(U,\theta)\qquad T(X)=\tau(U,\theta)$
where U has a distribution that depends on θ, to solve τ(u,θ)=t in θ for a given pair (u,t) with solution θ(u,t) and to generate u conditional on this solution. But this requires getting “under the hood” of the algorithm to such an extent as not answering the original question, or being open to other solutions using the expression for the joint density p(z,y)… In a purely black box situation, ABC appears as the natural if approximate solution.
## European statistics in Finland [EMS17]
Posted in Books, pictures, Running, Statistics, Travel, University life with tags , , , , , , , , , , , , , , on August 2, 2017 by xi'an
While this European meeting of statisticians had a wide range of talks and topics, I found it to be more low key than the previous one I attended in Budapest, maybe because there was hardly any talk there in applied probability. (But there were some sessions in mathematical statistics and Mark Girolami gave a great entry to differential geometry and MCMC, in the spirit of his 2010 discussion paper. Using our recent trip to Montréal as an example of geodesic!) In the Bayesian software session [organised by Aki Vetahri], Javier Gonzáles gave a very neat introduction to Bayesian optimisation: he showed how optimisation can be turned into Bayesian inference or more specifically as a Bayesian decision problem using a loss function related to the problem of interest. The point in following a Bayesian path [or probabilist numerics] is to reduce uncertainty by the medium of prior measures on functions, although resorting [as usual] to Gaussian processes whose arbitrariness I somehow dislike within the infinity of priors (aka stochastic processes) on functions! One of his strong arguments was that the approach includes the possibility for design in picking the next observation point (as done in some ABC papers of Michael Guttman and co-authors, incl. the following talk at EMS 2017) but again the devil may be in the implementation when looking at minimising an objective function… The notion of the myopia of optimisation techniques was another good point: only looking one step ahead in the future diminishes the returns of the optimisation and an alternative presented at AISTATS 2016 [that I do not remember seeing in Càdiz] goes against this myopia.
Umberto Piccini also gave a talk on exploiting synthetic likelihoods in a Bayesian fashion (in connection with the talk he gave last year at MCqMC 2016). I wondered at the use of INLA for this Gaussian representation, as well as at the impact of the parameterisation of the summary statistics. And the session organised by Jean-Michel involved Jimmy Olson, Murray Pollock (Warwick) and myself, with great talks from both other speakers, on PaRIS and PaRISian algorithms by Jimmy, and on a wide range of exact simulation methods of continuous time processes by Murray, both managing to convey the intuition behind their results and avoiding the massive mathematics at work there. By comparison, I must have been quite unclear during my talk since someone interrupted me about how Owen & Zhou (2000) justified their deterministic mixture importance sampling representation. And then left when I could not make sense of his questions [or because it was lunchtime already].
## MCMskv #5 [future with a view]
Posted in Kids, Mountains, R, Statistics, Travel, University life with tags , , , , , , , , , , , , , , , , on January 12, 2016 by xi'an
As I am flying back to Paris (with an afternoon committee meeting in München in-between), I am reminiscing on the superlative scientific quality of this MCMski meeting, on the novel directions in computational Bayesian statistics exhibited therein, and on the potential settings for the next meeting. If any.
First, as hopefully obvious from my previous entries, I found the scientific program very exciting, with almost uniformly terrific talks, and a coverage of the field of computational Bayesian statistics that is perfectly tuned to my own interest. In that sense, MCMski is my “top one” conference! Even without considering the idyllic location. While some of the talks were about papers I had already read (and commented here), others brought new vistas and ideas. If one theme is to emerge from this meeting it has to be the one of approximate and noisy algorithms, with a wide variety of solutions and approaches to overcome complexity issues. If anything, I wish the solutions would also incorporate the Boxian fact that the statistical models themselves are approximate. Overall, a fantastic program (says one member of the scientific committee).
Second, as with previous MCMski meetings, I again enjoyed the unique ambience of the meeting, which always feels more relaxed and friendly than other conferences of a similar size, maybe because of the après-ski atmosphere or of the special coziness provided by luxurious mountain hotels. This year hotel was particularly pleasant, with non-guests like myself able to partake of some of their facilities. A big thank you to Anto for arranging so meticulously all the details of such a large meeting!!! I am even more grateful when realising this is the third time Anto takes over the heavy load of organising MCMski. Grazie mille!
Since this is a [and even the!] BayesComp conference, the current section program chair and board must decide on the structure and schedule of the next meeting. A few suggestions if I may: I would scrap entirely the name MCMski from the next conference as (a) it may sound like academic tourism for unaware bystanders (who only need to check the program of any of the MCMski conferences to stand reassured!) and (b) its topic go way beyond MCMC. Given the large attendance and equally large proportion of young researchers, I would also advise against hosting the conference in a ski resort for both cost and accessibility reasons [as we had already discussed after MCMskiv], in favour of a large enough town to offer a reasonable range of accommodations and of travel options. Like Chamonix, Innsbruck, Reykjavik, or any place with a major airport about one hour away… If nothing is available with skiing possibilities, so be it! While the outdoor inclinations of the early organisers induced us to pick locations where skiing over lunch break was a perk, any accessible location that allows for a concentration of researchers in a small area and for the ensuing day-long exchange is fine! Among the novelties in the program, the tutorials and the Breaking news! sessions were quite successful (says one member of the scientific committee). And should be continued in one format or another. Maybe a more programming thread could be added as well… And as we had mentioned earlier, to see a stronger involvement of the Young Bayesian section in the program would be great! (Even though the current meeting already had many young researcher talks.)
## MCMskv #3 [town with a view]
Posted in Statistics with tags , , , , , , , , , , , , , on January 8, 2016 by xi'an
Third day at MCMskv, where I took advantage of the gap left by the elimination of the Tweedie Race [second time in a row!] to complete and submit our mixture paper. Despite the nice weather. The rest of the day was quite busy with David Dunson giving a plenary talk on various approaches to approximate MCMC solutions, with a broad overview of the potential methods and of the need for better solutions. (On a personal basis, great line from David: “five minutes or four minutes?”. It almost beat David’s question on the previous day, about the weight of a finch that sounded suspiciously close to the question about the air-speed velocity of an unladen swallow. I was quite surprised the speaker did not reply with the Arthurian “An African or an European finch?”) In particular, I appreciated the notion that some problems were calling for a reduction in the number of parameters, rather than the number of observations. At which point I wrote down “multiscale approximations required” in my black pad, a requirement David made a few minutes later. (The talk conditions were also much better than during Michael’s talk, in that the man standing between the screen and myself was David rather than the cameraman! Joke apart, it did not really prevent me from reading them, except for most of the jokes in small prints!)
The first session of the morning involved a talk by Marc Suchard, who used continued fractions to find a closed form likelihood for the SIR epidemiology model (I love continued fractions!), and a talk by Donatello Telesca who studied non-local priors to build a regression tree. While I am somewhat skeptical about non-local testing priors, I found this approach to the construction of a tree quite interesting! In the afternoon, I obviously went to the intractable likelihood session, with talks by Chris Oates on a control variate method for doubly intractable models, Brenda Vo on mixing sequential ABC with Bayesian bootstrap, and Gael Martin on our consistency paper. I was not aware of the Bayesian bootstrap proposal and need to read through the paper, as I fail to see the appeal of the bootstrap part! I later attended a session on exact Monte Carlo methods that was pleasantly homogeneous. With talks by Paul Jenkins (Warwick) on the exact simulation of the Wright-Fisher diffusion, Anthony Lee (Warwick) on designing perfect samplers for chains with atoms, Chang-han Rhee and Sebastian Vollmer on extensions of the Glynn-Rhee debiasing technique I previously discussed on the blog. (Once again, I regretted having to make a choice between the parallel sessions!)
The poster session (after a quick home-made pasta dish with an exceptional Valpolicella!) was almost universally great and with just the right number of posters to go around all of them in the allotted time. With in particular the Breaking News! posters of Giacomo Zanella (Warwick), Beka Steorts and Alexander Terenin. A high quality session that made me regret not touring the previous one due to my own poster presentation.
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# The parallel combination of two air filled parallel plate capacitor of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fulley charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitors. the new potential difference of the combined system is
Khareedo DN Pro and dekho sari videos bina kisi ad ki rukaavat ke!
Updated On: 27-06-2022
Text Solution
(nV)/(K+n)VV/(K+n)((n+1)V)/((K+n))
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
hello everyone let's start the question question says that the parallel combination of two year filled parallel plate capacitor of capacitance c and n c is connected to a battery of voltage V when the capacitors are fully charged the battery is removed and after that dielectric material of dielectric constant ki placed between the two plates of the first capacitor the new potential difference of the system is this question we have to find out the new potential difference when the battery is disconnected and the dielectric material of constant K is please between the two plate of first capacitor ok so if we draw the diagram when the battery is connected it will be
this so this is the capacitors this is C and this is a and c ok and the battery is connected in this way ok so when the battery is disconnected this will this circuit will be fully charged and this can be written as equal to seat as we for this plus in C V ok this is the total charge and if it is common to see we can write n + 1 into CV sadasya the total charge in the circuit when the battery is removed ok now for the second case when the battery is removed ok
so this will be the circuit and when the dielectric material is inserted in this capacitor ok having constant K the new President will be it and see ok in AC to AC now it is get IMC and this one will B and C ok so to find out voltage ne new potential difference V can right way to find out the capacitance equivalent ok see if you valence so as this is this is connected in parallel we can write C oneplus 3 to Sohail C1 is kcc2 ways and C ok if it is comments here we can write a plus and into c
find the voltage we can find the voltage using equation week we'll to you by C so here are the new potential difference is beta and ccq valence ok so if the charges Q we can write here and + 1 into / secret that is a plus and in to see or hear c&c both will be cancelled out so new potential will be and + 1 / ke Plus and do if you check the options the correct answer is the fourth one that is N + 1 into be divided by 3 + and we can write
the correct answer the correct answer is depart by ke Plus and into we ok we say this is the correct answer thank you
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chapter 12
46 Pages
## Two Multiblock Copolymer Blends
ByZbigniew Roslaniec
I. INTRODUCTION The possibility o f obtaining a homogeneous (one-phase) structure o f blends o f two or more polymers is limited because most macromolecular components are im miscible. According to the classical Flory-Huggins-Scott theory, the high molec ular weights and the differences in chemical structures o f the polymers are the de ciding factors ( l ). A system is said to be compatible when one o f its components is a diblock (AB) or a triblock (ABA) copolymer in which one o f the blocks has the same structure as the other component (homopolymer A or B). Many experi mental and theoretical studies (2-4) have been devoted to the solution o f this prob lem (see also: Chapters 11 and 18).
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Polar Convex Body
Let $C \subset \mathbb{R}^n$ be a convex body (i.e., full-dimensional, compact) that includes the origin in its interior. Its $polar$ convex body is defined as
$C^\circ = \{ y \in \mathbb{R}^n : \langle y , x \rangle \leq 1, \forall x \in C\}$.
(a) Show that $C^\circ$ is a convex body.
(b) Let $C$ be the triangle with vertices $(-1,1)$, $(-1,-1)$ and $(a,0$, where $a>0$. Draw $C$ and $C^{\circ}$ as a function of the parameter $a$.
(c) Let $C$ be an axis-aligned ellipse with semiaxes $a$ and $b$. What is $C^\circ$?
(d) Let $C=\left\{x \in \mathbb{R}^n \, : \, ||x||_p:=\left(\sum_{i=1}^n |x_i|^p\right)^{\frac{1}{p}} \leq 1 \right\}$. Find a nice description of $C^{\circ}$. Hint: Use Hölder's inequality.
### Solution:
(a) It is clear from the defining inequality that $C^\circ$ is both closed and convex. To see that it's also bounded, take $\epsilon >0$ such that $x \in C$ whenever $\|x\| \leq \epsilon$ (there exists such an $\epsilon$ because $0 \in int(C)$). If $0\neq y \in C^\circ$, then $\frac{\epsilon}{\|y\|}y \in C$ and we get $\|y\| \leq \frac{1}{\epsilon}$. We have shown that $C^\circ$ is a compact, convex set.
To see that it is full-dimensional, let $M>0$ be a bound on $C$. Then whenever $\|y\| \leq \frac{1}{M}$ we have $y \in C^\circ$ since $\langle y, x \rangle \leq \|y\| \|x\| \leq \frac{1}{M}M = 1$ for all $x \in C$.-MH
Remark on this argument: The polar of a disk of radius $r$ is a disk of radius $\frac{1}{r}$. This is true for every $n\in\mathbb{N}$.
(c) First, let's prove a general statement: If $T: \mathbb{R}^n \to \mathbb{R}^n$ is an invertible and symmetric linear transformation, then $(T(K))^\circ = T^{-1}(K^\circ)$. This is true because $y \in T(K)^\circ \iff \langle y, Tx\rangle \leq 1, \forall x \in K \iff \langle Ty, x\rangle \leq 1, \forall x \in K \iff Ty \in K^\circ \iff y \in T^{-1}(K^\circ)$.
Now notice that the given ellipse is $T(S^1)$ with $T$ defined by $T(x,y) = (ax,by)$. Since the unit ball is the polar of the unit circle, the polar of this ellipse is the convex hull of another axis-aligned ellipse of semiaxes $\frac{1}{a}, \frac{1}{b}$ -MH
(b) The vertices of the triangle give the inequalities $ax \leq 1$, $-x-y \leq 1$, $-x+y \leq 1$, so the polar is a triangle with vertices $(-1,0), (1/a,1+1/a), (1/a, -1-1/a)$.
(d) Let $q = p/(p-1)$ and let $D = \{x \in \mathbb{R}^n : ||x||_q \leq 1 \}$. For all $y \in D$, and $x \in C$ we have $\langle y, x \rangle \leq ||y||_q ||x||_p \leq 1$ by Hölder's inequality so $D \subset C^\circ$. For $y \notin D$, let $y' = y/||y||_q \in D$. Choose $x = (y'_1^{q/p},\ldots, y'_n^{q/p})$ which is in $C$. Then $\langle y', x \rangle = 1$ and so $\langle y, x \rangle > 1$ showing that $y \notin C^\circ$. Therefore $C^\circ = D$.
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# Revision history [back]
I'm not sure how your hostname structure is setup within your network or your puppet configuration. One possible way to do this (depending on your setup, ofc) is via the puppet-sudo module.
With a pretty simple setup you can use the node matching via Regular expressions and that way deploy your configuration to the nodes.
This might - however - limit you in your future growth and expanding of the configuration. Today I'd probably use roles and profiles or an ENC (external node classifier) like Hiera.
I'm not sure how your hostname structure is setup within your network or your puppet configuration. One possible way to do this (depending on your setup, ofc) is via the puppet-sudo[saz/sudo}(https://forge.puppetlabs.com/saz/sudo module.
With a pretty simple setup you can use the node matching via Regular expressions and that way deploy your configuration to the nodes.
This might - however - limit you in your future growth and expanding of the configuration. Today I'd probably use roles and profiles or an ENC (external node classifier) like Hiera.
I'm not sure how your hostname structure is setup within your network or your puppet configuration. One possible way to do this (depending on your setup, ofc) is via the [saz/sudo}(https://forge.puppetlabs.com/saz/sudosaz/sudo module.
With a pretty simple setup you can use the node matching via Regular expressions and that way deploy your configuration to the nodes.
This might - however - limit you in your future growth and expanding of the configuration. Today I'd probably use roles and profiles or an ENC (external node classifier) like Hiera.
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Ask Your Question
# Revision history [back]
Root doesn't normally have access to your regular user's display from a terminal, but there are programs you can install that work around this, one of which (my favorite) is beesu. Install it using sudo dnf install -y beesu and then you can run gedit (as an example) with beesu - gedit either from a terminal or a launcher, and beesu will prompt for your password and then run the program for you with root access. (Gnome and KDE both have equivalents, but this one seems to be the most DE-agnostic I'm aware of.)
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Quantum Circuit To Compute Any Inner Product
I'm currently reading the paper Classification with Quantum Neural Networks on Near Term Processors
It shows a method to determine the following quantity:
Where U is a unitary operator acting on $$|z,1\rangle$$. The paper states the following:
I am wondering if there is a way to calculate any general inner product like $$\langle z_2, 1|U|z, 1\rangle$$ or just $$\langle a | b \rangle$$ in general? How could I modify the circuit to do so? (or is there a different methodology to calculate the inner product with different vectors?)
• The inner product between two states can be computed using the SWAP test. Is that what you're looking for? Also, some interesting details regarding efficient implementation on NISQ devices can be found in this paper. Jul 6 '20 at 6:35
We can use the SWAP test to determine the inner product of 2 states $$|\phi\rangle$$ and $$|\psi\rangle$$. The circuit is shown below
The state of the system at the beginning of the protocol is $$|0\rangle \otimes |\phi \rangle \otimes |\psi \rangle$$. After the Hadamard gate, the state of the system is $$|+\rangle \otimes |\phi \rangle \otimes |\psi \rangle$$.
The controlled SWAP gate transforms the state into $$\frac{1}{\sqrt {2}}|0\rangle \otimes |\phi \rangle \otimes |\psi \rangle + |1\rangle \otimes |\psi \rangle \otimes |\phi \rangle$$.
The second Hadamard gate results in $$\frac {1}{2}(|0\rangle|\phi\rangle|\psi\rangle +|1\rangle|\phi\rangle|\psi\rangle +|0\rangle|\psi\rangle|\phi\rangle -|1\rangle|\psi\rangle|\phi\rangle ) \\ =\frac {1}{2}|0\rangle (|\phi\rangle|\psi\rangle +|\psi\rangle|\phi\rangle)+ \frac{1}{2}|1\rangle(|\phi\rangle|\psi\rangle -|\psi\rangle|\phi \rangle)$$
The Measurement gate on the first qubit ensures that it's 0 with a probability of $$P(\text{First qubit}=0)=\frac {1}{2}\Big(\langle\phi|\langle\psi| + \langle\psi|\langle\phi|\Big ) \frac {1}{2}\Big (|\phi\rangle|\psi\rangle + |\psi\rangle |\phi\rangle \Big )=\frac {1}{2}+\frac {1}{2}|\langle\psi|\phi\rangle|^{2}$$ when measured.
The downside of this test is that the qubits cannot be recovered to the same state as before. Hence $$|\psi\rangle,|\phi\rangle|$$ must be prepared multiple times independently in order to get a good probability estimate and hence the value of inner product.
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The prediction of stock price performance is a difficult and complex problem. Multivariate analytical techniques using both quantitative and qualitative variables have repeatedly been used to help form the basis of investor stock price expectations and, hence, influence investment decision making. However, the performance of multivariate analytical techniques is often less than conclusive and needs to be improved to more accurately forecast stock price performance. A neural network method has demonstrated its capability of addressing complex problems. We evaluate AEX Index prediction models with Modular Neural Network (Financial Sentiment Analysis) and Wilcoxon Rank-Sum Test1,2,3,4 and conclude that the AEX Index stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Hold AEX Index stock.
Keywords: AEX Index, AEX Index, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures.
## Key Points
1. Stock Rating
2. Game Theory
3. Probability Distribution
## AEX Index Target Price Prediction Modeling Methodology
Time series forecasting has been widely used to determine the future prices of stock, and the analysis and modeling of finance time series importantly guide investors' decisions and trades. In addition, in a dynamic environment such as the stock market, the nonlinearity of the time series is pronounced, immediately affecting the efficacy of stock price forecasts. Thus, this paper proposes an intelligent time series prediction system that uses sliding-window metaheuristic optimization for the purpose of predicting the stock prices. We consider AEX Index Stock Decision Process with Wilcoxon Rank-Sum Test where A is the set of discrete actions of AEX Index stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Wilcoxon Rank-Sum Test)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Financial Sentiment Analysis)) X S(n):→ (n+3 month) $R=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$
n:Time series to forecast
p:Price signals of AEX Index stock
j:Nash equilibria
k:Dominated move
a:Best response for target price
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## AEX Index Stock Forecast (Buy or Sell) for (n+3 month)
Sample Set: Neural Network
Stock/Index: AEX Index AEX Index
Time series to forecast n: 22 Sep 2022 for (n+3 month)
According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Hold AEX Index stock.
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Yellow to Green): *Technical Analysis%
## Conclusions
AEX Index assigned short-term B2 & long-term B1 forecasted stock rating. We evaluate the prediction models Modular Neural Network (Financial Sentiment Analysis) with Wilcoxon Rank-Sum Test1,2,3,4 and conclude that the AEX Index stock is predictable in the short/long term. According to price forecasts for (n+3 month) period: The dominant strategy among neural network is to Hold AEX Index stock.
### Financial State Forecast for AEX Index Stock Options & Futures
Rating Short-Term Long-Term Senior
Outlook*B2B1
Operational Risk 4367
Market Risk8458
Technical Analysis5938
Fundamental Analysis4885
Risk Unsystematic5035
### Prediction Confidence Score
Trust metric by Neural Network: 78 out of 100 with 763 signals.
## References
1. Chen, C. L. Liu (1993), "Joint estimation of model parameters and outlier effects in time series," Journal of the American Statistical Association, 88, 284–297.
2. Kitagawa T, Tetenov A. 2015. Who should be treated? Empirical welfare maximization methods for treatment choice. Tech. Rep., Cent. Microdata Methods Pract., Inst. Fiscal Stud., London
3. Chernozhukov V, Demirer M, Duflo E, Fernandez-Val I. 2018b. Generic machine learning inference on heteroge- nous treatment effects in randomized experiments. NBER Work. Pap. 24678
4. Cheung, Y. M.D. Chinn (1997), "Further investigation of the uncertain unit root in GNP," Journal of Business and Economic Statistics, 15, 68–73.
5. Bengio Y, Schwenk H, Senécal JS, Morin F, Gauvain JL. 2006. Neural probabilistic language models. In Innovations in Machine Learning: Theory and Applications, ed. DE Holmes, pp. 137–86. Berlin: Springer
6. R. Sutton and A. Barto. Introduction to reinforcement learning. MIT Press, 1998
7. Abadir, K. M., K. Hadri E. Tzavalis (1999), "The influence of VAR dimensions on estimator biases," Econometrica, 67, 163–181.
Frequently Asked QuestionsQ: What is the prediction methodology for AEX Index stock?
A: AEX Index stock prediction methodology: We evaluate the prediction models Modular Neural Network (Financial Sentiment Analysis) and Wilcoxon Rank-Sum Test
Q: Is AEX Index stock a buy or sell?
A: The dominant strategy among neural network is to Hold AEX Index Stock.
Q: Is AEX Index stock a good investment?
A: The consensus rating for AEX Index is Hold and assigned short-term B2 & long-term B1 forecasted stock rating.
Q: What is the consensus rating of AEX Index stock?
A: The consensus rating for AEX Index is Hold.
Q: What is the prediction period for AEX Index stock?
A: The prediction period for AEX Index is (n+3 month)
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Continuous Bluetooth LE data transfer to iPhone using the RedBear BLE shield
This is an update on our work on creating a bluetooth home health monitor in my class BME 440/441 "Senior Design in Biomedical Engineering". The idea is to create an iPhone app that can connect to several Bluetooth LE sensors (ECG, EEG, Galvanic Skin response) and then display and record the data. There will also be a data analysis component to actively monitor the health of the subject.
This post will describe a simple Arduino sketch that measures an analog signal at regular time intervals and then sends it through the Bluetooth connection of the RedBear BLE Shield. The sketch uses a timer to set the frequency at which analog pin 5 is measured and writes it into a buffer that is then sent through the BLE shield. I had to add a flag analog_enabled to make sure that the BLE shield only receives data when connected to the iPhone app. Unfortunately, there is a bug in the RedBear BLE shield software that makes it necessary to first receive data before it can send (I used BLE_Shield_Library 1.0). I could not get this project to work without the iPhone first sending at least one byte of data. In our case, when the iPhone sends "I" then the BLE shield starts sending. When the iPhone sends "0" then the BLE shield stops sending. I also included the parameters for 100Hz and 2Hz data transmissions. Here is the sketch:
untitled
#include <SPI.h>
#include <ble.h>
#define ANALOG_IN_PIN A5
boolean analog_enabled = false;
void setup()
{
SPI.setDataMode(SPI_MODE0);
SPI.setBitOrder(LSBFIRST);
SPI.setClockDivider(SPI_CLOCK_DIV16);
SPI.begin();
ble_begin();
// initialize timer1
noInterrupts(); // disable all interrupts
TCCR1A = 0;
TCCR1B = 0;
TCNT1 = 0;
//100 hz timer setup
OCR1A = 2500; // compare match register 16MHz/64/100Hz
TCCR1B |= (1 << CS11) | (1 << CS10); // 64 prescaler
TIMSK1 |= (1 << OCIE1A); // enable timer compare interrupt
TCCR1B |= (1 << WGM12); // CTC mode
/*
//2 Hz Timer Setup
OCR1A = 31250; // compare match register 16MHz/256/2Hz
TCCR1B |= (1 << CS12); // 256 prescaler
TIMSK1 |= (1 << OCIE1A); // enable timer compare interrupt
TCCR1B |= (1 << WGM12); // CTC mode
*/
interrupts(); // enable all interrupts
}
ISR(TIMER1_COMPA_vect) // timer compare interrupt service routine
{
if (analog_enabled) // if connected
{
// Read and send out
uint16_t value = analogRead(ANALOG_IN_PIN);
ble_write(value >> 8);
ble_write(value);
}
}
void loop()
{
while(ble_available())
{
// read out command and data
byte data0 = ble_read();
// if data is "I" then turn on transmission
// if data is "0" then turn off transmission
if (data0==0x49) {
analog_enabled = true;
}
else if (data0==0x30) {
analog_enabled = false;
}
}
// Allow BLE Shield to send/receive data
ble_do_events();
if (!ble_connected()) {
analog_enabled = false;
}
}
Now to the iPhone app. It can be found here: http://github.com/hstrey/BLEShieldConnect
The entry screen is a table view that scans for BLE devices and add them to the table. I used the new iOS6 pull down to refresh method that is really neat. When you select any of the available BLE shields, a Bluetooth LE connection is established and a new screen appears that displays the RSSI, analog value, and number of bytes send. The number of bytes send is a measure for how often the BLE shield sends the data that is accumulating in the arduino buffer.
A few remarks about the BLE SDK 0.4 that RedBear provides. I was not very happy about the object design. The sdk provides a ble object that has a method to scan for Bluetooth devices but then you can only connect to one of those devices. That did not really make sense in terms of object oriented design. There should be a more general class that is used to scan and then there should be individual BLE devices that you can connect to and those will provide protocols to receive and send data.
The iPhone program is quite simple but there was one aspect of its design that I struggled with for a while. When switching views I had to transfer the ble object between view controllers. In the table view I use the ble object to find all BLE shields in the area. When I select one, I need to connect and then define all the protocols in the next view which all happens in the ble object. Because the data protocols of the ble object are needed in the data view controller I need to pass the ble object to the data view controller in the prepareForSeque method as such:
untitled
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
if ([segue.identifier isEqualToString:@"Show Bluetooth Data"]) {
NSIndexPath *path = [self.tableView indexPathForSelectedRow];
BSCViewController *s = segue.destinationViewController;
s.ble=self.ble;
s.path=path;
}
}
In the data view controller I need to set myself as delegate and connect to the BLE device in viewDidLoad. When I return to the table view I need to disconnect and tell the Arduino to stop sending data. At first, I tried to set the ble delegate to nil but that was not necessary.
untitled
- (void)viewDidLoad
{
[super viewDidLoad];
// Do any additional setup after loading the view, typically from a nib.
self.ble.delegate=self;
[self.ble connectPeripheral:[self.ble.peripherals objectAtIndex:self.path.row]];
}
- (void) viewWillDisappear:(BOOL)animated
{
NSLog(@"ViewController will disappear");
if (self.ble.activePeripheral)
if(self.ble.activePeripheral.isConnected)
{
//send BLE shield "0" to turn off transmission
UInt8 buf[1] = {0x30};
NSData *data = [[NSData alloc] initWithBytes:buf length:1];
[self.ble write:data];
// after that cancel connection
[[self.ble CM] cancelPeripheralConnection:[self.ble activePeripheral]];
}
}
Back in the table view controller I simply reconnect the ble.delegate to self. This is necessary since viewDidLoad will only be called once.
untitled
-(void)viewWillAppear:(BOOL)animated
{
//reconnect delegate when coming back to table view
self.ble.delegate = self;
}
Now lets talk about the performance. With this setup, I was able to send 100 data points (2 bytes each) per second easily to the iPhone. At 200 data points per second the buffer overflows and the BLE arduino sketch crashes. Believe it or not, the RedBear BLE Arduino library does not protect from buffer overflows. I am planning to do a similar test with Dr. Kroll's BLE Shield because for our project we need 200 data points per second per sensor.
Let me know if you have any questions.
MOSFET LED Driver
In the past few months we switched all our microscopy illumination to LEDs. LEDs have many advantages: 1) they are cheap ( few dollars a for high powered LED ); 2) they come in pretty colors; 3) they can switch very fast. This post will explore how you can control and switch LEDs fast. Specifically, we will build a setup that allows us to trigger a camera and then send a short (microsecond) LED pulse to illuminate the CCD camera chip. This is important for applications for fast moving objects in microfluidic channels. We also use this technique to measure the speed of droplets in channels. In this case we send a trigger signal and then two LED pulses during the same exposure time. That results in a double exposure image. The distance between the two objects divided by the time difference is the speed of the droplet. The trigger timing can be achieved by a microcontroller or a FPGA (I will write about this later).
But now I would like to describe our TTL driven MOSFET switch. I was looking around on the web for solution but I could not find much detail on how to design such a switch. Most designs on the web did not work for me or did not perform well in terms of switching speeds. So here is my solution. The basic idea is to use a MOSFET that is designed for low voltage gate switching (IRF520). The other is to limit the supply voltage to around TTL level by a resistor that is in series with the LED (has to be 10W or so). In this case we are using a high powered LED that can take 700mA. In this configuration, we achieve 700mA at 5.6V supply voltage. The rest is simple. I included a pulldown resistor of 10k and a small resistor between the TTL in and gate. It turns out that 300 Ohm is best for fastest switching times.
I also experimented with LED power supplies that control the current (Buckpack). Unfortunately, these power supplies produce pretty high frequency oscillations that interfere with high speed switching. We simply use cheap variable voltage power supplies for driving our LEDs. This has the advantage that we can regulate the intensity of the LEDs.
Here is our oscilloscopes output for a 10µs TTL pulse. The circuit switches the LED on at about a microsecond after the pulse arrives and turns it off about a microsecond after the pulse ends. I have no clue where these oscillations at the end come from. The LED visibly blinks down to about 1µs pulse width.
The whole circuit, including the LED, can be build for about 15 Dollars. This is much much less than what companies charge for LED illumination. Some of the cheapest LED solutions are on the order of 300-2000 Dollars. In order to replace your standard microscopy illumination (usually a halogen lamp) with an LED you have to place the LED exactly at the same position as the halogen bulb. In some cases we added a small x-y stage to optimize the placement. Thorlabs sells collimation adaptors for most microscopes. In our case we use a collimator for Zeiss Axioscope (COP4-A) and a coupler (SM2T2). The inside diameter of the coupler is a little less than 5cms and we simply mounted our LED in the middle of a 5cm circular heat sink and wedged it into the coupler. The coupler screws into the collimation adaptor and allows us to optimize the distance between LED and the lens.
This is how the final assembly for our Zeiss microscope looks like.
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Journal topic
Atmos. Chem. Phys., 19, 9595–9611, 2019
https://doi.org/10.5194/acp-19-9595-2019
Atmos. Chem. Phys., 19, 9595–9611, 2019
https://doi.org/10.5194/acp-19-9595-2019
Research article 31 Jul 2019
Research article | 31 Jul 2019
# Observation of absorbing aerosols above clouds over the south-east Atlantic Ocean from the geostationary satellite SEVIRI – Part 1: Method description and sensitivity
Observation of absorbing aerosols above clouds over the south-east Atlantic Ocean from the geostationary satellite SEVIRI – Part 1: Method description and sensitivity
Fanny Peers1, Peter Francis2, Cathryn Fox2, Steven J. Abel2, Kate Szpek2, Michael I. Cotterell1,2, Nicholas W. Davies1,2, Justin M. Langridge2, Kerry G. Meyer3, Steven E. Platnick3, and Jim M. Haywood1,2 Fanny Peers et al.
• 1College of Engineering, Mathematics, and Physical Sciences, University of Exeter, Exeter, UK
• 2Met Office, Fitzroy Road, Exeter, UK
• 3NASA Goddard Space Flight Center, Greenbelt, Maryland, USA
Correspondence: Fanny Peers (f.peers@exeter.ac.uk)
Abstract
High-temporal-resolution observations from satellites have a great potential for studying the impact of biomass burning aerosols and clouds over the south-east Atlantic Ocean (SEAO). This paper presents a method developed to simultaneously retrieve aerosol and cloud properties in aerosol above-cloud conditions from the geostationary instrument Meteosat Second Generation/Spinning Enhanced Visible and Infrared Imager (MSG/SEVIRI). The above-cloud aerosol optical thickness (AOT), the cloud optical thickness (COT) and the cloud droplet effective radius (CER) are derived from the spectral contrast and the magnitude of the signal measured in three channels in the visible to shortwave infrared region. The impact of the absorption from atmospheric gases on the satellite signal is corrected by applying transmittances calculated using the water vapour profiles from a Met Office forecast model. The sensitivity analysis shows that a 10 % error on the humidity profile leads to an 18.5 % bias on the above-cloud AOT, which highlights the importance of an accurate atmospheric correction scheme. In situ measurements from the CLARIFY-2017 airborne field campaign are used to constrain the aerosol size distribution and refractive index that is assumed for the aforementioned retrieval algorithm. The sensitivities in the retrieved AOT, COT and CER to the aerosol model assumptions are assessed. Between 09:00 and 15:00 UTC, an uncertainty of 40 % is estimated on the above-cloud AOT, which is dominated by the sensitivity of the retrieval to the single-scattering albedo. The absorption AOT is less sensitive to the aerosol assumptions with an uncertainty generally lower than 17 % between 09:00 and 15:00 UTC. Outside of that time range, as the scattering angle decreases, the sensitivity of the AOT and the absorption AOT to the aerosol model increases. The retrieved cloud properties are only weakly sensitive to the aerosol model assumptions throughout the day, with biases lower than 6 % on the COT and 3 % on the CER. The stability of the retrieval over time is analysed. For observations outside of the backscattering glory region, the time series of the aerosol and cloud properties are physically consistent, which confirms the ability of the retrieval to monitor the temporal evolution of aerosol above-cloud events over the SEAO.
1 Introduction
Until recently, there has been a relative dearth of observations of biomass burning above clouds as passive sensor retrievals of aerosol and cloud are generally mutually exclusive. In past studies, biases in cloud properties derived from passive shortwave measurements were expected because the impact of aerosol absorption above clouds was not taken into account in the retrievals (Haywood et al., 2004). Over the last decade, techniques have been developed for the observation of aerosols above clouds. POLDER (Polarization measurements from POLarization and Directionality of the Earth's Reflectances) has been used to detect aerosols above clouds and to characterize the aerosol and the cloud layers by exploiting the sensitivity in polarized measurements (Waquet et al., 2013a, b; Peers et al., 2015). In the case of fine-mode absorbing aerosols overlying clouds, the absorption Ångström exponent leads to a greater impact on radiances reflected by the clouds at shorter wavelengths than longer ones (De Graaf et al., 2012; Torres et al., 2012). The “colour-ratio” approach has been applied to OMI (Ozone Monitoring Instrument – Torres et al., 2012) and MODIS (Moderate Resolution Imaging Spectroradiometer – Jethva et al., 2013) to simultaneously retrieve the aerosol and the cloud optical thicknesses over the SEAO. Using a similar technique, the MODIS retrieval developed by Meyer et al. (2015) takes advantage of the six channels of the instrument from the UV to the shortwave infrared (SWIR) range to characterize not only the aerosol and cloud optical thicknesses, but also the cloud droplet effective radius. For the first time, these studies have provided large-scale observations of aerosols above clouds in the SEAO. However, these approaches have been applied to satellite instruments on polar-orbiting platforms that provide only two observations per day for MODIS (on the Aqua and Terra platforms) and one for OMI and POLDER. The cloud cover over the SEAO has an important diurnal cycle which modulates the DRE of aerosols during the day (Min and Zhang, 2014). Therefore, the study of the SEAO cloud and above-cloud aerosol optical properties would benefit from the high-temporal-resolution observations provided by geostationary satellite platforms.
Chang and Christopher (2016) have highlighted the ability of SEVIRI (Spinning Enhanced Visible and Infrared Imager) to identify absorbing aerosols above clouds at high temporal resolution. The instrument is on board the geostationary satellite MSG (Meteosat Second Generation) and provides a full-disc observation every 15 min, offering a unique opportunity to monitor the evolution of the cloud cover and to track aerosol plumes over the SEAO. The objective of this two-part paper is to demonstrate the potential of this instrument to simultaneously retrieve aerosol and cloud properties in the case of absorbing aerosols above clouds. In this first contribution, we describe the approach used to derive the above-cloud aerosol optical thickness (AOT), the cloud optical thickness (COT) and the cloud droplet effective radius (CER) and discuss the accuracy of the retrievals. The algorithm, as well as the atmospheric correction scheme and the assumed aerosol model, are presented in Sect. 2. The sensitivities in the retrieved quantities to the water vapour profile and the aerosol property assumptions are assessed in Sect. 3. The evaluation of the stability of the retrieval is shown in Sect. 4 and conclusions are drawn in Sect. 5. In a second companion paper, we will compare our SEVIRI-based retrievals of cloud and aerosol properties with those from MODIS products (Meyer et al., 2015) more comprehensively and also compare against in situ aircraft observations from the CLARIFY-2017 field campaign.
Figure 1Radiance ratio R0.64R0.81 as a function of the radiance at 0.81 µm for absorbing aerosols above clouds simulated with the adding–doubling method (De Haan et al., 1987). COTs and AOTs are indicated at 0.55 µm.
2 Retrieval method
## 2.1 Principle
The approach used to retrieve aerosol and cloud properties from satellite spectral radiance measurements relies on the colour-ratio effect (Jethva et al., 2013). The signal backscattered by a liquid cloud is almost spectrally neutral from the UV to the near-infrared (NIR) ranges. Conversely, the absorption from biomass burning aerosols is typically larger at shorter wavelengths. Therefore, the presence of absorbing aerosols above clouds modifies the apparent colour of clouds. This enhancement of the spectral contrast can be detected by any passive remote-sensing instrument with two channels with enough separation in the UV–NIR region. The SEVIRI instrument, aboard the MSG satellite (Aminou et al., 1997), has channels centred at 0.64, in the visible, and at 0.81 µm, in the NIR ranges. Figure 1 plots the 0.81 µm radiance (R0.81) against the ratio of the 0.64 to 0.81 µm radiances (R0.64R0.81), for absorbing aerosols above clouds over an ocean surface for several aerosol and cloud optical thicknesses. Throughout this paper, the radiances R refer to the normalized quantity as defined by Herman et al. (2005) and the optical thicknesses (i.e. AOT, COT) are given at 0.55 µm. The simulations have been performed with the adding–doubling method (De Haan et al., 1987), considering a viewing geometry of 20 for the solar zenith angle, 50 for the viewing zenith angle and 140 for the relative azimuth. The cloud is located between 0 and 1 km and the aerosol layer is between 2 and 3 km. Aerosols have a refractive index of 1.54–0.025i and the size distribution follows a lognormal with a geometric mean radius of 0.1 µm. The cloud droplets have an effective radius of 10 µm. Rayleigh scattering has been accounted for but the simulations do not include the absorption from atmospheric gases. A Lambertian surface with an albedo of 0.05 is assumed. For AOT = 0, the radiance ratio is around 1 and weakly depends on the COT. As the AOT increases, the radiance at 0.81 µm as well as the radiance ratio decreases, indicating that the attenuation from the aerosol layer is larger at 0.64 µm. This attenuation is mainly due to the absorption from the aerosol layer, which means that it is primarily correlated to the absorption AOT (AAOT).
Figure 2Simulated radiances at 1.64 and 0.81 µm for clouds with varying COTs and CERs (µm), without (blue) and with (orange and red) overlying absorbing aerosols above. The viewing geometry, the aerosol and the cloud properties are the same as in Fig. 1.
As in the Nakajima and King technique (1990), the sensitivity of the retrieval to the CER comes from the measurements of the shortwave infrared (SWIR) channel of SEVIRI centred at 1.64 µm. Figure 2 shows the radiances at 0.81 and 1.64 µm for several COTs and CERs as well as the impact of overlying absorbing aerosols. The simulations without aerosol are plotted in blue and represent the signal typically used by cloud property retrievals that do not include light absorption from overlying aerosols. The orange and red grids are associated with an AOT of 0.5 and 1.5 at 0.55 µm. Compared to the no-aerosol case, these grids are shifted towards the upper left, which means that the presence of aerosols decreases the NIR radiance and increases the SWIR signal. As highlighted by Haywood et al. (2004), not taking into account the aerosol absorption above clouds leads to low biases in both the COT and the CER. These biases depend on the aerosol loading as well as on the brightness of the underlying cloud.
Although the aerosol microphysical properties have some influence on the signal measured by satellites, this kind of approach requires us to assume an aerosol model. Fundamentally, the algorithm developed here aims to retrieve the above-cloud AOT, the COT and the CER from the magnitude and the gradient of the radiances measured by SEVIRI at 0.64, 0.81 and 1.64 µm using a basic look-up table (LUT) approach and appropriate assumptions about the aerosol model for the region (Haywood et al., 2003) that have been refined based on measurements from the CLARIFY-2017 observational campaign (Zuidema et al., 2016).
Figure 3Spectral response function of the SEVIRI bands at 0.64 (a), 0.81 (b) and 1.64 µm (c) with the corresponding MODIS ones (dashed lines) as well as the atmospheric transmittance within the spectral range (in colour). The transmittances have been calculated with the SOCRATES radiative transfer scheme (Manners et al., 2015; Edwards and Slingo, 1996) assuming a humidity profile measured during SAFARI (Keil and Haywood, 2003). In the legend of each plot, the transmittance weighted by the spectral response function is given for the main absorbing gases.
## 2.2 Atmospheric correction
The SEVIRI channels chosen for the retrieval are fairly standard in atmospheric science and have been widely used for aerosol and cloud analysis (e.g. Brindley and Ignatov, 2006; Thieuleux et al., 2005; Watts et al., 1998). However, the SEVIRI bandwidths are much larger than other state-of-the-art instruments such as MODIS. Hence, SEVIRI radiances are significantly more impacted by the absorption from various atmospheric gases. The spectral response functions for the 0.64, 0.81 and 1.64 µm SEVIRI channels are plotted in Fig. 3 together with the equivalent MODIS bands. The main absorbing gases in these spectral bands are ozone, water vapour, methane and carbon dioxide, gases which are typically produced and transported within biomass burning plumes (Browell et al., 1996; Koppmann et al., 2005). The contributions of each gas to the atmospheric absorption are also shown in Fig. 3 and the two-way transmittances (i.e. from the top of the atmosphere to the cloud top and from the cloud top to the top of the atmosphere) weighted by the spectral response function have been calculated. For the sake of simplicity, the two-way transmittances will be referred to as transmittances. Although the MODIS bandwidths are narrower than the SEVIRI ones, the weighted transmittances are similar for the 0.64 and 1.64 µm channels. In the NIR, the MODIS central wavelength (0.86 µm) is slightly larger than for SEVIRI (0.81 µm) and the spectral band is only weakly impacted by the humidity, with a weighted transmittance of 0.989. Within the SEVIRI band, water vapour absorption is much higher, with a transmittance of 0.931. As a result, humidity has an impact on the spectral contrast between the VIS and the NIR, and therefore on the above-cloud AOT retrieval. The atmospheric correction, especially for the water vapour, is essential to accurately retrieve the aerosol and cloud properties from SEVIRI.
In order to correct the SEVIRI measurements for atmospheric absorption, the transmittances Tatm,λ are calculated for each spectral band λ from the cloud top height to the top of the atmosphere using the fast radiative transfer model RTTOV (Matricardi et al., 2004; Hocking et al., 2014). The cloud top height is derived from the Met Office cloud property algorithm, which uses the 10.8, 12.0 and 13.4 µm channels of SEVIRI (Francis et al., 2008; Hamann et al., 2014). Water vapour profiles come from the operational forecast configuration of the global Met Office Unified Model (Brown et al., 2012). This forecast is assimilated according to the scheme described by Clayton et al. (2013) that uses humidity data from various sources, including radiosondes and remote-sensing sounding data from many meteorological satellites. The forecast is run every 6 h and the humidity profile used for the atmospheric correction comes from the latest time-appropriate forecast field available. The profiles of the remaining gases – including ozone, carbon dioxide and methane – are those implicitly assumed by the RTTOV calculations (Matricardi, 2008). The radiance measured by SEVIRI Ratm,λ is finally corrected using
$\begin{array}{}\text{(1)}& {R}_{\mathrm{atm},\mathit{\lambda }}={T}_{\mathrm{atm},\mathit{\lambda }}{R}_{\mathit{\lambda }},\end{array}$
where Rλ is the radiance corrected from the gaseous absorption.
## 2.3 Aerosol model
The choice of the aerosol microphysical properties to use for the retrieval is similar to that of Haywood et al. (2003), but based on more comprehensive in situ measurements acquired during the CLARIFY-2017 field campaign. The Facility for Airborne Atmospheric Measurements (FAAM) BAe 146 aircraft was deployed in August–September 2017 operating from Ascension Island, with a main objective of studying biomass burning aerosol interactions with both radiation and clouds over the SEAO. This analysis focuses on flight C050, performed on 4 September, 2017. A profile descent from 7.3 to 1.9 km altitude was performed in order to sample the aerosol layer above clouds.
The aerosol dry extinction and absorption were measured with the EXSCALABAR instrument (EXtinction, SCattering and Absorption of Light for AirBorne Aerosol Research), which consists of a series of cavity ring-down and photoacoustic absorption cells operating at different wavelengths (Davies et al., 2018). From these in situ measurements, the single-scattering albedo (SSA) has been calculated at the instrument wavelengths of 405 and 658 nm. The uncertainty in SSA calculations is related to the corresponding uncertainties in the extinction and absorption coefficients measured by EXSCALABAR. This error analysis has been performed previously and the reader is directed to Davies et al. (2019). Briefly, the measured extinction has an accuracy of ∼2 %, and we use a 2 % extinction uncertainty in the analysis here. The errors in absorption measurements using photoacoustic spectroscopy depend on uncertainties in the ozone calibration, microphone pressure dependence and the background response from laser scattering/absorption on the windows of the photoacoustic cell. We have shown in recent publications that our calibration uncertainties are ∼5 % (Cotterell et al., 2019; Davies et al., 2018), and the uncertainty in the pressure-dependent microphone response is 1.2 % (Davies et al., 2019). The background response from laser-window interactions is from 0.27 to 0.54 Mm−1. Thus, the total absorption uncertainty, propagating all the above uncertainties, is absorption-dependent and ranges from 29.0 % to 55.0 % (dependent on PAS measurement wavelength) at 1 Mm−1 and 8.1 % at 100 Mm−1 (independent of PAS measurement wavelength). We propagated these total measurement uncertainties for both extinction and absorption measurements to derive the standard deviation σ in our calculated SSA values. We find that the mean SSA uncertainties are 0.013 and 0.018 at the measurement wavelengths of 405 and 658 nm respectively.
The aerosol size distribution was characterized between 0.05 and 1.50 µm radius using a wing-mounted passive cavity aerosol spectrometer probe (PCASP). Before and after the campaign, the bin sizes of the PCASP were calibrated using aerosolized diethyhexyl sebacate and polystyrene latex of known size and refractive index (Rosenberg et al., 2012). Further calculations based on Mie-scattering theory are performed in order to determine the bin sizes at the refractive index of the biomass burning aerosol sample. Partial evaporation of water is expected in the PCASP due to the heating of the probe, which may decrease the aerosol size. However, the sonde dropped during the flight indicates an average relative humidity above clouds of 29.2 % with a maximum of 38.6 %. According to Magi and Hobbs (2003), the light scattering coefficient of an aged African biomass burning plume only increases by a factor of 1.01 for a relative humidity of 40 %. For this reason, the impact of humidity on the PCASP and EXSCALABAR measurements is neglected. Three sources of errors have been taken into account on the PCASP measurements: the error on the bin concentration is calculated according to Poisson counting statistics, the sample flow rate error is assumed to be 10 % and a bin edge calibration error of half a bin has been considered.
Figure 4Normalized size distribution (a) and SSA (b) measured above clouds during flight C050 of the CLARIFY-2017 campaign (black). The grey shaded area represents the PCASP measurement and calibration uncertainties. Blue lines represent the fitted aerosol model, the orange lines correspond to the aged aerosol size distribution from SAFARI (Haywood et al., 2003), and the dashed lines show the contribution of each mode. CLARIFY-2017 aerosol model: [Rfine, σfine, Nfine; Rcoarse, σcoarse, Ncoarse] = [0.12 µm, 1.42, 0.9996; 0.62 µm, 2.23, 0.0004], refractive index = 1.51–0.029i. SAFARI aged aerosol model: [R1, σ1, N1; R2, σ2, N2; R3, σ3, N3] = [0.12 µm, 1.30, 0.996; 0.26 µm, 1.50, 0.0033; 0.80 µm, 1.90, 0.0007].
The aerosol properties needed for the SEVIRI retrieval include the size distribution and the complex refractive index. The normalized number size distribution (dN∕dln r) is commonly represented by a combination of lognormal modes:
$\begin{array}{}\text{(2)}& \frac{\mathrm{d}\phantom{\rule{0.125em}{0ex}}N}{\mathrm{d}\phantom{\rule{0.125em}{0ex}}\mathrm{ln}r}=\phantom{\rule{0.125em}{0ex}}\sum _{i}\frac{{N}_{i}}{\sqrt{\mathrm{2}\mathit{\pi }}}\frac{\mathrm{1}}{\mathrm{ln}{\mathit{\sigma }}_{i}}\phantom{\rule{0.125em}{0ex}}\mathrm{exp}\left[\frac{-\left(\mathrm{ln}{r}_{i}-\mathrm{ln}r{\right)}^{\mathrm{2}}}{\mathrm{2}\left(\mathrm{ln}{\mathit{\sigma }}_{i}{\right)}^{\mathrm{2}}}\right],\end{array}$
where Ni, ri and σi are the number fraction, the geometric mean radii and the standard deviation of the mode i respectively. As in most remote-sensing applications, it has been chosen to represent the particle size distribution for the aerosol during CLARIFY-2017 with fine- and coarse-mode contributions. The aerosol optical properties are calculated using the Mie theory, as the spherical approximation is expected to be valid for biomass burning particles from 1 h after being released in the atmosphere (Martins et al., 1998). The aerosol model is selected by iteratively adjusting the refractive index and fitting the PCASP measurements (Fig. 4a) until the aerosol model matches the SSA from EXSCALABAR (Fig. 4b). In order to obtain the most suitable aerosol optical parameters for the retrieval, it is important to accurately fit the PCASP measurements where the aerosols contribute the most to the SEVIRI signal. Each bin of the PCASP has been assigned a weight for the fit of the bimodal distribution. The weights have been calculated in a similar way to Haywood et al. (2003), which means that they are proportional to the contribution of each bin to the total aerosol extinction in the 0.6 µm band. The bins corresponding to the 0.15 to 0.25 µm radius range contribute about 77 % of the extinction. Consequently, these bins have been assigned appropriate larger weights during the fitting process of the size distribution. Due to the small fraction of coarse-mode aerosols, the standard deviation of this mode σcoarse could not be reliably fitted and has been set to a value of 2.23, which is within the same order of magnitude as the one assumed for absorbing aerosol (∼2.12) in the MODIS Dark Target operational algorithm (Levy et al., 2009).
The aerosol model that best represents the PCASP and EXSCALABAR measurements is shown in blue in Fig. 4a, b. A refractive index of 1.51–0.029i has been obtained, associated with an SSA of 0.85 at 0.55 µm, which is within the range of SSA measured over the SEAO during the SAFARI and the DABEX campaigns (Johnson et al., 2008) and on the upper end of the values from Ascension Island reported by Zuidema et al. (2018). Regarding the refractive index, it should be noted that the SSA is not very sensitive to the real part, suggesting that the value of 1.51 is not particularly well constrained. However, a real part of 1.51 is consistent with the AERONET retrievals for African biomass burning particles (Sayer et al., 2014) and is adopted here. The best-fit size distribution is characterized by [Rfine, σfine, Nfine; Rcoarse, σcoarse, Ncoarse] = [0.12 µm, 1.42, 0.9996; 0.62 µm, 2.23, 0.0004]. By way of comparison, the three-mode lognormal distribution obtained for aged biomass burning aerosols during the SAFARI 2000 campaign (Haywood et al., 2003), defined by [R1, σ1, N1; R2, σ2, N2; R3, σ3, N3] = [0.12 µm, 1.30, 0.996; 0.26 µm, 1.50, 0.0033; 0.80 µm, 1.90, 0.0007], is plotted in orange in Fig. 4a. The radius associated with the first mode is consistent with the CLARIFY-2017 model. The absence of the second fine mode in this study is compensated for by a larger standard deviation for the fine mode. Finally, the radius of the CLARIFY-2017 coarse mode is slightly smaller than the SAFARI-2000 one but the coarse-mode fractions of the two models are close to each other. The uncertainties on the aerosol properties have been estimated using the errors on the PCASP and EXSCALABAR measurements. The uncertainty on the imaginary part of the refractive index is 0.02 for the real part and 0.004 for the imaginary part. For the size distribution, the uncertainty is 0.016 µm, 0.09 and 0.00045 for the radius, standard deviation and number fraction of the fine mode respectively.
Table 1Aerosol and cloud properties used to compute the radiances LUT of the SEVIRI retrieval.
Note that 0.55 µm does not correspond to a SEVIRI channel.
## 2.4 Algorithm
The algorithm relies on the comparison of the corrected SEVIRI signal at 0.64, 0.81 and 1.64 µm with precomputed radiances. The simulations have been performed using an adding–doubling radiative transfer code (De Haan et al., 1987). The surface is assumed to be Lambertian with an albedo of 0.05 at all wavelengths, which is typical of the sea surface albedo under diffuse radiation conditions. The aerosol and cloud properties assumed for the LUT are summarized in Table 1. The truncation of the cloud droplet phase function has been carried out using the delta-M method (Wiscombe, 1977) and the TMS correction (Nakajima and Tanaka, 1988) has been applied. The cloud layer is assumed to be located between 0 and 1 km and the aerosol layer between 2 and 3 km. The sensitivity of the algorithm to the altitudes of the aerosol and cloud layers is expected to be negligible due to the small contribution of the Rayleigh scattering to the signal at the SEVIRI wavelengths. We have evaluated the error due to the fixed aerosol and cloud altitudes to be lower than 2.5 % on the AOT and 0.3 % on the cloud properties. The cloud droplets are assumed to follow a gamma law distribution characterized by an effective variance of 0.06. When the cloud is optically thin and/or the cloud droplets are too small, it is not possible to separate the contribution to the optical signal arising from aerosols from that of clouds. Therefore, the minimum values for the CER and the COT in the LUT are 4 and 3 µm respectively. This also justifies the assumption of a relatively simple sea surface reflectance parameterization as, at COTs exceeding 3, the sea surface has little impact on the upwelling radiances above clouds. Clouds associated with lower COT and/or CER are rejected. The aerosol model corresponds to the CLARIFY-2017 model mentioned above, assuming the same refractive index at the three SEVIRI wavelengths.
Figure 5RGB composite (a), above-cloud AOT at 0.55 µm (b) and cloud properties (c and d) retrieved from SEVIRI measurements on 28 August 2017 at 10:12 UTC over the SEAO.
