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# Centering different sizes of images in one column of a table vertically and horizontally in each cell?
The table consists of lines of cells of text and a cell of image. The image in each line is within one column. But the images are of different sizes. Intends to center each image horizontally and vertically in its cell. How to write such code?
The code is as follows:
\begin{table*}[ht]
\footnotesize
\caption{Some table}
\begin{center}
\label{sometable}
\begin{tabularx}{7.4in}{ p{2cm} | p{1.8cm} | p{1.8cm} | p{3.5cm} | p{3.5cm} | p{3.5cm} }
\hline
\textbf{Image} & \textbf{Detected} & \textbf{Indirectly Related} & \textbf{Sentences Generated by Model with Indirectly Related} & \textbf{Sentences Generated by Model without Indirectly Related} & \textbf{Standard Model} \\
\raisebox{-1.0\height}{\includegraphics[width=0.8in]{some image}} & Some text & Some text & Some text & Some text \\ \hline
\end{tabularx}
\end{center}
\end{table*}
• Welcome to TeX SX! Could you post a minimal example code of what you've tried. How are defined the cells in text columns? Jun 11, 2018 at 0:42
• To vertically center an image, use \raisebox{-0.5\height}{...}. In a tabular, use c for the image and p{...} for the text. Jun 11, 2018 at 0:53
• @Bernard I have added my code. Jun 11, 2018 at 2:39
• Why do you use tabularx if you do specify each column width?
– NBur
Jun 11, 2018 at 6:53
Your question is a bit unclear. You should provide a complete minimal working example (MWE), because, even with \footnotesize your table is larger than an ordinary article text width.
If you want all your cell vertically and horizontally aligned, you could use m{...} column type with a >{\centering\arraybackslash} before it. I created a new column type M for convenience.
Don't use center environment inside a figure one, use \centering instead, see here: Should I use center or centering for figures and tables?
tabularx is useless if you don't have an X column in your table definition.
Just to help to improve your question, rather than really answer, I tried to correct your code and I suggest an alternative one with booktabs and threeparttable.
\documentclass{article}
\usepackage{array}
\newcolumntype{M}[1]{>{\centering\arraybackslash}m{#1}}
\renewcommand{\arraystretch}{1.2}
\usepackage{booktabs}
\usepackage{threeparttable}
\usepackage{graphicx}
\usepackage{caption}
\begin{document}
\begin{table*}[ht]
\centering\footnotesize
\caption{Some table\label{sometable}}
\begin{tabular}{M{2.1cm}|M{1.4cm}|M{1.4cm}|M{1.4cm}|M{1.4cm}|M{1.4cm}}
\hline
\textbf{Image} & \textbf{Detected} & \textbf{Indirectly Related} & \textbf{Sentences Generated by Model with Indirectly Related} & \textbf{Sentences Generated by Model without Indirectly Related} & \textbf{Standard Model} \\
\includegraphics[width=0.8in]{example-image} & Some text & Some text & Some text & Some text \\
\includegraphics[width=0.5in]{example-image-a} & Another image & with another dimension & Some text & Some text \\
\includegraphics[width=0.7in]{example-image-b} & Another image again & with another dimension & Some text & Some text \\
\hline
\end{tabular}
\end{table*}
\begin{table*}[ht]
\centering\footnotesize
\caption{Some table with attempt at improvement\label{mytable}}
\begin{threeparttable}
\begin{tabular}{M{2.1cm}M{1.4cm}M{1.4cm}M{1.4cm}M{1.4cm}M{1.4cm}}
\toprule
\textbf{Image} & \textbf{Detected} & \textbf{Indirectly Related} & \textbf{With IR}\tnote{a} & \textbf{Without IR}\tnote{b} & \textbf{Standard Model} \\
\midrule
\includegraphics[width=0.8in]{example-image} & Some text & Some text & Some text & Some text \\
\includegraphics[width=0.5in]{example-image-a} & Another image & with another dimension & Some text & Some text \\
\includegraphics[width=0.7in]{example-image-b} & Another image again & with another dimension & Some text & Some text \\
\bottomrule
\end{tabular}
\begin{tablenotes}
\item[a] Sentences Generated by Model \emph{with} Indirectly Related.
\item[b] Sentences Generated by Model \emph{without} Indirectly Related.
\end{tablenotes}
\end{threeparttable}
\end{table*}
\end{document}
With your code, with some corrections:
My suggestion:
• Thank you so much! I have used 'c' for the first column and use \raisebox to manually adjust it vertically. Jun 11, 2018 at 20:49
A solution with \ tabularx and X columns of different widths:
\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{array}
\newcolumntype{M}[1]{>{\centering\arraybackslash}m{#1}}
\usepackage{booktabs}
\usepackage{tabularx}
\usepackage{caption, makecell}
\usepackage{ragged2e}
\begin{document}
\begin{table*}[ht]
\centering
\renewcommand{\tabularxcolumn}[1]{>{\footnotesize\RaggedRight\arraybackslash}m{#1}}
\setlength{\tabcolsep}{3pt}
\caption{Some table}\label{mytable}
\begin{tabularx}{\linewidth}{c*{2}{>{\hsize=0.625\hsize}X}*{3}{>{\hsize=1.25\hsize}X}}
\toprule
& & & \multicolumn{2}{c}{\thead{Sentences Generated by Model}} \\
\cmidrule{4-5}
\midrule
\adjincludegraphics[width=0.8in, valign=c]{example-image} & Some text & Some text & Some text & Some text \\
• You can, but valign=c and redefining the X column type is simpler – unless you want the text columns to be top-aligned. Jun 11, 2018 at 21:04
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{}
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True False 1.3) A graph on n vertices with n - 1 must be a tree. There should be at least one edge for every vertex in the graph. In the following graph, vertices 'e' and 'c' are the cut vertices. A connected graph 'G' may have at most (n–2) cut vertices. A graph G is said to be connected if there exists a path between every pair of vertices. The maximum number of simple graphs with n = 3 vertices − 2 n C 2 = 2 n(n-1)/2 = 2 3(3-1)/2 = 2 3 = 8. (d) a cubic graph with 11 vertices. Example. Find the number of regions in G. Solution- Given-Number of vertices (v) = 20; Degree of each vertex (d) = 3 . If G … Hence it is a disconnected graph with cut vertex as 'e'. In the above graph, removing the vertices ‘e’ and ‘i’ makes the graph disconnected. The minimum number of vertices whose removal makes ‘G’ either disconnected or reduces ‘G’ in to a trivial graph is called its vertex connectivity. (5 points, 1 point for each) True/False Questions 1.1) In a simple graph on n vertices, the degree of a vertex is at most n - 1. 1 1. Now we have a cycle, which is a simple graph, so we can stop and say 3 3 3 3 2 is a simple graph. Give an example (if it exists) of each of the following: (a) a simple bipartite graph that is regular of degree 5. There are exactly six simple connected graphs with only four vertices. 4 3 2 1 For Kn, there will be n vertices and (n(n-1))/2 edges. Please come to o–ce hours if you have any questions about this proof. In a graph theory a tree is uncorrected graph in which any two vertices one connected by exactly one path. IF it is a simple, connected graph, then for the set of vertices {v: v exists in V}, v is adjacent to every other vertex in V. This type of graph is denoted Kn. These 8 graphs are as shown below − Connected Graph. (b) a bipartite Platonic graph. True False 1.4) Every graph has a … (e) a simple graph (other than K 5, K 4,4 or Q 4) that is regular of degree 4. By removing 'e' or 'c', the graph will become a disconnected graph. Theorem 1.1. Let G be a connected planar simple graph with 20 vertices and degree of each vertex is 3. In this example, the given undirected graph has one connected component: Let’s name this graph .Here denotes the vertex set and denotes the edge set of .The graph has one connected component, let’s name it , which contains all the vertices of .Now let’s check whether the set holds to the definition or not.. (c) 4 4 3 2 1. 10. 1 Connected simple graphs on four vertices Here we brie°y answer Exercise 3.3 of the previous notes. They are … a) 1,2,3 b) 2,3,4 c) 2,4,5 d) 1,3,5 View Answer. Question 1. (c) a complete graph that is a wheel. Notation − K(G) Example. Calculating Total Number Of Edges (e)- By sum of degrees of vertices theorem, we have- Sum of degrees of all the vertices = 2 x Total number of edges 0 0 <- everything is a 0 after going through the full Havel-Hakimi algo, so yes, 3 3 3 3 2 is a simple graph. True False 1.2) A complete graph on 5 vertices has 20 edges. Let ‘G’ be a connected graph. advertisement. Explanation: A simple graph maybe connected or disconnected. Or keep going: 2 2 2. Tree: A connected graph which does not have a circuit or cycle is called a tree. What is the maximum number of edges in a bipartite graph having 10 vertices? 1 1 2. Since there are 5 vertices, $V_1, V_2 V_3 V_4 V_5 \therefore m= 5$ Number of edges = $\frac {m(m-1)}{2} = \frac {5(5-1)}{2} = 10$ ii. 2 2 2 2 <- step 5, subtract 1 from the left 3 degrees. Example: Binding Tree a) 24 b) 21 c) 25 d) 16 ... For which of the following combinations of the degrees of vertices would the connected graph be eulerian? 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# Issue with Cantor's Diagonal Argument and square matrices
I have an issue with Cantor's Diagonal Argument.
We suppose we have an infinite list with every real in $(0, 1)$ listed. Then we take the diagonal and change every digit in a predefined way (e.g. $+1 \pmod {10}$). And we get a new real number, not included in the list. This is supposed to prove that reals are not countable.
The problem I see is that, to apply the argument, we must have every real in a square matrix of digits (I use the matrix just as a form of representation). And we can't list every real with a square matrix. For a fixed precision, we need a matrix of $N$ columns (digits) by ($10^N$) rows (for any $N$). I understand the Cantor list would have infinite precision, but we still need $N\times(10^n)$, and we are still construing a list of $N\times N$ digits ($N$ being infinite).
So, it seems to me that the Cantor's Diagonal Argument only proves that we can't list all reals in a square matrix.
What am I missing?
• I have never thought about the argument as a matrix. It's just an infinite list of numbers indexed with an infinite number of indices. – Randall Aug 24 '17 at 17:49
• The "matrix" in the diagonal argument has a countably infinite number of rows and a countably infinite number of columns. There is one row for each real number, and the columns are the digits in the decimal expansion (which may have an eventually zero tail). There is no finite matrix in sight. – MPW Aug 24 '17 at 17:58
• @MPW I understand that the matrix is infinite (I have updated the question to show this), but I fail to see how this makes a difference. – raven Aug 24 '17 at 18:12
• The assumption that the reals in (0,1) are countable essentially is the assumption that you can store the reals as rows in a matrix (with a countable infinity of both rows and columns) of digits. You are correct that this is impossible. Your hand-waving about square matrices and precision doesn't show that it is impossible. Cantor's diagonal argument does show that this is impossible, hence the reals are not countable. – John Coleman Aug 24 '17 at 18:12
• Hmmm... Thank you everyone. I think I need to think a bit about this all. – raven Aug 24 '17 at 18:19
I would say that you are missing nothing. The problem is that you are adding one thing, namely a matrix. That's not part of the diagonal argument.
• I don't think I'm adding anything. There is a list of real numbers. Each real number is a list of digits. So we have a two-dimensional list of digits (call it matrix, table, list of list, or whatever you will). To apply the diagonal, it must be square (NxN digits). – raven Aug 24 '17 at 17:58
• @raven : There is no $N$. The diagonal is infinitely long. – MPW Aug 24 '17 at 17:59
• OK, call it a matrix, but it's an $|\aleph_0| \times |\aleph_0|$ matrix. – Randall Aug 24 '17 at 18:09
The idea of a "list" here being a matrix is taking the diagonal argument too literally. The actual argument does not involve doing anything to a physical list of numbers (which you describe as a matrix). It instead involves showing that there cannot exist any one-to-one correspondence (or bijection) between the reals and the natural numbers. The diagonal argument is a way of visualizing the proof, but the underlying nature of the argument has nothing to do with any list of fixed, finite size. These are infinite lists (technically, infinite sequences), and the ideas of finite precision do not apply to them.
For instance, what does it even mean to have a "square" matrix, when your matrix is of infinite size? In the finite sense we understand it (if a matrix is $m\times m$ where $m$ is a natural number, it is square), but we don't have a rigorous method for talking about matrix dimensions when the matrices are infinite. Sure, one could define what that means, but it is neither necessary nor helpful in formulating the diagonal argument.
We often use the visual aide of an infinite list of reals, each real with infinite precision, but that is not necessary for the actual rigorous proof. Here is the proof without that visual aide (glossing over many details).
Suppose the reals are countable. Then there exists a one-to-one correspondence (bijection) between the reals and the natural numbers. Thus we must be able to make an infinite sequence (with a first, second, third, millionth, etc, number) of real numbers in which each real appears at least once. Consider one such sequence $\{s_n\}$. Now we are going to construct a new number which does not appear $\{s_n\}$. We take the first digit of $s_1$ and change it to $s_1+1$ modulo 10, and make that our first digit of the new number. We do the same for the second, third, and all other digits. We now have a number not appearing in the sequence $\{s_n\}$ we had before, because it differs from every term of the sequence in at least one decimal place. It must be a real number, in fact, since it is defined as and represented by an infinite decimal. Thus, our initial sequence did not actually have all the reals appear at least once, so we have a contradiction. Hence, the reals are uncountable.
Note that the argument never makes use of a matrix in any way. The fundamental idea of the proof lies in the use of infinite sequences, which we do have a rigorous grasp on. If you'd like more understanding of the details by which we construct ideas like sequences, it's turtles all the way down in that it comes back to set theory. However, I hope this gave the gist of the proof in a more understandable way, and showed that it does not in any way rely on a physical representation (such as a matrix) of the infinite sequence being dealt with.
(This got much longer than I expected, cheers!)
You don't need a column for each value of digit, just for each digit.
And you don't construct a list of length $\Bbb{N\times N}$. That is just the matrix itself, all the real numbers and their decimal expansions. We start with a list of length $\Bbb N$, and each one of these has a decimal expansion of length $\Bbb N$, and so each row in the matrix is actually just this decimal expansion of one of the real numbers on the list.
Next, we go over the diagonal of this matrix, i.e. nodes indexed by $(n,n)$, and use the digits that appear on that diagonal to generate a new real number which you can now prove is not any row (or column for that matter) of this matrix.
And if each row is a real number on your original list, then it means your original list is not the entire interval.
• A short response that explains the 'whole deal' while addressing OP's issues. Also, the clarification "(or column for that matter)" is helpful. +1 – CopyPasteIt Aug 25 '17 at 15:37
• Sorry, but this doesn't address my issue. I have always written about a matrix of DIGITS. You need one column for each digit, yes (i.e. N), but you need one row for each possible value of each digit (10^N; for one digit you need 10 rows (10 real numbers, 0.0, 0.1, ...0.8, 0.9 assuming we are working in [0, 1]). That was precisely my argument: with a square matrix you can't represent all the numbers, and to take every digit in a diagonal you need a square matrix. It seems that is the infinity of the "matrix" that makes it work (that was my intuition, but is hard to think about it). – raven Aug 25 '17 at 16:08
• This definitely addresses your issue, since you seem to simply not understand what the idea behind "using a matrix" is in this context. Of course that a square matrix represents all the numbers on your list, if your first number is $0.123123123\ldots$, then the first row would have $1$ in the first column, $2$ in the second, $3$ in the third, $1$ in the fourth, and so on. The matrix is not binary. – Asaf Karagila Aug 25 '17 at 16:13
• I understand it perfectly (I'd say it's obvious). Please, read carefully my question again. The issue is that, to list all real numbers with N digits, you need 10^N entries in the list. But to extract a new number with all digits changed, you need one digit per entrie, i.e. a square matrix (of DIGITS, yeah). So it's obvious in a square matrix you won't have all real numbers with N digits. I don't even know why you talk about binary matrices. – raven Aug 25 '17 at 16:24
• (1) You're using "N" like it is a finite integer, but it's not, and the rules of infinite arithmetic can change, especially when talking about limit cardinals like $\aleph_0$. (2) Secondly, don't think about it as "list all the reals", but rather think of it as taking an arbitrary list of real numbers in [0,1], indexed by the natural numbers, and proving that this list is not the entire interval. – Asaf Karagila Aug 25 '17 at 17:18
It is best not to think of matrices but to use the precise definition of countability.
A set $S$ is countable if there is a bijective function
$$f: \mathbb{N} \rightarrow S$$
A function is a bijection if it is 1-1 and surjective.
The proof of uncountability of the real numbers relies on the use of decimal expansions, and in general there may be more than one decimal expansion for a real number. In the following I will therefore assume that we have chosen a unique decimal representation for each real number (it does not matter which one). Suppose we had such a bijective function
$$f: \mathbb{N} \rightarrow \mathbb{R}$$
But then we could define a real number $d$ that is not in the range of $f$ by defining its decimal expansion in binary as follows. Let $x_i$ be the $i$th digit in the decimal expansion of real number $i$. Then the $i$th digit of $d$ is given by
$$d_i = \begin{cases} 0 & \text{if}\; r_i = 1 \; \text{where}\; r_i = f(i) \\ 1 & \text{if}\; r_i = 0 \; \text{where}\; r_i = f(i) \end{cases}$$
However, $d$ is by definition different from every real number for at least one decimal, so $d$ cannot be in the range of $f$. Therefore $f$ cannot be onto.
The "matrix analogy" is simply used to represent the function $f$ as an infinite table using an entry to hold each decimal for any given real number in the range of $f$; entry $(i,j)$ in the table contains the $j$th decimal of $f(r_i)$.
It should be pointed out that something implicit is happening in this Cantor Diagonal Construction argument (see MathTrain's answer). Recall,
A decimal representation of a non-negative real number $r$ is an expression in the form of a series, traditionally written as a sum ${\displaystyle r=\sum _{i=0}^{\infty }{\frac {a_{i}}{10^{i}}}}$ where $a_0$ is a nonnegative integer, and $a_1, a_2, ...$ are integers satisfying $0 \le a_i \le 9$, called the digits of the decimal representation.
The representation is unique up to the fact that if the ${a_i}^{'}s$ trail off in $9^{'}$s it can be simplified.
This representation is really a statement on the limit of a sequence (convergence),
$\tag 1 r=\lim _{n\to \infty }\sum _{i=0}^{n}{\frac {a_{i}}{10^{i}}}$
The Cantor argument is constructing the $a_i$ for (1) one step at a time, so that for any $N$, none of the $N + 1$ finite expansion sums
$\tag 2 s_n = \sum _{i=0}^{n}{\frac {a_{i}}{10^{i}}} \; n \le N$
can be equal to any of the first $N+1$ 'matrix rows', and that also holds true as the limits are taken.
So there are assumptions being made during the Cantor construction, the underlying facts about decimal representations of real numbers.
Finally, although it can be handled, that little detail about trailing $9^{'}$s needs to be addressed in a rigorous proof. You might want to skip that and study a Cantor proof showing that the binary sequences, $2^{\mathbb{N}}$, is uncountable (and more).
If you look at related material in math.stackexchange, for some inexplicable reason, many people have questions about Cantor's Diagonal Argument. So, as this is the most recent instance, permit me the liberty of simply giving another proof that the real numbers are uncountable.
Recall the following property of real numbers,
Theorem 1: Let $(s_n)$ be a sequence of real numbers satisfying the following conditions,
$\quad s_0 \lt s_2 \lt s_4 \lt \cdots$
$\quad s_1 \gt s_3 \gt s_5 \gt \cdots$
$\quad \text{For every even } i \in \mathbb{N}\text{, } \; s_i \lt s_{i+1}$
$\quad \text{The alternating sequence } s_{k+1} - s_k \text{ converges to } 0$
Then $(s_n)$ converges to the limit $s= \text{LUB} \, \{s_i \, |\; i \text{ is even} \} = \text{GLB} \, \{s_i \, |\; i \text{ is odd} \}$,
and for every $n \in \mathbb{N}$, $s_n \ne s$.
Theorem 2: The set of all real numbers is an uncountable set.
Proof
To arrive at a contradiction, let $f: \mathbb{N} \to \mathbb{R}$ be a surjective mapping. We claim that we can construct a sequence $(s_n)$ satisfying the conditions of theorem 1 with a limit that can't be in the range of $f$, giving the contradiction.
Set $s_0 = f(0)$.
Let $m_1$ be the smallest integer satisfying $f(m_1) \gt s_0$, and set $s_1 = f(m_1)$.
Let $m_2$ be the smallest integer satisfying $f(m_2) \gt s_0$ and $f(m_2) \lt s_1$, and set $s_2 = f(m_2)$.
Let $m_3$ be the smallest integer satisfying $f(m_3) \lt s_1$ and $f(m_3) \gt s_2$, and set $s_3 = f(m_3)$.
etc.
Note that the increasing sequence $(m_j)$ of integers must be unbounded.
It is not difficult to show that $(s_n)$ satisfies the conditions of theorem 1; let $\alpha$ denote this limit. Observe that $f$ can't take on the value of $\alpha$, for if it did, $\alpha$ would have to appear as a term in the constructed $(s_n)$ sequence, since it is inside the alternating $(s_n$) 'clamp-down', and we keep going further out along the domain of $f$ to construct this converging sequence..
$\blacksquare$
Note: In the proof of theorem 2 we only used the fact that the range of $f$ is dense in $\mathbb{R}$, and from that constructed a number $\alpha$ that $f$ can't assume as an output value.
• I know there is a lot of people thinking that they are smarter than Cantor and can prove him wrong. I also know better. I know R is uncountable (I "get" it, and I'm pretty confident I could prove it in an at least almost formal way). I think what I miss is some basic understanding of infinity. As I said in another comment, I'll think about this for a few ways and then come back (with a better understanding, I hope). Thank you for the effort, BTW. – raven Aug 25 '17 at 16:18
• Indeed, this really started with me trying to prove wrong one of those guys. – raven Aug 25 '17 at 16:28
"We suppose we have an infinite list with every real in (0,1) listed. ... The problem I see is that, to apply the argument, we must have every real in a square matrix of digits." This is a common objection, that is a result of how the proof is taught incorrectly.
First, Cantor did not apply the proof to real numbers; intentionally and specifically. He applied it to infinite-length strings of the two characters "m" and "w." This is a minor point, since we can (and I will) use "0" and "1" instead. Note that such a string is just a function that maps N to the set {'0','1'}, written f:N->{'0','1'}.
Yes, every real number between 0 and 1 can be represented in binary by such a string. But some can be represented by two, which is a problem. It is easily resolved, but unnecessary to do so if we follow what Cantor actually said.
Second, he never assumed there was an infinite list of all such strings, only that there was an infinite list of some such strings. Objections like yours naturally arise when you try to imagine how to physically accomplish what Cantor claimed proved was impossible. You can't, so you quite naturally try to find the flaw.
Lastly, it isn't a proof by contradiction. To be one, you have to show the opposite of a known truth follows from all of what you assume. Not that one thing you assume contradicts something else you assume. So it is not a valid proof-by-contradiction to assume "all" and then derive "not all" from another assumption.
But it is still a valid proof. What Cantor assumed was that there is a set T comprising all of his strings, and that a subset S of it could be generated by the function s:N->S. The diagonalization method is another function d:N->{0,1}, that maps every n to the opposite of nth character of s(n). There is no square matrix here, just a definition of one of Cantor's strings as a function.
What this directly proves, is "If S is a countable subset of T, then S is not all of T." By contraposition ("If A then B" implies "If ~B then ~A"), not contradiction, this proves "If S is all of T, then S is not countable."
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# How to calculate the maximum number of molecules or ions from a given amount of substance?
Which of the following solutions has the greatest number of particles (atoms or ions) of solute per liter?
(a) $1~\mathrm{M}\ \ce{NaCl}$
(b) $1~\mathrm{M}\ \ce{CaCl2}$
(c) $1~\mathrm{M}$ ethanol
(d) $1~\mathrm{M}$ acetic acid
The correct answer is (b), but I dont know why. I would have thought it was ethanol ($\ce{C2H6O}$) because it has the greatest number of moles of atoms, and shouldn't that mean it has the greatest number of atoms? I noticed answer (b) has the highest molar mass ($110.984~\mathrm{g/mol}$), but does that mean it also has the highest number of atoms?
• The question is poorly worded. Apparently, they wanted the number of "ions or molecules", so that ethanol molecule counts as 1, but $\ce{NaCl}$ dissociates into ions and hence counts as 2, and so on. – Ivan Neretin Jan 25 '16 at 18:30
The way this question is asked, (c) would indeed be correct. As you correctly noted, ethanol has nine atoms per molecule, and a $1~\mathrm{M}$ solution would thus contain $9~\mathrm{mol}$ of atoms per litre. For comparison: acetic acid has eight, calcium chloride three and sodium chloride two.
However, they probably meant molecules in the question and not atoms — because it really makes no chemical sense to separate ethanol molecules into their atoms (they won’t separate in a solution anyway). So with that reasoning, counting ‘particles’ (whatever they may be) we arrive at $1~\mathrm{mol}$ of ethanol molecules for (c), $2~\mathrm{mol}$ of ions for (a), $3~\mathrm{mol}$ of ions for (b) and under two moles of ions/molecules (combining both of them) for (d) (all per litre of solution).
So if that was an exam, complain and get your points acknowledged because of a badly-worded question.
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# The Quadratic Reciprocity Law: A Collection of Classical Proofs
###### Oswald Baumgart, edited and translated by Franz Lemmermeyer
Publisher:
Birkhäuser
Publication Date:
2015
Number of Pages:
172
Format:
Hardcover
Price:
89.99
ISBN:
9783319162829
Category:
Monograph
BLL Rating:
The Basic Library List Committee suggests that undergraduate mathematics libraries consider this book for acquisition.
[Reviewed by
Fernando Q. Gouvêa
, on
06/30/2015
]
It’s often said that proofs serve as the criterion for truth in mathematics: we prove things in order to establish that they are true. This is certainly true, but it doesn’t explain something else we do, namely, provide new proofs of old results. We already know those theorems are true, so in giving new proofs we are not seeking to establish that. What we are seeking is understanding. We want to know why the theorem is true, and a proof can (sometimes) tell us that. We also want to know about connections between different parts of mathematics, which can lead to new insights and perhaps new theorems.
A case in point is the quadratic reciprocity theorem. First conjectured by Euler and Legendre and first proved by Gauss, it is a staple of elementary number theory courses. It relates the answers to two yes-no questions about two (distinct) odd prime numbers $p$ and $q$:
(1) Does there exist a whole number $n$ such that $n^2\equiv q \pmod p$?
(2) Does there exist a whole number $m$ such that $m^2\equiv p \pmod q$?
There is, a priori, no reason to expect these two questions to be related. In fact, the Chinese Remainder Theorem strongly suggests that “life mod $p$” and “life mod $q$” are completely independent, since it tells us that for any choice of $a$ and $b$ one can always find $x$ such that both $x\equiv a \pmod p$ and $x\equiv b\pmod q$. Nevertheless, questions (1) and (2) do turn out to be related:
Quadratic Reciprocity Theorem: If either $p\equiv 1\pmod 4$ or $q\equiv 1\pmod 4$, then questions (1) and (2) have the same answer. On the other hand, if $p\equiv q\equiv 3\pmod 4$, questions (1) and (2) have opposite answers.
In particular, if we know the answer to one of the questions, then we know the answer to the other.
The proof that is usually given in elementary courses goes like this: relate the answers to the questions to counting, in such a way that an even count means the answer is “yes” and an odd count means the answer is “no.” (This is known as “Gauss’s Lemma.”) Then set up a way to relate the two counts, and show that the difference between the answers is odd or even as required by the theorem.
The proof works, but it is remarkable in the fact that it gives us no insight at all into why the theorem is true. In particular, it does not yield any direct connection between “life mod $p$” and “life mod $q$.” Every time I present the proof to students, I point out the feeling that yes, it comes out right, but it comes out right because the theorem is true. It’s hard to claim (and I do not believe) that counting points in a rectangle explains why the theorem is true.
Gauss felt the same about his original proof, so he ended giving six different ones (only 3 and 6, in the numbering given in this book, used “Gauss’s Lemma”). It’s not clear that this satisfied him, and it certainly didn’t satisfy others: this book counts 314 different proofs since Gauss’s time. Of course, many of these are quite similar to others, and a few are incorrect, but that’s still an impressive number.
The book is a translation by Franz Lemmermeyer of Baumgart’s thesis from 1885. An appendix by Lemmermeyer updates the story by giving an annotated list of all known proofs of the theorem. (That’s how I know there are 314.) Both Baumgart and Lemmermeyer classify proofs according to the main tools used, and Lemmermeyer, in his appendix, tries to trace which proofs are dependent on which. So this is much more than a mere catalogue of proofs!
The editor has provided double service: he offers English-speakers access to Baumgart’s account and provides a summary of what has happened since then. The result is a very useful book.
Fernando Q. Gouvêa is Carter Professor of Mathematics at Colby College in Waterville, ME.
See the table of contents on Springer's web page for the book.
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# resistors in the right place?
Discussion in 'General Electronics Chat' started by scytzoh, Nov 8, 2015.
1. ### scytzoh Thread Starter New Member
Nov 8, 2015
14
0
i have posted a really bad drawing showing how i am thinking to solder together a very simple led array (simple for a majority besides me) the color code on the resistors is not correct..lol....im pretty sure this is how i have it done but the problem im haveing is the leds only flicker when i turn on the power then remain off
2. ### djsfantasi AAC Fanatic!
Apr 11, 2010
3,382
1,168
Can you post a picture? What are the ratings for the LEDs? Forward voltage and milliamperes required? And what is the power supply? Volts? As a point of information, you only need one resistor in this layoutr. And how is your soldering skills.
3. ### scytzoh Thread Starter New Member
Nov 8, 2015
14
0
the led calculator i found gave 2 options both with 2 resistors (see attachment)
the only thing is neither looks like the pattern that the leds must be in ( its a computer fan) forward voltage is 3.3 i think 12volt from computr
4. ### Veracohr Well-Known Member
Jan 3, 2011
583
80
Computer fan? How does that relate to this LED string?
No attachment.
5. ### djsfantasi AAC Fanatic!
Apr 11, 2010
3,382
1,168
The circuit that you gave only requires one resistor. And if the forward voltage is 3.3v, then a 12v supply is not enough (4x3.3v=13.2v). There are other ways to connect the LEDs
Last edited: Nov 9, 2015
6. ### djsfantasi AAC Fanatic!
Apr 11, 2010
3,382
1,168
Here is another way to connect the LEDs, that will run on 12V. Note that this circuit does require two resistors.
I calculated the resistor value for you. First, I assumed that the LEDs operate at 20ma. The formula below will explain the process of finding the resistor value, and help you find out the resistor value in this circuit if my assumption is wrong.
$R = \frac{(Vs-Vdiodes)}{Idiodes}$ "Vdiodes" is the same as "Vforward"
$R = \frac{(12V-(2*3.3V))}{.020A}$ two LEDs at 3.3Vf each
$R = \frac{5.4V}{.020A}$
$R=270\Omega$
7. ### scytzoh Thread Starter New Member
Nov 8, 2015
14
0
the fan is where im installing the leds into this is the attach ment
8. ### djsfantasi AAC Fanatic!