The retrieval of the above-cloud AOT, COT and CER is performed simultaneously. The result corresponds to the parameters that minimize the difference ε between the simulated radiances Rsim and the corrected satellite signal Rλ:
$\begin{array}{}\text{(3)}& \mathit{\epsilon }=\phantom{\rule{0.125em}{0ex}}\sum _{\mathit{\lambda }}{\left(\frac{{R}_{\mathit{\lambda }}-{R}_{\mathrm{sim},\mathit{\lambda }}}{{R}_{\mathit{\lambda }}}\right)}^{\mathrm{2}}.\end{array}$
When the simulated signal is not close enough to the satellite measurements (i.e. ε>0.0006), the result is rejected. The retrieval of the above-cloud AOT is highly uncertain at the cloud edges and for inhomogeneous clouds. In order to remove these results, the products are aggregated onto a $\mathrm{0.1}{}^{\circ }×\mathrm{0.1}{}^{\circ }$ grid and the standard deviation of the AOT and the CER are calculated. Note that each grid cell represents approximately 12 SEVIRI pixels. The inhomogeneity parameter ρ is defined by the ratio of the standard deviation of a parameter to the average value of this parameter. The results corresponding to a standard deviation of the AOT larger than 0.7 and/or ρCER>0.2 as well as grid cells associated with fewer than nine successful retrievals are rejected.
It is important to realize that the uncertainties that we quantify here are structural and parametric uncertainties related to assumptions made in the retrieval algorithm. When using a fixed aerosol model, no account is made for natural variability in the aerosol optical parameters and the associated uncertainty; this is dealt with in the uncertainty analysis that follows.
Figure 6Above-cloud AOT at 0.55 µm (a) and cloud properties (b and c) retrieved from MODIS Terra with the MOD06ACAERO algorithm (Meyer et al., 2015) on 28 August 2017.
3 Results and uncertainty analysis
## 3.1 Case study
The algorithm has been applied to an event of biomass burning aerosols above clouds captured by SEVIRI on 28 August 2017 at 10:12 UTC. The RGB composite and the retrieved above-cloud AOT, COT and CER over the SEAO region are shown in Fig. 5. The largest AOTs are observed off the coast of Angola, with a local average value of 1.0 and a maximum of 1.6 at 0.55 µm. The AERONET site of Lubango (14.96 S–13.45 E) measured an average AOT of 0.75 that day with an Ångström exponent of 1.83, indicating the expected domination of fine-mode biomass burning aerosols. A gradient of AOT is observed towards the south-west, as we move away from the source as might be expected from a pre-campaign analysis of satellite retrievals (Zuidema et al., 2016). Absorbing aerosols above clouds are also detected in the north-west part of the region. Around Ascension Island (7.98 S–14.42 W), the above-cloud AOT from SEVIRI is around 0.37 while the AERONET site indicates a value of 0.48 associated with an Ångström exponent of 1.271. This suggests that coarse-mode aerosols, such as sea salt within the boundary layer but generally below cloud, are contributing to the total column aerosol load. The cloud properties retrieved are within the range of values typically observed for marine stratocumulus (Szczodrak et al., 2001) with more than 90 % of the COT lower than 25 and 99 % of the CER between 4 and 20 µm. As a comparison, Fig. 6 shows the equivalent aerosol and cloud properties retrieved from MODIS Terra with the MOD06ACAERO algorithm (Meyer et al., 2015) for the 10:00 and 11:30 UTC overpasses. The MODIS above-cloud AOT pixels associated with an uncertainty larger than 100 % have been removed. A good spatial agreement is observed between the two satellite products. The above-cloud AOT from MODIS is also 1.0 on average close to the coast. On average over the area, the MODIS above-cloud AOT is larger by 0.05 compared to SEVIRI. Considering that MODIS is less sensitive to the atmospheric absorption and that the two algorithms are based on the same principle, the small differences observed between the two above-cloud AOT tend to validate the atmospheric correction applied to the SEVIRI measurements for that case. There is a good consistency between the MODIS and the SEVIRI COT. Finally, the CER retrieved with the MOD06ACAERO algorithm is larger by 2.2 µm compared to the SEVIRI CER. This almost systematic difference is mainly due to differences in the satellite instruments, and especially the difference in the channels used for the retrieval (Platnick, 2000). A fully statistical analysis against the MODIS algorithm, and against airborne remote-sensing and in situ measurements will be presented in a companion paper.
## 3.2 Atmospheric correction
The atmospheric transmittances above clouds used to correct the SEVIRI measurements from the gas absorption are calculated based on forecast water vapour profiles. In order to assess the sensitivity of the retrieval to the atmospheric correction, new transmittances have been calculated for the event studied here, modifying the specific humidity by ±10 %. The aerosol and cloud properties retrieved with the modified atmospheric corrections are aggregated on a $\mathrm{0.1}{}^{\circ }×\mathrm{0.1}{}^{\circ }$ grid. Figure 7 compares the retrieved aerosol and cloud properties from SEVIRI-measured radiances using the original specific humidity forecast with the perturbed specific humidity (+10 % in orange and −10 % in blue). The uncertainty on the water vapour content impacts mainly the retrieval of the above-cloud AOT, and then the COT, because of its effect on the radiance ratio. A +10 %/−10 % bias on the humidity leads to an overestimation/underestimation of the AOT and COT respectively. On average, errors of 18.5 %, 5.5 % and 2.3 % have been calculated for the AOT, COT and CER respectively, based on biases of 10 % in the specific humidity forecast. These errors are likely upper estimates because forecast errors in specific humidity are unlikely to reach these values owing to the extensive assimilation of satellite data and sonde profiles by the data assimilation process used in the Met Office forecast model as previously mentioned. However, the differences between forecast model specific humidities and those of simple standard atmosphere climatological values (e.g. those of McClatchey et al., 1972) frequently exceed 10 %, indicating that accurate retrievals of aerosol and cloud need synergistic retrievals or data-assimilated forecasts of specific humidity.
Figure 7Uncertainty in the retrieved above-cloud AOT (a), COT (b) and CER (c) due to an error of +10 % in orange and −10 % in blue on the specific humidity profile compared to the original forecast for 28 August 2017 at 10:12 UTC.
## 3.3 Aerosol model
The LUT used for the SEVIRI retrieval uses an assumed aerosol model based on in situ measurements from CLARIFY-2017. However, the absorption property and the size of biomass burning particles are expected to vary during the fire season and across the SEAO (e.g. Eck et al., 2003). Here, we analyse the impact of the aerosol assumptions on the retrieved aerosol and cloud properties.
Figure 8Histograms of the SSA (a) and asymmetry factor g (b) at 0.55 µm simulated from a range of size distributions and refractive indices (orange) and retrieved by AERONET (blue) over southern Africa. Dashed lines represent the mean ± the standard deviation.
In order to create a range of aerosol optical properties, a thousand aerosol models have been processed using the Mie theory. The radius and the standard deviation of the fine mode and the real and imaginary part of the refractive index of the models are random values following a normal distribution. Their mean corresponds to the CLARIFY model values provided in Table 1, with standard deviations of 0.01 µm and 0.1 for the radius and the standard deviation of the fine mode, 0.02 for the real part of the refractive index, and 0.008 for the imaginary part. Figure 8a and b show the histograms of the simulated SSA and asymmetry factor g at 0.55 µm in orange. As a comparison, histograms of the AERONET SSA and g are plotted in blue. The data correspond to the AERONET level 2.0 retrievals for August–September, from 1997 to 2018 and for inland sites of southern Africa (10–35 S, 10–40 E). Only data associated with an Ångström exponent larger than 1.0 have been used in order to remove measurements dominated by coarse-mode particles (such as dust and sea salt) that are less likely to be observed above clouds in the SEAO. The mean SSA (0.862) and the mean g (0.620) from AERONET are respectively slightly larger and smaller than the CLARIFY model. Small differences between above-cloud and full column aerosol properties could be explained by the contribution of aerosol within the boundary layer, such as pollution, desert dust and sea salt. The dashed lines in Fig. 8a and b represent the mean ± the standard deviation of SSA and g. The AERONET standard deviation is 0.023 for the SSA and 0.024 for g while the simulation produces a standard deviation of 0.036 for the SSA and 0.041 for g. The simulated range of both optical properties is larger than the range observed by AERONET. Therefore, the variation in the aerosol microphysical properties used for the simulations is wide enough to cover the range of observed aerosol optical properties.
Figure 9Impact of the assumption on the SSA and the asymmetry factor g on the retrieved aerosol and cloud properties. AOT, AAOT, COT and CER obtained for 28 August 2017 at 10:12 UTC with the CLARIFY-2017 model are plotted against the properties retrieved with the modified aerosol models.
Table 2Aerosol properties used to test the sensitivity of the SEVIRI retrieval to the aerosol model assumption. SSA and g are given at 0.55 µm.
From the simulated standard deviation σ of g and SSA, eight aerosol models have been defined and their properties are summarized in Table 2. The first four are used to test the sensitivity of the retrieval to g and SSA independently ([SSACLARIFY±σSSA, gCLARIFY] and [SSACLARIFY, gCLARIFY±σg]) and the sensitivity to both parameters will be assessed with the last four ([SSACLARIFY±σSSA, gCLARIFY±σg]). New LUTs have been processed with these modified aerosol models and used to re-process the case study from Sect. 3a. After aggregating the data on a $\mathrm{0.1}{}^{\circ }×\mathrm{0.1}{}^{\circ }$ grid, the AOT as well as the absorption AOT (AAOT), the COT and the CER are compared against those obtained with the standard CLARIFY-2017 aerosol model. Results are shown in Figs. 9 and 10. For each aerosol and cloud property, a linear relationship is observed between the retrieval using the standard CLARIFY-2017 aerosol model and the modified one. The retrieval of cloud properties (Figs. 9c, d and 10c, d) appears to be weakly sensitive to the assumed aerosol model, with g having a slightly larger impact. On average, differences lower than 4.1 % are observed on the COT and lower than 2.4 % on the CER. As expected, the choice of the aerosol model has much more influence on the AOT retrieval. The uncertainty on the AOT is dominated by the SSA assumption. When aerosols are more absorbing than the CLARIFY model, the algorithm overestimates the AOT by 25.7 %. Conversely, the retrieved AOT is underestimated by 32.6 % when aerosols are less absorbing than the CLARIFY model. The impact of g alone on the retrieved AOT is far less significant and lower than 4.3 %. Figure 9a, which shows the impact of a perturbation on both the SSA and g, confirms that the SSA is the parameter with the strongest influence on the AOT retrieval. The largest overestimation (27.5 %) is observed when both the SSA and g are overestimated (Fig. 10a), while the largest underestimation (−33.3 %) is obtained when the SSA is underestimated and g is overestimated. The retrieval of the above-cloud AOT depends mostly on the aerosol absorption of the light reflected by the cloud. Therefore, it is expected that the retrieved AAOT is less sensitive to the absorbing property of the aerosol than the AOT. The sensitivity of the AAOT to the assumed aerosol properties is shown in Figs. 9b and 10b. The uncertainty in the AAOT due to an error in g is similar to the uncertainty in the AOT (<5 %). However, the influence of the SSA assumption alone on the AAOT is smaller than the influence on the AOT, with differences of 1.9 % and −8.7 %. This means that a perturbation of the SSA primarily impacts the scattering AOT. The largest overestimation of the AAOT (2.7 %) is obtained when the assumed aerosol model overestimates g. An underestimation of the SSA and an overestimation of g lead to the largest underestimation of the AAOT (−5.1 %).
Figure 10Similar to Fig. 9 but for the combined impact and the SSA.
The variation in the solar zenith angle, and therefore in the satellite observation geometry during the day, can impact the sensitivity of the retrieval to the aerosol assumptions. Therefore, the 15 min SEVIRI observations for 28 August have been processed using the eight aerosol models described above and compared to the aerosol and cloud properties retrieved with the CLARIFY aerosol model. The difference Δxi of a product x is defined as
$\mathrm{\Delta }{x}_{i}=\phantom{\rule{0.125em}{0ex}}\left({x}_{\mathrm{CLARIFY}}-{x}_{i}\right)/{x}_{i}×\mathrm{100}\phantom{\rule{0.125em}{0ex}}\mathit{%},$
where xCLARIFY and xi is the mean product x retrieved over the SEVIRI slot with the aerosol CLARIFY model and the modified model i respectively. Figure 11 shows the time series of ΔAOT (a), ΔAAOT (b), ΔCOT (c) and ΔCER (d) obtained with the modified aerosol models. The sensitivity of the retrieved cloud properties to the aerosol model assumptions remains small (lower than 5.6 % for the COT and 2.6 % for the CER) and dominated by the sensitivity to g. Apart from a small decrease in ΔCOT at midday when g is overestimated (solid blue line) and an increase in ΔCOT in late afternoon when the SSA is underestimated (solid red line), no significant trend is observed in the cloud property sensitivities. As observed previously, the uncertainty on the AOT is led by the SSA assumption, with the AOT being overestimated (respectively underestimated) when the assumed SSA is overestimated (respectively underestimated). Until 15:00, ΔAOT stays within ±40 %, with the sensitivity to the SSA being slightly larger at midday. Then it increases up to 60 % when the SSA is overestimated and g is underestimated (dashed blue line). Similar trends are observed in ΔAAOT, with generally lower values than ΔAOT. An increase in the uncertainty is observed on the AAOT after 15:00, which reaches up to 27 % at 16:30. Before 15:00, there is a larger AAOT sensitivity to the SSA around midday (+8.9 % −15.2 %), but there is no evident evolution of the sensitivity to g with time. The case that leads to the largest biases on the AAOT is when the SSA is underestimated and g overestimated (dashed green lines), with an underestimation of up to 23 %. However, it should be noted that 0 % of the AERONET observations used in Fig. 8 are associated with an SSA lower than SSACLARIFYσSSA and a g larger than gCLARIFYσg. Otherwise, the sensitivity of the AAOT to the aerosol property assumptions stays between −16.6 and +9 % before 15:00.
Figure 11Time series (UTC) of the difference Δ (%) of the above-cloud AOT (a), AAOT (b), COT (c) and CER (d) retrieved with the CLARIFY model and the modified aerosol models for 28 August 2017.
In conclusion, the retrieved AOT is less sensitive to the aerosol property assumption before 15:00, with an uncertainty of 40 %. This uncertainty is dominated by the sensitivity of the retrieval to the SSA. An overestimation (respectively underestimation) of the AOT is expected when the observed aerosols are more (respectively less) absorbing than the aerosol model assumed for the retrieval. A better accuracy is obtained on the retrieved AAOT, with an uncertainty generally lower than 17 % before 15:00. The sensitivity of the cloud properties to the aerosol model assumption remains small all day long, with an uncertainty of 5.6 % on the COT and 2.6 % on the CER.
Figure 12Above-cloud AOT retrieved on 5 September 2017 at 11:42, 12:12 and 12:42 UTC. The red square represents the area over which the SEVIRI products have been averaged.
4 Assessing the stability of the retrieval
One of the major benefits from using SEVIRI is the ability to track both aerosol and cloud events at high temporal resolution. Therefore, it is important to evaluate how consistent the retrieval is over time. For that purpose, 2 d of continuous observations (i.e. 5 and 6 September 2017) have been analysed and the retrieved properties have been averaged over 20 and 10 S and 5 and 15 E, which correspond to the red square on the maps in Fig. 12. The above-cloud AOT, COT and CER time series are presented in Fig. 13a, b, c. The studied area is located next to the coast, where the AOT is typically the highest. The above-cloud AOT is around 0.66 and 0.72 for 5 and 6 September respectively. As expected, the transport of the aerosol plume from east to west is slow, resulting in a small evolution of the above-cloud AOT. On both days, a peak is observed at 12:12 with an anomaly larger than the AOT variability. This localized discontinuity in the above-cloud AOT is shown in the 11:42, 12:12 and 12:42 UTC maps for 5 September 2017 in Fig. 12. The evolution of the cloud properties is slightly more complex. A small decrease is observed in both the COT and CER until 14:00. After 15:00, both properties sharply increase. The clouds are strongly affected by the diurnal cycle and a shoaling of the cloud cover is expected from early morning to late afternoon. As the thinnest clouds vanish, the cloud fraction decreases together with the number of retrievals in the area. This results in a larger contribution of the thickest clouds to the mean value in the late afternoon. As for the above-cloud AOT, large variations in the CER are observed around noon. At that time, the sun and the satellite are almost aligned and the scattering angle (Fig. 13d) reaches values larger than 175, which corresponds to the region where the glory phenomenon is typically observed. Several reasons can explain why the retrieval does not perform well in backscattering direction. The first one is the uncertainty in the LUT due to the truncation of the cloud phase function. Although the TMS correction gives good results, biases still remain in the glory aureole (Iwabushi and Suzuki, 2009). Also, the radiances in the glory are more sensitive to the cloud droplet microphysics (Mayer et al., 2004). The assumption on the variance of the droplet size distribution may induce biases in the retrieval. Therefore, the accuracy of the retrieval cannot be guaranteed within the glory aureole and these observations should be discarded. In Fig. 13, the time spans corresponding to the MODIS Aqua and Terra overpasses in the region are highlighted in orange. This shows that MODIS measurements are typically performed before and after SEVIRI observes the glory backscattering over the SEAO, usually allowing comparisons between these instruments.
Figure 13Time series (UTC) of the above-cloud AOT (a), COT (b), CER (c) and scattering angle (d) averaged between 20 and 10 S and 5 and 15 E for 5 and 6 September 2017. The grey area represents scattering angles larger than 175 and the orange areas show the typical overpass times of MODIS Aqua and Terra over the region.
The performance of the algorithm is further assessed by evaluating the stability of the retrieved above-cloud AOT at pixel level. As noted by Chang and Christopher (2016), in this region over these scales, aerosols are expected to have a limited temporal variability and the variation in the above-cloud AOT is expected to be small between t=0 and t±15 min. The differences between the AOT retrieved at t=0 and the running mean estimated between t−15 and t+15 min have been calculated at pixel level for observations between 09:00 and 15:00 UTC, removing measurements within the glory backscattering region. Figure 14 shows the histogram of the AOT differences calculated over a 12 d period (1 to 12 September 2017). The differences follow a normal distribution centred around 0.0 with a standard deviation of 0.1. This short-term variability can be attributed to several sources of uncertainties, such as the total amount of water vapour, its vertical distribution, the retrieved cloud top height and the numerical fitting procedure. This analysis indicates that the retrieval of the above-cloud AOT remains relatively stable, with an observed variability of ±0.1 between consecutive observations. Except for the glory backscattering, the stability observed on the retrieved aerosol and cloud properties reinforces the reliability of the algorithm.
Figure 14Histogram of the difference between AOT retrieved at t=0 and the running mean calculated between t−15 and t+15 min from 1 to 12 September 2017. Observations within the glory region have been removed. Dashed lines represent the mean ± the standard deviation.
5 Conclusions
Recently, progress has been made in the remote-sensing field in order to fill the lack of aerosol above-cloud observations. Techniques have been developed to retrieve aerosol and cloud properties over the SEAO from passive remote-sensing instruments. These algorithms take advantage of the colour-ratio effect (Jethva et al., 2013), which is the spectral contrast produced by the aerosol absorption above clouds. Although OMI (Torres et al., 2012), MODIS (Jethva et al., 2013; Meyer et al., 2015) and POLDER (Peers et al., 2015) already provide useful information about aerosols above clouds, these instruments are on polar-orbiting satellites and their low temporal resolutions prevent monitoring the diurnal variation in the cloud cover and in the DRE of aerosols over the SEAO. For the first time, we have applied a similar algorithm to geostationary measurements from the SEVIRI instrument, which has a repeat cycle of 15 min. The method consists of a LUT approach, using the channels at 0.64, 0.81 and 1.64 µm in order to simultaneously retrieve the above-cloud AOT, COT and CER.
Compared to other satellite instruments, the SEVIRI measurements are more sensitive to the absorption from atmospheric gases because of their wider spectral bands. Therefore, an efficient atmospheric correction scheme is essential in order to separate the absorption from the aerosols and from the atmospheric gases. Atmospheric transmittances are calculated with the fast radiative transfer model RTTOV based on the cloud top height observed by SEVIRI and the forecasted water vapour profiles from the Met Office Unified Model. The water vapour correction has the largest impact on the above-cloud aerosol retrieval. The impact of errors in the atmospheric correction has been evaluated by modulating the humidity profile for a case study. A positive bias of both the AOT and the COT is observed when the water vapour is overestimated, and vice versa. On average, an 18.5 % bias on the AOT and a 5.5 % bias on the COT are expected for a 10 % error on the water vapour profile. Although a good accuracy is expected from the forecast model, this limitation should be kept in mind when utilizing or further developing SEVIRI products. In the companion paper, the humidity from the forecast will be compared against the dropsonde measurements from the CLARIFY-2017 campaign.
The choice of the aerosol model used to produce the LUT is also a key feature of the method. In situ measurements of aerosols above clouds have been performed off the coast of Ascension Island during the CLARIFY-2017 field campaign. An aerosol model optimized for the SEVIRI spectral bands has been obtained by analysing the vertical profiles of extinction and absorption from EXSCALABAR together with the size distribution from a PCASP. A bimodal lognormal distribution has shown to adequately reproduce the observations. A fine-mode radius of 0.12 µm has been obtained, which is in good agreement with the biomass burning measured over the SEAO during SAFARI 2000 (Haywood et al., 2003). The refractive index has been evaluated at 1.51–0.029i. The corresponding SSA of 0.85 at 0.55 µm is consistent with both in situ and remote-sensing observations of African biomass burning aerosols (Johnson et al., 2008; Sayer et al., 2014). In addition to the uncertainty associated with the estimation of the aerosol model, a seasonal dependence is expected in the biomass burning properties as well as modifications due to ageing processes during their transport over the SEAO. We have evaluated the impact of applying a single model assumption on both aerosol and cloud properties. Retrievals have been performed considering aerosol models with modified SSA and asymmetry factor g. It has been shown that the sensitivity of the retrieved cloud properties to the aerosol model assumption is small, with errors lower than 5.6 % on the COT and 2.6 % on the CER. As expected the impact of the assumed aerosol properties is much larger on the above-cloud AOT, with an uncertainty estimated at 40 % before 15:00 UTC. This uncertainty is led by the sensitivity of the retrieval to the SSA. Because the method relies on the impact of the aerosol absorption on the light reflected by the clouds, the perturbation of the SSA has primarily an impact on the scattering contribution of the AOT. Therefore, a better accuracy is obtained on the retrieved AAOT, with biases generally lower than 17 % before 15:00 UTC. After that time, an increase in the uncertainty on both the AOT and the AAOT has been observed, and users are advised to be careful when using the late afternoon aerosol product. For any satellite retrievals based on the colour-ratio technique, aerosol properties, including the SSA, have to be assumed and the same order of magnitude can be expected on the sensitivity of their AOT. This analysis highlights the importance of a suitable constraint on the SSA.
Despite the wider channels and the narrower spectral range of SEVIRI, it has been demonstrated that the geostationary instrument has the potential to detect and quantify the absorbing aerosol plumes transported above the clouds of the SEAO. Except from observations within the glory backscattering for which the retrieval has shown to be unstable, a good consistency has been observed on the aerosol and cloud properties. The stability of the results during the day is promising for future uses of the SEVIRI algorithm. In the companion paper, the reliability of the retrieved aerosol and cloud properties will be further assessed by analysing the consistency with the MODIS retrievals and comparing with direct measurements from the CLARIFY-2017 field campaign. The potential of such a retrieval is obvious. The 15 min resolution will aid in tracking the fate of above-cloud biomass burning aerosol and will prove invaluable for assessing models of the emission, transport and deposition of biomass burning aerosol, with implications for accurate determination of the direct radiative effects of biomass burning aerosol at high temporal resolution.
Data availability
Data availability.
The data used for this study are available from the corresponding author, FP, upon reasonable request.
Author contributions
Author contributions.
FP, PF and JMH developed the concept and the ideas for this paper. PF implemented the atmospheric correction scheme and FP the retrieval algorithm. CF, SJA, KS, MIC, NWD and JMH operated, calibrated and prepared the in situ measurements from EXSCALABAR and the PCASP. The reliability of the retrieved products was analysed throughout the development of the algorithm with the help of KGM and SEP. FP carried out the analysis and prepared the paper with contributions from all co-authors.
Competing interests
Competing interests.
The authors declare that they have no conflict of interest.
Special issue statement
Special issue statement.
This article is part of the special issue “New observations and related modelling studies of the aerosol–cloud–climate system in the Southeast Atlantic and southern Africa regions (ACP/AMT inter-journal SI)”. It is not associated with a conference.
Acknowledgements
Acknowledgements.
We thank the Natural Environment Research Council (NERC) and the Norwegian Research Council for their financial support. Airborne data were obtained using the BAe-146 Atmospheric Research Aircraft operated by Directflight Ltd. and managed by Facility for Airborne Atmospheric Measurements (FAAM), which is jointly supported by NERC and the Met Office. The authors acknowledge the dedicated work of FAAM and Directflight during the aircraft campaign. We thank the AERONET PIs, Paola Formenti, Derek Griffith, Brent Holben, Nichola Knox, Gillian Maggs-Kölling, Stuart Piketh, Carlos Ribeiro, Venkataram Sivakumar and Rick Wagener for their efforts in establishing and maintaining the Ascension Island, Bethlehem, Bonanza, DRAGON Henties, Durban UKZN, Elandsfontein, Etosha Pan, Gobabeb, Gorongosa, Henties Bay, HESS, Huambo, Inhaca, Joberg, Kaoma, Loskop Dam, Lubango, Maun Tower, Mongu, Mwinilunga, Namibe, Ndola, Paardefontein, Pietersburg, Possession Island, Potchefstroom, Pretoria CSIR-DPSS, Senanga, Sesheke, Skukuza, Solwezi, Swakopmund, Tsumkwe, Upington, Walvis Bay airport, Windhoek-NUST, Windpoort and Wits University sites.
Financial support
Financial support.
This research has been supported by the Natural Environment Research Council (NERC) via the CLARIFY project (grant no. NE/L013479/1) and the Research Council of Norway via the projects AC/BC (grant no. 240372) and NetBC (grant no. 244141).
Review statement
Review statement.
This paper was edited by Nikolaos Mihalopoulos and reviewed by Ian Chang and one anonymous referee.
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Watts, P., Mutlow, C., Baran, A., and Zavody, A.: Study on cloud properties derived from Meteosat Second Generation observations, EUMETSAT ITT, no 97/181, 344 pp., 1998.
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# Sub-center ArcFace: Boosting Face Recognition by Large-scale Noisy Web Faces
Margin-based deep face recognition methods (e.g. SphereFace, CosFace, and ArcFace) have achieved remarkable success in unconstrained face recognition. However, these methods are susceptible to the massive label noise in the training data and thus require laborious human effort to clean the datasets. In this paper, we relax the intra-class constraint of ArcFace to improve the robustness to label noise. More specifically, we design $K$ sub-centers for each class and the training sample only needs to be close to any of the $K$ positive sub-centers instead of the only one positive center. The proposed sub-center ArcFace encourages one dominant sub-class that contains the majority of clean faces and non-dominant sub-classes that include hard or noisy faces. Extensive experiments confirm the robustness of sub-center ArcFace under massive real-world noise. After the model achieves enough discriminative power, we directly drop non-dominant sub-centers and high-confident noisy samples, which helps recapture intra-compactness, decrease the influence from noise, and achieve comparable performance compared to ArcFace trained on the manually cleaned dataset. By taking advantage of the large-scale raw web faces (Celeb500K), sub-center Arcface achieves state-of-the-art performance on IJB-B, IJB-C, MegaFace, and FRVT.
PDF Abstract
## Results from the Paper Add Remove
Submit results from this paper to get state-of-the-art GitHub badges and help the community compare results to other papers.
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# What are the meanings of theta and gamma in this equation?
I'm having a bit of trouble understanding the notation from this paper: http://graphics.cs.aueb.gr/graphics/docs/papers/GraphiCon09_PapadopoulosPapaioannou.pdf
float3 surface = exp(dot(-y, d)) * fromSurface;
float3 viewer = exp(dot(-y, d)) * fromViewer;
float3 final = dot(photonIntensity, dot(mie(0), dot(viewer, surface));
I know I = intensity, mie is mie scattering, but what does the wierd 0 stand for? And what is y? Is d = distance?
Second go at it:
float Attenuation = 130.0;
float MieAnisotropy = 0.7;
float MiePhase(float CosTheta, float Anisotropy)
{
const float F = 1.0 / (4.0 * PI);
return F * ((1.0 - pow(Anisotropy, 2.0)) / pow(1.0 - 2.0 * Anisotropy * CosTheta + pow(Anisotropy, 2.0), 1.5));
}
float3 ViewPosition = normalize(WorldPosition - CameraPosition);
float CosViewSunAngle = dot(ViewPosition, SunDirection);
float Mie = MiePhase(CosViewSunAngle, MieAnisotropy);
float FromSurface = exp(dot(-Attenuation, DistanceFromSurface));
float FromViewer = exp(dot(-Attenuation, DistanceFromViewer));
float PhotonIntensity = 1.0;
float Final = dot(PhotonIntensity, dot(Mie, dot(FromViewer, FromSurface));
• The weird 0 is a Greek letter theta. The weird y is a Greek letter gamma. The paper should tell you what they are but I'm not going to read it.... – immibis Jan 30 '18 at 2:17
Immediately after the first appearance of the equation in the paper, the use of $\theta$ is described as
The use of $\gamma$ is first described just above; it
"Medium" in this sense means the substance through which the photon is traveling (not "medium" as in "low, medium, high"), so this is basically a constant for tweaking. Later in the paper, an example value for $\gamma$ is given (in the "Implementation Details" section).
And yes, $d$ is distance from the respective points.
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# Chapter 9 Magnetic Fields Due to Currents
Learning Objectives:
In this chapter you will basically learn:
$$\bullet$$ Learn that for a given point near a wire and a given current-element in the wire, how to determine the magnitude and direction of the magnetic field due to that element.
$$\bullet$$ Learn the the direction of the field vector using a right-hand rule for a point to one side of a long straight wire carrying current.
$$\bullet$$ Learn the Biot–Savart law to find the magnetic field set up at the point by the current.
$$\bullet$$ Learn how parallel currents attract each other, and antiparallel currents repel each other.
$$\bullet$$ Learn how to apply Ampere’s law, for determining the algebraic sign of an encircled current using a right-hand rule .
$$\bullet$$ Learn how to apply Ampere’s law to a long straight wire with current, to find the magnetic field magnitude inside and outside the wire.
$$\bullet$$ Learn how to apply Ampere’s law to find the magnetic field inside a solenoid.
$$\bullet$$ Learn what is a solenoid and how its internal magnetic field B, the current i, and the number of turns per unit length n of the solenoid are related.
$$\bullet$$ Learn how to apply Ampere’s law to find the magnetic field inside a toroid.
$$\bullet$$ Learn what is a toroid and how its internal magnetic field B, the current i, the radius r, and the total number of turns N are related.
$$\bullet$$ Learn how in a current-carrying coil, the dipole moment magnitude $$\mu$$ and the coil’s current i, number of turns N, and area per turn A are related.
$$\bullet$$ Learn how in a point along the central axis, magnetic field magnitude B, the magnetic dipole moment $$\mu$$, and the distance z from the center of the coil are related.
## 9.1 Magnetic Fields Due to Currents:
The magnitude of the field $$d\vec B$$ produced at point P at distance r by a current-length element $$id\vec s$$ turns out to be
$$$dB = \frac{\mu_0}{4\pi}\frac{ids~sin\theta}{r^2} \tag{9.1}$$$
where $$\theta$$ is the angle between the directions of $$d\vec s$$ and $$\hat r$$, a unit vector that points from ds toward P. Symbol $$\mu_0$$ is a constant, called the permeability constant, whose value is defined to be exactly
$$$\mu_0=4\pi\times10^{-7}~\frac{T\cdot m}{A}\approx1.26\times10^{-6}\frac{T\cdot m}{A} \tag{9.2}$$$
The direction of $$d\vec B$$, shown as being into the page in Fig. 9-1, is that of the cross product $$d\vec s\times \hat r$$.We can therefore write Eq. 9-1 in vector form as
$$$d\vec B =\frac{\mu_0}{4\pi}\frac{id\vec s\times\hat r}{r^2}~~~(Biot–Savart~law) \tag{9.3}$$$
This vector equation and its scalar form, Eq. 9-1, are known as the law of Biot and Savart. The law, which is experimentally deduced, is an inverse-square law. We shall use this law to calculate the net magnetic field produced at a point by various distributions of current.
## 9.2 Magnetic Field Due to a Current in a Long Straight Wire:
Figure 9-2, shows a straight wire of infinite length, for which we want to find the field $$\vec B$$ at point P, a perpendicular distance R from the wire. The magnitude of the differential magnetic field produced at P by the current-length element $$id\vec s$$ located a distance r from P is given by Eq. 9-1:
$$$dB = \frac{\mu_0}{4\pi}\frac{ids~sin\theta}{r^2} \tag{9.4}$$$
The direction of in Fig. 9-2 is that of the vector $$(d\vec s\times \vec r)$$, which is directly into the page. We can find the magnitude of the magnetic field produced at P by the current-length elements in the upper half of the infinitely long wire by integrating dB in Eq. 9-1 from 0 to $$\infty$$. Similarly, the magnetic field produced by the lower half of the wire is exactly the same as that produced by the upper half. To find the magnitude of the total magnetic field $$\vec B$$ at P, we need only multiply the result of our integration by 2. We get
$$$B=2\int_{0}^{\infty}dB=\frac{\mu_0i}{2\pi}\int_{0}^{\infty}\frac{sin\theta~ds}{r^2} \tag{9.5}$$$
The variables $$\theta$$, s, and r in this equation are not independent; Fig. 9-2 shows that they are related by
$$r=\sqrt {s^2+R^2}$$ and $$sin\theta=sin(\pi-\theta)=\frac{R}{\sqrt {s^2+R^2}}$$.
Using standard integral $$\int\frac{dx}{({x^2+a^2})^\frac{3}{2}}=\frac{x}{ a^2\sqrt {(x^2+a^2)}}$$, we can get
$$$B =\frac{\mu_0i}{2\pi}\int_{0}^{\infty}\frac{R~ds}{(s^2+R^2)^{3/2}}=\frac{\mu_0i}{2\pi R} \left[\frac{s}{\sqrt{s^2+R^2}}\right]_{0}^{\infty}=\frac{\mu_0i}{2\pi R} \tag{9.6}$$$
Note that the magnetic field at P due to either the lower half or the upper half of the infinite wire in Fig. 9-2 is half this value; that is,
$$$B =\frac{\mu_0i}{4\pi R}~~~(semi-infinite~straight~wire). \tag{9.7}$$$
Magnetic Field Due to a Current in a Circular Arc of Wire: Figure 9-3a shows such an arc-shaped wire with central angle $$\phi$$, radius R, and center C, carrying current i. At C, each current-length element $$id\vec s$$ of the wire produces a magnetic field of magnitude dB given by Eq. 9-4. Moreover, as Fig. 9-3b shows, no matter where the element is located on the wire, the angle $$\theta$$ between the vectors $$d\vec s$$ and $$\vec r$$is $$90^0$$; also, r = R. Thus, by substituting R for r and $$90^0$$ for $$\theta$$ in Eq. 9-4, we obtain
$$$dB = \frac{\mu_0}{4\pi}\frac{ids}{r^2} \tag{9.8}$$$
Now in Eq. 9-8 we substitute $$ds = Rd\phi$$ and then by integrating we get an expression for an arbitrary arc of a circular current carrying ring as follows:
$B=\int dB=\int_{0}^{\phi}\frac{\mu_0}{4\pi}\frac{iRd\phi}{R^2}= \frac{\mu_0i}{4\pi R}\int_{0}^{\phi}d\phi$
Integrating over the circular arc, we get
$$$B=\frac{\mu_0i}{4\pi R}\int_{0}^{\phi}d\phi=\frac{\mu_0i\phi}{4\pi R}~~~(at~center~of~ circular~arc). \tag{9.9}$$$
As an example, for a full circular current carrying conductor we can find the magnetic field at the center by substituting $$2\pi$$ rad for $$\phi$$ in Eq. 9-9, we get
$$$B=\frac{\mu_0i2\pi}{4\pi R}= \frac{\mu_0i}{2R}~~~(at~center~of~ full~circle). \tag{9.10}$$$
## 9.3 Force Between Two Parallel Currents:
The force between two long, straight, and parallel conductors separated by a distance d can be found by applying the result in Eq. 8-20 $$(\vec F_B = i\vec L \times \vec B)$$. Figure 9.4 shows the wires, their currents, the magnetic field created by wire 1, and the force it exerts on wire 2 (call the force $$\vec F_{21}$$. The field due to $$i_1$$ at a distance d is
$$$\vec F_{21}= i_2\vec L \times \vec B_1 \tag{9.11}$$$
$$$\vec F_{21}= i_2LB_1\sin90^0=\frac{\mu_0Li_1i_2}{2\pi d} \tag{9.12}$$$
The direction of $$\vec F_{21}$$ is the direction ofthe cross-product of $$\vec L\times \vec B_1$$ and is directed towards the wire 1, as shown in Fig. 9-4.
## 9.4 Ampere’s Law:
Over an arbitrary closed path,
$$$\oint\vec{\mathbf{B}}\cdot d\vec{\mathbf{s}}=\mu_0i_{enc}~~~(Ampere’s~Law) \tag{9.13}$$$
The line integral in this equation is evaluated around a closed loop called an Amperian loop. The current i on the right side is the net current encircled by the loop.
EXAMPLE 9.1 Using Ampère’s law to calculate the magnetic field due to a steady current i in an infinitely long, thin, straight wire as shown in Figure 9.5.
Figures shows an infinitely long, thin, straight wire with the current directed into the page. The radial and polar components of magnetic field, $$B_r$$ and $$B_{\Theta}$$ respectively, are shown at arbitrary points on a circle of radius r centered on the wire. The possible components of the magnetic field B due to a current i, which is directed into the page. The radial component is zero because the angle between the magnetic field and the path is at a right angle.
Solution: Over this path $$\vec B$$ is constant and is parallel to $$d\vec s$$, so
$\oint\vec B\cdot d\vec s = B_{\theta}\oint ds = B_{\theta}(2\pi r)$
Thus Ampère’s law reduces to
$B_{\theta}(2\pi r)=\mu_0i$
Finally, since $$B_{\theta}$$ is the only component of $$\vec B$$ we can drop the subscript and write
$B=\frac{\mu_0i}{2\pi r}~~~(Answer)$
This agrees with the result we got using Biot-Savart law in Eq.9-6.
Magnetic Field Inside a Long Straight Wire with Current: Figure 9-6 shows the cross section of a long straight wire of radius R that carries a uniformly distributed current i directly out of the page.
The magnetic field created by the uniform current must be cylindrically symmetrical. To calculate the magnetic field at points inside the wire, we can apply an Amperian loop of radius r, as shown in Fig. 9-6, where now $$r < R$$. Magnetic field $$\vec B$$ (Eq. 9-13) is tangent to the loop, as shown; so the left side of Ampere’s law yields
$$$\oint\vec B\cdot d\vec s = \oint B~cos\theta~ ds=B\oint ds=B(2\pi r) \tag{9.14}$$$
The current $$i_{enc}$$ encircled by the loop is proportional to the area encircled by the loop; that is,
$$$i_{enc}=\frac{\pi r^2}{\pi R^2} \tag{9.15}$$$
Now applying Ampere’s law we can find the magnetic inside the conductor
$B(2\pi r) = \mu_0 i\frac{\pi r^2}{\pi R^2}$
$$$B = \left(\frac{\mu_0 i}{2\pi R^2}\right)r~~~(magnetic~field~inside~the~straight~wire). \tag{9.16}$$$
Thus, inside the wire, the magnitude B of the magnetic field is proportional to r , is zero at the center, and is maximum at r = R (at the surface).
## 9.5 Solenoids:
Figure 9-7-a shows the lines of for a real solenoid and Figure 9-7-b shows an ideal solenoid. The spacing of these lines in the central region shows that the field inside the coil is fairly strong and uniform over the cross section of the coil. The external field, however, is relatively weak.
Let us now apply Ampere’s law,
$$$\oint\vec{\mathbf{B}}\cdot d\vec{\mathbf{s}}=\mu_0i_{enc}~~~(Ampere’s~Law) \tag{9.17}$$$
to the ideal solenoid of Fig. 9-7 (b), where $$\vec B$$is uniform within the solenoid and zero outside it, using the rectangular Amperian loop abcda. We write $$\oint\vec{\mathbf{B}}\cdot d\vec{\mathbf{s}}$$as the sum of four integrals, one for each loop segment:
$$$\oint\vec{\mathbf{B}}\cdot d\vec{\mathbf{s}}= \int_{a}^{b}\vec{\mathbf{B}}\cdot d\vec{\mathbf{s}} + \int_{b}^{c}\vec{\mathbf{B}}\cdot d\vec{\mathbf{s}} +\int_{c}^{d}\vec{\mathbf{B}}\cdot d\vec{\mathbf{s}} +\int_{d}^{a}\vec{\mathbf{B}}\cdot d\vec{\mathbf{s}} \tag{9.18}$$$
The first integral on the right of Eq. 9-18 is Bh, where B is the magnitude of the uniform field $$\vec B$$inside the solenoid and h is the (arbitrary) length of the segment from a to b. The second and fourth integrals are zero because for every element ds of these segments $$\vec B$$, either is perpendicular to ds or is zero, and thus the product $$\vec B.d\vec s$$is zero.The third integral, which is taken along a segment that lies outside the solenoid, is zero because B = 0 at all external points. Thus, $$\vec{\mathbf{B}}\cdot d\vec{\mathbf{s}}$$for the entire rectangular loop has the value Bh.
The net current $$i_{enc}$$ encircled by the rectangular Amperian loop in Fig. 9-7 is not the same as the current i in the solenoid windings because the windings pass more than once through this loop. Let n be the number of turns per unit length of the solenoid; then the loop encloses nh turns and $i_{enc}=i(nh)$ Ampere’s law then gives us $Bh=\mu_0 inh$ $$$B = \mu_0 in~~~(ideal~solenoid). \tag{9.19}$$$
An application of solenoid is the a magnetic resonance imaging (MRI). In an MRI scan, the person lies down on a table that is moved into the center of a large solenoid that can generate very high magnetic fields that is being created by passing high currents flowing through the superconducting wires. The large magnetic field is used to change the spin of protons in the patient’s body. The time it takes for the spins to align or relax (return to original orientation) is a signature of different tissues that can be analyzed to see if the structures of the tissues is normal (Figure 9.8).
EXAMPLE 9.2
A solenoid has 300 turns wound around a cylinder of diameter $$1.20~\mathrm{cm}$$ and length $$14.0~\mathrm{cm}$$. If the current through the coils is $$0.410~\mathrm{A}$$, what is the magnitude of the magnetic field inside and near the middle of the solenoid?
Solution: The number of turns per unit length is
$n=\frac{300~\mathrm{turns}}{0.140~\mathrm{m}}=2.14\times10^3~\mathrm{turns/m}.$
The magnetic field produced inside the solenoid is
$\begin{eqnarray*}B&=&\mu_0nI=(4\pi\times10^{-7}~\mathrm{T}\cdot\mathrm{m/A})(2.14\times10^3~\mathrm{turns/m})(0.410~\mathrm{A})\\&=&1.10\times10^{-3}~\mathrm{T}.\end{eqnarray*}$
## 9.6 Toroids:
Figure 9-9a shows a toroid, which we may describe as a (hollow) solenoid that has been curved until its two ends meet, forming a sort of hollow bracelet.What magnetic field is set up inside the toroid (inside the hollow of the bracelet)? We can find out from Ampere’s law and the symmetry of the bracelet.
From the symmetry, we see that the lines of $$\vec B$$ form concentric circles inside the toroid, directed as shown in Fig. 9-9b. Let us choose a concentric circle of radius r as an Amperian loop and traverse it in the clockwise direction. Using Ampere’s law we get
$(B)(2\pi r)=\mu_0iN$ $$$B = \frac{\mu_0 iN}{2\pi}\frac{1}{r}~~~(Toroid). \tag{9.20}$$$
In Eq. 9-20,one can see that B is not constant over the cross section of a toroid.
## 9.7 A Current-Carrying Coil as a Magnetic Dipole:
In the last few sections we studied the magnetic fields produced by current in a long straight wire, a solenoid, and a toroid. We found how magnetic dipole moment resulted in a current carrying loop in section 8.9, Eq. 8-24. We also briefly discussed in Example 8.2 how torque $$\vec\tau$$ tau can be created if we place it in an external magnetic field $$\vec B$$,
$$$\vec\tau = \vec\mu \times \vec B \tag{9.21}$$$
In Eq. 9-21, $$\vec\mu$$ is the magnetic dipole moment of the coil and has the magnitude NiA, where N is the number of turns, i is the current in each turn, and A is the area enclosed by each turn. The direction of the magnetic dipole moment $$\vec\mu$$is shown in Figure 9-10, can be found by applying right-handed rule placing your fingers of your right hand curl around it in the direction of the current; with your extended thumb then points in the direction of the dipole moment $$\vec\mu$$ as shown in Figure 9-10.
The magnetic field element $$d\vec B$$ has two components, $$dB_{\|}$$ along the axis of the loop and $$dB_{\perp}$$ perpendicular to this axis. From the symmetry, the vector sum of all the perpendicular components $$dB_{\perp}$$ due to all the loop elements ds is zero. This leaves only the axial (parallel) components $$dB_{\|}$$, and we have
$B=\int dB_{\|}$
The element $$d\vec s$$ as shown in in Fig. 9-11, the law of Biot and Savart (Eq. 9-3) tells us that the magnetic field at distance r is
$dB = \frac{\mu_0}{4\pi}\frac{ids\sin90^0}{r^2}$
$dB_{\|} = dB\cos\alpha$
$$$dB_{\|}=\frac{\mu_0i~\cos\alpha~ ds}{4~\pi~r^2} \tag{9.22}$$$
Figure 9-11 shows that r and $$\alpha$$ are related to each other. Let us express each in terms of the variable z, the distance between point P and the center of the loop. The relations are
$$$r=\sqrt{R^2+z^2} \tag{9.23}$$$
and
$$$\cos\alpha=\frac{R}{r}=\frac{R}{\sqrt{R^2+z^2}} \tag{9.24}$$$
Substituting Eqs. 9-23 and 9-24 into Eq. 9-22, we find
$dB_{\|}=\frac{\mu_0iR}{4\pi(R^2+z^2)^{3/2}}$ $B=\int dB_{\|}=\frac{\mu_0iR}{4\pi(R^2+z^2)^{3/2}}\int ds$ or, because $$\int~ds$$is simply the circumference $$2\pi R$$ of the loop,
$$$B(z)=\frac{\mu_0iR^2}{2(R^2+z^2)^{3/2}}. \tag{9.25}$$$
## Solved Problems Magnetic Fields Due to Currents
1. [7] In Fig. 29-39, two circular arcs have radii a = 13.5 cm and b = 10.7 cm, subtend angle $$\theta=74^\circ$$, carry current i = 0.411 A, and share the same center of curvature P. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at P?
2. [9] Two long straight wires are parallel and 8.0 cm apart. They are to carry equal currents such that the magnetic field at a point halfway between them has magnitude 300 mT. (a) Should the currents be in the same or opposite directions? (b) How much current is needed?
3. [11] In Fig. 29-42, two long straight wires are perpendicular to the page and separated by distance $$d_1$$ = 0.75 cm. Wire 1 carries 6.5 A into the page. What are the (a) magnitude and (b) direction (into or out of the page) of the current in wire 2 if the net magnetic field due to the two currents is zero at point P located at distance $$d2$$ = 1.50 cm from wire 2?
4. [13] In Fig. 29-44, point $$P_1$$ is at distance R = 13.1 cm on the perpendicular bisector of a straight wire of length L = 18.0 cm carrying current i = 58.2 mA. (Note that the wire is not long.) What is the magnitude of the magnetic field at $$P_1$$ due to i?
5. [21] Figure 29-49 shows two very long straight wires (in cross section) that each carry a current of 4.00 A directly out of the page. Distance $$d_1$$ = 6.00 m and distance $$d_2$$ = 4.00 m. What is the magnitude of the net magnetic field at point P, which lies on a perpendicular bisector to the wires?
6. [31] In Fig. 29-59, length a is 4.7 cm (short) and current i is 13 A. What are the (a) magnitude and (b) direction (into or out of the page) of the magnetic field at point P?
Section 9-2 Force Between Two Parallel Currents
1. [35] Figure 29-63 shows wire 1 in cross section; the wire is long and straight, carries a current of 4.00 mA out of the page, and is at distance $$d_1$$ = 2.40 cm from a surface. Wire 2, which is parallel to wire 1 and also long, is at horizontal distance $$d_2$$ = 5.00 cm from wire 1 and carries a current of 6.80 mA into the page. What is the x component of the magnetic force per unit length on wire 2 due to wire 1?
Section 9-3 Ampere’s Law
1. [43] Figure 29-67 shows a cross section across a diameter of a long cylindrical conductor of radius a = 2.00 cm carrying uniform current 170 A.What is the magnitude of the current’s magnetic field at radial distance (a) 0, (b) 1.00 cm, (c) 2.00 cm (wire’s surface), and (d) 4.00 cm?
Section 9-4 Solenoids and Toroids
1. [49] A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of 0.800 A. (It is made up of a square solenoid—instead of a round one as in Fig. 29-17—bent into a doughnut shape.) What is the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius?
2. [51] A 200-turn solenoid having a length of 25 cm and a diameter of 10 cm carries a current of 0.29 A. Calculate the magnitude of the magnetic field inside the solenoid.
3. [53] A long solenoid has 100 turns/cm and carries current i.An electron moves within the solenoid in a circle of radius 2.30 cm perpendicular to the solenoid axis. The speed of the electron is 0.0460c (c = speed of light). Find the current i in the solenoid.
4. [55] A long solenoid with 10.0 turns/cm and a radius of 7.00 cm carries a current of 20.0 mA. A current of 6.00A exists in a straight conductor located along the central axis of the solenoid. (a) At what radial distance from the axis will the direction of the resulting magnetic field be at $$45^\circ$$ to the axial direction? (b) What is the magnitude of the magnetic field there?
Section 9-5 A Current-Carrying Coil as a Magnetic Dipole
1. [62] In Fig. 29-75, current i = 56.2 mA is set up in a loop having two radial lengths and two semicircles of radii a = 5.72 cm and b ! 9.36 cm with a common center P. What are the (a) magnitude and (b) direction (into or out of the page) of the magnetic field at P and the (c) magnitude and (d) direction of the loop’s magnetic dipole moment?
## Problems Magnetic Fields Due to Currents
1. [8] In Fig. 29-40, two semicircular arcs have radii $$R_2$$ = 7.80 cm and $$R_1$$ = 3.15 cm, carry current i = 0.281 A, and have the same center of curvature C. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at C?
2. [10] In Fig. 29-41, a wire forms a semicircle of radius R = 9.26 cm and two (radial) straight segments each of length L = 13.1 cm. The wire carries current i = 34.8 mA. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at the semicircle’s center of curvature C?
3. [12] In Fig. 29-43, two long straight wires at separation d = 16.0 cm carry currents $$i_1$$ = 3.61 mA and $$i_2$$ = 3.00$$i_1$$ out of the page. (a) Where on the x axis is the net magnetic field equal to zero?
1. If the two currents are doubled, is the zero-field point shifted toward wire 1, shifted toward wire 2, or unchanged?
1. [24] Figure 29-52 shows, in cross section, four thin wires that are parallel, straight, and very long. They carry identical currents in the directions indicated. Initially all four wires are at distance d = 15.0 cm from the origin of the coordinate system, where they create a net magnetic field $$\vec B$$. (a) To what value of x must you move wire 1 along the x axis in order to rotate counterclockwise by $$30^\circ$$? (b) With wire 1 in that new position, to what value of x must you move wire 3 along the x axis to rotate by $$30^\circ$$ back to its initial orientation?
2. [33] Figure 29-61 shows a cross section of a long thin ribbon of width w = 4.91 cm that is carrying a uniformly distributed total current i = 4.61 $$\mu A$$ into the page. In unit-vector notation, what is the magnetic field $$\vec B$$ at a point P in the plane of the ribbon at a distance d = 2.16 cm from its edge? (Hint: Imagine the ribbon as being constructed from many long, thin, parallel wires.)
Section 9-2 Force Between Two Parallel Currents
1. [41] In Fig. 29-66, a long straight wire carries a current $$i_1$$ = 30.0 A and a rectangular loop carries current $$i_2$$ = 20.0 A. Take the dimensions to be a = 1.00 cm, b = 8.00 cm, and L = 30.0 cm. In unitvector notation, what is the net force on the loop due to $$i_1$$?
Section 9-3 Ampere’s Law
1. [48] In Fig. 29-71, a long circular pipe with outside radius R ! 2.6 cm carries a (uniformly distributed) current i = 8.00 mA into the page. A wire runs parallel to the pipe at a distance of 3.00R from center to center. Find the (a) magnitude and (b) direction (into or out of the page) of the current in the wire such that the net magnetic field at point P has the same magnitude as the net magnetic field at the center of the pipe but is in the opposite direction.
Section 9-4 Solenoids and Toroids
1. [50] A solenoid that is 95.0 cm long has a radius of 2.00 cm and a winding of 1200 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.
2. [52] A solenoid 1.30 m long and 2.60 cm in diameter carries a current of 18.0 A. The magnetic field inside the solenoid is 23.0 mT. Find the length of the wire forming the solenoid.
3. [54] An electron is shot into one end of a solenoid. As it enters the uniform magnetic field within the solenoid, its speed is 800 m/s and its velocity vector makes an angle of $$30^\circ$$ with the central axis of the solenoid. The solenoid carries 4.0 A and has 8000 turns along its length. How many revolutions does the electron make along its helical path within the solenoid by the time it emerges from the solenoid’s opposite end? (In a real solenoid, where the field is not uniform at the two ends, the number of revolutions would be slightly less than the answer here.)
Section 9-5 A Current-Carrying Coil as a Magnetic Dipole
1. [56] Figure 29-72 shows an arrangement known as a Helmholtz coil. It consists of two circular coaxial coils, each of 200 turns and radius R = 25.0 cm, separated by a distance s = R.The two coils carry equal currents i = 12.2 mA in the same direction. Find the magnitude of the net magnetic field at P, midway between the coils.
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# Equivalence of definitions of cartesian morphisms
Let $p: C\to D$ be a functor, and let $f:y\to x$ be a morphism of $C$. We say that $f$ is cartesian if the canonical map $Q:(C\downarrow f) \to P:=(C\downarrow x)\times_{(D\downarrow p(x)} (D\downarrow p(f))$ is a surjective (on objects) equivalence of categories. However, if we write out what the (strict 2-) pullback means, the objects are precisely the pairs of morphisms $g: z\to x$ and $h:p(z)\to p(y)$ such that $p(g)=p(f) \circ h$. If we look at the fibres of $Q$ over objects of $A$, we see that that they are contractible groupoids.
Using the more common definition of a cartesian morphism, we must show that any pair of morphisms $(g, h)$ as above uniquely determines an arrow $\ell:z\to y$ such that $f\circ \ell= g$ and $p(\ell)=h$.
I see how the first definition implies the existence of such a map, but how does it determine the map's uniqueness (up to more than a contractible space of choices)?
-
The only morphisms in the fibers of $Q$ are identity maps, so it is actually an isomorphism of categories. To see this, suppose $\ell,\ell'\colon z\to y$ both induce $g\colon z\to y$. What would a morphism from $\ell$ to $\ell'$ in the fiber of $Q$ be? It would be a morphism $\varphi\colon z\to z$ over $y$ (the first $z$ is over $y$ via $\ell$ and the second via $\ell'$) which induces the identity morphism on $g$ in $(C\downarrow x)$. But $\varphi$ induces the morphism
φ
z ----> z
\ /
g\ /g
v v
x
The only way this is the identity morphism of $g$ is if $\varphi=id_z$.
-
That's what I thought as well, but the nLab says that it only needs to be a surjective equivalence (a trivial fibration in the natural model structure). ncatlab.org/nlab/show/Cartesian+morphism . In fact, I asked a similar question earlier on the nForum where I raised the same issue math.ntnu.no/~stacey/Vanilla/nForum/… . I agree with you, but maybe someone can explain why the definition on the nForum is so strange? Does it give a good definition for weakly cartesian morphisms? – Harry Gindi Aug 29 '10 at 8:38
I think, as you say, the motivation for the strange phrasing is to weaken well. As Anton shows, the two versions are equivalent in this setting — “surjective equivalence” implies “isomorphism” — so just looking at this case, it'd seem more natural to use the simpler phrasing of “isomorphism”. But if in higher-dimensional/weakened generalisations the general concept needed is “trivial fibration”, that would be an argument for using it here, slightly harder to understand but more principled. – Peter LeFanu Lumsdaine Aug 29 '10 at 17:58
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## Data Presentation
Data is only as good as how it is presented. How do you take hundreds or thousands of data points and create something a human can understand? Check out the world of charts, graphs, and more.
# Stem-Leaf Plot
How many observations are recorded in the following stem and leaf plot?
$\begin{array} { r | l } 14 & 4 2 13 3 \\ 15 & 7 4 \\ 16 & 13 32 7 9 \\ 17 & 7 19 4 \\ \end{array}$
Victor measured the heights of his students, who ranged from 110 cm to 150 cm. He recorded the data in the following ordered stem and leaf plot.
$\begin{array} { r | l } 11 & 0 1 3 5 8 9 \\ 12 & 1 2 4 5 8 \\ 13 & 2 5 5 7 9 \\ 14 & 0 1 2 6 8 \\ \end{array}$
What is the median height in class?
Victor measured the heights of his students, who ranged from 80 cm to 120 cm. He recorded the data in the following stem and leaf plot.
$\begin{array} { r | l } 8 & 8 5 10 \\ 9 & 12 0 10 \\ 10 & 10 8 12 2 \\ 11 & 12 1 8 \\ \end{array}$
What is the height of the tallest student in class?
Victor measured the heights of his students, who ranged from 80 cm to 120 cm. He recorded the data in the following stem and leaf plot.
$\begin{array} { r | l } 8 & 8 5 4 \\ 9 & 16 0 4 \\ 10 & 4 8 16 2 \\ 11 & 16 1 8 \\ \end{array}$
What is the height of the shortest student in class?
Which of the following statements is always true of Stemplots?
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## § Lie bracket versus torsion
This picture finally made the difference between these two things clear. The lie bracket moves along the flow , while the torsion moves along parallel transport . This is why the sides of the parallelogram that measure torsion form, well, a parallelogram: we set them up using parallel transport. On the other hand, the lie bracket measures the actual failure of the parallelogram from being formed.
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# 02 In-Class Assignment: Vectors¶
What can you solve with $$Ax=b$$?
Image from http://wikipedia.org/
Vector mathematics are used in physics all of the time. Consider the picture at the beginning of this notebook. While the space shuttle is attached to the Airplane they have the same velocity vector. This vector likely has three components; the velocity in the x direction, y direction and z direction. In fact this velocity is just a combination of multiple components such as thrust from the engines, wind speed and drag.
In this notebook we will be discussing basic vector mathematics and practicing our Python. We will be using the commands in this notebook all semester so make sure you have have some mastery before moving on.
## 1. Example linear System:¶
Suppose that we have three objects on a balanced beam. Also suppose we know that one has a mass of 2 kg, and we want to find the two unknown masses. Experimentation with a (assume weightless) meter stick produces these two balances. (diagram not to scale)
For the masses to balance we must have the sum of the moments on the left equal to the sum of the moments on the right, where the moment of an object is its mass times its distance from the balance point. That gives a system of two equations:
$40A + 15B = 50 \times 2$
$25B = 25 \times 2 + 50A$
Do This: Find a solution for the above systems of equations and place your solution in the following cell. Make sure you delete the instructional text in the cell first.
# Put your answer to the above question here
Do This: Using Python as a calculator, verify that the solution you have found is correct.
# Put your answer to the above question here
Do This: Now lets consider a system where we have three unknown masses instead of two. Experimentation with a meter stick produces the two balanced states shown below (diagram not to scale). Write the equations for this system.
Do This: Find a solution to the second set of equations and report the mass for objects A, B and C.
# Put your answer to the above question here
Do This: Using Python as a calculator, verify that the solution you have found is correct.
# Put your answer to the above question here
## 3. Pre-class assignment review¶
For any vectors x, y, and z of the same size/dimension, we have the following properties:
1. Vector addition is commutative: x + y = y + x.
2. Vector addition is associative: (x + y) + z = x + (y + z). We can therefore write both as x + y + z.
3. Adding the zero vector to a vector has no effect: x + 0 = 0 + x = x. (This is an example where the size of the zero vector follows from the context, i.e., its size must be the same as the size of x)
4. x − x = 0. Subtracting a vector from itself yields the zero vector. (Here too the size of 0 is the size of a.)
### Scalar-Vector Multiply Properties¶
For any vectors x, y, and scalars a, b, we have the following properties
• Scalar-vector multiply is commutative: ax = x * a; This means that scalar-vector multiplication can be written in either order.
• Scalar-vector multiply is associative: (ab)x = a(bx)
• Scalar-vector multiply is distributive: a(x + y) = ax + ay, (x+y)a = xa + ya, and (a+b)x = ax + bx.
## 4. Vectors in Python¶
For those who are new to Python, there are many common mistakes happen in this course. Try to fix the following codes.
### SyntaxError¶
It means that the code does not make sense in Python. We would like to define a vector with four numbers.
DO THIS: Fix the following code to create three vectors with four numbers.
x = [1 2 3.4 4]
y = [1, 2, 3, 5]]
z = [[1, 2, 3, 6.3]
File "<ipython-input-5-cd07b469c255>", line 1
x = [1 2 3.4 4]
^
SyntaxError: invalid syntax
Although you may have been able to get rid of the error messages the answer to you problem may still not be correct. Throughout the semester we will be using a python program called answercheck to allow you to check some of your answers. This program doesn’t tell you the right answer but it is intended to be used as a way to get immediate feedback and accelerate learning.
DO THIS: First we will need to download answercheck.py to your current working directory. You only really need to do this once. However, if you delete this file by mistake sometime during the semester, you can come back to this notebook and download it again by running the following cell:
from urllib.request import urlretrieve
DO THIS: How just run the following command to see if you got x, y and z correct when you fixed the code above.
from answercheck import checkanswer
NOTE make sure you do not change the checkanswer commands. The long string with numbers and letters is the secret code that encodes the true answer. This code is also called the HASH. Feel free to look at the answercheck.py code and see if you can figure out how it works?
### Numpy¶
Numpy is a common way to represent vectors, and you are suggested to use numpy unless otherwise specified. The benefit of numpy is that it can perform the linear algebra operations listed in the previous section.
For example, the following code uses numpy.array to define a vector of four elements.
import numpy as np
x_np = np.array([-1, 0, 2, 3.1])
x_np
#### Scalars versus 1-vectors¶
In mathematics, 1-vector is considered as a scalar. But in Python, they are not the same.
x = 2.4
y = [2.4]
x == y
x == y[0]
#### Lists of vectors¶
We have a list of numpy arrays or a list of list. In this case, the vectors can have different dimensions.
DO THIS: Modify the print statement using indexing to only print the value 3 from the list_of_vectors defined below.
x_np = np.array([-1,0, 2 , 3.1])
y_np = np.array([1,-1,3])
z_np = np.array([0,1])
list_of_vectors = [x_np,y_np,z_np]
print(list_of_vectors)
#### Indexing¶
The index of a vector runs from 0 to $$n-1$$ for a $$n$$-vector.
DO THIS: The following code tries to get the third element of x_np, which is the number 2.0. Fix this code to provide the correct answer.
print(x_np(3))
DO THIS: Replace only the third element of x_np with the number 20.0 such that the new values of x_np is [-1, 0, 20., 3.1]
# Replace the third element using 20.0, then the resulting element is
print(x_np)
from answercheck import checkanswer
There is a special index -1, which represents the last element in an array. There are several ways to get more than one consecutive elements.
• x_np[1:3] gives the 2nd and 3rd elements only. It starts with the first index and ends before the second index. So the number of element is just the difference between these two numbers.
• x_np[1:-1] is the same as x_np[1:3] for a 4-vector.
• If you want the last element also, then you do not need to put the index, e.g., x_n[1:] gives all elements except the first one. You can do the same thing as the first one.
DO THIS: you are given a vector (x_np) of $$n$$ elements, define a new vector (d) of size $$n-1$$ such that $$d_i = x_{i+1}-x_i$$ for $$i=1,\dots,n-1$$. Hint try doing this without writing your own loop. You should be able to use simple numpy indexing as described above.
x_np = np.array([1,8,3,2,1,9,7])
from answercheck import checkanswer
### Assignment versus copying¶
Take a look at the following code.
• we create one numpy array x_np
• we let y_np = x_np
• we change the third element of y_np
• The third element of x_np is also changed
This looks weired and may not make sense for those uses other languages such as MATLAB.
The reason for this is that we are not creating a copy of x_np and name it as y_np. What we did is that we give a new name y_np to the same array x_np. Therefore, if one is changed, and the other one is also changed, because they refer to the same array.
x_np = np.array([-1, 0, 2, 3.1])
y_np = x_np
y_np[2] = 20.0
x_np
DO THIS: There is a method named copy that can be used to create a new array. You can search how it works and fix the code below. If this is done correctly the x_np vector should stay the same and the y_np you now be [-1 0 2 3.1].
## modify the following code to copy the x_np instead of just giving it a new name
x_np = np.array([-1, 0, 2, 3.1])
y_np = x_np
y_np[2] = 20.0
print(x_np)
from answercheck import checkanswer
from answercheck import checkanswer
### Vector equality in numpy and list¶
The relational operator (==, <, >, !=, etc.) can be used to check whether the vectors are same or not. However, they will act differently if the code is comparing numpy.array objects or a list. In numpy, In numpy relational operators checks the equality for each element in the array. For list, relational operators check all elements.
x = [-1, 0, 2, 3.1]
y = x.copy()
y[2] = 20.2
x_np = np.array(x)
y_np = np.array(y)
x == y
np.array(x_np) == np.array(y_np)
### Zero vectors and Ones vectors in numpy¶
• zeros(n) creates a vector with all 0s
• ones(n) creates a vector with all 1s
DO THIS: Create a zero vector (called zero_np) with the same dimension as vector x_np. Create a ones vector (called ones+np) also with the same dimension as vector x_np.
x_np = np.array([-1, 0, 2, 3.1])
### Define zero_np and ones_np here
from answercheck import checkanswer
### Random vectors¶
• random.random(n) creates a random vector with dimension $$n$$.
random_np = np.random.random(2)
print(random_np)
In this section, you will understand why we use numpy for linear algebra opeartions. If x and y are numpy arrays of the same size, we can have x + y and x-y for their addition and subtraction, respectively.
x_np = np.array([1,2,3])
y_np = np.array([100,200,300])
v_sum = x_np + y_np
v_diff = x_np - y_np
print (f'Sum of vectors: {v_sum}')
print (f'Difference of vectors: {v_diff}')
For comparison, we also put the addition of two lists below. Recall from the pre-class assignment, we have to define a function to add two lists for linear algebra.
DO THIS: Modify the following code to properly add and subtract the two lists. HINT it is perfectly okay NOT to write your own function try you should be able to cast the lists as arrays:
x = [1,2,3]
y = [100,200,300]
v_sum = x + y
v_diff = x - y
print (f'Sum of vectors: {v_sum}')
print (f'Difference of vectors: {v_diff}')
A scalar-vector addition means that the scalar (or a 1-vector) is added to all elements of the vector.
DO THIS: Add a scalar 20.20 to all elements of the following vector x_np and store teh result back into x_np
x_np = np.array([1.0,2.0,3.0])
from answercheck import checkanswer
### Scalar-vector multiplication and division¶
When a is a scalar and x is numpy array. We can express the scalar-vector multiplication as a*x or x*a.
We can also do scalar-vector division for x/a or a/x. (note that x/a and a/x are different)
DO THIS: Divide all elements of the following vector x_np by 20.20 and put it into y_np
x_np = np.array([1,2,3])
y_np =
print(y_np)
from answercheck import checkanswer
### Element-wise operations.¶
As stated above relational operations on numpy arrays are performed element-wise. Examples we mentioned before are
• The == operator
• The addition + and subtraction -
Note for this to work the two vectors have to be the same dimensions.
If they are not have the same dimension, such as a scalar and a vector, we can think about expanding the scalar to have the same dimension as the vector and perform the operations. For example.
• Vector-scalar multiplication and division
DO THIS: Assume that you invested three assets with initial values stored in p_initial, and after one week, their values are stored in p_final. Then what are the asset return ratio (r) for these three assets (i.e. price change over the initial value).
p_initial = np.array([22.15, 89.32, 56.77])
p_final = np.array([23.05, 87.32, 53.13])
from answercheck import checkanswer
### Linear combination¶
We have two vectors $$x$$ and $$y$$ we can get the linear combination of these two vectors as $$ax + by$$ where $$a$$ and $$b$$ are scalar coefficients.
In the following example, we are given two vectors (x_np and y_np), and two scalars (alpha and beta), we obtain the linear combination alpha*x_np + beta*y_np.
x_np = np.array([1,2])
y_np = np.array([3,4])
alpha = 0.5
beta = -0.8
c = alpha*x_np + beta*y_np
print(c)
We can also define a function lincomb to performn the linear combination.
DO THIS: Finish the following code for lincomb and compare the results we just get.
def lincomb(coef, vectors):
n = len(vectors[0]) # get the dimension of the vectors. note they have to be of the same dimension
comb = np.zeros(n) # initial the value with all zeros.
### Add code here to calculate the linear combination of the input vecotrs and the coefficients.
return comb
from answercheck import checkanswer
combination = lincomb([alpha, beta], [x_np,y_np])
We can also test the functions ourselves by using values for which we know the answer. For example, the following tests are multiplying and adding by zero we know what these answers should be and can check them.
combination = lincomb([0, 0], [x_np,y_np])
combination == np.zeros(combination.shape)
combination = lincomb([2, 2], [combination,np.zeros(combination.shape)])
combination == 2*combination
If you want to check that all values in a numpy.array are the same you could convert it to a list or there is a method called alltrue which checks if everything is true. It is a good idea to use this method if vectors get big.
combination = lincomb([2, 2], [combination,np.zeros(combination.shape)])
np.alltrue(combination == 2*combination)
Written by Dr. Dirk Colbry, Michigan State University and Dr. Ming Yan, Michigan State University
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#### Archived
This topic is now archived and is closed to further replies.
# Circular Depandancies
This topic is 5332 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I could not get rid of them on my exam and I cant get rid of them now. I am having endless problems with circular dependancies with .h files including other .h files that have already been included if you catch me? The result being things such as global variables being declared twice. A friend of mine said you have to use some #ifdef or something? Can someone explain this to me and how to use it (assuming my friend was right)? Thanks as always.
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Run a search for "inclusion gaurds".
But, how it works is like this:
#ifndef MY_BoxClass_H
#define MY_BoxClass_H
... Put includes here ...
... Put classes here ...
#endif
Basically, this keeps the compiler from reading and defining the same code multiple times.
Also, do not define global variables in a header like such:
int g_Handle;
handle.h
extern int g_Handle;
handle.cpp
int g_Handle;
That way it is not defined multiple times, but other source file know of it''s existance...
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Thanks, but just one thing...what goes where the MY_BoxClass_H is? I dont understand what this is...
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That''s a FAQ.
“Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.”
— Brian W. Kernighan (C programming language co-inventor)
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Yeah I just found that, thanks anyway
I cant believe we were not taught inclusion guards this semester in software development principals...would have saved me alot of trouble.
Anyhow, thanks again.
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Watch out for class circular dependency as well (for example, when two classes have members that are of the other class).
Inclusion guards don't entirely prevent them. You need an empty declaration of a class in order for it to work.
class A;
class B
{
A data;
};
class A
{
int value;
B member;
};
I think that works.
[edited by - Waverider on January 12, 2004 12:53:11 PM]
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You can also use '#pragma once' if you only want a file included once during compilation (if your compiler supports it).
[edited by - arm on January 12, 2004 2:36:56 PM]
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Hmm for some reason I still cant get rid of some of them. This is what I have in my OpenGL.h file (included in 2 .c files). hDC, hRC, hWnd and hInstance are still causing problems. Actually ifndef didnt seem to solve anything, declaring keys, active and fullscreen, then defining them in the .c file solved the problem with them, but the others are no good. The functions i never had a problem with.
#ifndef INC_OPENGL_H#define INC_OPENGL_H#include <windows.h>#include <gl\gl.h>#include <gl\glu.h> //#include <gl\glaux.h>extern bool keys[256];extern bool active;extern bool fullscreen;extern HDC hDC;extern HGLRC hRC;extern HWND hWnd;extern HINSTANCE hInstance;GLvoid ReSizeGLScene(GLsizei width, GLsizei height);int InitGL(GLvoid);int DrawGLScene(GLvoid);GLvoid KillGLWindow(GLvoid);BOOL CreateGLWindow(char* title, int width, int height, int bits, bool fullscreenflag);LRESULT CALLBACK WndProc(HWND, UINT, WPARAM, LPARAM);#endif /*INC_OPENGL_H*/
The problem with the global variables was that i was defining them in the header...not just declaring (i am a bit rusty). However, the other 4 globals are not working, still causing the double defining.
I was also taught that extern actually does nothing as all functions and global variables are globals in C anyway, unless declared static, making them local to the source file...
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If you post the exact error messages, it will be easier to help solve your specific error.
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quote:
Original post by Waverider
Watch out for class circular dependency as well (for example, when two classes have members that are of the other class).
Inclusion guards don''t entirely prevent them. You need an empty declaration of a class in order for it to work.
class A;
class B
{
A data;
};
class A
{
int value;
B member;
};
I think that works.
[edited by - Waverider on January 12, 2004 12:53:11 PM]
That doesn''t work. This works.
class Bar;
class Foo
{
Bar * pBar;
};
class Bar
{
Foo * pFoo;
};
You have to use pointers (at least one has to be a pointer) or it won''t compile. It''s not even logical if you really think about it.
-----------
VenDrake
"My stupid jar is full. I can''t talk to you anymore."
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# Information Security and Cryptography Research Group
## One-and-a-half quantum de Finetti theorems
### Matthias Christandl, Robert Koenig, Graeme Mitchison, and Renato Renner
Feb 2006, Available at http://arxiv.org/abs/quant-ph/0602130.
We prove a new kind of quantum de Finetti theorem for representations of the unitary group U(d). Consider a pure state that lies in the irreducible representation U_{mu+nu} for Young diagrams mu and nu. U_{mu+nu} is contained in the tensor product of U_mu and U_nu; let xi be the state obtained by tracing out U_nu. We show that xi is close to a convex combination of states Uv, where U is in U(d) and v is the highest weight vector in U_mu. When U_{mu+nu} is the symmetric representation, this yields the conventional quantum de Finetti theorem for symmetric states, and our method of proof gives near-optimal bounds for the approximation of xi by a convex combination of product states. For the class of symmetric Werner states, we give a second de Finetti-style theorem (our 'half' theorem); the de Finetti-approximation in this case takes a particularly simple form, involving only product states with a fixed spectrum. Our proof uses purely group theoretic methods, and makes a link with the shifted Schur functions. It also provides some useful examples, and gives some insight into the structure of the set of convex combinations of product states.
## BibTeX Citation
@unpublished{CKMR06,
author = {Matthias Christandl and Robert Koenig and Graeme Mitchison and Renato Renner},
title = {One-and-a-half quantum de {F}inetti theorems},
year = 2006,
month = 2,
note = {Available at http://arxiv.org/abs/quant-ph/0602130},
}
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# Hackerrank Solution: Even Odd Query
Original Problem
You are given an array $$A$$ of size $$N$$. You are also given an integer $$Q$$. Can you figure out the answer to each of the $$Q$$ queries?
Each query contains 2 integers $$x$$ and $$y$$, and you need to find whether the value find(x, y) is Odd or Even:
find(int x,int y) {
if(x>y) return 1;
ans = pow(A[x],find(x+1,y))
return ans
}
Note: pow(a, b) = $$a^b$$.
## Input Format
The first line of the input contains an integer $$N$$. The next line contains $$N$$ space separated non-negative integers (whole numbers less than or equal to 9).The line after that contains a positive integer, $$Q$$ , the denotes the number of queries to follow. $$Q$$ lines follow, each line contains two positive integer $$x$$ and $$y$$ separated by a single space.
## Output Format
For each query, display 'Even' if the value returned is Even, otherwise display 'Odd'.
## Constraints
• 2 ≤ N ≤ 105
• 2 ≤ Q ≤ 105
• 1 ≤ x,y ≤ N
• x ≤ y
The array is 1-indexed.
No 2 consecutive entries in the array will be zero.
## Solution
When analyzing the recurrence, we can see that the function calculates the continues exponentiation of the array elements, beginning with $$x$$ up to $$y$$:
$a_x^{⋰^{{a_{y-k}^{⋰^{a_{y-1}^{a_y}}}}}}$
From it, it is clear that we can state the algorithm in a non-recursive manner:
function find(arr, x, y) {
let e = 1;
while (x <= y) {
e = Math.pow(arr[y], e);
y--;
}
return e;
}
But looping is not necessary, since we want to know if the result is even or odd, which is
$a^b\bmod{2} \equiv \begin{cases} 1 &\text{if } b=0 \\a\bmod{ 2} & \text{else} \end{cases}$
That means that except the special case $$b=0$$, we can ignore the exponent completely. With this knowledge, the only remaining part is $$a_x \bmod{2}$$.
What we still have to investigate on is the case that $$b=0$$. $$b$$ can only be zero if $$a_{x+1}=0$$ and $$x\neq y$$. If $$a_{x+2}$$ would be $$0$$, $$a_{x+1}$$ would be $$1$$, so we don't have to go further. But of course, if $$x=y$$, we have $$a_x^{a_x}$$ and it does not matter if the next array element is zero for that query. Case $$0^0$$ can be ignored, since the problem statement promises no 2 consecutive zeros in the array.
The problem statement says $$x\leq y < N$$, but we require $$x\neq y$$, from which follows that $$x<y$$. Another thing we should keep track of is that $$x + 1 < N$$, to avoid out of bound index access. Collecting all these information leads to
$x + 1 < N \land x < y \land x \leq y < N \Rightarrow x + 1 \leq y \Leftrightarrow x<y$
The final solution is then
function find(A, x, y) {
if (x < y && A[x + 1] == 0)
return 1;
return A[x] % 2;
}
function solve(arr, queries) {
return queries.map(([x, y]) => find(arr, x - 1, y - 1) ? "Odd" : "Even")
}
« Back to problem overview
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# Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises - Page 43: 15
$\frac{2x+1}{2x-6}$
#### Work Step by Step
First, cancel out like terms and simplify: from: $\frac{5(x-3)(2x+1)}{10(x-3)^{2}}$ to: $\frac{(2x+1)}{2(x-3)}$ then, expand the bracket: $\frac{2x+1}{2x-6}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Fourier transformation - connection between exponential and trigonomeric forms
On Wikipedia i have come across a Fourier transformation equation in exponential form and its inverse (Wiki):
$$\begin{split} \mathcal{F}(x) &= \int\limits^{\infty}_{-\infty}\mathcal{f}(k) \, e^{2 \pi i kx} \, \textrm{d} k\\ \mathcal{f}(k) &= \int\limits^{\infty}_{-\infty}\mathcal{F}(x) \, e^{2 \pi i kx} \, \textrm{d} x \end{split}$$
but i allso found that there is a trigonometric form of Fourier transformation (PDF, page 2)
$$\begin{split} \mathcal{F}(x) &= \int\limits^{\infty}_{-\infty} f(k) \cos(kx) \, \textrm{d}k\\ \mathcal{f}(k) &= \int\limits^{\infty}_{-\infty} \mathcal{F}(x) \cos(kx) \, \textrm{d}x \end{split}$$
MAIN QUESTION:
Could someone show me, how these pairs of equations are connected?
SUB QUESTION:
(i) I think that $\textrm{d}x$ is used for spatial integration (please correct me if i am wrong).
(ii) I think that $\textrm{d}k$ is used for integration over wave vector (please correct me if i am wrong) .
-
## migrated from physics.stackexchange.comFeb 26 at 0:05
This question came from our site for active researchers, academics and students of physics.
Do you know about Euler's Formula?
$e^{ix} = cos(x) + isin(x)$
I think that's a hint...
Also, see this formula http://en.wikipedia.org/wiki/Sine_and_cosine_transforms#Fourier_inversion The first one in this column is simply the fourier integral applied over the above relation. Everything else is just a simplification into cosine
And this is even better...a whole justification of the connection http://en.wikipedia.org/wiki/Sine_and_cosine_transforms#Relation_with_complex_exponentials
-
Second link you provided is suplying link between time dependant and frequency dependant fourier. I have wave vector ependant and spatial dependant fourier. – 71GA Feb 25 at 23:53
Call it time or frequency or wave or space --- the formulas are the same, no? – Gerry Myerson Feb 26 at 0:17
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# Wednesday, May 14, 2014¶
Tried to solve docs/tickets/97. A real solution would require changes in store.py and elems.py, take much more time than I currently have and would bring unstability.
I finally ended up by simply extending the workaround: the GFK_HACK regular expression used by lino.utils.instantiator.GenericForeignKeyConverter didn’t work when the value had been rendered by a request with a known requesting panel.
The “Manage addresses” button is cool, but (1) it should be next to the address, not after the eid_info. And (2) we want it for every site where lino.modlib.addresses is installed. So I moved this code from lino_welfare.modlib.pcsw to lino.modlib.contacts.
The test suite had some minor failures.
Deleting a primary address now will clear the address fields of the partner.
## First session with Taavi¶
Getting Lino development version to run on Windows.
• The first thing he did was Installing a Lino development environment. Main difficulties where pip itself, pycrypto and pylibtidy. The latter produced an OSError: Could not libtidy using any of these names: libtidy,libtidy.so,libtidy-0.99.so.0,cygtidy-0-99-0,tidylib,libtidy.dylib,tidy traceback.
• The Lino test suite then failed because atelier.utils.SubProcessParent had from __future__ import unicode_literals, which caused a traceback TypeError: environment can only contain strings in the Python subprocess module.
• Updated The Lino Polls tutorial.
• lino.modlib.humanlinks requires ml.contacts.Person to be lino.mixins.human.Born. But it wasn’t. Because at some moment I thought that the default ml.contacts.Person should be minimal. But meanwhile I’d say that indeed most contact managers have a birth date for their persons.
• Taavi’s first Lino application will have 3 models: Song, Concert and Performance
## Continued on docs/tickets/93¶
• The DELETE key failed to invoke the DeleteSelected action (after converting this to a customizable action). Now it works again. And Lino is now able for the first time to handle custom hotkey definitions using Lino.GridPanel.handle_key_event(). The system is not yet stable and cannot yet handle modifiers, but I am glad that a beginning is finally there.
• The IBAN number of a ml.sepa.Account was indeed not validated. Simply because I forgot to call super() in its dd.Model.full_clean() method.
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cupcake_queen 2 years ago 7a-3=3-2a solve for a
Bring a's to one side and constants to the other.
2. ChristianGeek
$7a - 3 = 3 - 2a$$7a - 3 + 2a = 3 - 2a + 2a$$9a - 3 = 3$$9a - 3 + 3 = 3 + 3$$9a = 6$$\frac{9a}{9} = \frac{6}{9}$$a = \frac{6}{9}$
3. cupcake_queen
9a-3=3
4. cupcake_queen
oh ok
5. zordoloom
BTW, 6/9 can be further simplified, but it looks good @ChristianGeek
6. ChristianGeek
True, sorry...the final answer should be: $a = \frac{ 2 }{ 3 }$
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2. ## Re: Range
What have you tried? For example: What's the range of the sinus? and cosinus? ...
3. ## Re: Range
I transcribe it correctly: Find the range of the function
$f(x)=\cos\left(\sin \left(\log\frac{e^{2}+x^{2}}{x^{2}}+1\right)\right )+\sin \left(\cos \left(\log\frac{e^{2}+x^{2}}{x^{2}}+1\right)\right )$
4. ## Re: Range
sir sin and cos both are taking values from [-1,1] but next
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Algebra with fractions is a representation of algebraic expression in simple fraction form as numerator and denominator. In an algebra fraction, when we are adding or subtracting, one thing we need to keep in mind is that it should have a common denominator by cross multiplying.
In this article we will be solving few problems related to Algebra with fractions in order to get better understanding about it.
Examples 1: Simplify
\frac{1}{(x + 2)}+ \frac{6}{(x + 10)}
Solution:
We have,\frac{1}{(x + 2)}+ \frac{6}{(x + 10)} \\ =\frac{ 1(x + 10) + 6(x + 2)}{(x + 2)(x + 10)} (by Cross Multiplying)\\ =\frac{x + 10 + 6x + 12}{(x + 2)(x + 10)}\\ = \frac{7x + 22}{(x + 2)(x + 10)} Hence, after simplification we get \frac{7x + 22}{(x + 2)(x + 10)}
Example 2: Simplify
\frac{5x+10}{5}
Solution:
We have,\frac{5x+10}{5} Here, the common factor of 5 =\frac{5(x+2)}{5} =x+2 So, after simplification we get x+2
Example 3: Reduce
\frac{4x^4}{2x^2}
Solution:
We have, \frac{4x^4}{2x^2} =\frac{2x^2(2x^2)}{2x^2} (after cancelling) =2x^2 Therefore, after reducing the answer is 2x^2
## Solve Equation of Algebra with fractions
Example 1: Solve
\frac{1}{x+4}=\frac{1}{2}
Solution:
\frac{1}{x+4}=\frac{1}{2} (1)(2)=(1)(x+4) 2=x+4 x= -2 As a result, the value of x= -2
Example 2: Solve
\frac{1}{(x + 2)}+ \frac{6}{(x + 10)} =\frac{4}{3}\\
Solution:
\frac{ 1(x + 10) + 6(x + 2)}{(x + 2)(x + 10)}=\frac{4}{3}\\ \frac{x + 10 + 6x + 12}{(x + 2)(x + 10)}=\frac{4}{3}\\ 3 (7x + 22)=4(x + 2)(x + 10)\\ 21x+66=4(x^2+10x+2x+20)\\ 4x^2+48x-21x+80-66=0\\ 4x^2+27x+14=0\\ x=\frac{-b±\sqrt{b^2-4ac}}{2a}\\ =\frac{-27±\sqrt{27^2-4(4)(14)}}{2(4)}\\ =\frac{-27±\sqrt{729-224}}{8}\\ =\frac{-27±\sqrt{505}}{8}\\ =\frac{-27±22\sqrt{21}}{8}
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# X-axis homing failed with odd results on Tevo Tarantula
I recently finished building my Tevo Tarantula but now I have an issue with homing the X-axis.
I have uploaded MarlinTarantula EasyConfig 2.0.x firmware with Arduino Version 1.8.15 on to the Tevo Tarantula. Whenever I press Home X on the LCD, or by sending the g code command G28 X0, the X carriage moves to activate the end stop, it activates the end stop, then the carriage moves to the right, or away from the end stop, then it moves back and jams in to the end stop.
When I physically press the X end stop and check the state with the G-code command M119 the end does register as 'triggered' so I don't think the end stop is the problem. However after reading some other threads I'm not sure anymore. I will link a video to show how the problem I have just described.
This is the link to the video: https://streamable.com/kbdnwc
Edit: When I do press the X end stop manually with my hand during homing of the X axis, the carriage has very similar behavior. It stops, then it goes to the right, which is away from the X end stop then it comes back to the left and I get the message of "Homing Failed. Please Reset."
• Just to confirm, you hear the impact/stalling motor sound when the carrage moves towards the end stop for the 2nd time during the X home? Aug 9 at 3:19
• Yes, when the X carriage moves to the end stop for the second time the X motor makes a weird stalling noise like it's being jammed and then I get the error message that the Homing was Failed. Aug 9 at 3:42
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CGAL 5.3 - 2D Arrangements
ArrTraits::Intersect_2 Concept Reference
## Operations
A model of this concept must provide:
OutputIterator operator() (ArrTraits::X_monotone_curve_2 xc1, ArrTraits::X_monotone_curve_2 xc2, Output_iterator &oi)
computes the intersections of xc1 and xc2 and inserts them in an ascending lexicographic $$xy$$-order into a range begining at oi. More...
## ◆ operator()()
OutputIterator ArrTraits::Intersect_2::operator() ( ArrTraits::X_monotone_curve_2 xc1, ArrTraits::X_monotone_curve_2 xc2, Output_iterator & oi )
computes the intersections of xc1 and xc2 and inserts them in an ascending lexicographic $$xy$$-order into a range begining at oi.
The type OutputIterator dereferences a boost::variant of either the type pair<ArrTraits::Point_2,ArrTraits::Multiplicity> or the type ArrTraits::X_monotone_curve_2. An object of the former type represents an intersection point with its multiplicity (in case the multiplicity is undefined or unknown, it should be set to $$0$$). An object of the latter type representing an overlapping subcurve of xc1 and xc2. The operator returns a past-the-end iterator of the destination range.
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# Q4a) Find the QR factorization the tatrixb) Test Aing the spectral method suitable watrix HOFLS . the guaranteed COMVCTECICE Gauss
###### Question:
Q4a) Find the QR factorization the tatrix b) Test Aing the spectral method suitable watrix HOFLS . the guaranteed COMVCTECICE Gauss Jacobi method for the following *ystem Ir +4 = 4v+2:= 9 4 +2v _ 2: = 10
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##### Evaluate 3 integral 8 52(3x)dx
Evaluate 3 integral 8 52(3x)dx...
##### Determine whether the following series converge. $\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k^{2}+10}$
Determine whether the following series converge. $\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k^{2}+10}$...
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# Why is the number of components of Lie group finite?
The following quote is from An Introduction to Lie Group and Lie Algebra by Alexander Kirillov. By "discrete" he means that $|G/G^0|$ is finite. However, according to Introduction to Smooth Manifolds by J. Lee, a topological manifold has countable (not finite in general) components. So, I'd like to know what condition makes lie group to have a finite components.
• It isn't. ${}{}$ – Mariano Suárez-Álvarez Oct 4 '15 at 3:20
• For example, one wants that a closed subgroup of a Lie group itself, but that is false if we insisted in that the number of components be finite, as the group of integers inside the additive group of real number shows. – Mariano Suárez-Álvarez Oct 4 '15 at 3:23
• Thanks for your answer! That solved my question. – Math.StackExchange Oct 4 '15 at 3:24
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# Check My Beginner's Guide to Shells, Energy Levels, Subshells, and Orbitals [closed]
I am a beginner trying to summarize what I've learned about:
• Shells
• Energy Levels
• Subshells and
• Orbitals
all at the high school level.
Unlike related threads, I'm going to try to:
1. Define each word and state a few key properties of each.
2. Highlight what is most important for beginners.
3. Summarize all of that in a table.
Please let me know how I've done!
Thanks!
tl;dr Summary: Electrons move within orbitals which are in subshells which are in shells which are in atoms.
## Shell aka Energy Level
Below atoms, shells are the largest grouping of electrons beginner chemists should care about, and among those, the innermost two shells are most important to beginners.
They correspond to periods (rows) on the periodic table of the element.
• In the 3rd period, all elements have 3 shells.
• In the 5th period, all elements have 5 shells.
• In the $$n$$th period, all elements have $$n$$ shells.
Generally speaking, the larger $$n$$ means:
• The shells are farther away from the nucleus and thus the electrons in it are less attracted to the nucleus.
• The shells can hold more electrons, i.e. an atom of that element requires more valence electrons to be stable.
## Subshell
Each shell consists of at least one subshell. Below atoms, this is the second largest grouping of electrons beginner chemists should care about.
There are 4 types of subshells: $$s$$, $$p$$, $$d$$, and $$f$$. For beginners, the $$s$$ and $$p$$ subshells are the most important.
The notation for shells and subshells:
• $$1s^2$$ means the 1st shell's $$s$$ subshell has 2 electrons in it.
• $$4d^5$$ means the 4th shell's $$d$$ subshell has 5 electrons in it
• $$AB^X$$ means the $$A$$th shell's $$B$$th subshell has $$X$$ electrons in it.
Note: This notation does not specify which orbital those electrons are in... but that info does not matter much to beginners.
## Orbital
An orbital is an actual path (probability distribution function, PDF) that either 1 or 2 electrons can be whizzing through at a time. This is the smallest grouping of electrons beginner chemists should care about. The math underlying that PDF is not critical for beginners.
Each subshell consists of at least one orbital.
# THE BIG TABLE
Feel free to edit the table. :)
• This could be a great blog post. However, Chemistry.SE is a Q&A site meaning it'a place where a well-defined question receives a concise answer. The main Chemistry.SE site is not designed for discussions, polling, collecting feedback or peer reviews. If you can re-formulate your post into a question (see Help Center — Asking), it can be re-opened. – andselisk Apr 3 '20 at 5:44
All looks good at a school level, but I would suggest or convince you that we should not attach any physical pictures or meaning to all these atomic concepts. If you pursue science in future, you will have to unlearn a lot of these stories ingrained by school or even university teachers. It is okay to acknowledge and say that we do not fully understand everything as yet. Nature is very very complex. Our only effort is to make simple models of the phenomena we observe in the Nature.
The shell model is outdated and obsolete yet it is good for educational purposes. It was inspired by the planets rotating around the Sun, so learning about it from a historical perspective is good. For example, you wrote "The shells are farther away from the nucleus and thus the electrons in it are less attracted to the nucleus." It may not be a correct picture, because in reality there are no circular shells per se, like an onion ring around a nucleus. Brilliant physicists spent their life time in understanding the atom so don't be surprised if you find atomic structure overwhelming. Indeed it is.
More importantly just delete this interpretation from your mind "An orbital is an actual path (probability distribution function, PDF) that either 1 or 2 electrons can be whizzing through at a time."
An orbital is just a mathematical construct with no physical reality as textbooks show them as balloon models and so on. The rigorous and correction version of an orbital is "Wavefunction depending explicitly on the spatial coordinates of only one electron" (IUPAC)
Their shape depends on the coordinate system being used to solve the Schrodinger's equation. Can you believe that Schrodinger never showed or drew a single picture of orbitals in his four famous papers?
The typical shapes of s, p, d, f orbital are shown in Cartesian coordinate system, but physicists and mathematicians have plenty of other coordinate systems to solve differential equations. Do the electrons care what coordinate system a human chose to represent them or calculate their energies? The answer is no. Their energy turns out to be the same regardless of the coordinate system. Nobody teaches that and I also recently learned this from a physicist who has published a lot on orbital shapes. That gentleman is a big opponent to the teaching of orbitals and the myths associated with them in undergraduate curriculum. Not all agree with him including Linus Pauling.
The atomic structures "stories" as taught to chemistry students as facts is one of the worst things in modern educational system. In my humble opinion, chemists should take the help of modern physicists and ask them how to represent the latest views of atomic structure right from beginning classes. It does not need to have crazy mathematics, but good scientists can explain and elaborate things in words as well. As someone said, "If you can't explain it simply, you don't understand it well enough."
Book recommendation: Hydrogen: The essential element by J. Rigden (from Harvard). A thin book which is very illuminating. Chemical educators should adapt this style. The last chapter or so talks about anti-hydrogen.
• "chemists should take the help of modern physicists and ask them how to represent the latest views of atomic structure right from beginning classes..." Where might I find such representations that are intelligible to teenagers? If they don't exist, might that suggest nobody has found a way to do it? If they do, I'd love to read them! – WeCanLearnAnything Apr 3 '20 at 5:40
• The quotes by Einstein, on simplicity, connect our ability to explain theories to their quality, not to how well we as scientists understand the underlying concepts: "All physical theories, their mathematical expressions notwithstanding, ought to lend themselves to so simple a description that even a child could understand them." By contrast, the idea contained within your quotation, which connects our ability to explain theories to our own understanding of them, famously comes from Feynman: kottke.org/17/06/… – theorist Apr 3 '20 at 6:08
• Dear @WeCanLearnAnything, I added the book recommendation. I liked it a lot (perhaps the only book I read from start to end) and you can easily follow it as well. – M. Farooq Apr 3 '20 at 12:59
• Cool. Hopefully my library has it! – WeCanLearnAnything Apr 4 '20 at 18:11
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##Introduction Problem. Suppose you are in a room with $n$ lightbulbs and $k$ switches, where $1 \leq n$ and $1 \leq k$. Suppose that any particular switch turns a subset of light bulbs on. Find the minimum number of switches such that flipping them on turns on all of the . . .
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University Physics Volume 2
# Challenge Problems
University Physics Volume 2Challenge Problems
### Challenge Problems
123.
A pendulum is made of a rod of length L and negligible mass, but capable of thermal expansion, and a weight of negligible size. (a) Show that when the temperature increases by dT, the period of the pendulum increases by a fraction $αLdT/2αLdT/2$. (b) A clock controlled by a brass pendulum keeps time correctly at $10°C10°C$. If the room temperature is $30°C30°C$, does the clock run faster or slower? What is its error in seconds per day?
124.
At temperatures of a few hundred kelvins the specific heat capacity of copper approximately follows the empirical formula $c=α+βT+δT−2,c=α+βT+δT−2,$ where $α=349J/kg·K,α=349J/kg·K,$ $β=0.107J/kg·K2,β=0.107J/kg·K2,$ and $δ=4.58×105J·kg·K.δ=4.58×105J·kg·K.$ How much heat is needed to raise the temperature of a 2.00-kg piece of copper from $20°C20°C$ to $250°C250°C$?
125.
In a calorimeter of negligible heat capacity, 200 g of steam at $150°C150°C$ and 100 g of ice at $−40°C−40°C$ are mixed. The pressure is maintained at 1 atm. What is the final temperature, and how much steam, ice, and water are present?
126.
An astronaut performing an extra-vehicular activity (space walk) shaded from the Sun is wearing a spacesuit that can be approximated as perfectly white $(e=0)(e=0)$ except for a $5cm×8cm5cm×8cm$ patch in the form of the astronaut’s national flag. The patch has emissivity 0.300. The spacesuit under the patch is 0.500 cm thick, with a thermal conductivity $k=0.0600W/m°Ck=0.0600W/m°C$, and its inner surface is at a temperature of $20.0°C20.0°C$. What is the temperature of the patch, and what is the rate of heat loss through it? Assume the patch is so thin that its outer surface is at the same temperature as the outer surface of the spacesuit under it. Also assume the temperature of outer space is 0 K. You will get an equation that is very hard to solve in closed form, so you can solve it numerically with a graphing calculator, with software, or even by trial and error with a calculator.
127.
Find the growth of an ice layer as a function of time in a Dewar flask as seen in Exercise 1.120. Call the thickness of the ice layer L. (a) Derive an equation for dL/dt in terms of L , the temperature T above the ice, and the properties of ice (which you can leave in symbolic form instead of substituting the numbers). (b) Solve this differential equation assuming that at $t=0t=0$, you have $L=0.L=0.$ If you have studied differential equations, you will know a technique for solving equations of this type: manipulate the equation to get dL/dt multiplied by a (very simple) function of L on one side, and integrate both sides with respect to time. Alternatively, you may be able to use your knowledge of the derivatives of various functions to guess the solution, which has a simple dependence on t. (c) Will the water eventually freeze to the bottom of the flask?
128.
As the very first rudiment of climatology, estimate the temperature of Earth. Assume it is a perfect sphere and its temperature is uniform. Ignore the greenhouse effect. Thermal radiation from the Sun has an intensity (the “solar constant” S) of about $1370W/m21370W/m2$ at the radius of Earth’s orbit. (a) Assuming the Sun’s rays are parallel, what area must S be multiplied by to get the total radiation intercepted by Earth? It will be easiest to answer in terms of Earth’s radius, R. (b) Assume that Earth reflects about 30% of the solar energy it intercepts. In other words, Earth has an albedo with a value of $A=0.3A=0.3$. In terms of S, A, and R, what is the rate at which Earth absorbs energy from the Sun? (c) Find the temperature at which Earth radiates energy at the same rate. Assume that at the infrared wavelengths where it radiates, the emissivity e is 1. Does your result show that the greenhouse effect is important? (d) How does your answer depend on the the area of Earth?
129.
Let’s stop ignoring the greenhouse effect and incorporate it into the previous problem in a very rough way. Assume the atmosphere is a single layer, a spherical shell around Earth, with an emissivity $e=0.77e=0.77$ (chosen simply to give the right answer) at infrared wavelengths emitted by Earth and by the atmosphere. However, the atmosphere is transparent to the Sun’s radiation (that is, assume the radiation is at visible wavelengths with no infrared), so the Sun’s radiation reaches the surface. The greenhouse effect comes from the difference between the atmosphere’s transmission of visible light and its rather strong absorption of infrared. Note that the atmosphere’s radius is not significantly different from Earth’s, but since the atmosphere is a layer above Earth, it emits radiation both upward and downward, so it has twice Earth’s area. There are three radiative energy transfers in this problem: solar radiation absorbed by Earth’s surface; infrared radiation from the surface, which is absorbed by the atmosphere according to its emissivity; and infrared radiation from the atmosphere, half of which is absorbed by Earth and half of which goes out into space. Apply the method of the previous problem to get an equation for Earth’s surface and one for the atmosphere, and solve them for the two unknown temperatures, surface and atmosphere.
1. In terms of Earth’s radius, the constant $σσ$, and the unknown temperature $TsTs$ of the surface, what is the power of the infrared radiation from the surface?
2. What is the power of Earth’s radiation absorbed by the atmosphere?
3. In terms of the unknown temperature $TeTe$ of the atmosphere, what is the power radiated from the atmosphere?
4. Write an equation that says the power of the radiation the atmosphere absorbs from Earth equals the power of the radiation it emits.
5. Half of the power radiated by the atmosphere hits Earth. Write an equation that says that the power Earth absorbs from the atmosphere and the Sun equals the power that it emits.
6. Solve your two equations for the unknown temperature of Earth.
For steps that make this model less crude, see for example the lectures by Paul O’Gorman.