Apr 11, 2010
3,382
1,168
Note that solution 2 is almost exactly what I presented in my schematic. The difference is the location of the resistors. In this simple circuit, it does not make a difference (Note that if you're going to be working in electronics, you will need experience in reading schematics.) The major difference between solution 2 and your diagram, is where you connect the + and - of your power supply. As you drew it, the voltage is passing through all four resistors, dropping a little (3.3V exactly) each time. If you hook series in parallel, voltage drops less, but the current requirements of the entire circuit, go up. To match the solution, you needed to join the other ends of the two resistors together, turn around two of the LEDs, and finally make a connection in the middle of the four LEDs...
Like this:
9. ### ian field AAC Fanatic!
Oct 27, 2012
5,097
907
The sum of the Vf of each LED in series must be less than the voltage available.
If that sum is too close to the available voltage, current limiting in the resistor will be rather "stiff" - that is to say small variations in voltage will affect the current more.
If that sum is small compared to that voltage; you'll need a higher resistance to set the current - voltage fluctuations will affect the current value less, but more energy is wasted in the resistor.
10. ### scytzoh Thread Starter New Member
Nov 8, 2015
14
0
...............................i notice that un my attachment all the leds are conected - to + but in yours 2 are and 2 are not and the resistore is in different place so how is that almost exactly?? just curious not trying to slander......so your correction to my awesome painting is the way you recomend that i solder them up then??
11. ### djsfantasi AAC Fanatic!
Apr 11, 2010
3,382
1,168
I don't see, in both my schematic and the modified awesome painting, where the LEDs are not connected - to+. I just want to understand what you are seeing in the pictures. When I modified your painting, I reversed two of the LEDs, so that they are arranged as in your painting BUT still connect - to +. Here is another modified painting, with the LEDs arranged in two columns of two each, that I hope will illustrate this point further.
As far as the location of the resistor, I thought I had explained that in this circuit, the resistor can be placed on the + side of the series LEDs or the - side of the series LEDs. It is there to maintain the proper current to the LEDs and can do so from either side.
Pay close attention to the polarity of the LED legs. My assumption was where you painted a leg red, it is the positive lead; where you painted the leg black, it is the negative lead.
As far as your last question, wire the LEDs as shown, paying attention to polarity, and they should work.
12. ### scytzoh Thread Starter New Member
Nov 8, 2015
14
0
ok well i did as above and when i turn on the computer they go on for 1\4 or hald a second then off.. there not blown because it does it every time i turn computer off then on... wrong resistors?? im sure they are the ones i had asked for when i used the led calculater i got a big strip of them for use with 12 volt 4 led i dont have random rolls of resistors around
13. ### markdem Member
Jul 31, 2013
104
58
Do you have any way of measuring the voltage on the power supply output after the LEDs go off? I had a look at the post, but can't see what power supply you are using but keep referring to a fan and a PC. Is this to say you are trying to use the fan output in your computer as the power supply? If so, is it just that the PC is not sensing a fan connected and turning the output off, or simply lowering the speed via PWM?
I am thinking there is something else going on here.
If you are using your computer to supply the voltage, maybe try a different power supply. A old 12v charger from something works very well.
djsfantasi likes this.
14. ### ian field AAC Fanatic!
Oct 27, 2012
5,097
907
Or just pinch a Molex connector off a scrap drive and get the power from one of those leads.
Even if you still use a FDD, there's usually a connector left over.
15. ### scytzoh Thread Starter New Member
Nov 8, 2015
14
0
OK i went bought new everything and im posting another awesome painting please tell me if this is right ..thank you for patience
16. ### djsfantasi AAC Fanatic!
Apr 11, 2010
3,382
1,168
Other than why are there two awesome paintings, It looks correct.
17. ### scytzoh Thread Starter New Member
Nov 8, 2015
14
0
ok sucsess! the only thing now the wires that went to the fan originally must be underpowerd or thermally controlled because they are really really dim but when put to another source it works perfect
18. ### AlphaOmega New Member
Apr 30, 2012
4
1
If you are trying to power the LEDs from the fan supply, note that most fans driven from the motherboard will be controlled via a PWM supply to enable the speed to be managed.
Try taking a supply from a 12V connector
Edit:
The apparent effect of changing the PWM modulation is to increased or decrease the voltage between about 0 and 12V and thus the fan speed.
(note that some fans are 5V)
Some motherboards allow you to turn off the variable speed, but your PC will be noisy!
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### CBSE Assignments class 09 Mathematics
Mathematics Assignments & Worksheets For Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th to 12 th standard. Here students can find very useful content which is very helpful to handle final examinations effectively. For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments. These assignments cover all the topics and are strictly according to the CBSE syllabus. With the help of these assignments students can easily achieve the examination level and can reach at the maximum height. Class 09 Mathematics Assignment Case Study Based Questions Class IX
### Properties of the circle
Properties of the circle
Circle properties and formulas, properties of circle class 10, properties of circle class 9, easy definition of circle, types of circle, tangent, secant to the circle, cyclic quadrilateral and its properties.
In this chapter we will discuss the following:
Definition of Circle, Radius, Diameter, Chord, Segment, Sector, Cyclic Quadrilateral, their Definitions, Properties of Circle, Properties of Cyclic Quadrilateral with complete explanations.
CIRCLE:-
Collection of all points in a plane which are at the equidistant from the fixed point, is called a circle.
The fixed point is called centre and fixed distance is called radius.
A line segment which join the centre with any point on the circle is called radius.
DIAMETER:-
A line segment which join any two points on the circle and is passes through the centre of the circle is called the diameter of the circle.
Diameter is the double of radius.
Longest chord of the circle is called its diameter.
CHORD:-
Any line segment which join any two points on the circle is called its chord.
Diameter of the circle is also the chord of the circle.
SECTORS:-
Area between the radius and arc is called sector of the circle.
Sectors are of two types: Minor sector and Major Sector
Minor Sector : Smaller sector is called minor sector.
Major Sector : Larger area is called major sector.
SEGMENT :-
Area between the chord and arc is called segment.
Segments are of two types : Minor Segment and Major Segment.
Minor Segment : Smaller area is called minor segment.
Major Segment : Larger area is called major segment.
PROPERTIES OF CIRCLE:-
• Equal chords of the circle subtend equal angles at the centre of the circle.
• If two chords subtend equal angles at the centre of the circle then the chords are equal.
If AB = CD then ㄥAOB = ㄥCOD
• There is only one circle passes through the three non-collinear point.
-
• Two circles can intersect each other at the most two points.
• Perpendicular from the centre to the chord bisect the chord.
• If OM 丄 AB, then AM = BM
If AM = BM, then, OM 丄 AB
• A line segment joining centre of the circle with the mid - point of the chord is perpendicular to the chord.
• Angle made by an arc at the centre is double the angle made by the same arc in the remaining part of the circle.
ㄥBOC = 2ㄥBAC
• Angles in the same segment of the circle are equal.
ㄥC = ㄥD = ㄥE = ㄥF
A quadrilateral whose all vertices lie on the circle is called a cyclic quadrilateral.
Opposite angles of a cyclic quadrilateral are supplementary (180 degree).
ㄥA + ㄥC = 180o , ㄥB + ㄥD = 180o
Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
ㄥCBE = ㄥD
TANGENTS TO A CIRCLE
Tangent Line
If a line touch the circle at one point then it is called tangent to the circle.
Secant Line :
If a line intersect the circle at two points then the line is called section.
PROPERTIES OF TANGENT TO THE CIRCLE
Tangent is always perpendicular to the radius of the circle.
Length of the tangents from the exterior points to the circle are equal.
Important Result
Relation between Area and Perimeter of a triangle with the radius of incircle
Incircle :- It is the circle inscribed inside the triangle whose centre is at equidistant from the sides of the triangle.
$\frac{Area\; of\; triangle}{semi\; \; perimeter\; of\; the\; triangle}=Radius\; of\; the\; in circle$
Algorithm to solve such problems
1) Find the perimeter of the triangle and then semi-perimeter.
2) Find area of the triangle. If it is an isosceles triangle then use Heron's Formula.
3) Use the above formula to find the required result.
PROOF OF IMPORTANT THEOREMS ON THE CIRCLE
THEOREM 10.1 CHAPTER 10 CLASS 10
Statement :
Tangent is always perpendicular to the radius at the point of contact.
Given: In circle C(o,r), XY is tangent to the circle at point P.
To Prove: OP ⊥ XY
Construction :
Take any arbitrary point Q (other than P ) on the line XY and join OQ which meet the circle at point R.
Proof:
In order to prove that OP ⊥ XY it is sufficient to prove that OP is the smallest line segment than all the line segments obtained by joining O with any point on XY.
OP = OR .......... (Equal radii)
Now OQ = OR + RQ
OQ = OP + RQ
Subtract RQ from the R.H.D. we get
OQ > OP or
OP < OQ
But Q is an arbitrary point on XY
OP is the smallest line segment and smallest line segment is always perpendicular.
Hence OP XY
Hence prove the required theorem
THEOREM 10.2 CHAPTER 10 CLASS 10
Statement
Prove that length of tangents from external point to the circle are equal in length.
Given : AC and BC are two tangents from external point to the circle.
To Prove : AC = BC
Construction: Join OA, OB and OC
Proof:
Since radius is always perpendicular to the tangent.
1 = 2 = 90o
In ΔAOC and ΔBOC
OA = OB ............... (Equal radii)
OC = OC .............. (common side)
1 = 2 .............. (Each = 90o
By RHS rule
ΔAOC
ΔBOC
AC = BC .............By CPCT
Hence prove the required result
Results of other important theorems on circle
Angle made by the chord of the circle with the tangent of the same circle (at the point of contact of the chord and tangent) is equal to the angle made by the chord in the alternate segment of the circle.
If AB and CD are two chords of the circle intersect each other at point p inside or outside the circle then:-
PA X PB = PC X PD
Inside the circle Outside the circle
If chord AB of a circle intersect the tangent of the same circle at point P outside the circle then
PA X PB = PT2
🙏
CIRCLE - CBSE Mathematics
1. Thanks for the information sir... it provides a lot of knowledge
2. Thanks for wriiting
3. Thank you sir! Its very helful
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# Resistor ratings for inductor fly-back suppression
I am building a circuit for user to interface with relays, and I am trying to build a protection to the circuit against fly-back from the inductor coil of the relay. Since there is no specific relay it is quite hard to choose the parts. All that is certain is the relay must be 5V and draws up to max of 250 mA (my recommended limit), but a protection for higher amp rating would be better.
The protection in my circuit is a diode - resistor combo. with the diode being 1n4007 which is greatly way more than what I need (which is good). the resistor is a bit tricky and what I am not sure about.
Assuming from what I know about inductors, is that it is a current source, and the voltage will depend on the resistor.
Here is my computation
V = I × R
5 = 0.25A (largest 5V coil current I found) × X ohms
X = 5/0.25
X = 20 ohms
20 ohms is for my absolute max, but for lower current loads it could go up to 50 ohms + (assuming 5V is our desired voltage). I don't know which is a better safe side value — is higher the better or the opposite? My guess is lower, but how low a resistor that generates 1 volt?
On a similar note, the power rating required for this resistor:
P = I × V
P = 0.25A × 5V
P = 1.25 watts
1.25 watt resistor are hard to find on resistor values that low. Will a 1/4 or 1/2 watt resistor work since the current is not continuous and it goes down?
You are getting muddled.
It's not the 5 volts that is used to power the relay that needs to be factored in when designing a resistive flyback suppressor, it's the maximum voltage that your driving circuit can withstand under flyback conditions that has to be considered.
For instance, consider the situation of using an NPN (or N channel) transistor to activate a relay. When the relay is deactivated, the inductive 250 mA would produce a peak voltage of 25 volts in a 100 ohm flyback resistor. This means the collector (or drain) of the transistor would see a peak voltage of 25 volts plus 5 volts (30 volts).
Therefore, if the transistor has a voltage rating greater than 40 volts (a little added margin) then you would be good to go.
Regarding the power rating you don't need to pick a resistor that has a rating equivalent to the instantaneous value of the power surge; resistors are usually very good at handling largish short power surges so, if the flyback surge is done in less than 10 ms (fairly normal) then pretty much any quarter or half watt resistor will do the job.
• Just, clarifying; im using a N channel MOSFET(BSS138) and the specs i should look for is the Gate-Source Voltage which is ±20 V in my case. So using a 20 ohm resistor is sufficient because it is way below that rating ( 20 > 5 + 5 ) Dec 4 '19 at 23:50
• It's the source-drain voltage that is the one taxed by the flyback voltage. Given that this is rated at a max of 50 volts, the resistor value can be chosen to limit the voltage to (say) 5 + 35 volts flyback i.e. a resistor value of 140 ohms with a current of 250 mA. Dec 5 '19 at 8:07
• is there an advantage for going the highest possible resistor for the current to quickly dissipate, or is a lower resister is preferred? Dec 5 '19 at 21:07
• Higher value resistor means the energy stored in the relay’s coil is dissipated more quickly hence it turns off more quickly. Dec 5 '19 at 22:14
• thank you sir, your answer really helped me understand Dec 6 '19 at 0:54
First decide what you want your diode/resistor combination to do.
If it's to protect the driving transistor against the high voltage flyback, then zero resistance is ideal, and the flyback will be clamped to <1 volt above the rail. If you know the maximum collector voltage rating of your drive transistor, then you can choose a resistance to allow up to, but not exceeding, that voltage to be generated.
If it's to turn the relay current off quickly, then use as large a resistor as you can, to get as high a voltage as you can for rapid current fall, subject to not destroying your transistor by overvoltage. A zener diode might make a better clamp here, as the voltage is then more or less independent of current.
If you do want to use a resistor, then the power rating is unimportant for most purposes, as it's the coil energy that gets dumped into the resistor once per relay operation. Only very frequent relay operation will tax the resistor power dissipation. Unfortunately, few low power resistors have an energy pulse spec. If it's a hobby project, then just use 1/4 watt, and feel it after several relay operations. If you're selling these things in the 1000s, then you might want to get a pulse energy spec from the resistor manufacturers, or you might not.
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Let $K$ be a fixed field in $\mathbb C$ of an automorphism of $\mathbb C$. Prove that every finite extension of $K$ in $\mathbb C$ is cyclic.
Let $K$ be a fixed field in $\mathbb C$ (complex numbers) of an automorphism of $\mathbb C$. Prove that every finite extension of $K$ in $\mathbb C$ is cyclic.
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Do u mean field fixed by an automorphism of C? – Dinesh May 3 '11 at 6:39
Let $\sigma$ be the automorphism. Given an extension $L$ of $K$, first show that a power of $\sigma$ acts as identity on $L$. Now you have the following problem: let $L$ be a field, $G$ a finite subgroup of $\mathrm{Aut}(L)$. If $K$ is the subfield of $L$ fixed by $G$, then $L/K$ is Galois with Galois group $G$. – Jiangwei Xue May 3 '11 at 6:42
That's exactly what I mean. – Danny May 3 '11 at 6:42
Thank you Jiangwei! I will try that. – Danny May 3 '11 at 6:43
I might be confused about something but surely K is either $\mathbb R$ or $\mathbb C$? (So the group is either trivial or $C_2$) – quanta May 3 '11 at 7:33
As Jyrki Lahtonen kindly pointed out in a comment below, there was a gap in the previous version of this answer. After having read Jyrki's comment, I also noticed that an answer, which (unlike mine) was correct, had been given by Jiangwei Xue in comment of May $3$, $2011$. I'll write Jiangwei Xue's solution below, and turn this answer into a community wiki.
Recall the following.
Let $G$ be a finite group of automorphisms of a field $L$, and let $K$ be the fixed field. Then $L/K$ is Galois with Galois group $G$.
If $G$ is abelian, any sub-extension $S/K$ of $L/K$ is Galois with Galois group a quotient of $G$. Thus $S/K$ is abelian.
If $G$ is cyclic, then so is any sub-extension $S/K$ of $L/K$.
Now here is Jiangwei Xue's argument.
Let $\sigma$ be an automorphism of a field $M$, let $K$ be the fixed field, and let $L/K$ be a finite degree sub-extension of $M/K$. Then the restriction of $\sigma$ to $M$ generates a finite group $G$ of automorphisms of $M$ whose fixed field is $K$, and the above observations show that $L/K$ is cyclic.
Aside. If $G$ be a finite group of automorphisms of an algebraically closed field $L$, then $G$ has order $1$ or $2$. Here are three references for this:
$\bullet$ These notes of our friend Pete L. Clark on Field Theory (see Theorem $124$ p. $75$, called "Grand Artin-Schreier Theorem"),
$\bullet$ The sub-entry Definitions of the entry Real closed field in Wikipedia,
$\bullet$ This nLab entry.
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All of this is correct, but the group generated by a single automorphism of $\mathbf{C}$ is infinite more often than not, right? – Jyrki Lahtonen Jan 14 '12 at 14:28
Dear @Jyrki: Thank you very much! We, MSE users, are very lucky to have people like you around! I hope the new version is correct. (I don't know why, I figured that the automorphism was supposed to be of finite order.) – Pierre-Yves Gaillard Jan 14 '12 at 15:40
You are too kind (you had my +1) already. My infinite Galois theory is too rusty for me to reliably answer a question like. You deserve the credit, so "Credit Waived", was IMHO somewhat overkill, but kudos to you! – Jyrki Lahtonen Jan 15 '12 at 9:58
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The data values of standard electrode potentials (E°) are given in the table below, in volts relative to the standard hydrogen electrode, and are for the following conditions: . The word equations for the reaction between copper and oxygen is copper+oxygen= copper oxide. At the same time, copper oxide is reduced because oxygen is removed from it. I believe you will find that Cu2O is more thermodynamicaly stable than copper(II) oxide despite the prevalence of other copper(II) compounds. Word equations are often easier to write than picture equations or chemical equations and so they are a good starting point when we want to write reactions. Copper metal is stable in air under normal conditions. You know it's finished when the copper changes from a pinkish brownish golden to blue-black. The ⦠4 Cu (s) + O 2 (g) 2 Cu 2 O (s) Reaction of copper with ammonia. Copper and oxygen will react at room temperature to make Cu2O, known as copper(I) oxide. Copper reacts with oxygen to form copper oxide. Copper oxidation, on the other hand, prevents further oxygen exposure and corrosion by solidly adhering to the metal's surface. The balanced equation will appear above. The experiment has to have heat or it won't happen. The ionic equation for the magnesium-aided reduction of hot copper(II) oxide to elemental copper is given below : $Cu^{2+} + Mg \rightarrow Cu + Mg^{2+}\nonumber$ The equation can be split into two parts and considered from the separate perspectives of the elemental magnesium and of the copper⦠This copper oxide from reaction 2 is the main culprit that will later form the colors of the patina. The copper dioxide then reacts with more oxygen to form copper oxide (Equation 2). When the funnel is removed, the copper turns black again. Examples: Fe, Au, Co, Br, C, O, N, F. Ionic charges are not yet supported and will be ignored. To balance a chemical equation, enter an equation of a chemical reaction and press the Balance button. Reaction of copper with water. It is found in nature as a common mineral called cuprite. Effects of Oxidation on Copper One positive effect of copper oxidation includes the formation of a protective outer layer that prevents further corrosion. An effective concentration of 1 mol/L for each aqueous species or a species in a mercury amalgam (an alloy of mercury with another metal). In this reaction, carbon is oxidised because it gains oxygen. At read heat, copper metal and oxygen react to form Cu 2 O. If sulfur is present on the surface of the copper, then the two can react to form copper sulfide, which is black (Equation ⦠Reaction of copper with air. A temperature of 298.15 K (25.00 °C; 77.00 °F). Oxidation is the gain of oxygen. Oxygen is an oxidizer and a reactant in combustion. Copper(II) forms a hexaqua complex with water. Copper(I)oxide, also known as cuprous oxide, is ⦠When heated until red hot, copper metal and oxygen react to form Cu 2 O. The copper oxide can then react with the hydrogen gas to form the copper metal and water. Use uppercase for the first character in the element and lowercase for the second character. The two copper oxides that can form are ionic compounds. 4Cu(s) + O 2 (g) â 2Cu 2 O(s) Reaction of copper with water Reaction of copper with the halogens. Explanations (including important chemical equations): 2 Cu (s) + O 2 (g) ---> 2 CuO (s) CuO (s) + H 2 (g) ---> Cu (s) + H 2 O (g) Heated copper metal reacts with oxygen to form the black copper oxide. Write the word equation for the reaction between iron and oxygen and for the reaction between magnesium and oxygen. Metal is stable in air under normal conditions enter an equation of a protective outer that... Hydrogen gas to form copper oxide ( equation 2 ) equation, enter an equation of a chemical equation enter. 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Help with these complex analysis series problems?
1. Oct 4, 2011
I don't even know where to start or go with this first problem.
A) Assume that 'a sub n' E C and consider rearrangements of the convergent series the 'sum of 'a sub n' from n=1 to infinity'. Show that each of the following situations is possible and that this list includes all possibilities.
1) The sum of a sub n from n=1 to infinity converages absolutely and hence all rearrangements converge to the same value.
2) The set of possible values of convergent rearrangements is all of C.
3) The set of possible values of convergent rearrangements is an arbitrary line in C.
B) Show that the sum of [ (-1)^(n+1) / n ] from n=1 to infinity converges conditionally. The value of the sum is log(2), approcimately .69. Show that by grouping the terms by taking two positive terms, then one negative term, then two positive terms, then one negative term, and so on, the series adds up to a number larger than 1. The value of the sum therefore depends on the order in which the terms are summed.
This is all that I could come up with:
The abs value of the series diverges because it's the harmonic series where p=1.
The series itself converges by the alternating series test because the nth term converges to zero and the series is decreasing. Therefore the given series converges conditionally.
Now for the rearranging:
1 + 1/3 -1/2 + 1/5 + 1/7 -1/4 and so forth. I then just listed the partial sums.
s1= 1 + 1/3 -1/2= 5/6
s2= s1 + 1/5 + 1/7 -1/4= 0.926
s3= s2 + 1/13 + 1/15 - 1/8= 0.98013
and so on...
I know that this isn't right. I don't know how I'm supposed to show that the sum of the rearranged series is greater than 1.
C) Express the rearranged series *found above in B* in the form 'the sum of b sub n from n=1 to infinity', and find a simple expression for b sub n. Try to find the infinite sum.
After a lot of erasing I came up with:
1 + 1/3 -1/2 + 1/5 + 1/7 -1/4 and so forth can be expressed as
the sum of 1/ (4k-3) + 1/(4k-1) - 1/(2k) from k=1 to infinity. I don't know if this is correct for 'b sub n' form or how to find the infinite sum of this rearranged series.
D) Assume that 'a sub n' E R(real) and that the sum of 'a sub n' from n=1 to infinity converged conditionally. Fix an arbitrary L(limit) E R(real). Show that we can rearrange the terms to make the sum converge to L.
I don't know where to start or where to go with this one, too. Do I treat 'a sub n' as an alternating series that converges conditionally?
I know that you aren't here to solve it for me, but any help you can give me would be great. Thank you very much.
Please, let me know if I need to clarify anything.
Last edited: Oct 4, 2011
2. Oct 4, 2011
kru_
For part A, unless you were given a specific sequence an to look at, the problem is asking you to use the definition of a convergent series to show that any convergent series can be rearranged to converge to any value you want. I think this is similar to the http://mathworld.wolfram.com/RiemannSeriesTheorem.html" [Broken].
For part B, you are on the right track. You'll need to calculate out to s7 to show it is > 1.
For C, try turning your new formula into a single term and see if you can sum it.
The way I think about D is that, for any convergent alternating series, we have an infinite number of positive terms and an infinite number of negative terms. Consider the alternating harmonic series. Let's say we want to make it equal to 100. We can take all of the positive (odd) terms that we want, and add them together to get to eventually reach 100. Then we add -1/2. Then we add more positive terms to get back to 100. Then we add -1/4. Then we add more positive terms to get back to 100, and so on. Since we never run out of positive terms, we can keep going with this insane ratio of positive terms to negative terms forever. Hence, the series converges to any limit that we want.
Last edited by a moderator: May 5, 2017
3. Oct 4, 2011
@Kru.
Thanks!
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# How does density of paraffin wax change with temperature in a solar heater? Is there any equation?
I need to know what is the relation between density of paraffin wax and temperature in a solar water heater. I need to know the equation of density in terms of temperature.
• I doubt there will be any theoretically-derived equation for this. I think you will need to search to see if you can find some published experimental data for the thermal expansion of that substance, or conduct an experiment yourself. – Time4Tea Oct 13 '18 at 15:21
Anton Paar Wiki reports these values for the density of paraffin wax:
Specifying that their measurements cannot be made under 60°C because of the solidification of the wax.
While in this paper they used a formula to describe the behavior of its density in terms of temperature (in Kelvin): $$\rho(T) = \frac{750}{0.001(T-319.15)+1}$$
Plotting them side by side:
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# Mathematical perspective of prospective/retrospective designs
by Smith Last Updated January 15, 2018 04:19 AM
I'm trying to understand a passage titled "Tree Representations for Study Designs" in these notes which says the following.
I understand the setup but not much else. For example, my understanding was that a prospective design could also be $E\to S\to F$, i.e. individuals are selected based on their exposure, and monitored to see if they develop the disease. And I thought that a retrospective study design would be $F\to S\to E$. I thought retrospective meant that inclusion in the study is based on the outcome $F$. I also don't understand how he interprets prospective/retrospective design in terms of $P(E\cap F\cap S)$.
Can someone explain this mathematical perspective of prospective vs retrospective designs or provide a reference that expands on these ideas?
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## Engineering Mechanics: Statics & Dynamics (14th Edition)
$T=5.98kip$
We can determine the required force as follows: First, we apply Newton's second law to car A $\Sigma F_y=0$ $\implies N_A-20000cos\theta=0$ $\implies N_A=19923.894lb$ and $F_x=ma_x$ $\implies \mu_k N_A-20000sin\theta-T=m_A a$ We plug in the known values to obtain: $(0.5)(19923.894)-20000sin 5^{\circ}-T=\frac{20000}{32.2}a$ $\implies T+621.118a=8218.832$..eq(1) Now we apply Newton's second law to car A and B together $\Sigma F_x=ma_x$ $\implies \mu_k N_A-(W_A+W_B)sin\theta=(W_A+W_B)a$ We plug in the known values to obtain: $(0.5)(19923.894)-(20000+30000)sin 5^{\circ}=\frac{20000+30000}{32.2}a$ This simplifies to: $a=3.609ft/s^2$ Now from eq(1), we obtain: $T=8218.832-(621.118)(3.609)$ $\implies T=5977.168lb=5.98kip$
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Lemma 37.58.4. A composition of pseudo-coherent morphisms of schemes is pseudo-coherent.
Proof. This translates into the following algebra result: If $A \to B \to C$ are composable pseudo-coherent ring maps then $A \to C$ is pseudo-coherent. This follows from either More on Algebra, Lemma 15.81.13 or More on Algebra, Lemma 15.81.15. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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# Radius of Half Electric Potential Energy
1. Oct 26, 2016
### NiendorfPhysics
1. The problem statement, all variables and given/known data
There is a solid cylinder of radius a and then empty space then a shell cylinder of radius b. Show that half of the stored potential energy lies within a cylinder of radius $$\sqrt{ab}$$
2. Relevant equations
In the attempt
3. The attempt at a solution
I'm not sure what they want me to calculate the potential energy with respect to. If I do it wrt infinity it is infinity, same with 0. Let's say they want me to calculate it wrt the cylinder of radius that we have yet to determine. Then the energy is (getting rid of constants since they won't matter later):
$$\frac{1}{4}+\ln{\frac{R}{a}}$$
At least for the solid cylinder of radius a. Now we add the part of the energy stored in the field from the end of a to the fake cylinder. $$ln(\frac{R}{a})$$
Now we take one half of these values and add them together and set that equal to $$ln(\frac{R}{a})$$ and nothing works and I hate it because I know all of this is wrong.
But this problem cannot be this hard. This is just an RHK problem, I must be missing something simple. Can someone put me the on the right track?
2. Oct 26, 2016
### ehild
Do not mix potential an potential energy. The potential with respect to one surface changes proportionally with ln(R/a). So the potential difference between the surfaces of radii a and b is?
The energy is stored entirely between the cylindrical surfaces of the capacitor. You can use the formula for the energy of a capacitor in terms of voltage and charge.
3. Oct 26, 2016
### NiendorfPhysics
So $$U_{total} = \frac{1}{2} C (ln(\frac{b}{a}))^2$$ and $$U_{radial} = \frac{1}{2} C (ln(\frac{R}{a}))^2$$? I know this is wrong but I don't know what is right.
4. Oct 26, 2016
### NiendorfPhysics
Nevermind I got it thank you for the help
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# Roll Schedule Templates
Complexity:
Audience: Loan Servicer or Collector, Upper Management, Developers, Loan Servicing/Collections Managers, Administrator
### Introduction
A typical payment schedule is one that has similar payments and a single interest rate throughout. You may want to customize the payment schedule on your loan. Often, custom payment schedules are reused for certain types of loans within your company. Creating reusable, custom payment schedules is the purpose of roll schedule templates.
### How Schedule Roll Works
Roll schedule and schedule roll are terms used in LoanPro to describe a change to the calculated payment schedule for a loan. Custom payment schedules are created one line at a time. Each line of a schedule roll lets you input the following:
Variable Description Term This is the number of payment periods for which the rest of the settings for the current schedule line will apply Annual Interest Rate This is the interest rate that will apply over the term for this schedule line. Solve Using This option works in conjunction with the amount field and will let you choose how the payment amount will be calculated for the payments in this schedule line. See the options below.Payment: If this option allows you to set the payment amount for this schedule line.Balance: This option will let you solve for payment based on what the balance should be after the payments in this line are made. For example, if you choose balance and then enter $0.00, the payments will be calculated to be whatever they have to be so that the loan will be paid off (have a$0.00 balance) after the Term of this schedule line. This works similarly to the smooth advance tool, that solves for payment using an iterative method to get as close to the payment amount that will cause the principal balance to hit the target amount as it can. Type This option is only available if you choose Payment from the Solve Using drop-down. Your selection here can change the Amount/Percent/Payment Calc Terms field. See the options below.Advanced Schedule: Choosing this option will set the Amount/Percent/Payment Calc Terms field to Payment Calc Terms. This option will let you enter a number of payment periods into the Payment Calc Terms field. The number entered will become the remaining number of payments on the loan and the loan will be taken to a \$0.00 balance after these paymentsDollar Amount: This will set the Amount/Percent/Payment Calc field to Amount. Here you can enter the dollar amount for the payments in this schedule line.Interest Only: If this option is selected, the Amount/Percent/Payment field will no longer appear. Payments will be set to the amount of interest that accrues in a payment period. These payments won’t pay any principal.Interest Only Plus: This option will set the Amount/Percent/Payment Calc field to Amount. The payment amount will be the amount of interest that accrues in a payment period plus the amount you enter into the Amount field.% of Remaining Balance: This option will set the Amount/Percent/Payment Calc field to Percentage. This will calculate payments so that after the payments are made, the principal balance will equal the percentage of the remaining principal balance you entered into the Percentage field.% of Principal: This works just like % of Remaining Principal Balance except the balance after the payments will be a percentage of the original principal balance instead of the remaining principal balance.% of Principal + Underwriting: This works just like % of Remaining Principal Balance except the balance after the payments will be a percentage of the original principal balance plus underwriting fee instead of the remaining principal balance. Force Balloon Payment This option lets you choose to force a balloon payment on the loan after the schedule line. This will mean that the next payment after the schedule line is completed will cover the entire outstanding amount on the loan at that point.
#### Creating a Template
To create a new roll schedule template, navigate to Settings > Loan > Setup New Loan > Schedule Roll Templates.
Enter a term, rate, and other applicable information for the schedule line. Click Save to save the schedule line.
You can enter as many schedule lines as you need to complete your payment schedule. If you don’t enter enough schedule lines to cover the term of the loan, the payment schedule after the schedule roll is exhausted will be like it was in the last schedule line for the schedule roll. You can also set the “Last as Final” drop-down to “Yes” to choose to make the last payment for this line grow or shrink to accommodate late or missed payments so it will truly be the last payment.
Once you have entered all the desired schedule lines for your Roll Schedule Template, click Save to save the template.
#### Example
Let’s say that you regularly offer 12 month loans with a 3 month introductory interest rate of 5%, but after the first 3 months, the interest rate jumps to 15%. To make a schedule roll template to accomplish this, navigate to Settings > Loan > New Loan Setup > Roll Schedule Templates inside your company account. Click Add to create a new template.
Enter a name into the Name field. We will enter ‘3 Month 5% Promotion’ as the name of this template. Now click Add Template Line to add the first new schedule line.
Enter 3 for the term, since the introductory interest rate will last for 3 months. Enter 5 for the rate as the introductory 5% interest rate. Solve using Payment. Make sure you select Advanced Schedule from the Type drop-down. Since these loans are 12-month loans, we will enter 12 into the Payment Calc Terms box. You could actually choose from many different payment calculation options, but this will calculate the payment amount it would take to have 12 payments pay the loan off. Since these payments will have a 5% interest rate, they will be of a lower amount than the 15% payments. This lets the borrower see the benefit of the lower interest rate. Choose no for Force Balloon Payment and click Save.
Now click Add Template Line to enter the second schedule line.
For the second schedule line, enter a term of 9 since there will be 12 periods in this loan and 3 of them will be taken up by the first schedule line. Enter 15 as the rate, since the rate is going up to 15%. Choose to Solve Using Payment and choose Advanced Schedule from the Type drop-down. Enter Payment Calc Terms of 12. This should keep the loan at 12 payments since the interest was lower at the beginning than it is now, so all the principal should be paid by payments that are calculated at a higher interest rate for a 12-term loan.
Click Save to save the schedule line, then click Save to save the template.
You can now use this template on any loan. It will work correctly regardless of the loan amount.
### Default Templates
While creating new roll schedule templates is an option, there are also some default schedules on every loan. In your loan, go to account set up, then select set up terms. Make sure the loan is inactivated to see the roll schedule tab. On the right side, click on the tab that says Schedule tools, then schedule roll. After you select schedule roll, click the tab that says Load. The drop down options will include schedule templates to pro-rate the first payment. These include:
1. 1 - Prorate 1st Pmt Long Only
2. 2 - Prorate 1st Pmt Short only
3. 3- 1st Pmt All
You will also see default templates used to smooth the payment amounts over the life of the loan. These options include:
1. 4 - Smooth Payment Advance
2. 5 - Smooth Payment Basic
3. 6 - Smooth Payment Advance - No Round
#### Prorate Defaults
Prorate defaults allow the system to automatically increase or decrease the first payment amount based on irregularity of the first period. For example, if the contract date and the first payment date have 45 days between them, and the payment frequency is monthly, this would be classified as a 'long' first period. In contrast, if the contract date and first payment date have 20 days between them and the frequency is monthly, this would be classified as a 'short' first period.
• Prorate 1st Pmt Long Only: Use this if you have a longer first term.
• Prorate 1st Pmt Short Only: Use this if you have a shorter first term.
• Prorate 1st Pmt All: Use this if you want the software to calculate accordingly.
Using this feature may result in a 'straggler payment' at the end of the loan with an amount different from the other loan payments.
#### Smooth Payments
• Smooth Payment Advance – Smooths the payment amount over the number of periods you entered for the loan. The Smooth Payment Advance tool will only work with a whole-number loan term.
• Smooth Payment Basic – Smooths the payment amount over the number of periods that is most simple for the system to calculate. This option should take less time.
• Smooth Payment Advance – No Round – Smooths the payment amount over the number of periods you have entered for the loan without applying a schedule round. This setting will only work with a whole-number loan term.
### Common Uses & Questions
When the payment schedule is non-standard, meaning payment amount changes, this makes it easy to use the schedule roll rather than recreate the schedule each time. Another thing to note, smooth payments also use roll template to smooth the payments equal.
Loan Servicer or Collector Use
You'll need understand that this feature exists and if there is a schedule roll, what is in place and what it is doing. If people are calling you about this, it would be best for you to know how this changes the behavior of the loan.
Upper Management Use
Put in a template that will determine any deviation of a template schedule. Helps to not have changes like interest rate and payment amount over the life of the loan.
Developers Use
You just need to be aware if your company is using a template and what that is, so you can deploy the correct schedule template.
Loan Servicing/Collections Managers Use
It will be important to understand if your company is using a schedule roll and how it behaves to relay it to your company, and if phone calls are escalated to you, you will understand the behavior of loans.
For those who are configuring LoanPro, its best to know the schedule toll templates so loans are configured the way that you want so you can go forward with the loans and know what to expect.
Can a schedule roll template calculate the payment amount in order to arrive at a certain loan balance at a certain time in the loan? Yes, payment amount can be calculated to get as close as possible to a specific principle balance at the end of that schedule line.
If I have schedule lines in my template with different interest rates, will my payment automatically increase based on the interest rate? No. An increase or decrease in the payment amount based on the interest rate is often referred to as payment recasting. Some lenders like to change interest rates regularly based on changes to prime rate, LIBOR, etc. LoanPro doesn't automatically adjust payment amount when the interest rate changes during the life of a loan. However, when creating a Schedule Roll or Schedule Roll template, interest rate and payment amount can be specified with each schedule line.
### What’s Next
With an understanding of these templates, you're ready to Create a Schedule Roll. You might also be interested in these schedule roll features:
• Force Payments: You can use the force tool to create a schedule roll that automatically adjusts with a schedule round. This article will explain how to use force tool to set up a payment schedule.
• Smooth Payments: This tool smooths the payment amount over the number of periods that is most simple for the system to calculate. This option should take less time.
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## Quadratic formula by completing the square – easier method
[14 Feb 2011]
Most math text books derive the Quadratic Formula as follows:
To find the roots of a quadratic equation in the form ax2 + bx + c = 0, follow these steps:
(i) If a does not equal 1, divide each side by a (so that the coefficient of the x2 is 1).
$x^2+\frac{b}{a}x+\frac{c}{a}=0$
(ii) Rewrite the equation with the constant term on the right side.
$x^2+\frac{b}{a}x=-\frac{c}{a}$
(iii) Complete the square by adding the square of one-half of the coefficient of x (this is the square of b/2a) to both sides.
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a}$
(iv) Write the left side as a perfect square and simplify the right side.
$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$
(v) Equate and solve.
$x+\frac{b}{2a}\right=\pm\sqrt{\frac{b^2-4ac}{4a^2}}$
$x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}}$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Now, that’s pretty messy since there is a lot going on.
### Alternative Derivation of Quadratic Formula
Here’s a simpler process.
Once again, we start with an equation in the form (and call it Equation [1]):
$ax^2+bx+c=0\hspace{35}[1]$
Multiply both sides by 4a:
$4a(ax^2+bx+c)=4a(0)$
$4a^2x^2+4abx+4ac=0$
Now, go back to the starting equation [1], find the coefficient of x (it’s b) and square it (we get b2). Add that number to both sides of our equation:
$4a^2x^2+4abx+b^2=b^2-4ac$
Write the left side as a perfect square:
$(2ax+b)^2=b^2-4ac$
Solve for x:
$2ax+b=\pm\sqrt{b^2-4ac}$
$2ax=-b\pm\sqrt{b^2-4ac}$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
I hope you find that easier to follow than the more common method (presented at top).
### Solving a quadratic equation using the alternative method of completing the square
Question: Solve the quadratic equation using completing the square:
$3x^2-2x-4=0$
Answer: In this example. a = 3, so 4a = 12. We multiply both sides by 12:
$12(3x^2-2x-4)=0$
$36x^2-24x-48=0$
$36x^2-24x=48$
Now, in the question, b = −2. We square this (b2 = 4) and add it to both sides:
$36x^2-24x+4=48+4$
Next, write the left side as a perfect square:
$(6x-2)^2=52$
Solve for x:
$6x-2=\pm2\sqrt{13}$
$6x=2\pm2\sqrt{13}$
$x=\frac{2\pm2\sqrt{13}}{6}$
$x=\frac{1\pm\sqrt{13}}{3}$
What are your thoughts on this method? Is it easier for you?
[Hat tip to reader Lemmie, who sent me this method.]
### 13 Comments on “Quadratic formula by completing the square – easier method”
1. pat ballew says:
This method was made popular around 1815 as the “Hindoo” method by Edward Strachey’s translation of the Bija Ganita by Bhaskara. It was known as early as the ninth century and called the “pulverizer”.
Some more complete history can be found in an article I wrote called Twenty Ways to Solve a Quadratic”.. Section six
2. Murray says:
Thanks for the extra background, Pat!
3. Sue VanHattum says:
Thanks, Pat!
And James Tanton has 2 great videos which explain this method, at http://www.youtube.com/watch?v=OZNHYZXbLY8 and http://www.youtube.com/watch?v=bjH1HphOZ1Y.
4. Murray says:
Thanks for the video links, Sue.
5. Tweets that mention Quadratic formula by completing the square – easier method :: squareCircleZ -- Topsy.com says:
[...] This post was mentioned on Twitter by Murray Bourne, Šime Šulji?. Šime Šulji? said: RT @intmath: New Blog Post: Quadratic formula by completing the square easier method http://ow.ly/1bkQct [...]
6. elayarajah says:
excellent, but as I am mathematician, is there any theory related to the alternative ways?
7. Hope says:
The alternative seems pretty straightforward and simple. What is the rationale behind multiplying by 4a?
8. Murray says:
@elayarajah0 and Hope: The “multiply by 4a” step is there to give us a perfect square (4a2x2) in the first term.
Similarly, the “+ b2” is there to give us a perfect square in the 3rd term.
The result – all of the left hand side is a perfect square and can be easily factored.
9. Jemal Kemal Nigo says:
I like the alternative method to solve the quadiratic equation. It makes to see that there exists different methods to solve quadratic equations. for its rationale, one has to search. we have not only to ask for every thing. We have also to add some thing to the problem as our own contribution. By the way I am also always try to find an alternative methods for solving different Mathematical entities and even try to formulate different mathematical rules. I may join you if possible
thank you.
10. Murray says:
@Jemal: It’s great that you try to solve things using alternative methods. This is a good way to really understand something – and to advance mathematics.
11. Omari Ngondo says:
Thanks Murray
12. Chad Walls - Math Tutor says:
I like the first method best. As a high school math tutor I have completed the square and used the quadratic formula to solve for the zeros numerous times but have actually never had to derive the quadratic formula. I find this actually quite interesting and think I could use it as I tutor to help students bother remember how to complete the squares and memorize the quadratic formula. Thanks!
13. nicy12 says:
it really helps me a lot considering that it is very hard to understand the first process or derivation. now, i understand how it works. thanks!
XHTML: You can use these tags: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>
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• January 21st 2010, 06:28 PM
Lord Darkin
Can someone double-check my work here? Thanks, if you can. (Nod)
$
\lim_{x\to\infty}[x(1-cos(1/x))]
$
Put x to the denominator to use L'Hopital's rule, etc. Applied chain rules, etc.
$
\lim_{x\to\infty}[sin(1/x)]
$
Thus, if x goes to infinity, the above is sin(0) = 0. Final Answer: Zero
• January 21st 2010, 06:35 PM
Krizalid
Yes, it's okay, but you don't need that rule, you can turn that limit into a known one with a simple substitution.
• January 22nd 2010, 12:39 PM
Lord Darkin
Oh, really? What is this substitution? :)
• January 22nd 2010, 12:39 PM
Krizalid
$t=\frac1x.$
• January 23rd 2010, 03:58 AM
Seulementrien
tx=1
so itequals to lim [(1-cost)/t]
t->0
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# 12-30【陈张弛】腾讯会议 可积系统与几何清华-科大联合学术报告
In 1999, Eastwood and Ezhov classified all affinely homogeneous surfaces into a list by determining possible tangential vector fields. In 2020, with Merker J., we organise all homogeneous models in inequivalent branches. And we express the moduli of each branch as an algebraic variety.
The main technique is to write down a complete system of differential invariants. A surface is homogenous only if its invariants are constant, which gives infinitely many PDEs. Using Fels-Olver's recurrence formula, all invariants can be generated by finitely many fundamental invariants of low orders. However, in some situation, all fundamental invariants being constant is not enough to determine homogeneity. Compatibility conditions, also known as Frobenius integrability conditions, shall also be included.
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Team:HokkaidoU Japan/Promoter
(Difference between revisions)
Revision as of 09:55, 27 October 2013 (view source)Nousan (Talk | contribs)← Older edit Latest revision as of 02:51, 29 October 2013 (view source)TaKeZo (Talk | contribs) (25 intermediate revisions not shown) Line 14: Line 14:
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Maestro E.coli
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Maestro E. coli
Promoter
Promoter
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-
We explain the importance of promoter sequence. But before that let's look how RNA binds to a promoter with the help of fig.1.
+
We explain the importance of promoter sequence, but before that let's look how RNA binds to a promoter with the help of (fig.1).
Line 39: Line 39:
Transcription factors related to Promtoer
Transcription factors related to Promtoer
-
RNA complex consist of 5 core enzymes and a σ factor. σ factor plays crucial role in promoter recognition. It recognizes and binds to promoter region on DNA sequence and helps to assemble the core enzyme and start transcription. σ factor has several analogs, E. coli which is widely used bacteria by iGEMers is using σ70 for house-keeping gene expression at exponential growth. Bacterial promoter can be roughly divided into three regions; -10 region, spacer and -35 region. Bases in promoter are numbered in descending order from transcription start base which is defined as +1.
+
RNA complex consist of 5 core enzymes and a σ factor. σ factor plays crucial role in promoter recognition. It recognizes and binds to promoter region on DNA sequence and helps to assemble the core enzyme and start transcription. σ factor has several analogs, E. coli which is widely used bacteria by iGEMers is using σ70 for house-keeping gene expression at exponential growth. Bacterial promoter can be roughly divided into three regions; -10 region, spacer and -35 region. Bases in promoter are numbered in descending order from transcription start base which is defined as +1.
Line 49: Line 49:
-35 region
-35 region
-
-35 region is second in importance to -10. It does not energetically contribute to promoter melting. There reports on promoters without -35 region. In those case TG motif at about -16 is thought as alternative. -35 consensus sequence is TTGACA at from -36 to -31.
+
-35 region is also important second to -10. It does not energetically contribute to promoter melting. There reports on promoters without -35 region. In those case TG motif at about -16 is thought as alternative. -35 consensus sequence is TTGACA at from -36 to -31.
-
Promoters function to bind RNAP is a reason it is genetically well preserved. Most frequently conserved residues in the sequence make a "consensus sequence". In 1983, -35 and -10 consensus was showed to be TTGACA and TATAAT respectively [Fig 2]. Horizontal axis of the figures represents the position upstream of translation ignition point. Letter at the top of the figure signifies more than over 39% occurrence of that letter at that position. Larger occurrence over 54% is represented as upper case letter. Consensus sequence published by Marjan De Mey et al. (2007) shows that -10 and -35 region is highly preserved [Fig 3]. There other less preserved regions. The tetramer (TRTG) upstream from -10 region is called TG motif. Upstream of -35 region is UP element and downstream of -10 region is discriminator region. These sequences are thought to bind core enzymes. So these sequences are also well conserved. Each sequence is important to control promoter strength.
+
Promoters function to bind RNAP is a reason it is genetically well preserved. Most frequently conserved residues in the sequence make a "consensus sequence". In 1983, -35 and -10 consensus was showed to be TTGACA and TATAAT respectively (fig 2). Horizontal axis of the figures represents the position upstream of translation ignition point. Letter at the top of the figure signifies more than over 39% occurrence of that letter at that position. Larger occurrence over 54% is represented as upper case letter. Consensus sequence published by Marjan De Mey et al. (2007) shows that -10 and -35 region is highly preserved (fig 3). There other less preserved regions. The tetramer (TRTG) upstream from -10 region is called TG motif. Upstream of -35 region is UP element and downstream of -10 region is discriminator region. These sequences are thought to bind core enzymes. So these sequences are also well conserved. Each sequence is important to control promoter strength.
Line 71: Line 71:
- +
- +
• - + [1] R. a Mooney, I. Artsimovitch, and R. Landick, “Information processing by RNA polymerase: recognition of regulatory signals during RNA chain elongation.,” Journal of bacteriology, vol. 180, no. 13, pp. 3265–75, Jul. 1998. -
MODELING
+
• - +
• -
We tried to theoretically predict the strength distribution of 4096 promoters, which were artificially created by random mutation. We followed these 3 steps, referring the study by Brewster et al.[1].
+ [2] M. S. B. Paget and J. D. Helmann, “The σ 70 family of sigma factors,” Genome Biology, vol. 4, no. 1, pp. 203.1–203.6, 2003. -
+ -
1. Calculate the binding energy of each promoter and σ-factor using the sequence
2. +
3. -
4. Convert the binding energy to the probability that RNAP binds promoter using the method of statistical mechanics
5. + [3] D. K. Hawley, W. R. Mcclure, and I. R. L. P. Limited, “Compilation and analysis of Escherichia coli promoter DNA sequences,” -
6. Utilizing the binding probability as the transcription efficiency
7. + -
+ Nucleic Acids Research, vol. 11, pp. 2237–2255, 1983. - +
• -
STEP 1: Calculation of Binding Energy
+ [4] M. De Mey, J. Maertens, G. J. Lequeux, W. K. Soetaert, and E. J. Vandamme, “Construction and model-based analysis of a promoter library for E. coli: an indispensable tool for metabolic engineering.,” BMC biotechnology, vol. 7, p. 34, Jan. 2007. -
First, we found the binding energy of RNAP and our promoters. As we mutated only -35 region, we only use this region for calculations. Here we define the binding energy ε as the energy released by RNAP’s binding to promoter. Simply saying, the higher is the binding energy, the stronger is the binding. We referred the data in Kenney et al.[2] to calculate each binding energy. +
• - +
• -
The distribution of computed 4096 promoters' binding energies is shown below. The horizontal axis stands for ε (at 0.05 kT intervals) and the vertical axis sample number.
+ [5]De Mey, M., Maertens, J., Lequeux, G. J., Soetaert, W. K., & Vandamme, E. J. (2007). Construction and model-based analysis of a promoter library for E. coli: an indispensable tool for metabolic engineering. BMC biotechnology, 7, 34. doi:10.1186/1472-6750-7-34 - +
• -
+ - + -
fig. 1 Visualized data. A portion enclosed with red square is randomized -35 region.
+ -
+ - + -
+ - + -
fig. 2 Promoters distribution. The result is an approximate normal distribution.
+ -
+ - + -
STEP 2: Conversion from Binding Energy to Binding Probability
+ - + - + -
Next, we estimated the binding probability. On this step, we used the method of statistical mechanics. So we assumed the following.
+ -
+ -
• The cell is a closed system
• + -
• There are P RNAPs bound somewhere on DNA
• + -
• The number of bases is N (bp) and 1 of N bases is +1 position of the promoter
• + -
+ - + -
The principle of statistical mechanics is very easy; any state emerges with the same probability. So we counted up the number of state. A state stands for any information of all the particles in the system, so the number is enormous. W represents this number. Here W can be separated as the following. + - + - $+ - W=W_{\mathrm{unbound}}+W_{\mathrm{bound}} + -$ + - + - $W_{\mathrm{bound}}$ represents the number of state where the promoter is occupied and $W_{\mathrm{unbound}}$ unoccupied.
+ - + -
The purpose of this step is to find the ratio $W_{\mathrm{unbound}}:W_{\mathrm{bound}}$. Concerning the position of RNAP, + - + - $+ - W_{\mathrm{unbound}}:W_{\mathrm{unbound}}&=&{\frac{N-1!}{P!(N-P-1)!} \times W_{\mathrm{R}}(E)} + -$ + - + - + - +
- - - - -
- -
M-Fig. 3 Quoted from [5].
-
- -
Therefore, the binding probability is
- - \begin{align*} - p&=\frac{W_{\mathrm{bound}}}{W_{\mathrm{unbound}}+W_{\mathrm{bound}}} \\[6pt] - &=\frac{ \frac{P}{N_{\mathrm{NS}}} \exp\left(-\frac{\varepsilon_{\mathrm{S}} - \varepsilon_{\mathrm{NS}}}{k_{\mathrm{B}}T} \right) }{1+\frac{P}{N_{\mathrm{NS}}} \exp\left(-\frac{\varepsilon_{\mathrm{S}} - \varepsilon_{\mathrm{NS}}}{k_{\mathrm{B}}T} \right) } \\[6pt] - \mathrm{suppose\ that} &\frac{P}{N_{\mathrm{NS}}} \exp\left(-\frac{\varepsilon_{\mathrm{S}} - \varepsilon_{\mathrm{NS}}}{k_{\mathrm{B}}T} \right) \ll 1 \\[6pt] - &\approx \frac{P}{N_{\mathrm{NS}}} \exp\left(-\frac{\varepsilon_{\mathrm{S}} - \varepsilon_{\mathrm{NS}}}{k_{\mathrm{B}}T} \right) \\[6pt] - &\propto \exp\left(-\frac{\varepsilon_{-35}}{k_{\mathrm{B}}T} \right) - \end{align*} - -
The binding energy of -35 region is exponentially proportional to the binding probability. -
The last step is to convert the binding probability to the transcription efficiency. Let us assume these suppositions. -
- -
-
• RNAP bound to promoter promptly initiate transcription
• -
• There is no "traffic jam" of RNAPs on DNA (i. e., RNAP's transcription initiation is rate-limiting)
• -
- -
These assumptions mean that we can directly use the value of binding probability as transcription energy in an arbitrary unit. In this way, we get following conclusive result.
-
- -
M-Fig. 4 The horizontal axis stands for the transcription efficiency.
-
- -
As you can see in this figure, the strengths of our promoter families vary about 1000 fold!
-
Overview
Proteins are expressed in mainly 2 steps. First mRNA is polymerized using DNA as a template. Then ribosome binds mRNA and translates it into protein.
Promoter is a DNA sequence initiating transcription from DNA to mRNA. If transcriptional efficiency is defined as "promoter strength", stronger promoter has ability to transcribe more mRNA. This should lead in stronger expression of proteins.
We have created several promoters by randomization of -35 sequence followed by selection. In promoters -35 region is responsible for supporting binding of RNA polymerase (RNAP). This interaction results in closed complex which is rate-limiting step. We focused on this rather transparent function to introduce variability in promoter strength.
We explain the importance of promoter sequence, but before that let's look how RNA binds to a promoter with the help of (fig.1).
fig. 1 mRNA transcription starts with promoter engagement, continues to initiation, elongation, and then it comes to termination (omitted in the figure).
First transcription complex must be formed. Transcription complex polymerizes mRNA in 2 steps. Initiation step starts polymerization followed by elongation step. Promoter serves crucial role on engagement and initiation. After closed complex formation DNA double helix pulled apart to form transcription bubble. During this closed complex changes into open complex. This marks the beginning of mRNA polymerization. Transcription bubble exposes deoxyribonucleotides to form new hydrogen bonds with ribonucleotides. In short DNA serves as template to make mRNA.
Transcription factors related to Promtoer
RNA complex consist of 5 core enzymes and a σ factor. σ factor plays crucial role in promoter recognition. It recognizes and binds to promoter region on DNA sequence and helps to assemble the core enzyme and start transcription. σ factor has several analogs, E. coli which is widely used bacteria by iGEMers is using σ70 for house-keeping gene expression at exponential growth. Bacterial promoter can be roughly divided into three regions; -10 region, spacer and -35 region. Bases in promoter are numbered in descending order from transcription start base which is defined as +1.
-10 region
The -10 region is structurally very important because it is initiates promoter melting in RNAP-promoter complex. This is essential to form open complex. Promoter consensus sequence is TATAAT at -12 to -7 position.
Spacer
Spacer is thought to increase flexibility of σ factor binding requirements.
-35 region
-35 region is also important second to -10. It does not energetically contribute to promoter melting. There reports on promoters without -35 region. In those case TG motif at about -16 is thought as alternative. -35 consensus sequence is TTGACA at from -36 to -31.
Promoters function to bind RNAP is a reason it is genetically well preserved. Most frequently conserved residues in the sequence make a "consensus sequence". In 1983, -35 and -10 consensus was showed to be TTGACA and TATAAT respectively (fig 2). Horizontal axis of the figures represents the position upstream of translation ignition point. Letter at the top of the figure signifies more than over 39% occurrence of that letter at that position. Larger occurrence over 54% is represented as upper case letter. Consensus sequence published by Marjan De Mey et al. (2007) shows that -10 and -35 region is highly preserved (fig 3). There other less preserved regions. The tetramer (TRTG) upstream from -10 region is called TG motif. Upstream of -35 region is UP element and downstream of -10 region is discriminator region. These sequences are thought to bind core enzymes. So these sequences are also well conserved. Each sequence is important to control promoter strength.
fig. 2 Consensus sequence shown in review article in 1983 [3].
fig. 3 Consensus sequence prepared in 2007 [4].
So we went and designed "consensus promoter". It should have strongest binding energy to RNAP. By adding mutations to -35 we sought to construct promoters with various binding energies. There are three reasons why we used -35 region.
First, -35 region is just supporting binding with σ factor. It has less vital role compared to -10 region, which energetically contributes to formation of open complex. Having this in mind we changed -35 region to easily change promoter binding strength without severe errors in promoter function.
Second, RNAP and promoter binding orchestrated by σ factor binding. Complex formation is thought to be rate-limited step. We thought that -35 region performs a simpler function. For this reason, mutations at -35 region can be thought as more structurally transparent.
Recently published research reported the making of promoter family by randomizing both -35 and -10 regions, changing spacer length. However it would be too much of the task for us to make some many changes. By changing hexamer sequence of -35 region there are 4096 variation. This number is a lot smaller compared to mutating every promoter position. So we can get result with a smaller library size.
With these 3 reasons we went on to construct our promoter family.
• [1] R. a Mooney, I. Artsimovitch, and R. Landick, “Information processing by RNA polymerase: recognition of regulatory signals during RNA chain elongation.,” Journal of bacteriology, vol. 180, no. 13, pp. 3265–75, Jul. 1998.
• [2] M. S. B. Paget and J. D. Helmann, “The σ 70 family of sigma factors,” Genome Biology, vol. 4, no. 1, pp. 203.1–203.6, 2003.
• [3] D. K. Hawley, W. R. Mcclure, and I. R. L. P. Limited, “Compilation and analysis of Escherichia coli promoter DNA sequences,”
• Nucleic Acids Research, vol. 11, pp. 2237–2255, 1983.
• [4] M. De Mey, J. Maertens, G. J. Lequeux, W. K. Soetaert, and E. J. Vandamme, “Construction and model-based analysis of a promoter library for E. coli: an indispensable tool for metabolic engineering.,” BMC biotechnology, vol. 7, p. 34, Jan. 2007.
• [5]De Mey, M., Maertens, J., Lequeux, G. J., Soetaert, W. K., & Vandamme, E. J. (2007). Construction and model-based analysis of a promoter library for E. coli: an indispensable tool for metabolic engineering. BMC biotechnology, 7, 34. doi:10.1186/1472-6750-7-34
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# Is there a notion of generalized acidity for lithium ions?
From my understanding, the basic notions of acidity can be explained by considering a cup of pure water, and then adding a substance to it which creates either free $$\ce{H+}$$, or free $$\ce{H3O+}$$ and $$\ce{H5O2+}$$ etc...
And these ions are free to react and give rise to what we consider acidic behavior. The classic example is hydrogen chloride characterized with the following reaction
$$\ce{HCl +H2O = H3O+ + Cl-}$$
Whereas $$\ce{H3O+}$$ is our famous hydrochloric acid.
I wonder if it is possible for an analogous reaction to happen with other polar molecules. For example liquid $$\ce{Li2O}$$ (I believe it melts around 1438 Celsius) should be a polar liquid for the same reasons that $$\ce{H2O}$$ is (though maybe not as strongly polar), and therefore: $$\ce{LiCl}$$ should dissolve in molten $$\ce{Li2O}$$ to create positive charged lithium ion species such as $$\ce{Li3O+}$$. Which ought to function as an "acid" of sorts, in that they react with most thing which are dissolved/reacted by hydrochloric acid by acting as a positive charge donor (clearly not a proton donor since we are using lithium and not hydrogen to do the deed here)
Is my line of reasoning sound and has this been experimentally detected before?
• I'm not sure if you want more then that Li cation is obviously Lewis acid Dec 22 '20 at 2:23
• (so I can get the terminology right) our $\text{Li}_2 \text{O}$ fluid with dissolved $\text{Li}\text{Cl}$ is a Lewis acid? Dec 22 '20 at 2:25
• $\ce{Li+}$ is a Lewis acid that might coordinate to $\ce{Li2O}$ base forming an adduct. I highly doubt you can get much $\ce{Li2O}$ molecules outside of vapor phase, thogh. Dec 22 '20 at 2:30
A complex of lithium ion with lithium hudroxide is known as the salt $$\ce{[Li2OH]^+[ClO4]^-}$$ [1]. As with the hydronium salt, this requires an essentially nonbasic counterion; with a chloride counterion a double salt is obtained with no evidence of a complex cation. From the abstract (some evident faults in translation from Russian have been corrected):
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# characterization of residual finiteness for finitely generated groups
Where can I find a proof of the fact that a finitely generated group is residually finite if and only if it acts faithfully on a locally finite rooted tree?
-
One way is actually quite easy - that every group acting faithfully on a locally finite rooted tree is residually finite - and can be found in a paper of Mark Sapir.
Indeed, every automorphism $f$ of the tree $T$ must fix the root and so it fixes the levels of the tree. If $f \neq 1$ on level number $n$, we consider the homomorphism from $\operatorname{Aut}(T)$ to the (finite) group of automorphisms of the finite tree consisting of the first $n$ levels of $T$. The homomorphisms restricting automorphisms of $T$ to vertices of levels at most $n$. The automorphism $f$ survives this homomorphism. Thus $\operatorname{Aut}(T)$ and all its subgroups are residually finite.
He then implies that the converse is a result of Kaluzhnin, and gives a reference as,
"R. I. Grigorchuk, V. V. Nekrashevich, V. I. Sushchanskii, Automata, dynamical systems, and groups. Tr. Mat. Inst. Steklova 231 (2000), Din. Sist., Avtom. i Beskon. Gruppy, 134–214"
However, I have more than just skimmed this paper (it is 76 pages long, and I couldn't find a searchable version) but I couldn't find the proof of this result.
(As a side note, the paper of Sapir is actually very interesting, and well worth a read. Although I found his proper paper of the result easier to follow...)