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yscale(value, **kwargs) [source] ¶ Set the y-axis scale. text () is used to place text on the graph. It is better to exchange the X/Y. Base class of all Geoms. Plotting x and y points. Pastebin is a website where you can store text online for a set period of time. Here we'll create a 2 × 3 grid of subplots, where all axes in the same row share their y-axis scale, and all axes in the same column share their x-axis scale: In [6]: fig, ax = plt. invert_yaxis () in Python. Set heatmap y-axis label. Use this option to show two Y-axes (left and right) with two or more different scales. x = scale(x) y = scale(y). If you want the axes to choose the appropriate limits, set the limits mode back to automatic. Most of the time when one creates a plot in matplotlib, whether it being a line plot using plot(), scatter plot using scatter(), 2D plot using imshow() or contour(), matplotlib will automatically tick the x- and y- axes and add the tick labels at reasonable intervals. The axis line acts as a legend; it explains the mapping between locations and values. How To Create Scatterplots in Python Using Matplotlib. You can manually set the Minimum, Maximum, Interval, IntervalOffset, IntervalType, and IntervalOffsetType properties for each axis. This can also be achieved using. In other words, I want the y-axis values shown in the above plot to be 0%, 5%, 10%, 15%, 20%, 25%, and 30%. Now we run filters in the order they are presented on the command line, whether lua or JSON. Here’s an example how to access an attribute of a particular axis: import linuxcnc s = linuxcnc. Next, right click the y-axis. In this article we'll demonstrate that using a few examples. plot (a, b) i ax. js methods that will allow you to. For further examples also see the Scales section of the gallery. y = (x - min) / (max - min) 1. To plot logarithmic Y-axis bins in Python, we can take the following steps − Create x and y points using numpy. figure() ax = fig. If you’ve ever used MATLAB, Matplotlib might feel a bit familiar as that’s where it drew its inspiration from. 04 is the variance) The code has been implemented in Google colab with Python 3. line(y=[1, 0]) fig. For x and y aesthetics, plotnine does not create a legend, but it creates an axis line with tick marks and a label. import matplotlib. We can customize the horizontal boxplot further as we can see the horizontal boxplot is dominated by the outlier salaries. Except for the trans argument any of the arguments can be set to derive () which would result in the secondary axis inheriting the settings from the primary axis. GitHub Gist: instantly share code, notes, and snippets. I set the Y axis scale as follows: Min = 8 PM (0. XY and Bubble charts¶. set_ylabel('Yearly Hours') Finally, we set the position of the yearly scale to the far right, and scale it based on the axis limits to be an annual total: yy. min(x, axis), np. set_xscale() function in axes module of matplotlib library is used to set the x-axis scale. And this is how you set the x and y limit in matplotlib with Python. You can use the plot (x,y) method to create a line chart. To start, let's specify n_neighbors = 1: model = KNeighborsClassifier(n_neighbors = 1). We then plot graphs of the x and y values on each of the axes. The scatter() function plots one dot for each observation. Set the y-axis scale. In many applications, we need the axis of subplots to be aligned with each other. Python Coding for Minecraft: This Instructable shows how to install and use a mod I wrote that lets you control Minecraft with python scripts. The subgraph which terminates with the scale tensor must be a build-time constant. normal(0,1,1000) numBins = 50 ax. You can manually set the Minimum, Maximum, Interval, IntervalOffset, IntervalType, and IntervalOffsetType properties for each axis. There’s no hard rule about which coordinate system is better. append_axes function. I have some y -axis values in thousands example:1000,30,000 and so on and due to this the bar graphs plotted have very small bars and user has requested we increase the bar height. 251-255 of "Introduction to Statistical Learning with Applications in R" by Gareth James, Daniela Witten, Trevor Hastie and Robert Tibshirani. The default x- (and y-) axes scales are scale_x_continuous and scale_y_continuous, but other options include scale_x_sqrt and scale_x_reverse. The building blocks of Matplotlib library is 2-D NumPy Arrays. barh instead of plt. In total there is 4 coordinate system in Matplotlib, which are data, axes, figures, and display. tight: This parameter is used to expand the axis limits using the margins. Eu estou tentando criar uma trama matplotlib com um exponenciais (?) Eixo Y como o falso eu zombou-se abaixo. When we need to plot data in symmetric logarithmic form, then we can use an inbuilt defined function matplotlib. Here is a solution. Follow along, if you want to understand what's the simple idea behind the implementation. Show zoom slider. This is called the Matplotlib set axis range. scatter(x,y) plt. How to Randomly Select From or Shuffle a List in Python. Also, we set font size as 2, according to your requirements you can set it. One is a simple MATLAB-style API (Application. We have six different categories. This section shows how to build a barplot with Python, using Matplotlib and Seaborn. rand(4, 6) >>> heat_map = sb. aft") Using the Fault Data File Chooser An example of a data chooser for selecting one file: chooser = FaultDataChooser() faultData = chooser. PyChart is a Python library for creating high quality Encapsulated Postscript, PDF, PNG, or SVG charts. The actual Y-axis is divided by the actual X-axis and that refers to the method set_aspect() i. autoScale()). However, there might be times when you want to alter some specific tick labels, with a different numerical value. So for this particular case, the aspect ratio becomes the ratio of the Y-axis to the X-axis. set_yscale('log') plt. Figure constitutes of subplots, sub axis, titles, subtitles, legends, everything inside the plot but an overview. answered Apr 7 '17 at 21:08. It consists of pyplot (in the code often shortened by "plt"), which is an object oriented interface to the plotting. setGrid(100) # the alpha value of grids on x-axis y_axis = pg. arange(-3, 3, 0. I set the Y axis scale as follows: Min = 8 PM (0. Next, right click the y-axis. Voila, now the x-axis tick label text is Female and Male as we wanted. xlim () : (left, right). A better way to make the scatter plot is to change the scale of the x-axis to log scale. La classe Axes contient la plupart des éléments de la figure: Axis, Tick, Line2D, Text, Polygon, etc. Two y axises (code below) * reinit grads and turns off grid lines and grads labels. from pylab import figure, show, legend, ylabel # create the general figure fig1 = figure () # and the first axes using subplot. Graphing Library Plotly. The axis limits to be set. This cluster plot uses the ‘murder’ and ‘assault’ columns as X and Y axis. Ahora, tengo que trazar dos Xejes en el gráfico, pero necesito las marcas de tiempo a ser el mismo para cada gráfica, es decir, todos deben empezar y terminar en el mismo punto en el tiempo, incluso si no hay ningún evento en aquellos tiempos para algunos de los gráficos (por ejemplo, en el gráfico # 1 comienza a las 15:50 del 05/10/2015 y. Use log scale. Next, we're going to talk about how we can have multiple Y axis. This code is based on the following websites:. Valid types are: None, 'none', or 'off' : no axis decoration is displayed. how to flip or modify something, or add a line or add more labels on Y-axis to show different labels for different ranges?. Basic Plotting with Python and Matplotlib This guide assumes that you have already installed NumPy and Matplotlib for your Python distribution. 3D Axes in Python How to format axes of 3d plots in Python with Plotly. To run the app below, run pip install dash, click "Download" to get the code and run python app. array(data)) gl = op. ticker import NullFormatter # useful for logit scale # Fixing random state. The scale command is used to change the sizes of geometric objects. How to save a matplotlib plot as an image in Python. For example, for a dataset, we could guesstimate the min and max observable values as 30 and -10. Image matching using sift python. Question or problem about Python programming: Say I have some input data: data = np. Python Matplotlib. Takes a set of vertices that have been created around the origin and translates them to the be centered around the x, y, z coordinates supplied in center. If you want the axes to choose the appropriate limits, set the limits mode back to automatic. For a boxplot, which works at the Axes level, you'll need to make sure to assign your boxplot to a variable ax, which will be a matplotlib. yscale ('log'). 5153740Z ##[section]Starting: Initialize job 2021-06-08T06. Python y axis scale. Y range is chosen with respect to the y values of the whole x range and does not change after zooming-in. Simple matplotlib Histogram Example. Esto pondrá marcas X de garrapatas, y etiquetarlos correctamente, por lo [1, 10, 100, 1000, 10000, 100000]. The axis scale type to apply. Start The Ply2vrmesh Executable From C:\Program Files\Chaos Group\V-Ray For 3ds Max\bin\ Usage. """ @property def category_type (self): """ A member of :ref:XlCategoryType specifying the scale type of this axis. pyplot as plt plt. In this tutorial, I'll walk you through how one can scale and rotate a contour based on OpenCV Python API. I wanna modify the calculation of the graph to show that a lower Y-axis (lower trust score) is a better value. If both the lines have the same scale you can simply plot the second line on the same plot as per your formatting. 0, x_test / 255. Matplotlib is a library in Python and it is numerical – mathematical extension for NumPy library. three-dimensional plots are enabled by importing the mplot3d toolkit. I do not want to only display share of women but also the median salary on the chart, so I wondered how to add a secondary axis to this bar chart so that I could have the median salary on another axis, and then display the median salary above each bar in the form of a line?. Setting axis range in matplotlib using Python. set_yscale (log) ax. Change a ggplot theme and modify the background color. 1) Change the scale of the y-axis by multiplying its values by 10000 and, if it's possible, add a % sign to the numbers. Notice how the zoom box is constrained to prevent the distortion of the shape of the line plot. pyplot as plt # Prepare data. If that is not what you want, but you want to rescale only some axis, I would make this a parameter of the scale function to give the user full control: def scale(x, out_range=(-1, 1), axis=None): domain = np. Dash is the best way to build analytical apps in Python using Plotly figures. How can I plot the x-axis to scale? 0. Python Matplotlib (pyplot), a step-by-step Tutorial. To plot multiple sets of coordinates on the same set of axes, specify at least one of X or Y as a matrix. An example can be seen in the figure above. Previously we ran all lua filters before JSON filters. figure() ax = fig. Python y axis scale. Note that the x-axis is the exact same, but the y-axis is now on a log scale. xlim () : (left, right). get_ylim() [0]*12, mm. axis [ 1 ][ 'homed' ]. set_ylim (0,100). from bokeh. The size of the scale must match the size of the zeroPt. To do this, we have built an object-oriented visualization framework in Python. Dask and XGBoost can work together to train gradient boosted trees in parallel. matplotlib plot without a y axis. The Department of Transportation publicly released a dataset that lists flights that occurred in 2015, along with specificities such as delays, flight time and other information. E as instâncias de Axes suportam callbacks por meio de um atributo callbacks. #!/usr/local/bin/python import sys import math import putil import numpy as np import matplotlib. Algunas parcelas sólo contienen los valores de 0 a 1 y algunas contienen valores en los millones. However, you can customize the scale to better meet your needs. Getting started. csv') fig = plt. This tutorial will cover the basics of how to use three Python plotting libraries — Matplotlib, Seaborn, and Plotly. Related Resources. Plot two lines in two different Y axes (secondary axis) Sometimes the the data you want to plot, may not be on the same scale. sec_axis () is used to create the specifications for a secondary axis. The function semilogy() from matplotlib. To change the X axis scale range: Click the Rescale Tool button and press Z, or simply press Z. However, there are many cases, particularly in science and engineering, where the independent variable is a continuous value, such as temperature or frequency. import numpy as np import matplotlib. set_ylabel("y");. Two y axises (code below) * reinit grads and turns off grid lines and grads labels. loglog() function which returns the base 10 log scaling x-axis. set_yscale(self, value, **kwargs) ¶ Set the y-axis scale. Axis auto-range improvements plotly/plotly. 11900420156 Azienda certificata UNI EN ISO 9001:2015 - Certificato No. Inputs to this class can be convergent objects. new_train = new_train/255 x_train,x_test,y_train,y_test = train_test_split(new_train,clearalllabels,test_size=0. How can I control scale range so that bars are of good size even when the values are. fig, ax = plt. It plots the power of each frequency component on the y-axis and the frequency on the x-axis. 7 (not yet tested with Python 3 but it should work) • ROOTwith PyROOTenabled • matplotlib • numpy • scipy mATLASplotlib is developed and tested on Linux and Mac. Python: Altair - Setting the range of Date values for an axis In my continued experiments with the Altair visualisation library , I wanted to set a custom range of data values on the x axis of a chart. 6388901Z ##[section]Starting: Onnxruntime_Linux_GPU_ORTModule_Test 2021-06-08T06:25:30. #!/usr/local/bin/python import sys import math import putil import numpy as np import matplotlib. Matplotlib: Multiple Y-Axis Scales. To set logarithmic values along both axes, we could use both semilogx () and semilogy () functions: Python. legend Control the legend placement options. To install in your home directory: pip install--user mATLASplotlib. In general, the argument of the xticks command is a list or array with the desired numerical labels, starting at the left end of the axis and ending at the right end. Documentation. The building blocks of Matplotlib library is 2-D NumPy Arrays. To generate the DataFrame bar plot, we have specified the kind parameter value as ‘bar’. Jordan Crouser at Smith College for SDS293: Machine Learning. 83333), Max = 10 AM (1. import matplotlib. Matplotlib is showing them as, for example, 1,000,000,000. plot(x, x**2, x, np. geom_* classes determine the kind of geometric objects and every plot must have at least one geom added to it. Matplotlib was created as a plotting tool to rival those found in other software packages, such as MATLAB. However, there might be times when you want to alter some specific tick labels, with a different numerical value. Parameters: vertices ( list of lists ( e. Running the code, you’ll see the following graphic: This graphic shows the same information you saw in the previous plot, but by flipping the axes you may find it easier to understand and compare different bars. If empty placement use WCS. The following are 30 code examples for showing how to use matplotlib. ylim manual hold on y2 = 2*sin (x); plot (x,y2) hold off. csv') fig = plt. prop(obj, 'scale', index=2, text Scale (s) # scale all three axes the same Scale (x, y, z) # scale the axes independently Deprecated since version 1. To improve the model accuracy we'll scale both x and y data then, split them into train and test parts. The size of the 1-D scale tensor must match the size of the quantization axis. AFM micrograph with added scale bar. text () is used to place text on the graph. This import random from time import sleep from oscilloscope import Osc # adjust window_sec and intensity to improve visibility osc = Osc ( window_sec = 10 , intensity = 1 ) @osc. formatting of faceting values. 1) In this example we have modify the y value, the result look as following. I wanna modify the calculation of the graph to show that a lower Y-axis (lower trust score) is a better value. Circle((0, 0), radius=1, fc='w') plt. add_subplot () Function for Plotting Seaborn Subplots in Python. In the dsize, we will keep the width same as that of original image but change the height. Here, sales data could be in 10's whereas the number of sales enquiries could be in 100's. By default, the plot () function draws a line from point to point. skewtest (a[, axis, nan_policy]) Test whether the skew is different from the normal distribution. Plot the graph of y = g(x + k) with the text "g(x+k)' at the top middle inside each panel 1 plt. CHAPTER 3 Total luminosity inspector Brilview application has a total luminosity inspector component for plotting total luminosity values. twinx method creates a new y-axis that shares the same x-axis. To do this, first we need to define a new axis, but this axis will be a "twin" of the ax2 x axis. ylabel Give the Y-axis a title. waitKey(0) # waits until a key is pressed cv2. set_xscale (1, 'linear') Here's the documentation for that function. Going back to our earlier example, below is the function y=x with the y-axis on a logarithmic scale. Parameters. 5153740Z ##[section]Starting: Initialize job 2021-06-08T06. 99261715, 0. matplotlib is not doing this, and all I can find when I do research is how to plot data points using different y axis scales (which is. In matplotlib, a secondary y-axis sharing the same x-axis with another one is called a twin axis, and can be created using: twinax = ax. This can also be achieved using. With Pandas plot(), labelling of the axis is achieved using the Matplotlib syntax on the “plt” object imported from pyplot. In the same dialog, in the "Regularly spaced ticks" section, choose the option "Antilog" in the Format dropdown. Therefore, Matplotlib has a defined function for this operation matplotlib. First, you need to import the style package from python matplotlib library and then use styling functions as shown in below code: plt. Our previous post detailed the best practices to manipulate data. Dash and Dot. Just like in the viewport, when you move an object pressing "g" and then pressing the x-axis key (or y / z) twice. We then modify the y-coordinates from 0 to 25. import numpy as np import matplotlib. Syntax : matplotlib. CHAPTER 3 Total luminosity inspector Brilview application has a total luminosity inspector component for plotting total luminosity values. ObjectPlacement attribute definition from IFC documentation. Change the appearance - color, size and face - of titles. pyplot; matplotlib graph; python plot multiple lines in same figure; legend of colorbar python; python hello world; sleep function python; python iterate through dictionary; how to make a python list. The charts so far use a discrete set of values for the independent variable (the X axis, roughly speaking). Hello, I am quite new in matplotlib, I am now facing a quite simple problem but have no idea to solve. It needs two arrays of the same length, one for the values of the x-axis, and one for values on the y-axis:. And we also set the x and y-axis labels by updating the axis object. axis_items = {'left': y_axis, 'bottom': x_axis} # add plot self. subplots (2, 3, sharex=True, sharey=True). plot ( x , y1 ) plt. 416667), and Major Unit = 2 hours (0. xticks()-> plt. The axis line acts as a legend; it explains the mapping between locations and values. Geeksforgeeks. Let’s try the support vector machine, with a grid search over a few choices of the C parameter:. Feature Matching using OpenCV Sep 24, 2012 · KeyPoints Matching using FLANN: After successfully finding SIFT keypoints in an image, we can use these keypoints to match the keypoints from other image. And this is how you set the x and y limit in matplotlib with Python. Here we'll create a $2 \times 3$ grid of subplots, where all axes in the same row share their y-axis scale, and all axes in the same column share their x-axis scale:. scale_integer (b) → Rotator ¶ Returns rotator representing rotator A scaled by B. It selects a reasonable scale to use with the aesthetic, and it constructs a legend that explains the mapping between levels and values. Returns : y: Gray-scale (uint8 or uint16) or binary image. See the respective class keyword arguments: matplotlib. Making Everything Easieri HTML5 and CSS3 i L ' I N ' O N E FOR _ A Wiley Brand BOOKS IN 1 • Creating the HTML Foundation • Styling with CSS • Building Layouts with CSS • C. How can I get my graph to look the same as his? How do I re-scale my x axis so that the hours are stretched out? I tried to. So say we have an x-axis where the range is from 0 to 10. A solution is to scale salary values the x-axis to log-scale using scale_y_log10() in ggplot2. Kite is a free autocomplete for Python developers. To find value of x-axis, we can use get_x () and get_width () function. Dask uses existing Python APIs and data structures to make it easy to switch between NumPy, pandas, scikit-learn to their Dask-powered equivalents. 11900420156 Azienda certificata UNI EN ISO 9001:2015 - Certificato No. I will introduce some of the many D3. Forums Selected forums Clear. Now we'll see how to save this plot. Adapted by R. How To Create Scatterplots in Python Using Matplotlib. Step 2: Create variables for the chart. If you’ve ever used MATLAB, Matplotlib might feel a bit familiar as that’s where it drew its inspiration from. The length of the scale widget. Line Graph. Rescale y-axis for graphs with x-axis rangeslider (Python) #932. Creating Scatter Plots. How to modify the below code to get this done. Question or problem about Python programming: Say I have some input data: data = np. ax (Axes): matplotlib Axes, optional; The sns. X-axis scale factor for start of the B-Bone, adjusts thickness (for tapering effects) Type. Use the mouse wheel or +/-keys to change the horizontal axis scale range. waitKey(0) # waits until a key is pressed cv2. Ich würde gerne wissen, was der Unterschied ist zwischen: ax. Add second x-axis at top of figure using Python and matplotlib. The multi-bar chart is made by drawing four separate bar charts on the same axes—offset each bar chart by a certain amount, so they appear side-by-side. If we use "equal" as an aspect ratio in the function, we get a plot with the same scaling from data points to plot units for X-axis and Y-axis. Line Graph. kurtosistest (a[, axis, nan_policy]) Test whether a dataset has normal kurtosis. Use hold on to add a second plot to the axes. (In this case, it sends an email every Monday at 5 PM). In many applications, we need the axis of subplots to be aligned with each other. You can control the limits of X and Y axis of your plots using matplotlib function plt. In a more targeted evaluation, NetCheck correctly detects over 95% of the network problems we found from bug trackers of projects like Python, Apache, and Ruby. The Axes Class contains most of the figure elements: Axis, Tick, Line2D, Text, Polygon, etc. Create plots on different scales. from pylab import figure, show, legend, ylabel # create the general figure fig1 = figure () # and the first axes using subplot. Let us modify the minimum and maximum limit on each axis, by calling the set_xlim, set_ylim, and set_zlim methods. In the above example, basex = 10 and basey = 2 is passed as arguments to the plt. ax is simply the 3D axes which can be obtained with ax = fig. properties ( height = 100) & map. Commented: Walter Roberson on 13 Sep 2020 Accepted Answer: Walter Roberson. Basic Plotting with Pylab ¶. Rohit Bhoi on 15 Apr 2016. The simplified formats of the functions are : scale_x_continuous(name, breaks, labels, limits, trans) scale_y_continuous(name, breaks, labels, limits, trans) name : x or y axis labels. set_xlim() will be used to set the limit of units of x-axis and ax. Scaling Y and X axis python graph. You will need to copy the Axes() and Point() classes below, then all you will need is the following. Feature Matching using OpenCV Sep 24, 2012 · KeyPoints Matching using FLANN: After successfully finding SIFT keypoints in an image, we can use these keypoints to match the keypoints from other image. Y-axis log scale. A picture is worth a thousand words. Otherwise use an Action Control to triger the script once happy with the slider value selection: from Spotfire. set_xscale () in Python. This is called the Matplotlib set axis range. Subplots are useful if you want to show the same data on different scales. By default, the minimum and maximum scale values of each axis in a chart are calculated automatically. The size of the bar represents its numeric value. 8 # If we were to simply plot pts, we'd lose most of the interesting. Unlike the previous method, this function can be used to create subplots dynamically. Below is an example that shows how to do it:-. PyChart is a Python library for creating high quality Encapsulated Postscript, PDF, PNG, or SVG charts. The following three basic rotation matrices rotate vectors by an angle θ about the x-, y-, or z-axis, in three dimensions, using the right-hand rule—which codifies their alternating signs. how to flip or modify something, or add a line or add more labels on Y-axis to show different labels for different ranges?. Dask uses existing Python APIs and data structures to make it easy to switch between NumPy, pandas, scikit-learn to their Dask-powered equivalents. To create scatterplots in matplotlib, we use its scatter function, which requires two arguments: x: The horizontal values of the scatterplot data points. Learn About Dask APIs ». A label is simply a string of text. The power can be plotted in linear scale or in log scale. To do this, we have built an object-oriented visualization framework in Python. ticker import NullFormatter # useful for logit scale # Fixing random state. It will return two values – the keypoints and the descriptors. How can I get my graph to look the same as his? How do I re-scale my x axis so that the hours are stretched out? I tried to. Python program using seaborn for line chart plotting. The arguments are the number of rows and number of columns, along with optional keywords sharex and sharey, which allow you to specify the relationships between different axes. grid(axis = 'y') plt. sec_axis () is used to create the specifications for a secondary axis. To find value of x-axis, we can use get_x () and get_width () function. pandas is a fast, powerful, flexible and easy to use open source data analysis and manipulation tool, built on top of the Python programming language. Then a scale division is calculated (QwtScaleEngine. In such case, instead of manually computing the x and y positions for each axes, you can specify the x and y values in relation to the axes (instead of x and y axis values). Multiple axes in Dash¶. Here, in this tutorial we will see a few examples of python bar plots using matplotlib package. Change Y-Axis Options - Single Scale: This IronPython script shows how to change the following options, roughly in order as they appear on the y-axis tab when the "One axis with a single scale" option is selected. yscale ('symlog'). So essentially I just stored the 'reasons' column values and its frequencies as a series in p and then converted them to another dataframe k but this new countplot doesn't have the right range of Y-axis for the given values. Just copy/paste the following code into a file (say. pyplot as plt plt. xlim () and plt. 1) Change the scale of the y-axis by multiplying its values by 10000 and, if it's possible, add a % sign to the numbers. Sample rate used to determine time scale in x-axis. js methods that will allow you to. add_subplot. Let's say you now want to plot two bar charts in the same figure. With the object at Axes level, you can make use of the set() function to set xlim, ylim,… Just like in the following example:. In case of autoscaling the boundaries of a scale are calculated from the bounding rectangles of all plot items, having the QwtPlotItem. Luckily, matplotlib provides functionality to change the format of a date on a plot axis using the DateFormatter module, so that you can customize the look of your labels without having to rotate them. Total luminosity component is meant for querying total luminosity from brilcalc and visualising it in 4 types of charts:. 0 release, some three-dimensional plotting utilities were built on top of Matplotlib's two-dimensional display, and the result is a convenient (if somewhat limited) set of tools for three-dimensional data visualization. Image matching using sift python. The x-axis should have tick marks every π/2 and the y-axis should have tick marks at -1 and 1. plotting import figure from bokeh. Jeff Forcier Paul Bissex Wesiey Chun A Covers DJango 1. Set the Y-axis scale using the yscale () method. Eu estou tentando criar uma trama matplotlib com um exponenciais (?) Eixo Y como o falso eu zombou-se abaixo. One reason you might want to turn off the Y-axis, is to save space for more data. Application. accepts: [ True | False]. I want to set only y-axis as log scale and x-axis as linear?. how to flip or modify something, or add a line or add more labels on Y-axis to show different labels for different ranges?. In the dsize, we will keep the width same as that of original image but change the height. 416667), and Major Unit = 2 hours (0. Each of the axes' scales are set seperately using set_xscale and set_yscale methods which accept one parameter (with the value "log" in this case): In [1]:. pandas is a fast, powerful, flexible and easy to use open source data analysis and manipulation tool, built on top of the Python programming language. Given an object that is rotated and an example vector (1,2,3). Syntax: Axes. So this is how we can use the axis() provided by Matplotlib to change xxes size of our output graph plot. Jeff Forcier Paul Bissex Wesiey Chun A Covers DJango 1. Sometimes for quick data analysis, it is required to create a single graph having two data variables with different scales. Set the intercept of x and y axes at zero (0,0). By default the seaborn displaces the X axis ranges from -5 to 35 in distplots. pyplot as plt from matplotlib. 8776244Z ##[section]Starting: Initialize job. Next, it will take input for the x and y axes labels. Output:Ideal Normal curveThe points on the x-axis are the observations and the y-axis is the likelihood of each. I just want set the y axis scale to value 100, which means in the image, the y axis is always of scale 100, because the points in my image indicates the percentage value(for example, 20%, 87%) which will never exceed 100. Making the bar plots. A graph should appear with a line that animates much faster than in the previous example (i. Going back to our earlier example, below is the function y=x with the y-axis on a logarithmic scale. Matplotlib: Multiple Y-Axis Scales. Frequency types: 'linear', 'fft', 'hz' : frequency range is determined by the FFT window and sampling rate. Python’s x % y returns a result with the sign of y instead, and may not be exactly computable for float arguments. Python has a number of powerful plotting libraries to choose from. breaks: One of: NULL for no breaks. In our example, we will use a for loop to create an axes object with the subplots. set_title method from Matplotlib, can be leveraged to modify your title position in Seaborn. Application. The following plot parameters can be used : xlim: the limit of x axis; format : xlim = c (min, max) ylim: the limit of y axis; format: ylim = c (min, max) Transformation to log scale: log = “x”. This IronPython script shows how to change the following options, in order from top-left to bottom-right on the x-axis tab in the Spotfire user interface: Change the x-axis expression. Scale the x and y axis values so the plot will have an appealing aspect ratio. Return value. waitKey(0) # waits until a key is pressed cv2. Change the legend title and position, as well, as the color and the size. enableAutoSIPrefix(enable=False) # Prevent automatic SI # prefix scaling on this axis. python - How to scale Seaborn's y-axis with a bar plot? - Stack Overflow. add_patch(circle) plt. Unlike X/Y/Z Scale , this value can be negative if the object is flipped by negative scaling. First, we'll generate random regression data with make_regression () function. User guide. Y-axis log scale. To plot logarithmic Y-axis bins in Python, we can take the following steps − Create x and y points using numpy. For example, 0, 0. axis('scaled') for this to work. I have some y -axis values in thousands example:1000,30,000 and so on and due to this the bar graphs plotted have very small bars and user has requested we increase the bar height. xticks() doesn't really work. How To Create Scatterplots in Python Using Matplotlib. Feature Matching using OpenCV Sep 24, 2012 · KeyPoints Matching using FLANN: After successfully finding SIFT keypoints in an image, we can use these keypoints to match the keypoints from other image. Set the intercept of x and y axes at zero (0,0). Next, let's create an instance of the KNeighborsClassifier class and assign it to a variable named model. It needs two arrays of the same length, one for the values of the x-axis, and one for values on the y-axis:. 2006003002, 2: 0. In this Python Matplotlib tutorial series, you will learn how to create and improve a plot in Python using pyplot. The Y-axis labels are displayed to the left by default. However, if you look at the scatter plot most of the points are clumped in a small region of x-axis and the pattern we see is dominated by the outliers. Here is the matplotlib histogram demo. It selects a reasonable scale to use with the aesthetic, and it constructs a legend that explains the mapping between levels and values. Broken Axis. How can I do a translation on the local axis, so the object is moved 1 on it's local x-axis, 2 on the local y-axis, 3 on the local z-axis. rotation can either be horizontal or vertical. What must I change to get a uniform scale of 1 on the left hand side (starting from 5 and ending at 10)? Same for the x-axis. Example 3: Log Scale for Both Axes. Esto pondrá marcas X de garrapatas, y etiquetarlos correctamente, por lo [1, 10, 100, 1000, 10000, 100000]. This is just a short introduction to the matplotlib plotting package. get_ymajorticklabels(), fontsize = 18) Note: to control the labels rotation there is the option "rotation":. add_subplot. sin (x) z = np. Format the text color, size, and font: Color: Select black. To scale the image vertically using OpenCV, scale the image only along the y-axis or vertical axis, and keep the width of the image unchanged. The functions scale_x_continuous() and scale_y_continuous() are used to customize continuous x and y axis, respectively. org DA: 21 PA: 50 MOZ Rank: 74. ylabel ( 'Function' ) plt. Styles de codage « pyplot » et « OO » [modifier | modifier le wikicode] On voit qu'il y a aussi deux styles de codage, deux manières de rédiger. To create a new instance of this class, use NXOpen. Convenience method to get or set some axis properties. The yticks command works similarly for the y axis. Finally, you enter a title for your graph. First we need to find out the position where we need to show the labels. 2 is at the same distance as 1. loglog() function which returns the base 10 log scaling x-axis. The tick_params() function of matplotlib makes it possible to customize x and y axis ticks. The matplotlib subplots () method accepts two more arguments namely sharex and sharey so that all the subplots axis have similar scale. I have some y -axis values in thousands example:1000,30,000 and so on and due to this the bar graphs plotted have very small bars and user has requested we increase the bar height. Takes a set of vertices that have been created around the origin and translates them to the be centered around the x, y, z coordinates supplied in center. Ich würde gerne wissen, was der Unterschied ist zwischen: ax. axes (returns integer) - reflects [TRAJ] AXES ini value. Questions: I am trying to fix how python plots my data. For example, large ticks can indicate each new week day and small ticks can indicate each day. Dash is the best way to build analytical apps in Python using Plotly figures. Hello,I have a combination chart and my Y axis values are typically in millions and my x-axis is a date. When Mike plots ‘viewers_hour vs hours’ his x axis is scaled differently to his y axis like so: Whereas when I post the exact same code my x and y axes are scaled in a weird way. Data can be retrieved from Zetane through Form retrievals. scale_y_continuous(breaks = seq(0,350,50)) manually overrides the numbers on the y-axis to start from zero and end at 350 with 50 unit increments. Unlike the previous method, this function can be used to create subplots dynamically. plot(x,y) plt. Setting axis limits, adding a legend, configuring marker size, and other custom configurations are effective ways to improve the readability of a plot in Pyt. If the x-axis is scaled logarithmically, negative values in the domain will be discarded. The building blocks of Matplotlib library is 2-D NumPy Arrays. set_ylim(mm. The yticks command works similarly for the y axis. Similarly, to change the y-axis we can use the scale_y_continuous option. Cluster Plot canbe used to demarcate points that belong to the same cluster. Here we have used the range [0, 50] for X-Axis and [-1, 0] for Y-Axis. The picture is here. set_yscale('log') plt. Anyway, here is the updated code…. fig, ax = plt. If you’ve ever used MATLAB, Matplotlib might feel a bit familiar as that’s where it drew its inspiration from. ; To change the Y axis scale range:. Python Seaborn module is used to visualize the data and explore various aspects of the data in a graphical format. Analogous to the binwidth of a histogram, a density plot has a parameter called the bandwidth that changes the individual kernels and significantly affects the final result of the plot. xscale(‘log’). How can I do a translation on the local axis, so the object is moved 1 on it's local x-axis, 2 on the local y-axis, 3 on the local z-axis. semilogy (X,Y) plots x - and y -coordinates using a linear scale on the x -axis and a base-10 logarithmic scale on the y -axis. In general, the argument of the xticks command is a list or array with the desired numerical labels, starting at the left end of the axis and ending at the right end. Matplotlib’s Split Personality ¶. xticks()-> plt. axes (returns integer) - reflects [TRAJ] AXES ini value. Set heatmap y-axis label. You will need to copy the Axes() and Point() classes below, then all you will need is the following. Familiar for Python users and easy to get started. Basic Plotting with Pylab ¶. x y z k; 0: 466: 948: 1: male: 1: 832: 481: 0: male: 2: 978: 465: 0: male: 3: 510: 206: 1: female: 4: 848: 357: 0: female. labels the x axis with the numerical values 5, 10, 15, 20, 25. pyplot The result is: This page shows how to add second x-axis at the top of the figure using python and matplotlib. Python Tutorial Python HOME Python Percentile Data Distribution Normal Data Distribution Scatter Plot Linear Regression Polynomial Regression Multiple Regression Scale Train/Test Decision Tree Python MySQL plt. Use hold on to add a second plot to the axes. Matplotlib's plt. As the above graph is limited to [0,50] in x-Axis and [-1,0] in y Axis. , and sets the coordinate system. Just like in the viewport, when you move an object pressing "g" and then pressing the x-axis key (or y / z) twice. add_subplot () Function for Plotting Seaborn Subplots in Python. I then found a cool trick to label the axis using exponents rather than decimals or the lame default scientific notation. Use this option to show two Y-axes (left and right) with two or more different scales. Change axis limits. Follow along, if you want to understand what's the simple idea behind the implementation. Plot direction field, eigenvectors, and some solutions¶. You can use one of the following two methods to do so using only ggplot2: 1. For example, X Location, X Rotation, X Scale. Parameters. contour-lines-dataviz. sin (x) z = np. 0 than it is 1. plot(x,y) matplotlib. Specify Which Grid Lines to Display. Determine optimal k. pyplot as plt iplevel = pd. The following three basic rotation matrices rotate vectors by an angle θ about the x-, y-, or z-axis, in three dimensions, using the right-hand rule—which codifies their alternating signs. As we are aware of the fact that Matplotlib is the plotting library of Python. The y axis is compsed of the values 12-144. Thank you!Is there a way for a Spotfire. prop(obj, 'scale', index=2, text Scale (s) # scale all three axes the same Scale (x, y, z) # scale the axes independently Deprecated since version 1. MultipleLocator () places ticks on multiples of some base. There's no hard rule about which coordinate system is better. to the usual figure plotting script. py package on PyPi for import. Change the axis expression Use one axis with a single scale. Now we run filters in the order they are presented on the command line, whether lua or JSON. The size of the scale must match the size of the zeroPt. A basic rotation (also called elemental rotation) is a rotation about one of the axes of a coordinate system. Scale builder. How can I do a translation on the local axis, so the object is moved 1 on it's local x-axis, 2 on the local y-axis, 3 on the local z-axis. All data visible in Zetane is represented using hierarchical tree-like data structures called Forms. xticks() doesn't really work. And, the last section will focus on handling timezone in Python. I used two StandardScaler X_scaler and y_scaler to scale X and y respectively. plot (x,y) We can use the. Si necesita más etiquetas, cambiar el 6. 83333), Max = 10 AM (1. Now add the bar charts themselves. We will mainly use two libraries for audio acquisition and playback: 1. The axis line acts as a legend; it explains the mapping between locations and values. This is not unique but seems to work with matplotlib 1. Builder" means the Builder class in the sphinx. yscale ('log'). To increase the size of the labels on the y-axis just add the following line: res. The axes renders human-readable reference marks for scales. import matplotlib. CIAA 1514165 - REA 1507781 - Capitale Sociale € 100. To create scatterplots in matplotlib, we use its scatter function, which requires two arguments: x: The horizontal values of the scatterplot data points. This tutorial aims at showing good practices to visualize data using Python's most popular libraries. Here, sales data could be in 10's whereas the number of sales enquiries could be in 100's. It consists of pyplot (in the code often shortened by “plt”), which is an object oriented interface to the plotting. Sign up to +=1 for access to these, video downloads, and no ads. This is not unique but seems to work with matplotlib 1. Python: Altair - Setting the range of Date values for an axis In my continued experiments with the Altair visualisation library , I wanted to set a custom range of data values on the x axis of a chart. Sample program:. For starters, we will place sepalLength on the x-axis and petalLength on the y. “ax can be either a single matplotlib. Running the code, you’ll see the following graphic: This graphic shows the same information you saw in the previous plot, but by flipping the axes you may find it easier to understand and compare different bars. Note: To set a different scale type to the top X axis, you can enter the “identity function” to the link formulae for the second layer, that is: X1=X1, X2=X2. Use this option to show two Y-axes (left and right) with two or more different scales. Image matching using sift python. You can use the plot (x,y) method to create a line chart. Next, it will take input for the x and y axes labels. In above example if k=3 then new point will be in class B but if k=6 then it will in class A. The problem. Change x axes scale in matplotlib. Scaling up by 2. It is also possible to use the functions scale_x_continuous () and scale_y_continuous () to change x and y axis limits, respectively. 7 (not yet tested with Python 3 but it should work) • ROOTwith PyROOTenabled • matplotlib • numpy • scipy mATLASplotlib is developed and tested on Linux and Mac. 04 is the variance) The code has been implemented in Google colab with Python 3. xticks() doesn't really work. Python Tutorial Python HOME Scatter Plot Linear Regression Polynomial Regression Multiple Regression Scale Train/Test a label for the x- and y-axis. get_ymajorticklabels(), fontsize = 18) Note: to control the labels rotation there is the option "rotation":. Let’s try the support vector machine, with a grid search over a few choices of the C parameter:. November 13, 2017 Pan. This is just a short introduction to the matplotlib plotting package. Here a linear, a logarithmic, a symmetric logarithmic and a logit scale are shown. Dash is the best way to build analytical apps in Python using Plotly figures. One of the nice features in Altair is to be able to build, as in Tableau, nice looking Dashboards. The parameter breaks controls the split of the axis. pandas is a fast, powerful, flexible and easy to use open source data analysis and manipulation tool, built on top of the Python programming language. The combination chart y-axis tab layout in the Spotfire user interface changes quite a bit depending on whether the "One axis with a single scale" or "Multiple scales" option is selected. After integrating the d3. set_title method from Matplotlib, can be leveraged to modify your title position in Seaborn. Making Plots With plotnine (aka ggplot) Introduction. I'm particularly interested in showing the data in intervals of 200. To make a Matlabplot figure axis have integer-only labels, use method like: A complete standalone example follows: If too few ticks are displayed, as per the Matplotlib MaxNLocator , you must have "at least min_n_ticks integers…found within the view limits. Logarithmic Scaling¶. Note: Throughout this post we'll just focus on the x-axis but know you can do the exact same thing for the y-axis as well - just substitute x for y (e. 2) Add more values to the x-axis. get_yscale¶ Axes. py (matplotlib-3. I wanna modify the calculation of the graph to show that a lower Y-axis (lower trust score) is a better value. fig, ax = plt. Et les instances d’Axes prennent en charge les callbacks via un attribut. 18ACM4964Q. Python’s x % y returns a result with the sign of y instead, and may not be exactly computable for float arguments. To change the Y axis scale range: Click the Rescale Tool button , press Z and hold the Shift key down, or simply press Z+. yscale (value, **kwargs). 'set grid off'. There is a single x axis, and there are multiple columns that represent the points on the y axis. The scale command is used to change the sizes of geometric objects.
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# Demand Elasticity, Factor Substitution: Independent?
Given $$Y=f(K,L;\sigma)$$, the effect on labor from a change in the price of capital can be gauged through a substitution effect and a scale effect: \begin{align*} \frac{\partial L}{\partial r} & =\underset{Sub}{\underbrace{\left. \frac{\partial L^{c}}{\partial r}\right\vert _{Y}}}% +\underset{Scale}{\underbrace{\frac{\partial L^{c}}{\partial Y}\frac{\partial Y}{\partial r}}}\\ \\ & =SH_{_{Labor}}\left( \sigma+\eta\right) \end{align*} where $$L^c$$ is the conditional demand (the smooth movement along the isoquant).
This can then - as far as I can see - be also expressed as the sum of the factor substitution elasticity and the firm's elasticity of demand, weighted by the labor cost share.
But are $$\eta$$ and $$\sigma$$ independent of one another? If a firm can readily substitute factors in response to factor price changes, this will presumably impact its price elasticity of demand. Any thoughts?
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# Is $\sqrt{x/a}+\sqrt{y/b}=1$ the equation of a parabola tangent to the coordinate axes?
Is the below equation represents a parabola that touches the axes of coordinates? $$\sqrt{x/a}+\sqrt{y/b}=1$$
I know it is very stupid to ask this type of easy question here in the forum, but I'm very curious to know. I have searched many places and found nothing. My professor is not here, so I can't ask him. Suspense would have killed me.
Note from @Blue. Months later, I have edited the original problem to move the "$a$" and "$b$" under the radical signs. (This is because a duplicate problem recently appeared and I wanted to minimize confusion.) Most answers assumed this was the intention and proceeded accordingly. Those answers that use "$\sqrt{x}/a$" and "$\sqrt{y}/b$" should not be penalized for this after-the-fact notational change.
• Do you know how to identify conics (in particular, parabolas) from the general second-degree form $A x^2 + B x y + C y^2 + D x + E y + F = 0$? Can you see how to transform your equation into second-degree form? – Blue Apr 9 '17 at 6:08
• See revised solution using the Bezier Curve formula, which is much shorter and more direct. – hypergeometric Apr 17 '17 at 17:27
I'll take a different approach, describing a parabola that satisfies the equation. (More precisely, "a parabola with an arc that satisfies the equation", since, as @Alex notes, the equation's solution set must be bounded and therefore cannot include a complete parabola.)
Your original problem statement seemed a little unclear as to whether $a$ and $b$ belong inside the square roots. The first TeX edit of your question assumed they don't, and I preserved that assumption in my own edit. Here, however, I make the other call, so that the target is ... $$\sqrt{\frac{x}{a}} + \sqrt{\frac{y}{b}} = 1 \tag{1}$$
where I'll take $a > 0$ and $b > 0$ (and therefore also $x > 0$ and $y > 0$). With that aside ...
My parabola is tangent to points $A=(a,0)$ and $B=(0,b)$. Its directrix, $\ell$, is perpendicular to diagonal $\overline{OC}$ of the rectangle $\square OACB$, and its focus, $F$, is the foot of the perpendicular from $O$ to $\overline{AB}$. Without too much trouble, we determine that the directrix has equation $$\ell : a x + b y = 0 \qquad\text{and}\qquad F = \frac{ab}{c^2}\left(b,a\right)$$ where $c := |\overline{OC}| = \sqrt{a^2+b^2}$. A point $(x,y)$ on the parabola must be equidistant to $F$ and $\ell$; invoking the corresponding distance formulas, we have ... $$\sqrt{\left(x-\frac{a b^2}{c^2}\right)^2 + \left(y-\frac{a^2 b}{c^2}\right)^2} = \frac{|a x + b y|}{c} \tag{2}$$ Squaring, clearing fractions, and expanding $c^2$ as $a^2 + b^2$, and then dividing-through by $a^2 b^2$, we can ultimately re-write the above as ... $$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 + \left(1\right)^2 - 2 \left(\frac{x}{a}\right)\left(\frac{y}{b}\right) - 2 \left(1\right) \left(\frac{x}{a}\right) - 2 \left(1\right) \left(\frac{y}{b}\right) = 0 \tag{3}$$ I've made various factors (and powers!) of $1$ conspicuous to put the reader in the mind of the expanded form of Heron's formula for the area of a triangle. Specifically, $(3)$ represents ($16$-times) the square of the area of a triangle with side-lengths $\sqrt{\frac{x}{a}}$, $\sqrt{\frac{y}{b}}$, $\sqrt{1}$. Since the area vanishes, we must have a degenerate "flat" triangle: two side-lengths must equal the third. The target equation $(1)$ represents one of the three ways this can happen, and its solution set is arc $\stackrel{\frown}{AB}$ of the parabola. The other cases, $$\sqrt{\frac{y}{b}} + 1 = \sqrt{\frac{x}{a}} \qquad\text{and}\qquad \sqrt{\frac{x}{a}} + 1 = \sqrt{\frac{y}{b}}$$ correspond to the unbounded "arms" attached at points $A$ and $B$, respectively. $\square$
It seems like it should be possible to make the degenerate triangle interpretation of $(3)$ "visible" in the diagram, but I have not yet found a good way to do this.
• Should $F$ be $\displaystyle \frac {ab}{c^{\color{red}2}}\; (b,a)$? – hypergeometric Apr 17 '17 at 18:15
• @hypergeometric: Indeed! Good eye ... Thanks. – Blue Apr 17 '17 at 22:41
(New Solution - much shorter and more direct!)
Here we adopt the form of the equation used in @Blue's solution, i.e. $$\boxed{\qquad \sqrt{\frac xa}+\sqrt{\frac yb}=1\qquad}$$ Converting to parametric form by putting $\displaystyle \sqrt{\frac xa}=t$ gives \begin{align} \left[x\atop y\right] &=\left[t^2 a\atop (1-t)^2 b\right]\\ &=(1-t)^2\left[0\atop b\right] +2(1-t)t\left[0\atop 0\right]+ t^2\left[a\atop 0\right] \end{align} which is in the form of a Quadratic Bezier Curve (which is a parabola) with control points $B(0,b), \;O(0,0) ,\; A(a,0)$ where tangents to the curve at $B, A$ intersect at $O$.
Hence the equation represents (part of) a parabola which touches the $x-$axis and $y-$axis at points $A, B$ respectively. $\blacksquare$
(Previous Solution - much longer)
Taking the form used in @Blue's solution, we have
\begin{align} \sqrt{\frac xa}+\sqrt{\frac yb}&=1\tag{1}\\ \sqrt{bx}+\sqrt{ay}&=\sqrt{ab}\\ bx+ay+2\sqrt{abxy}&=ab\\ 4abxy&=\big[ab-(bx+ay)\big]^2\\ &=a^2b^2-2ab(bx+ay)+(bx+ay)^2\\ 0&=a^2b^2-2ab(bx+ay)+(bx-ay)^2\\ (bx-ay)^2&=2ab\left(bx+ay-\frac {ab}2\right)\tag{2}\\ (bx-ay)^2-2ab^2x-2a^2by+a^2b^2&=0\tag{3} \end{align} As ($3$) is of the form $(Ax+Cy)^2+Dx+Ey+F=0$, it must be a parabola. See this.
Hence the curve ($1$) is part of the same parabola.
Note that ($2$) can also be written as
$$\left(\frac xa-\frac yb\right)^2=2\left(\frac xa+\frac yb\right)-1\tag{2a}$$ or $$\left(\frac xa-\frac yb-1\right)^2=\frac {4y}b\tag{2b}$$ It can also be worked out that the parabola touches the $x$ and $y$ axes at $A(a,0)$ and $B(0,b)$ respectively. Setting $x=0$ in ($2$) gives $(y-b)^2=0$ i.e. coincident roots at $y=b$. Similarly, setting $y=0$ in ($2$) gives $(x-a)^2=0$ i.e. coincident roots at $x=a$. Hence the coordinate axes are tangent to the parabola. $\blacksquare$
Using information from the solutions here and here we can work out the following easily:
\begin{align} &\text{Parameter t:} &&t=\frac {ab(a^2-b^2)}{a^2+b^2}\\ &\text{Axis of symmetry:} &&bx-ay+\frac {ab(a^2-b^2)}{a^2+b^2}=0 &&\left[\frac xa-\frac yb+\frac {a^2-b^2}{a^2+b^2}=0\right]\\ &\text{Vertex, V, of parabola: } &&\left(\frac {ab^4}{(a^2+b^2)^2},\frac {a^4b}{(a^2+b^2)^2}\right)\\ &\text{Tangent at vertex:} &&ax+by-\frac{a^2b^2}{a^2+b^2}=0 &&\left[\frac xb+\frac ya-\frac {ab}{a^2+b^2}=0\right]\\ &\text{Directrix of parabola:} &&ax+by=0 &&\left[\frac xb+\frac ya=0\right]\\ &\text{Focus, F:} &&\left(\frac {ab^2}{a^2+b^2},\frac {a^2b}{a^2+b^2}\right)\\ &\text{Centre of Directrix*, M:} &&\left(\frac {ab^2(b^2-a^2)}{(a^2+b^2)^2},\frac {a^2b(a^2-b^2)}{(a^2+b^2)^2} \right)\\ &\text{Focal length, z:} &&\frac {a^2b^2}{(a^2+b^2)^{3/2}}=\frac {a^2b^2}{r^3} \end{align}
Note the following points:
• *The centre of directrix, $M$, is the intersection between the axis of symmetry and the directrix. By definition, $FV=VM$.
• The directrix is parallel to the tangent at the vertex.
• $O$ lies on the directrix of the parabola. This is a standard property of the parabola - the intersection point of two perpendicular tangents to the parabola lies on its directrix.
• The focus of the parabola, $F$, lies on the line $AB$ as well as the axis of symmetry.
See graphical implementation here.
Note the following:
Using $r=\sqrt{a^2+b^2}$, and dividing the equations above by $r$,
the Axis of Symmetry (the "$Y$" axis) can also be written as $$\overbrace{\frac {bx-ay}{r}+\frac {ab(a^2-b^2)}{r^3}}^X=0$$ and the Tangent at the Vertex (the "$X$" axis) can also be written as
$$\overbrace{\frac {ax+by}r-\frac {a^2b^2}{r^3}}^Y=0$$
Using focal length $z=\dfrac{a^2b^2}r^3$, the equation of the parabola can then be written as
$$X^2=4zY\\ \color{red}{\left[\frac {bx-ay}{r}+\frac {ab(a^2-b^2)}{r^3}\right]^2=4\left(\frac{a^2b^2}{r^3}\right)\left[\frac {ax+by}r-\frac {a^2b^2}{r^3}\right]\tag{4}}$$ It can be shown that equation $(4)$ is equivalent to equations $(2), (2a), (2b), (3)$, and hence the complete parabola for $(1)$.
(Relationship with standard rotated parabola form)
Let vertex $\displaystyle V=(h,k)=\left(\frac{ab^4}{r^2},\frac{a^4b}{r^2}\right)$ where $r^2=a^2+b^2$ and $\displaystyle\tan\theta=\frac ba$.
Some pre-processing. Note that $$\color{orange}{\frac ha-\frac kb=\frac {b^4-a^4}{r^r}=\frac {(b^2-a^2)(b^2+a^2)}{(a^2+b^2)^2}=\frac {b^2-a^2}{r^2}}$$ and $$\color{green}{\frac hb+\frac ka=\frac {ab(a^2+b^2)}{(a^2+b^2)^2}=\frac {ab}{a^2+b^2}=\frac {ab}{r^2}}$$ Also, $$\color{blue}{-\frac {4a^2b^2}{r^4}-\left(\frac {b^2-a^2}{r^2}\right)^2=\frac {-4a^2b^2-(b^4-2a^2b^2+a^4)}{r^4}=\frac {-(b^2+a^2)^2}{r^4}=-1}$$. A parabola with focal length $\displaystyle z=\frac {a^2b^2}{r^3}$ with vertex at $V$ and axis of symmetry rotated by $\theta$ clockwise from the vertical is given by \begin{align} (x-h)\cos\theta+(h-k)\sin\theta &=\frac 1{4a}\big[(x-h)\sin\theta-(h-k)\cos\theta)\big]^2\\ (x-h)\frac ar+(y-k)\frac br &=\frac {r^3}{4a^2b^2}\big[(x-h)\frac br-(y-k)\frac ar\bigg]^2\\ \frac {ab}r\bigg[\frac {x-h}b+\frac {y-k}a\bigg] &=\frac {r^3}{4a^2b^2}\cdot \frac {a^2b^2}{r^2}\bigg[\frac {x-h}a-\frac {h-k}b\bigg]^2\\ \frac {4ab}{r^2}\bigg[\left(\frac xb+\frac ya\right)-\left(\color{orange}{\frac hb+\frac ka}\right)\bigg] &=\bigg[\left(\frac xa-\frac yb\right)-\left(\color{green}{\frac ha-\frac kb}\right)\bigg]^2\\ \frac {4ab}{r^2}\bigg[\left(\frac xb+\frac ya\right)-\color{orange}{\frac {ab}{r^2}}\bigg] &=\bigg[\left(\frac xa-\frac yb\right)-\color{green}{\frac {b^2-a^2}{r^2}}\bigg]^2\\ \frac {4ab}{r^2}\left(\frac xb+\frac ya\right)-\color{blue}{\frac {4a^2b^2}{r^4}} &=\left(\frac xa-\frac yb\right)^2-2\left(\frac {b^2-a^2}{r^2}\right)\left(\frac xa-\frac yb\right)+\color{blue}{\left(\frac {b^2-a^2}{r^2}\right)^2}\\ \left(\frac xa-\frac yb\right)^2 &=\frac {4ab}{r^2}\left(\frac xb+\frac ya\right)+2\left(\frac {b^2-a^2}{r^2}\right)\left(\frac xa-\frac yb\right)\color{blue}{-1}\\ &=\bigg[\frac {4a}{r^2}+\frac 2a\left(\frac{b^2-a^2}{r^2}\right)\bigg]x+\bigg[\frac {4b}{r^2}-\frac 2b\left(\frac{b^2-a^2}{r^2}\right)\bigg]y-1\\ &=\frac 2a\bigg[\frac {2a^2+b^2-a^2}{r^2}\bigg]x+\frac 2b\bigg[\frac {2b^2-(b^2-a^2)}{r^2}\bigg]y-1\\ &=\frac 2a\left(\frac {a^2+b^2}{\\ r^2}\right)x+\frac 2b\left(\frac{a^2+b^2}{r^2}\right)y-1\\ &=2\left(\frac xa+\frac yb\right)-1 \end{align} which is effectively equation ($2a$) as derived from the original equation.
However, from the above, it can be seen working backwards from equation ($2a$) to the standard rotated/translated form is not quite so straightforward.
(From First Principles)
Start with the general equation for parabola, specify that it passes through and are tangential to the axes at $(a,0),(0,b)$.
General equation for parabola: $$(Ax+Cy)^2+Dx+Ey+F=0\tag{1}$$ At $(a,0):$ $A^2a^2+Da+F=0\tag{2}$ At $(0,b):$ $C^2b^2+Eb+F=0\tag{3}$ Differentiating $(1)$ and rearranging: $$\frac{dy}{dx}=-\frac {D+2A(Ax+CY)}{E+2C(aAx+Cy)}$$ At $(a,0)$, $\dfrac {dy}{dx}=0$ $\Rightarrow \quad D=-2A^2a\tag{4}$ At $(0,b)$, $\dfrac {dy}{dx}=\infty$ $\Rightarrow \quad E=-2C^2b\tag{5}$ Putting $(4),(5)$ in $(2),(3)$ gives $$F=A^2a^2=C^2b^2 \\ \Rightarrow {C=\pm \frac ab A\tag{6}}$$ Putting $(4),(5),(6)$ into $(1)$, diving by $A^2$ and rearranging: $$\left(\frac xa\pm\frac yb\right)^2-2\left(\frac xa+\frac yb\right)+1=0\tag{7}$$ Taking the $+$ sign in $\pm$ gives $$\left(\frac xa+\frac yb-1\right)^2=0$$ which graphs as two parallel lines.
Taking the $-$ sign in $\pm$ gives $$\left(\frac xa-\frac yb\right)^2=2\left(\frac xa+\frac yb\right)-1$$ which is the same as equation $(2a)$ derived from the original equation.
Hence the equation in the question represents part of a parabola to which the coordinate axes are tangential at $(a,0),(0,b)$ respectively.
(ANOTHER METHOD)
Some further thoughts based on a refreshing method by a friend of mine who is an excellent mathematician.
First note that in parametric form the curve is $$\left(x\atop y\right)=\left(at^2\atop b(1-t)^2\right)$$ Apply the rotation matrix $\dfrac 1{\sqrt{a^2+b^2}}\left(\begin{array} \ b&-a\\a&\;\;b\end{array}\right)$ to get rid of $t^2$ in the $x$-component, i.e. rotating clockwise by $\arctan \left(\frac ab\right)$ about the origin: \begin{align} \left(X\atop Y\right) &=\frac 1{\sqrt{a^2+b^2}}\left(\begin{array} \ b&-a\\a&\;\;b\end{array}\right) \left(at^2\atop b(1-t)^2\right)\\ &=\frac 1{\sqrt{a^2+b^2}}\left(ab(2t-1)\atop (a^2+b^2)t^2-2b^2t+b^2\right)\qquad {\leftarrow \text{linear in t}\quad\;\atop {\leftarrow \text{quadratic in t}}}\\ &=\frac 1r\left(ab(2t-1)\atop r^2t^2-2b2t+b^2\right) \qquad\qquad\qquad\text{(where r^2=a^2+b^2)}\\ &=\frac 1r\left(ab(2t-1)\atop r^2\left(t-\frac {b^2}{r^2}\right)^2+\frac {a^2b^2}{a^2+b^2}\right) \end{align} i.e. $Y=AX^2+BX+C$ which is a parabola. Hence the original curve is also a parabola. $\blacksquare$
By simple differentiation it can be shown that the the axes are tangent to the original parabola at $(a,0)$ and $(0,b)$. $\blacksquare$
Note that at $t=\frac {b^2}{r^2}$, $Y=Y_{\text{min}}=\frac {a^2b^2}{r^3}$ and $X=-\frac {ab(a^2-b^2)}{r^3}$, which is also the equation of the axis of symmetry.
Using the fact that two perpendicular tangents (the coordinate axes in this case) to a parabola intersect at the directrix, we conclude that the origin $O$ lies on the directrix of the original parabola. Since $O$ is invariant under the applied (as the rotation is about $O$), therefore $O$ also lies on the directrix of the rotated parabola. Also, since the rotated parabola is upright, its directrix must be the $x-$axis itself. As such the focal length of the parabola must be $Y_{\text{min}}$, i.e. $\dfrac {a^2b^2}{r^3}=\dfrac {a^2b^2}{(a^2+b^2)^{3/2}}$.
Applying the reverse rotation matrix $\displaystyle\frac 1r\left(\;\;b\;\;a\atop -a\;\;b\right)$ to the vertex, axis of symmetry and directrix of the rotated parabola, it can be easily shown that, for the original parabola:
Vertex is $$\frac 1{(a^2+b^2)^2}\left(ab^4\atop a^4b\right)$$ Axis of symmetry is $$\left(\frac ar\left(\frac {b^2(a^2-b^2)}{r^3}+Y\right)\atop \frac br\left(-\frac {a^2(a^2-b^2)}{r^3}+Y\right)\right)\quad\Longrightarrow\quad \frac xa-\frac yb=\frac {a^2-b^2}{a^2+b^2}$$ Directrix is $$\frac 1r\left(\;\;bX\atop -aX\right)\quad\Longrightarrow\quad \frac xb+\frac ya=0$$
(Special Note)
You can see right away that your equation definitely does not represent an entire parabola.
The branches of a parabola go to infinity, whereas in your equation both $x$ and $y$ are bounded: $$0 \le x \le a^2, \qquad 0 \le y \le b^2.$$
However, if we transform the equation by squaring it, isolating the radical and squaring again (thus introducing into the picture infinitely many additional points $(x,y)$ that were not solutions of the original equation), then we would see that e.g. for $a=b=1$ your equation does represent a (bounded) subset of points of a parabola, $$4xy=(1-x-y)^2.\tag{1}$$
An easy way to see that the above equation describes a parabola is a linear transformation of coordinates: $x=v+u, \ y=v-u$; so $v={1\over2}(x+y), \ u={1\over2}(x-y)$. In the transformed coordinates, equation $(1)$ reduces to $$v = u^2 + {1\over4}, \tag{2}$$ so $v$ is a simple quadratic function of $u$.
In our $a=b=1$ example, the (arc of) parabola indeed touches the $x$ and $y$ axes at the points $(1,0)$ and $(0,1)$ because these two points correspond to the slope ${dv\over du}=\pm1$ at $u=\pm{1\over2}$ in the transformed coordinates $(u,v)$.
In the general case, for arbitrary positive $a$ and $b$, the touch points are $(a^2,0)$ and $(0,b^2)$. (We can go from the particular case $a=b=1$ to the general case simply by rescaling the coordinate axes. Such rescaling preserves the type of conic, so a parabola in rescaled axes remains a parabola.)
• In the equation $$\frac{\sqrt{x}}{a}+\frac{\sqrt{y}}{b}=1,$$ which value of $\sqrt{x}$ should you take in order to get $\sqrt{y}=2b$? We must allow negative values of $\sqrt{\cdot}$ in order to get the entire parabola. My remark in the beginning should be understood "with positive (arithmetic) square root values". – Alex May 31 '17 at 2:20
• In my answer I included a generalized ellipse. Only by squaring the entire parabola comes in. the given form only convex portion wrt arc comes in, as stated in the last line of my answer. – Narasimham May 31 '17 at 3:15
A lot of physical punch is lost from the parabola equation if we do not stick to dimensional agreement to see $a,b$ as segments made on the axes.
$$\sqrt{x/a} + \sqrt{y/b} = 1\tag1$$
is a subset of generalized ellipse family
$$\left(\frac{x}{a} \right)^n + \left(\frac{y}{b}\right)^n = 1 \tag2$$
When $\pm$ sign is respected we appreciate that it could belong to any of the four quadrants.
$$\pm \sqrt{x/a} +\pm \sqrt{y/b} = 1\tag3$$
An equation of $\frac12$ order is not a conic. To get it to a classic conic form we need to massage it a bit.. to remove the $\pm$ in front of radicals we square two times, getting
$$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2+1 =\pm 2 \frac{x}{a} \pm2 \frac{y}{b}\pm 2 \frac{xy}{ab} \tag4$$
The determinant $$(2/ab)^2 - 4(1/a^2)(1/b)^2= 0$$ so they are all parabolas symmetrical to coordinate axes.
They are plotted below for values $a= 3, b=2.$ It can be seen that the conics plot beyond tangent points which could not be enabled by the equation in radicals form.
To confirm tangency of the four parabolas to coordinate axes set $x=0$ or $y=0$ we see that $x=a,y=b$ have double roots where the quadratic equations have zero discriminant and so are tangential to the $x,y$ coordinate axes.
Let it's possible for $x$-axes.
Hence, we have: $$\left(\frac{\sqrt{x}}{a}+\frac{\sqrt{y}}{b}\right)'=0$$ or $$\frac{1}{a\sqrt{x}}+\frac{y'}{b\sqrt{y}}=0,$$ which is impossible for $y'=0$.
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# Math Help - Help please
Show that 1+tan^2x = (cosx)^-2 where defined.
Don't have a clue what to do.
2. Originally Posted by jumba
Show that 1+tan^2x = (cosx)^-2 where defined.
Don't have a clue what to do.
To prove whether or not
$1+tan^2x=\frac{1}{cos^2x}$
then... either express tanx in terms of cosx, or vice versa
Simplest is to express tan in terms of cos
$1+\frac{sin^2x}{cos^2x}=\frac{cos^2x}{cos^2x}+\fra c{sin^2x}{cos^2x}$
A very important trigonometric identity completes the proof after you express that as a single fraction
3. Don't make it complicated -_-
$1+tan^2(x)=sec^2(x)=\frac{1}{cos^2(x)}$
4. Excellent response, Miss !!
However, $\frac{1}{cosx}$ is "called" $sec(x)$
so we must prove that
$1+tan^2(x)=sec^2(x)=\frac{1}{cos^2(x)}$
without swopping between the alternative descriptions of the same trigonometric ratio.
You have highlighted a situation where marks could be easily lost in an exam!
5. Originally Posted by Miss
Don't make it complicated -_-
$1+tan^2(x)=sec^2(x)=\frac{1}{cos^2(x)}$
I don't know if you understood, jumba,
the above is not a proof,
it's only a statement.
Whether you choose to call $\frac{1}{cosx}$ as $sec(x)$ or not, you still need to prove that it is $1+tan^2(x)$
6. Originally Posted by Archie Meade
I don't know if you understood, jumba,
the above is not a proof,
it's only a statement.
Whether you choose to call $\frac{1}{cosx}$ as $sec(x)$ or not, you still need to prove that it is $1+tan^2(x)$
Its an acceptable proof, not a statement !
Its a standard identity.
If you think like this, then you need to prove that $tan^2(x)=\frac{sin^2(x)}{cos^2(x)}$.
7. $sec(x)=\frac{1}{cos(x)}$
is a definition.
Please don't try to confuse the student.