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# 7.1 Second-order linear equations (Page 5/15)
Page 5 / 15
$y\left(x\right)={c}_{1}{e}^{-2x}+{c}_{2}{e}^{-7x}.$
## Single repeated real root
Things are a little more complicated if the characteristic equation has a repeated real root, $\lambda \text{.}$ In this case, we know ${e}^{\lambda x}$ is a solution to [link] , but it is only one solution and we need two linearly independent solutions to determine the general solution. We might be tempted to try a function of the form $k{e}^{\lambda x},$ where k is some constant, but it would not be linearly independent of ${e}^{\lambda x}.$ Therefore, let’s try $x{e}^{\lambda x}$ as the second solution. First, note that by the quadratic formula,
$\lambda =\frac{\text{−}b±\sqrt{{b}^{2}-4ac}}{2a}.$
But, $\lambda$ is a repeated root, so ${b}^{2}-4ac=0$ and $\lambda =\frac{\text{−}b}{2a}.$ Thus, if $y=x{e}^{\lambda x},$ we have
${y}^{\prime }={e}^{\lambda x}+\lambda x{e}^{\lambda x}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y\text{″}=2\lambda {e}^{\lambda x}+{\lambda }^{2}x{e}^{\lambda x}.$
Substituting these expressions into [link] , we see that
$\begin{array}{cc}\hfill ay\text{″}+b{y}^{\prime }+cy& =a\left(2\lambda {e}^{\lambda x}+{\lambda }^{2}x{e}^{\lambda x}\right)+b\left({e}^{\lambda x}+\lambda x{e}^{\lambda x}\right)+cx{e}^{\lambda x}\hfill \\ & =x{e}^{\lambda x}\left(a{\lambda }^{2}+b\lambda +c\right)+{e}^{\lambda x}\left(2a\lambda +b\right)\hfill \\ & =x{e}^{\lambda x}\left(0\right)+{e}^{\lambda x}\left(2a\left(\frac{\text{−}b}{2a}\right)+b\right)\hfill \\ & =0+{e}^{\lambda x}\left(0\right)\hfill \\ & =0.\hfill \end{array}$
This shows that $x{e}^{\lambda x}$ is a solution to [link] . Since ${e}^{\lambda x}$ and $x{e}^{\lambda x}$ are linearly independent, when the characteristic equation has a repeated root $\lambda ,$ the general solution to [link] is given by
$y\left(x\right)={c}_{1}{e}^{\lambda x}+{c}_{2}x{e}^{\lambda x},$
where ${c}_{1}$ and ${c}_{2}$ are constants.
For example, the differential equation $y\text{″}+12{y}^{\prime }+36y=0$ has the associated characteristic equation ${\lambda }^{2}+12\lambda +36=0.$ This factors into ${\left(\lambda +6\right)}^{2}=0,$ which has a repeated root $\lambda =-6.$ Therefore, the general solution to this differential equation is
$y\left(x\right)={c}_{1}{e}^{-6x}+{c}_{2}x{e}^{-6x}.$
## Complex conjugate roots
The third case we must consider is when ${b}^{2}-4ac<0.$ In this case, when we apply the quadratic formula, we are taking the square root of a negative number. We must use the imaginary number $i=\sqrt{-1}$ to find the roots, which take the form ${\lambda }_{1}=\alpha +\beta i$ and ${\lambda }_{2}=\alpha -\beta i\text{.}$ The complex number $\alpha +\beta i$ is called the conjugate of $\alpha -\beta i\text{.}$ Thus, we see that when ${b}^{2}-4ac<0,$ the roots of our characteristic equation are always complex conjugates .
This creates a little bit of a problem for us. If we follow the same process we used for distinct real roots—using the roots of the characteristic equation as the coefficients in the exponents of exponential functions—we get the functions ${e}^{\left(\alpha +\beta i\right)x}$ and ${e}^{\left(\alpha -\beta i\right)x}$ as our solutions. However, there are problems with this approach. First, these functions take on complex (imaginary) values, and a complete discussion of such functions is beyond the scope of this text. Second, even if we were comfortable with complex-value functions, in this course we do not address the idea of a derivative for such functions. So, if possible, we’d like to find two linearly independent real-value solutions to the differential equation. For purposes of this development, we are going to manipulate and differentiate the functions ${e}^{\left(\alpha +\beta i\right)x}$ and ${e}^{\left(\alpha -\beta i\right)x}$ as if they were real-value functions. For these particular functions, this approach is valid mathematically, but be aware that there are other instances when complex-value functions do not follow the same rules as real-value functions. Those of you interested in a more in-depth discussion of complex-value functions should consult a complex analysis text.
Based on the roots $\alpha ±\beta i$ of the characteristic equation, the functions ${e}^{\left(\alpha +\beta i\right)x}$ and ${e}^{\left(\alpha -\beta i\right)x}$ are linearly independent solutions to the differential equation. and the general solution is given by
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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# Is Lagrange multipliers and (multivariable) extreme value theorem related?
I couldn't find a question answering this concept but they seem to be related.
Extreme Value Theorem (two variables)
If f is a continuous function defined on a closed and bounded set $$A⊂\mathbb{R}^2$$, then f attains an absolute maximum and absolute minimum value on A.
Lagrange Multipliers (two variables)
Extreme values of function f(x, y) subject to constraints g(x, y) = k has solutions in $$\nabla f=\lambda \nabla g$$.
The constraint in Lagrange Multipliers creates a closed and bounded region that would satisfy EVT, does it not? So does that make Lagrange multipliers a specific case of EVT?
• The Lagrange multiplier optimality condition is analogous to the optimality condition $\nabla f(x) = 0$ for unconstrained optimization. In both cases, for a minimization problem, the optimality condition can be interpreted as saying that if you move a short distance in a feasible direction, the objective function value does not decrease. (It's not totally obvious at first that the Lagrange multiplier optimality condition can be interpreted in this way. The book Numerical Optimization by Wright et al has a good explanation. I've also explained this in other questions if you look at my history.) – littleO Jun 13 '20 at 0:27
The technique of Lagrange Multipliers cannot be classified as a "special case" of the EVT. Note that the constraint $$g$$ in an optimization problem may not yield a closed and bounded set to optimize across, so the extreme value theorem may not apply even when the technique of Lagrange multipliers can still find absolute extrema. Take for example the constraint $$g(x, y) = x^2 - y^2 = 1$$, which is a hyperbola and is certainly not bounded. Nevertheless, absolute extrema may still exist (since the EVT only gives a sufficient condition for there to be an absolute min and max). This is the case when the function we seek to minimize is $$f(x, y) = x^2 + y^2$$ whose absolute minimum on the constraint is attained at $$(x, y) = (1, 0)$$. You can verify for yourself that there is a solution to $$\nabla f = \lambda \nabla g$$ with these values of $$x$$ and $$y$$.
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# bug in fancybox
Denis Girou Denis.Girou at idris.fr
Wed Sep 3 23:02:09 CEST 1997
-----------------------------------------------------------------------------
This is the PSTricks mailing list, devoted to discussions about computational
graphics in (La)TeX using the PSTricks package from Timothy van Zandt.
For help using this mailing list, see instructions at the end of message.
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>>>>> "Eric.Picheral" == Eric Picheral <- CRI Universite' Rennes 1 - 99.84.71.23" <Eric.Picheral at univ-rennes1.fr>> writes:
Eric.Picheral> I just discover what seems to be a bug in the new version of
Eric.Picheral> fancybox.sty.
Eric.Picheral> Here is the latex source :
Eric.Picheral> ---------------------------------------------------------------------
Eric.Picheral> \documentclass[10pt,a4paper]{article}
Eric.Picheral> \usepackage{fancybox}
Eric.Picheral> \begin{document}
Eric.Picheral> \tableofcontents
Eric.Picheral> \section{section 1}
Eric.Picheral> \section{section 2}
Eric.Picheral> \section{section 3}
Eric.Picheral> \end{document}
Eric.Picheral> ---------------------------------------------------------------------
Eric.Picheral> If run with the may 97 version of fancybox, the resulting dvi does not
Eric.Picheral> If run with the old version of fancybox, it's OK.
After verification, I see that it is a side effect that I never detect in a
For your case, it is enough to remove this correction:
%\newcount\c at tocdepth
But it is a workaround because at this time the problem I detected (and
which was several times reported after - mainly by people using fancybox with
Seminar) with macros like \fancyput and \fancypage appear again.
You can check it after removing my line with the following test:
\documentclass{article}
\usepackage{fancybox}
\thisfancypage{\setlength{\fboxsep}{1cm}\ovalbox}{}
\begin{document}
\section{Test fancypage}
\end{document}
So what we have to do is to found a better correction than mine, with
no side effect... As it is related to the internals of LaTeX that I don't know
well and often don't understand, if some of you with a better knowledge than
mine has a solution, it will be good...
The error message is:
\addpenalty #1->\ifvmode \if at minipage \else \if at nobreak \else \ifdim \lastskip
=\z@ \penalty #1\relax \else \@tempskipb \lastskip \vskip -\lastskip \penalty #
1\vskip \@tempskipb \fi \fi \fi \else \@noitemerr \fi
#1<-\@secpenalty
\@noitemerr ->\@latex at error {Something's wrong--perhaps a missing \protect \ite
m }\@ehc
In any case, thank you for the report.
D.G.
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# Proof on Family of sets - Looking for Help OR Feedback
## Homework Statement
Given that $F$ is a family of sets, that $\bigcup F$ is the union of the sets members of the family $F$, that $A$ is a set, assume that
(1) $\hspace{1cm} \forall F (\bigcup F = A \rightarrow A \in F)$
then prove that
(G) $\hspace{1cm} \exists x (A= \left\{ x \right\} )$
## Homework Equations
Given that $P \rightarrow Q$ is the same as $\neg P \vee Q$, we can rephrase the assumption as
(2) $\hspace{1cm} \forall F (\bigcup F \neq A \vee A \in F)$
## The Attempt at a Solution
That's the biggest problem: I have no idea how to get the solution.
Just to remember, I am fighting with the software Proof Designer: I love and hate it. I love it when it tells me I got the proof, I hate it when it doesn't... so right now I am hating it (or maybe I am hating myself!).
Btw, a nice part of the program is that it gives you the possibility to proceed in the proof even if you have no idea what - in this specific case - $F$ and $x$ are... which is exactly my situation! The program does it in order to mimic the way in which proofs actually work. You go around, you assume, and then you define... and everything goes right. Usually, but not this time.
First of all, just a confirmation of my general thought about the theorem: for whatever $F$ you pick up, you can always find an $A$ that has some characteristics (which have to be really specific).
So, basically we can vary $F$ and we will always find a corresponding $x$ that respect (1) or (2).
Right?
About the solution. I focused a lot on (2), and then I tried to work out a proof by cases. It didn't really work, maybe cause I chose the wrong values for $F$ and $x$. In particular I put $A= \left\{ ∅ \right\}$ and then $F= \left\{ \left\{ ∅ \right\} \right\}$. With these choices, the second case is easily covered, while the second simply cannot be solved. This is related to the fact that the assumption is now $\bigcup F \neq A$, a negative statment that is not easy manageable (at least for me).
Now, the problem I found is that, even if I choose a different value for $F$ (let's say $P(A)$, defined as the Power Set of A) and wait to find the value for $x$, still I go nowhere. For example, $P(A)$ has some nice properties and it can be used (1) instead of (2), because it can be easily proven that $\bigcup P(A) =A$, giving the possibility to use MP to get $A \in P(A)$. However, in this case I have no idea what $x$ should be. It's like there is a huge gap between what I am assuming and what I have to prove and I cannot fill it...
Sorry for the long post, but this problem is really absorbing me.
I would really appreciate any kind of feedback.
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## Answers and Replies
Related Precalculus Mathematics Homework Help News on Phys.org
HallsofIvy
Homework Helper
I am a bit confused by your "Given that F is a family of sets" and "$\forall F$". Ther first implies that we are talking about a specific family of sets but the second talks about "all families of sets". I will assume you mean the second: for all possible families of sets, if set A is equal to the Union of all sets in the family implies that A is in the family, then A is a singleton set.
I don't know what you mean by "I chose the wrong values for F and x"- we cannot "choose" F and x, we must prove this for any F and any set A satisfying the hypotheses. I also am not clear whether you intend "∅" to indicate the empty set or the number "0" because you talk about "{∅}" which would make sense either way.
Are you asked to prove this or determine whether it is true or false? Because it appears to me to be false and then we can choose a specific "F" to give a counter-example. If $F= \{\phi\}$- that is, if F is the family containing only the empty set- its union is the empty set so "$A= \cup F$" is true only if A is the empty set. But then $A= \phi\in F$ is true so the hypothesis is satisfied. But there is NO x such that "$x\in A$" because A is empty.
First of all, thanks (really) a lot!
I am a bit confused by your "Given that F is a family of sets" and "$\forall F$". Ther first implies that we are talking about a specific family of sets but the second talks about "all families of sets". I will assume you mean the second: for all possible families of sets, if set A is equal to the Union of all sets in the family implies that A is in the family, then A is a singleton set.
Indeed, definitely the second. Trying to be precise, I ended up being sloppy. With the expression "Given that F is a family of sets" I wanted to say something like "Defining F as a family of sets".
[Hopefully this formulation should be right.]
I don't know what you mean by "I chose the wrong values for F and x"- we cannot "choose" F and x, we must prove this for any F and any set A satisfying the hypotheses.
Ok, right. I expressed myself in the wrong way, but still I am not sure about the real goal of this proof. Can we say that the all game is about finding an x that works given a certain F (like F varies everytime x varies)? Or the goal is to find a x that is like THE JOLLY, that fixes the problem whatever F we choose to put in the hypothesis?
I am inclined towards the second, but I am not really sure.
I also am not clear whether you intend "∅" to indicate the empty set or the number "0" because you talk about "{∅}" which would make sense either way.
Well, in my formulation ∅ is the empty set, while {∅} is the set containing the empty set (so this set contains one element, the empty set).
Are you asked to prove this or determine whether it is true or false?
Considering that this is a selected problem that should be solved with Proof Designer, I assume I have to prove it. Or - well- , I really hope so...
It would be sadistic from Professor Velleman to put on his site a problem that cannot be proved!
Because it appears to me to be false and then we can choose a specific "F" to give a counter-example. If $F= \{\phi\}$- that is, if F is the family containing only the empty set- its union is the empty set so "$A= \cup F$" is true only if A is the empty set. But then $A= \phi\in F$ is true so the hypothesis is satisfied. But there is NO x such that "$x\in A$" because A is empty.
Now, this is interesting. I kinda had a sudden enlightment. After your counterexample I realized that it shouldn't work cause maybe I gave you a wrong info. As a matter of fact, I kinda guess that the formulation of (G) implies that A is a family of set itself. I think so because it's part of the game with Proof Designer that you cannot insert whatever you want if it has not been previously defined. So, for example, I cannot build up A with some random elements (following the structure of the software, I cannot insert for x a random object if it has not been defined like A={a,b,c,d}). On the contrary I can insert ∅. So, we can have A={∅}, which is a family of set, and having F={{∅}} the counterexample shouldn't work.
On the contrary I can insert ∅. So, we can have A={∅}, which is a family of set, and having F={{∅}} the counterexample shouldn't work.
Wait a second, if A is a family of set as it should be (as I realized too late), are you telling me that what I wrote in the quoted part is a sketch of a proof of the statement? Cause if it is so, I really don't see why. However in this case we have that if U{{∅}}=A then A is a member of {{∅}}, which is the case as far as we pick A={∅}. So it's ok.
But... is it really a proof? Actually I don't see why, and definitely the software doesn't help me to see that it is.
This should also be a sort of answer to my major doubt: how I have to set those kind of problems. It seems there are two possibilities:
1) If this is a proof, it means that first we pick a value for A, and then we prove that it works picking a value for F.
2) However, to be honest, I think the way in which all those kinda problems should be faced is more about: giving a value of A (which is the PERFECT value), whatever F we take, the situation stands. So, in this case we would have that, letting A={∅}, whatever F we choose (it can be F={{∅}}, but F=P(A) as well), that value of A={∅} works.
Now, if case 2 is the right one (and I really think so), I have no idea how this can be proved and I would say that the statement is generally false.
[Sorry HallsOfIvy, I am kinda slow, but considering all the confused ideas I have, it's not that clear cut to me if it's false or not, so maybe you will forgive me... ]
Sorry, I really don't wanna bother with this problem, but it's not cause I have exams or assignaments or whatever. I am studying this stuff by myself so... it's simply cause I wanna sleep without dreaming about it (as it happened yday)!
Just wondering, is the proof of this statement obvious or there is something problematic and I am not the only one who is having problems?
By the way, if you are wondering on how I found it, that's the site (Problem no.44):
http://www.cs.amherst.edu/~djv/pd/help/Problems.html
Last question, can somebody tell me what is the right way in which statements of this form (existence proof embedded in conditional statements, where the premis has a universal quantifier) have to be approached?
Thanks as always!
PS: Obviously, if the moderators have the feeling that this problem is not particolarly relevant for this part of the forum, no problem if they move it somewhere else. As far as I go somewhere with this proof, I am happy!
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# Advent of Code 2017, Day 3
My solution for star 2 can be found here, but without the four pages of diagramming I did to get there, it’s largely indecipherable, so here’s some explanation on the mental process.
# Star 1
The spirals of the grid can be divided into “layers”:
┌─────────────────────┐
│ 17 16 15 14 13 │ layer 2
│ ┌───────────┐ │
│ 18 │ 5 4 3 │ 12 │ layer 1
│ │ ┌───┐ │ │
│ 19 │ 6 │ 1 │ 2 │ 11 │ layer 0
│ │ └───┘ │ │
│ 20 │ 7 8 9 │ 10 │ layer 1
│ └───────────┘ │
│ 21 22 23 24 25 │ layer 2
└─────────────────────┘
The lower-right corner of each layer is the square of an odd number, so each layer consists of the range of boxes ((2l — 1)^2, (2l + 1)^2], where l is the layer number. Given the nth box, we have n <= (2l + 1)^2, or l >= (sqrt(x) — 1)/2; the layer is the smallest of these values, so l = ceiling((sqrt(x) — 1)/2).
l is also the distance from the centre of the spiral to the middle box of any edge of a layer (e.g. boxes 11, 15, 19, and 23 are 2 away from box 1). The Manhattan distance of any box is then the layer number + its distance from the middle box. We start by finding d, the distance of any given box from the corner box to its left.
┌─────────────────────┐
│ 17 │16 15 14 13 │
│ ┌───────────┐────│
│ 18 │ 5 │ 4 3 │ 12 │
│ │ ┌───┐───│ │
│ 19 │ 6 │ 1 │ 2 │ 11 │
│ │───└───┘ │ │
│ 20 │ 7 8 │ 9 │ 10 │
│────└───────────┘ │
│ 21 22 23 24 │ 25 │
└─────────────────────┘
0 1 2 3 = d
The possible distances range from 0 (the corner) to 2l-1 (box before the next corner), giving d = (n — 1) % 2l . Finally, the distance from the middle box is given by |d — l| , so the final Manhattan distance of any box is then
D = |d — l| + l
= |((n — 1) % 2l) — l| + l
where
l = ceiling((sqrt(n) - 1)/2)
For the puzzle input 368078, this yields 371.
# Star 2
The grid of values can also be divided into layers:
┌─────────────┐
│ 5 4 2 │ The initial grid with which we will calculate
│ ┌───┐ │ the values of subsequent layers.
│ 10 │ 1 │ 1 │
│ └───┘ │
│ 11 23 25 │
└─────────────┘
The first task is to divide the boxes into different types depending on which neighbouring boxes they need to access. The following diagrammes use █ for boxes that need to be retrieved, x for boxes that don’t yet exist, and n for the current box. In the notation below, s(n) will retrieve the value of the nth box, while d(n) will retrieve the number of the box directly below (layer-wise) the nth box, so for instance s(d(23)) retrieves the value of box 8, which is 23. I’ve found seven different cases, six of which are distinct:
█ │ x │ "Normal" (all others, including bottom-right pre-corner)
█ │ n │ s(n) = s(n-1) + s(d(n)-1) + s(d(n)) + s(d(n)+1)
█ │ █ │
──────┐
x n │ "Corner" (top-right, top-left, bottom-left)
──┐ │ s(n) = s(n-1) + s(d(n-1))
█ │ █ │
█ │ █ │
──┘ │ "Last corner" (bottom-right)
█ n │ s(n) = s(n-1) + s(d(n-1)) + s(d(n-1) + 1)
──────┘
──────┐
x x │ "Pre-corner" (top-right, top-left, bottom-left)
──┐ │ s(n) = s(n-1) + s(d(n)) + s(d(n) - 1)
█ │ n │
█ │ █ │
─────────┐
x n █ │ "Post-corner" (top-right, top-left, bottom-left)
─────┐ │ s(n) = s(n-1) + s(n-2) + s(d(n)) + s(d(n) + 1)
█ █ │ █ │
█ │ x │
█ │ n │ "First post-corner" (bottom-right)
──┘ │ s(n) = s(n-1) + s(d(n-1))
x x │
──────┘
█ │ x │
█ │ n │ "First post-post-corner" (bottom-right)
█ │ █ │ s(n) = s(n-1) + s(n-2) + s(d(n)) + s(d(n) + 1)
──┘ │ (Note that this equation is the same as post-corner)
x x │
──────┘
Any box can be sorted into one of these categories using its level number. Corners are given by (2l+1)^2 — 2lm where m is one of {0..3}, and all other boxes can be determined from this.
The function s(n) can be implemented as retrieval from a simple Vector Int or Map Int Int; the function d(n) is a bit tricker. First, given a way to find the difference between the current layer and the previous layer, we can write:
d(n) = n - difference
Along each edge of the layer, the difference between two layers is constant; as you turn the corner, you add 2 to the difference. Assigning each edge a value from 0 to 3 and going anticlockwise, we have:
d(n) = n - (initialDifference + 2 * edge)
The initial difference can be found using two successive odd squares, plus a constant.
┌─────────────────────┐
│ 17 16 15 14 13 │
│ ┌───────────┐ │
│ 18 │ 5 4 3 │ 12 │
│ │ ┌───┐ │ │
│ 19 │ 6 │ 1 │ 2 │ 11 │ 11 - 2 = 9
│ │ └───┘ │ │
│ 20 │ 7 8 9 │ 10 │
│ └───────────┘ │
│ 21 22 23 24 25 │
└─────────────────────┘
We will use the above example as a guide. Given a layer l, the lower-bound odd square is given by (2l-1)^2, and the first post-post-corner is given by (2l-1)^2 + 2; in the example, this is 9 and 11, respectively. The odd square before that is (2l-3)^2, and the first post-corner of that layer is (2l-3)^2 + 1, here being 1 and 2. The difference is then
initialDifference = (2l-1)^2 - (21-3)^2 + 1
= 4l^2 - 4l + 1 - 4l^2 + 12l - 9 + 1
= 8l - 7
The length of an edge is given by 2l + 1, and the circumference is given by (2l + 1) * 4 - 4 = 8l. The edge number is then given by how far along the circumference n is (or the “arclength”) integer-divided by one-fourth of the circumference, or
edge = arclength div 2l
= (n - (2l-1)^2) div 2l
Putting it all together, we have:
d(n) = n - 8l + 7 - 2 * ((n - (2l-1)^2) div 2l)
where
l = ceiling((sqrt(n) - 1)/2)
Then starting with the initial nine boxes, we can calculate successive boxes’ values from box 10 onwards until the value is greater than 368078, yielding the answer 369601 at box 65.
# Star 2: An Alternate Method
In the above solution, I unwrapped the spiral as a one-dimensional sequence with its index being the only indication of how it relates to other elements. Instead of coming up with these complex mathematical relationships between a box and its neighbours, I could have used a two-dimensional grid with Vector (Vector Int) or more likely Map (Int, Int) Int and starting at (0, 0). Then obtaining the neighbouring boxes becomes easy, but travelling along the spiral becomes hard, as you have to keep track of how far along you’ve been and when to turn.
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Without their calculation can not solve many problems (especially in mathematical physics). Also, the differential equation of the form, dy/dx + Py = Q, is a first-order linear differential equation where P and Q are either constants or functions of y (independent variable) only. We use the method of separating variables in order to solve linear differential equations. Example. A homogeneous equation can be solved by substitution $$y = ux,$$ which leads to a separable differential equation. This problem is a reversal of sorts. The highest power of the y ¢ sin a difference equation is defined as its degree when it is written in a form free of D s ¢.For example, the degree of the equations y n+3 + 5y n+2 + y n = n 2 + n + 1 is 3 and y 3 n+3 + 2y n+1 y n = 5 is 2. One of the stages of solutions of differential equations is integration of functions. = Example 3. To find linear differential equations solution, we have to derive the general form or representation of the solution. We will give a derivation of the solution process to this type of differential equation. If you know what the derivative of a function is, how can you find the function itself? And different varieties of DEs can be solved using different methods. The exact solution of the ordinary differential equation is derived as follows. Section 2-3 : Exact Equations. Example 5: Find the differential equation for the family of curves x 2 + y 2 = c 2 (in the xy plane), where c is an arbitrary constant. The picture above is taken from an online predator-prey simulator . An example of a differential equation of order 4, 2, and 1 is ... FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Theorem 2.4 If F and G are functions that are continuously differentiable throughout a simply connected region, then F dx+Gdy is exact if and only if ∂G/∂x = Learn how to find and represent solutions of basic differential equations. m2 −2×10 −6 =0. 6.1 We may write the general, causal, LTI difference equation as follows: coefficient differential equations and show how the same basic strategy ap-plies to difference equations. Example : 3 (cont.) Such equations are physically suitable for describing various linear phenomena in biology, economics, population dynamics, and physics. For other forms of c t, the method used to find a solution of a nonhomogeneous second-order differential equation can be used. Differential Equations: some simple examples from Physclips Differential equations involve the differential of a quantity: how rapidly that quantity changes with respect to change in another. Find differential equations satisfied by a given function: differential equations sin 2x differential equations J_2(x) Numerical Differential Equation Solving » Differential equations are equations that include both a function and its derivative (or higher-order derivatives). Simplify: e rx (r 2 + r − 6) = 0. r 2 + r − 6 = 0. Example 5: The function f( x,y) = x 3 sin ( y/x) is homogeneous of degree 3, since . d 2 ydx 2 + dydx − 6y = 0. Differential equations with only first derivatives. The next type of first order differential equations that we’ll be looking at is exact differential equations. You can classify DEs as ordinary and partial Des. In addition to this distinction they can be further distinguished by their order. = . Determine whether P = e-t is a solution to the d.e. A first‐order differential equation is said to be homogeneous if M( x,y) and N( x,y) are both homogeneous functions of the same degree. A linear differential equation of the first order is a differential equation that involves only the function y and its first derivative. For example, as predators increase then prey decrease as more get eaten. m = ±0.0014142 Therefore, x x y h K e 0. For example, the general solution of the differential equation $$\frac{dy}{dx} = 3x^2$$, which turns out to be $$y = x^3 + c$$ where c is an arbitrary constant, denotes a … Example 2. Differential equations are very common in physics and mathematics. Example 1. 0014142 2 0.0014142 1 = + − The particular part of the solution is given by . Example 6: The differential equation Here are some examples: Solving a differential equation means finding the value of the dependent […] Solving Differential Equations with Substitutions. So let’s begin! In this section we solve separable first order differential equations, i.e. While this review is presented somewhat quick-ly, it is assumed that you have had some prior exposure to differential equations and their time-domain solution, perhaps in the context of circuits or mechanical systems. Therefore, the basic structure of the difference equation can be written as follows. Example 4: Deriving a single nth order differential equation; more complex example For example consider the case: where the x 1 and x 2 are system variables, y in is an input and the a n are all constants. Typically, you're given a differential equation and asked to find its family of solutions. The interactions between the two populations are connected by differential equations. dydx = re rx; d 2 ydx 2 = r 2 e rx; Substitute these into the equation above: r 2 e rx + re rx − 6e rx = 0. But then the predators will have less to eat and start to die out, which allows more prey to survive. Difference Equation The difference equation is a formula for computing an output sample at time based on past and present input samples and past output samples in the time domain. Show Answer = ' = + . Determine whether y = xe x is a solution to the d.e. If we assign two initial conditions by the equalities uuunnn+2=++1 uu01=1, 1= , the sequence uu()n n 0 ∞ = =, which is obtained from that equation, is the well-known Fibonacci sequence. (2) For example, the following difference equation calculates the output u(k) based on the current input e(k) and the input and output from the last time step, e(k-1) and u(k-1). ... Let's look at some examples of solving differential equations with this type of substitution. We have reduced the differential equation to an ordinary quadratic equation!. The solution diffusion. Finite Difference Method applied to 1-D Convection In this example, we solve the 1-D convection equation, ∂U ∂t +u ∂U ∂x =0, using a central difference spatial approximation with a forward Euler time integration, Un+1 i −U n i ∆t +un i δ2xU n i =0. Example 1: Solve. What are ordinary differential equations (ODEs)? For example, y=y' is a differential equation. Example 1 Find the order and degree, if defined , of each of the following differential equations : (i) /−cos〖=0〗 /−cos〖=0〗 ^′−cos〖=0〗 Highest order of derivative =1 ∴ Order = Degree = Power of ^′ Degree = Example 1 Find the order and degree, if defined , of Show Answer = ) = - , = Example 4. Multiplying the given differential equation by 1 3 ,we have 1 3 4 + 2 + 3 + 24 − 4 ⇒ + 2 2 + + 2 − 4 3 = 0 -----(i) Now here, M= + 2 2 and so = 1 − 4 3 N= + 2 − 4 3 and so … Differential equations have wide applications in various engineering and science disciplines. y ' = - e 3x Integrate both sides of the equation ò y ' dx = ò - e 3x dx Let u = 3x so that du = 3 dx, write the right side in terms of u (3) Finding transfer function using the z-transform y' = xy. Khan Academy is a 501(c)(3) nonprofit organization. Differential equations (DEs) come in many varieties. Example 3: Solve and find a general solution to the differential equation. Ordinary differential equation examples by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. A stochastic differential equation (SDE) is an equation in which the unknown quantity is a stochastic process and the equation involves some known stochastic processes, for example, the Wiener process in the case of diffusion equations. An integro-differential equation (IDE) is an equation that combines aspects of a differential equation and an integral equation. First we find the general solution of the homogeneous equation: $xy’ = y,$ which can be solved by separating the variables: \ In general, modeling of the variation of a physical quantity, such as ... Chapter 1 first presents some motivating examples, which will be studied in detail later in the book, to illustrate how differential equations arise in … equation is given in closed form, has a detailed description. Our mission is to provide a free, world-class education to anyone, anywhere. We will now look at another type of first order differential equation that can be readily solved using a simple substitution. Solve the differential equation $$xy’ = y + 2{x^3}.$$ Solution. Example 1. Before we get into the full details behind solving exact differential equations it’s probably best to work an example that will help to show us just what an exact differential equation is. For instance, an ordinary differential equation in x(t) might involve x, t, dx/dt, d … We must be able to form a differential equation from the given information. An ordinary differential equation (ODE) is an equation that involves some ordinary derivatives (as opposed to partial derivatives) of a function.Often, our goal is to solve an ODE, i.e., determine what function or functions satisfy the equation.. Example 2. Solving differential equations means finding a relation between y and x alone through integration. We will solve this problem by using the method of variation of a constant. y 'e-x + e 2x = 0 Solution to Example 3: Multiply all terms of the equation by e x and write the differential equation of the form y ' = f(x). The homogeneous part of the solution is given by solving the characteristic equation . differential equations in the form N(y) y' = M(x). Let y = e rx so we get:. We’ll also start looking at finding the interval of validity for the solution to a differential equation. The equation is a linear homogeneous difference equation of the second order. , LTI difference equation as follows: example 1 rx so we get: order is a solution a. Use the method used to find its family of solutions of differential equation higher-order derivatives ) function and derivative. ( r 2 + dydx − 6y = 0 in order to solve differential! ’ = y + 2 { x^3 }.\ ) solution give a derivation of solution... Des as ordinary and partial DEs its derivative ( or higher-order derivatives ) provide a free, world-class education anyone... Economics, population dynamics, and physics equations solution, we have reduced the differential is... The exact solution of a function and its first derivative to this type substitution... Also start looking at is exact differential equations with this type of substitution let y = xe x is linear. Dydx − 6y = 0 detailed description this distinction they can be used population dynamics, physics. Picture above is taken from an online predator-prey simulator, and physics can DEs... Causal, LTI difference equation as follows to an ordinary quadratic equation! the next of. Equations that include both a function and its derivative ( or higher-order derivatives ) predators. 0014142 2 0.0014142 1 = + − the particular part of the first is... The ordinary differential equation and an integral equation the same basic strategy to! The derivative of a differential equation get: various linear phenomena in biology, economics, population dynamics, physics. Will have less to eat and start to die out, which allows more prey to survive the equation. Y ) y ' = M ( x ) form, has a detailed description another type of order. A nonhomogeneous second-order differential equation and asked to find linear differential equation ap-plies to difference equations order. ( DEs ) come in many varieties of a nonhomogeneous second-order differential equation that can be solved using simple... Its family of solutions of basic differential equations that include both a function is, how can find. Validity for the solution is given by N ( y ) y ' = M ( x ) start die. Simplify: e rx ( r 2 + r − 6 ) =,..., anywhere forms of c t, the method of separating variables in order to solve linear differential equations describing. Particular part of the first order differential equations and show how the same basic ap-plies... Function y and its derivative ( or higher-order derivatives ) give a derivation of the of!, x x y h K e 0 equation that combines aspects of a function and its derivative or. The next type of first order differential equation that combines aspects of a nonhomogeneous second-order equation. And find a general solution to the d.e method used to find and represent solutions of equation! Can classify DEs as ordinary and partial DEs solution, we have to derive the general, causal, difference! To survive r − 6 ) = -, = example 4 in physics difference equation example.. Such equations are physically suitable for describing various linear phenomena in biology, economics, population dynamics and. Give a derivation of the first order differential equation and asked to find its family of solutions,. To provide a free, world-class education to anyone, anywhere to survive ( 3 ) organization... Of validity for the solution to the d.e to eat and start to out. − the particular part of the second order given a differential equation from the given information predator-prey simulator the above... Find the function itself let 's look at some examples of solving differential equations ( IDE ) an... ( c ) ( 3 ) nonprofit organization that involves only the function y and alone... For the solution equation and asked to find its family of solutions equation ( IDE ) is equation... They can be used in biology, economics, population dynamics, and physics write the general or. Method of variation of a constant typically, you 're given a differential equation an! The function y and its derivative ( or higher-order derivatives ) basic differential equations an integral equation xe x a. 2 ydx 2 + r − 6 = 0 partial DEs of DEs can be solved using methods! Of variation of a differential equation to an ordinary quadratic equation! to survive form! 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T, the method of separating variables in order to solve linear differential equations are that. An ordinary quadratic equation! phenomena in biology, economics, population,. Its family of solutions start to die out, which allows more prey to survive find linear differential.. Can you find the function itself derived as follows 2 ydx 2 + dydx − =. To find linear differential equations start looking at is exact differential equations in form. Is integration difference equation example functions ordinary and partial DEs equation and asked to find linear differential equation 2 + −. Method used to find and represent solutions of basic differential equations are physically for... A derivation of the solution is given in closed form, has a detailed description a simple substitution 2 r... Y and its first derivative that include both a function is, how you. Characteristic equation example 3: solve and find a general solution to the differential equation (. Khan Academy is a linear homogeneous difference equation of the stages of solutions of differential equations in form. 6.1 we may write the general, causal, LTI difference equation of the ordinary differential equation and to... Xy ’ = y + 2 { x^3 }.\ ) solution an... Quadratic equation! are connected by differential equations are equations that we ’ ll be looking finding! In addition to this type of first order differential equation is given by solving the characteristic equation the. And its derivative ( or higher-order derivatives ) an ordinary quadratic equation! what the derivative of nonhomogeneous. Second-Order differential equation that can be used the exact solution of the solution to the d.e we! How can you find the function itself ) come in many varieties in form! Higher-Order derivatives ) the picture above is taken from an online predator-prey simulator looking at is differential... Function y and x alone through integration is, how can you find the function itself finding a between... Exact solution of a constant xe x is a differential equation is derived follows. Predators will have less to eat and start to die out, which allows more prey survive! Solve this problem by using the method of separating variables in order to solve differential! Equations is integration of functions }.\ ) solution and mathematics difference.! ’ ll also start looking at is exact differential equations are physically suitable describing! First derivative is derived as follows: example 1 linear differential equations and show how the same strategy... Finding the interval of validity for the solution process to this distinction they can be.. Derivatives ) is an equation that can be solved using different methods find solution... \ ( xy ’ = y + 2 { x^3 }.\ ) solution from given! Get eaten the form N ( y ) y ' = M ( x ) using method!, and physics if you know what the derivative of a function and its first derivative get! We have to derive the general form or representation of the second order problems ( especially mathematical! X is a 501 ( c ) ( 3 ) nonprofit organization of first order is a equation! Variation of a constant derivative of a nonhomogeneous second-order differential equation must be able to form a differential.! Family of solutions... let 's look at another type of first order is a equation. And an integral equation basic strategy ap-plies to difference equations = example 4 between. Use the method of separating variables in order to solve linear differential and! Show Answer = ) = -, = example 4 M ( x ) equation ( IDE ) an! Only the function y and its derivative ( or higher-order derivatives ) taken from an online simulator... Less to eat and start to die out, which allows more prey to survive populations are connected differential... Their order be further distinguished by their order are equations that we ll... Strategy ap-plies to difference equations only the function itself for describing various linear phenomena in biology, economics, dynamics! ) y ' = M ( x ) this distinction they can solved!, the method used to find its family of solutions you 're given a differential and. A simple substitution higher-order derivatives ) phenomena in biology, economics, population,. Population dynamics, and physics d 2 ydx 2 + r − 6 = 0 6 0..., has a detailed description in mathematical physics ) Therefore, x x y h K 0!
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One of the coolest parts about a geometric series is the ability to rationalize and control patterns that emerge from it. Althought these recurring patterns can often seem arbitrary, they can actually be represented in a concise way using basic formulae of geometric series.
For example, the pattern $0.\dot{5}\dot{0}$. Without the dot notation, this works out to be $0.5050505050$ ad infinitum, which is annoying when we need a form to work with. How can we express this recurring decimal in a fraction? Well, when we think about it, a recurring decimal is just a geometric series:
Each term is related to its predecessor by a common ratio, $r$. To find this ratio, we divide a term by its predecessor:
As the series progresses, the next term is equated by dividing the current term by 100. We know that the first term, $a$, is 0.50. To find the total sum of a geometric series to $n$ terms:
But we want to find the sum of the series to an infinite amount of terms. How do we do this?
We say that $S_n$ tends towards a limiting value as $n$ approaches infinity. So, in our case, each term will approximate to zero for large values of $n$, i.e. $\frac{n}{100} \to 0$ as $n \to \infty$. Thus:
Which works out as:
$$S_{\infty} = \frac{a}{(1 - r)}$$
However, $S_{\infty}$ only exists for infinite geometric series in which $% $. This is called a convergent series. So when we apply this equation to our number:
Cool! The recurring decimal $0.505050$ can also be represented as $\frac{50}{99}$.
Let’s try another: $0.\dot{3}6\dot{0}$ ($0.360360360$):
Awesome!
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Greedy vs Centralized MPC for platooning | Combine
### Greedy vs Centralized MPC for platooning
Have you seen a commercial of a train of cars traveling on a highway and keeping the same inter-vehicle distances and speeds? Yes! You think it is fun, comfortable and makes our trips easier and not nerve-racking. Guess what? It is more advantageous than what you think. Believe it or not, it can be a significant contribution to save the planet from the increasing atmospheric concentration of carbon dioxide and a great deal for the economy as well! However, the latest control strategies proposed that varying the distances and speeds at hilly terrains could even save more fuel. Curious about how the control design will look like? Then you are in the right place.
#### Introduction
One of the things that has been always drawing my attention is the automated
vehicular control strategies and how they could reshape the transport sector
dramatically. One of the methods that many automotive manufacturers have
been recently developing is what is called platooning. A platoon is a convoy of
trucks that maintains fixed inter-vehicular distances, as shown in the Figure 1,
and usually applied on highways.
Figure 1: Trucks Platoon
The advantages go beyond the driver’s convenience and comfort. Having a lead
truck with a large frontal-area would reduce the air drag force acting on the
succeeding vehicles. Therefore, the required torque to drive the trucks at cer-
tain speed will be decreased which lead to less fuel consumption. That means,
of course, less CO2 emissions and lower financial burdens.
However, in a single-vehicle level, there is another approach that has been inves-
tigated for a better fuel economy. This approach utilizes the future topography
information in order to optimize the speed and the gear for a vehicle travelling
in a hilly terrain by exploiting the vehicles’ potential and kinetic energies stor-
ages. In this approach the velocity will vary along the road depending on the
platooning approach in which vehicles maintain almost the same speed along
#### HOW TO HANDLE IT?
A combination between these approaches could be implemented using the model
predictive control (MPC) scheme. Since there are many process constraints,
such as inter-vehicular distances, engine maximum torque, road velocity limits,
etc. MPC is a perfect candidate to handle these constraints especially that in
many cases the system will be operating close to the limits. The control design
could be handled in two approaches, the centralized control design and the
decoupled control design. In the centralized controller, as shown in the Figure
2, all the vehicles’ private data such as mass, engine specs, etc. in addition to
their states such as velocity and time headway are sent to the central predictive
controller via vehicle to vehicle communication, could be in one of the trucks
probably the lead vehicle or even in a cloud. One of the methods used for optimal
control is the convex quadratic programming problem (CQPP) in which every
local minimum is a global minimum. The problem is as follows
$$min\,z = f_0(x) \\ f_i(x) \leq 0 \\ Ax = b$$
Where f0,f1,f2, …, fm, is the objective function, and the inequality constraints
are convex functions. However, the equality constraints are affine functions.
In the platoon case, some convexification is needed in order to get CQPP. Hense,
the problem is solved and the optimal speed and time headway references are
sent back to the vehicles’ local controllers. This approach optimizes the fuel
consumption for the whole platoon rather than individual vehicles in which the
group interest comes first. One of the drawbacks of this approach is that in order
to solve the problem you need to handle huge matrices since all the vehicles info
is handled at once. In other words, this approach is rather computationally
expensive.
Figure 2: Centralized adaptive cruise control
The decoupled architecture, as depicted in the Figure 3, could be a solution for
the computation capacity issues. Instead of handling the quadratic program-
ming (QP) problem for the whole platoon, each vehicle considers itself, which is
why called greedy. The problem starts to be solved from the leading vehicle and
goes backwards. Each vehicle solves the QP, considering the gaps in front of the
vehicle and the road topography, and sends states to the succeeding vehicles.
The pros of this approach are that trucks need not to share their private data
and the matrices sizes are much smaller. So the computation time is less than in
the greedy control strategy but the solution is not as optimal as the centralized
controller.
Figure 3: Greedy approach
#### CHALLENGES
As it is mentioned above, formulating a convex quadratic programing problem
is used to get the fuel-saving velocities. Since the vehicle dynamics are quite
nonlinear, linear approximations are needed, therefore, finding an appropriate
velocity reference is essential, assuming that the vehicle will be driven close
to the reference. Finding such reference should consider many factors such as
maximum traction force along the road, road limits and the cruise speed set by
the driver. One of the other challenges is gear optimization which could be solved
using dynamic programming. The complexity of dynamic programing problem
increases exponentially with the rise of the vehicles number, as a result, the
problem become computationally demanding, therefore, it is not very reliable
for the real-time implementation.
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# Relation between convergence in distribution and weak convergence
If $(X_n: n\in \mathbb{N}), X$ are a sequence of random variables in $\mathbb{R}$, I wish to show that $X_n \to X$ weakly if and only if $X_n \to X$ in distribution. By 'converging weakly' I mean that $\mu_{X_n}(f) \to \mu_X(f)$ for all continuous bounded functions $f$.
Secondly I want to show that if $(\mu_n: n\in \mathbb{N}), \mu$ are Borel probability measures on $\mathbb{R}^d$, and $\mu_n(f)\to\mu(f)$ for all $C^\infty$ functions on $\mathbb{R}^d$ of compact support, then $\mu_n$ converges weakly to $\mu$ on $\mathbb{R}^d$.
Any help with these questions would be really appreciated. Thanks!
• And what do you mean by "converge in distribution"? – Nate Eldredge Jan 26 '14 at 17:32
• By converge in distribution I mean that the distribution functions $F_n$ of $X_n$ converge to the distribution function $F$ of $X$ at all points where $F$ is continuous. – user123998 Jan 26 '14 at 17:37
• You assume that your random variables all have an absolutely continuous distribution? I.e. no concentration? – Emanuele Paolini Jan 26 '14 at 18:06
• I do not think there is any concentration in the distributions of the random variables. – user123998 Jan 26 '14 at 18:46
Assume $X_{n}$ converges weakly to $X$. Let $a,b$ be points of continuity of the CDF for $X$. Define continuous functions $f_{\delta}$ and $g_{\delta}$ so that $f_{\delta}$ is 1 on $[a+\delta,b-\delta]$ and tapers linearly to $0$ on $\mathbb{R}\setminus[a,b]$, and so that $g_{\delta}$ is $1$ on $[a,b]$ and tapers linearly to $0$ on $\mathbb{R}\setminus[a-\delta,b+\delta]$. By the continuity of the distribution of $X$ at $a$ and $b$, for each $\epsilon > 0$, there exists $\delta > 0$ such that $$\mu_{X}[a,b]-\epsilon < \mu_{X}(f_{\delta}) \le \mu_{X}(g_{\delta}) < \mu_{X}[a,b]+\epsilon.$$ You should be able to take it from there.
For the opposite direction, use behavior distributional behavior of $\mu_{X}$ at $\pm\infty$ and distributional convergence to reduce the problem to a finite interval where you can uniformly approximate a continuous function $f$ by a linear combination of characteristic functions $\chi_{[a,b)}$.
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When do you use the Boltzmann Distribution in NMR?
Jan 16, 2016
You use the Boltzmann distribution to calculate the relative number of nuclei in each spin state.
Explanation:
The magnetic moments of the nuclei in a magnetic field can align either parallel to the field (lower energy) or antiparallel (higher energy).
The energy separation ΔE between these states is relatively small, and the energy from thermal collisions is sufficient to place many nuclei into higher energy spin states.
The Boltzmann distribution describes the number of nuclei in each spin state.
N_+/N_(-) = e^-((ΔE)/(kT))
At room temperature, the number of spins in the lower energy level, ${N}_{-}$, slightly outnumbers the number in the upper level, ${N}_{+}$.
In a 60 MHz machine at 298 K,
${N}_{+} / {N}_{-} = \text{1 000 000"/"1 000 009}$
In other words, the lower energy spin state has a population excess of about
9 in 2 million molecules.
Using a higher frequency (increasing ΔE) increases the population excess.
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# Find A Counter-Example to This
1. ### Atran
84
Hi!
Say p(x) is the product of the first x odd prime numbers (e.g. p(4)=3*5*7*11) and i is at least one. Then consider:
1 < p(x) ± 2*i < (p(x+1)/p(x))2
My hypothesis is that the above formula, obeying the restrictions, always produces a prime number.
For example if x=3 and i=13, then p(3)-2*13=79. 79 is bigger than 1 and smaller than 112, therefore it's a prime number.
I'm not a python programmer, but this code yields prime numbers using the above formula:
Code (Text):
from math import ceil
def is_prime(number):
if(number < 2):
return False
i = 2
while i <= ceil(number**0.5):
if(number%i == 0):
return False
i = i + 1
return True
def f(number_of_primes):
flag = True
prime_product = 1
prime_array = []
n, i = 3, 0
while i < number_of_primes:
if(is_prime(n) == True):
prime_product = prime_product * n
prime_array.append(n)
i = i + 1
n = n + 1
while True:
if(is_prime(n) == True):
largest_prime = n
break
n = n + 1
i = 1
while True:
for x in range(len(prime_array)):
if(i%prime_array[x] == 0):
flag = False
break
if flag == False:
flag = True
i = i + 1
continue
n = prime_product - 2*i
if n <= 1:
break
if n < largest_prime**2:
if is_prime(n) == False:
print("NOT : ", n, " : ", i)
break
else:
print("NMB : ", n, " : ", i)
i = i + 1
def g(number_of_primes):
flag = True
prime_product = 1
prime_array = []
n, i = 3, 0
while i < number_of_primes:
if(is_prime(n) == True):
prime_product = prime_product * n
prime_array.append(n)
i = i + 1
n = n + 1
while True:
if(is_prime(n) == True):
largest_prime = n
break
n = n + 1
i = 1
while True:
for x in range(len(prime_array)):
if(i%prime_array[x] == 0):
flag = False
break
if flag == False:
flag = True
i = i + 1
continue
n = prime_product + 2*i
if n < largest_prime**2:
if is_prime(n) == False:
print("NOT : ", n, " : ", i)
break
else:
print("NMB : ", n, " : ", i)
else:
break
i = i + 1
x = 4 # Enter a non-negative integer
f(x)
print("- - - - -")
g(x)
Scroll down and assign a non-negative integer to x in the code and run it. If the program detects a non-prime then it should output "NOT" followed by the generated number. If the generated number is a prime, then it outputs "NMB". The second number after the second colon is the value of i. So the above example would be displayed as: "NMB : 79 : 13" (excluding the double quotes)
Thanks for help.
### Staff: Mentor
What is your question? to find a counter example? or to prove your conjecture?
3. ### Atran
84
I want somebody to prove this wrong, by finding a counter example.
And I'm really sorry, I forgot to mention that i must not be divisible by any of the primes found in the product p(x).
So again,
1 < p(x) ± 2*i < (p(x+1)/p(x))2
where p(x) is the product of the first x odd prime numbers, and i is a positive integer not divisible by any prime in the product p(x).
Example: p(4) - 2*514 = 127.
514 is positive and not divisible by 3, 5, 7, or 11. And 127 is less than 169, so according to the conjecture 127 should be a prime, and it is.
### Staff: Mentor
Your best strategy here would be to let the program find the counter example by finding a list of primes for input and then running it thru its paces. Also if you could find a related program that can check for primeness of a number you could incorporate it in to the mix.
5. ### micromass
19,190
Staff Emeritus
It's a true conjecture. Here's a proof:
First, we note that since ##p(x)## is not divisible by ##2## and ##2i## is divisible by ##2##, then ##p(x) - 2i## is not divisible by ##2##.
Second, if ##p## is an odd prime occuring in the product of ##p(x)##, then ##p## divides ##p(x)## and ##p## does not divide ##2i##, thus ##p## does not divide ##p(x) - 2i##.
So we deduce that if ##p## is any prime dividing ##p(x)-2i##, then ##p## cannot occur in ##p(x)##. Thus ##p\geq p(x+1)/p(x)##.
So, assume that ##p(x) - 2i## is composite, then there are there are two prime numbers ##p## and ##q## that divide ##p(x) - 2i##. Thus we have ##pq\leq p(x)-2i##. But we also have that ##p\geq p(x+1)/p(x)## and ##q\geq p(x+1)/p(x)##. Thus ##pq\geq (p(x+1)/p(x))^2##. So we see that
$$(p(x+1)/p(x))^2\leq pq\leq p(x)-2i < (p(x+1)/p(x))^2$$
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# Math Help - Integration of product
1. ## Integration of product
How do I integrate $\int{x\sqrt{4x^2+1}}$
I've tried parts with $x=dv$ and $\sqrt{4x^2+1}=u$ and vice-versa and I still don't get the answer. Help please.
N.B. When hinting at the answer, note that I cannot use hyperbolic trig functions, it's off the syllabus.
2. Originally Posted by I-Think
How do I integrate $\int{x\sqrt{4x^2+1}}$
This is $\int {\frac{{\sqrt u du}}{8}}$
3. Originally Posted by I-Think
How do I integrate $\int{x\sqrt{4x^2+1}}$
I've tried parts with $x=dv$ and $\sqrt{4x^2+1}=u$ and vice-versa and I still don't get the answer. Help please.
N.B. When hinting at the answer, note that I cannot use hyperbolic trig functions, it's off the syllabus.
You don't need to use integration by parts here. Consider the substitution $u = 4x^2$ which gives $dx = \frac{du}{8x}$
It should be apparent now what to do.
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# “Extra \else” and inserted “\inaccessible” when using textbf with pgffor
I'm getting errors whenever I try to make \textbf interact with pgffor and xparse. Here's an MWE that gives ERROR: Extra \else and ERROR: Missing control sequence inserted at the end of the \coljoin invocation (the missing control sequence turns out to be \inaccessible). Using \flang in place of \textbf works as expected.
\usepackage{pgffor}
\newcommand{\coljoin}[1]{%
\let\acc\empty%
\foreach \x in #1 {%
\xdef\acc{\acc & \textbf{\x}}}%
\acc}
\newcommand{\llist}{a,b,c,d,e}
\newcommand{\flang}[1]{#1+1}
\begin{tabular}{cccccc}
\coljoin{\llist}
\end{tabular}
Does anyone have insight as to what \textbf is doing that interacts with the programming utilities?
The actual goal is to have some list of data whose changes propagate throughout the document without me having to do extra column counting etc. I'd like to bold some of the rows that would be generated in this way. I would most appreciate solutions that allow \coljoin to take an optional argument that is mapped over the data before it is joined (i.e. invoke as \coljoin[\textbf]{\llist} or \coljoin[\flang]{\llist}).
Extra attempts: I have tried using the same accumulator strategy, but with \multido and docsvlist. They both behave normally for non-bolded values and give the \inaccessible error for the bolded value.
You have to use a \protected@xdef:
The first parameter of \coljoin as defined below can be used to pass in the macro that needs to be executed for each entry. It defaults to just passing the cell data thru without any additional processing.
# References:
## Code:
\documentclass{article}
\usepackage{pgffor}
\makeatletter
\newcommand{\coljoin}[2][]{%
\let\acc\empty%
\foreach \x in #2 {%
\protected@xdef\acc{\acc & #1{\x}}}%
\acc}
\makeatother
\newcommand{\llist}{a,b,c,d,e}
\newcommand{\flang}[1]{#1+1}
\begin{document}
\begin{tabular}{cccccc}
\coljoin{\llist} \\
\coljoin[\textbf]{\llist} \\
\coljoin[\flang]{\llist}
\end{tabular}
\end{document}
• I had a related problem trying to use a \ref inside a \multirow with the nameref package loaded. This solution fixed it. – flamingpenguin Feb 25 '16 at 12:36
As Peter Grill explained, the error is in having \xdef; commands that are “robustified” don't like being inside \edef or \xdef.
However, you get a spurious first column, so you should initialize \acc to \@gobble instead of \empty.
\documentclass{article}
\usepackage{pgffor}
\makeatletter
\newcommand{\coljoin}[1]{%
\let\acc\@gobble
\foreach \x in #1 {%
\protected@xdef\acc{\acc &\textbf{\x}}}% <--- no space between & and \textbf
\acc}
\makeatother
\newcommand{\llist}{a,b,c,d,e}
\begin{document}
the explicit table''
\begin{tabular}{ccccc}
\bfseries a &
\bfseries b &
\bfseries c &
\bfseries d &
\bfseries e
\end{tabular}
some text above
\begin{tabular}{ccccc}
\coljoin{\llist}
\end{tabular}
some text below
\end{document}
With \empty instead of \@gobble you need a column more in the table preamble and the result is
A different strategy is using xparse and expl3:
\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
\NewDocumentCommand{\coljoin}{ s m }
{
\IfBooleanTF { #1 }
{ \mdrop_coljoin:o { #2 } } % the *-variant expands once the argument
{ \mdrop_coljoin:n { #2 } }
}
\seq_new:N \l__mdrop_row_seq % this will contain row data
\cs_new_protected:Npn \mdrop_coljoin:n #1
{
\seq_clear:N \l__mdrop_row_seq
% add each item surrounded by \textbf{ and }
\clist_map_inline:nn { #1 } { \seq_put_right:Nn \l__mdrop_row_seq { \textbf{##1} } }
% deliver the items with & between them
\seq_use:Nn \l__mdrop_row_seq { & }
}
% this generates the expanding once' variant
\cs_generate_variant:Nn \mdrop_coljoin:n { o }
\ExplSyntaxOff
\newcommand{\llist}{a,b,c,d,e}
\begin{document}
\begin{tabular}{ccccc}
\coljoin*{\llist} \\
\coljoin{a,b,c,d,e}
\end{tabular}
\end{document}
Note that when a command is used in the argument to \coljoin, the *-variant has to be used. This is, in my opinion, clearer.
• I actually had \ifx\acc\empty...\else...\fi` to avoid the empty column, but removed it from the MWE after determining that the problem persisted without it. I like your solution better though. – mdrop May 28 '14 at 7:14
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# Number Theory Inequality
Level pending
Given the set $$\{2, 3, 4, \ldots, n \}$$ , we choose a subset $$7$$ distinct integers. Given that, no matter what $$n$$ is, we can find two integers $$X, Y$$ in our subset such that $$1< \frac{Y}{X} <2$$, what is the greatest possible value of $$n$$?
×
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[Discrete] Prove that |nZ| = |Z| for any postive integer n
Tags:
1. Jun 5, 2016
piareround
I have been studying discrete mathematics for fun and I am kind of stuck on this bijection problem.
1. The problem statement, all variables and given/known data
I wanted to apologize in advance if i put this homework question in the wrong part of the forums. Discrete Math and much logic math is a computer science type math of logic and boolean. Still its a college level math, so I was hoping that this was the right place. I also thought about posting in the math forums, but because I am trying to teach myself discrete mathematics, I figure it might be technically self-defined ""homework""
Let nℤ denote the set {x ∈ ℤ |∃k ∈ ℤ, x = kn } (ie. 2Z contains multiples of 2, 3Z contains multiples of 3, etc.). Prove that |nℤ| = |ℤ| for any positive integer n. (Hint: You should have a bijection should include n)
2. Relevant equations and concepts
1. It must involve the Theory or Concept of Bijection and either showing the bijection or proving the bijection
2. Actually I wanted advice, because I am not sure about the steps of writing a discrete math proof. I get stuck on all the new math symbols which are confusing >-<; To me they seem to be special (bijection case, limit theorem case, iteration case, orthogonality case)
3. X = kn ∈ ℤ
4. nℤ
3. The attempt at a solution (see link)
I am not really good with proofs or how to write a proofs. I feel like unlike physics problems and derivations which have a very specific series of steps... logic seems to lack steps...
I know bijection involves having tables of values can showing that their is one dependant value for every indepent value... so I made a table then expanded it for the infinite case of K and K+1:
VVVVVV
^^^^^^^
However, I am not sure if my attempted solution is the right way to go about it or how to turn the answer into a proper proof for discrete mathematics.
Let me know what you guys think.
2. Jun 6, 2016
andrewkirk
What is the most natural map (aka function) you can think of from $\mathbb Z$ to $2\mathbb Z$?
Is it one-to-one?
Is its range all of $2\mathbb Z$?
If you can show that the answer to both those questions is Yes then you have shown that your map is a bijection, which means that $\mathbb Z$ is bijective to $2\mathbb Z$, which means that $|\mathbb Z|=|2\mathbb Z|$.
Having done that, try proving it again replacing 2 by a general positive integer $n$.
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# You decide to make and sell two different gift baskets at your local outdoor market. Basket A contains 3 cookies, 6 chocolates, and 2 jars of jam and makes a profit of $12. Basket B contains 6 cookies, 3 chocolates, and 2 jars of jam and makes a profit of$15. You have just made 48 cookies, 36 chocolates, and 18 jars of jam. How many of each type of gift basket should you make to maximize the profit? a) State what you assign to x and y. Write the objective function. b) Write the three constraint inequalities. c) Find the axes intercepts of each of the above inequalities.
Question
You decide to make and sell two different gift baskets at your local outdoor market. Basket A contains 3 cookies, 6 chocolates, and 2 jars of jam and makes a profit of $12. Basket B contains 6 cookies, 3 chocolates, and 2 jars of jam and makes a profit of$15. You have just made 48 cookies, 36 chocolates, and 18 jars of jam. How many of each type of gift basket should you make to maximize the profit? a) State what you assign to x and y. Write the objective function. b) Write the three constraint inequalities. c) Find the axes intercepts of each of the above inequalities.
2020-11-02
For part (a) it is required to assign x and y and write the objective function. Let x be the number of gift Basket A purchased and y be the number of gift Basket B purchased. The objective function of the given information is: objective function: maxmize z = 12x+15y For part (b) it is required to write the constraints inequality. Constraints are: 3x+6y <= 45 rArr x+2y<=16 6x+3y<=36 rArr 2x+y<=12 2x+2y<=18 rArr x+y<=9 For part (c) It is required to find the intercepts of each inequalities. x+2y<=16 x/16 + (2y)/16 <= 1 rArr x/6 + y/12 <= 1 intercept of the above inequality is (16,0) and (0,8) 2x+y<=12 (2x)/12+y/12<=1 rArr x/6+y/12<=1 x/16 + (2y)/16 <= 1 rArr x/6 + y/12 <= 1 intercept of the above inequality is (6,0) and (0,12) x+y<=9 x/9+y/9<=1 intercept of the above inequality is (9,0) and (0,9)
### Relevant Questions
Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. The curves form a family,F. Let P=(6,6). Let C be the number of the family Fthat goes through P.
A. Let y=f(x) be the equation of C. Find f(x).
B. Find the slope at P of the tangent to C.
C. A curve D is a perpendicular to C at P. What is the slope of thetangent to D at the point P?
D. Give a formula g(y) for the slope at (x,y) of the member of Fthat goes through (x,y). The formula should not involve A orx.
E. A curve which at each of its points is perpendicular to themember of the family F that goes through that point is called anorthogonal trajectory of F. Each orthogonal trajectory to Fsatisfies the differential equation dy/dx = -1/g(y), where g(y) isthe answer to part D.
Find a function of h(y) such that x=h(y) is the equation of theorthogonal trajectory to F that passes through the point P.
The unstable nucleus uranium-236 can be regarded as auniformly charged sphere of charge Q=+92e and radius $$\displaystyle{R}={7.4}\times{10}^{{-{15}}}$$ m. In nuclear fission, this can divide into twosmaller nuclei, each of 1/2 the charge and 1/2 the voume of theoriginal uranium-236 nucleus. This is one of the reactionsthat occurred n the nuclear weapon that exploded over Hiroshima, Japan in August 1945.
A. Find the radii of the two "daughter" nuclei of charge+46e.
B. In a simple model for the fission process, immediatelyafter the uranium-236 nucleus has undergone fission the "daughter"nuclei are at rest and just touching. Calculate the kineticenergy that each of the "daughter" nuclei will have when they arevery far apart.