8. Originally Posted by Archie Meade
$sec(x)=\frac{1}{cos(x)}$
is a definition.
Please don't try to confuse the student.
I did not confuse him
But what I wrote is an accpetable proof.
Am sure
9. It's not a proof.
$sec(x)$ is the name given to $\frac{1}{cos(x)}$
The question may be written
Prove $1+tan^2(x)=\frac{1}{cos^2(x)}$
Or..
Prove $1+tan^2(x)=sec^2(x)$
It's the same question!
Interposing the alternative name does not outline a proof.
10. Originally Posted by Archie Meade
It's not a proof.
$sec(x)$ is the name given to $\frac{1}{cos(x)}$
The question may be written
Prove $1+tan^2(x)=\frac{1}{cos(x)}$
Or..
Prove $1+tan^2(x)=sec(x)$
It's the same question!
Interposing the alternative name does not outline a proof.
Well, then $\frac{sin^2(x)}{cos^2(x)}$ is a name given to $tan^2(x)$. So prove that $tan^2(x)=\frac{sin^2(x)}{cos^2(x)}$.
and $\frac{cos(x)}{cos(x)}$ is a name given to 1.
....etc etc
You are moving around a circle.
11. Are you serious????
12. Originally Posted by Archie Meade
Are you serious????
Sure.
13. I reckon it's a case for a moderator then
14. Originally Posted by Archie Meade
I reckon it's a case for a moderator then
OK. Do it.
Question: Do you know how many proofs have trigonometric identities in Earth?
According to your posts: they are not proofs, they are just statements!
15. How do you show that
$sec(x)=\frac{1}{cos(x)}$
without only stating that
$sec(x)=\frac{1}{cos(x)}$ ?
or without just calling it an identity.
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## Class files
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Impulse response corresponding to differential equation always real?
An important class of linear and time-invariant systems are modeled by linear constant-coefficient differential equations (LCCDEs), i.e.
dnydtn+cn−1dn−1ydtn−1+…+c2d2ydt2+c1dydt=x(t),dnydtn+cn−1dn−1ydtn−1+…+c2d2ydt2+c1dydt=x(t),\frac{d^n y}{dt^n}+c_{n-1}\frac{d^{n-1} y}{dt^{n-1}}+…+c_2\frac{d^2 y}{dt^2}+c_1\frac{d y}{dt}=x(t),
where y(t)y(t) is the output and x(t)x(t) is the input.
When the input is a Dirac delta, the output is called the impulse response. If some of the coefficients cnc_n’s are complex, then it’s possible for the impulse response to be complex. But if coefficients are all real, is the impulse response of such a system always purely real? If not, please provide a counterexample.
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1
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For this system to represent an LTI, you have to specify that all initial conditions must be zero. If Y(s)Y(s) is the Laplace transform of y(t)y(t) and x(t)=δ(t−a)x(t) = \delta(t-a) then we have
Y(s)=e−assn+cn−1sn−1+⋯c1s. Y(s) = \frac{e^{-as}}{s^n + c_{n-1}s^{n-1} + \cdots c_1s}.
Now we can factor the denominator into terms that look like (s−r)(s-r) for real roots or (s−α)2+β2(s-\alpha)^2 + \beta^2 if α+iβ\alpha + i \beta is a complex root.
The inverse Laplace transforms of functions of the type
1s−r\frac{1}{s-r} and
1(s−α)2+β2 \frac{1}{(s-\alpha)^2+\beta^2}
are well known to be real functions (see any table of inverse transforms). So we can write Y(s)Y(s) as a product of functions with real inverse Laplace transforms, and response y(t)y(t) is a convolution of these functions, which is real-valued.
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Archive for the ‘polymath’ Category
Frankl’s union-closed conjecture — a possible Polymath project?
January 21, 2016
Although it was from only a couple of people, I had an enthusiastic response to a very tentative suggestion that it might be rewarding to see whether a polymath project could say anything useful about Frankl’s union-closed conjecture. A potentially worrying aspect of the idea is that the problem is extremely elementary to state, does not seem to yield to any standard techniques, and is rather notorious. But, as one of the commenters said, that is not necessarily an argument against trying it. A notable feature of the polymath experiment has been that it throws up surprises, so while I wouldn’t expect a polymath project to solve Frankl’s union-closed conjecture, I also know that I need to be rather cautious about my expectations — which in this case is an argument in favour of giving it a try.
A less serious problem is what acronym one would use for the project. For the density Hales-Jewett problem we went for DHJ, and for the Erdős discrepancy problem we used EDP. That general approach runs into difficulties with Frankl’s union-closed conjecture, so I suggest FUNC. This post, if the project were to go ahead, could be FUNC0; in general I like the idea that we would be engaged in a funky line of research.
(more…)
Gil Kalai starts Polymath10
November 5, 2015
I’ve already mentioned this on Google Plus, but my readership there is different from my readership here, so it seems a good idea to write a similar post here, to give what publicity I can to the fact that Gil Kalai has started a new Polymath project. The problem he would like solved (and I would very much like to see it solved too) is the famous delta-systems or sunflower conjecture of Erdős and Rado.
The problem, as with many problems in combinatorics, is easy to state, but fascinatingly hard to solve. It is a classic extremal problem, in that it asks how big some combinatorial object needs to be before it is guaranteed to contain a subobject of some particular kind. In this case, the object is a $k$uniform hypergraph, which just means a collection of sets of size $k$. The subobject one would like to find is a sunflower of size $r$, which means a collection of sets $A_1,\dots,A_r$ such that we can find disjoint sets $H, P_1,\dots,P_r$ with the $P_i$ disjoint and with $A_i=H\cup P_i$ for each $i$. I have used the letters $H$ and $P$ to stand for “head” and “petal” — $H$ is the head of the sunflower and $P_1,\dots,P_r$ are the petals. (more…)
Whither Polymath?
February 28, 2013
Over at the Polymath blog, Gil Kalai recently proposed a discussion about possible future Polymath projects. This post is partly to direct you to that discussion in case you haven’t noticed it and might have ideas to contribute, and partly to start a specific Polymathematical conversation. I don’t call it a Polymath project, but rather an idea I’d like to discuss that might or might not become the basis for a nice project. One thing that Gil and others have said is that it would be a good idea to experiment with various different levels of difficulty and importance of problem. Perhaps one way of getting a Polymath project to take off is to tackle a problem that isn’t necessarily all that hard or important, but is nevertheless sufficiently interesting to appeal to a critical mass of people. That is very much the spirit of this post.
Before I go any further, I should say that the topic in question is one about which I am not an expert, so it may well be that the answer to the question I’m about to ask is already known. I could I suppose try to find out on Mathoverflow, but I’m not sure I can formulate the question precisely enough to make a suitable Mathoverflow question, so instead I’m doing it here. This has the added advantage that if the question does seem suitable, then any discussion of it that there might be will take place where I would want any continuation of the discussion to take place.
(more…)
Polymath6: A is to B as C is to ???
February 5, 2011
The Polymath experiment is still very much in its infancy, with the result that we still have only a rather hazy idea of what the advantages and disadvantages are of open online multiple collaboration. It is easy to think of potential advantages and disadvantages, but the more actual experience we can draw on, the more we will get a feel for which of these plausible speculations are correct.
In my last post but one I outlined a few thoughts about the cap-set problem. Although this wasn’t meant as the beginning of a Polymath project (as I said in the post, it was modelled more on Gil Kalai’s notion of an “open discussion”), it had something in common with such projects, to the point where one of the worries people have about the Polymath approach actually occurred: I suggested a line of attack that was, unbeknownst to me, actively being pursued by somebody.
I do not know exactly what calculations went on in the minds of Michael Bateman and Nets Katz when they decided to go public with their ideas before they had fully worked them out, but my guess is that they wanted to establish that they had got there first. This they did by the very effective method of explaining that the simple observations that I had made were much weaker than what they already knew how to prove. As it happens, the story has had a happy ending for them, since they managed, soon after posting their comments, to push their ideas to the point where they have obtained the first improvement to the Roth/Meshulam bound, something that many people have wanted to do for a long time.
(more…)
Polymath3 now active
September 30, 2010
After a long initial discussion period, Polymath3, a project on Gil Kalai’s blog that aims to solve the polynomial Hirsch conjecture, has now started as an active research project. There is already quite a lot of material on his blog, and soon some of it should have migrated to a wiki, which will be a good place to get up to speed on the basics. I will update this post from time to time with news of how the project is going.
Something that is maybe worth pointing out is that although the problem looks at first as though you need to know all sorts of facts about convexity, there is a very nice purely combinatorial statement (about set systems) that would imply the conjecture. So there is no excuse not to think about it …
Polymath news
July 13, 2010
If you haven’t already spotted this, you might like to know that Scott Aaronson has just posed a very interesting unsolved combinatorial problem and invited Polymath-style thoughts on it. I’ll give a quick and I hope appetite-whetting account of the problem here, but you could skip it if you want and jump straight to his posts on his blog and at Mathoverflow.
The problem is completely elementary, but it has consequences of great interest to theoretical computer scientists. As Scott says, it also seems to be at a realistic level of difficulty and highly suitable for a Polymath approach. (The first assertion is a coded way of saying that it isn’t a notorious “impossible” complexity assertion in disguise.) It asks this. Suppose you colour the points of $\mathbb{Z}^d$ with two colours in such a way that the origin is red and each of the coordinate axes contains at least one blue point. Do there exist constants $c,\alpha>0$, independent of $d$, such that there must be a point $x$ that has at least $cd^\alpha$ neighbours that are coloured differently from $x$? At the moment, it seems that there isn’t even a proof that for sufficiently large $d$ there must be a point with 100 neighbours of the other colour. There is a non-trivial but not hard example that shows that $\alpha$ cannot be more than $1/2$. (Scott presents this in his Mathoverflow post.) A very brief remark is that if you just assume that not all points are coloured the same way, then the assertion is false: a counterexample is obtained if you colour a point red if its first coordinate is positive and blue otherwise.
Erdős’s discrepancy problem as a forthcoming Polymath project
January 6, 2010
This is an emergency post, since the number of comments on the previous post about Erdős’s discrepancy problem has become unwieldy while I’ve been away enjoying a bit of sunshine in Luxor. My head is full of amazing details from the walls and columns of ancient Egyptian tombs and temples. The main thing I didn’t see, or at least not until I finally saw them sticking up through the mist from my plane window yesterday morning, was the pyramids, since those are near Cairo. However, I didn’t feel too bad about that as my guide book, the Lonely Planet guide to Egypt, assured me that the pyramids never fail to disappoint (which was just one of many little gems of bad writing in that book).
For the first time for ages, I was completely away from all email and internet for a week, but just before I left, the results of the polls, as they then stood, about the next Polymath project were already suggesting that the Erdős discrepancy problem was the clear favourite, so I couldn’t help thinking about it a certain amount while I was in Egypt. I said that the process of choosing the next problem was not going to be fully democratic, but the fact is that I love the Erdős discrepancy problem and am currently somewhat grabbed by it, so I see no reason to go against such a clear majority, especially as it also has a clear majority of people who say that they are ready to work on it. (more…)
The next Polymath project on this blog: some polls
December 28, 2009
For reasons that I have already gone into, I do not think that the origin-of-life project is suitable as the next Polymath project on this blog, though there seems to be enough enthusiasm for it that I am quite serious about giving it a try at some point in the not too distant future. The complexity-related project also no longer seems such a great idea for now. That leaves three candidates from amongst the problems that I have posted about recently: the project related to the polynomial-DHJ problem, the project related to the Littlewood problem, and the project to solve Erdős’s discrepancy problem. My impression is that for each of these three projects there is already a small group of highly interested people, and certainly my level of enthusiasm for each of these three problems is enough for me to be ready to devote plenty of time to it. (A theory I want to test is that if I post regularly and am not completely stuck, then that will be enough to keep the project feeling active and attracting contributions from other people as well, even if a proof does not appear to be round the corner.)
To help me decide which of the above three problems to go for, here are four polls. (more…)
Erdős’s discrepancy problem
December 17, 2009
I’ve been pretty busy over the last month or so — hence my blog silence — and continue to be busy. But here is one more discussion of a problem that was on my earlier list of possible future Polymath projects. The problem in question is sometimes known as Erdős’s discrepancy problem. This will be a shortish post, because I don’t have much of a programme for solving the problem: if it were adopted, then we would be starting almost from scratch, but that could be an interesting experiment.
Here is the question.
Problem. Is it possible to find a $\pm 1$-valued sequence $x_1,x_2,\dots$ and a constant $C$ such that $|\sum_{i=1}^nx_{id}|\leq C$ for every $n$ and every $d$?
With the help of a little terminology we can ask this question in a slightly different way. If $\mathcal{A}$ is a collection of subsets of a set $X$ and $f:X\rightarrow\{-1,1\}$, define the discrepancy of $f$ to be the maximum value of $|\sum_{a\in A}f(a)|$ over all $A\in\mathcal{A}$. In our case, $\mathcal{A}$ is the collection of all arithmetic progressions of the special form $\{d,2d,3d,\dots,nd\}$, and the question is whether there is any function that has bounded discrepancy for this $\mathcal{A}$. I say “bounded” rather than “finite” because one can define the function $\delta(N,f)$ to be the discrepancy of $f$ with respect to all those sets in $\mathcal{A}$ that are subsets of $\{1,2,\dots,N\}$. Then the question is equivalent to asking whether there is any function $f$ for which $\delta(N,f)$ is a bounded function of $N$. (more…)
Problems related to Littlewood’s conjecture
November 17, 2009
This is the third in a series of posts in which I discuss problems that could perhaps form Polymath projects. Again, I am elaborating on a brief discussion that appeared in an earlier post on possible future projects. [Not for the first time in my experience, WordPress’s dates have gone funny and this was posted not on the 17th as it says above but on the 20th.]
An obvious objection to the Littlewood conjecture as a Polymath project is that it is notoriously hard. On its own that might not necessarily be a convincing argument, since part of the point of Polymath is to attempt to succeed where individual mathematicians have failed. However, a second objection is that the best results in the direction of the Littlewood conjecture, due to Einsiedler, Katok and Lindenstrauss, use methods that are far from elementary (and far from understood by me). I envisage this project as an elementary one, at least to begin with, so does that make it completely unrealistic? I shall try to argue in this post that there is plenty that could potentially be done by elementary methods, even if attempting to prove the conjecture itself is probably too ambitious.
Another advantage of tackling the conjecture by elementary means is that if we find ourselves forced to reintroduce the non-elementary methods that have led to the very interesting results of Einsiedler, Katok and Lindenstrauss, we will have a deeper understanding of those methods than if we had just passively learnt about them. I myself prefer to rediscover things than to learn them: it isn’t always practical, but it’s easier if you half bear in mind that they are there and have a vague idea about them. (more…)
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## Precalculus (10th Edition)
$9.56$ square units
By Heron's theorem $K=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}.$ Hence, $s=\frac{9+6+4}{2}=9.5$. Thus, $K=\sqrt{9.5(9.5-9)(9.5-6)(9.5-4)}\approx9.56.$
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# How can I remember this definition of gene? (Like hints) A sequence of nucleotides at a given locus along a chromosome.
###### Question:
How can I remember this definition of gene? (Like hints)
A sequence of nucleotides at a given locus along a chromosome.
### Question 12* Which pair of decimals has a sum that is greater than 1 and a difference that is less than 1? A. B. C. D. 0.1, 0.1 0.65, 0.45 0.7, 0.09 0.9, 0.1 A B с D
Question 12* Which pair of decimals has a sum that is greater than 1 and a difference that is less than 1? A. B. C. D. 0.1, 0.1 0.65, 0.45 0.7, 0.09 0.9, 0.1 A B с D...
### Tina is looking out a window and throws marble straight downward toward the sidewalk below at a speed of 5.00 m/s and the window is 39.0 meters above the sidewalk. Answer the two questions below, using three significant digits. Part A: What is the speed of the ball (11) when it hits the ground? Number m/s (Remember: enter a magnitude.) Part B: After 1.90 s, how far down has the marble travelled?
Tina is looking out a window and throws marble straight downward toward the sidewalk below at a speed of 5.00 m/s and the window is 39.0 meters above the sidewalk. Answer the two questions below, using three significant digits. Part A: What is the speed of the ball (11) when it hits the ground? Numb...
### Find the point on the terminal side of θ = three pi divided by four that has a y coordinate of 1. Show your work for full credit.
Find the point on the terminal side of θ = three pi divided by four that has a y coordinate of 1. Show your work for full credit....
### After the completion of the railroad, a major industry in the West was Oranching. banking. milling. O fishing
After the completion of the railroad, a major industry in the West was Oranching. banking. milling. O fishing...
### Find the next number in the sequence 6,11,16,21...
Find the next number in the sequence 6,11,16,21......
### Select the correct comparison set a set b O A the typical value is greater is set a the spread is greater in set b
Select the correct comparison set a set b O A the typical value is greater is set a the spread is greater in set b...
### PLS HELP I WILL GIVE BRAINLY
PLS HELP I WILL GIVE BRAINLY...
### Today Ian wants to run less than 7/2 mile write a fraction with the denominator of four to represent a distance that is less than 7/2 mile
Today Ian wants to run less than 7/2 mile write a fraction with the denominator of four to represent a distance that is less than 7/2 mile...
### In triangle XYZ, the measure of angle Z is 42°, and the measure of angle Y is 68.4°. What is the measure of angle X?
In triangle XYZ, the measure of angle Z is 42°, and the measure of angle Y is 68.4°. What is the measure of angle X?...
### Help would be amazing right now!
Help would be amazing right now!...
### Which of the following statements about the physical geography of North America is true? A. The Mississippi drainage basin is the fifth largest in the world. B. The largest tributary of the Mississippi is the Arkansas River. C. The most geologically active region of North America is found along the West Coast. D. The ancient rock of the Canadian Shield covers about one-fourth of Canada.
Which of the following statements about the physical geography of North America is true? A. The Mississippi drainage basin is the fifth largest in the world. B. The largest tributary of the Mississippi is the Arkansas River. C. The most geologically active region of North America is found along the...
### Which scale is used to describe the intensity of an earthquake? a. richter scale b. mercalli scale c. intensity scale d. destruction scale
Which scale is used to describe the intensity of an earthquake? a. richter scale b. mercalli scale c. intensity scale d. destruction scale...
### 7. You own a portfolio that has $1,750 invested in Stock A and$3,950 invested in Stock B. If the expected returns on these stocks are 9% and 14%, respectively, what is the expected return on the portfolio?
7. You own a portfolio that has $1,750 invested in Stock A and$3,950 invested in Stock B. If the expected returns on these stocks are 9% and 14%, respectively, what is the expected return on the portfolio?...
### Death Valley, located in California, is the lowest place in North America. It is 282 feet below sea level, or -282 feet. Mount McKinley in Colorado is the highest place in North America. Its elevation is approximately -71.76 times the elevation of Death Valley. What is the approximate elevation of the Mount McKinley? -20,236.32 feet 20,236.32 feet -3.93 feet 3.93 feet
Death Valley, located in California, is the lowest place in North America. It is 282 feet below sea level, or -282 feet. Mount McKinley in Colorado is the highest place in North America. Its elevation is approximately -71.76 times the elevation of Death Valley. What is the approximate elevation of t...
### Help please it due in 5 mins and i’m almost finshed
help please it due in 5 mins and i’m almost finshed...
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## Unambiguous “weak” vector valued $L^{+\infty}$ spaces?
For some time, I have been stuck to the problem to be described as follows. The (perhaps not so commonly known) facts given here are taken from R. E. Edwards' Functional Analysis (Holt, Rinehart and Winston 1965) pp. 578−589. Let $I=[0,1]$ with the Lebesgue measure, and consider real separable (not necessarily reflexive) Banach spaces $E$ with strong dual $F=E'_\beta$. A certain kind of "$L^{+\infty}(I,F)$", $\Lambda(I,E')$, representing the dual of $L^1(I,E)$ is constructed as follows. Let $Y$ be the vector space of a.e. bounded (i.e. bounded outside some set of measure zero) functions $g:I\to F$ such that ${\rm ev}_x\circ g:I\to\mathbb R$ given by $t\mapsto(g(t))(x)$ is measurable for all $x\in E$. Letting $N_0$ be the subspace formed by $g\in Y$ vanishing a.e., then $\Lambda(I,E')=Y/N_0$ becomes a Banach space when we equip it with the essential supremum norm of representatives $g$ of the equivalence classes $[g]=\{g+h:h\in N_0\}$. Then a linear homeomorphism $\Lambda(I,E')\to(L^1(I,E))'_\beta$ is given by $[g]\mapsto\ell:[f]\mapsto\int_I(g\ .f)$ where $(g\ .f)(t)=(g(t))(f(t))$.
The problem is now the following. Since generally preduals are not unique, there may be different (separable) spaces $E$ having linearly homeomorphic duals $F$. So, at least a priori, we cannot invariantly define some space "$\Lambda(I,F)$" as a certain kind of "$L^{+\infty}(I,F)$". According to this Philip Brooker's answer, there are nonisomorphic separable spaces $C(S)$ having isomorphic duals. One may then ask, whether (1) the corresponding spaces $\Lambda(I,(C(S))')$ are (isometrically) isomorpic or linearly homeomorphic, under the associated "natural" maps. Further, the dual of $L^1(I\times I)$ is represented by $L^{+\infty}(I\times I)$. Since $L^1(I\times I)$ is isomorphic to $L^1(I,L^1(I))$, we see that $L^{+\infty}(I\times I)$ is isomorphic to $\Lambda(I,(L^1(I))')$. One may then ask, whether (2) there are separable Banach spaces $E$ not linearly homeomorphic to $L^1(I)$, but having dual linearly homeomorphic to $L^{+\infty}(I)$ and $\Lambda(I,E')$ not linearly homeomorphic to $L^{+\infty}(I\times I)$ under the associated natural maps.
So, there are two concrete questions (1) and (2) above.
As an explanation of the phrase "natural map" above, I add the following. If $\ell_0$ is a linear homeomorphism $(C(S_1))'_\beta\to(C(S_2))'_\beta$, then the question is about whether a linear homeomorphism $\Lambda(I,(C(S_1))')\to\Lambda(I,(C(S_2))')$ is given by $[g]\mapsto[\ell_0\circ g]$. For the second question, if $\ell_0$ is a linear homeomorphism $E'_\beta\to L^{+\infty}(I)$, then the question is about whether a linear homeomorphism $\Lambda(I,E')\to L^{+\infty}(I\times I)$ is given by $[g]\mapsto[\hat g]$ where $[\hat g(t,\cdot)]=\ell_0(g(t))$ for suitably chosen $\hat g$.
I have above taken the attitude that the isomophism (or linear homeomorphism) class the space $\Lambda(I,E')$ is not solely determined by that of $E'_\beta$ but depends also on $E$. If someone knows the contrary to be true, I am gratefull for a reference or a proof. Also possible couterexamples where for separable Banach spaces $E,E_1$ with $E'_\beta$ and $(E_1)'_\beta$ linearly homeomorphic but $\Lambda(I,E')$ and $\Lambda(I,E_1')$ not, where the spaces $E,E_1$ are not some $C(S)$ or $L^1(I)$ as I suggested above, are wellcome.
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# Python etc / zen of python (PEP-20)
Published: 27 August 2020, 18:00
The famous “Zen of Python” was introduced in PEP-20. This is 19 aphorisms authored by Tim Peters. Do import this in the Python interpreter to see them:
>>> import this
The Zen of Python, by Tim Peters
Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than *right* now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!
The fun thing is how this module looks like. The original text is encoded by ROT13 algorithm and is decoded on the fly:
s = """Gur Mra bs Clguba, ol Gvz Crgref
...
"""
d = {}
for c in (65, 97):
for i in range(26):
d[chr(i+c)] = chr((i+13) % 26 + c)
print("".join([d.get(c, c) for c in s]))
Some say it violates almost all the principles that it contains.
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# A weaker form of Zariski's connectedness principle
Let $A$ be a complete regular local noetherian ring of dimension $d>1$ and $B$ an $A$-algebra, finite and free as $A$-module. Assume moreover that there exists an open subset $U$ of $\textrm{Spec}\ A$ of primes of height $d-1$ such that there is exactly one prime in $\textrm{Spec}\ B$ over a prime of $U$. Is it then true that $\textrm{Spec}\ B$ is an irreducible topological space?
Under various choices of ancillary hypotheses, results like this one follow from Zariski's connectedness principle; see e.g A useful form of principle of connectedness.
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Yes. Assume not. Since $B$ is finite flat over $A$, each irreducible component of $\mathrm{Spec}(B)$ is pure $d$-dimensional, finite and surjective over $\mathrm{Spec}(A)$. If $Y\neq Z$ are two of them, then the image of $Y\cap Z$ in $\mathrm{Spec}(A)$ is a proper closed subset, hence does not contain $U$. Every point of $U$ has a preimage in $X$ and one in $Y$.
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# Find the limit as x goes to 0 of [sin^2 Ax]/[cos Ax - 1] *using L'Hopital's Rule?
• Find the limit as x goes to 0 of [sin^2 Ax]/[cos Ax - 1] *using L'Hopital's Rule?
Example 1: Find the limit lim x→0 sin x / x Solution to Example 1: ... are both equal to zero, we can use L'hopital's rule to calculate limit lim x→1 ...
Positive: 63 %
... , which handles the indeterminate form 0 / 0 at x = 0: → () ... as x goes to infinity; with this ... is using L'Hôpital's rule with some ...
Positive: 60 %
### More resources
The following problems require the algebraic computation of limits of functions as x approaches ... of L'Hopital's Rule. ... by using algebraic ...
Positive: 63 %
x→0 x In order to compute ... θ→0(sin(θ)/θ) = 1. We switch from using x to using θ because we want to start thinking about ... Limit sin(x)/x = 1 ...
Positive: 58 %
The limit of a function f(x) ... and also for infinite limits using the rules. ... This rule uses derivatives to find limits of indeterminate forms 0/0 or ...
... I still can't take the limit as x ... to \ \dfrac{0}{0}$$Now use L'Hopital's Rule.$$\lim_{x\to 0 ... Using L'Hôpital's rule to find \$\lim_{x\to ...
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### Home > CALC > Chapter 7 > Lesson 7.3.4 > Problem7-151
7-151.
1. Multiple Choice: If y = x + sin(xy), then =
Homework Help ✎
1. 1 + cos(xy)
2. 1 + y cos(xy)
Implicit differentiation with the Chain Rule can be messy.
$y'=1+\text{cos}(xy)\left ( y\frac{dx}{dx}+x\frac{dy}{dx} \right )$
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# zbMATH — the first resource for mathematics
On the convergence of a four-node plate bending element based on Mindlin/Reissner plate theory and a mixed interpolation. (English) Zbl 0589.73068
The mathematics of finite elements and applications V, MAFELAP 1984, Proc. 5th Conf., Uxbridge/Engl. 1984, 491-503 (1985).
[For the entire collection see Zbl 0566.00028.]
The paper aims at analysing, from a mathematical point of view, the convergence of a four-node plate bending element based on Mindlin/Reissner plate theory and a mixed interpolation. A sequence of problems (for different values of thickness parameter t) is constructed and the behaviour of the corresponding solutions is studied as thickness becomes ”too small”, to test for the stability of discretization of the four-node plate element. It is proved that, at least for particular cases (like a uniform rectangular mesh considered by the authors), the element is uniformly stable with respect to the thickness parameter and convergent with optimal rate. The paper does not establish the stability for shear strains nor the uniform convergence. The authors suggest the application of filtering procedure for shear strains to have an $$L^ 2$$- stability and optimal rate of convergence uniformly in t.
The proofs are based on sound mathematical analysis and involve the use of several theorems and lemmas. The paper is a good contribution to the mathematical theory of finite elements.
Reviewer: V.K.Arya
##### MSC:
74S05 Finite element methods applied to problems in solid mechanics 74K20 Plates
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# Find the program that prints this integer sequence (Cops' thread)
• The cop's task is to write a program/function that takes a positive (or non-negative) integer and outputs/returns another integer.
• The cop must state it if the program is intolerant towards zero.
• The cop will provide at least 2 sample inputs/outputs.
• For example, if I have decided to write the Fibonacci sequence, I would write this in my answer:
a(0) returns 0
a(3) returns 2
• The number of examples is up to the cop's discretion.
• However, the sequence must actually exist in the On-Line Encyclopedia of Integer Sequences®, so no pseudo-random number generator for you. :(
• The cop can hide as many characters as is so wished.
• For example, if my program is:
function a(n)
if n>2 then
return n
else
return a(n-1) + a(n-2)
end
end
• Then I would hide these characters as I want:
function a(n)
if ### then
########
else
######################
end
end
• is obviously to find the original source code.
• However, any proposed source code that produces the same set of output also counts as valid, as long as it is also found in OEIS.
## Tips for the cops
• The search function in the OEIS only works for consecutive terms, so if you want to hide your sequence, then just leave a hole anywhere.
• Apparently there is no way to hide the sequence. Put this in mind when you choose the sequence.
The winner will be the submission with the lowest score that hasn't been cracked in 7 days.
Only submissions that are posted in 2016 April are eligible for the win. Submissions that are posted later than this are welcome, but cannot win.
In order to claim the win you need to reveal the full code and the OEIS sequence (after 7 days).
Your post should be formatted like this (NN is the number of characters):
# Lua, 98 bytes
Output:
a(0) returns 0
a(3) returns 2
Code (# marks unrevealed characters):
function a(n)
if ### then
########
else
######################
end
end
If the code is cracked, insert [Cracked](link to cracker) in the header. If the submission is safe, insert "Safe" in the header and reveal the full code in your answer. Only answers that have revealed the full code will be eligible for the win.
• Also, OEIS search can have blanks with _, fyi – Sp3000 Apr 9 '16 at 0:59
• It may be too late to change, but allowing sequences that are multiples of an OEIS sequence, and/or only include every nth term would have made this challenge much better. sandbox, hint, hint – Nathan Merrill Apr 9 '16 at 2:37
• Can I, for example, pick the Fibonacci sequence and provide only a(1000)? (which is part of the sequence, but too large to be searchable on OEIS) – Sp3000 Apr 9 '16 at 6:36
• I'd say the values have to actually be searchable on OEIS, so that it can easily be verified that the values are correct for the chosen sequence. – Mego Apr 9 '16 at 6:38
• "Intolerant towards zero" makes no sense. What is that supposed to mean? – feersum Apr 10 '16 at 23:28
## 05AB1E, 3 bytes, Cracked
Output:
a(3) = 2
a(4) = 6
Code:
___
_ (underscore) indicates an obfuscated character
Try it online
Original Code:
D;^
Which is the sequence A003188 (Decimal equivalent of Gray code for n)
• Cracked. That was a nice puzzle! :) – Adnan Apr 28 '16 at 22:54
## Python 3, 58 Bytes, Cracked
# marks unrevealed chars.
import math;x=int(input());print(int(x/math.####(#)*#**#))
If x < 1, it will not work.
in out
1 0
4 457
7 2578097
10 44721359549
13 1760848250285208
16 131994155879539032064
19 16810747184697114703691776
## Pyth, 51 bytes, Safe
# denote hidden characters:
K"#"D#d=#:d"#|#"1Rs#?qd"#"####"##"b#W<#K+#1=KhK;@KQ
K"0"Dhd=b:d"0|1"1Rsm?qd"0""01""10"b;W<lK+Q1=KhK;@KQ
## Outputs
a(0) -> 0
a(1) -> 1
a(2) -> 1
a(4) -> 0
This is a fair one.
Thue-morse sequence, A010060.
# Python, 17 bytes, Cracked
####a(n):########
Test cases:
a(1) -> 1
a(2) -> 2
• Cracked – Blue Apr 12 '16 at 17:40
# Python, 60 bytes, Cracked
def a(n):
if n<#:
return #
else:
return a(n-#)+a(n-#)
Test cases:
a(0) -> 1
a(1) -> 1
a(3) -> 2
a(6) -> 6
• Cracked – Blue Apr 12 '16 at 17:46
# Jelly, 5 bytes (safe)
### Code
??Œ??
Each ? denotes a missing character.
### Output
n a(n)
0 1
1 3
2 9
3 21
4 45
5 93
6 189
7 381
8 765
9 1533
This is OEIS entry A068156.
### Solution
Ḷ¡ŒṘL
Try it online!
### How it works
As noted on the OEIS page, this sequence obeys the recursive formula a(n + 1) = 2a(n) + 3, with base cases a(0) = 1, a(1) = 3.
The code begins with the integer n and applies Ḷ (unlength) n times, which convert all integers k in the previous array into the range [0, ... k - 1]. The output after all n steps is similar to the set-theoretic definition of n.
For n = 0, 1, 2, 3, 4, the outputs of Ḷ¡ŒṘ (unlength n times, then generate Python's string representation) is as follows.
0: 0
1: [0]
2: [[], [0]]
3: [[], [[]], [[], [0]]]
4: [[], [[]], [[], [[]]], [[], [[]], [[], [0]]]]
If we examine the outputs for 3 and 4 closely, we can note the following pattern.
[[], [[]], [[], [0]]]XX[[], [[]], [[], [0]]]X
[[], [[]], [[], [[]]], [[], [[]], [[], [0]]]]
The output for 4 consists of two copies of the output for 3 (the actual characters differ, but we're only interested in the length), and three additional characters (X marks the spots). Thus, if we take the length with L, our program follows the desired recursive formula.
## Python 3, 108 bytes (Cracked)
Warning (and hint): my implementation is dreadfully slow...
>>> a(0)
0
>>> a(4)
2
>>> a(16)
8
>>> a(20)
42
a(0) is the first element of the series exactly as defined in OEIS. # is used to hide chars.
def a(n):
###n####0######n#0
f=a#######
return f #f#####a###### f ####a(##f###i#i###a####n##else a#######
• Cracked – FliiFe Apr 10 '16 at 20:58
## Python 3, 50 bytes (Cracked)
Another one, because I like this sequence
>>> a(1)
0
>>> a(10)
2303
The first number from the sequence is a(0). # used to obfuscate characters.
from #### import ####
a=lambda n:####(###########)
• Cracked – Mego Apr 11 '16 at 1:44
# Python 3, 58 bytes (Cracked)
One last series
>>> a(0)
0
>>> a(4)
8
>>> a(5)
17
Sequence starts with a(0).
a=lambda n:###(###(###)## if ###### else #####(#######)##)
Original was different than the cracked, because I can't golf Python :P
a=lambda n:int(3**(n/2)-1 if n%2==0 else 2*3**(n/2-1/2)-1)
## Pyth, 29 bytes
W##YQ#t^#Z=#h#I#_#=#+Y##;@#tQ
Probably not an easy one. Begins at 1.
a(1) -> 2
a(2) -> 3
a(3) -> 5
a(4) -> 7
a(5) -> 13
# Python 2, 72 bytes, cracked
print####((#y##########.####c###6#####16#.######)#131963###input()#3###)
Outputs:
a(1) = 1
a(2) = 2
a(3) = 3
a(4) = 6
a(5) = 9
a(6) = 18
a(1) is the first element of the sequence. Tested on my personal computer, on repl.it and on ideone (which have versions 2.7.11, 2.7.2 and 2.7.10 respectively).
The sequence is finite, and all terms are listed - it's just the divisors of 18. However, I overlooked the idea that #s could be substituted for newlines...
The original one-liner was:
print len((BytesWarning.__doc__*6)[8::16].split()[1319648>>input()*3&7])
## Pyth, 4 bytes, Cracked
# denote hidden characters:
###Q
## Output
a(1) -> 1
a(2) -> 2
a(3) -> 3
a(10) -> 42
• Is this offset at all? – Blue Apr 11 '16 at 17:22
• @muddyfish ... I posted this answer without double-checking. I updated the examples. – FliiFe Apr 11 '16 at 17:30
• @muddyfish To answer your question, a(0) is the first term of the sequence. I'm not sure if that's what you wanted to know, but at least you know that. – FliiFe Apr 11 '16 at 17:32
• – Adnan Apr 11 '16 at 18:31
# Molecule (v5.5), 25 bytes (UTF-8)
"ᘰᜟᘄఄሳÝ̲"Cn
No characters are hidden here ;)
An input of 0 is allowed.
Output of 1: 16
Output of 2: 21
• You need to hide some characters to make it a valid submission – TuxCrafting Oct 22 '16 at 21:37
• The worst thing: It's safe xD – TuxCrafting Oct 22 '16 at 21:37
## Python, 42 bytes Cracked
def f(x):return # if x<#else ######+######
Where
f(0) -> 0
f(1) -> 1
f(7) -> 13
f(15) -> 610
• Cracked. – Dennis Oct 4 '16 at 15:40
# PHP, 118 Bytes, Safe
I hope that I'm understand the rules right
function f($n){for($a=[],$i=0,$n++;$n!=@@@@@@($a);$i++)if(@@@@@@@)if(@@@@@@@@(@@@@@@@@@@@@@@))$a[]=$i;return end($a);}
echo f(6); # 16
echo f(13);# 32
echo f(0); #2
echo f(1); #4
echo f(54); #128
Original code
function f($n){ for($a=[],$i=0,$n++;$n!=sizeof($a);$i++){ if($t=$i%7)if(in_array($t,array(2,4,6)))$a[]=$i;
}
return end(\$a);
}
`
generates this sequence
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Miktex amsfont problem: \scriptfont is undefined [duplicate]
Today, After updating Miktex system, I have trouble with error \scriptfont is undefined. So I tried to reinstall Miktex 2.9.4813 (32 bit version). Everything OK, but again after updating, problem is back. I see, that this update related for example file fontspec, amsmath and so on. Does anyone have a similar problem as me?
Texlipse Console:
running: C:\Program Files (x86)\MiKTeX 2.9\miktex\bin\xelatex.exe -synctex=1 - interaction=nonstopmode --src-specials main.tex --enable-write18
xelatex.exe> This is XeTeX, Version 3.1415926-2.5-0.9999.3 (MiKTeX 2.9)
...
xelatex.exe> ! \textfont 173 is undefined (character ???).
xelatex.exe> l.126 \item $V_{OH}(3.3V)>V_{IH}(5V)$
xelatex.exe> ,
xelatex.exe> ! \textfont 53 is undefined (character ???).
xelatex.exe> l.126 \item $V_{OH}(3.3V)>V_{IH}(5V)$
xelatex.exe> ,
xelatex.exe> [395]
xelatex.exe> ! \textfont 117 is undefined (character ???).
xelatex.exe> l.127 \item $V_{OL}(3.3V)<V_{IL}(5V)$
xelatex.exe> .
xelatex.exe> ! \textfont 176 is undefined (character ???).
xelatex.exe> l.127 \item $V_{OL}(3.3V)<V_{IL}(5V)$
xelatex.exe> .
xelatex.exe> <use "../src/CES/img/level_shifting_MOSFET.pdf" >
xelatex.exe> ! \textfont 100 is undefined (character ???).
xelatex.exe> l.153 $V_{OH}(3.3V)<V_{IH}(5V)$
...
xelatex.exe>
xelatex.exe> ! \textfont 52 is undefined (character ???).
xelatex.exe> l.153 $V_{OH}(3.3V)<V_{IH}(5V)$
...
xelatex.exe>
xelatex.exe> ! \scriptscriptfont 108 is undefined (character ???).
xelatex.exe> l.153 ...tÃm $$V_{GS_{th_{max}}}<V_{OH_{min}}.$$
MWE: I came to the conclusion that the problem is with the font in mathematical environment.
\documentclass{book}
\usepackage[log-declarations=false]{xparse}
\usepackage{xltxtra}
\usepackage{fontspec}
\setmainfont[Mapping=tex-text,Numbers={OldStyle,Proportional}]{Calibri}
\setsansfont[Mapping=tex-text,Numbers={OldStyle,Proportional}]{Candara}
\setmonofont[Numbers=OldStyle]{Consolas}
\usepackage{ifthen}
\usepackage{amsmath, amsthm, amssymb, amsfonts, amsbsy}
\usepackage{accents}
\usepackage{unicode-math}
\setmathfont{Cambria Math}
\begin{document}
\chapter{Chapter 1}
\section{Section 1}
$b_{i,j}$
\end{document}
I see another suspicious warning:
• Couldn't open font map file "kanjix.map"
This warning came after the update.
Same difficulties are discussed in this post
• I think you are not the only one having this problem... See Updating MiKTeX: again errors. – karlkoeller Jun 30 '13 at 6:03
• I can reproduce the problem. TL13 doesn't give an error but a suspicious missing character message. It looks like an engine bug. @KhaledHosny could probably say more. I have sent a message to the xetex mailing list. – Ulrike Fischer Jun 30 '13 at 10:08
• It indeed seems to be the same problem as discussed in Updating MiKTeX: again errors. In the preamble of the MWE you just need the \usepackage{unicode-math} and \setmathfont{Cambria Math} to reproduce the problem. It also does not work with lualatex, so I'd rather suspect it to be a problem with unicode-math!? – dpprdan Jul 1 '13 at 8:17
• FWIW: I can also reproduce the problem with XITS Math. – dpprdan Jul 1 '13 at 8:30
• @dapperdan: Are you sure that you get the same error with lualatex? Or only an error about Umathcode? In the second case: check the archive of the miktex mailing. I have described there what to do to get around this problem. – Ulrike Fischer Jul 2 '13 at 12:04
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# Microsoft, Now Loving Hadoop
Last week, OStatic noted the rumor, first reported by VentureBeat, that Microsoft intended to buy Silicon Valley semantic search engine Powerset for $100 million. Lo and behold, Microsoft and Powerset are confirming today that an acquisition agreement has been signed. The terms of the deal have not been disclosed, but the rumored$100 million figure was in line with valuations put on Powerset based on its early financing.
Powerset’s search technology uses the open-source, cluster-based technology Hadoop, which provides fast answers to queries by using the resources of many computers. Hadoop, a project from the Apache Software Foundation, is also behind Yahoo’s search.
Natural language search got a bad rap early on in the rise of the web as players such as AskJeeves stumbled, but clustered query technology like Hadoop’s may represent a game-changer. Microsoft, of course, has been desperately trying to catch up in search, where it is a distant third to Google and Yahoo. It won’t be surprising to see large portions of Microsoft’s LiveSearch start to depend on Powerset, and in so doing, depend on open-source upstart Hadoop.
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# Prove following points lie on a circle.
I found this in a textbook without a solution and I wasnt able to solve it myself.
Let ABCD be a tetrahedron with all faces acute. Let E be the mid point of the longer arc AB on a circle ABD. Let F be the mid point of the longer arc BC on a circle BCD. Let G be the mid point of the longer arc AC on a circle ACD.
Show that points D,E,F,G lie on a circle.
My approach to that was to try to show these point were co-planear. Since they all lie on one sphere (the one with inscribed tetrahedron ABCD) that would solve the problem. Needles to say I failed at that.
Let $$\vec{a} = \vec{DA}, \vec{b} = \vec{DB}, \vec{c} = \vec{DC}$$ and $$a,b,c$$ be corresponding magnitudes.
Let us look at what happens on the plane holding circle $$ABD$$. Let $$X$$ be the circle's center and $$E'$$ be the mid point of the shorter arc $$AB$$. It is not hard to see
$$\angle E'DB = \frac12 \angle E'XB = \frac12 \angle AXE' = \angle ADE'$$
This implies $$DE'$$ is the angular bisector of $$\angle ADB$$ and $$\vec{DE'}$$ is pointing along the direction $$\frac{\vec{a}}{a} + \frac{\vec{b}}{b}$$. Since $$DE$$ is perpendicular to $$DE'$$, $$\vec{DE}$$ is pointing along the direction $$\frac{\vec{a}}{a} - \frac{\vec{b}}{b}$$.
To proceed, we will re-express this fact in terms of barycentric coordinates.
For any $$P \in \mathbb{R}^3$$, the barycentric coorindates of $$P$$ with respect to tetrahedron $$ABCD$$ is a 4-tuple $$(\alpha_P, \beta_P, \gamma_P, \delta_P)$$ which satisfies: $$\alpha_P + \beta_P + \gamma_P + \delta_P = 1\quad\text{ and }\quad \vec{P} = \alpha_P \vec{A} + \beta_P \vec{B} + \gamma_P \vec{C} + \delta_P\vec{D}$$
In particular, the barycentric coordinates for $$D$$ is $$(0,0,0,1)$$.
Let's look at point $$E$$. Since $$E$$ lies on the plane holding $$ABD$$, $$\gamma_E = 0$$. Since $$DE$$ is pointing along the direction $$\frac{\vec{a}}{a} - \frac{\vec{b}}{b}$$, we find $$\alpha_E : \beta_E = \frac1a : -\frac1b$$. From this, we can deduce there is a $$\lambda_E$$ such that
$$(\alpha_E, \beta_E, \gamma_E, \delta_E) = \left(\frac{\lambda_E}{a}, -\frac{\lambda_E}{b}, 0, 1 + \lambda_E\frac{a - b}{ab}\right)$$
By a similar argument, we can find $$\lambda_F$$ and $$\lambda_G$$ such that
$$(\alpha_F, \beta_F, \gamma_F, \delta_F) = \left(0,\frac{\lambda_F}{b}, -\frac{\lambda_F}{c}, 1 + \lambda_F\frac{b - c}{bc}\right)\\ (\alpha_G, \beta_G, \gamma_G, \delta_G) = \left(-\frac{\lambda_G}{a},0,\frac{\lambda_G}{c}, 1 + \lambda_G\frac{c - a}{ca}\right)$$ In terms of barycentric coordinates, $$D,E,F,G$$ are coplanar when and only when following determinant evaluates to zero.
$$\mathcal{D}\stackrel{def}{=}\left|\begin{matrix} \alpha_E & \beta_E & \gamma_E & \delta_E\\ \alpha_F & \beta_F & \gamma_F & \delta_F\\ \alpha_G & \beta_G & \gamma_G & \delta_G\\ \alpha_D & \beta_D & \gamma_D & \delta_D \end{matrix}\right| = \left|\begin{matrix} \alpha_E & \beta_E & \gamma_E & \delta_E\\ \alpha_F & \beta_F & \gamma_F & \delta_F\\ \alpha_G & \beta_G & \gamma_G & \delta_G\\ 0 & 0 & 0 & 1 \end{matrix}\right| = \left|\begin{matrix} \alpha_E & \beta_E & \gamma_E\\ \alpha_F & \beta_F & \gamma_F\\ \alpha_G & \beta_G & \gamma_G\\ \end{matrix}\right|$$ Substitute above expression of barycentric coordinates of $$E,F,G$$ into last determinant, we find $$\mathcal{D} = \lambda_E\lambda_F\lambda_G \left| \begin{matrix} \frac1a & -\frac1b & 0\\ 0 & \frac1b & -\frac1c\\ -\frac1a & 0 &\frac1c \end{matrix} \right| = 0$$ as the rows of determinant on RHS sum to zero.
From this, we can conclude $$D, E, F, G$$ are coplanar. Since $$D, E, F, G$$ lie on the intersection of a sphere and a plane, they lie on a circle.
• Sorry for the ignorance but how does the direction of DE tell us anything about the ratio αE:βE. Could you explain that part? – Dood Dec 3 '18 at 19:49
• Also does the fact that D,E,F,G are coplanar if and only if the determinant D=0 come from the formula for volume of tetrahedron using its verticies in barycentric coordinates? If so could you show the whole formula or at least link to it? I cant find it anywhere but i assume it's simmilar to the formula for a triangle which I did find. Many thanks for your contribution anyway. – Dood Dec 3 '18 at 19:59
• @Dood, yes, it is related to the formula of volume of tetrahedron. No. I don't have a link. This is a well known result and generalise to any finite dimensional simplex. – achille hui Dec 4 '18 at 0:30
• For the first question, $\vec{DE} \propto \frac{\vec{a}}{a} - \frac{\vec{b}}{b}$ implies existance of $\lambda_E$ such that $$\vec{E}-\vec{D} = \lambda_E\left(\frac{\vec{a}}{a} -\frac{\vec{b}} {b}\right) = \lambda_E\left(\frac{\vec{A}-\vec{D}}{a} - \frac{\vec{B}-\vec{D}}{a}\right)\\ \iff \vec{E} = \frac{\lambda_E}{a}\vec{A} - \frac{\lambda_E}{b}\vec{B} + \left(1 - \frac{\lambda_E}{a} + \frac{\lambda_E}{b}\right)\vec{D}$$ – achille hui Dec 4 '18 at 0:38
I'll write the tetrahedron as $$OABC$$, with $$O$$ at the origin, and I'll let $$D$$, $$E$$, $$F$$ be the new points associated with faces $$\triangle OBC$$, $$\triangle OCA$$, $$\triangle OAB$$. Define $$a := |OA| \qquad b := |OB| \qquad c := |OC| \qquad \alpha := \angle BOC \qquad \beta := \angle COA \qquad \gamma = \angle AOB$$ and recall that, for instance, $$B\cdot C = b c \cos\alpha \qquad |B\times C| = b c \sin\alpha$$
Consider the situation with $$\triangle OBC$$. The defining arc property for $$D$$ indicates that this point lies on the perpendicular bisector of $$\overline{BC}$$ within the plane of $$\triangle OBC$$. Thus, $$\overrightarrow{DD^\prime}$$ is perpendicular to both $$\overrightarrow{BC}$$ and the normal to the plane (that is, $$B\times C$$). we can write
$$D=D^\prime + |DD^\prime| \frac{( C-B )\times ( B \times C )}{|BC|\,|B\times C|} \tag{1}$$ where $$D^\prime := \frac12(B+C)$$ is the midpoint of $$\overline{BC}$$.
Further, by the Inscribed Angle Theorem, $$\angle BOC\cong\angle BDC$$, and we conclude that $$\overline{DD^\prime}$$ is the altitude of an isosceles triangle with vertex angle $$\alpha$$ and base $$|BC|$$. Therefore, $$|DD^\prime| = \frac12|BC|\,\cot\frac12\alpha$$, so we have $$\frac{|DD^\prime|}{|BC|\,|B\times C|} = \frac{\cot\frac12\alpha}{2bc\sin\alpha} = \frac{\cos\frac12\alpha\,/\,\sin\frac12\alpha}{4bc\sin\frac12\alpha\cos\frac12\alpha}=\frac{1}{4bc\sin^2\frac12\alpha} = \frac{1}{2bc(1-\cos\alpha)} \tag{2}$$ Further, via a cross product identity, \begin{align} (C-B)\times(B\times C) &= \phantom{-}B\,((C-B)\cdot B) - C\,((C-B)\cdot C) \\ &=\phantom{-}B\left(B\cdot C - |B|^2\right)-C\left(|C|^2 - B\cdot C\right) \\ &=-B\,b(b-c\cos\alpha) - C\,c(c-b\cos\alpha) \end{align}\tag{3} Altogether, this gives us \begin{align}D\;2bc(1-\cos\alpha) &= (B+C)bc(1-\cos\alpha)-B\,b(b-c\cos\alpha)-C\,c(c-b\cos\alpha) \\ &=Bb(c-b)+Cc(b-c) \\ &=(c-b)(B b-C c) \tag{4a} \end{align} (Note that, if $$b=c$$, then $$D=0$$, as we would expect. This confirms that we got our cross-product vector directions correct in $$(1)$$.) Likewise, \begin{align} E\,2ca(1-\cos\beta) &= (a-c)(C c-A a) \tag{4b} \\ F\,2ab(1-\cos\gamma) &= (b-a)(A a-B b) \tag{4c} \end{align}
Finally, observe that, for $$a$$, $$b$$, $$c$$ not all equal (the only case with which we need concern ourselves), $$(\text{eq }4a)\;(a-c)(b-a) + (\text{eq }4b)\;(c-b)(b-a)+(\text{eq }4c)\;(a-c)(c-b) \tag{5}$$ gives a non-trivial linear combination of $$D$$, $$E$$, $$F$$ that vanishes. Consequently, $$D$$, $$E$$, $$F$$ are linearly dependent; that is, they lie on a common plane through $$O$$, and the result follows. $$\square$$
• Pardon my ignorance but I'm confused as to what is being written as soon as equality (1). How can a point be a sum of other points and lenghts. – Dood Dec 3 '18 at 11:35
• Think of the points as position vectors, which add coordinate-wise. The lengths act as scalar multipliers. – Blue Dec 3 '18 at 11:38
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# Forum
Help with beta extr...
Clear all
# Help with beta extraction
Posts: 8
Registered
Topic starter
(@bdalcinbalda)
Active Member
Joined: 2 months ago
Hello,
I have a question about the beta coefficients shown in the output.
I used a GLM with the Gaussian basis function, and I now have 6 different beta coefficients, which I read that are one for each basis function. How do I choose between those beta? (see picture) They seem quite different from one another. How do I get to the final beta?
Thank you very much,
Bruna
4 Replies
Posts: 168
Registered
(@dboas)
Estimable Member
Joined: 1 year ago
We use consecutive gaussian basis functions to model the HRF to allow for more flexiblity in the estimated shape of the HRF. This is quite different than using, for instance, a gamma function with specified width and delay parameters and then you estimate the beta which is just the amplitude. In the gamma function case, you get a single beta coefficient for the amplitude that you can do your statistics on. In the gaussian basis functions case, you need to multiple the series of beta coefficients by the basis functions to restore the shape of the HRF. Then you can measure the mean across a chosen time range of that HRF. Then you can look at the statistics of that mean across subjects.
Posts: 8
Registered
Topic starter
(@bdalcinbalda)
Active Member
Joined: 2 months ago
Hello David,
Thank you very much, that makes sense. Can you just explain how to do this on Matlab for someone with very basic calculus background and little matlab skills?
"you need to multiple the series of beta coefficients by the basis functions to restore the shape of the HRF. Then you can measure the mean across a chosen time range of that HRF. Then you can look at the statistics of that mean across subjects."
Also, if I choose to use the gamma function, do you recommend using the default parameters of [0.1 3.0 10.0 1.8 3.0 10.0]? Sorry I am not very familiar with this function.
Posts: 105
Moderator
(@mayucel)
Joined: 2 years ago
Hi Bruna,
That multiplication results in HRF, and Homer already outputs HRF with the output name "dcAvg". Please check out the link below. You can also export HRF.
https://github.com/BUNPC/Homer3/wiki/Output-Files
Meryem
Posts: 8
Registered
Topic starter
(@bdalcinbalda)
Active Member
Joined: 2 months ago
Hi @dboas,
I ended up using the gamma function like you had mentioned.
I was just wondering if there is a better way of exporting the betas in Matlab? I am only able to see them when I go in each subject's .mat file. Is there a way to get all betas for all subjects and conditions?
Thank you!
Share:
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*These starred week numbers are one behind CIT’s week numbers. This is because of the snow.
## Linear Algebra: 20% Test
Will now take place Wednesday 21 March, in Week 7* [21 March].
The test will take place from 19:00-20:30 but most students should be able to complete the test in about an hour. It has about 35 Marks worth of questions: five in all (with one very short, and three shortened versions of longer questions).
Anything done in the first five weeks is examinable (see “Independent Learning” below) and it is recommended that you understand what is going on with the summaries of p. 57-59.
The nine questions from p. 60 on are a good revision but not every possible question is listed there. In next week’s Maple you will get a chance to revise these questions.
## Week 5* [7 March]
We saw how linear systems can be written as matrix equations, and (sometimes) solved using matrix inverses. Then we spoke about determinants, and their use in figuring out if homogeneous linear systems have non-zero solutions. Finally we looked at Cramer’s Rule.
## Week 6* [14 March]
We will start the class with one more example of Cramer’s Rule, and then start pushing into statistics.
In Maple, we will do Lab 3, which is really revision for the Linear Algebra Test.
## Week 7* [21 March]
The test is going to begin at 19:00 sharp and run until 20:30. Class will resume at 20:35 sharp. This seems a very short break but the test is designed so that it shouldn’t take much longer than an hour to complete, so almost everyone should have a solid enough break.
At 20:35 we will continue working on statistics.
## Week 8* [28 March/4 April]
This may or may not be a Maple night (it depend on how far we get in the previous week).
It appears that at most one student will miss the class, which isn’t too bad. So now we now go back to the poll to pick between the two nights.