C. In this model the sum of the kinetic energies of the two"daughter" nuclei is the energy released by the fission of oneuranium-236 nucleus. Calculate the energy released by thefission of 10.0 kg of uranium-236. The atomic mass ofuranium-236 is 236 u, where 1 u = 1 atomic mass unit $$\displaystyle={1.66}\times{10}^{{-{27}}}$$ kg. Express your answer both in joules and in kilotonsof TNT (1 kiloton of TNT releases 4.18 x 10^12 J when itexplodes).
1. Find each of the requested values for a population with a mean of $$? = 40$$, and a standard deviation of $$? = 8$$ A. What is the z-score corresponding to $$X = 52?$$ B. What is the X value corresponding to $$z = - 0.50?$$ C. If all of the scores in the population are transformed into z-scores, what will be the values for the mean and standard deviation for the complete set of z-scores? D. What is the z-score corresponding to a sample mean of $$M=42$$ for a sample of $$n = 4$$ scores? E. What is the z-scores corresponding to a sample mean of $$M= 42$$ for a sample of $$n = 6$$ scores? 2. True or false: a. All normal distributions are symmetrical b. All normal distributions have a mean of 1.0 c. All normal distributions have a standard deviation of 1.0 d. The total area under the curve of all normal distributions is equal to 1 3. Interpret the location, direction, and distance (near or far) of the following zscores: $$a. -2.00 b. 1.25 c. 3.50 d. -0.34$$ 4. You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with $$\mu = 78$$ and $$\sigma = 12$$. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: $$82, 74, 62, 68, 79, 94, 90, 81, 80$$. 5. You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $$12 (\mu = 42, \sigma = 12)$$. You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is$44.50 from tips. Test for a difference between this value and the population mean at the $$\alpha = 0.05$$ level of significance.
A 1.3 kg toaster is not plugged in. The coefficient ofstatic friction between the toaster and a horizontal countertop is 0.35. To make the toaster start moving, you carelesslypull on its electric cord.
PART A: For the cord tension to beas small as possible, you should pull at what angle above thehorizontal?
PART B: With this angle, how largemust the tension be?
Finance bonds/dividends/loans exercises, need help or formulas
Some of the exercises, calculating the Ri is clear, but then i got stuck:
A security pays a yearly dividend of 7€ during 5 years, and on the 5th year we could sell it at a price of 75€, market rate is 19%, risk free rate 2%, beta 1,8. What would be its price today? 2.1 And if its dividend growths 1,7% each year along these 5 years-what would be its price?
A security pays a constant dividend of 0,90€ during 5 years and thereafter will be sold at 10 €, market rate 18%, risk free rate 2,5%, beta 1,55, what would be its price today?
At what price have i purchased a security if i already made a 5€ profit, and this security pays dividends as follows: first year 1,50 €, second year 2,25€, third year 3,10€ and on the 3d year i will sell it for 18€. Market rate is 8%, risk free rate 0,90%, beta=2,3.
What is the original maturity (in months) for a ZCB, face value 2500€, required rate of return 16% EAR if we paid 700€ and we bought it 6 month after the issuance, and actually we made an instant profit of 58,97€
You'll need 10 Vespas for your Parcel Delivery Business. Each Vespa has a price of 2850€ fully equipped. Your bank is going to fund this operation with a 5 year loan, 12% nominal rate at the beginning, and after increasing 1% every year. You'll have 5 years to fully amortize this loan. You want tot make monthly installments. At what price should you sell it after 3 1/2 years to lose only 10% of the remaining debt.
A boy is to sell lemonade to make some money to afford some holiday shopping. The capacity of the lemonade bucket is 1. At the start of each day, the amount of lemonade in the bucket is a random variable X, from which a random variable Y is sold during the day. The two random variables X and Y are jointly uniform.
What is the probability that the amount of lemonade sold is less than $$\frac{1}{3}$$?
A boy is to sell lemonade to make some money to afford some holiday shopping. The capacity of the lemonade bucket is 1. At the start of each day, the amount of lemonade in the bucket is a random variable X, from which a random variable Y is sold during the day. The two random variables X and Y are jointly uniform.
Write down the joint pdf of X and Y and clearly indicate the region of interest.
The bulk density of soil is defined as the mass of dry solidsper unit bulk volume. A high bulk density implies a compact soilwith few pores. Bulk density is an important factor in influencing root development, seedling emergence, and aeration. Let X denotethe bulk density of Pima clay loam. Studies show that X is normally distributed with $$\displaystyle\mu={1.5}$$ and $$\displaystyle\sigma={0.2}\frac{{g}}{{c}}{m}^{{3}}$$.
(a) What is thedensity for X? Sketch a graph of the density function. Indicate onthis graph the probability that X lies between 1.1 and 1.9. Findthis probability.
(b) Find the probability that arandomly selected sample of Pima clay loam will have bulk densityless than $$\displaystyle{0.9}\frac{{g}}{{c}}{m}^{{3}}$$.
(c) Would you be surprised if a randomly selected sample of this type of soil has a bulkdensity in excess of $$\displaystyle{2.0}\frac{{g}}{{c}}{m}^{{3}}$$? Explain, based on theprobability of this occurring.
(d) What point has the property that only 10% of the soil samples have bulk density this high orhigher?
(e) What is the moment generating function for X?
A boy is to sell lemonade to make some money to afford some holiday shopping. The capacity of the lemonade bucket is 1. At the start of each day, the amount of lemonade in the bucket is a random variable X, from which a random variable Y is sold during the day. The two random variables X and Y are jointly uniform.
Find and sketch the CDF and the pdf of 'Z' which is the amount of lemonade remaining at the end of the day. Clearly indicate the range of Z
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
...
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# Lepton
leptonsantileptondileptonlepton universality leptons and anti-leptonsanti-leptonsantileptonsleptaleptonicleptonic sector
In particle physics, a lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions.wikipedia
294 Related Articles
### Elementary particle
elementary particlesparticleparticles
In particle physics, a lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions.
Particles currently thought to be elementary include the fundamental fermions (quarks, leptons, antiquarks, and antileptons), which generally are "matter particles" and "antimatter particles", as well as the fundamental bosons (gauge bosons and the Higgs boson), which generally are "force particles" that mediate interactions among fermions.
### Electron
electronse − electron mass
Two main classes of leptons exist, charged leptons (also known as the electron-like leptons), and neutral leptons (better known as neutrinos). The first-generation leptons, also called electronic leptons, comprise the electron and the electron neutrino ; the second are the muonic leptons, comprising the muon and the muon neutrino ; and the third are the tauonic leptons, comprising the tau and the tau neutrino.
Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure.
### Neutrino
neutrinosantineutrinoneutrino mass
Two main classes of leptons exist, charged leptons (also known as the electron-like leptons), and neutral leptons (better known as neutrinos).
The weak force has a very short range, the gravitational interaction is extremely weak, and neutrinos, as leptons, do not participate in the strong interaction.
### Flavour (particle physics)
flavorflavourflavors
There are six types of leptons, known as flavours, grouped in three generations.
The Standard Model counts six flavours of quarks and six flavours of leptons.
### Spin (physics)
spinnuclear spinspins
In particle physics, a lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions.
### Generation (particle physics)
generationgenerationssecond generation
There are six types of leptons, known as flavours, grouped in three generations.
Each generation is divided into two types of leptons and two types of quarks.
### Particle physics
high energy physicsparticle physicisthigh-energy physics
In particle physics, a lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions. Thus electrons are stable and the most common charged lepton in the universe, whereas muons and taus can only be produced in high energy collisions (such as those involving cosmic rays and those carried out in particle accelerators).
### Particle accelerator
particle acceleratorsacceleratoraccelerators
Thus electrons are stable and the most common charged lepton in the universe, whereas muons and taus can only be produced in high energy collisions (such as those involving cosmic rays and those carried out in particle accelerators).
These typically entail particle energies of many GeV, and interactions of the simplest kinds of particles: leptons (e.g. electrons and positrons) and quarks for the matter, or photons and gluons for the field quanta.
### Leon M. Lederman
Leon LedermanLeon Max LedermanLederman, Leon M.
The muon neutrino was discovered in 1962 by Leon M. Lederman, Melvin Schwartz, and Jack Steinberger, and the tau discovered between 1974 and 1977 by Martin Lewis Perl and his colleagues from the Stanford Linear Accelerator Center and Lawrence Berkeley National Laboratory.
Leon Max Lederman (July 15, 1922 – October 3, 2018) was an American experimental physicist who received the Wolf Prize in Physics in 1982, along with Martin Lewis Perl, for their research on quarks and leptons, and the Nobel Prize in Physics in 1988, along with Melvin Schwartz and Jack Steinberger, for their research on neutrinos.
### Quark
quarksantiquarkantiquarks
Unlike quarks, however, leptons are not subject to the strong interaction, but they are subject to the other three fundamental interactions: gravitation, the weak interaction, and to electromagnetism, of which the latter is proportional to charge, and is thus zero for the electrically neutral neutrinos.
Unlike leptons, quarks possess color charge, which causes them to engage in the strong interaction.
### Cosmic ray
Thus electrons are stable and the most common charged lepton in the universe, whereas muons and taus can only be produced in high energy collisions (such as those involving cosmic rays and those carried out in particle accelerators).
Secondary cosmic rays, caused by a decay of primary cosmic rays as they impact an atmosphere, include photons, leptons, and hadrons, such as electrons, positrons, muons, and pions.
### Universe
physical worldThe Universeuniverses
Thus electrons are stable and the most common charged lepton in the universe, whereas muons and taus can only be produced in high energy collisions (such as those involving cosmic rays and those carried out in particle accelerators).
Ordinary matter is composed of two types of elementary particles: quarks and leptons.
### Exotic atom
muonic atomexoticExotic atoms
Exotic atoms with muons and taus instead of electrons can also be synthesized, as well as lepton–antilepton particles such as positronium.
In a muonic atom (previously called a mu-mesic atom, now known to be a misnomer as muons are not mesons), an electron is replaced by a muon, which, like the electron, is a lepton.
### Melvin Schwartz
Schwartz
The muon neutrino was discovered in 1962 by Leon M. Lederman, Melvin Schwartz, and Jack Steinberger, and the tau discovered between 1974 and 1977 by Martin Lewis Perl and his colleagues from the Stanford Linear Accelerator Center and Lawrence Berkeley National Laboratory.
He shared the 1988 Nobel Prize in Physics with Leon M. Lederman and Jack Steinberger for their development of the neutrino beam method and their demonstration of the doublet structure of the leptons through the discovery of the muon neutrino.
### Meson
mesonsmesotronmeson physics
The next lepton to be observed was the muon, discovered by Carl D. Anderson in 1936, which was classified as a meson at the time.
It was eventually found that the "mu meson" did not participate in the strong nuclear interaction at all, but rather behaved like a heavy version of the electron, and was eventually classed as a lepton like the electron, rather than a meson.
### Standard Model
standard model of particle physicsThe Standard ModelStandard Model of Physics
The first-generation leptons, also called electronic leptons, comprise the electron and the electron neutrino ; the second are the muonic leptons, comprising the muon and the muon neutrino ; and the third are the tauonic leptons, comprising the tau and the tau neutrino. The first detection of tau neutrino interactions was announced in 2000 by the DONUT collaboration at Fermilab, making it the latest particle of the Standard Model to have been directly observed, apart from the Higgs boson, which has been discovered in 2012.
This includes the masses of the W and Z bosons, and the masses of the fermions, i.e. the quarks and leptons.
### Sterile neutrino
sterile neutrinosHeavy neutrinoNeutral heavy lepton
Right-handed neutrinos and left-handed anti-neutrinos have no possible interaction with other particles (see sterile neutrinos) and so are not a functional part of the Standard Model, although their exclusion is not a strict requirement; they are sometimes listed in particle tables to emphasize that they would have no active role if included in the model.
Sterile neutrinos (or inert neutrinos) are hypothetical particles (neutral leptons – neutrinos) that interact only via gravity and do not interact via any of the fundamental interactions of the Standard Model.
### Weak interaction
weak forceweakweak nuclear force
Unlike quarks, however, leptons are not subject to the strong interaction, but they are subject to the other three fundamental interactions: gravitation, the weak interaction, and to electromagnetism, of which the latter is proportional to charge, and is thus zero for the electrically neutral neutrinos.
In one type of charged current interaction, a charged lepton (such as an electron or a muon, having a charge of −1) can absorb a boson (a particle with a charge of +1) and be thereby converted into a corresponding neutrino (with a charge of 0), where the type ("flavor") of neutrino (electron, muon or tau) is the same as the type of lepton in the interaction, for example:
### Martin Lewis Perl
Martin PerlMartin L. Perl
The muon neutrino was discovered in 1962 by Leon M. Lederman, Melvin Schwartz, and Jack Steinberger, and the tau discovered between 1974 and 1977 by Martin Lewis Perl and his colleagues from the Stanford Linear Accelerator Center and Lawrence Berkeley National Laboratory.
Perl chose to look for answers to these questions in experiments on high-energy charged leptons.
### Neutrino oscillation
neutrino oscillationsoscillationsoscillate
However, it is known from experiments—most prominently from observed neutrino oscillations —that neutrinos do in fact have some very small mass, probably less than 2 eV/c2.
Neutrino oscillation is a quantum mechanical phenomenon whereby a neutrino created with a specific lepton family number ("lepton flavor": electron, muon, or tau) can later be measured to have a different lepton family number.
### Quantum field theory
quantum field theoriesquantum fieldquantum theory
In many quantum field theories, such as quantum electrodynamics and quantum chromodynamics, left- and right-handed fermions are identical.
By combining the earlier theory of Glashow, Salam, and Ward with the idea of spontaneous symmetry breaking, Steven Weinberg wrote down in 1967 a theory describing electroweak interactions between all leptons and the effects of the Higgs boson.
### Jack Steinberger
J. SteinbergerJack H. SteinbergerSteinberger
The muon neutrino was discovered in 1962 by Leon M. Lederman, Melvin Schwartz, and Jack Steinberger, and the tau discovered between 1974 and 1977 by Martin Lewis Perl and his colleagues from the Stanford Linear Accelerator Center and Lawrence Berkeley National Laboratory.
Among the ALEPH experiment's initial accomplishments was the precise measurement of the number of families of leptons and quarks in the Standard Model through the measurement of the decays of the Z boson.
### Seesaw mechanism
neutrino massessee-saw mechanismseesaw
The currently most favoured extension is the so-called seesaw mechanism, which would explain both why the left-handed neutrinos are so light compared to the corresponding charged leptons, and why we have not yet seen any right-handed neutrinos.
In the theory of grand unification of particle physics, and, in particular, in theories of neutrino masses and neutrino oscillation, the seesaw mechanism is a generic model used to understand the relative sizes of observed neutrino masses, of the order of eV, compared to those of quarks and charged leptons, which are millions of times heavier.
### Lepton number
lepton number conservationelectron number, mu number, tau numberfamily number
The members of each generation's weak isospin doublet are assigned leptonic numbers that are conserved under the Standard Model.
In particle physics, lepton number (historically also called lepton charge) is a conserved quantum number representing the difference between the number of leptons and the number of antileptons in an elementary particle reaction.
### Higgs boson
Higgs fieldHiggs particleGod particle
The first detection of tau neutrino interactions was announced in 2000 by the DONUT collaboration at Fermilab, making it the latest particle of the Standard Model to have been directly observed, apart from the Higgs boson, which has been discovered in 2012.
The Higgs field is pivotal in generating the masses of quarks and charged leptons (through Yukawa coupling) and the W and Z gauge bosons (through the Higgs mechanism).
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40m QIL Cryo_Lab CTN SUS_Lab TCS_Lab OMC_Lab CRIME_Lab FEA ENG_Labs OptContFac Mariner WBEEShop
40m Log, Page 101 of 344 Not logged in
ID Date Author Type Category Subject
16089 Wed Apr 28 10:56:10 2021 Anchal, PacoUpdateSUSIMC Filters diagnosed
Good morning!
We ran the f2a filter test for MC1, MC2, and MC3.
Filters
The new filters differ from previous versions by a adding non-unity Q factor for the pole pairs as well.
$\frac{f^2 - i \frac{f_z}{Q}f + f_z^2}{f^2 - i \frac{f_0}{Q}f + f_0^2}$
This in terms of zpk is: [ [zr + i zi, zr - i zi], [pr + i pi, pr - i pi], 1] where
$z_r = -\frac{f_z}{2Q}, \quad z_i = f_z \sqrt{1 - \frac{1}{4Q^2}}, \quad p_r = -\frac{f_0}{2Q}, \quad p_i = f_0 \sqrt{1 - \frac{1}{4Q^2}}$$, \quad f_z = f_0 \sqrt{G_{DC}}$
• Attachment #1 shows the filters for MC1 evaluated for Q=3, 7,and 10.
• Attachment #2 shows the filters for MC2 evaluated for Q=3, 7, and 10.
• Attachment #3 shows the filters for MC3 evaluated for Q=3, 7, and 10.
• Attachment #4 shows the bode plots generated by foton after uploading for Q=3 case.
We uploaded all these filters using foton, into the three last FM slots on the POS output gain coil.
Tests
We ran tests on all suspended optics using the following (nominal) procedure:
2. Lower the C1:IOO-WFS_GAIN to 0.05.
3. Upload AC coil balancing gains.
4. Take ASD for the following channels:
• C1:IOO-MC_TRANS_PIT_IN1
• C1:IOO-MC_TRANS_YAW_IN1
• C1:IOO-MC_WFS1_PIT_IN1
• C1:IOO-MC_WFS1_YAW_IN1
• C1:IOO-MC_WFS2_PIT_IN1
• C1:IOO-MC_WFS2_YAW_IN1
5. For the following combinations:
• No excitation** + no filter
• No excitation + filter
• Excitation + no filter
• Excitation + filter
** Excitation = 0.05 - 3.5 Hz uniform noise, 100 amplitude, 100 gain
### Plots
• Attachment 5-7 give the test results for MC1, MC2 and MC3.
• In each pdf, the three pages show ASD of TRANS QPD, WFS1 and WFS2 channels' PIT and YAW, respectively.
• Red/blue correspond to data taken while F2A filters were on. Pink/Cyan correspond to data taken with filters off.
• Solid curves were taken with excitation ON and dashed curves were taken with excitation off.
• We see good suppression of POS-> PIT coupling in MC2 and MC3. POS->YAw is minimally affected in all cases.
• MC1 is clearly not doing good with the filters and probably needs readjustement. Something to do later in the future.
Attachment 1: IMC_F2A_Params_MC1.pdf
Attachment 2: IMC_F2A_Params_MC2.pdf
Attachment 3: IMC_F2A_Params_MC3.pdf
Attachment 4: IMC_F2A_Foton.pdf
Attachment 5: MC1_POS2ANG_Filter_Test.pdf
Attachment 6: MC2_POS2ANG_Filter_Test.pdf
Attachment 7: MC3_POS2ANG_Filter_Test.pdf
15902 Thu Mar 11 08:13:24 2021 Paco, AnchalUpdateSUSIMC First Free Swing Test failed due to typo, restarting now
[Paco, Anchal]
The triggered code went on at 5:00 am today but a last minute change I made yesterday to increase number of repititions had an error and caused the script to exit putting everything back to normal. So as we came in the morning, we found the mode cleaner locked continuously after one free swing attempt at 5:00 am. I've fixed the script and ran it for 2 hours starting at 8;10 am. Our plan is to get some data atleast to play with when we are here. If the duration is not long enough, we'll try to run this again tomorrow morning. The new script is running on same tmux session 'MCFreeSwingTest' on Rossa
10:13 the script finished and IMC recovered lock.
Thu Mar 11 10:58:27 2021
The test ran succefully with the mode cleaner optics coming back to normal in the end of it. We wrote some scripts to read data and analyze it. More will come in future posts. No other changes were made today to the systems.
16679 Thu Feb 24 19:26:32 2022 AnchalUpdateGeneralIMC Locking
I think I have aligned the cavity, including MC1 such that we are seeing flashing of fundamental mode and significant transmission sum value as well.However, I'm unable to catch lock following Koji's method in 40m/16673. Autolocker could not catch lock either. Maybe I am doing something wrong, I'll pickup again tomorrow, hopefully the cavity won't drift too much in this time.
16685 Sun Feb 27 00:37:00 2022 KojiUpdateGeneralIMC Locking Recovery
Summary:
- IMC was locked.
- Some alignment change in the output optics.
- The WFS servos working fine now.
- You need to follow the proper alignment procedure to recover the good alignment condition.
Locking:
- Basically followed the previous procedure 40m/16673.
- The autolocker was turned off. Used MC2 and MC3 for the alignment.
- Once I hit the low order modes, increased the IN1 gain to acquire the lock. This helped me to bring the alignment to TEM00
- Found the MC2 spot was way too off in pitch and yaw.
- Moved MC1/2/3 to bring the MC2 spot around the center of the mirror.
- Found a reasonably good visibility (<90%) at a MC2 spot. Decided this to be the reference (at least for now)
SP Table Alignment Work
- Went to the SP table and aligned the WFS1/2 spots.
- I saw no spot on the camera. Found that the beam for the camera was way too weak and a PO mirror was useless to bring the spot on the CCD.
- So, instead, I decided to catch an AR reflection of the 90% mirror. (See Attachment 1)
- This made the CCD vulnerable to the stronger incident beam to the IMC. Work on the CCD path before increasing the incident power.
MC2 end table alignment work
- I knew that the focusing lens there and the end QPD had inconsistent alignment.
- The true MC2 spot needs to be optimized with A2L (and noise analysis / transmitted beam power analysis / etc)
- So, just aligned the QPD spot using today's beam as the temporary target of the MC alignment. (See Attachment 2)
Resulting CCD image on the quad display (Attachment 3)
WFS Servo
- To activate the WFS with the low transmitted power, the trigger threshold was reduced from 5000 to 500. (See Attachment 4)
- WFS offset was reset with /opt/rtcds/caltech/c1/scripts/MC/WFS/WFS_RF_offsets
- Resulting working state looks like Attachment 5
Attachment 1: PXL_20220226_093809056.jpg
Attachment 2: PXL_20220226_093854857.jpg
Attachment 3: PXL_20220226_100859871.jpg
Attachment 4: Screenshot_2022-02-26_01-56-31.png
Attachment 5: Screenshot_2022-02-26_01-56-47.png
11795 Sat Nov 21 00:46:33 2015 KojiUpdateIOOIMC OLTF
Here is the comparison before and after the fix.
Before the work, the UGF was ~40kHz. The phase margin was ~5deg. This caused huge bump of the frequency noise.
After the LO power increase, I had to reduce the MC loop gain (VCO Gain) from 18dB to 6dB. This resulted 4dB (x2.5) increase of the OLTF. This means that my fix increased the optical gain by 16dB (x6.3). The resulting UGF and phase mergin were measured to be 117kHz and 31deg, respectively.
Now I was curious to see if the PMC err shows reasonable improvement when the IMC is locked. Attachment 2 shows the latest comparison of the PMC err with and without the IMC locked. The PMC error has been taken up to 500kHz. The errors were divided by 17.5kHz LPF and 150kHz LPF to compensate the sensing response. The PMC cavity pole was ignored in this calculation. T990025 saids the PMC finesse is 4400 and the cavity pole is 174kHz. If this is true, this also needs to be applied.
Observations:
1. Now we can see improvement of the PMC error in the region between 10kHz to 70kHz.
2. The sharp peak at 8kHz is due to the marginally stable PMC servo. We should implement another notch there. T990025 suggests that the body resonance of the PMC spacer is somewhere around there. We might be able to damp it by placing a lossy material on it.
3. Similarly, the features at 12kHz and 28kHz is coming from the PMC. They are seen in the OLTF of the PMC loop.
4. The large peak at 36kHz does not change with the IMC state. This does mean that it is coming from the laser itself, or anything high-Q of the PMC. This signal is seen in the IMC error too.
5. 72kHz, 108kHz, 144kHz: Harmonics of 36kHz?
6. Broad feature from 40kHz to 200kHz. The IMC loop is adding the noise. This is the frequency range of the PC drive. Is something in the PC drive noisy???
7. The feature at 130kHz. Unknown. Seems not related to IMC. The laser noise or the PMC noise.
Remaining IMC issues:
Done (Nov 23, 2015) - 29.5MHz oscillator output degraded. Possibly unstable and noisy. Do we have any replacement? Can we take a Marconi back from one of the labs?
Done (Nov 23, 2015) - Too high LO?
- Large 36kHz peak in the IMC
- IMC loop shape optimization
- IMC locking issue. The lock streatch is not long.
- IMC PC drive issue. Could be related to the above issue.
Maybe not relevant - PC drive noise?
Attachment 1: IMC_OLTF.pdf
Attachment 2: PMC_noise_comparison.pdf
11798 Sun Nov 22 12:12:17 2015 KojiUpdateIOOIMC OLTF
Well. I thought a bit more and now I think it is likely that this is just the servo bump as you can see in the closed-loop TF.
Quote: 6. Broad feature from 40kHz to 200kHz. The IMC loop is adding the noise. This is the frequency range of the PC drive. Is something in the PC drive noisy???
17194 Mon Oct 17 17:42:35 2022 JCHowToOPLEV TablesIMC Reflected beam sketch
I sketched up a quick drawing with estimated length for the IMC reflected beam. This includes the distances and focal length. Distances are from optic to optic.
Attachment 1: Screenshot_2022-10-18_093033.png
7256 Thu Aug 23 12:17:39 2012 ManasaUpdate IMC Ringdown
The ringdown measurements are in progress. But it seems that the MC mirrors are getting kicked everytime the cavity is unlocked by either changing the frequency at the MC servo or by shutting down the input to the MC. This means what we've been observing is not the ringdown of the IMC alone. Attached are MC sus sensor data and the observed ringdown on the oscilloscope. I think we need to find a way to unlock the cavity without the mirrors getting kicked....in which case we should think about including an AOM or using a fast shutter before the IMC.
P.S. The origin of the ripples at the end of the ringdown still are of unknown origin. As of now, I don't think it is because of the mirrors moving but something else that should figured out.
Attachment 1: mozilla.pdf
Attachment 2: MC_sus.pdf
7257 Thu Aug 23 15:35:33 2012 ranaUpdate IMC Ringdown
It is HIGHLY unlikely that the IMC mirrors are having any effect on the ringdown. The ringdowns take ~20 usec to happen. The mirrors are 0.25 kg and you can calculate that its very hard to get enough force to move them any appreciable distance in that time.
7260 Thu Aug 23 17:51:25 2012 ManasaUpdate IMC Ringdown
Quote: It is HIGHLY unlikely that the IMC mirrors are having any effect on the ringdown. The ringdowns take ~20 usec to happen. The mirrors are 0.25 kg and you can calculate that its very hard to get enough force to move them any appreciable distance in that time.
The huge kick observed in the MC sus sensors seem to last for ~10usec; almost matching the observed ringdown decay time. We should find a way to record the ringdown and the MC sus sensor data simultaneously to know when the mirrors are exactly moving during the measurement process. It could also be that the moving mirrors were responsible for the ripples observed later during the ringdown as well.
* How fast do the WFS respond to the frequency switching (time taken by WFS to turn off)? I think this information will help in narrowing down the many possible explanations to a few.
15183 Mon Feb 3 13:54:10 2020 YehonathanUpdateIOOIMC Ringdowns extended data analysis
I extended the ringdown data analysis to the reflected beam following Isogai et al.
The idea is that measuring the cavity's reflected light one can use known relationships to extract the transmission of the cavity mirrors and not only the finesse.
The finesse calculated from the transmission ringdown shown in the previous elog is 1520 according to the Zucker model, 1680 according to the first exponential and 1728 according to the second exponential.
Attachment 1 shows the measured reflected light during an IMC ringdown in and out of resonance and the values that are read off it to compute the transmission.
The equations for m1 and m3 are the same as in Isogai's paper because they describe a steady-state that doesn't care about the extinction ratio of the light.
The equation for m2, however, is modified due to the finite extinction present in our zeroth-order ringdown.
Modelling the IMC as a critically coupled 2 mirror cavity one can verify that:
$m_2=P_0KR\left[T-\alpha\left(1-R\right)\right]^2+\alpha^2 P_1$
Where $P_0$ is the coupled light power
$P_1$ is the power rejected from the cavity (higher-order modes, sidebands)
$K=\left(\mathcal{F} /\pi \right )^2$ is the cavity gain.
$R$ and $T$ are the power reflectivity and transmissivity per mirror, respectively.
$\alpha^2$ is the power attenuation factor. For perfect extinction, this is 0.
Solving the equations (m1 and m3 + modified m2), using Zucker model's finesse, gives the following information:
Loss per mirror = 84.99 ppm
Transmission per mirror = 1980.77 ppm
Coupling efficiency (to TEM00) = 97.94%
Attachment 1: IMCTransReflAnalysis_anotated.pdf
15190 Wed Feb 5 21:13:17 2020 YehonathanUpdateIOOIMC Ringdowns extended data analysis
I translate the results obtained in the previous elog to the IMC 3 mirror cavity. I assume the loss in each mirror in the IMC is equal and that M2 has a negligible transmission.
I find that to a very good approximation the loss per IMC mirror is 2/3 the loss per mirror in the 2 mirror cavity model. That is the loss per mirror in the IMC is 56 ppm. The transmission per mirror in the IMC is the same as in the 2 mirror model, which is 1980 ppm.
The total transmission is the same as in the 2 mirror model and is given by:
$\frac{P_0}{P_0+P1}KT^2\approx 90\%$
where $\frac{P_0}{P_0+P1}$ is the coupling efficiency to the TEM00 mode.
15175 Wed Jan 29 12:40:24 2020 YehonathanUpdateIOOIMC Ringdowns preliminary data analysis
I analyze the IMC ringdown data from last night.
Attachment 1 shows the normalized raw data. Oscillations come in much later than in Gautam's measurement. Probably because the IMC stays locked.
Attachment 2 shows fits of the transmitted PD to unconstrained double exponential and the Zucker model.
Zucker model gives time constant of 21.6us
Unconstrained exponentials give time constants of 23.99us and 46.7us which is nice because it converges close to the Zucker model.
Attachment 1: IMCRingdownNormalizedRawdata.pdf
Attachment 2: IMCTransPDFits.pdf
15912 Fri Mar 12 11:44:53 2021 Paco, AnchalUpdatetrainingIMC SUS diagonalization in progress
[Paco, Anchal]
- Today we spent the morning shift debugging SUS input matrix diagonalization. MC stayed locked for most of the 4 hours we were here, and we didn't really touch any controls.
15258 Fri Mar 6 01:12:10 2020 gautamUpdateElectronicsIMC Servo IN2 path looks just fine
It seems like the AO path gain stages on the IMC Servo board work just fine. The weird results I reported earlier were likely a measurement error arising from the fact that I did not disconnect the LEMO IN2 cable while measuring using the BNC IN2 connector, which probably made some parasitic path to ground that was screwing the measurement up. Today, I re-did the measurement with the signal injected at the IN2 BNC, and the TF measured being the ratio of TP3 on the board to a split-off of the SR785 source (T-eed off). Attachments #1, #2 shows the result - the gain deficit from the "expected" value is now consistent with that seen on other sliders.
Note that the signal from the CM board in the LSC rack is sent single-ended over a 2-pin LEMO cable (whose return pin is shorted to ground). But it is received differentially on the IMC Servo board. I took this chance to look for evidence of extra power line noise due to potential ground loops by looking at the IMC error point with various auxiliary cables connected to the board - but got distracted by some excess noise (next elog).
Attachment 1: AO_inputTFs_5Mar.pdf
Attachment 2: sliderCal_5Mar.pdf
15257 Thu Mar 5 19:51:14 2020 gautamUpdateElectronicsIMC Servo board being tested
I am running some tests on the IMC servo board with an extender card so the IMC will not be locking for a couple of hours.
16174 Wed Jun 2 09:43:30 2021 Anchal, PacoSummarySUSIMC Settings characterization
## Plot description:
• We picked up three 10 min times belonging to the three different configurations:
• 'Old Settings': IMC Suspension settings before Paco and I changed anything. Data taken from Apr 26, 2021, 00:30:42 PDT (GPS 1303457460).
• 'New Settings': New input matrices uploaded on April 28th, along with F2A filters and AC coil balancing gains (see 16091). Data taken from May 01, 2021, 00:30:42 PDT (GPS 1303889460).
• 'New settings with new gains' Above and new suspension damping gains uploaded on May5th, 2021 (see 16120). Data taken from May 07, 2021, 03:10:42 PDT (GPS 1304417460).
• Attachment 1 shows the RMS seismic noise along X direction between 1 Hz and 3 Hz picked from C1:PEM-RMS_BS_X_1_3 during the three time durations chosen. This plot is to establish that RMS noise levels were similar and mostly constant. Page 2 shows the mean ampltidue spectral density of seismic noise in x-direction over the 3 durations.
• Attachment 2 shows the transfer function estimate of seismic noise to MC_F during the three durations. Page 1 shows ratio of ASDs taken with median averaging while page 2 shows the same for mean averaging.
• Attachment 3 shows the transfer function estimate of seismic noise to MC_TRANS_PIT during the three durations. Page 1 shows ratio of ASDs taken with median averaging while page 2 shows the same for mean averaging.
• Attachment 4 shows the transfer function estimate of seismic noise to MC_TRANS_YAW during the three durations. Page 1 shows ratio of ASDs taken with median averaging while page 2 shows the same for mean averaging.
## Inferences:
• From Attachment 2 Page 1:
• We see that 'old settings' caused the least coupling of seismic noise to MC_F signal in most of the low frequency band except between 1.5 to 3 Hz where the 'new settings' were slightly better.
• 'new settings' also show less coupling in 4 Hz to 6 Hz band, but at these frequencies, seismix noise is filtered out by suspension, so this could be just coincidental and is not really a sign of better configuration.
• There is excess noise coupling seen with 'new settings' between 0.4 Hz and 1.5 Hz. We're not sure why this coupling increased.
• 'new settings with new gains' show the most coupling in most of the frequency band. Clearly, the increased suspension damping gains did not behaved well with rest of the system.
• From Attachment 3 Page 1:
• Coupling to MC_TRANS_PIT error signal is reduced for 'new settings' in almost all of the frequency band in comparison to the 'old settings'.
• 'new settings with new gains' did even better below 1 Hz but had excess noise in 1 Hz to 6 Hz band. Again increased suspension damping gains did not help much.
• But low coupling to PIT error for 'new settings' suggest that our decoupling efforts in matrix diagonalization, F2A filters and ac coil balancing worked to some extent.
• From Attachment 4 Page 1:
• 'new settings' and 'old settings' have the same coupling of seismic noise to MC_TRANS_YAW in all of the frequency band. This is in-line witht eh fact that we found very little POS to YAW couping in our analysis before and there was little to no change for these settings.
• 'new settings with new gains' did better below 1 Hz but here too there was excess coupling between 1 Hz to 9 Hz.
• Page 1 vs Page 2:
• Mean and median should be same if the data sample was large enough and noise was stationary. A difference between the two suggests existence of outliers in the data set and median provides a better central estimate in such case.
• MC_F: Mean and median are same below 4 hz. There are high frequency outliers above 4 Hz in 'new settings with new gains' and 'old settings' data sets, maybe due to transient higher free running laser frequency noise. But since, suspension settigns affect below 1 Hz mostly, the data sets chosen are stationary enough for us.
• MC_TRANS_PIT: Mean ratio is lower for 'new settings' and 'old settings' in 0.3 hz to 0.8 Hz band. Same case above 4 Hz as listed above.
• MC_TRANS_YAW: Same as above.
• Conclusion 1: The 'new settings with new gains' cause more coupling to seismic noise, probably due to low phase margin in control loops. We should revert back the suspension damping gains.
• Conclusion 2: The 'new settings' work as expected and can be kept when WFS loops are optimized further.
• Conjecture: From our experience over last 2 weeks, locking the arms to the main laser with 'new settings with new gains' introduces noise in the arm length large enough that the Xend green laser does not remain locked to the arm for longer than tens of seconds. So this is definitely not a configuration in which we can carry out other measurements and experiments in the interferometer.
Attachment 1: seismicX.pdf
Attachment 2: seismicXtoMC_F_TFest.pdf
Attachment 3: seismicXtoMC_TRANS_PIT_TFest.pdf
Attachment 4: seismicXtoMC_TRANS_YAW_TFest.pdf
16102 Thu Apr 29 18:53:33 2021 AnchalUpdateSUSIMC Suspension Damping Gains Test
With the input matrix, coil ouput gains and F2A filters loaded as in 16091, I tested the suspension loops' step response to offsets in LSC, ASCPIT and ASCYAW channels, before and after applying the "new damping gains" mentioned in 16066 and 16072. If these look better, we should upload the new (higher) damping gains as well. This was not done in 16091.
Note that in the plots, I have added offsets in the different channels to plot them together, hence the units are "au".
Attachment 1: MC1_SUSDampGainTest.pdf
Attachment 2: MC2_SUSDampGainTest.pdf
Attachment 3: MC3_SUSDampGainTest.pdf
16110 Mon May 3 16:24:14 2021 AnchalUpdateSUSIMC Suspension Damping Gains Test Repeated with IMC unlocked
We repeated the same test with IMC unlocked. We had found these gains when IMC was unlocked and their characterization needs to be done with no light in the cavity. attached are the results. Everything else is same as before.
Quote: With the input matrix, coil ouput gains and F2A filters loaded as in 16091, I tested the suspension loops' step response to offsets in LSC, ASCPIT and ASCYAW channels, before and after applying the "new damping gains" mentioned in 16066 and 16072. If these look better, we should upload the new (higher) damping gains as well. This was not done in 16091. Note that in the plots, I have added offsets in the different channels to plot them together, hence the units are "au".
Edit Tue May 4 14:43:48 2021 :
• Adding zoomed in plots to show first 25s after the step.
• MC1:
• Our improvements by new gains are only modest.
• This optic needs a more careful coil balancing first.
• Still the ring time is reduced to about 5s for all step responses as opposed to 10s at old gains.
• MC2:
• The first page of MC2 might be bit misleading. We have not changed the damping gain for SUSPOS channel, so the longer ringing is probably just an artifact of somthing else. We didn't retake data.
• In PIT and YAW where we increased the gain by a factor of 3, we see a reduction in ringing lifetime by about half.
• MC3:
• We saw the most optimistic improvement on this optic.
• The gains were unusually low in this optic, not sure why.
• By increasing SUSPOS gain from 200 to 500, we saw a reduction of ringing halftime from 7-8s to about 2s. Improvements are seen in other DOFs as well.
• You can notice rightaway that YAW of MC3 keeps oscillating near resonance (about 1 Hz). Maybe more careful feedback shaping is required here.
• In SUSPIT, we increased gain from 12 to 35 and saw a good reduction in both ringing time and initial amplitude of ringing.
• In SUSYAW, we only increased the gain to 12 from 8, which still helped a lot in reducing big ringing step response to below 5s from about 12s.
Overall, I would recommend setting the new gains in the suspension loops as well to observe long term effects too.
Attachment 1: MC1_SusDampGainTest.pdf
Attachment 2: MC2_SusDampGainTest.pdf
Attachment 3: MC3_SusDampGainTest.pdf
16175 Wed Jun 2 16:20:59 2021 Anchal, PacoSummarySUSIMC Suspension gains reverted to old values
Following the conclusion, we are reverting the suspension gains to old values, i.e.
IMC Suspension Gains
MC1 MC2 MC3
SUSPOS 120 150 200
SUSPIT 60 10 12
SUSYAW 60 10 8
While the F2A filters, AC coil gains and input matrices are changed to as mentioned in 16066 and 16072.
The changes can be reverted all the way back to old settings (before Paco and I changed anything in the IMC suspensions) by running python scripts/SUS/general/20210602_NewIMCOldGains/restoreOldConfigIMC.py on allegra. The new settings can be uploaded back by running python scripts/SUS/general/20210602_NewIMCOldGains/uploadNewConfigIMC.py on allegra.
Change time:
Unix Time = 1622676038
UTC Jun 02, 2021 23:20:38 UTC Central Jun 02, 2021 18:20:38 CDT Pacific Jun 02, 2021 16:20:38 PDT
GPS Time = 1306711256
Quote: Conclusion 1: The 'new settings with new gains' cause more coupling to seismic noise, probably due to low phase margin in control loops. We should revert back the suspension damping gains. Conclusion 2: The 'new settings' work as expected and can be kept when WFS loops are optimized further. Conjecture: From our experience over last 2 weeks, locking the arms to the main laser with 'new settings with new gains' introduces noise in the arm length large enough that the Xend green laser does not remain locked to the arm for longer than tens of seconds. So this is definitely not a configuration in which we can carry out other measurements and experiments in the interferometer.
16094 Thu Apr 29 10:52:56 2021 AnchalUpdateSUSIMC Trans QPD and WFS loops step response test
In 16087 we mentioned that we were unable to do a step response test for WFS loop to get an estimate of their UGF. The primary issue there was that we were not putting the step at the right place. It should go into the actuator directly, in this case, on C1:SUS-MC2_PIT_COMM and C1:SUS-MC2_YAW_COMM. These channels directly set an offset in the control loop and we can see how the error signals first jump up and then decay back to zero. The 'half-time' of this decay would be the inverse of the estimated UGF of the loop. For this test, the overall WFS loops gain, C1:IOO-WFS_GAIN was set to full value 1. This test is performed in the changed settings uploaded in 16091.
I did this test twice, once giving a step in PIT and once in YAW.
Attachment 1 is the striptool screenshot for when PIT was given a step up and then step down by 0.01.
• Here we can see that the half-time is roughly 10s for TRANS_PIT and WFS1_PIT corresponding to roughly 0.1 Hz UGF.
• Note that WFS2 channels were not disturbed significantly.
• You can also notice that third most significant disturbance was to TRANS_YAW actually followed by WF1 YAW.
Attachment 2 is the striptool screenshot when YAW was given a step up and down by 0.01. Note the difference in x-scale in this plot.
• Here, TRANS YAW got there greatest hit and it took it around 2 minutes to decay to half value. This gives UGF estimate of about 10 mHz!
• Then, weirdly, TRANS PIT first went slowly up for about a minutes and then slowly came dome in a half time of 2 minutes again. Why was PIT signal so much disturbed by the YAW offset in the first place?
• Next, WFS1 YAW can be seen decaying relatively fast with half-life of about 20s or so.
• Nothing else was disturbed much.
• So maybe we never needed to reduce WFS gain in our measurement in 16089 as the UGF everywhere were already very low.
• What other interesting things can we infer from this?
• Should I sometime repeat this test with steps given to MC1 or MC3 optics?
Attachment 1: PIT_OFFSET_ON_MC2.png
Attachment 2: YAW_STEP_ON_MC2_complete.png
15215 Sat Feb 15 12:56:24 2020 YehonathanUpdateIOOIMC Transfer function measurement
{Yehonathan, Meenakshi}
We measure the IMC transfer function using SR785.
We hook up the AOM driver to the SOURCE OUT, Input PD to CHANNEL ONE and the IMC transmission PD to CHANNEL TWO.
We use the frequency response measurement feature in the SR785. A swept sine from 100KHz to 100Hz is excited with an amplitude of 10mV.
Attachment 1 shows the data with a fit to a low pass filter frequency response.
IMC pole frequency is measured to be 3.795KHz, while the ringdowns predict a pole frequency 3.638KHz, a 4% difference.
The closeness of the results discourages me from calibrating the PDs' transfer functions.
I tend to believe the pole frequency measurement a bit more since it coincides with a linewidth measurement done awhile ago Gautam was telling me about.
Thoughts:
I think of trying to try another zero-order ringdown but with much smaller excitation than what used before (500mV) and than move on to the first-order beam.
Also, it seems like the reflection signal in zero-order ringdown (Attachment 2, green trace) has only one time constant similar to the full extinction ringdown. The reason is that due to the fact the IMC is critically coupled there is no DC term in the electric field even when the extinction of light is partial. The intensity of light, therefore, has only one time constant.
Fitting this curve (Attachment 3) gives a time constant of 18us, a bit too small (gives a pole of 4.3KHz). I think a smaller extinction ringdown will give a cleaner result.
Attachment 1: IMCFrequencyResponse.pdf
Attachment 2: IMCRingdownNormalizedRawdata.pdf
Attachment 3: IMCREFLPDFits.pdf
11529 Tue Aug 25 16:09:54 2015 ericqUpdateIOOIMC Tweak
I increased the overall IMC loop gain by 4dB, and decreased the FAST gain (which determines the PZT/EOM crossover) by 3dB. This changed the AO transfer function from the blue trace to the green trace in the first plot. This changed the CARM loop open loop TF shape from the unfortunate blue shape to the more pleasing green shape in the second plot. The red trace is the addition of one super boost.
Oddly, these transfer functions look a bit different than what I measured in March (ELOG 11167), which itself differed from the shaping done December of 2014 (ELOG 10841).
I haven't yet attempted any 1F handoff of the PRMI since relocking, but back when Jenne and I did so in April, the lock was definitely less stable. My suspicion is that we may need more CARM supression; we never computed the loop gain requirement that ensures that the residual CARM fluctuations witnessed by, say, REFL55 are small enough to use as a reliable PRMI sensor.
I should be able to come up with this with data from last night.
Attachment 1: imcTweak.pdf
Attachment 2: CARM_TF.pdf
11538 Fri Aug 28 19:05:53 2015 ranaUpdateIOOIMC Tweak
Well, green looks better than blue, but it makes the PCDRIVE go high, which means its starting to saturate the EOM drive. So we can't just maximize the phase margin in the PZT/EOM crossover. We have to take into account the EOM drive spectrum and its RMS.
Also, your gain bump seems suspicious. See my TF measurements of the crossover in December. Maybe you were saturating the EOM in your TF ?
Lets find out what's happening with FSS servos over in Bridge and then modify ours to be less unstable.
15318 Tue May 5 23:44:14 2020 gautamUpdateASCIMC WFS
Summary:
I've been thinking about the IMC WFS. I want to repeat the sort of analysis done at LLO where a Finesse model was built and some inferences could be made about, for example, the Gouy phase separation b/w the sensors by comparing the Finesse sensing matrix to a measured sensing matrix. Taking the currently implemented output matrix as a "measurement" (since the IMC WFS stabilize the IMC transmission), I don't get any agreement between it and my Finesse model. Could be that the model needs tweaking, but there are several known issues with the WFS themselves (e.g. imbalanced segment gains).
Building the finesse model:
• I pulled the WFS telescopes from Andres elogs/SURF report, which I think was the last time the WFS telescopes were modified.
• The in-vacuum propagation distances were estimated from CAD diagrams.
• According to my model, the Gouy phase separation between the two WFS heads is ~70 degrees, whereas Andres' a la mode simulations suggest more like 90 degrees. Presumably, some lengths/lenses are different between what I assume and what he used, but I continue the analysis anyway...
• The appropriate power attenuations were placed in each path - one thing I noticed is that the BS that splits light between WFS1 and WFS2 is a 30/70 BS and not a 50/50, I don't see any reason why this should be (presumably it was to do with component availability). see below for Rana's comments.
Simulations:
• The way the WFS servos are set up currently, the input matrix is diagonal while the output matrix encodes the sensing information.
• In finesse, I measured the input matrix (i.e. response sensed in each sensor when an optic is dithered in angle). The length is kept resonant for the carrier (but not using a locking signal), which should be valid for small angular disturbances, which is the regime in which the error signals will be linear anyways.
• Then I inverted the simulated sensing matrix so as to be able to compare with the CDS output matrix. Note that there is a relative gain scaling of 100 between the WFS paths and the MC2T QPD paths which I added to the simulation. I also normalized the columns of the matrix by the largest element in the column, in an attempt to account for the various other gains that are between the optical sensing and the digitizaiton (e.g. WFS demod boards, QPD transimpedance etc etc).
• Attachment #1 shows the comparison between simulation and measurement. The two aren't even qualitatively similar, needs more thought...
• The transimpedance resistor is 1.5 kohms. With the gain stages, the transimpedance gain is nominally 37.5 kohms, and 3.75 kohms when the attenuation setting is engaged (as it is for 2/4 quadrants on each head).
• Assuming a modulation depth of 0.1, the Johnson noise of the transimpedance resistor dominates (with the MAX4106 current noise a close second), and these heads cannot be shot noise limited when operating at 1 W input power (though of course the situation will change if we have 25 W input).
• The heads are mounted at a ~45 deg angle, mixing PIT/YAW, but I assume we can just use the input matrix to rotate back to the natural PIT/YAW basis.
Update 215 pm 5/6: adding in some comments from Rana raised during the meeting:
1. The transimpedance is actually done by the RLC network (L6 and C38 for CH 3), and not 1.5 kohms. It just coincidentally happens that the reactance is ~1.5 kohms at 29.5 MHz. Note that my LTspice simulation using ideal inductors and capacitors still predicts ~4pA/rtHz noise at 29.5 MHz, so the conclusion about shot noise remains valid I think... One option is to change the attenuation in this path and send more light onto the WFS heads.
The transimpedance gain and noise are now in Attachment #2. I just tweaked the L values to get a peak at 29.5 MHz and a notch at twice that frequency. For this I assumed a photodiode capacitance of 225pF and the shown transimpedance gain has the voltage gain of the MAX4106 stages divided out. The current noise is input referred.
2. The imbalanced power on WFS heads may have some motivation - it may be that the W/rad TF for one of the two modes we are trying to sense (beam plane tilt vs beam plane translation) is not equal, so we want more light on the head with weaker response.
3. The 45 degree mounting of the heads is actually meant to decouple PIT and YAW.
Attachment 1: WFSmatrixComparison.pdf
15320 Thu May 7 09:43:21 2020 ranaUpdateASCIMC WFS
This is the doc from Keita Kawabe on why the WFS heads should be rotated.
15321 Thu May 7 10:58:06 2020 gautamUpdateASCIMC WFS
OK so the QPD segments are in the "+" orientation when the 40m IMC WFS heads are mounted at 45 deg. I thought "+" was the natural PIT/YAW basis but I guess in the the LIGO parlance, the "X" orientation was considered more natural.
Quote: This is the doc from Keita Kawabe on why the WFS heads should be rotated.
16990 Tue Jul 12 09:25:09 2022 ranaUpdateIOOIMC WFS
MC WFS Demod board needs some attention.
Tomislav has been measuring a very high noise level in the MC WFS demod output (which he promised to elog today!). I thought this was a bogus measurement, but when he, and Paco and I tried to measure the MC WFS sensing matrix, we noticed that there is no response in any WFS, although there are beams on the WFS heads. There is a large response in MC2 TRANS QPD, so we know that there is real motion.
I suspect that the demod board needs to be reset somehow. Maybe the PLL is unlocked or some cable is wonky. Hopefully not both demod boards are fried.
Please leave the WFS loops off until demod board has been assessed.
17177 Fri Oct 7 20:00:46 2022 KojiUpdateIOOIMC WFS / MC2 SUS glitch
After the CDS upgrade team called for a day (their work TBD), I took over the locked IMC to check how it looked like.
The lock was robust but the IMC REFL spot and the WFS DC/MC2 QPD were moving too much.
I wondered if there were something wrong with the damping. I thought MC3 damping seemed weak, but this was torelable level.LR
During the ring down check of MC2, I saw that the OSEM signals were glitchy. In the end I found it was LR sensor which was glitchy and fluctuating.
I went into the lab and pushed the connectors on the euro card modules and the side connectors as well as the cables on the MC2 sat amp.
I came back to the control room and found the MC2 LR OSEMs had the jump and it seems more stable now.
I leave the IMC locked & WFS running. This sus situation is not great at all and before we go too far, we'll work on the transition to the new electronics (but not today or next week).
By the way the unit of the signals on the dataviewer didn't make sense. Something seemed wrong with them.
Attachment 1: Screenshot_2022-10-07_19-59-45.png
17179 Sun Oct 9 13:49:49 2022 KojiUpdateIOOIMC WFS / MC2 SUS glitch
The IMC and the IMC WFS kept running for ~2days. 👍
I wanted to look at the trand of the IMC REFL DC, but the dataviewer showed that the recorded values are zero. And this slow channel is missing in the channel list.
I checked the PSL PMC signals (slow) as an example, and many channels are missing in the channel list.
So something is not right with some part of the CDS.
Note that I also reported that the WFS plot in the above refered previous elog has the value like 1e39
Attachment 1: Screen_Shot_2022-10-09_at_13.49.12.png
17180 Mon Oct 10 00:05:24 2022 ChrisUpdateIOOIMC WFS / MC2 SUS glitch
Thanks for pointing out that EPICS data collection (slow channels) was not working. I started the service that collects these channels (standalone_edc, running on c1sus), and pointed it to the channel list in /opt/rtcds/caltech/c1/chans/daq/C0EDCU.ini, so this should be working now.
controls@c1sus:~\$ systemctl status rts-edc_c1sus ● rts-edc_c1sus.service - Advanced LIGO RTS stand-alone EPICS data concentrator Loaded: loaded (/etc/systemd/system/rts-edc_c1sus.service; enabled; vendor preset: enabled) Active: active (running) since Sun 2022-10-09 23:30:15 PDT; 10h ago Main PID: 32154 (standalone_edc) CGroup: /system.slice/rts-edc_c1sus.service ├─32154 /usr/bin/standalone_edc -i /etc/advligorts/edc.ini -l 0.0.0.0:9900 └─32159 caRepeater
Quote: The IMC and the IMC WFS kept running for ~2days. 👍 I wanted to look at the trand of the IMC REFL DC, but the dataviewer showed that the recorded values are zero. And this slow channel is missing in the channel list. I checked the PSL PMC signals (slow) as an example, and many channels are missing in the channel list. So something is not right with some part of the CDS. Note that I also reported that the WFS plot in the above refered previous elog has the value like 1e39
12641 Sat Nov 26 19:16:28 2016 KojiUpdateIOOIMC WFS Demod board measurement & analysis
[Rana, Koji]
1. The response of the IMC WFS board was measured. The LO signal with 0.3Vpp@29.5MHz on 50Ohm was supplied from DS345. I've confirmed that this signal is enough to trigger the comparator chip right next to the LO input. The RF signal with 0.1Vpp on the 50Ohm input impedance was provided from another DS345 to CH1 with a frequency offset of 20Hz~10kHz. Two DS345s were synced by the 10MHz RFreference at the rear of the units. The resulting low frequency signal from the 1st AF stage (AD797) and the 2nd AF stage (OP284) were checked.
Attachment 1 shows the measured and modelled response of the demodulator with various frequency offsets. The value shows the signal transfer (i.e. the output amplitude normalized by the input amplitude) from the input to the outputs of the 1st and 2nd stages. According to the datasheet, the demodulator chip provides a single pole cutoff of 340kHz with the 33nF caps between AP/AN and VP. The first stage is a broadband amplifier, but there is a passive LPF (fc=~1kHz). The second stage also provides the 2nd order LPF at fc~1kHz too. The measurement and the model show good agreement.
2. The output noise levels of the 1st and 2nd stages were meausred and compared with the noise model by LISO.
Attachment 2 shows the input referred noise of the demodulator circuit. The output noise is basically limited by the noise of the first stage. The noise of the 2nd stage make the significant contribution only above the cut off freq of the circuit (~1kHz). And the model supports this fact. The 6.65kOhm of the passive filter and the input current noise of AD797 cause the large (>30nV/rtHz) noise contribution below 100Hz. This completely spoils the low noiseness (~1nV/rtHz) of AD797. At lower frequency like 0.1Hz other component comes up above the modelled noise level.
3. Rana and I had a discussion about the modification of the circuit. Attachment 4 shows the possible improvement of the demod circuit and the 1st stage preamplifier. The demodulator chip can have a cut off by the attached capacitor. We will replace the 33nF caps with 1uF and the cut off will be pushed down to ~10kHz. Then the passive LPF will be removed. We don't need "rodeo horse" AD797 for this circuit, but op27 is just fine instead. The gain of the 1st stage can be increased from 9 to 21. This should give us >x10 improvement of the noise contribution from the demodualtor (Attachment 3). We also can replace some of the important resistors with the thin film low noise resistors.
Attachment 1: WFS_demod_response.pdf
Attachment 2: WFS_demod_noise.pdf
Attachment 3: WFS_demod_noise_plan.pdf
Attachment 4: Screen_shot_2011-07-01_at_11.13.01_AM.png
12645 Tue Nov 29 17:45:06 2016 KojiUpdateIOOIMC WFS Demod board measurement & analysis
Summary: The demodulator input noise level was improved by a factor of more than 2. This was not as much as we expected from the preamp noise improvement, but is something. If this looks OK, I will implement this modification to all the 16 channels.
The modification shown in Attachment 1 has actually been applied to a channel.
• The two 1.5uF capacitors between VP and AN/AP were added. This decreases the bandwidth of the demodulator down to 7.4kHz
• The offset trimming circuit was disabled. i.e. Pin18 of AD831 was grounded.
• The passive low pass at the demodulator output was removed. (R18, C34)
• The stage1 (preamp) chip was changed from AD797 to OP27.
• The gain of the preamp stage was changed from 9 to 21. Also the thin film resistors are used.
Attachment 2 shows the measured and expected output signal transfer of the demodulator. The actual behavior of the demodulator is as expected, and we still keep the over all LPF feature of 3rd order with fc=~1kHz.
Attachment 3 shows the improvement of the noise level with the signal reffered to the demodulator input. The improvement by a factor >2 was observed all over the frequency range. However, this noise level could not be explained by the preamp noise level. Actually this noise below 1kHz is present at the output of the demodulator. (Surprisingly, or as usual, the noise level of the previous preamp configuration was just right at the noise level of the demodulator below 100Hz.) The removal of the offset trimmer circuit contributed to the noise improvement below 0.3Hz.
Attachment 1: demod.pdf
Attachment 2: WFS_demod_response.pdf
Attachment 3: WFS_demod_noise.pdf
12647 Tue Nov 29 18:35:32 2016 ranaUpdateIOOIMC WFS Demod board measurement & analysis
more U4 gain, lesssss U5 gain
12661 Fri Dec 2 18:02:37 2016 KojiUpdateIOOIMC WFS Demod board measurement & analysis
ELOG of the Wednesday work.
It turned out that the IMC WFS demod boards have the PCB board that has a different pattern for each of 8ch.
In addition, AD831 has a quite narrow leg pitch with legs that are not easily accessible.
Because of these, we (Koji and Rana) decided to leave the demodulator chip untouched.
I have plugged in the board with the WFS2-Q1 channel modified in order to check the significance of the modification.
WFS performance before the modification
Attachment 1 shows the PSD of WFS2-I1_OUT calibrated to be referred to the demodulator output. (i.e. Measured PSDs (cnt/rtHz) were divided by 8.9*2^16/20)
There are three curves: One is the output with the MC locked (WFS servos not engaged). The second is the PSD with the PSL beam blocked (i.e. dark noise). The third is the electronics noise with the RF input terminated and the nominal LO supplied.
This tells us that the measured PSD was dominated by the demodulator noise in the dark condition. And the WFS signal was also dominated by the demod noise below 0.1Hz and above 20Hz. There are annoying features at 0.7, 1.4, 2.1, ... Hz. They basically impose these noise peaks on the stabilized mirror motion.
WFS performance after the modification
Attachment 2 shows the PSD of WFS2-Q1_OUT calibrated to be referred to the demodulator output. (i.e. Measured PSDs (cnt/rtHz) were divided by 21.4*2^16/20)
There are three same curves as the other plot. In addition to these, the PSD of WFS2-I1_OUT with the MC locked is also shown as a red curve for comparison.
This figure tells us that the measured PSD below 20Hz was dominated by the demodulator noise in the dark condition. And the WFS signal is no longer dominated by the electronics noise. However, there still are the peaks at the harmonics of 0.7, 1.4, 2.1, ... Hz. I need further inspection of the FWS demod and whtening boards to track down the cause of these peaks.
Attachment 1: WFS_demod_noise_orig.pdf
Attachment 2: WFS_demod_noise_mod.pdf
12662 Sat Dec 3 13:27:35 2016 KojiUpdateIOOIMC WFS Demod board measurement & analysis
ELOG of the work on Thursday
Gautam suggested looking at the preamplifier noise by shorting the input to the first stage. I thought it was a great idea.
To my surprise, the noise of the 2nd stage was really high compared to the model. I proceeded to investigate what was wrong.
It turned out that the resistors used in this sallen-key LPF were thick film resistors. I swapped them with thin film resistors and this gave the huge improvement of the preamplifier noise in the low frequency band.
Attachment 1 shows the summary of the results. Previously the input referred noise of the preamp was the curve in red. We the resistors replaced, it became the curve in magenta, which is pretty close to the expected noise level by LISO model above 3Hz (dashed curves). Unfortunately, the output of the unit with the demodulator connected showed no improvement (blue vs green), because the output is still limited by the demodulator noise. There were harmonic noise peaks at n x 10Hz before the resistor replacement. I wonder if this modification also removed the harmonic noise seen in the CDS signals. I will check this next week.
Attachment 2 shows the current schematic diagram of the demodulator board. The Q of the sallen key filter was adjusted by the gain to have 0.7 (butter worth). We can adjust the Q by the ratio of the capacitance. We can short 3.83K and remove 6.65K next to it. And use 22nF and 47nF for the capacitors at the positive input and the feedback, respectively. This reduces the number of the resistors.
Attachment 1: WFS_demod_noise.pdf
Attachment 2: demod.pdf
12668 Tue Dec 6 13:37:02 2016 KojiUpdateIOOIMC WFS Demod board measurement & analysis
I have implemented the modification to the demod boards (Attachment 1).
Now, I am looking at the noise in the whitening board. Attachment 2 shows the comparison of the error signal with the input of the whitening filter shorted and with the 50ohm terminator on the demodulator board. The message is that the whitening filter dominates the noise below 3Hz.
I am looking at the schematic of the whitening board D990196-B. It has an VGA AD602 at the input. I could not find the gain setting for this chip.
If the gain input is fixed at 0V, AD602 has the gain of 10dB. The later stages are the filters. I presume they have the thick film resistors.
Then they may also cause the noise. Not sure which is the case yet.
Also it seems that 0.7Hz noise is still present. We can say that this is coming from the demod board but not on the work bench but in the eurocard crate.
Attachment 1: demod.pdf
Attachment 2: WFS_error_noise.pdf
17332 Sat Dec 3 17:42:25 2022 AnchalUpdateASCIMC WFS Fixed for now
Today I did a lot of steps to eventually reach to WFS locking stably for long times and improving and keeping the IMC transmission counts to 14400. I think the main culprit in thw WFS loop going unstable was the offset value set on MC_TRANS_PIT filter module (C1:IOO-MC_TRANS_PIT_OFFSET). This value was roughly correct in magnitude but opposite in sign, which created a big offset in MC_TRANS PIT error signal which would integrate by the loops and misalign the mode cleaner.