## Maple Catch Up
If you have missed a lab you have two options: either download Maple onto your own machine (instructions may be found here) or come into CIT at another time to use Maple.
Go through the missed lab on your own, doing all the exercises in Maple. Save the worksheet and email it to me.
## Independent Learning
Questions you can do include:
• After Week 5: P.44, Q. 1-3, Q. 4-5 more abstract. P.47, Q. 1-3, Q. 4 more abstract. P.56, Q. 1-3, Q. 4 more abstract. P.69, Q. 9 is an important question. A $2\times 2$ version might be
Use only determinants to determine if the following homogeneous system of linear equations has non-zero solutions:
$2x+y=0$
$-6x-3y=0$
• After Week 4: P. 41, Q. 1-4
• After Week 3: P. 28, Q. 1-5, 6-9 have answers with Q. 7 a harder question. P. 34 exercises.
• After Week 2: P. 18 Q. 2
• After Week 1: P. 18 Q. 1, 3 – 6. Harder questions are 7 and 8. For those who do not yet have the manual, see here.
I am not suggesting you should do all of these. It is recommended by the module descriptor that you do two hours of independent and directed learning every week but of course this isn’t feasible for everyone.
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# Is coding multi-line curly braces in text mode possible?
I need to typeset the following document:
I have coded the following
\documentclass[french, a4paper, 10pt]{book}
\usepackage[french]{babel}
\usepackage{fontspec}
\begin{document}
\setlength{\tabcolsep}{1em}
\scriptsize
\section*{\centering \small\sc Tableau des principaux champignons comestibles de france}
\begin{center}
\rule{2cm}{0.4pt}
\end{center}
\subsection*{\centering \small\sc Basidiomycètes}
\bigskip
\bigskip
\begin{tabular}{llll}
{\sc Hyménomycètes}\dotfill &
\begin{tabular}{lll}
\textit{Agaricinées} &
\begin{tabular}{ll}
Spores blanches\dotfill &
\begin{tabular}{l}
Amanita \\
Lepiota \\
Armilliaria \\
Tricholoma \\
Clitocybe \\
Collybia \\
Pleurotus \\
Hygrophorus \\
Cantharellus\\
Lactarius \\
\\
\end{tabular} \\[3em]
Spores roses \dotfill &
\begin{tabular}{l}
Volvaria \\
Pluteus \\
Entoloma \\
Clitopilus \\
\\
\end{tabular} \\[2em]
Spores ocracées\dotfill &
\begin{tabular}{l}
Cortinarius \\
Paxillus \\
\\
\end{tabular} \\[2em]
Spores pourpres\dotfill &
\begin{tabular}{l}
Psalliota \\
\\
\end{tabular} \\[2em]
Spores noires\dotfill &
\begin{tabular}{l}
Coprinum \\
\\
\end{tabular} \\[2em]
\end{tabular} \\
\textit{Polyporées} & \begin{tabular}{ll}
....... ....... & ...... ....... \\
....... ....... & ...... ....... \\
....... ....... & ...... ....... \\
\end{tabular} \\
\textit{Hydnées} & \begin{tabular}{ll}
....... ....... & ...... ....... \\
....... ....... & ...... ....... \\
....... ....... & ...... ....... \\
\end{tabular} \\
\end{tabular} \\
\end{tabular}
\end{document}
The above code compiles without errors using xelatex and correctly outputs an abridged version of the page I am trying to typeset but it misses the curly braces.
I have been trying to add
$\left\{ ... \right.$
and many variations thereof in many different places without success.
Is typesetting this via latex at all possible? Is switching to math mode imperative or is there any way I could create the curly braces without switching to math mode? I anything wrong with the method I adopted? Is there a "better way"?
• Welcome to TeX.SE. – Sebastiano Mar 18 at 21:39
Taking a fresh look at this. I followed advice given by other posters and tried my luck with cases and the schemata package without much luck. I eventually managed to get the solution I had initially tried with nested tables and a switch to math mode to compile without errors but the result is pretty awful where just about everything is misaligned.
Here is an as yet incomplete but very simple and robust solution:
It is missing the braces but I might use it as a replacement since it reflects the original's data content and its structure.
Now I was wondering if I could persuade latex (xelatex) to draw left braces that would scale to occupy the narrow cells numbered 1-8.
One quick DIY style solution consists in drawing them manually using the U+23a7, U+23a8, U+23a9 and U+23aa unicode characters and switching to a font that has the glyphs such as DejaVu.
The problem with this approach is that there will some horizontal alignment issues with the lists that have an even number of items (such as in cells #3 and #4 for instance.Not to mention that the curly braces will not be scaled - ie. they will just be extended by additional U+23aa's but their thickness/weight will all be the same.
Is there any way I can write some code that simply tells Latex: please draw a left/opening curly brace that exactly fills each of these cells and centers it vertically (meaning that the little tit in the middle of the curly brace will be exactly centered) and scales the curly braces in termes of thickness to match the sample I initially posted?
Thanks,
RSF
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## Nihonkai Mathematical Journal
### Simultaneous extensions of Selberg and Buzano inequalities
#### Abstract
We give a simultaneous extension of Selberg and Buzano inequalities: If $y1, y2$ and nonzero vectors $\{ z_i; i = 1, 2, \dots , n \}$ in a Hilbert space $\mathscr{H}$ satisfy the orthogonality condition $\langle y_k; z_i \rangle = 0$ for $i = 1, 2, \dots , n$ and $k = 1, 2,$ then $| \langle x, y_1 \rangle \langle x, y_2 \rangle | + \mathit{B} (y_1, y_2) \sum_i \frac{| \langle x, z_i \rangle |^2}{\sum_j | \langle z_i, Z-j \rangle |} \leq \mathit{B} (y_1, y_2) \|x\|^2$ holds for all $x \in \mathscr{H}$, where $\mathit{B} (y1; y2) = \frac{1}{2} (\|y_1\| \|y_2\| + | \langle y_1, y_2 \rangle |)$. As an application, we discuss some refinements of the Heinz-Kato-Furuta inequality and the Bernstein inequality.
#### Article information
Source
Nihonkai Math. J., Volume 25, Number 1 (2014), 45-63.
Dates
First available in Project Euclid: 17 October 2014
https://projecteuclid.org/euclid.nihmj/1413555412
Mathematical Reviews number (MathSciNet)
MR3270971
Zentralblatt MATH identifier
1311.47020
Subjects
Primary: 47A63: Operator inequalities
#### Citation
Fujii, Masatoshi; Matsumoto, Akemi; Tominaga, Masaru. Simultaneous extensions of Selberg and Buzano inequalities. Nihonkai Math. J. 25 (2014), no. 1, 45--63. https://projecteuclid.org/euclid.nihmj/1413555412
#### References
• H.J. Bernstein, An inequality for selfadjoint operators on a Hilbert space, Proc. Amer. Math. Soc., 100 (1987), 319–321.
• M.L. Buzano, Generalizzazione della diseguaglianza di Cauchy-Schwarz, Rend. Sem. Mat. Univ. e Politech. Torino, 31 (1971-73). (1974), 405–409.
• M. Fujii, Furuta's inequality and its mean theoretic approach, J. Operator Theory, 23 (1990), 67–72.
• M. Fujii, T. Furuta and E. Kamei, Furuta's inequality and its application to Ando's theorem, Linear Algebra Appl., 179 (1993), 161–169.
• M. Fujii, T. Furuta and Y. Seo, An inequality for some nonnormal operators –Extension to normal approximate eigenvalues, Proc. Amer. Math. Soc., 118 (1993), 899–902.
• M. Fujii, J.-F. Jiang and E. Kamei, Characterization of chaotic order and its application to Furuta inequality, Proc. Amer. Math. Soc., 125 (1997), 3655–3658.
• M. Fujii, J.-F. Jiang, E. Kamei and K. Takahashi, A characterization of chaotic order and a problem, J. Inequal. Appl., 2 (1998), 149–156.
• M. Fujii and E. Kamei, Furuta's inequality and a generalization of Ando's theorem, Proc. Amer. Math. Soc., 115 (1992), 409–413.
• M. Fujii, K. Kubo and S. Otani, A graph theoretic observation on the Selberg inequality, Math. Japon., 35 (1990), 381–385.
• M. Fujii, C.-S. Lin and R. Nakamoto, Bessel type extension of the Bernstein inequality, Sci. Math., 3 (2000), 95–98.
• M. Fujii and R. Nakamoto, Simultaneous extensions of Selberg inequality and Heinz-Kato-Furuta inequality, Nihonkai Math. J., 9 (1998), 219–225.
• M. Fujii and R. Nakamoto, Extensions of Heinz-Kato-Furuta inequality, Proc. Amer. Math. Soc., 128 (2000), 223–228.
• M. Fujii and R. Nakamoto, Extensions of Heinz-Kato-Furuta inequality, II, J. Inequal. Appl., 3 (1999), 293–302.
• M. Fujii, R. Nakamoto and Y. Seo, Covariance in Bernstein's inequality for operators, Nihonkai Math. J., 8 (1997), 1–6.
• T. Furuta, $A \ge B \ge 0$ assures $(B^rA^pB^r)^{1/q} \ge B^{(p+2r)/q}$ for $r \ge 0, p \ge 0, q \ge 1$ with $(1+2r)q \ge p+2r$, Proc. Amer. Math. Soc., 101 (1987), 85–88.
• T. Furuta, An inequality for some nonnormal operators, Proc. Amer. Math. Soc., 104 (1988), 1216–1217.
• T. Furuta, Elementary proof of an order preserving inequality, Proc. Japan Acad., 65 (1989), 126.
• T. Furuta, When does the equality of a generalized Selberg inequality hold?, Nihonkai Math. J., 2 (1991), 25–29.
• T. Furuta, Determinant type generalizations of the Heinz-Kato theorem via the Furuta inequality, Proc. Amer. Math. Soc., 120 (1994), 223–231.
• E. Kamei, A satellite to Furuta's inequality, Math. Japon., 33 (1988), 883–886.
• K. Kubo and F. Kubo, Diagonal matrix dominates a positive semidefinite matrix and Selberg's inequality, preprint.
• C.-S. Lin, Heinz's inequality and Bernstein's inequality, Proc. Amer. Math. Soc., 125 (1997), 2319–2325.
• G.K. Pedersen, Some operator monotone functions, Proc. Amer. Math. Soc., 36 (1972), 309–310.
• K. Tanahashi, Best possibility of the Furuta inequality, Proc. Amer. Math. Soc., 124 (1996), 141–146.
• M.Uchiyama, Some exponential operator inequalities, Math. Inequal. Appl., 2 (1999), 469–471.
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Related Searches
Definitions
# Marsh test
The Marsh test is a highly sensitive method in the detection of arsenic, especially useful in the field of forensic toxicology when arsenic was used as a poison. It was developed by the chemist James Marsh and first published 1836.
Arsenic, in the form of white arsenic trioxide As2O3, was a highly favored poison, for it is odorless, easily incorporated into food and drink, and before the advent of the Marsh test, untraceable in the body. For the untrained, arsenic poisoning would have symptoms similar to cholera. Indeed, in France it came to be known as poudre de succession, "inheritance powder" for obvious reasons.
## Precursor methods
The first breakthrough in the detection of arsenic poisoning was in 1775 when Carl Wilhelm Scheele discovered a way to change arsenic trioxide to arsine gas (AsH3), a garlic-smelling gas by treating it with nitric acid (HNO3) and combining it with zinc.
As2O3 + 6Zn + 12HNO3 → 2AsH3 + 6Zn(NO3)2 + 3H2O
In 1787, Johann Metzger discovered that if arsenic trioxide was heated in the presence of charcoal, a shiny black powder (arsenic mirror) would be formed over it. This is the reduction of As2O3 by carbon:
2As2O3 + 3C → 3CO2 + 4As
In 1806, Valentine Rose took the stomach of a victim suspected of being poisoned and treated it with potassium carbonate (K2CO3), calcium oxide (CaO) and nitric acid. Any arsenic present would appear as arsenic trioxide and then could be subjected to Metzger's test.
However, the most common test (and used even today in water test kits) was discovered by Samuel Hahnemann. It would involve combining a sample fluid with hydrogen sulfide (H2S) in the presence of hydrochloric acid (HCl). A yellow precipitate, arsenic trisulfide (AsS3) would be formed if arsenic were present.
## Circumstances and methodology behind the Marsh test
Even so, these tests have proven not to be sensitive enough. In 1832, a certain John Bodle was brought to trial for poisoning his grandfather by putting arsenic in his coffee. James Marsh, a chemist working at the Royal Arsenal in Woolwich was called by the prosecution to try to detect its presence. He performed the standard test by passing hydrogen sulfide through the suspect fluid. While Marsh was able to detect arsenic, the yellow precipitate did not keep very well, and by the time it was presented to the jury it deteriorated. The jury was not convinced, and John Bodle was acquitted.
Angered and frustrated by this, especially when John Bodle confessed later that he indeed killed his grandfather, Marsh decided to devise a better test to demonstrate the presence of arsenic. Taking Scheele's method as a basis, he constructed a simple glass apparatus capable of not only detecting minute traces of it but also measure its quantity. While the Scheele test used nitric acid, in Marsh's case the suspect fluid would be mixed with sulfuric acid $\left(mathrm\left\{H\right\}_2mathrm\left\{SO\right\}_4 \right)$ and passed through a U-shaped tube with a piece of arsenic-free zinc at the end. If even a trace of arsenic was present, arsine gas would result. When he ignited this gas, it would decompose into arsenic and hydrogen. When he held a cold ceramic bowl, the arsenic would form a silvery-black deposit on the bowl, a result similar to that of Metzger's test. Not only could minute amounts of arsenic be detected (for as little as 0.02 mg), the test was very specific for arsenic. Although antimony $\left(mathrm\left\{Sb\right\}\right)$could give a false-positive test by forming a similar black deposit, it would not react with sodium hypochlorite $\left(mathrm\left\{NaOCl\right\}\right)$, while arsenic would.
## Specific reactions involved with the Marsh test
The Marsh test treats the sample with sulfuric acid and arsenic-free zinc. Even if there are minute amounts of arsenic present, the zinc reduces the trivalent arsenic (As3+ ). Here are the two half-reactions:
Oxidation: Zn → Zn2+ + 2e
Reduction: As2O3 + 12e + 6H+ → 2As3− + 3H2O
Overall, we have this reaction:
As2O3 + 6Zn + 6H + → 2As3− + 6Zn2+ + 3H2O
But In an acidic medium, As3+ actually forms arsine gas (AsH3), so adding sulfuric acid (H2SO4) to each side of the equation we get:
As2O3 + 6Zn + 6H+ + 6H2SO4 → 2As3 − + 6H2SO4 + 6Zn2+ + 3H2O,
or As the As3 − , combines with the H+ to form arsine:
As2O3 + 6Zn + 6H+ + 6H2SO4 → 2AsH3 + 6ZnSO4 + 3H2O + 6H+ ,
or by eliminating the common ions:
As2O3 + 6Zn + 6H2SO4 → 2AsH3 + 6ZnSO4 + 3H2O
## First notable application
Although the Marsh test was efficacious, its first publicly documented use — as the matter of fact the first time evidence from forensic toxicology was introduced — was in Tulle, France in 1840 with the celebrated LaFarge poisoning case. Charles LaFarge, a foundry owner, was suspected of being poisoned with arsenic by his wife Marie. The circumstantial evidence was great: it was shown that she brought arsenic trioxide from a local chemist, supposedly to kill rats which infested their home. In addition, their maid swore that she had mixed a white powder into his drink. Although the food was found to be positive for the poison using the old methods as well as the Marsh test, when the husband's body was exhumed and tested, the chemists assigned to the case were not able to do so. Mathieu Orfila, the renowned toxicologist retained by the defence and an acknowledged authority of the Marsh test examined the results. He performed the test again and demonstrated that the Marsh test was not at fault for the misleading results but rather those who performed it did it incorrectly. Orfila thus proved the presence of arsenic in LaFarge's body using the test. As a result of this, Marie was found guilty and sentenced to life imprisonment.
## Effects of the Marsh test
The case proved to be controversial, for it divided the country into factions who were convinced or otherwise of Mme. LaFarge's guilt; nevertheless, the impact of the Marsh test was great. The French press covered the trial and gave the test the publicity it needed to give the field of forensic toxicology the legitimacy it deserved, although in some ways it trivialized it: Marsh test assays were actually done in salons, public lectures and even in some plays that recreated the LaFarge case.
The existence of the Marsh test also served a deterrent effect: deliberate arsenic poisonings became rarer because of the fear of discovery became more present.
## References
• Marsh J. (1836). "Account of a method of separating small quantities of arsenic from substances with which it may be mixed". Eddinburgh New Philosophical Journal 21 229–236..
• Marsh J. (1837). "Arsenic; nouveau procédé pour le découvrir dans les substances auxquelles il est mêlé". Journal de Pharmacie 23 553–562.
• Marsh, James (1837). "Beschreibung eines neuen Verfahrens, um kleine Quantitäten Arsenik von den Substanzen abzuscheiden, womit er gemischt ist". Liebigs Annalen der Chemie 23 207.
• Mohr C. F. (1837). "Zusätze zu der von Marsh angegebenen Methode, den Arsenik unmittelbar im regulinischen Zustande aus jeder Flüssigkeit auszuscheiden". Annalen der Pharmacie und Chemie 23 217–225.
• Lockemann, Georg (1905). "Über den Arsennachweis mit dem Marshschen Apparate". Angewante Chemie 18 416.
• Harkins, W. D. (1910). "The Marsh test and Excess Potential (First Paper.1) The Quantitative Determination of Arsenic". Journal of the American Chemical Society 32 518–530.
• Campbell W. A. (1965). "Some landmarks in the history of arsenic testing". Chemistry in Britain 1 198–202.
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$$\require{cancel}$$
# 12.4: Refraction
[ "article:topic", "Snell\u2019s law of refraction", "authorname:crowellb", "index of refraction", "Lens-Maker\u2019s Equation" ]
Economists normally consider free markets to be the natural way of judging the monetary value of something, but social scientists also use questionnaires to gauge the relative value of privileges, disadvantages, or possessions that cannot be bought or sold. They ask people to imagine that they could trade one thing for another and ask which they would choose. One interesting result is that the average light-skinned person in the U.S. would rather lose an arm than suffer the racist treatment routinely endured by African-Americans. Even more impressive is the value of sight. Many prospective parents can imagine without too much fear having a deaf child, but would have a far more difficult time coping with raising a blind one.
So great is the value attached to sight that some have imbued it with mystical aspects. Joan of Arc saw visions, and my college has a “vision statement.” Christian fundamentalists who perceive a conflict between evolution and their religion have claimed that the eye is such a perfect device that it could never have arisen through a process as helter-skelter as evolution, or that it could not have evolved because half of an eye would be useless. In fact, the structure of an eye is fundamentally dictated by physics, and it has arisen separately by evolution somewhere between eight and 40 times, depending on which biologist you ask. We humans have a version of the eye that can be traced back to the evolution of a light-sensitive “eye spot” on the head of an ancient invertebrate. A sunken pit then developed so that the eye would only receive light from one direction, allowing the organism to tell where the light was coming from. (Modern flatworms have this type of eye.) The top of the pit then became partially covered, leaving a hole, for even greater directionality (as in the nautilus). At some point the cavity became filled with jelly, and this jelly finally became a lens, resulting in the general type of eye that we share with the bony fishes and other vertebrates. Far from being a perfect device, the vertebrate eye is marred by a serious design flaw due to the lack of planning or intelligent design in evolution: the nerve cells of the retina and the blood vessels that serve them are all in front of the light-sensitive cells, blocking part of the light. Squids and other molluscs, whose eyes evolved on a separate branch of the evolutionary tree, have a more sensible arrangement, with the light-sensitive cells out in front.
### 12.4.1 Refraction
The fundamental physical phenomenon at work in the eye is that when light crosses a boundary between two media (such as air and the eye's jelly), part of its energy is reflected, but part passes into the new medium. In the ray model of light, we describe the original ray as splitting into a reflected ray and a transmitted one (the one that gets through the boundary). Of course the reflected ray goes in a direction that is different from that of the original one, according to the rules of reflection we have already studied. More surprisingly --- and this is the crucial point for making your eye focus light --- the transmitted ray is bent somewhat as well. This bending phenomenon is called refraction. The origin of the word is the same as that of the word “fracture,” i.e., the ray is bent or “broken.” (Keep in mind, however, that light rays are not physical objects that can really be “broken.”) Refraction occurs with all waves, not just light waves.
Figure $$\PageIndex{1}$$: A human eye. Figure $$\PageIndex{2}$$: The anatomy of the eye.
The actual anatomy of the eye, Figure $$\PageIndex{2}$$, is quite complex, but in essence it is very much like every other optical device based on refraction. The rays are bent when they pass through the front surface of the eye, Figure $$\PageIndex{3}$$. Rays that enter farther from the central axis are bent more, with the result that an image is formed on the retina. There is only one slightly novel aspect of the situation. In most human-built optical devices, such as a movie projector, the light is bent as it passes into a lens, bent again as it reemerges, and then reaches a focus beyond the lens. In the eye, however, the “screen” is inside the eye, so the rays are only refracted once, on entering the jelly, and never emerge again.
Figure $$\PageIndex{3}$$: A simplified optical diagram of the eye. Light rays are bent when they cross from the air into the eye. (A little of the incident rays' energy goes into the reflected rays rather than the ones transmitted into the eye.)
A common misconception is that the “lens” of the eye is what does the focusing. All the transparent parts of the eye are made of fairly similar stuff, so the dramatic change in medium is when a ray crosses from the air into the eye (at the outside surface of the cornea). This is where nearly all the refraction takes place. The lens medium differs only slightly in its optical properties from the rest of the eye, so very little refraction occurs as light enters and exits the lens. The lens, whose shape is adjusted by muscles attached to it, is only meant for fine-tuning the focus to form images of near or far objects.
#### Refractive Properties of Media
What are the rules governing refraction? The first thing to observe is that just as with reflection, the new, bent part of the ray lies in the same plane as the normal (perpendicular) and the incident ray, Figure $$\PageIndex{4}$$.
Figure $$\PageIndex{4}$$: The incident, reflected, and transmitted (refracted) rays all lie in a plane that includes the normal (dashed line). Figure $$\PageIndex{5}$$: The angles $$\theta_1$$ and $$\theta_2$$ are related to each other, and also depend on the properties of the two media. Because refraction is time-reversal symmetric, there is no need to label the rays with arrowheads.
If you try shooting a beam of light at the boundary between two substances, say water and air, you'll find that regardless of the angle at which you send in the beam, the part of the beam in the water is always closer to the normal line, Figure $$\PageIndex{5}$$.
It doesn't matter if the ray is entering the water or leaving, so refraction is symmetric with respect to time-reversal, Figure $$\PageIndex{6}$$ .
Figure $$\PageIndex{6}$$: Refraction has time-reversal symmetry. Regardless of whether the light is going into or out of the water, the relationship between the two angles is the same, and the ray is closer to the normal while in the water.
If, instead of water and air, you try another combination of substances, say plastic and gasoline, again you'll find that the ray's angle with respect to the normal is consistently smaller in one and larger in the other. Also, we find that if substance A has rays closer to normal than in B, and B has rays closer to normal than in C, then A has rays closer to normal than C. This means that we can rank-order all materials according to their refractive properties. Isaac Newton did so, including in his list many amusing substances, such as “Danzig vitriol” and “a pseudo-topazius, being a natural, pellucid, brittle, hairy stone, of a yellow color.” Several general rules can be inferred from such a list:
• Vacuum lies at one end of the list. In refraction across the interface between vacuum and any other medium, the other medium has rays closer to the normal.
• Among gases, the ray gets closer to the normal if you increase the density of the gas by pressurizing it more.
• The refractive properties of liquid mixtures and solutions vary in a smooth and systematic manner as the proportions of the mixture are changed.
• Denser substances usually, but not always, have rays closer to the normal.
The second and third rules provide us with a method for measuring the density of an unknown sample of gas, or the concentration of a solution. The latter technique is very commonly used, and the CRC Handbook of Physics and Chemistry, for instance, contains extensive tables of the refractive properties of sugar solutions, cat urine, and so on.
### Snell's Law
The numerical rule governing refraction was discovered by Snell, who must have collected experimental data something like what is shown on this graph and then attempted by trial and error to find the right equation. The equation he came up with was
$\frac{\sin\theta_1}{\sin\theta_2} = \text{constant} .$
The value of the constant would depend on the combination of media used. For instance, any one of the data points in the graph would have sufficed to show that the constant was 1.3 for an air-water interface (taking air to be substance 1 and water to be substance 2).
Figure $$\PageIndex{7}$$: The relationship between the angles in refraction.
Snell further found that if media A and B gave a constant $$K_{AB}$$ and media B and C gave a constant $$K_{BC}$$, then refraction at an interface between A and C would be described by a constant equal to the product, $$K_{AC}=K_{AB}K_{BC}$$. This is exactly what one would expect if the constant depended on the ratio of some number characterizing one medium to the number characteristic of the second medium. This number is called the index of refraction of the medium, written as $$n$$ in equations. Since measuring the angles would only allow him to determine the ratio of the indices of refraction of two media, Snell had to pick some medium and define it as having $$n=1$$. He chose to define vacuum as having $$n=1$$. (The index of refraction of air at normal atmospheric pressure is 1.0003, so for most purposes it is a good approximation to assume that air has $$n=1$$.) He also had to decide which way to define the ratio, and he chose to define it so that media with their rays closer to the normal would have larger indices of refraction. This had the advantage that denser media would typically have higher indices of refraction, and for this reason the index of refraction is also referred to as the optical density. Written in terms of indices of refraction, Snell's equation becomes
$\frac{\sin\theta_1}{\sin\theta_2} = \frac{n_2}{n_1} ,$
but rewriting it in the form
$n_1 \sin \theta_1=n_2 \sin \theta_2$
[relationship between angles of rays at the interface between media with indices of refraction $$n_1$$ and $$n_2$$; angles are defined with respect to the normal] makes us less likely to get the 1's and 2's mixed up, so this the way most people remember Snell's law. A few indices of refraction are given in the back of the book.
Exercise $$\PageIndex{1}$$
1. What would the graph look like for two substances with the same index of refraction?
2. Based on the graph, when does refraction at an air-water interface change the direction of a ray most strongly?
Example $$\PageIndex{1}$$: Finding an angle using Snell's law
A submarine shines its searchlight up toward the surface of the water. What is the angle $$\alpha$$ shown in Figure $$\PageIndex{8}$$?
Figure $$\PageIndex{8}$$: Example 10.
Solution
The tricky part is that Snell's law refers to the angles with respect to the normal. Forgetting this is a very common mistake. The beam is at an angle of $$30°$$ with respect to the normal in the water. Let's refer to the air as medium 1 and the water as 2. Solving Snell's law for $$\theta_1$$, we find
$\theta_1 = \sin^{-1}\left(\frac{n_2}{n_1}\sin\theta_2\right) .$
As mentioned above, air has an index of refraction very close to 1, and water's is about 1.3, so we find $$\theta_1=40°$$. The angle $$\alpha$$ is therefore $$50°$$.
What neither Snell nor Newton knew was that there is a very simple interpretation of the index of refraction. This may come as a relief to the reader who is taken aback by the complex reasoning involving proportionalities that led to its definition. Later experiments showed that the index of refraction of a medium was inversely proportional to the speed of light in that medium. Since $$c$$ is defined as the speed of light in vacuum, and $$n=1$$ is defined as the index of refraction of vacuum, we have
$n=\frac{c}{v} .$
[$$n=$$ medium's index of refraction, $$v=$$ speed of light in that medium, $$c=$$ speed of light in a vacuum]
Many textbooks start with this as the definition of the index of refraction, although that approach makes the quantity's name somewhat of a mystery, and leaves students wondering why $$c/v$$ was used rather than $$v/c$$. It should also be noted that measuring angles of refraction is a far more practical method for determining $$n$$ than direct measurement of the speed of light in the substance of interest.
A mechanical model of Snell's law
Why should refraction be related to the speed of light? The mechanical model shown in the figure may help to make this more plausible. Suppose medium 2 is thick, sticky mud, which slows down the car. The car's right wheel hits the mud first, causing the right side of the car to slow down. This will cause the car to turn to the right until is moves far enough forward for the left wheel to cross into the mud. After that, the two sides of the car will once again be moving at the same speed, and the car will go straight.
Figure $$\PageIndex{8}$$: A mechanical model of refraction.
Of course, light isn't a car. Why should a beam of light have anything resembling a “left wheel” and “right wheel?” After all, the mechanical model would predict that a motorcycle would go straight, and a motorcycle seems like a better approximation to a ray of light than a car. The whole thing is just a model, not a description of physical reality.
#### A Derivation of Snell's law
However intuitively appealing the mechanical model may be, light is a wave, and we should be using wave models to describe refraction. In fact Snell's law can be derived quite simply from wave concepts. Figure $$\PageIndex{9}$$ shows the refraction of a water wave. The water in the upper left part of the tank is shallower, so the speed of the waves is slower there, and their wavelengths is shorter. The reflected part of the wave is also very faintly visible.
Figure $$\PageIndex{9}$$: A derivation of Snell's law.
In the close-up view on the right, the dashed lines are normals to the interface. The two marked angles on the right side are both equal to $$\theta_1$$, and the two on the left to $$\theta_2$$.
Trigonometry gives
\begin{align*} \sin \theta_1 &= \lambda_1/h \text{and} \\ \sin \theta_2 &= \lambda_2/h . \end{align*}
Eliminating $$h$$ by dividing the equations, we find
$\frac{\sin\theta_1}{\sin\theta_2}=\frac{\lambda_1}{\lambda_2} .$
The frequencies of the two waves must be equal or else they would get out of step, so by $$v=f\lambda$$ we know that their wavelengths are proportional to their velocities. Combining $$\lambda\propto v$$ with $$v\propto 1/n$$ gives $$\lambda\propto 1/n$$, so we find
$\frac{\sin\theta_1}{\sin\theta_2}=\frac{n_2}{n_1} ,$
which is one form of Snell's law.
Example $$\PageIndex{1}$$: Ocean waves near and far from shore
Ocean waves are formed by winds, typically on the open sea, and the wavefronts are perpendicular to the direction of the wind that formed them. At the beach, however, you have undoubtedly observed that waves tend come in with their wavefronts very nearly (but not exactly) parallel to the shoreline. This is because the speed of water waves in shallow water depends on depth: the shallower the water, the slower the wave. Although the change from the fast-wave region to the slow-wave region is gradual rather than abrupt, there is still refraction, and the wave motion is nearly perpendicular to the normal in the slow region.
#### Color and Refraction
In general, the speed of light in a medium depends both on the medium and on the wavelength of the light. Another way of saying it is that a medium's index of refraction varies with wavelength. This is why a prism can be used to split up a beam of white light into a rainbow. Each wavelength of light is refracted through a different angle.
#### How much light is reflected, and how much is transmitted?
In section 6.2 we developed an equation for the percentage of the wave energy that is transmitted and the percentage reflected at a boundary between media. This was only done in the case of waves in one dimension, however, and rather than discuss the full three dimensional generalization it will be more useful to go into some qualitative observations about what happens. First, reflection happens only at the interface between two media, and two media with the same index of refraction act as if they were a single medium. Thus, at the interface between media with the same index of refraction, there is no reflection, and the ray keeps going straight. Continuing this line of thought, it is not surprising that we observe very little reflection at an interface between media with similar indices of refraction.
The next thing to note is that it is possible to have situations where no possible angle for the refracted ray can satisfy Snell's law. Solving Snell's law for $$\theta_2$$, we find
$\theta_2 = \sin^{-1}\left(\frac{n_1}{n_2}\sin\theta_1\right) ,$
and if $$n_1$$ is greater than $$n_2$$, then there will be large values of $$\theta_1$$ for which the quantity $$(n_1/n_2)\sin\theta$$ is greater than one, meaning that your calculator will flash an error message at you when you try to take the inverse sine. What can happen physically in such a situation? The answer is that all the light is reflected, so there is no refracted ray. This phenomenon is known as total internal reflection, and is used in the fiber-optic cables that nowadays carry almost all long-distance telephone calls.
Figure $$\PageIndex{10}$$: Total internal reflection in a fiber-optic cable.
The electrical signals from your phone travel to a switching center, where they are converted from electricity into light. From there, the light is sent across the country in a thin transparent fiber. The light is aimed straight into the end of the fiber, and as long as the fiber never goes through any turns that are too sharp, the light will always encounter the edge of the fiber at an angle sufficiently oblique to give total internal reflection. If the fiber-optic cable is thick enough, one can see an image at one end of whatever the other end is pointed at.
Figure $$\PageIndex{11}$$: A simplified drawing of a surgical endoscope. The first lens forms a real image at one end of a bundle of optical fibers. The light is transmitted through the bundle, and is finally magnified by the eyepiece.
Alternatively, a bundle of cables can be used, since a single thick cable is too hard to bend. This technique for seeing around corners is useful for making surgery less traumatic. Instead of cutting a person wide open, a surgeon can make a small “keyhole” incision and insert a bundle of fiber-optic cable (known as an endoscope) into the body.
Figure $$\PageIndex{12}$$: Endoscopic images of a duodenal ulcer.
Since rays at sufficiently large angles with respect to the normal may be completely reflected, it is not surprising that the relative amount of reflection changes depending on the angle of incidence, and is greatest for large angles of incidence.
##### Discussion Questions
◊ What index of refraction should a fish have in order to be invisible to other fish?
◊ Does a surgeon using an endoscope need a source of light inside the body cavity? If so, how could this be done without inserting a light bulb through the incision?
◊ A denser sample of a gas has a higher index of refraction than a less dense sample (i.e., a sample under lower pressure), but why would it not make sense for the index of refraction of a gas to be proportional to density?
◊ The earth's atmosphere gets thinner and thinner as you go higher in altitude. If a ray of light comes from a star that is below the zenith, what will happen to it as it comes into the earth's atmosphere?
◊ Does total internal reflection occur when light in a denser medium encounters a less dense medium, or the other way around? Or can it occur in either case?
### 12.4.2 Lenses
Figures n/1 and n/2 show examples of lenses forming images. There is essentially nothing for you to learn about imaging with lenses that is truly new. You already know how to construct and use ray diagrams, and you know about real and virtual images. The concept of the focal length of a lens is the same as for a curved mirror. The equations for locating images and determining magnifications are of the same form. It's really just a question of flexing your mental muscles on a few examples. The following self-checks and discussion questions will get you started.
Figure $$\PageIndex{13}$$: 1. A converging lens forms an image of a candle flame. 2. A diverging lens.
Exercise $$\PageIndex{1}$$
1. In figures Figure $$\PageIndex{13; part 1}$$ and Figure $$\PageIndex{13; part 2}$$, classify the images as real or virtual.
2. Glass has an index of refraction that is greater than that of air. Consider the topmost ray in Figure $$\PageIndex{13; part 1}$$. Explain why the ray makes a slight left turn upon entering the lens, and another left turn when it exits.
3. If the flame in Figure $$\PageIndex{13; part 2}$$ were moved closer to the lens, what would happen to the location of the image?
### Discussion Questions
◊ In figures n/1 and n/2, the front and back surfaces are parallel to each other at the center of the lens. What will happen to a ray that enters near the center, but not necessarily along the axis of the lens? Draw a BIG ray diagram, and show a ray that comes from off axis.
In discussion questions B-F, don't draw ultra-detailed ray diagrams as in A.
◊ Suppose you wanted to change the setup in figure n/1 so that the location of the actual flame in the figure would instead be occupied by an image of a flame. Where would you have to move the candle to achieve this? What about in n/2?
◊ There are three qualitatively different types of image formation that can occur with lenses, of which figures n/1 and n/2 exhaust only two. Figure out what the third possibility is. Which of the three possibilities can result in a magnification greater than one? Cf. problem 10, p. 797.
◊ Classify the examples shown in figure o according to the types of images delineated in discussion question C.
◊ In figures n/1 and n/2, the only rays drawn were those that happened to enter the lenses. Discuss this in relation to figure o.
◊ In the right-hand side of figure o, the image viewed through the lens is in focus, but the side of the rose that sticks out from behind the lens is not. Why?
Figure $$\PageIndex{14}$$: Two images of a rose created by the same lens and recorded with the same camera.
#### The Lensmaker's Equation
Figure $$\PageIndex{15}$$: The radii of curvature appearing in the lensmaker's equation.
The focal length of a spherical mirror is simply $$r/2$$, but we cannot expect the focal length of a lens to be given by pure geometry, since it also depends on the index of refraction of the lens. Suppose we have a lens whose front and back surfaces are both spherical. (This is no great loss of generality, since any surface with a sufficiently shallow curvature can be approximated with a sphere.) Then if the lens is immersed in a medium with an index of refraction of 1, its focal length is given approximately by
$f = \dfrac{1}{(n-1) \left|\dfrac{1}{r_1}\pm\dfrac{1}{r_2}\right|} \label{LM}$
where $$n$$ is the index of refraction and $$r_1$$ and $$r_2$$ are the radii of curvature of the two surfaces of the lens. Equation \ref{LM} is known as the lensmaker's equation. In my opinion it is not particularly worthy of memorization. The positive sign is used when both surfaces are curved outward or both are curved inward; otherwise a negative sign applies. The proof of this equation is left as an exercise to those readers who are sufficiently brave and motivated.
### 12.4.3 Dispersion
For most materials, we observe that the index of refraction depends slightly on wavelength, being highest at the blue end of the visible spectrum and lowest at the red. For example, white light disperses into a rainbow when it passes through a prism, q.
Figure $$\PageIndex{16}$$: Dispersion of white light by a prism. White light is a mixture of all the wavelengths of the visible spectrum. Waves of different wavelengths undergo different amounts of refraction.
Even when the waves involved aren't light waves, and even when refraction isn't of interest, the dependence of wave speed on wavelength is referred to as dispersion. Dispersion inside spherical raindrops is responsible for the creation of rainbows in the sky, and in an optical instrument such as the eye or a camera it is responsible for a type of aberration called chromatic aberration (subsection 12.3.3 and problem 28). As we'll see in subsection 13.3.2, dispersion causes a wave that is not a pure sine wave to have its shape distorted as it travels, and also causes the speed at which energy and information are transported by the wave to be different from what one might expect from a naive calculation. The microscopic reasons for dispersion of light in matter are discussed in optional subsection 12.4.6.
#### The principle of least time for refraction
We have seen previously how the rules governing straight-line motion of light and reflection of light can be derived from the principle of least time. What about refraction? In the figure, it is indeed plausible that the bending of the ray serves to minimize the time required to get from a point A to point B. If the ray followed the unbent path shown with a dashed line, it would have to travel a longer distance in the medium in which its speed is slower. By bending the correct amount, it can reduce the distance it has to cover in the slower medium without going too far out of its way. It is true that Snell's law gives exactly the set of angles that minimizes the time required for light to get from one point to another. The proof of this fact is left as an exercise (problem 38, p. 802).
Figure $$\PageIndex{17}$$: The principle of least time applied to refraction.
#### Microscopic description of refraction
Given that the speed of light is different in different media, we've seen two different explanations (on p. 774 and in subsection 12.4.5 above) of why refraction must occur. What we haven't yet explained is why the speed of light does depend on the medium.
Figure $$\PageIndex{18}$$: Index of refraction of silica glass, redrawn from Kitamura, Pilon, and Jonasz, Applied Optics 46 (2007) 8118, reprinted online at http://www.seas.ucla.edu/~pilon/Publications/AO2007-1.pdf.
A good clue as to what's going on comes from the figure s. The relatively minor variation of the index of refraction within the visible spectrum was misleading. At certain specific frequencies, $$n$$ exhibits wild swings in the positive and negative directions. After each such swing, we reach a new, lower plateau on the graph. These frequencies are resonances. For example, the visible part of the spectrum lies on the left-hand tail of a resonance at about $$2\times10^{15}\ \text{Hz}$$, corresponding to the ultraviolet part of the spectrum. This resonance arises from the vibration of the electrons, which are bound to the nuclei as if by little springs. Because this resonance is narrow, the effect on visible-light frequencies is relatively small, but it is stronger at the blue end of the spectrum than at the red end. Near each resonance, not only does the index of refraction fluctuate wildly, but the glass becomes nearly opaque; this is because the vibration becomes very strong, causing energy to be dissipated as heat. The “staircase” effect is the same one visible in any resonance, e.g., figure k on p. 180: oscillators have a finite response for $$f \ll f_0$$, but the response approaches zero for $$f \gg f_0$$.
So far, we have a qualitative explanation of the frequency-variation of the loosely defined “strength” of the glass's effect on a light wave, but we haven't explained why the effect is observed as a change in speed, or why each resonance is an up-down swing rather than a single positive peak. To understand these effects in more detail, we need to consider the phase response of the oscillator. As shown in the bottom panel of figure j on p. 181, the phase response reverses itself as we pass through a resonance.
Figure $$\PageIndex{19}$$: 1. A wave incident on a sheet of glass excites current in the glass, which produce a secondary wave. 2. The secondary wave superposes with the original wave, as represented in the complex-number representation introduced in subsection 10.5.7.
Suppose that a plane wave is normally incident on the left side of a thin sheet of glass, t/1, at $$f \ll f_0$$. The light wave observed on the right side consists of a superposition of the incident wave consisting of $$\mathbf{E}_0$$ and $$\mathbf{B}_0$$ with a secondary wave $$\mathbf{E}^*$$ and $$\mathbf{B}^*$$ generated by the oscillating charges in the glass. Since the frequency is far below resonance, the response $$q\mathbf{x}$$ of a vibrating charge $$q$$ is in phase with the driving force $$\mathbf{E}_0$$. The current is the derivative of this quantity, and therefore 90 degrees ahead of it in phase. The magnetic field generated by a sheet of current has been analyzed in subsection 11.2.1, and the result, shown in figure e on p. 664, is just what we would expect from the right-hand rule. We find, t/1, that the secondary wave is 90 degrees ahead of the incident one in phase. The incident wave still exists on the right side of the sheet, but it is superposed with the secondary one. Their addition is shown in t/2 using the complex number representation introduced in subsection 10.5.7. The superposition of the two fields lags behind the incident wave, which is the effect we would expect if the wave had traveled more slowly through the glass.
In the case $$f \gg 0$$, the same analysis applies except that the phase of the secondary wave is reversed. The transmitted wave is advanced rather than retarded in phase. This explains the dip observed in figure s after each spike.
All of this is in accord with our understanding of relativity, ch. 7, in which we saw that the universal speed $$c$$ was to be understood fundamentally as a conversion factor between the units used to measure time and space --- not as the speed of light. Since $$c$$ isn't defined as the speed of light, it's of no fundamental importance whether light has a different speed in matter than it does in vacuum. In fact, the picture we've built up here is one in which all of our electromagnetic waves travel at $$c$$; propagation at some other speed is only what appears to happen because of the superposition of the $$(\mathbf{E}_0,\mathbf{B}_0)$$ and $$(\mathbf{E}^*,\mathbf{B}^*)$$ waves, both of which move at $$c$$.
But it is worrisome that at the frequencies where $$n\lt1$$, the speed of the wave is greater than $$c$$. According to special relativity, information is never supposed to be transmitted at speeds greater than $$c$$, since this would produce situations in which a signal could be received before it was transmitted! This difficulty is resolved in subsection 13.3.2, where we show that there are two different velocities that can be defined for a wave in a dispersive medium, the phase velocity and the group velocity. The group velocity is the velocity at which information is transmitted, and it is always less than $$c$$.
### Contributors
Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.
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# Place landscape figure on page and in maximum available size
I need to put landscape figure on a separate page of my appendix.
The Code i tried looks like this:
\chapter{Appendix 1}
\begin{figure}[h]
\centering
\rotatebox{90}{\includegraphics[width=\textwidth,height=\textheight,keepaspectratio]{Figure 1}}
\caption{Figure 1} \cite{Test} \label{fig:1}
\end{figure}
\myfigures{Fig. 1}
\clearpage
But Latex will throw the figure on a separate page, leaving the page with the chapter heading empty. Is there a way to automatically scale the figure, so it used all space available without being pushed on the next page?
you presumably know how much space your chapter heading takes up so just change height=\textheight, to be a suitable smaller value, but you seem to be using the wrong values, you scale the height of the image to \textheight but then rotate it so that length is trying to fit into the width of the page, I assume you want
\rotatebox{90}{\includegraphics[width=.7\textheight]{Figure 1}}
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# Centroid Problems
To demonstrate this concept, I’ll review a simple example of K-Means Clustering in Python. The centroid of the half-ellipse-shaped-cross-sectional area shown is given by: ус 4. (1) Define. Assignment Problems Notice. It seeks to identify the origin of a problem. The centroid of a shape is the arithmetic mean (i. 7 Part A Locate the centroid x of the parabolic area. Centroid Design Pvt. B it could be used to spy on people. Centroid of a triangle is the point of intersection of all its three medianscomplete information about the centroid, definition of an centroid, examples of an centroid, step by step solution of problems. Notice there’s one centroid for every hidden node and that each centroid has the same number of values as an input vector. SPOJ has a rapidly growing problem set/tasks available for practice 24 hours/day, including many original tasks prepared by the community of expert problem. Problem-solving is human. SOLUTION: • Divide the area into a triangle, rectangle, and semicircle with a circular cutout. x y 10 in 20 in Solution: The strategy is to find the centroid for the half circle area, and use the result in the composite algorithm. Centroids do not change location any more) OR 4. It is also known as "Separator Decomposition". Both centroid-linkage and Ward's method are applicable to the clutter removal problem because the overlap of representative icons is checked based on the distance between cluster centroids. Z-Z axis is called polar axis and is known as the polar moment of inertia. Games and puzzles. A simple example of application for these relationships is given below, Lets try a more practical problem using the centroid, ([Hibbeler, 1992], prob. The centroid is computed using the following formula, where μ ( x i ) is the membership value for point x i in the universe of discourse. In this case the whole complex is a word combination "adjective. Calculate the centroid for the entire lamina. Hi,these are the some problems on Centroid decomposition. In effect, this makes it similar to the label updating phase of the sklearn. Get Started. CoCreate User Forum > Applications > CoCreate Modeling: Centroid !. Equation of a line passing through the two given points. Using the moment-of-a-force analogy, the numerator in Eq. NOC:Artificial Intelligence Search Methods For Problem Solving. Note that there is a problem with linear combinations: As previously stated, no set of variables can add to a constant, yet the sum of the values for each plot is 1. This is not necessarilly true for the mean. mec 109 notes 3. Locate the centroid of the plane area shown. The centroid of a triangle is the point through which all the mass of a triangular plate seems to act. For example, if a person finds themselves sitting in the same. The display reads limit switch faults on the z and x axes even though the axes are in the middle of their travel. Rectangle Centroid x̅ and y̅ Right Triangle Centroid x̅ I and y̅ about x Semi-circle Centroid δ L Deformation: Axial δ deformation F = axial force 0 = original length = cross-sectional area E = modulus of elasticity POE 5 AE 4 CEA 4 xx b Moment of Inertia = moment of inertia of a rectangular section -x axis x h x. Service Delivery Manager at Centroid Systems Lake In The Hills, Illinois 500+ connections. Anchor: #VHVWXHXH; Determine the potential head from the centroid of the culvert opening, which is approximated as the sum of the invert elevation and one half the rise of the culvert. Centroids (grey) and shrunken centroids (red) for the SRBCT dataset. problem (Aloise et al. Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels. For example, in the triangle shown below, length of AG is twice the length of GD, while length of BG is twice the length of GE. First moments, centroids Papus' theorem. Introduction K-means clustering is one of the most widely used unsupervised machine learning algorithms that forms clusters of data based on the similarity between data instances. How to find centroids of clusters created with hclust using R program? any suggestions please? A centroid is always part of the given dataset (vectors). 25 Problem 5. Introduction : Centroid Decomposition is just a divide and conquer technique which is used on trees and turns out useful in many problems. In other words, for moment calculations you use the centroid position to calculate lever arms. This post will approach our task as follows: prepare the data; prepare our sample; perform centroid initialization search iterations to determine our “best” collection of initial centroids. By: Jaimin Pandya Centroid and Centre of Gravity 2. So this whole thing right over here is going to be equal to 15. Collinearity of points in a projective setting. Solution RP 5 } 2 3. Click here for a printable parallel line construction worksheet containing two problems to try. Center of Mass & Centroid Problems - Calculus. We conducted experiments with various Centroid based numbers of summarization approaches and obtain effective classification results. The centroid coincides with the center of mass or the center of gravity only if the material of the body is homogenous (density or specific weight is constant throughout the body). $\begingroup$ @dkr, You might want to ask this as a new question to get more (and more in-depth) responses. A median is the segment from a vertex of a triangle to the midpoint of the opposite side. Problems by Year. For HW i /D > 3. Let be I the incenter and G the centroid. Locate the centroid of the plane area shown. 102154 1 r 4 29 54 38. > Centroid Class Problem. We calculate the average pressure at the centroid. The following practice questions ask you to find the coordinates of a centroid in a triangle and to find the distance from one of the vertices to the centroid, given the median length. Common centroid constraint is not the same as 2-D symmetry constraint. Centroid = weighted average of centroids of component clusters. The task is to implement the K-means++ algorithm. Get Started. Problems on centroid, circumcentre and orthocentre - example If the circumcentre of the triangle lies at (0, 0) and centroid is middle point of (a 2 + 1, a 2 + 1) and (2 a, − 2 a) then the orthocentre lies on the line? Given coordinates of circumcentre is (0, 0). Besides being interesting in its own right, it is a good lead-in to Ellipsoid, since it gives some. The problem was: my data was scaled but my centroids were unscaled. Personal Web Page | Information Technology | Drexel University. Question 1 : Find the centroid of triangle whose vertices are (1, 10) (-7, 2) and (-3, 7). In other words, for moment calculations you use the centroid position to calculate lever arms. Where medians cross, the point common to all three medians is called the centroid. examples randomly, and set the cluster centroids to be equal to the values of these k examples. It has recently been shown that classification by nearest centroids provides an accurate predictor that may outperform much more complicated methods. Centroid initialization is given by *cluster_init*, the only available options: are ‘sample’ and ‘kmeans’. Introduction to centroids. We can solve above problem in O(N Log N) time using Centroid Decomposition. Worksheet - Centroid, Circumcenter, Orthocenter Author: 20619 Created Date: 11/22/2013 12:48:36 AM. 727418 1 r 1 20 36 20. Therefore, the. From symmetry, we can see that the centroid lies along the y-axis, which acts like a mirror to the quarter circles either side. 5 40 4 Total -----12---- ----42 12 xc =∑ xi⋅Ai A. Consequently we shall devote the next group of frames to the determination of centroidal moments of inertia. This will serve as an avenue to further enhance the CE Students’ skills in writing various issues and informations that will be beneficial to the ICE community; more updated. There are serious issues to address in Planning, Execution and Controls to secure successful deliveries. Determine tile centroid (X. We have step-by-step solutions for your textbooks written by Bartleby experts! Find the centroid of the region bounded by the given curves. return getCentroid( id, u, n ); } inline int Centroid ( int u ). Calculating the centroids requires efficiently evaluating the integrals in equation. When a cross-section of a beam is under bending from above, everything above the centroid is in compression, and everything below the centroid is in tension. Get Started. 8 Q S G F H B 4) Find QJ if WQ = 5 L J W K X Y Q 5) Find DY if DR = 16. ANO TliE CENTROIO OF A Booy 4 61 FUNDAMENTAL PROBLEMS F'l-I. Centroid and center of gravity are two important concepts in statics. Jump to Navigation Abaqus element centroid coordinates. Nadia, if this is in a loop over frames, just extract the centroids (x_centroids and y_centroids) from the structure. Do many runs of k-Means, each with different initial centroids. 5 Centroid • Centroid for multiple points • Centroid about x-axis 1 1 n i i i n i i m x x 9. You may need to use the parallel axis theorem to determine the Moment of Inertia of an I-Beam around it's centroid because the top and bottom flange will not be acting through the centroid of the. 1 For the plane area shown, determine the first moments with respect to the x and y axes. I will discuss a few problems that can be efficiently solved using Centroid Decomposition to help get a better feel of the technique and its usage. In this step we assign each input value to closest center. Related course: Complete Machine Learning Course with Python. For example, in the triangle shown below, length of AG is twice the length of GD, while length of BG is twice the length of GE. Step 8: Once the cluster becomes static, the k-means algorithm is said to be converged. Centroid The Triangle Inequality Theorem Inequalities in one triangle. Each client chooses the nearest facility as his supplier. In applying Eq. buffer problems are expected. Look a little closer and life is just one of those big problems full of little, unavoidable. cov_list: list. Calculating the centroids requires efficiently evaluating the integrals in equation. 411 Chapel Drive Durham, NC 27708 (919) 660-5870 Perkins Library Service. Typical (straight sided) Problem. A fascinating fact is that the centroid is the point where the triangle's medians intersect. 475,010 -267,590 12,723. In the above triangle , AD, BE and CF are called medians. k-means clustering centroid problem. Illinois Tractors, Inc. So it's equal to 2/3 times 15, which is equal to 10. 6 Moments, Centers of Mass, and Centroids 497 Center of Mass in a One-Dimensional System You will now consider two types of moments of a mass—the moment about a point and the moment about a line. Centroids and centre of gravity. Find another word for problem at YourDictionary. Surprisingly, many people find problems while working with them, silly errors due to how they work most of the time. Color each point according to the nearest centroid. Here's a Quick Look at the kind of Problems which have been solved in the Tutorial document at the end : Using integration find the centroid of the parabolic area OAB as shown in the figure below. This is not the case for composite beams and is one of the main difficulties in solving for the bending stress. A simple example of application for these relationships is given below, Lets try a more practical problem using the centroid, ([Hibbeler, 1992], prob. MOTIVATION: Classification of biological samples by microarrays is a topic of much interest. Centroid fuel senders determine fuel level by measuring the capacitance between their metal inner and outer tubes. Correct answers: 1 question: In the diagram, which must be true for point P to be the centroid of the triangle? LN ⊥ JK, JO ⊥ LK, and JL ⊥ MK. All points (except for Australia and Italy) shown on the map are based on geographic centroids, and are not Australian and Italian dots are located at the centroid of the largest city in each state. For initialization, sample 16 colors randomly from the original small picture. Rectangle Centroid x̅ and y̅ Right Triangle Centroid x̅ I and y̅ about x Semi-circle Centroid δ L Deformation: Axial δ deformation F = axial force 0 = original length = cross-sectional area E = modulus of elasticity POE 5 AE 4 CEA 4 xx b Moment of Inertia = moment of inertia of a rectangular section -x axis x h x. What is the location of its centroid from the line x = b?. X is the given data-point from which we have to determine the Euclidean Distance to the centroid. arg min c i ∈ C d i s t (c i, x) 2 \arg \min_{c_i \in C} dist(c_i, x)^2 ar g c i ∈ C min d i s t (c i , x) 2. But in reality, they are unavoidable. khubeb Mar 26, 2011 1:26 PM I made 10,000 k-means cluster in ODM 10g, version 10. The high speed image processor may calculate centroids of feet of the leads, average height of the feet and border violation. The denominator of the centroid is transformed as follows:. Problems seem like something to avoid. Problem on Centroid of I Section Video Lecture from Chapter Centroid and Centre of Gravity in Engineering Mechanics for First. Practice Problem Centroid of a circular arc: Locate the centroid of the circular arc as shown in the figure. P is the centroid of TQRS and RP 5 10. R is the region bounded by y = x +x, y = 0, and x = 2. In machine learning, a nearest centroid classifier or nearest prototype classifier is a classification model that An extended version of the nearest centroid classifier has found applications in the medical. Can anyone suggest how to integrate this problem? maybe even just the top would be great help. ‘sample’ selects random samples as centroids. Especially in the early days of teaching, figuring out what is going to be problematic for students is one of the more challenging. 'random': choose k observations (rows) at random from data for the initial centroids. Service Delivery Manager at Centroid Systems Lake In The Hills, Illinois 500+ connections. So how would I go about solving a problem like this: Find the y-coordinate of the centroid oft he curve given by the parametric equations: x=(3^(1/2))t^2 y=t-t^3 t= [0, 1] I can figure out most of it, I'm just stuck trying to figure out how I would find the area of the centroid Thanks. Use the words to complete an if-sentence about solutions to environmental problems. * First it will deal with the centroids of simple geometric shapes. GROUP PROBLEM SOLVING. Centroid The Triangle Inequality Theorem Inequalities in one triangle. Then find the exact coordinates of the centroid. It has recently been shown that classification by nearest centroids provides an accurate predictor that may outperform much more complicated methods. The k-means algorithm is a very useful clustering tool. The centroid is computed using the following formula, where μ ( x i ) is the membership value for point x i in the universe of discourse. In the case of finding initial centroids using Lloyd’s algorithm for K-Means clustering, we were using randomization. Problem-solving is human. To show how to determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape. In general, Centroid formula is a point on a given body or shape at which the entire mass of the body acts (center of gravity of the mass), it might also be the center of area for certain shapes. y-centroid in. Introduction K-means clustering is one of the most widely used unsupervised machine learning algorithms that forms clusters of data based on the similarity between data instances. P(-6, 9), Q(6, 1), R(-6, -7) Solution (-2, 1). pappus centroid theorem problems The surface of revolution generated by a smooth curve γ in the xz-plane with x. All of its centroids are stored in the attribute cluster_centers. In centroid mode, the backbone remains fully atomic, but the representation of each side chain is Centroid scorefunctions are kind of vague, in the same way that the protein representation is kind of. Hello, There is a problem with the command 'centroid' (in Dutch 'zwaartepunt'). In this page you can discover 38 synonyms, antonyms, idiomatic expressions, and related words for problem , like. The centroid of a triangle is just going to be the average of the coordinates of the vertices. If not converged, go back to 2 d cluster distance ( x, m. In this paper, we propose. A manufacturing facility is being planned to provide fabricated components to three tractor assembly facilities. Close Popup. Sketch the region bounded by the curves, and visually estimate the location of the centroid. As one can see, the bottom left image is a more accurate representation. The Gergonne point, triangle centroid , and mittenpunkt are collinear, with. Current Interns (Fall 2019) Tabitha Covey. I tried to cluster data points by k-medoids where k-centroids/medoids are randomly picked from (X), how those centroids are recalculated when there are some clusters changing by the updating x1? wi. A median of a triangle is the segment from a vertex to the midpoint of the opposite side. Two players, the leader and his competitor, open facilities, striving to capture the largest market share. cov_list: list. A condition under which the centroid must be inside the figure is when the figure is convex. Well they show it as the probe and software so likely yes there is an unlock code. All of its centroids are stored in the attribute cluster_centers. Title: 5-Medians Author: Mike Created Date: 7/12/2012 1:40:46 PM. y = x 3 , x + y = 2 , y = 0 | bartleby. Make x_centroids and y_centroids arrays that depend on the frame. Area Area Centroid Area Momets Volume Volume Centroid Volume Moments Hidrostatics. Initialize a set of centroids randomly 2. (Other initialization methods are also possible. Triangle, Centroid, Incenter, Parallel, Proportions. in fact we replace that all of these small forces by a single equivalent force (the resultant force). Problem – E – Codeforces Let ans[i] denote the min distance to a red node for the centroid “i” in it’s corresponding part. RFC 1913 Architecture of the Whois++ Index Service February 1996 3. Next you break your object down into polygons and find the centroids of these polygons. Powered by Create your own unique website with customizable templates. SPOJ has a rapidly growing problem set/tasks available for practice 24 hours/day, including many original tasks prepared by the community of expert problem. Check out this awesome Centroid Computer Corporations: The New Sales Manager Case Studies Examples for writing techniques and actionable ideas. Home About Algebra 1 > > > > > > > Principles of Engineering >. Integration formulas for calculating the Centroid are:. The final cluster with centroids c1 and c2 is as shown below:. Centroid Problems If this is your first visit, be sure to check out the FAQ by clicking the link above. 1 , Article 8. The geographic center of the USA was found this. P is the centroid of TQRS and RP 5 10. Define centroid. Aug 30, 2020 vector mechanics for engineers statics and dynamics 8th eigth edition Posted By John CreaseyLtd TEXT ID 66911762 Online PDF Ebook Epub Library eighth vector mechanics for engineers dynamics edition 9 8 sample problem 91 determine the moment of inertia of a triangle with respect to its base solution o a differential strip parallel to the x axis is. Problems seem like something to avoid. Visit BYJU'S to learn definitions, properties and centroid formulas for different geometrical shapes with examples. C is the number of centroids and j is the number of clusters. K]) for each % example, and K, the number of centroids. You can browse the problem sets from each year of the. Arm Type ATC Position Drum Problem (Legacy) 05-10-2007. In k-means clustering, each group is defined by creating a centroid for each group. Problems on differentiation of trigonometric functions. SPOJ has a rapidly growing problem set/tasks available for practice 24 hours/day, including many original tasks prepared by the community of expert problem. Each client chooses the nearest facility as his supplier. Common centroid constraint is not the same as 2-D symmetry constraint. Reposition the random centroid to the actual centroid. Centroid & Centre of Gravity 1. : Returns : center_list: list. If we restrict the concept of center of gravity or center of mass to a closed plane curve we obtain the idea of "centroid". The first result relates the centroid of a plane region with the volume of the. Given: A plate as shown. Various innovative solutions to these problems are also shown with examples. The centroid of plane geometric area can be located by one of the following methods a) Graphical methods b) Geometric consideration c) Method of moments The centroid of simple elementary areas can be located by geometric consideration. It is an iterative procedure that starts with guessing the initial centroids. See center of mass. 1 Centroids by Integration Problem Statement for Example 1 1. To solve this problem, several sentence-level text similarity methods have recently been established [1, 2, 3, 4, 5, 6, 7, 8, 9. Summations. To show how to determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape. Set ϵ to a small value as the terminating threshold. Centroid of a Triangle The centroid of a triangle is the point where the three medians coincide. It is formed by the intersection of the medians. Sample data matrix Consider the following matrix: {\bf X} = \left[ \begin{array}{ccc} 4. for example in lights: $JOB/tex/bathroom_4k. Consequently we shall devote the next group of frames to the determination of centroidal moments of inertia. In this page you can discover 38 synonyms, antonyms, idiomatic expressions, and related words for problem , like. Implementation: #find new centroid by taking the centroid of the points in the cluster class for cluster_index in self. BYU ScholarsArchive Citation. But what is a centroid? Well the centroid, we said was a point of concurrency of a three medians. Some people believe the solution to these problems is everyone accepts a simpler way of life, while others say that technology can solve these. A condition under which the centroid must be inside the figure is when the figure is convex. Transform stiffness involves an important distinction in which the line that represents the position of a frame object may be different from the line that represents the object centroid and its physical attributes. paper, we propose new efficient methods for centroid clustering problems. Using Geometry Calculation. But I have no idea how to calculate 8. 1 Centroids by Integration Problem Statement for Example 1 1. More formally, if c i is the collection of centroids in set C, then each data point x is assigned to a cluster based on. Solutions for Water Scarcity What is the Global Problem? Education Recycle water Advance Technology Related to Water Conservation Top 10 Global Problems of the world 1. Therefore, point J is the centroid of and, according to the Centroid Theorem, UJ is of UY. (a) circumcenter (b) incenter (c) centroid (d) orthocenter 10. SPOJ has a rapidly growing problem set/tasks available for practice 24 hours/day, including many original tasks prepared by the community of expert problem. Determine The Location Y Of The Centroid C Of The Beam Having The Cross Sectional Area Shown. It is % given a dataset X where each row is a single data point, a vector % idx of centroid assignments (i. Definition 1. Finding the new centroid from the clustered group of points: S i is the set of all points assigned to the ith cluster. The centroid method helps businesses identify the location of facilities used for manufacturing and other production-related aspects of the business. C C C is the set of all centroids. This means that any point inside a hexagon is closer to the centroid of the hexagon than any given point in an equal-area square or triangle would be (this is due. Seamlessly integrated into the CENTROID CNC control, digitzing parts is simple with the fill in the blank menus. K are the number of clusters and n are the number of cases. It is a centroid-based algorithm meaning that the goal is to locate the center points of each Each black dot represents the centroid of a sliding window and each gray dot is a data point. (The centroid is also known as the "center of gravity. If a test data is more closer to , then that data is labelled with ‘0’. Sample Problem 5. return getCentroid( id, u, n ); } inline int Centroid ( int u ). y = 49 − x 2, y = 0 I have the graph, i just dont understand how to get the centroid pointI''ve tried to follow other probems while doing this, but i just cant get this answer. More formally, if c i is the collection of centroids in set C, then each data point x is assigned to a cluster based on. These are your means. Show Edges Split Edge Merge Edge Join 2 Naket Edge Rebuild Edges. 5 2 2 8 2 1 x 8 8 5 0. This algorithm can be used to find groups within unlabeled data. Problems on centroid, circumcentre and orthocentre - example If the circumcentre of the triangle lies at (0, 0) and centroid is middle point of (a 2 + 1, a 2 + 1) and (2 a, − 2 a) then the orthocentre lies on the line? Given coordinates of circumcentre is (0, 0). Using intersecting tool to get the intersecting point of the line segments. the problem comes from the scaling the chord of the wing is supposed to be 1 cm but i am meshing a geometry of chord 100 units (as far as i know Gridgen has no If dimensions are far away from this range Prostar will create problem. Centroid and Center of Gravity. Nearest shrunken centroids (NSC) is a popular classification method for microarray data. Suppose, for example, that an area A consists of two parts A 1 and A 2, with centroids at and respectively. Surprisingly, many people find problems while working with them, silly errors due to how they work most of the time. mec 109 notes 3. 1/22/20 Jure Leskovec, Stanford CS246: Mining Massive Datasets ¡. Problem Solution Sample Essay. In this paper, we propose. Parallel Axis Theroem. 9—6, or the analogous centroid equations. a) When R is the area under : ∫ ∫ ∫ The density cancels out, so the centroid is: ̅ ̅ Formulas: b) When R is the area bounded above by and below by :. Proof in the style of Descartes Direct observation of a few examples suggests that the medians of a triangle not only meet at the same point, but that this point is two-thirds of the way from the vertex to the midpoint of the opposite side on each median. Two players, the leader and his competitor, open facilities, striving to capture the largest market share. Multiproduct Scheduling Problem Benchmark library was created thanks to financial support by Russian Foundation for Basic Research (grants NN 98-07-90259, 01-07-90212, 04-07-90096, 08-07-00037, 11-07-00474). This can be carried out only by phonological analysis based on phonological rules. A centroid is the point of intersection of the medians of the triangle. In fact, you only need to know depth first search to understand it. Find the Centroid of an Irregular Shape (Example #11) Statics. Games and puzzles. Make x_centroids and y_centroids arrays that depend on the frame. Various innovative solutions to these problems are also shown with examples. I am in Grade 12 and I am stuck on a problem dealing with Geometry. The area: The element of area is a vertical strip y high and dx wide. Centroid definition is - center of mass. Centroid M400 control with 15” Touch Screen LCD 24 tools ATC arm type 25 HP spindle motor,. The location of the centroid is often denoted with a 'C' with the coordinates being x̄ and ȳ, denoting that they are the average x and y coordinate for the area. The centroid of a triangle is a point, where the three medians intersect. The centroid will lie on an axis of symmetry however. 814-353-9290 M-F 8:30-5:00 159. Symbols If P is the centroid of TABC , then AP 5} 2 3}AD , BP 5} 2 3}BF , and CP 5} 2 3}CE. The centroid for components in each dimension is defined as follows: The centroid of points is the arithmetic mean of the input coordinates. The centroid C is the mathematical center of the plate’s area. The centroid is also the center of balance of a triangle. constant density. This problem is similar to problem 49, except that the gate extends above the top of the water. It is assumed that the student is already familiar with the following concepts. Repeat steps 1-5 until the centroids no longer move. A height is each of the perpendicular lines drawn from one vertex to the opposite side (or its extension). 1 , Article 8. The fundamentals of centroid-based object tracking. Centroids do not change location any more) OR 4. Polar moment of inertia is also denoted as. Without the diagram, this problem is rather difficult. SOLUTION : • Divide the area into a triangle, rectangle, and semicircle with a circular cutout. https://www. What do you mean by the Centroid of a Triangle? In order to understand the term centroid, we first need to know what do we mean by a median. Definitions. 1 Centroids. We need to choose p facilities of the leader in such a way as to maximize his market share. Furthermore, we have also shown that if the cost is less than n/2, then the centroid is unique. 8) y R = y c + I xc y cA (3. (The centroid does not have to be in the figure, however. Sketch the region bounded by the curves, and visually estimate the location of the centroid. In k-means clustering, each group is defined by creating a centroid for each group. This post will approach our task as follows: prepare the data; prepare our sample; perform centroid initialization search iterations to determine our “best” collection of initial centroids. While there are no best solutions for the problem of determining the number of clusters to extract # Centroid Plot against 1st 2 discriminant functions library(fpc) plotcluster(mydata, fit$cluster). Finding the Centroid of Two Dimensional Shapes Using Calculus. Determine The Location Y Of The Centroid C Of The Beam Having The Cross Sectional Area Shown. The main element of the algorithm The figure below shows the centroids and SSE updating through the first five iterations from two. The centroid is the location that we can model the shape as if the entire weight acts through this point. This post will approach our task as follows: prepare the data prepare our sample perform centroid initialization search iterations to determine our "best" collection of initial centroids. Lower the centroid (or center of mass) of a system more stable the system will be. The Centroid of a parallelogram is the Intersection point of its diagonals In a coordinate plane, the center of mass of a parallelogram with vertices P=P(x1,y1), Q=Q(x2,y2), R=R(x3,y3) and S=S(x4,y4) is the point with the coordinates = , =. I am trying to find the centroid of a contour but am having trouble implementing the example code in C++ (OpenCV 2. 150729 1 r 2 28 30 14. Here are some examples of obtaining a centroid. The centroid C is the mathematical center of the plate’s area. The centroid of a triangle is the point through which all the mass of a triangular plate seems to act. Calculating the centroids requires efficiently evaluating the integrals in equation. Problems on moment, mass, center of mass, and centroid. Whether you're dealing with your dad's decade-old computer or your own custom-built gaming rig, troubleshooting PC problems is a part of everyday life. For floating bodies the stability problem is more complicated, since as the body rotates the location of the center of buoyancy (which passes through the centroid of the displaced volume) may change. How long you have been experiencing problems opening Yandex Browser. edu † Georgia Institute of Technology, Atlanta, Georgia 30332–0250 Email: matzi. The problem I am having, is that the route algorithm obviously searches for the route that costs the least (black line below) causing the route to traverse multiple centroid connectors and avoiding the network all together. Determine the location of the centroid 'c' the beam's cross section and the moment of inertia about both centroidal axis. 44% of the radius. Each Chapter Begins With A Quick Discussion Of The Basic Concepts And Principles. How To Calibrate Centroid Fuel Sender. But in reality, they are unavoidable. The fundamentals of centroid-based object tracking. each entry in range [1. If an object has an axis of symmetry, then the centroid of object lies on that axis. The moment of inertia of a triangle with respect to an axis passing through its centroid, parallel to its base, is given by the following expression: I = \frac{b h^3}{36} where b is the base width, and specifically the triangle side parallel to the axis, and h is the triangle height (perpendicular to the axis and the base). 2 • Public • Published 4 years ago. There is no net internal axial force. Weight of component times distance from its centroid to new centroid. The centroid is 2/3 of the way from a given vertex to the opposite midpoint. Use fc’ = 4000 psi, and fy = 60,000 p si. The first problem of phonological analysis is to establish the phonemes in a definite language. Nanotechnology: technology on a small scale. The centroid of an area is similar to the center of mass of a body. Textbook solution for Calculus (MindTap Course List) 8th Edition James Stewart Chapter 8. The first result relates the centroid of a plane region with the volume of the. The main element of the algorithm The figure below shows the centroids and SSE updating through the first five iterations from two. BYU ScholarsArchive Citation. In this case, the Do not Transform Stiffness for Offsets from. As one can see, the bottom left image is a more accurate representation. in houdini -> add file node. Engineering. In this case the whole complex is a word combination "adjective. Anyway I hope this helps someone I wanted to do a cap type sender but was afraid of the wiring and as it turns out it is nothing at all. More formally, if c i is the collection of centroids in set C, then each data point x is assigned to a cluster based on. Centroids (grey) and shrunken centroids (red) for the SRBCT dataset. The 2nd moment of area, also known as moment of inertia of plane area, area moment of inertia, or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Circle Word Problems Exercise 1Anne is riding a horse which is tied to a pole with a 3. Solutions for the example problem from the topic of Centroid of Composite Bodies for the Statics course. https://clnk. They are the Incenter, Orthocenter, Centroid and Circumcenter. Each centroid is an existing data point in the given input data set, picked at random, such that all centroids are unique (that is, for all centroids ci and cj, ci ≠ cj). Solution to Problem 1 The x and y coordinate of the centroid are given by: x = (- 2 + 4 + 1) / 3 = 1 y = (0 + 3 + 6) / 3 = 3 Below is shown the graphical solution including triangle, the medians of the three vertices and their point of intersection O called the centroid. The K-means algorithm starts by randomly choosing a centroid value for each cluster. Service Delivery Manager at Centroid Systems Lake In The Hills, Illinois 500+ connections. Note that this could be done using other techniques too, and even faster than centroid decomposition, but this will serve as a nice example for the basics. Properties of the Centroid. P is the centroid of TQRS and RP 5 10. The normal planes remain normal for pure bending. K are the number of clusters and n are the number of cases. Update Centroid We use the following equation to calculate the n dimensionalWe use the following equation to calculate the n dimensional centroid point amid k n-dimensional points Example: Find the centroid of 3 2D points, (2,4), (5,2) and (8,9)and (8,9) Example of K-means Select three initial centroids 1 1. The centroid is equal to the centroid of the set of component Geometries of highest dimension SELECT ST_AsText(ST_centroid(g)) FROM ST_GeomFromText('CIRCULARSTRING(0 2, -1 1,0 0. CoCreate User Forum > Applications > CoCreate Modeling: Centroid !. a) When R is the area under : ∫ ∫ ∫ The density cancels out, so the centroid is: ̅ ̅ Formulas: b) When R is the area bounded above by and below by :. Definition of Centroid - Centroid and Centre of Gravity - Engineering Mechanics. This is done by calculating Euclidean(L2) distance between the point and the each centroid. In this page you can discover 38 synonyms, antonyms, idiomatic expressions, and related words for problem , like. Solutions to this problem. Step 7: Reposition the random centroid to the actual centroid. Locate the centroid \\bar{y} of the shaded area. Plots 1, 3 and 10 are on granite; 2, 4 and 5 are on limestone, and 6, 7, 8 and 9 are on basalt. If you like motor controllers, drives, and power electronics; they Nothing is a problem until it is, and once it's brought to attention it's expected to be handled, yesterday. Examples for Moment of Inertia 14 8” 1” Typ. Each Chapter Begins With A Quick Discussion Of The Basic Concepts And Principles. I am trying to find the centroid of a contour but am having trouble implementing the example code in C++ (OpenCV 2. Each client chooses the nearest facility as his supplier. Assume that the density is 1. ly/2kNxjuH Other First. I have some pictures but darned if I can figure out how to post them. I ordered another and installed it. SOLUTION: •Divide the area into a triangle, rectangle, and semicircle with a circular cutout. It is the "center of mass". NSC calculates centroids for each class and "shrinks" the centroids toward 0 using soft thresholding. It is important to realize that the choice of the initial centroid has a huge effect on the final result, for instance the figure below shows the resulting clusters based on two different randomly selected centroids. The article examines various global environmental problems and their causes. distortion [Returned Value] [float] The distortion between the observations passed and the centroids generated. The pieces of a unicentroidal tree T having centroid z are the components of T −z; when we know T and z, the neighbors of z in the pieces are the roots of the pieces. The centroid is the location that we can model the shape as if the entire weight acts through this point. Can anyone help me out?. A median is the line joining the mid-points of the sides and the opposite vertices. 2441 Component Quarter circle. Here are three theorems involving centroid, orthocenter, and circumcenter of a triangle. The second develops the use of integration to find the center of mass of a planar lamina, and with that, the centroid of a region. So the centroid is located at Xc = 183. com/matlabcentral/answers/62938-problem-with-centroid-of. I will try to show you the way of efficiently working with them, avoid this errors and. cal 2 centroid problem? Find the centroid of the region bounded by the given curves. Therefore U = (U1, U2, …, UN) denotes the membership value matrix. Chapter 5 already introduced a centroid function st_centroid(), but this example highlights how One of the most intriguing things about spatial data problems is that things which appear to be trivially. 251 yV9 – yż dy Is Calculate ya • Rectangular Method (Left) • Rectangular Method (Right) • Midpoint For each integration method use a spacing of 0. Course Links. centroids 2. The elevation is derived from the pline or, if a set, the mean elevation of the points. 'random': choose k observations (rows) at random from data for the initial centroids. X is the given data-point from which we have to determine the Euclidean Distance to the centroid. Circle Word Problems Exercise 1Anne is riding a horse which is tied to a pole with a 3. Create points at centroids of plines and sets. 5 3 y Iteration 1. The centroid C is the mathematical center of the plate’s area. So the centroid is located at Xc = 183. Centroid is the first and official newsletter of the Institute of Civil Engineering. Harbor Freight inverter generator. Center of Mass and Centroids: Composite Bodies and Figures Integration vs Appx Summation: Irregular Volume Reduce the problem to one of locating the centroid of area Appx Summation may be used instead of integration Divide the area into several strips Volume of each strip = A∆x Plot all such A against x. edu is a platform for academics to share research papers. : Returns : center_list: list. buffer problems are expected. khubeb Mar 26, 2011 1:26 PM I made 10,000 k-means cluster in ODM 10g, version 10. Seamlessly integrated into the CENTROID CNC control, digitzing parts is simple with the fill in the blank menus. Say you have a 12"x12"x10'-0" wood column supporting a platform. In this formulation, the path integral centroid density occupies the same role as the Boltzmann density in classical statistical mechanics. Find the position of the neutral axis for combined stress situations. A condition under which the centroid must be inside the figure is when the figure is convex. It is a new project pioneered by this year's’ Civil Engineering Representatives. the average) of all the points in a shape. Install it yourself or we can professionally install, on-site, in just days. Anchor: #VHVWXHXH; Determine the potential head from the centroid of the culvert opening, which is approximated as the sum of the invert elevation and one half the rise of the culvert. Solution Practice Problem Centroid of a triangular area. 005) from (select. In-vitro fertilization (IVF), as the most common fertility treatment, has never reached its maximum potentials. The task is to implement the K-means++ algorithm. The centroid is a point of concurrency. Anyone have any ideas or suggestions? I'm using Python for this, but I can adapt examples from other languages. The centroid of a section is not always within the area or material of the section. Finding the Area and individual centroid dimensions of each of these sub shapes will allow you to find the overall centroid. Triangle centroid. Visualizing K-Means Clustering. x i is the distance from the axis to the centroid of the simple shape, A i is the area of the simple shape. And AG is going to be 2/3 of AD. She specializes in Spatial Analysis/spatial problem solving, Python scripting, mobile data collection, and the ArcGIS Desktop suite. Centroid Products 2104 Hibiscus Dr. Determine by direct integration the location of the centroid of a parabolic spandrel. More precisely, we are going to apply our methods to problems of the following type: given entities with weights ( =1,…, ) it is searched centres ( =1,…, ) minimizing , where ( , ) measures the dissimilarity between and However, the methods are very general. from solving statics problems in matlab - sample problems from solving statics problems in matlab by School of. I am in Grade 12 and I am stuck on a problem dealing with Geometry. Centroid is the first and official newsletter of the Institute of Civil Engineering. Since the distance along the gate where water is present is 6 ft, the centroid of the area where there is water pressure is yc = 3 ft. In this section we are going to find the center of mass or centroid of a thin plate with uniform density $$\rho$$. Arm Type ATC Position Drum Problem (Legacy) 05-10-2007. Motivation problem : We have a tree consisting of n(<=10^5) nodes. 5 3 y Iteration 1. Centroid: Centroid is the point of intersection of the three medians of a triangle. The problem is that when the centroid of the mpolygon is outside the outline, my text inserts outside, too. Both centroid-linkage and Ward's method are applicable to the clutter removal problem because the overlap of representative icons is checked based on the distance between cluster centroids. A centroid divides the median in the ratio 2:1. 1 5 - 16 For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid. applying the center of gravity equations, Determine i, j, Eqs. 'random': choose k observations (rows) at random from data for the initial centroids. 131: Problems: p. Definition 1. Centroid Draw a line (called a "median") from each corner to the midpoint of the opposite side. Note that there is a problem with linear combinations: As previously stated, no set of variables can add to a constant, yet the sum of the values for each plot is 1. A median of a triangle is the segment from a vertex to the midpoint of the opposite side. ” The same formula, with y i substituting for x i, gives us the y coordinate of the centroid. The centroid divides each median into a piece one-third the length of the median and two-thirds the length. Define centroid. x_centroid = [(Area of each piece)(Centroid of each piece)]/(Area of each piece) The Attempt at a Solution This seems like a really basic problem, I am just reviewing for my statics final on saturday. Abaqus element centroid coordinates. 43: Centroid: Investigate, justify, and apply theorems about the centroid of a triangle, dividing each median into segments whose lengths are in the ratio 2:1 1 In the diagram below of ACE, medians AD, EB, and CF intersect at G. The first thing that comes to mind for calculating such a center is the polygon centroid. Systematic selection of embryos with the highest implementation potentials is a necessary step toward enhancing the effectiveness of IVF. k-means clustering is a method of vector quantization, originally from signal processing, that aims to partition n observations into k clusters in which each observation belongs to the cluster with the nearest mean (cluster centers or cluster centroid), serving as a prototype of the cluster. Learn vocabulary, terms and more with flashcards, games and other study tools. The Gergonne point, triangle centroid , and mittenpunkt are collinear, with. The centroid decomposition is a very simple idea that is able to solve some really scary problems. The center of gravity will equal the centroid if the body is homogenous i. Centroid and Centre of Gravity Centroid Center of Gravity • It is defined as a point about which the entire line, area or volume is assumed to be concentrated. The second develops the use of integration to find the center of mass of a planar lamina, and with that, the centroid of a region. Assumptions made for this problem. No change between iterations 3 and 4 has been noted. Sample Problem 9. 1 Centroids by Integration Problem Statement for Example 2 2. Triangle Inequality Problem… 1. Lloyd’s algorithm (Lloyd, 1982) uses a greedy strategy to approxi-mate the Eq. In case of ties, the leader’s. Each centroid is an existing data point in the given input data set, picked at random, such that all centroids are unique (that is, for all centroids ci and cj, ci ≠ cj). https://www. I am in Grade 12 and I am stuck on a problem dealing with Geometry. But I have no idea how to calculate 8. It doesn't rot, like paper or food, so. Centroids are data points representing the center of a cluster. Centroids and Their Application to Some Mechanical Problems 1. It divides each median into two sections at a 2:1 ratio. Solutions for Water Scarcity What is the Global Problem? Education Recycle water Advance Technology Related to Water Conservation Top 10 Global Problems of the world 1. January 19, 2014. 1 cluster function to cluster 1000 pdbs of a homodimer that were created by Symmdock of rosetta3. Common centroid constraint is not the same as 2-D symmetry constraint. I am trying to find the centroid of a contour but am having trouble implementing the example code in C++ (OpenCV 2. In this model the number of clusters required at the end is known in prior. By viewing the training of classifiers as an optimisation problem, we have developed a method in this paper to train A Stepwise Multi-centroid Classification Learning Algorithm with GPU Implementation. Iteration-1, determine centroids : Knowing the members of each group, now compute the new centroid of each group based on these new memberships. Here is the K-means algorithm as it applies to this problem: K-means algorithm. P-708 bounded by the x-axis, the line x = b, and the curve y = kxn where n ≥ 0. Centroid and center of gravity are two important concepts in statics. Moments of Inertia explains how to find the resistance of a rotating body. It picks the parameter that gives the largest separation between the centroids, in kernel feature space, of two classes of data. In machine learning, a nearest centroid classifier or nearest prototype classifier is a classification model that An extended version of the nearest centroid classifier has found applications in the medical. Determine the moment of inertia and radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the section. R: calculate the area of any convex polygon. So this whole thing right over here is going to be equal to 15. A simple example of application for these relationships is given below, Lets try a more practical problem using the centroid, ([Hibbeler, 1992], prob. CoCreate User Forum > Applications > CoCreate Modeling: Centroid !. Users seeking the latest gridded population data are advised to use Gridded Population of the World, Version 4 (GPWv4), which supersedes GPWv3 and the Global Rural-Urban Mapping Project, Version 1 (GRUMPv1). C is the number of centroids and j is the number of clusters. I tried to cluster data points by k-medoids where k-centroids/medoids are randomly picked from (X), how those centroids are recalculated when there are some clusters changing by the updating x1? wi. Zero stress exists at the centroid and the line of centroids is the neutral axis (n. Assign all points to the closest centroid. Only one integration point at the centroid. Triangle centroid. 102 the machine locate the x coordinate of the center Of gravity. What Is the Centroid Method in Business?. The centroid of an area can be thought of as the geometric center of that area. No problem what so ever. Transform stiffness involves an important distinction in which the line that represents the position of a frame object may be different from the line that represents the object centroid and its physical attributes. Problem of estimation of economic, social and cultural, social and ecological consequences of technics development. This calculus video tutorial provides a basic introduction into the center of mass of a system also known as the centroid. 1 5 - 16 For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid. Question 1 : Find the centroid of triangle whose vertices are (1, 10) (-7, 2) and (-3, 7). However for an object such as a beam, the centroid is only represented by the geometric center if the material is uniform or homogenous. Repeat steps 2 and 3 until the centroids don’t change. Centroid: Centroid is the point of intersection of the three medians of a triangle. Centroid = weighted average of centroids of component clusters. Centroids are data points representing the center of a cluster. Centroid of Composite Body/Figure Irregular volume:: Integration vs Approximate Summation - Reduce the problem to one of locating the centroid of area-Approximate summation instead of integration Divide the area into several strips Volume of each strip = AΔx Plot all such A against x. Yaskawa Sigma I and II with Centroid CNC11 based CNC controls (Oak) 06-25-2015. KMeans algorithm. It is the center of mass (center of gravity) and therefore is always located within the triangle. Initialize m, M, and initial cluster centroids C0. The second develops the use of integration to find the center of mass of a planar lamina, and with that, the centroid of a region. 25 Problem 5. Mechanical Engineering: Centroids & Center of Gravity (5 of 35) Center of Gravity of a 1/4 Circle. Indeed such claims and. In this section we are going to find the center of mass or centroid of a thin plate with uniform density $$\rho$$. R: calculate the area of any convex polygon. In this tutorial, we will tell you about the steps to generate BOM(Bill of Materials) and CPL(component placement list, as known as a Centroid file/Pick and pla. The centroid divides the length of each median in 2:1 ratio. The centroid of a triangle is the center of the triangle, which can be determined as the point of intersection of all the three medians of a triangle.
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## Performance Measures in Electric Power Networks under Line Contingencies
14 May 2019 · Coletta Tommaso, Jacquod Philippe ·
Classes of performance measures expressed in terms of ${\cal H}_2$-norms have been recently introduced to quantify the response of coupled dynamical systems to external perturbations. So far, investigations of these performance measures have been restricted to nodal perturbations... Here, we go beyond these earlier works and consider the equally important, but so far neglected case of line perturbations. We consider a network-reduced power system, where a Kron reduction has eliminated passive buses. Identifying the effect that a line fault in the physical network has on the Kron-reduced network, we find that performance measures depend on whether the faulted line connects two passive, two active buses or one active to one passive bus. In all cases, performance measures depend quadratically on the original load on the faulted line times a topology dependent factor. Our theoretical formalism being restricted to Dirac-$\delta$ perturbations, we investigate numerically the validity of our results for finite-time line faults. We find good agreement with theoretical predictions for longer fault durations in systems with more inertia. read more
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# Categories
Systems and Control Adaptation and Self-Organizing Systems
# Datasets
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# Munn tree
Let $X$ be a finite set, and $\left(X\amalg X^{-1}\right)^{\ast}$ the free monoid with involution on $X$. It is well known that the elements of $\left(X\amalg X^{-1}\right)^{\ast}$ can be viewed as words on the alphabet $\left(X\amalg X^{-1}\right)$, i.e. as elements of the free monod on $\left(X\amalg X^{-1}\right)$.
The Munn tree of the word $w\in\left(X\amalg X^{-1}\right)^{\ast}$ is the $X$-inverse word graph $\mathrm{MT}(w)$ (or $\mathrm{MT}_{X}(w)$ if $X$ needs to be specified) with vertex and edge set respectively
$\mathrm{V}(\mathrm{MT}(w))=\mathrm{red}(\mathrm{pref}(w))=\left\{\mathrm{red}(% v)\,|\,v\in\mathrm{pref}(w)\right\},$
$\mathrm{E}(\mathrm{MT}(w))=\left\{(v,x,\mathrm{red}(vx))\in\mathrm{V}(\mathrm{% MT}(w))\times\left(X\amalg X^{-1}\right)\times\mathrm{V}(\mathrm{MT}(w))\right\}.$
The concept of Munn tree was created to investigate the structure of the free inverse monoid. The main result about it says that it “recognize” whether or not two different word in $\left(X\amalg X^{-1}\right)^{\ast}$ belong to the same $\rho_{X}$-class, where $\rho_{X}$ is the Wagner congruence on $X$. We recall that if $w\in\left(X\amalg X^{-1}\right)^{\ast}$ [resp. $w\in\left(X\amalg X^{-1}\right)^{+}$], then $[w]_{\rho_{X}}\in\mathrm{FIM}(X)$ [resp. $[w]_{\rho_{X}}\in\mathrm{FIS}(X)$].
###### Theorem 1 (Munn)
Let $v,w\in\left(X\amalg X^{-1}\right)^{\ast}$ (or $v,w\in\left(X\amalg X^{-1}\right)^{+}$). Then $[v]_{\rho_{X}}=[w]_{\rho_{X}}$ if and only if $\mathrm{MT}(v)=\mathrm{MT}(w)$
As an immediate corollary of this result we obtain that the word problem in the free inverse monoid (and in the free inverse semigroup) is decidable. In fact, we can effectively build the Munn tree of an arbitrary word in $\left(X\amalg X^{-1}\right)^{\ast}$, and this suffice to prove wheter or not two words belong to the same $\rho_{X}$-class.
The Munn tree reveals also some property of the $\mathcal{R}$-classes of elements of the free inverse monoid, where $\mathcal{R}$ is the right Green relation. In fact, the following result says that “essentially” the Munn tree of $w\in\left(X\amalg X^{-1}\right)^{\ast}$ is the Schützenberger graph of the $\mathcal{R}$-class of $[w]_{\rho_{X}}$.
###### Theorem 2
Let $w\in\left(X\amalg X^{-1}\right)^{\ast}$. There exists an isomorphism (in the category of $X$-inverse word graphs) $\Phi:\mathrm{MT}(w)\rightarrow\mathcal{S}\Gamma(X;\varnothing;[w]_{\rho_{X}})$ between the Munn tree $\mathrm{MT}(w)$ and the Schützenberger graph $\mathcal{S}\Gamma(X;\varnothing;[w]_{\rho_{X}})$ given by
$\Phi_{\mathrm{V}}(v)=[v]_{\rho_{X}},\ \ \forall v\in\mathrm{V}(\mathrm{MT}(w))% =\mathrm{red}(\mathrm{pref}(w)),$
$\Phi_{\mathrm{E}}((v,x,\mathrm{red}(vx)))=([v]_{\rho_{X}},x,[vx]_{\rho_{X}}),% \ \ \forall(v,x,\mathrm{red}(vx))\in\mathrm{E}(\mathrm{MT}(w)).$
## References
• 1 W.D. Munn, Free inverse semigroups, Proc. London Math. Soc. 30 (1974) 385-404.
• 2 N. Petrich, Inverse Semigroups, Wiley, New York, 1984.
• 3 J.B. Stephen, Presentation of inverse monoids, J. Pure Appl. Algebra 63 (1990) 81-112.
Title Munn tree MunnTree 2013-03-22 16:11:59 2013-03-22 16:11:59 Mazzu (14365) Mazzu (14365) 20 Mazzu (14365) Definition msc 20M05 msc 20M18 SchutzenbergerGraph Munn tree
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# Help in factorization of a third degree polynomial
Gold Member
## Homework Statement
Looking to factor $-2x^3-3$ and having an issue. To my understanding, the Fundamental Theorem of Algebra tells us that it is at least theoretically possible to factor any polynomial of degree n.
## The Attempt at a Solution
So my first step to factor this was to use the Rational Zeros Theorem, and synthetically divide $-2x^3-3$ by these possible rational zeros. After multiple attempts, no possible rational zeros were found to correspond to any factors of $-2x^3-3$. So this tells me that we either have real but irrational zeros, or non-real zeros.
Where can I go from here to factor this thing out in knowing this?
## Answers and Replies
Related Precalculus Mathematics Homework Help News on Phys.org
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
## Homework Statement
Looking to factor $-2x^3-3$ and having an issue. To my understanding, the Fundamental Theorem of Algebra tells us that it is at least theoretically possible to factor any polynomial of degree n.
## The Attempt at a Solution
So my first step to factor this was to use the Rational Zeros Theorem, and synthetically divide $-2x^3-3$ by these possible rational zeros. After multiple attempts, no possible rational zeros were found to correspond to any factors of $-2x^3-3$. So this tells me that we either have real but irrational zeros, or non-real zeros.
Where can I go from here to factor this thing out in knowing this?
Solve $\ -2x^3-3=0\$ and use the Factor Theorem.
Mark44
Mentor
## Homework Statement
Looking to factor $-2x^3-3$ and having an issue. To my understanding, the Fundamental Theorem of Algebra tells us that it is at least theoretically possible to factor any polynomial of degree n.
That's not what it says.
To quote the wiki article, https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra,
The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.
In fact, Galois provided that there is no general factorization formula for fifth-degree and higher polynomials. That's not to say that you can't factor some fifth-degree or higher polynomials.
opus said:
## The Attempt at a Solution
So my first step to factor this was to use the Rational Zeros Theorem, and synthetically divide $-2x^3-3$ by these possible rational zeros. After multiple attempts, no possible rational zeros were found to correspond to any factors of $-2x^3-3$. So this tells me that we either have real but irrational zeros, or non-real zeros.
Where can I go from here to factor this thing out in knowing this?
Write it as $-2(x^3 + \frac 3 2)$. Do you have any formulas for factoring cubics of the form $x^3 + a^3$ or $x^3 - a^3$?
Gold Member
Are you referring to the Zero Factor Theorem? I thought that was used for quadratics, and the reason for using things like the Rational Zeros Theorem, Descartes Rule of Signs, Intermediate Value Theorem, Upper and Lower Bounds, etc, was to drop a polynomial down to a quadratic so that we can use the Zero Factor Property?
Gold Member
That's not what it says.
To quote the wiki article, https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra,
Write it as $-2(x^3 + \frac 3 2)$. Do you have any formulas for factoring cubics of the form $x^3 + a^3$ or $x^3 - a^3$?
But isn't that article quote saying the same thing? If it does have at least one complex root, then it can be factored, can't it?
Ok I do know the sum and difference of cubes formulas. Are you suggesting that I go with solving something like $-2(x^3 + 1.44714243)$ with the second term in the parentheses being equal to $3/2$ when cubed?
Mark44
Mentor
But isn't that article quote saying the same thing? If it does have at least one complex root, then it can be factored, can't it?
No, they're not saying the same thing. The Fund. Thm. of Algebra says that at least one complex root exists, but it doesn't tell you how to find it. You can't factor a polynomial if you don't know the roots.
opus said:
Ok I do know the sum and difference of cubes formulas. Are you suggesting that I go with solving something like $-2(x^3 + 1.44714243)$ with the second term in the parentheses being equal to $3/2$ when cubed?
No, it would be $-2(x^3 + 1.5)$, so you should be able to use one of your formulas to get a linear factor and a quadratic (which will have complex roots only).
Gold Member
Could you please explain the 1.5? My train of thought here is that, by your recommendation, I use the sum of cubes formulas which is $a^3+b^3=(a+b)(a^2-ab+b^2)$.
But to use this, I need $-2(x^3+\frac{3}{2})$ to be in the form $-2(a^3+b^3)$. To do so, I would take the cube root of the 1.5, which will give me 1.44714243 which, when cubed, is equivalent to 1.5
Mark44
Mentor
Could you please explain the 1.5? My train of thought here is that, by your recommendation, I use the sum of cubes formulas which is $a^3+b^3=(a+b)(a^2-ab+b^2)$.
But to use this, I need $-2(x^3+\frac{3}{2})$ to be in the form $-2(a^3+b^3)$. To do so, I would take the cube root of the 1.5, which will give me 1.44714243 which, when cubed, is equivalent to 1.5
$-2(x^3 + (\sqrt[3]{1.5})^3)$ -- Now the part in parentheses fits your sum of cubes formula. It's best to leave the terms exact for now, not as decimal approximations.
In what you wrote in post #5, you had $-2(x^3 + 1.44714243)$, which is incorrect. The decimal approximation should have been to the power 3.
Gold Member
Ok that makes sense.
Now as a more general question, which is why I was asking this question in the first place:
Consider a rational function $f(x)=\frac{2x^3+2x^2-4x+2}{x^3-27}$
If I want to graph this, there are a number of things I need to find. In all cases, I'll need to find the x-intercepts (if any) and the vertical asymptotes (if any).
Now in a rational function $f(x)=\frac{P(x)}{q(x)}$, our zeros, and x-intercepts are determined by $P(x)$. This is because we have a fraction, and for a fraction to equal zero, the numerator has to equal zero. Now for the vertical asymptotes, these are determined by $q(x)$, because any values that make the denominator equal to zero and are thus undefined and the vertical asymptote is at the zero x=c.
Now to find the zeros and vertical asymptotes, I need to be able to factor both the numerator and the denominator.
Given this problem, $f(x)=\frac{2x^3+2x^2-4x+2}{x^3-27}$, my text states that the fraction is already simplified. Which I take it to mean, that the numerator and the denominator have no common factors other than 1. Now to determine if this fraction is simplified, we would need to factor the numerator and denominator. But the text did no such factoring.
So, is there a way to notice that the numerator and denominator here have no common factors without factoring them? I'm finding a little bit of difficulty in this step. The prior section went over finding all of the zeros for higher degree polynomials, and that was easy with the use of the Rational Zeros Theorem, and accompanying theorems. However, in these graphing problems, they aren't nearly as neat or straightforward and I'm having a hard time factoring these. I'm starting to think that the last section had cherry-picked, very easy polynomials to factor. Whereas in reality, they aren't that straight forward.
StoneTemplePython
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Gold Member
2019 Award
Given this problem, $f(x)=\frac{2x^3+2x^2-4x+2}{x^3-27}$, my text states that the fraction is already simplified. Which I take it to mean, that the numerator and the denominator have no common factors other than 1. Now to determine if this fraction is simplified, we would need to factor the numerator and denominator. But the text did no such factoring.
So, is there a way to notice that the numerator and denominator here have no common factors without factoring them? I'm finding a little bit of difficulty in this step.
What I've written below is more of a point of interest -- something to think about for the future-- as I don't think it will directly help you now. Again emphasizing monic polynomials, you have
$f(x) = 2\frac{p(x)}{q(x)} = 2 \frac{x^3 + x^2 -2x + 1}{x^3-27}$
if hypothetically I told you that $p(x)$ has $k$ distinct roots (over $\mathbb C$) and $q(x)$ had $r$ distinct roots, then there are no common roots if and only if the 'combined' polynomial given by $g(x) = p(x)q(x)$ has $(k + r)$ distinct roots. Why?
There's another, more direct approach that is harder to describe that uses resultants.
In both cases, the underlying approaches come from Sylvester and require linear algebra to interpret and solve, I'm afraid.
I'm starting to think that the last section had cherry-picked, very easy polynomials to factor. Whereas in reality, they aren't that straight forward.
I think a lot of introductory textbook problems are this way. Solve easy stuff first... hopefully you can return to these questions and ideas you have at some future point once you have more powerful tools at your disposal.
Last edited:
epenguin
Homework Helper
Gold Member
Given this problem, $f(x)=\frac{2x^3+2x^2-4x+2}{x^3-27}$, my text states that the fraction is already simplified. Which I take it to mean, that the numerator and the denominator have no common factors other than 1. Now to determine if this fraction is simplified, we would need to factor the numerator and denominator. But the text did no such factoring.
So, is there a way to notice that the numerator and denominator here have no common factors without factoring them? I'm finding a little bit of difficulty in this step. The prior section went over finding all of the zeros for higher degree polynomials, and that was easy with the use of the Rational Zeros Theorem, and accompanying theorems. However, in these graphing problems, they aren't nearly as neat or straightforward and I'm having a hard time factoring these. I'm starting to think that the last section had cherry-picked, very easy polynomials to factor. Whereas in reality, they aren't that straight forward.
Perhaps the most obvious and elementary way to test whether or not you have a common factor in this, as you say, made easy case, is to factorise the denominator. It fairly obviously has the factor (x - 3). So you could divide the numerator by that and see whether there is any remainder, or easier just substitute 3 for x in it and see whether the result is 0 - if it is then (x - 3) is a factor of it.
To get the other factor of the denominator you could divide it by (x - 3). You get a real quadratic (one with nonreal roots). You can divide the numerator by that and see whether you get a nonzero remainder.
The Sylvester method is not very far from these things in spirit. I think what you have done has well prepared and given you an advantage for what you need to do beyond now, i.e. read at the next chapters in your algebra book which might be about complex numbers (in connection with polynomial equations in particular) or cube roots of unity, or cubic equations etc.
scottdave
Science Advisor
Homework Helper
Great point @epenguin just divide the polynomials and see if there is a remainder, without having to factor.
Gold Member
Thanks everyone. It appears these things aren't quite as simple as I originally thought! But I think I've got a better idea now. So can I expect to see these again down the road, and be introduced to new methods of solving these?
Gold Member
Can I summarize by saying that if have a similar rational, with say $P(x)$ is of degree 6 and $q(x)$ is of degree 5, and neither has obvious factors, and I divide $\frac{P(x)}{q(x)}$ and a remainder results, then $q(x)$ doesn't divide evenly into $P(x)$, and there are no common factors?
StoneTemplePython
Science Advisor
Gold Member
2019 Award
Can I summarize by saying that if have a similar rational, with say $P(x)$ is of degree 6 and $q(x)$ is of degree 5, and neither has obvious factors, and I divide $\frac{P(x)}{q(x)}$ and a remainder results, then $q(x)$ doesn't divide evenly into $P(x)$, and there are no common factors?
Have you tried testing this idea over a simple example using what you know about synthetic division? For example, consider
$\frac{p(x)}{q(x)} = \frac{(x-3)(x-5)(x-6)}{(x-3)(x+3)} = \frac{x^3 - 14x^2 + 63x - 90}{x^2 - 9}$
so doing synthetic division, you can split the division into two steps. What you want is:
$(x^2-9)g(x) = p(x)$
but do the easy part first, i.e. first synthetically divide out $(x-3)$. What do you get? Is their a remainder? Then from that result synthetically divide out $(x+3)$. Now what happens?
- - - -
Coming up with simple examples to test your own ideas on... is an important habit to develop.
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(50g) Euler Transform - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Software Libraries (/forum-10.html) +--- Forum: General Software Library (/forum-13.html) +--- Thread: (50g) Euler Transform (/thread-12610.html) (50g) Euler Transform - John Keith - 03-12-2019 10:17 PM Following are programs for computing the Euler transform and its inverse for sequences of integers. Both require the ListExt Library. The inverse transform also requires Gerald Hillier's MOB program which computes the Moebius Mu function. Euler transform: Code: \<< DUP SIZE R\->I \-> n \<< DUP HEAD SWAP 2 n FOR k k DIVIS DUP2 LPICK * LSUM SWAP NEXT DROP n \->LIST DUP 1. 1. SUB 2 n FOR j OVER 1 j 1 - SUB OVER REV * LSUM PICK3 j GET + j / + NEXT NIP \>> \>> Inverse Euler transform: Code: \<< DUP SIZE R\->I \-> n \<< DUP 1. 1. SUB 2 n FOR j OVER 1 j 1 - SUB OVER REV * LSUM PICK3 j GET j * SWAP - + NEXT NIP 1 n FOR k k DIVIS DUP2 LPICK SWAP REV MOB * LSUM k / SWAP NEXT DROP n \->LIST \>> \>> RE: (50g) Euler Transform - Thomas Klemm - 03-14-2019 07:49 PM Your link to the Binomial transform made me write this implementation of the binomial transform T: Code: \<< { } SWAP WHILE DUP SIZE 1 > REPEAT SWAP OVER HEAD + SWAP \GDLIST NEG END + \>> This can now be used to define a function ΣL to create the partial sum of a list: Code: \<< T 0 SWAP + NEG T \>> And ΔL is just the transformation of TAIL negated: Code: \<< T TAIL NEG T \>> Of course ΔL and the built-in ΔLIST are the same. But since T is an involution we can see that « ΣL ΔL » is the identity: Code: \<< T 0 SWAP + NEG T T TAIL NEG T \>> = Code: \<< T 0 SWAP + NEG TAIL NEG T \>> = Code: \<< T NEG NEG T \>> = Code: \<< T T \>> = Code: \<< \>> We notice that the binomial transform of a polynomial is 0 after a while. E.g. in case of the cubes of the natural numbers we get: [0 1 8 27 64 125 216 343 512 729] T [0 -1 6 -6 0 0 0 0 0 0] So if we want to calculate the partial sum of this list we negate it and add 0 at its head: [0 0 1 -6 6 0 0 0 0 0 0] T [0 0 1 9 36 100 225 441 784 1296 2025] We might try to figure out the pattern of $$T(n^k)$$ for $$k \in \mathbb{N}$$: [1 0 0 0 0 0 0 0 0 0] [0 -1 0 0 0 0 0 0 0 0] [0 -1 2 0 0 0 0 0 0 0] [0 -1 6 -6 0 0 0 0 0 0] [0 -1 14 -36 24 0 0 0 0 0] [0 -1 30 -150 240 -120 0 0 0 0] [0 -1 62 -540 1560 -1800 720 0 0 0] [0 -1 126 -1806 8400 -16800 15120 -5040 0 0] [0 -1 254 -5796 40824 -126000 191520 -141120 40320 0] [0 -1 510 -18150 186480 -834120 1905120 -2328480 1451520 -362880] Or then check the powers of 2: [1 2 4 8 16 32 64 128 256 512] T [1 -1 1 -1 1 -1 1 -1 1 -1] What about Fibonacci? [0 1 1 2 3 5 8 13 21 34 55 89] T [0 -1 -1 -2 -3 -5 -8 -13 -21 -34 -55 -89] What else can you come up with? Cheers Thomas RE: (50g) Euler Transform - John Keith - 03-16-2019 09:12 PM Thanks for another enlightening post, Thomas. Your program T is similar to the second program in my post here. The difference is, of course the negation that happens after the ΔLIST. Your other programs ΣL and ΔL do not return the same results without the negation. My programs are those described in the last paragraph of the "Definitions" section of the Wikipedia page you linked to, which are not self-inverse. I have linked my binomial transform thread to this one as it seems your version would be of interest to anyone reading that thread. RE: (50g) Euler Transform - John Keith - 08-07-2019 11:32 AM I just updated post #1 to fix an erroneous program listing for the inverse transform and to replace both programs with shorter, faster versions. Please delete previous versions if you have them.
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# Similarity-first search: A new algorithm with application to Robinsonian matrix recognition
Monique Laurent, Matteo Seminaroti
Research output: Contribution to journalArticleScientificpeer-review
6 Citations (Scopus)
## Abstract
We present a new efficient combinatorial algorithm for recognizing if a given symmetric matrix is Robinsonian, i.e., if its rows and columns can be simultaneously reordered so that entries are monotone nondecreasing in rows and columns when moving toward the diagonal. As the main ingredient we introduce a new algorithm, named Similarity-First Search (SFS), which extends lexicographic breadth-first search (Lex-BFS) to weighted graphs and which we use in a multisweep algorithm to recognize Robinsonian matrices. Since Robinsonian binary matrices correspond to unit interval graphs, our algorithm can be seen as a generalization to weighted graphs of the 3-sweep Lex-BFS algorithm of Corneil for recognizing unit interval graphs. This new recognition algorithm is extremely simple and it exploits new insight on the combinatorial structure of Robinsonian matrices. For an $n\times n$ nonnegative matrix with $m$ nonzero entries, it terminates in $n-1$ SFS sweeps, with overall running time $O(n^2 +nm\log n)$.
Original language English 1765–1800 SIAM Journal on Discrete Mathematics 31 3 https://doi.org/10.1137/16M1056791 Published - 2017
## Keywords
• Robinson (dis)similarity
• seriation
• similarity search
• Lex-BFS
• LBFS
• partition refinement
## Fingerprint
Dive into the research topics of 'Similarity-first search: A new algorithm with application to Robinsonian matrix recognition'. Together they form a unique fingerprint.
• ### Similarity-First Search: A New Algorithm With Application to Robinsonian Matrix Recognition
Laurent, M. & Seminaroti, M., Jan 2016, Ithaca: Cornell University Library, 35 p. (arXiv; vol. 1601.03521).
Research output: Working paperOther research output
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# gsubfn v0.7
0
0th
Percentile
## Utilities for Strings and Function Arguments
The gsubfn function is like gsub but can take a replacement function or certain other objects instead of the replacement string. Matches and back references are input to the replacement function and replaced by the function output. gsubfn can be used to split strings based on content rather than delimiters and for quasi-perl-style string interpolation. The package also has facilities for translating formulas to functions and allowing such formulas in function calls instead of functions. This can be used with R functions such as apply, sapply, lapply, optim, integrate, xyplot, Filter and any other function that expects another function as an input argument or functions like cat or sql calls that may involve strings where substitution is desirable. There is also a facility for returning multiple objects from functions and a version of transform that allows the RHS to refer to LHS used in the same transform.
## Functions in gsubfn
Name Description transform2 Like transform but allows right hand sides to refer to left hand sides. list Multiple value assignment. gsubfn-package gsubfn match.funfn Generic extended version of R match.fun gsubfn Pattern Matching and Replacement strapply Apply a function over a string or strings. read.pattern Read file or text string using a regular expression to separate fields. as.function.formula Make a one-line function from a formula. fn Transform formula arguments to functions. No Results!
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# Calculate the amount of heat evolved in the complete oxidation of "8.17 g" of "Al" (M_M = "26.98 g/mol") at 25^@"C" and "1 atm" pressure?
## a 101 kJ b 127 kJ c 203 kJ d 237 kJ e 254 kJ DeltaH_f^@("Al"_2"O"_3(s)) = -"1676 kJ/mol" $4 {\text{Al"(s) + 3"O"_2(g) -> 2"Al"_2"O}}_{3} \left(s\right)$
Apr 3, 2018
$\text{254 kJ}$
#### Explanation:
The key here is the standard enthalpy change of formation, $\Delta {H}_{f}^{\circ}$, of aluminium oxide.
$\Delta {H}_{f}^{\circ} = - {\text{1676 kJ mol}}^{- 1}$
Note that the minus sign is used here to illustrate that the heat is being given off.
This value tells you that when $1$ mole of aluminium oxide is formed at standard conditions from its constituent elements in their most stable form, $\text{1676 kJ}$ of heat are being given off.
In other words, you know that standard enthalpy change for this reaction
$2 {\text{Al"_ ((s)) + 3/2 "O"_ (2(g)) -> "Al"_ 2"O}}_{3 \left(s\right)}$
is equal to
$\Delta {H}_{f}^{\circ} = - \text{1676 kJ}$
This implies that this reaction
$\left(\textcolor{red}{2} \cdot 2\right) {\text{Al"_ ((s)) + (color(red)(2) * 3/2) "O"_ (2(g)) -> color(red)(2)"Al"_ 2"O}}_{3 \left(s\right)}$
which can be written as
$4 {\text{Al"_ ((s)) + 3"O"_ (2(g)) -> 2"Al"_ 2"O}}_{3 \left(s\right)}$
will have
$\Delta {H}_{\text{rxn}}^{\circ} = \textcolor{red}{2} \cdot \Delta {H}_{f}^{\circ}$
$\Delta {H}_{\text{rxn"^@ = - "3352 kJ}}$
So, when $2$ moles of aluminium oxide are formed under standard conditions, the reaction gives off $\text{3352 kJ}$ of heat.
Use the molar mass of aluminium to calculate the number of moles present in your sample
8.17 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.30282 moles Al"
According to the balanced chemical equation, the reaction will produce
0.30282 color(red)(cancel(color(black)("moles Al"))) * ("2 moles Al"_2"O"_3)/(4color(red)(cancel(color(black)("moles Al")))) = "0.15141 moles Al"_2"O"_3
You can thus say that when $\text{8.17 g}$ of aluminium undergo complete combustion, the reaction will give off
$0.15141 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Al"_2"O"_3))) * "3352 kJ kJ"/(2 color(red)(cancel(color(black)("moles Al"_2"O"_3)))) = color(darkgreen)(ul(color(black)("254 kJ}}}}$
The answer is rounded to three sig figs, the number of sig figs you have for the mass of aluminium.
In other words, you have
DeltaH_ ("rxn for 8.17 g Al")^@ = - "254 kJ"
Once again, the minus sign is used to show that the heat is being given off.
Notice that you can get the same result by using the standard enthalpy change of formation of aluminium oxide
$0.15141 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Al"_2"O"_3))) * "1676 kJ kJ"/(1 color(red)(cancel(color(black)("mole Al"_2"O"_3)))) = color(darkgreen)(ul(color(black)("254 kJ}}}}$
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# Kinematic problem (initial acceleration) [closed]
A ball is rolling over a soccer field with constant velocity $V_{b} > 0$ at an angle of 45° to the goal line. It starts at the corner of the field (distance $d$ beside the middle of the goal line) at the time $t_0 =0$. At the same time a player starts heading towards the goal on a path perpendicular to the goal line. He starts from a position at distance 2d in front of the middle of the goal line) with $a_p (t) = a_{0}(1-{t\over ß})$
I've solved the problems but the left thing to do is just the initial acceleration. I use integration to find these solutions:
• Distance covered by the ball $s_b = V_b t$
• Time when ball is in front of the middle of the goal line $t_A = {\sqrt2 d\over V_b}$ (with distance $s_A={d\over cos 45°}$)
• Velocity of player $v_p = a_0t(1-{t\over 2ß})$
• Distance covered by the player $s_p = {a_0t^2\over 2} (1-{t\over 3ß})$
• Velocity of the player when he reach the point (the point after the ball reach distance $s_A$) $v_{pA} = {a_{0}d\over v_b} (\sqrt2-{d\over v_bß})$ *I'm inserting $t_A$ into $v_p$, am I right?
• Initial acceleration of the player in order to reach the point (the point after the ball reach distance $s_A$) at the same time as the ball $a_{0}$ = ??
Any ideas would be very appreciated.
edited: sorry I mixed up the calculations.
## closed as unclear what you're asking by Gert, honeste_vivere, ACuriousMind♦, sammy gerbil, knzhouJul 28 '16 at 18:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• $v_p=\int_0^ta_p(t)dt=\int_0^ta_0(1-\frac{t}{\beta})dt=a_0t-a_0\frac{t^2}{2\beta}=a_0t(1-\frac{t}{2\beta})$ – Gert Jul 27 '16 at 17:31
• "I'm trying to derivate $v_p$ over $t$". $a_p=\frac{dv_p}{dt}$. So you already have that. The acceleration at $t=0$ is $a_0$. – Gert Jul 27 '16 at 17:36
• @Gert but how can I define $a_0$? – hello Jul 27 '16 at 18:27
• you can't. It's a given, a known. It's the acceleration at $t=0$, by definition. – Gert Jul 27 '16 at 18:36
• Also:$s_p=\int_0^tv_pdt=\frac12 t^2(1-\frac{1}{3\beta}t)$ – Gert Jul 27 '16 at 18:41
I think some errors crept into your calculation. And at any rate - your confusion is around $a_0$ - which is the quantity you need to solve for (the initial acceleration needed to reach the ball at just the right moment).
The player has to cover a distance $d$ in the time that the ball covers a distance $\sqrt{2} d$. The velocity of the ball is $v_b$, so the mean velocity of the player must be $\frac{v_b}{\sqrt{2}}$.
Given the expression for the acceleration, the instantaneous velocity is
$$v(t) = \int_0^t a_0(1-\frac{\tau}{\beta})d\tau\\ = a_0t(1-\frac{t}{2\beta})$$
The average velocity is the total distance divided by the total time:
$$v_{av} = \frac{\int_0^t v(\tau)d\tau}{t}=\frac12 a_0 t\left(1- \frac{t}{3\beta}\right)$$
Which has to be equal to $\frac{v_b}{\sqrt{2}}$. Finally, we know the time it takes the ball to get to this point:
$$t = \frac{\sqrt{2} d}{v_b}$$
Putting it all together, we find an expression for $a_0$:
\begin{align} \frac{v_b}{\sqrt{2}} &= \frac12 a_0 \frac{\sqrt{2} d}{v_b}\left(1 - \frac{\sqrt{2} d}{3\beta v_b}\right) \\ a_0 &= \frac{v_b^2}{d\left(1 - \frac{\sqrt2 d}{3\beta v_b}\right)}\\ \end{align}
Dimensionally and directionally this makes sense. The faster the ball goes, the greater the required initial acceleration. Also, the more the player tires (small $\beta$), the greater the required initial acceleration. Also, for a sufficiently small value for $\beta < \frac{\sqrt2 d}{3v_b}$ the initial acceleration would have to be negative (the player starts by running away from the goal, and then quickly turns around and sprints to the ball...). And when $\beta$ becomes very large, the problem reduces to the usual constant acceleration, and you can trivially see that you get the expected answer.
Just to confirm this, I ran a simple numerical simulation in Python. Using a large value of beta (and the value of $a_0$ calculated with the equation above) I get the following plot (plotting the distance to the goal line) for $\beta=100$:
And this when I use $\beta=1$ - as predicted, the player would "run away" then turn around and accelerate towards the ball:
In both cases the intersection happens at 25.0 m (I set $d=25~\rm{m}$, arbitrarily) at $t=3.5~\rm{s}$ - just as you would expect for a ball velocity of $10~\rm{m/s}$
Code used:
# football problem
import numpy as np
import matplotlib.pyplot as plt
d = 25
vb = 10
a0 = 1
beta = 1
dt = 0.001
# equation says
a0 = vb*vb/(d*(1-np.sqrt(2)*d/(3*beta*vb)))
def acc(t):
return a0*(1-t/beta)
tv = np.arange(0,4,dt)
v = 0
x=2*d
X=[]
V=[]
for ti,t in enumerate(tv):
vt = v - dt*acc(t)
x = x + (v+vt)/2*dt
v = vt
V.append(v)
X.append(x)
plt.figure()
plt.plot(tv, X)
X2 = tv * vb / np.sqrt(2)
plt.plot(tv, X2)
plt.show()
• Sorry, isn't that should be $v(t) = a_0t(1-{t\over 2ß})$ ? – hello Jul 27 '16 at 21:10
• @hello - yes, good catch. I will have to edit... – Floris Jul 27 '16 at 21:11
• Why do I have to use average velocity (which is my $s_p$ divided by $t$) instead of instantaneous velocity (my $v_p$) ? – hello Jul 28 '16 at 12:58
• Because the average over the interval tells us when you will arrive... – Floris Jul 28 '16 at 12:59
• Okay, that's one thing that I don't know. Thank you so much for your kindly reply. – hello Jul 28 '16 at 13:01
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# How to calculate cubic spline coefficients from end slopes
I want to know how to calculate cubic spline interpolation coefficients, which uses end point slope constraint.
There are $N$ points $(x_0,y_0),(x_1,y_1),\dots,(x_{N-1},y_{N-1}) \in \mathbb{R}^2$ where $x_0 < x_1 < \cdots < x_{N-1}$. Cubic spline interpolation should give $N-1$ polinomials
$$S_j(x) = a_j+b_j(x-x_j)+c_j(x-x_j)^2+d_j(x-x_j)^3$$
where $j \in \{0,1,\dots,N-2\}$. Because there are $4(N-1)$ unknown variables $a_j,b_j,c_j,d_j$, $4(N-1)$ equations are required to solve.
Like natural cubic splines, we usually get the first $4(N-1)-2$ equations from the following conditions
• $S_i(x_i) = S_{i+1}(x_i)=y_i$ for $i \in \{0,1,\dots,N-1\}$
• $S_i(x_{i+1}) = S_{i+1}(x_{i+1})$ for $i \in \{0,1,\dots,N-1\}$
• $S_i^\prime(x_{i+1}) = S^\prime_{i+1}(x_{i+1})$ for $i \in \{0,1,\dots,N-2\}$
• $S_i^{\prime\prime}(x_{i+1}) = S^{\prime\prime}_{i+1}(x_{i+1})$ for $i \in \{0,1,\dots,N-2\}$
and so two other conditions are required. For natural cubic splines, conditions $$S_0^{\prime\prime}(x_0)=0 , S_{N-1}^{\prime\prime}(x_{N-1})=0$$ are used, but instead, I want to use edge slope conditions $$S_0^{\prime}(x_0)=v_{\rm first},S_{N-1}^{\prime}(x_{N-1})=v_{\rm last}$$ for my issue. How can it be solved?
The two conditions $S_0^{\prime}(x_0)=v_{\rm first}$ and $S_{N-1}^{\prime}(x_{N-1})=v_{\rm last}$ give you two more equations. So now you have a total of $4(N-1)$ linear equations for the $4(N-1)$ unknowns, just like in the "natural spline" case. You can solve this system of equations any way you like. The system has some special structure, and you can solve it most efficiently by taking advantage of this structure. But, any linear system solver will work, and the inefficiency probably won't matter unless $N$ is huge (in the thousands).
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## Define Depreciation
### Learning Outcomes
• Define accounting depreciation and differentiate from economic definition
In economics, the term depreciation represents a loss in value, usually due to either wear and tear or obsolescence (in contrast, the term appreciation refers to an increase in value). Usually, this decrease (or increase) is measured by fair market value, which is the price that a willing buyer would pay to a willing seller if both parties are aware of all the relevant facts and circumstances and neither party is under any undue influence.
For example, you buy a brand new car for $50,000 and drive it off the lot. According to bankrate.com, a new car will lose 15–20% of its value in that single moment. So, your car just went down in value by about$10,000. Presumably, because it is now a used car rather than a new car, if you went to trade it in the next day, or you tried to sell it to another individual, you’d only be able to get $40,000 for it. This is a classic example of economic depreciation, and a good way to contrast economic depreciation from accounting depreciation. Accounting depreciation is a systematic and rational method of allocating the cost of an asset to the revenues it indirectly generates. In other words, accounting depreciation is based on the matching principle and is not directly related to a loss in value. For instance, Baker, Co. buys a new delivery truck for$100,000 cash on July 1, and expects it to be of service for five years delivering baked goods to customers around the city, and therefore contributing to revenue indirectly. The cost of the truck ($100,000) is not a period cost for July, nor is it a product cost; so we accountants, in order to match the cost with the revenue (recognizing the expense as incurred), borrow the idea of economic depreciation, but we modify it slightly. First, we capitalize the cost of the truck as an asset. Then, instead of recognizing a$100,000 expense in July for the loss of value (economic depreciation), we recognize the amount of the total cost that is used up in the delivery process. We can only estimate the total cost, but we can make a reasonable estimate. For instance, if we expect the truck to last five years, which is 60 months, we could allocate the cost of $100,000 over 60 months, which would be$1,666.67 per month in depreciation, or \$20,000 per year for five years. The first year we only owned the truck for half a year, so we would only record depreciation expense for half the year.
Year Total Cost Depreciation 20X1 20X2 20X3 20X4 20X5 20X6 100,000 100,000 10,000 20,000 20,000 20,000 20,000 10,000 100,000
This is depreciation in its simplest form, but as you may have realized by now, there are usually factors that complicate things. Even so, if you go back to the matching principle, upon which depreciation is based, you will find that all the variations you are about to learn make sense. In this section, you’ll be learning three different methods of systematically and rationally allocating the cost of an asset over its useful life, as well as how to journalize and track depreciation over time.
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# Is there a reference for "computing $\pi$" using external rays of the Mandelbrot set?
I was recently reminded of the following cute fact which I will state as a proposition to fix notation:
Proposition Given $\epsilon > 0$, let $c = -3/4 + \epsilon i \in \mathbb{C}$ and $q_c(z) = z^2 + c$. Define the sequence of polynomials $q_c^n$ inductively by $q_c^0(z) = z$ and $$q_c^{n+1}(z) = q_c(q_c^n(z)).$$ Since $q_c^n(0) \to \infty$ as $n \to \infty$, $$N(\epsilon) = \min\{n ~:~ |q_c^n(0)| > 2 \}$$ is well-defined. The cute fact is that: $$\epsilon N(\epsilon) \to \pi,$$ as $\epsilon \to 0$.
The next-simplest example of this phenomenon uses $c = 1/4 + \epsilon$. If we define $N(\epsilon)$ similarly then this time it turns out that: $$\sqrt{\epsilon} N(\epsilon) \to \pi,$$ as $\epsilon \to 0$.
The points $-3/4$ and $1/4$ are the "neck" and "butt", respectively, of the Mandelbrot set and there are various other examples of these "$\pi$ paths": they are suitably-parameterised curves simultaneously tangent to two bulbs of the Mandelbrot set where they meet it. In fact I'd guess any part of the boundary where a bulb bubbles off works and so there are infinitely many such paths. External rays seem the obvious paths and they have a natural parameterisation. (Note that although $\epsilon \mapsto -3/4 + \epsilon i$ is not an external ray, it is asymptotic to one, as least set-wise, which is really all that matters.)
If true, I'm sure this is all in the literature (perhaps implicitly) but I don't know the field and after of searching through some lovely papers I couldn't find what I wanted. A naive guess is that something like the following might be true:
Half-serious conjecture Let $M \subset \mathbb{C}$ be the Mandelbrot set, $\overline D \subset \mathbb{C}$ the closed unit disc and $\Phi : \mathbb{C} - M \to \mathbb{C} - \overline D$ the unique conformal isomorphism such that $\Phi(c) \sim c$ as $c \to \infty$. Let $e^{i\theta} \in S^1$ (for $\theta$ rational) and let: $$\alpha : (1, \infty) \to \mathbb{C} - M\\ r \mapsto \Phi^{-1}(re^{i\theta})$$ Let $c = \alpha(r)$ and define $N(r)$ as above, then: $$N(r) \sim \pi / f_\theta(r)$$ for some function $f_\theta$, such that $f_\theta(r) \to 0$ as $r \to 1$ and in particular $f_0(r) \sim \sqrt{\alpha(r)-1/4}$ and $f_{1/3} \sim -i(\alpha(r)+3/4)$.
Here then, at last, is my question:
Question Is some suitably-modified form of the above conjecture correct and is there a proof in the literature toward which somebody could direct me?
Aside from general curiosity my motivation stems from the fact that it's not too hard to prove the stated results for the $c = -3/4 + \epsilon i$ and $c = 1/4 + \epsilon$ paths separately but I'd like to have a unified proof. E.g., to prove the result for the (easier) $c = 1/4 + \epsilon$ path we consider: $$z_{n+1} = z_n^2 + 1/4 + \epsilon\\ \Rightarrow z_{n+1} - z_n = (z_n - 1/2)^2 + \epsilon$$ Then approximate: $$\frac{dz}{dn} \simeq (z-1/2)^2 + \epsilon\\ \Rightarrow z(n) = 1/2 + \sqrt{\epsilon}\tan(\sqrt{\epsilon}n)\\ \Rightarrow \sqrt{\epsilon}N(\epsilon) \simeq \tan^{-1}(\frac{3}{2\sqrt{\epsilon}} ) - \tan^{-1}(-\frac{1}{2\sqrt{\epsilon}})$$ And so provided we can justify the approximation is accurate as $\epsilon \to 0$ (which is tricky but possible) the result follows.
• This doesn't answer your very interesting question (and surely you are already aware of what follows), but a good reference for this $\pi$ phenomenon is the paper " $\pi$ in the Mandelbrot set" by Klebanoff. In the conclusion, Klebanoff conjectures that there are infinitely many routes at each of the infinitely many pinches of the Mandelbrot set which lead to $\pi$ in this way. Aug 19, 2015 at 19:29
• Thank you @MalikYounsi! I was indeed aware of Klebanoff's paper but I had not looked at it in years and had forgotten his concluding remarks about infinitely-many such paths. Indeed I see he even gives a parameterisation of a path to -5/4. Aug 19, 2015 at 20:13
• Thinking about this a tiny bit more, I could believe the right conjecture is $N(r) \sim \pi / |\alpha(r) - \alpha(1)|$ (the square root for $c=1/4$ being absorbed by parameterisation). Note $\alpha(1)$ exists because rational rays land. This should be easy to decide one way or another by numerical calculation, if all else fails. Aug 20, 2015 at 1:05
These results (which are indeed cute - I hadn't seen them before) are well-explained by the theory of parabolic explosion, which is by now classical. Indeed, for the Mandelbrot set, I think that the relevant statements were known to the experts already in the 1980s; they may already be contained implicitly in the Orsay notes by Douady and Hubbard. For an introduction, see Shishikura, Bifurcation of parabolic fixed points, in The Mandelbrot set, Theme and Variations.
Let me focus on the case of "primitive" parabolics, i.e. the "cusps" of Mandelbrot copies, with $c=1/4$ being the simplest case.
Then (see Theorem 3.2.2 of Shishikura's article) - under perturbations as you describe - the number of iterates to escape through the "eggbeater" dynamics into a prescribed part of the basin of infinity is essentially $$N(c) \approx \frac{1}{|\alpha(c)|},$$ where $\lambda(c) = e^{2\pi i \alpha(c)}$ is the multiplier of one of the orbits bifurcating from the parabolic fixed point. (This multiplier is a holomorphic function defined on a double cover around the parabolic; hence the square root in the expression.)
So, for perturbations $c=1/4+\epsilon$, the repelling fixed point has multiplier $$\lambda(c) = 1 + 2\sqrt{-\epsilon}$$ by an elementary and well-known calculation. So $$N(c) \approx \frac{2\pi}{|\log \lambda(c)|}\approx \frac{2\pi}{2\sqrt{\epsilon}}=\frac{\pi}{\sqrt{\epsilon}}.$$
Hence indeed $$N(c)\cdot \sqrt{\epsilon}\to\pi.$$ For other parameters, you will get similar results, which will however depend on the derivative of the multiplier map with respect to the parameter in a suitable sense. Note that the above formulation (in terms of repelling orbits) is, in any case, the natural one, since it is independent of parameterisation.
For the satellite case, there are similar results that you can find in the literature. Again, for the bifurcation at -3/4, you get an explicit formula because you can compute the multipliers of the orbits in question directly.
• Marvellous, thank you. Most satisfying to get an idea of what's going on at last. I look forward to reading Shishikura's article in more detail over lunch. Aug 22, 2015 at 12:10
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# Hamiltonian simulation with complex coefficients
As part of a variational algorithm, I would like to construct a quantum circuit (ideally with pyQuil) that simulates a Hamiltonian of the form:
$$H = 0.3 \cdot Z_3Z_4 + 0.12\cdot Z_1Z_3 + [...] + - 11.03 \cdot Z_3 - 10.92 \cdot Z_4 + \mathbf{0.12i \cdot Z_1 Y_5 X_4}$$
When it comes to the last term, the problem is that pyQuil throws the following error:
TypeError: PauliTerm coefficient must be real
I started diving into the literature and it seems like a non-trivial problem. I came across this paper on universal quantum Hamiltonians where complex-to-real encodings as well as local encodings are discussed. However, it is still not clear to me how one would practically implement something like this. Can anyone give me some practical advice how to solve this problem?
• Does it throw an error when you replace that i with $S_j^2*(X_j S_j X_j)^2$? – AHusain Jul 24 '18 at 17:57
• Remember that a Hamiltonian should be Hermitian. That’s only true of the coefficients are real. – DaftWullie Jul 24 '18 at 18:15
• I might be using a different definition for $S$ than you are. But the point is you can find some combination that results in $i Id_2$. – AHusain Jul 24 '18 at 18:31
• Don't you have another term somewhere in those $\cdots$, that is the Hermitian conjugate? $H = i A B - i B^\dagger A^\dagger$ – AHusain Jul 24 '18 at 21:50
• Or are all the terms of the form such that those cancel out? – AHusain Jul 24 '18 at 21:53
A conventional Hamiltonian is Hermitian. Hence, if it contains a non-Hermitian term, it must either also contain its Hermitian conjuagte as another term, or have 0 weight. In this particular case, since $Z\otimes X\otimes Y$ is Hermitian itself, the coefficient would have to be 0. So, if you're talking about conventional Hamiltonians, you've probably made a mistake in your calculation. Note that if the Hermitian conjugate of the term is not present, you cannot simply fix things by adding it in; it will give you a completely different result.
On the other hand, you might be wanting to implement a non-Hermitian Hamiltonian. These things do exist, often for the description of noise processes, but are not nearly so widespread. You need to explicitly include the "non-Hermitian" terminology, otherwise everyone will just think that what you're doing is wrong because it's not Hermitian, and a Hamiltonian should be Hermitian. I'm not overly familiar with what capabilities the various simulators provide, but I'd be surprised if they have non-Hermiticity built in.
However, you can simulate it, at the cost of non-deterministic implementation. There will be more sophisticated methods than this (see the links in this answer), but let me describe a particularly simply one: I'm going to assume there's only one non-Hermitian component, which is $i\times$(a tensor product of Paulis). I'll call this tensor product of Paulis $K$. The rest of the Hamiltonian is $H$. You want to create the evolution $$e^{-iHt+Kt}$$ We start by Trotterising the evolution, $$e^{-iHt+Kt}= \prod_{i=1}^Ne^{-iH\delta t+K\delta t}$$ where $N\delta t=t$. Now we work on simulating an individual term $e^{-iH\delta t+K\delta t}\approx e^{-iH\delta t}e^{K\delta t}$ (which becomes more accurate at large $N$). You already know how to deal with the Hermitian part so, focus on $$e^{K\delta t}=\cosh(\delta t)\mathbb{I}+\sinh(\delta t)K.$$
We introduce an ancilla qubit in the state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$, and we use this as the control qubit in a controlled-$K$ gate. Then we measure the ancilla in the $\{|\psi\rangle,|\psi^\perp\rangle\}$ basis (where $\langle\psi|\psi^\perp\rangle=0$). If the outcome is $|\psi\rangle$, then on the target qubits we have implemented the operation $|\alpha|^2\mathbb{I}+|\beta|^2K$, up to normalisation. So, if you fix $(1-|\alpha|^2)/|\alpha|^2=\tanh(\delta t)$, you have perfectly implemented that operation. If the measurement fails, then it's up to you whether you want to try to recover (this may well not be possible) or start again.
This simple MATLAB/Octave code shows that $i0.12Z_1Y_2X_3$ is not Hermitian:
z=[1 0 ; 0 -1];
x=[0 1; 1 0];
y=[0 -1i; 1i 0];
z1 = kron(z,eye(4));
y2 = kron(kron(eye(2),y),eye(2));
x3 = kron(eye(4),x);
H=0.12*1i*z1*y2*x3
The output is H:
0 0 0 0.12 0 0 0 0
0 0 0.12 0 0 0 0 0
0 -0.12 0 0 0 0 0 0
-0.12 0 0 0 0 0 0 0
0 0 0 0 0 0 0 -0.12
0 0 0 0 0 0 -0.12 0
0 0 0 0 0 0.12 0 0
0 0 0 0 0.12 0 0 0
Since it's a real matrix, Hermitian means symmetric, but this is not symmetric and therefore not Hermitian. The top-right triangle isn't equal to the bottom-right triangle.
However the top-right triangle is the negative of the bottom-right triangle, so it is anti-Hermitian.
So doing AHussain's suggestion of adding the conjugate transpose, results in 0. Just run this command:
H + H'
and you will get an 8x8 matrix of 0's.
So when you make your Hamiltonian Hermitian by adding the conjugate transpose, you get 0 for this term, and therefore you do not need to have any imaginary coefficients.
• First of all, thanks for the thoughtful reply! However, I'm wondering about the implications of adding the conjugate transpose. The non-hermitian Hamiltonian that I'm facing is the mixer/driver Hamiltonian $H_M$ that I would like to use for QAOA (I generated it based on equations from a theoretical paper). If I know use $H_M + H_M'$ rather than $H_M$, doesn't that fundamentally change my results? – Mark Fingerhuth Jul 26 '18 at 15:49
• That's why @DaftWullie's comment is mistaken without further assumptions. – AHusain Jul 26 '18 at 20:37
• @MarkFingerhuth: Sorry for delay in the replay. I've been extremely busy during the days and have been getting home near midnight every day this month. If you can show me the paper where the equations come from, I can think about how your results get fundamentally different. I may change my answer to say "PyQuil does not support non-Hermitian matrices, but that doesn't mean a different program cannot". – user1271772 Jul 26 '18 at 21:55
• @MarkFingerhuth: you say "I generated it based on equations from a theoretical paper" which equations from which theoretical paper? The paper linked in the question is 82 pages long, can you not just show me which equations you used to generate this "Hamiltonian" ? – user1271772 Aug 2 '18 at 19:36
• @MarkFingerhuth, yes we can talk offline, however I won't get any points there for it. I only got 1 upvote for my effort here, so the incentive is low. – user1271772 Oct 22 '18 at 18:04
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D. Inversion Counting
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
A permutation of size n is an array of size n such that each integer from 1 to n occurs exactly once in this array. An inversion in a permutation p is a pair of indices (i, j) such that i > j and ai < aj. For example, a permutation [4, 1, 3, 2] contains 4 inversions: (2, 1), (3, 1), (4, 1), (4, 3).
You are given a permutation a of size n and m queries to it. Each query is represented by two indices l and r denoting that you have to reverse the segment [l, r] of the permutation. For example, if a = [1, 2, 3, 4] and a query l = 2, r = 4 is applied, then the resulting permutation is [1, 4, 3, 2].
After each query you have to determine whether the number of inversions is odd or even.
Input
The first line contains one integer n (1 ≤ n ≤ 1500) — the size of the permutation.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n) — the elements of the permutation. These integers are pairwise distinct.
The third line contains one integer m (1 ≤ m ≤ 2·105) — the number of queries to process.
Then m lines follow, i-th line containing two integers li, ri (1 ≤ li ≤ ri ≤ n) denoting that i-th query is to reverse a segment [li, ri] of the permutation. All queries are performed one after another.
Output
Print m lines. i-th of them must be equal to odd if the number of inversions in the permutation after i-th query is odd, and even otherwise.
Examples
Input
31 2 321 22 3
Output
oddeven
Input
41 2 4 341 11 41 42 3
Output
oddoddoddeven
Note
The first example:
1. after the first query a = [2, 1, 3], inversion: (2, 1);
2. after the second query a = [2, 3, 1], inversions: (3, 1), (3, 2).
The second example:
1. a = [1, 2, 4, 3], inversion: (4, 3);
2. a = [3, 4, 2, 1], inversions: (3, 1), (4, 1), (3, 2), (4, 2), (4, 3);
3. a = [1, 2, 4, 3], inversion: (4, 3);
4. a = [1, 4, 2, 3], inversions: (3, 2), (4, 2).
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## FANDOM
1,800 Pages
Governor's Residence
Function:
Mitigation of corruption
Requirements:
Expansion ()
Can only be built in a colony
Expansion requirements:
Use requirements:
None
## Building Information
A governor in your colony guarantees that all the daily administrative tasks are done properly and lowers the level of corruption in your colony.
The Governor's Residence costs as much as the Palace and does not require Bureaucracy nor Governor - you can build it on a normal building ground (with a red flag on it) in your colonies - but it does require Expansion as it is needed to be built in your colonies in order to control corruption.
• The Governor's Residence can also be upgraded to a Palace if you ever want to move your capital. However, if you change your capital by converting a Governor's Residence to a Palace, the Palace in your original Capital will be completely destroyed and you will then have to build and expand a Governor's Residence as a replacement for the Palace to remove the corruption that will be there.
• You cannot shorten the build time of the Governor's Residence.
## Expansion Details
The time (in seconds) it takes to upgrade to the next level is determined by the following formula: ${ \text{Building time (seconds)} = \left \lbrack 11,520 \times 1.4 ^ \text{Level} \right \rbrack }$
The accumulative time (in seconds) it takes to upgrade up to the next level is determined by the following formula: ${ \text{Accumulative building time (seconds)} = \left \lbrack 40,320 \times \left (\ 1.4^\text{Level} -\ 1\ \right ) \right \rbrack}$
## Other Standard Buildings
Community content is available under CC-BY-SA unless otherwise noted.
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# CAN bus reverse-engineering
I am struggling with CAN bus reverse-engineering. It might be a silly question, but it irritates me.
These are traffics that arise from pressing buttons A and B which is lifting up the 1. Axis of the robot at the end.
Pressing down button A changes the value "09" in 181h node and button B changes the value "C8" in 281h.
If I am not wrong, I have to feed RPDOs to replicate the lifting up action instead of sending the messages "0A 00 09 00 00 00 00 FF" to 181h and "00 00 00 C8 6F BD 00 FF" to 281h. So basically I resent the messages from RPDOs back to RPDOs.
Is there anything wrong until now? (It must be...otherwise it would have worked.)
As you can see from the above screenshot I manullay sent the messages by pressing each line with the space bar. And it seems like that between my Tx messages lots of Rx messages show up. Is this maybe the reason why the robot shows no reaction?
UPDATE:
The TPDO message "0A 00 09 00 00 00 00 FF" that arises from the pressing down button A gets ignored because whenever I write a TPDO message 0A 00 09 00 00 00 00 FF, it gets overwritten with its default value 0A 00 0A 00 00 00 00 FF so fast that my message is like "not arrived" at all.
The transmission type of TPDO is currently asynchronous with event timer of 50 ms. This leads that my TPDO somehow overwritten with default value 0A 00 0A every 50 ms. How do I handle this problem? I thought this asynchronous transmission type with 50 ms means that the TPDO has to be checked every 50 ms and if there was a change -> transmission. But where does this 0A 00 0A 00 00 00 00 FF default value come from then?
Another question: I thought the whole time that I have to write the RPDOs value back to RPDO to replicate the action. But it seems like that only from pressing down the button A the RPDOs does not change at all. Then how can I replicate the pressing button A action at all?
UPDATE 2: The bit rate is 125 kbit/s. I am using CANopen so that I can control the robot with my computer, instead of using the remote controller. The robot that I am using is Brokk 170. Below you can find an Excel file where the recorded CAN messages are. Those CAN messages arose when I powered up the robot using the robot controller.
I transmitted the messages until the message with number 107, since the value 0A 00 0A 00 00 00 00 FF indicates that the robot is now powered up. But somehow the transmitted sequence does not power up the robot. Now I am trying to find a way to block the messages from the remote controller.
• Can you provide more information about the system (by editing the question)? E.g. bit rate on the CAN bus and how many devices are on the CAN bus. What is the CAN bus used for? - is it only to communicate to/from the robot or are individual joints on the CAN bus? Is there some kind of central controller that is on the CAN bus? Following where the CAN bus cables are connected might give some clue. What is the manufacturer and model name/number of the robot? Oct 8 '18 at 18:08
• Can you provide a full CAN bus log file, over the entire time this is going on (where you are not sending out messages, but operate the robot with the buttons)? An external place is required (say, by a direct URL (public) on Dropbox) due to the size. Later, when we have narrowed down this problem we can post a reduced version in this question). Oct 8 '18 at 18:11
This looks like CANopen traffic (RPDO is also mentioned in the question). 0x80 is SYNC, and it seems it is send at regular intervals (about every 26 ms, 38-39 Hz). And some device responds to the SYNC messages by sending out messages with ID 0x181 and 0x281. But that is just a guess at this point.
It could also be that the content of ID 0x181 and 0x281 are set points to a servo (thus the same device sends out 0x80, 0x181, and 0x281) and that the feedback position is contained in the 0x301 messages.
It should be possible to correlate physical positions of the robot with the messages. A set point is (probably) set immediately and the actual positions are lacking behind.
Note: 181h is not a node, and you are not sending messages to it. 181h is a CAN message ID. As it is likely CANopen, 0x181 is the message "PDO1, transmit" for the device with ID 1. Note that it is not always clear if the device ID indicates what device sends it or if the device is the destination.
Type Function code: Device ID range:
Binary Decimal ID in Hex Decimal
CAN ID
--------------------------------------------------------------------
NMT 0000 0 No 0 - 0 0 - 0
SYNC 0001 1 No 0x80 - 0x80 128 - 128
Emergency 0001 1 Yes 0x81 - 0xFF 129 - 255
Time stamp 0010 2 No 0x100 - 0x100 256 - 256
PDO1, transmit 0011 3 Yes 0x181 - 0x1FF 385 - 511
PDO1, receive 0100 4 Yes 0x201 - 0x27F 513 - 639
PDO2, transmit 0101 5 Yes 0x281 - 0x2FF 641 - 767
PDO2, receive 0110 6 Yes 0x301 - 0x37F 769 - 895
PDO3, transmit 0111 7 Yes 0x381 - 0x3FF 897 - 1023
PDO3, receive 1000 8 Yes 0x401 - 0x47F 1025 - 1151
PDO4, transmit 1001 9 Yes 0x481 - 0x4FF 1153 - 1279
PDO4, receive 1010 10 Yes 0x501 - 0x57F 1281 - 1407
SDO, transmit 1011 11 Yes 0x581 - 0x5FF 1409 - 1535
SDO, receive 1100 12 Yes 0x601 - 0x67F 1537 - 1663
NMT error ctrl 1110 14 Yes 0x701 - 0x77F 1793 - 1919
• "It should be possible to correlate physical positions of the robot with the messages" Yes, this is what I am trying to achieve. Your guess that 0x281 is a set point to a servo sounds reasonable since the value is only changing when I actually try to change the physical position. 0x181 might be something else since it changes its value when I turn up the motor and pressing other buttons which do not correlate with physical position of the robot.
– Joe
Oct 8 '18 at 8:16
• So now back to the question, now I sent the messages in a certain intervall (50ms )over and over again. But the robot is not moving at all and there is this warning that the bus is now busy. It is as if something blocks that the robot is moved by messages. Any idea how to handle, debugg or analyze such a case ?
– Joe
Oct 8 '18 at 8:20
• @Joe: It is not a good idea to have two devices send a CAN message with the same CAN ID. This will result in CAN frame errors. The frequency of these CAN frame errors will depend on a lot of factors, including the bus load. As the CAN frame errors happens during transmit it only takes 16-32 CAN frame errors before both devices go into the bus off state. Some devices very stubbornly refuses to come out of busoff unless power cycled (I think this is misinterpretation of the CAN standard, but that is another story). Oct 8 '18 at 17:41
• I think I found the reason why my message gets ignored all the time. There is a remote controller for the robot. The robot brokk170 that I am using is designed to be controlled with this remote controller. So I have to turn on the robot via the remote controller. And this remote controller was sending his value to the robot. This was the reason why my messages gets overwritten the whole time.
– Joe
Oct 8 '18 at 19:50
• So I see two options here : 1) I find a way that the messages from the remote controller gets ignored by the robot. 2) I turn on the robot using the remote controller and capture the CAN message. For the first option I do not have any clue how to do it (although it would be more "elegant" way than the other). For the second option, I captured the sequence of the messages. Then I typed each CAN message manually and transmitted, but somehow the robot gets not started.
– Joe
Oct 8 '18 at 19:59
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# Testrunner binary not available
Hello,
when trying to run the public tests I encounter the following error.
I’ve already implented all functions and haven’t made any changes to the test files.
I also applied the patch only after completing the first few functions.
This is the output of make check?
Try to delete the ./bin folder and run make followed by make check again.
Do any unusual warnings or errors appear?
No, this is the output of ./run_tests.py, sorry. When running make check, apart from warning of unused variables, it says
and doesn’t run the tests at all.
Deleting bin folder didn’t help.
The python script does not build the project, it only uses the already built files to run the tests. If there are no files, it can not run the tests. If you want to run all tests, use make check.
Regarding the output of make check:
It seems that you use floor from the math library. But we do not link against this library in the Makefile. Therefore, these functions will not be available during execution:
Further note that using floor is likely not necessary.
floor takes a float, and gives you a new float, which is floored (i.e. it computes \lfloor x \rfloor given x).
Since you most likely convert your float to an integer later anyway, this is unnecessary. The code
float f = 3.5; //example value
int i = f;
performs an automatic conversion from float to int, which is defined to truncate the portion after the decimal point. In the above example, i has value 3 afterwards. If you had written i = floor(f);, the floor would be redundant.
PS: If your floats are negative, this does not apply, since floor rounds towards -\infty, while truncation is rounding towards 0. But I guess your floats are all non-negative .
1 Like
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## Thomas' Calculus 13th Edition
Published by Pearson
# Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 26
#### Answer
$$\int_{0}^{\pi} \sqrt{1-\cos ^{2} \theta} \mathrm{d} \theta=2$$
#### Work Step by Step
Given $$\int_{0}^{\pi} \sqrt{1-\cos ^{2} \theta} \mathrm{d} \theta$$ So, we have \begin{aligned} I&=\int_{0}^{\pi} \sqrt{1-\cos ^{2} \theta} \ d\theta\\ &=\int_{0}^{\pi}|\sin \theta| \ d \theta\\ &=\int_{0}^{\pi} \sin \theta \ d \theta\\ &=[-\cos \theta]_{0}^{\pi}\\ &=-\cos \pi+\cos0\\ &=1+1\\ &=2\\ \end{aligned}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Warning someone that they have been suckered into a scam
A friend of mine (let's call her Anne) was recently drawn into attending life-coaching sessions, through an organization which is a scam based on the Lifespring model (tl;dr they give you a psychological high, and then make you pay $to attend training sessions to get another whiff of the high). Anne has already spent thousands of dollars (which she cannot afford to spend) to attend these sessions, and is utterly convinced that they have changed her life for the better. She wants to attend the next "VIP" training session, but has run out of money, and resorted to trying to get other people to pay for it via begging family and friends and a GoFundMe campaign. I just found out about this today, and I want to help her get out, but I know I have to be very careful about how I go about it. She almost certainly won't be receptive to any criticism of the organization or their practices, and the last thing I want is to do something that drives her towards this group. How can I talk to her about my concern (hopefully to get her out of this scam)? In response to comments: I am deliberately obfuscating the details of our relationship, but suffice to say that I am close enough to Anne that she feels comfortable asking me for personal advice, but not so close that I know intimate details of her life. Our communication is primarily in person, secondarily Facebook/texts/etc. She told me about her involvement personally, including asking me for money. In other circumstances, if she was legitimately in trouble and in need of money, I would likely have given it to her, no strings attached. • Is she actually wanting help to get out of this? That would be an important thing considering the approach, or if you should approach at all. Also, we can't really tell you "How you can help her" as that's a sort of "What should I do?"-question which is off-topic here. You should instead consider, what message you want to transport to her. And ask us, how to communicate that to her. As, if she expressively asked you for help we aren't really well equiped for answering how to help her getting out of that. And if she didn't ask for it, you should rather consider what to say than what to do. – dhein Jun 19 at 5:46 • I'm wondering... has she already asked you personally about money, or did you just find the GoFundMe? If it's the first, where she asked you personally, have you already told her the reasons you're not willing to help out? Are you expecting her to come asking at some time, and are you willing to wait until that point to have that as a 'moment', a good point at which you can start getting involved in getting her out? – Tinkeringbell Jun 19 at 8:01 • Could you elaborate on "psychological high"? The article doesn't really make Lifespring out to be a scam (although I'd certainly believe it is one). – scatter Jun 19 at 13:02 • Also, could you describe a bit more about what your specific concerns are? Is it the financial burden (if Anne could easily afford the trainings, would you still be concerned)? Are there negative impacts to her life from these sessions outside of the cost? – Upper_Case Jun 19 at 13:22 • @scatter I haven't attended myself, so I can't comment on the particulars. It is a similar type of manipulation to psychics, faith healers, etc, but working with different emotional states. The scam is legal, incidentally, though there are many civil suits against its practitioners which have out-of-court settlements. – asgallant Jun 19 at 22:22 ## 1 Answer This is a tough one and the blow back from Anne might include her cutting you out of her life. My Story A little over a decade ago I was involved in a religious cult and I completely bought into it hook, line, and sinker. For the first year I showed positive results, impressing my family, friends, and other members in my religious community with my "special zeal" for my religion. In the second year, life took a turn. Everything that was good was soured, everything that was bad was even worse. My parents were constantly upset with me and started to worry about me. About 22 months into my experience, I traveled home and my parents decided to stage an intervention where they listed their complaints. I don't remember the details aside from a generic "You've been uncharacteristically disrespectful." However, what did stand out was my dad telling me that they think I've been brainwashed and that they didn't want me to return to the cult. I dismissed that claim. I felt that it was nearly impossible to be brainwashed and it's only done by sinister people with ill intentions (Nazis, KGB, etc). My cult had nothing but the best of intentions for me, or so I thought. I felt incredulous at their accusations of me and my organization. After our discussion we decided that I should return back to the cult because I was also going to college (on the side). Long story "short" I ended up leaving a month and a half later. On Leaving What got me to leave was a perfect storm. Before the intervention I started noticing the negatives in my life. Most of my support group within the cult left and I was feeling listless. However, everything was pretty subconscious so I didn't put enough thought into my situation. Then a couple of weeks after my parents confronted me, it finally hit me that I was wasting my time. I wasn't growing any, despite the numerous promises that I would. I left at the end of the year. Anne's Part I think that Anne probably needs to have some cynicism, which gets squelched in these kinds of programs (I too participated in a LifeSpring-like convention). For these programs to be considered effective there has to be a suspension of skepticism so that participants "can fully take advantage" of the lessons to be learned. Then, if one session goes well, whether there are breakthroughs or there's that high, then participant can determine that program is right for them. For them the buy-in is that much bigger. It would probably be best to approach Anne when she's "in a valley." When it's been some time since her last involvement and her high has worn off. Approach her and explain your concerns. Remind her that you are her friend and that you support her. Bring into question if these sessions are actually worth the money. One of my mom's criticisms was that my cult was$8k a year and my returns were definitely coming up short.
Perhaps there's been some good, don't throw the baby out with the bathwater as they say. Maybe she can take these breakthroughs and go through therapy, presumably much cheaper. Hopefully she'll come to realize that the appeal for these kinds of things is not the breakthroughs but the highs. Breaking through is just an added bonus.
Things to Remember
You will not convince Anne of anything. She'll be stubborn and set. She'll believe that what she's doing is perfectly healthy and that she's totally in the right. Part of that stems from our refusal to believe that we can be duped into scams/cults/etc. We know that it happens, but we always think it'll never happen to us. The goal here is to put a bug in Anne's ear so that she starts doubting, so that she starts looking at her situation in a different light.
If she's like me she might have some negative thoughts about the organization subconsciously. My parents were pretty much the catalyst to pull me out. You could be too, but the key is to keep things light. Don't come off preachy and let Anne come to the conclusions that she's in trouble herself.
The Aftermath
Like I said in the beginning, Anne might cut you out of her life. She may get so offended at what you have to say that she may deem you as toxic. She has a support group in this organization, as misguided as it may be, and Anne might realize that you don't mean much to her anymore. Outsiders just can't understand how good the organization is and should be ignored.
Try your best to not hold these actions against her. They come from a place of fear and if you, her friend, are scary, then she might feel that she'll have to get rid of what's scary. In the future, when she's wised up, hopefully she'll make amends. Humans are capable of doing crazy things when they drink the koolaid. It makes recovery all the much easier when friends and family are welcoming and supportive when we finally realize our mistakes.
• Thank you for sharing @LuxClaridge; it is helpful to hear the perspective of someone who has been on the other side of a problem like this. You affirmed a lot of the thoughts and concerns churning in my head. – asgallant Jun 19 at 22:40
• The Aftermath is very important. Even though you're seeing the problem, she doesn't. How many doctors treat patients that do not really want to believe they have a problem? If the doctor goes nuts and starts yelling at the patient, it stops mattering how right he is, because the threat is now the messenger, not the initial problem. – Nelson Jun 21 at 1:49
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# Tag Info
15
It probably is a dwarf planet. (It almost certainly is a dwarf planet.) The naming procedures at the IAU are that "Objects that have an absolute magnitude (H) less than +1 [...] are overseen by two naming committees, one for minor planets and one for planets. [...] All other bodies are named by the minor-planet naming committee alone." source—wikipedia ...
6
This has to do with how minor planets obtain provisional names (see also Wikipedia). The year is divided into 24 half months, with a particular letter associated with each one. Each of these half-months has a number of cycles of length 25 depending on how many minor planets are discovered; a minor planet is assigned a letter corresponding to its order in the ...
4
One of the talking points of people how think that the definition of a planet should include Pluto is that during the 2006 vote on the definition of a planet, only a small percentage of the members of the IAU actually voted. This argument is very disingenuous or rather naive, or both. One the most disingenuous of all is Dr. Alan Stern, the principal ...
4
No, only the Hubble law was recommended to have its name changed (I'm a member of the IAU, so hopefully I'd have known if there were more votings). However, several astronomers (including myself) found the voting a bit… weird; while acknowledging the work of George Lemaître is admirable, many more people than him and Edwin Hubble contributed to the ...
4
You use this from https://planetarynames.wr.usgs.gov/FeatureNameRequest. You need to follow the rules as set out in https://planetarynames.wr.usgs.gov/Page/Rules as well as following the themes for naming Moon based features https://planetarynames.wr.usgs.gov/Page/Categories.
3
All star names are unofficial. A few stars have ancient names (such as Sirius) all other stars are referred to by their position in a catalogue or star atlas. As well as I can tell, Al Sadira means "the ostrich", or perhaps in context "The (riverbank) ostrich" indicating a type of tree that grows by rivers (baby name sites offer "lotus tree"). There may be ...
3
"What" is straightforward: L4 and L5 are two points on an orbit that are 60 degrees ahead and 60 degrees behind the planet, and so move as the planet orbits. If the planet is less than about 1/25 of the mass of the central object, then a combination of centrifugal and Coriolis forces will cause the L4 and L5 points to be places where a third body ...
2
To directly answer your question, the other answer is correct: The United States Geologic Survey's Astrobiology office handles ad hoc US submissions to the International Astronomical Union's nomenclature committee, so their form is what you would need to fill out. Nomenclature across the solar system has been set by the IAU to follow themes based on ...
2
In the linked article by A. Bouchard and another by J. Daley, the words "in the journal Lunar and Planetary Science" link not to a journal article but to a poster in a K-12 education session at the 2017 Lunar and Planetary Science Conference. First author Kirby Runyon told Universe Today in 2017 that he would not submit this geophysical definition ...
2
Nomenclature on planetary bodies is meant to ease and standardize communication. If an object is referred to often, or if it is important for someone's research, then the scientist(s) involved can submit a name request to the International Astronomical Union (IAU), or through some other body that submits names to the IAU (such as the United States Geologic ...
1
I asked the at the minor planet center how the codes were decided, and the answer was Historically, the observatory codes were assigned ascending by longitude toward east (from prime meridian): 360 degrees were divided by numbers. When three digit numerical codes were not sufficient, letters plus two numbers were used again in bands toward the east. Some ...
1
The values $$\alpha_0 = 0.00 - 0.641 T$$ $$\delta_0 = 90.00 - 0.557 T$$ provide a first-order estimate of the movement of the right ascension ($\alpha_0$) and declination ($\delta_0$) of the direction of the Earth's north pole for short periods of time after the epoch, expressed to three digits of precision only. $T$ represents the time after epoch ...
1
As an example consider the prime meridian for Vesta. Based on Hubble images, the IAU had established a prime meridian based on the observation of the Obler's regio (a dark region). The prime meridian was defined as passes through the middle of this region. When the Dawn mission arrived at Vesta, it was discovered that the asteroid's pole was not in the ...
Only top voted, non community-wiki answers of a minimum length are eligible
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## Thursday, July 23, 2015
### Solve in positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$: First Solution (Continued)
I want to apology for posting the unfinished blog post on last Saturday (solve-in-positive-integers-equation.html), and I am truly sorry for that.
Here is the other part of the solution that I unwittingly left out.
[MATH]\color{black}\bbox[5px,orange]{(b-3)^4\lt (b-3)^4+120(b-3)\lt ((b-3)^2+1)^2}[/MATH] and determine the range of $b$ which it does not satisfy the set inequality, then we've succeeded half way.
Please bear in mind that $(b-3)^4+120(b-3)$ must be a perfect square!
We know there is no way $(b-3)^4+120(b-3)$ could be a perfect square for any $b$ positive integer values that satisfy the above inequality. Therefore, we need to find out that range of values of $b$ that satisfies the above inequality and then and only then we look for each case for the values of $b$ outside that range, since $(b-3)^4+120(b-3)$ must be a perfect square.
Solve for the LHS inequality, we have:
$(b-3)^4\lt (b-3)^4+120(b-3)$
$\cancel{(b-3)^4}\lt \cancel{(b-3)^4}+120(b-3)$
$3\lt b$
Solve for the RHS inequality, we got:
$(b-3)^4+120(b-3)\lt ((b-3)^2+1)^2$
$(b-3)^4+120(b-3)\lt (b-3)^4+2(b-3)^2+1$
$\cancel{(b-3)^4}+120(b-3)\lt \cancel{(b-3)^4}+2(b-3)^2+1$
$0\lt 2(b-3)^2-120(b-3)+1$
$2b^2-132b+379\gt 0$
Since $b$ is integer, so we obtain $b\le 3$ or $b\ge 63$.
Combining both results from the LHS and RHS ranges we got $3 \le b\gt 63$.
Thus, we need to consider the cases for which $3\le b\le 62$ so to determine all the positive integers the equation $ab(a+b-10)+21a-3a^2+16b-2b^2=60$.
Wow...that seems like there are a total of $63$ (including $0$) cases to look for...but then we know if we continue and checking out all cases from $b=0,\,1,\,2,\cdots,62$, we will reach to the correct answers for certain.
$(3-3)^4+120(3-3)=0$ gives $a=\text{complex solutions}$
$(4-3)^4+120(4-3)=121=11^2$ gives $a=-4,\,7$
$(5-3)^4+120(5-3)=121=16^2$ gives $a=-3,\,5$
$(6-3)^4+120(6-3)=21^2$ gives $a=-3,\,4$
$(7-3)^4+120(7-3)\ne\text{perfect square}$
$(8-3)^4+120(8-3)=35^2$ gives $a=-4,\,3$
$(9-3)^4+120(9-3)\ne\text{perfect square}$
$(10-3)^4+120(10-3)\ne\text{perfect square}$
$(11-3)^4+120(11-3)\ne\text{perfect square}$
$(12-3)^4+120(12-3)\ne\text{perfect square}$
$\,\,\,\,\,\,\,\vdots$
$(62-3)^4+120(62-3)\ne\text{perfect square}$
Therefore, the answers to this problem are:
$(a,\,b)=(3,\,8),\,(4,\,6),\,(5,\,5),\,(7,\,4)$
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# Coin weighing puzzle
We have 5 boxes with 20 coins each. And coins of three boxes weigh 10, one of them weighs 9 and the other weighs 11. How can we differentiate between them by using a scale(digital, not the one with two cups) only once, with 1 unit accuracy?
• Weigh them? shrugs This puzzle needs more restrictions to be meaningful. – mr23ceec Dec 11 '17 at 16:14
• @mr23ceec "using a scale only once" – Apep Dec 11 '17 at 16:15
• Is it a balancing scale (two cups and one side is equal to the other) or a spring scale (one cup and it gives you a numerical answer, usually in a certain range) – mr23ceec Dec 11 '17 at 16:23
• One cup digital scale – Nemexia Dec 11 '17 at 17:11
• each coin weigh 10 or whole box weigh 10? – Oray Dec 11 '17 at 18:07
This is pretty straight forward question. Just choose $0,1,3,7,12$ and this is also the minimum number of coins you can choose as well. You will get different $20$ results except the boxes with coins with the same weight coin.
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# What is the remainder when 1!+2!+3!+4!+5!+…+50! is divided by 5!?
What is the remainder when 1!+2!+3!+4!+5!+.......+50! is divided by 5!
My Approach
$1$+$2$+$6$+$24$+$5$!/$5$!+$6 . 5$!/$5$!+$7$ .$6$ . $5$!/$5$!....so on
$33$+$1$+$6$+$42$+......
I am not getting the correct answer as the solution is getting complex.
Can anyone guide me how to approach the problem?
• You asked almost the same question here...Using the same tools leads to a very similar solution... – user37238 Nov 16 '15 at 9:34
Hint: Terms of $5!$ onwards are divisble by $5!$, so you only need the remainer of $1! +2!+3!+4!$.
As Lee said $5!$ onwards all are divisible by $5!$ so we need remainder of($1!+2!+3!+4!$) 33 so remainder is 33.
• @ArchisWelankar: Would you care to repeat that comment in plain English? FYI $5!=120$. – Marc van Leeuwen Nov 16 '15 at 11:13
• Cant you write $33=6.5+3$ thats what i want to say let the sum be x .so $x+30+3$ whats the remainder??. Hope now you are clear on that. – Archis Welankar Nov 16 '15 at 11:19
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# Is “indeterminate” a better name than “indifferent” for neutral fixed-points?
let $f(t)=t$ be an analytic function which has a fixed-point at $t$. The multiplier $\lambda (t) = f'(t)$ is just the 1st derivative evaluated at the fixed-point. The standard nomenclature is that when $\| \lambda (t) \|=1$ the fixed-point is said to be indifferent, or neutral. However, it does not necessarily mean that trajectories can't be attracted or repelled by this point, only that its 1st derivative equals 1.
Wouldn't "indeterminate" be a better name than "indifferent" ?
The function $(1 / ( 1 - \| \lambda (t) \| ) )$ is ill-defined when $\lambda(t) = 1$
Perhaps a matter of opinion, but "indeterminate" should mean that its type cannot be determined, while "indifferent" should mean that its type can be any. Right?
From this point of view, although it is only language, it seems better to use "indifferent" since clearly the type of a fixed point will definitely be well defined, for each specific dynamics.
On the other hand, it also common to use "parabolic" (and in my opinion much better) in opposition to "hyperbolic". The latter (usually) means that $|f'(t)|\ne1$.
• I agree with your interpretation of indeterminate vs indifferent. I think that parabolic is typically reserved for the case where $f'(t)$ is a root of unity, though, which is a stronger assumption that $|f'(t)|=1$. As such, parabolic points are definitely not indeterminate, as the Leau-Fatou flower theorem describes their dynamics quite precisely. – Mark McClure Mar 4 '17 at 23:32
• I see, now that you put it that way, indifferent does make more sense. since if one is indifferent, they could go either way. indeterminate would imply it cant be determined as you say. on an indifferent fixed-point the trajectories at that point are only changing by at most a constant independent of t. at a superattractive point it isnt changing by anything since the derivative is exactly 0 – crow Mar 5 '17 at 0:09
• i understand the case of indifferent fixed-points is much more complicated than attractive or repulsive . – crow Mar 5 '17 at 0:12
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