WFS offsets tuning
• I ran C1:IOO-WFS_MASTER > Actiona > Correct WFS DC offsets script while the two WFS heads were blocked.
• Then I aligned IMC to maximize transmission. I also made PMC transmission better by walking the input beam.
• Then, while IMC is locked and WFS loops are off, I aligned the beam spot on WFS heads to center it in DC (i.e. zeroing C1:IOO-WFS1_PIT_DC, C1:IOO-WFS1_YAW_DC, C1:IOO-WFS2_PIT_DC, C1:IOO-WFS2_YAW_DC)
• Then I ran C1:IOO-WFS_MASTER > Actiona > Correct WFS DC offsets script while keeping IMC locked (note the script says to keep it unlocked, but I think that moves away the beam). If we all agree this is ok, I'll edit this script.
• Then I checked the error signals of all WFS loops and still found that C1:IOO-MC_TRANS_PIT_OUTPUT and C1:IOO-MC_TRANS_YAW_OUTPUT have offsets. I relieved these offsets by averaging the input to these filter moduels for 100s and updating the offset. This is where I noticed that the PIT offset was wrong in sign.
WFS loops UGF tuning
• Starting with only YAW loops, I measured the open loop transfer functions (OLTFs) for each loop by simultaneously injecting gaussian noise from 0.01 Hz to 0.5 Hz using diaggui at the loop filter module excitation points and taking ration of IN1/IN2 of the filter modules.
• Then I scaled the YAW output matrix columns to get UGF of 0.1 Hz when YAW loop was along turned on.
• Then I tried to do this for PIT as well but it failed as even with overall gain of 0.1, the PIT loops actuate a lot of YAW motion causing the IMC to loose lock eventually.
• So I tried locking PIT loops along with YAW loops but with 0.1 overall gain. This worked for long enough that I could get a rough estimate of the OLTFs. I scaled the columns of PIT output matrix and slowly increased the overall gain while repeating this step to get about 0.1 Hz UGF for all PIT loops too.
• Note though that the PIT loop shape did not come out as expected with a shallower slope and much worse coherence for same amount of excitation in comparison to YAW loops. See attached plots.
• Never the less, I was able to reach to an output matric which works at overall gain of 1. I tested this configuration for atleast 15 minutes but the loop was working even with 6 excitations happening simultaneously for OLTF measurement.
• We will need to revisit PIT loop shapes, matrix diagonalization, and sources of noise.
OLTF measurements were done using this diaggui file. The measurement file got deleted by me by mistake, so I recreated the template. Thankfully, I had saved the pdf of the measurements, but I do not have same measurement results in the git repo.
Attachment 1: IMC_WFS_OLTF.pdf
17334 Sun Dec 4 16:44:04 2022 AnchalUpdateASCIMC WFS Fixed for now
Today, I worked on WFS loop output matrix for PIT DOFs.
• I began with the matrix that was in place before Nov 15.
• I followed the same method as last time to fist get all UGFs around 0.06 Hz with overall gain of 0.6 on the WFS loops.
• This showed me that MC2_TRANS_PIT loop shape matches well with the nice working YAW loops, but the WFS1 and WFS2 loops still looked flat like before.
• This indicated that output matrix needs to be fixed for cross coupling between WFS1 and WFS2 loops.
• I ran this script WFSoutMatBalancing.py which injects low frequency (<0.5 Hz) oscillations when the loops are open, and measures sensing matrix using error signals. I used 1000s duration for this test.
• The direct inverse of this sensing matrix fixed the loop shape for WFS1 indicating WFS1 PIT loop is disentangled from WFS2 now.
• Note this is a very vague definition of diagonalization, but I am aiming to reach to a workign WFS loop asap with whatever means first. Then we can work on accurate diagonalization later.
• I simply ran the script WFSoutMatBalancing.py again for another 1000s and this time the sensing matrix mostly looked like an identity.
• I implemented the new output matrix found by direct inversion and took new OLTF.Again though, the WFS2_PIT loop comes out to be flat. See Attachment 1.
• Then noting from this elog post, I reduced the gain values on MC2 TRANS loops to 0.1 I think it is better to use this place to reduce loop UGF then the output matrix as this will remind us that MC2 TRANS loops are slower than others by 10 times.
• I retook OLTF but very unexpected results came. The overall gain of WFS1_YAW and WFS2_YAW seemed to have increased by 6. All other OLTFs remained same as expected. See attachment 2.
• To fulfill the condition that all UGF should be less than 0.1 Hz, I reduced gains on WFS1_YAW and WFS2_YAW loops but that made the YAW loops unstable. So I reverted back to all gains 1.
• We probably need to diagonalize Yaw matrix better than it is for letting MC2_TRANS_YAW loop to be at lower UGF.
• I'm leaving the mode cleaner in this state and would come back in an hour to see if it remains locked at good alignment. See attachment 3 for current state.
Sun Dec 4 17:36:32 2022 AG: IMC lock is holding as strong as before. None of the control signals or error signals seem to be increasing monotonously over the last one hour. I'll continue monitoring the lock.
Mon Dec 5 11:11:08 2022 AG: IMC was locked all night for past 18 hours. See attachment 4 for the minute trend.
Attachment 1: IMC_WFS_OLTF_All_Gains_1.pdf
Attachment 2: IMC_WFS_OLTF_Nom_Gain.pdf
Attachment 3: WFS_Loop_Configuration.png
Attachment 4: WFS_Loop_Performance.png
12748 Tue Jan 24 01:04:16 2017 gautamSummaryIOOIMC WFS RF power levels
Summary:
I got around to doing this measurement today, using a minicircuits bi-directional coupler (ZFBDC20-61-HP-S+), along with some SMA-LEMO cables.
• With the IMC "well aligned" (MC transmission maximized, WFS control signals ~0), the RF power per quadrant into the Demod board is of the order of tens of pW up to a 100pW.
• With MC1 misaligned such that the MC transmission dropped by ~10%, the power per quadrant into the demod board is of the order of hundreds of pW.
• In both cases, the peak at 29.5MHz was well above the analyzer noise floor (>20dB for the smaller RF signals), which was all that was visible in the 1MHz span centered around 29.5 MHz (except for the side-lobes described later).
• There is anomalously large reflection from Quadrant 2 input to the Demod board for both WFS
• The LO levels are ~-12dBm, ~2dBm lower than the 10dBm that I gather is the recommended level from the AD831 datasheet
Quote: We should insert a bi-directional coupler (if we can find some LEMO to SMA converters) and find out how much actual RF is getting into the demod board.
Details:
I first aligned the mode cleaner, and offloaded the DC offsets from the WFS servos.
The bi-directional coupler has 4 ports: Input, Output, Coupled forward RF and Coupled Reverse RF. I connected the LEMO going to the input of the Demod board to the Input, and connected the output of the coupler to the Demod board (via some SMA-LEMO adaptor cables). The two (20dB) coupled ports were connected to the Agilent spectrum analyzer, which have input impedance 50ohms and hence should be impedance matched to the coupled outputs. I set the analyzer to span 1MHz (29-30MHz), IF BW 30Hz, 0dB input attenuation. It was not necessary to turn on averaging to resolve the peaks at ~29.5MHz since the IF bandwidth was fine enough.
I took two sets of measurements, one with the IMC well aligned (I maximized the MC Trans as best as I could to ~15,000 cts), and one with a macroscopic misalignment to MC1 such that the MC Trans fell to 90% of its usual value (~13,500 cts). The peak function on the analyzer was used to read off the peak height in dBm. I then converted this to RF power, which is summarized in the table below. I did not account for the main line loss of the coupler, but according to the datasheet, the maximum value is 0.25dB so there numbers should be accurate to ~10% (so I'm really quoting more S.Fs than I should be).
WFS Quadrant Pin (pW) Preflected(pW) Pin-demod board (pW)
## IMC well aligned
1 1 50.1 12.6 37.5
2 20.0 199.5 -179.6
3 28.2 10.0 18.2
4 70.8 5.0
65.8
2 5 100 19.6 80.0
6 56.2 158.5 -102.3
7 125.9 6.3 11.5
8 17.8 6.3
119.6
WFS Quadrant Pin (pW) Preflected(pW) Pin-demod board (pW)
## MC1 Misaligned
1 1 501.2 5.0 496.2
2 630.6 208.9 422
3 871.0 5.0 866
4 407.4 16.6
190.8
2 5 407.4 28.2 379.2
6 316.2 141.3 175.0
7 199.5 15.8 183.7
8 446.7 10.0 436.7
For the well aligned measurement, there was ~0.4mW incident on WFS1, and ~0.3mW incident on WFS2 (measured with Ophir power meter, filter out).
I am not sure how to interpret the numbers for quadrants #2 and #6 in the first table, where the reverse coupled RF power was greater than the forward coupled RF power. But this measurement was repeatable, and even in the second table, the reverse coupled power from these quadrants are more than 10x the other quadrants. The peaks were also well above (>10dBm) the analyzer noise floor
I haven't gone through the full misalginment -> Power coupled to TEM10 mode algebra to see if these numbers make sense, but assuming a photodetector responsivity of 0.8A/W, the product (P1P2) of the powers of the beating modes works out to ~tens of pW (for the IMC well aligned case), which seems reasonable as something like P1~10uW, P2 ~ 5uW would lead to P1P2~50pW. This discussion was based on me wrongly looking at numbers for the aLIGO WFS heads, and Koji pointed out that we have a much older generation here. I will try and find numbers for the version we have and update this discussion.
Misc:
1. For the sake of completeness, the LO levels are ~ -12.1dBm for both WFS demod boards (reflected coupling was negligible)
2. In the input signal coupled spectrum, there were side lobes (about 10dB lower than the central peak) at 29.44875 MHz and 29.52125 MHz (central peak at 29.485MHz) for all of the quadrants. These were not seen for the LO spectra.
3. Attached is a plot of the OSEM sensor signals during the time I misaligned MC1 (in both pitch and yaw approximately by equal amounts). Assuming 2V/mm for the OSEM calibration, the approximate misalignment was by ~10urad in each direction.
4. No IMC suspension glitching the whole time I was working today
Attachment 1: MC1_misalignment.png
12759 Fri Jan 27 00:14:02 2017 gautamSummaryIOOIMC WFS RF power levels
It was raised at the Wednesday meeting that I did not check the RF pickup levels while measuring the RF error signal levels into the Demod board. So I closed the PSL shutter, and re-did the measurement with the same measurement scheme. The detailed power levels (with no light incident on the WFS, so all RF pickup) is reported in the table below.
IMC WFS RF Pickup levels @ 29.5MHz
1 1 0.21 10.
2 1.41 148
3 0.71 7.1
4 0.16 3.6
2 1 0.16 10.5
2 1.48 166
3 0.81 5.1
4 0.56 0.33
These numbers can be subtracted from the corresponding columns in the previous elog to get a more accurate estimate of the true RF error signal levels. Note that the abnormal behaviour of Quadrant #2 on both WFS demod boards persists.
14709 Sun Jun 30 19:47:09 2019 ranaUpdateIOOIMC WFS agenda
we are thinking of doing a sprucing up of the input mode cleaner WFS (sensors + electronics + feedback loops)
1. it has been known since ~2002 that the RF circuits in the heads oscillate.
2. in the attached PDF you can see that 2 opamps (U3 & U4; MAX4106) are used to amplify the tank circuit made up of the photodiode capacitance and L6.
3. due to poor PCB layout (the output of U4 runs close to the input of U3) the opamps oscillate if the if the Reed relay (RY2) is left open (not attenuating)
4. we need to remove/disable the relay
5. also remove U3 for each quadrant so that it has a fixed gain of (TBD) and a 50 Ohm output
6. also check that all the resonances are tuned to 1f, 2f, & 3f respectively
2. Demod boards
4. Whitening
5. Noise budget of sensors, including electronics chain
6. diagonalization of sensors / actuators
7. Requirements -
8. Optical Layout
9. What does the future hold ?
1. what is our preferred pin-for-pin replacement for the MAX4106/MAX4107? internet suggests AD9632. Anyone have any experience with it? The Rabbott uses LMH6642 in the aLIGO WFSs. It has a lower slew rate than 9632, but they both have the same distortion of ~ -60 dB for 29.5 MHz.
2. the whole DC current readout is weird. Should have a load resistor and go into the + input of the opamp, so as to decouple it from the RF stuff. Also why such a fast part? Should have used a OP27 equivalent or LT1124.
3. LEMO connectors for RF are bad. Wonder if we could remove them and put SMA panel mount on there.
4. as usual, makes me feel like replacing with better heads...and downstream electronics...
15747 Sun Jan 3 16:26:06 2021 KojiUpdateSUSIMC WFS check (Yet another round of Sat. Box. switcharoo)
I wanted to check the functionality of the IMC WFS. I just turned on the WFS servo loops as they were. For the past two hours, they didn't run away. The servo has been left turned on. I don't think there is no reason to keep it turned off.
Attachment 1: Screen_Shot_2021-01-03_at_17.14.57.png
10728 Thu Nov 20 22:43:15 2014 KojiUpdateIOOIMC WFS damping gain adjustment
From the measured OLTF, the dynamics of the damped suspension was inferred by calculating H_damped = H_pend / (1+OLTF).
Here H_pend is a pendulum transfer function. For simplicity, the DC gain of the unity is used. The resonant frequency of the mode
is estimated from the OLTF measurement. Because of inprecise resonant frequency for each mode, calculated damped pendulum
has glitches at the resonant frequency. In fact measurement of the OLTF at the resonant freq was not precise (of course). We can
just ignore this glitchiness (numerically I don't know how to do it particularly when the residual Q is high).
Here is my recommended values to have the residual Q of 3~5 for each mode.
MC1 SUS POS current 75 -> x3 = 225 MC1 SUS PIT current 7.5 -> x2 = 22.5 MC1 SUS YAW current 11 -> x2 = 22 MC1 SUS SD current 300 -> x2 = 600
MC2 SUS POS current 75 -> x3 = 225 MC2 SUS PIT current 20 -> x0.5 = 10 MC2 SUS YAW current 8 -> x1.5 = 12 MC2 SUS SD current 300 -> x2 = 600
MC3 SUS POS current 95 -> x3 = 300 MC3 SUS PIT current 9 -> x1.5 = 13.5 MC3 SUS YAW current 6 -> x1.5 = 9 MC3 SUS SD current 250 -> x3 = 750
This is the current setting in the end.
MC1 SUS POS 150 MC1 SUS PIT 15 MC1 SUS YAW 15 MC1 SUS SD 450
MC2 SUS POS 150 MC2 SUS PIT 10 MC2 SUS YAW 10 MC2 SUS SD 450
MC3 SUS POS 200 MC3 SUS PIT 12 MC3 SUS YAW 8 MC3 SUS SD 500
Attachment 1: MC_OLTF_CLTF.pdf
17337 Mon Dec 5 20:02:06 2022 AnchalUpdateASCIMC WFS heads electronic feasibility test for using for Arm ASC
I took transfer function measurement of WFS2 SEG4 photodiode between 1 MHz to 100 MHz in a linear sweep.
### Measurement details:
• The reincarnated Jenne laser head was used for this test. The laser diode is 950 nm though, which should just mean a different responsivity of the photodiode while we are mainly interested in relative response of the WFS heads at 11 MHz and 55 MHz with respect to 29.5 MHz.
• See attachment 2 for how the laser was placed on AP table.
• The beam was injected in between beam splitter for MC reflection camera and beam splitter for beam dump.
• The input was aligned such that all the light of the laser was falling on Segment 4 of WFS2.
• Using moku, I took RF transfer function from 1 MHz to 100 MHz, 512 points, linearly spaced, with excitation amplitude of 1 V and 100,000 cycles of averaging.
• Measurement data and settings are stored here.
### Results:
Relative to 29.5 MHz, teh photodiode response is:
• At 11 MHz: -20.4 dB
• At 55 MHz: -36.9 dB
• At 71.28 MHz: -5.9 dB
I'm throwing in an extra number at the end as I found a peak at this frequency as well. This means to use these WFS heads for arm ASC, we need to have 10 times more light for 11 MHz and roughly 100 times more light for 55 MHz. According to Gautam's thesis Table A.1 and this elog post, the modulation depth for 11 MHz is 0.193 and for 55 MHz is 0.243 in comparison to 0.1 for 29.5 MHz., so the sideband TEM00 light available for beating against carrier TEM01/TEM10 is roughly twice as much for single arm ASC. That would mean we would have 5 times less error signal for 11 MHz and 40 times less error signal for 55 MHz. These are rough calculations ofcourse.
Attachment 1: 20221205_193105_WFS2_SEG4_RF_TF_Screenshot.png
Attachment 2: PXL_20221206_033419110.jpg
10561 Thu Oct 2 20:54:45 2014 KojiUpdateIOOIMC WFS measurements
[Eric Koji]
We made sensing matrix measurements for the IMC WFS and the MC2 QPD.
The data is under further analysis but here is some record of the current state to show
IMC Trans RIN and the ASC error signals with/without IMC ASC loops
The measureents were done automatically running DTT. This can be done by
/users/Templates/MC/wfsTFs/run_measurements
The analysis is in preparation so that it provides us a diagnostic report in a PDF file.
Attachment 1: IMC_RIN_141002.pdf
Attachment 2: IMC_WFS_141002.pdf
10564 Fri Oct 3 13:03:05 2014 ericqUpdateIOOIMC WFS measurements
Yesterday, Koji and I measured the transfer function of pitch and yaw excitations of each MC mirror, directly to each quadrant of each WFS QPD.
When I last touched the WFS settings, I only used MC2 excitations to set the individual quadrant demodulation phases, but Koji pointed out that this could be incomplete, since motion of the curved MC2 mirror is qualitatively different than motion of the flat 1&3.
We set up a DTT file with twenty TFs (the excitation to I & Q of each WFS quadrant, and the MC2 trans quadrants), and then used some perl find and replace magic to create an xml file for each excitation. These are the files called by the measurement script Koji wrote.
I then wrote a MATLAB script that uses the magical new dttData function Koji and Nic have created, to extract the TF data at the excitation frequency, and build up the sensing elements. I broke the measurements down by detector and excitation coordinate (pitch or yaw).
The amplitudes of the sensing elements in the following plots are normalized to the single largest response of any of the QPD's quadrants to an excitation in the given coordinate, the angles are unchanged. From this, we should be able to read off the proper digital demodulation angles for each segment, confirm the signs of their combinations for pitch and yaw, and construct the sensing matrix elements of the properly rotated signals.
The axes of each quadrant look consistent across mirrors, which is good, as it nails down the proper demod angle.
The xml files and matlab script used to generate these plots is attached. (It requires the dttData functions however, which are in the svn (and the dttData functions require a MATLAB newer than 2012b))
Attachment 5: analyzeWfs.zip
10565 Sun Oct 5 10:09:49 2014 ranaUpdateIOOIMC WFS measurements
It seems clever, but I wonder why use DTT and command line perl, instead of using the FE lockins or just demod the offline data or all of the other sensing matrix scripts made for the LSC (at 40m) or ASC (at LLO) ?
10566 Sun Oct 5 23:43:08 2014 KojiUpdateIOOIMC WFS measurements
There are several non scientific reasons.
16108 Mon May 3 09:14:01 2021 Anchal, PacoUpdateLSCIMC WFS noise contribution in arm cavity length noise
Lock ARMs
• Try IFO Configure ! Restore Y Arm (POY) and saw XARM lock, not YARM. Looks like YARM biases on ITMY and ETMY are not optimal, so we slide C1:SUS-ETMY_OFF from 3.0 --> -14.0 and watch Y catch its lock.
• Run ASS scripts for both arms and get TRY/TRX ~ 0.95
• We ran X, then Y and noted that TRX dropped to ~0.8 so we ran it again and it was well after that. From now on, we will do Y, then X.
WFS1 noise injection
• Turn WFS limits off by running switchOffWFSlims.sh
• Inject broadband noise (80-90 Hz band) of varying amplitudes from 100 - 100000 counts on C1:IOO-WFS1_PIT_EXC
• After this we try to track its propagation through various channels, starting with
• C1:LSC-XARM_IN1_DQ / C1:LSC-YARM_IN1_DQ
• C1:SUS-ETMX_LSC_OUT_DQ / C1:SUS-ETMY_LSC_OUT_DQ
• C1:IOO-MC_F_DQ
• C1:SUS-MC1_**COIL_OUT / C1:SUS-MC2_**COIL_OUT / C1:SUS-MC3_**COIL_OUT
• C1:IOO-WFS1_PIT_ERR / C1:IOO-WFS1_YAW_ERR
• C1:IOO-WFS1_PIT_IN2
** denotes [UL, UR, LL, LR]; the output coils.
• Attachment 1 shows the power spectra with IMC unlocked
• Attachment 2 shows the power spectra with the ARMs (and IMC) locked
Attachment 1: WFS1_PIT_Noise_Inj_Test_IMC_unlocked.pdf
Attachment 2: WFS1_PIT_Noise_Inj_Test_ARM_locked.pdf
ELOG V3.1.3-
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In this example, the function is undefined where x is 0. g (0) = 24 (0) 7 6 (0) 5 = 0 0 u n d e f i n e d. Therefore, the domain consists of all real numbers x, where x ≠ 0. HA : approaches 0 as x increases. For example, a quadratic for the numerator and a cubic for the denominator is identified as a quadratic/cubic rational function. By comparing the interpolating accuracy, we can note that … Like logarithmic and exponential functions, rational functions may have asymptotes. All polynomials are rational functions. just create an account. Cancel common factors. But what if there are common factors between the numerator and denominator of a rational function? Let’s look at an example of a rational function that exhibits a “hole” at one of its restricted values. All rights reserved. Sal matches three graphs of rational functions to three formulas of such functions by considering asymptotes and intercepts. Since x^2 + 1 = 0 has no real solutions, the only vertical asymptote comes from x + 3 = 0. Quiz & Worksheet - Who is Judge Danforth in The Crucible? (+) Graph rational functions, identifying zeros and asymptotes when suitable factorizations are available, and showing end behavior. We highlight the first step. 's' : ''}}. The factor x+1 in the denominator does not cancel, so x+1=0 gives x=-1 as a vertical asymptote. Rational Functions A rational function is a function of the form where g (x) 0 Domain of a Rational Function The domain of a rational function is the set of real ... – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow.com - … They … Note that these look really difficult, but we’re just using a lot of steps of things we already know. The function R(x) = 1 / ((x - 1)(x^2 + 3)) is a rational function since the numerator, 1, is a polynomial (yes, a constant is still a polynomial) and the denominator, (x - 1)(x^2 + 3), is also a polynomial (it's just in a factored form). This is what we call a horizontal asymptote. Graph rational functions In Example 9, we see that the numerator of a rational function reveals the x -intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. Thanks to all of you who support me on Patreon. That is the case in this example, since both the numerator and denominator are cubic polynomials. Conflict Between Antigone & Creon in Sophocles' Antigone, Quiz & Worksheet - Metaphors in The Outsiders, Quiz & Worksheet - Desiree's Baby Time & Place, Quiz & Worksheet - The Handkerchief in Othello. Rational functions follow the form: In rational functions, P(x) and Q(x) are both polynomials, and Q(x) cannot equal 0. In a similar way, any polynomial is a rational function. where n n is the largest exponent in the numerator and m m is the largest exponent in the denominator. These are called the holes of rational functions. It is also known as a Rational Expression. Our vertical asymptotes exist at x = 0 and x = -1. A rational function is a function made up of a ratio of two polynomials. is the constant of variation. For example, the rational function R(x) = ((x+1)(x-1))/(x-1) has a common factor of x-1 in the numerator and denominator. Quiz & Worksheet - What is Abstract Conceptualization? For example, the following is a rational function: $$f(x)=\frac{4x+4}{6x-9}$$ How do we add or subtract them? 00:29. © copyright 2003-2021 Study.com. Since our denominator is x^2 + x, we'll set it equal to 0 and solve for x. Rational functions follow the form: In rational functions, P(x) and Q(x) are both polynomials, and Q(x) cannot equal 0. We have to find what values of x make our denominator equal to 0. A vertical asymptote at a value x is when the value of our function approaches either positive or negative infinity when we evaluate our function at values that approach x (but are not equal to x). Therefore, we have x^2 + x = 0. If there are more instances of the common factor in the denominator, the result is a vertical asymptote. She has over 10 years of teaching experience at high school and university level. Rational functions supply important examples and occur naturally in many contexts. Find Rational Functions (1) This is an analytical tutorial on rational functions to further understand the properties of the rational functions and their graphs. Equations and Inequalities. Integration of Rational Functions Recall that a rational function is a ratio of two polynomials $$\large{\frac{{P\left( x \right)}}{{Q\left( x \right)}}}\normalsize.$$ We will assume that we have a proper rational function in which the degree of the numerator is less than the degree of the denominator. If we were to cancel the common factors, R(x) would look like R(x)=x+1. 8.1 Model Inverse and Joint Variation. Is the integral of a rational function always a rational function? Try to picture an imaginary line y = 0. This method can also be used with rational … Polynomial and rational functions are examples of _____ functions. Exponential and Logarithmic Functions. ****UPDATED**** NOW WITH 11 THOUGHTFUL EXAMPLES This is a great one page cheat sheet on graphing rational functions. / (x 3 + ….) The function R(x) = (-2x^5 + 4x^2 - 1) / x^9 is a rational function since the numerator, -2x^5 + 4x^2 - 1, is a polynomial and the denominator, x^9, is also a polynomial. In other words, R(x) is a rational function if R(x) = p(x) / q(x) where p(x) and q(x) are both polynomials. Rational Function Holes – Explanation and Examples. Few examples of rational equations are given below: – Example No.1: Solve Some examples and diagrams are taken from the textbook. 2.For finding distance. At what points is the function y = \frac{x + 9}{x^2 - 14x + 45} continuous? Exponential and Logarithmic Functions. Evaluate the integral: integral fraction {x^2 - 48}{x+7}dx, Identify the extent, leading coefficient, and constant coefficient in the following polynomial functions: (a) f ( x ) = 9 x 2 ? This is because if x = 0, then the function would be undefined. Precalculus 10th. The control of a medical dosage is a great example of a scenario where rational functions can be used to ensure safety for patient, when anesthetic enters the consumers body, the amount of medicine must be measured to determine the … You will learn more about asymptotes later on. In particular: f(x) = (3x 3 + ….) 25 chapters | As you can see, is made up of two separate pieces. Analyze the function f(x)= (x^3-4x^2-31x+70)/(x^2-5x+6). Sal matches three graphs of rational functions to three formulas of such functions by considering asymptotes and intercepts. A rational function is one such that f(x)=P(x)Q(x)f(x)=P(x)Q(x), where Q(x)≠0Q(x)≠0; the domain of a rational function can be calculated. That’s the fun of math! In this lesson you will learn about rational functions, discontinuities and how we can use them to model real-life scenarios. Let f be a rational function given by f (x) = \dfrac {2x + 2} {x+1} . algebraic. The function R(x) = (x - 4) / x^(-2/3) + 4 is not a rational function since the denominator, x^(-2/3) + 4, is not a polynomial since the exponent of x is not a non-negative integer. I. Find the asymptotes. Rational Functions are used in real-life situations as well. This is simply a brief introduction to the topic. We have a hole at (5, 91/3). rational functions - functions which are ratios of polynomials. Example $$\PageIndex{2}$$ Sketch the graph of $f(x)=\frac{x-2}{x^{2}-4}$ Solution. Once again, that's great news because that means we can use our theorem! A rational function is a fraction of polynomials. (12.7.1) f (x) = x 2 − 4 2 x 2 + x − 3 Why study rational and radical functions? 1 Ex. In particular, they are quite good for describing distance-speed-time questions, and modeling multi-person work problems. The numerator is p(x)andthedenominator is q(x). 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The height of an object launched upward, ignoring the effects of air resistance, can be modeled with the following quadratic function: h (t) = − 1 2 g t 2 + v 0 t + s 0. Domain: Recall that for the … Therefore, we have (x + 3)(x^2 + 1) = 0. | PBL Ideas & Lesson Plans, SAT Subject Test Physics: Tutoring Solution, College Mathematics Syllabus Resource & Lesson Plans, Foundations of Education for Teachers: Professional Development, Social Psychology for Teachers: Professional Development, Reading Comprehension on the LSAT: Tutoring Solution, Understanding Function Operations: Tutoring Solution, Quiz & Worksheet - Effect of pH on LeChatelier's Principle. In Example$$\PageIndex{10}$$, we see that the numerator of a rational function reveals the x-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. Complex analysis. RATIONAL FUNCTIONS A rational function is a function of the form: () () ()xq xp xR = where p and q are polynomials 2. With this formula, the height h (t) can be calculated at any given time t after the object is launched. In complex analysis a rational function is the ratio of two polynomials with complex coefficients. Well, we technically did use the numerator since we had to make sure there were no common factors between the numerator and denominator. Find all vertical asymptotes of the function: Get access risk-free for 30 days, Rational Functions Here is a set of practice problems to accompany the Rational Functions section of the Common Graphs chapter of the notes for Paul Dawkins Algebra course at Lamar University. Rational Function with Removable Discontinuity And lastly, we plot points and test our regions in order to create our graph! You can test out of the Note that the numerator and denominator can be polynomials of any order, but the rational function is undefined when the denominator equals zero. f (x) = \dfrac {2 (x+1)} {x+1} = 2 , for x \ne -1 . Find all the vertical asymptotes of the function: First, we see that R(x) is indeed a rational function (because remember, a factored polynomial is still a polynomial) with no common factors between the numerator and denominator. . There are no common factors, so using the theorem from the lesson, we have vertical asymptotes when x+1=0 or x-2=0, so we have vertical asymptotes at x=-1 and x=2. First I'll find any vertical asymptotes, by setting the denominator equal to zero and solving: Since this equation has no solutions, then the denominator is never zero, and there are no vertical asymptotes. Because by definition a rational function may have a variable in its denominator, the domain and range of rational functions … Example. Richard Wright, Andrews Academy . Limit of a Rational Function, examples, solutions and important formulas. Examples Ex. Remember, a rational function is a function that is a fraction where both its numerator and denominator are polynomials. Working Scholars® Bringing Tuition-Free College to the Community, Find all holes or vertical asymptotes for the rational function R(x) = ((x^2+1)(x-5)(x+2)) / ((x-5)(x+1)), Find all holes or vertical asymptotes for the rational function R(x) = (x+3) / ((x+1)(x-2)). Kathryn earned her Ph.D. in Mathematics from UW-Milwaukee in 2019. x-1 =0 when x=1, so we have a hole at x=1. Runge's function is the classical example of a function which cannot be interpolated by a polynomial on an equidistant grid. Mathematical articles, tutorial, examples. Jenn, Founder Calcworkshop ®, 15+ Years Experience (Licensed & Certified Teacher) First, we need to make sure that our function is in it’s lowest … Try refreshing the page, or contact customer support. ., a-sub-n are all real numbers and the exponents of each x is a non-negative integer. Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher) Before we get ahead of ourselves, let’s first talk about what a Limit is. Then, the real values of x that make our denominator equal to 0 will have vertical asymptotes. De nition 4.1. We will assume that we have a proper rational function in which the degree of the numerator is less than the degree of the denominator. The holes in a rational function are the result of it sharing common factors shared by the numerator and denominator. Create an account to start this course today. Make a table of values around the vertical asymptotes. Wait, we didn't use the numerator! () () ()xq xp xR = What would the … That's great because that means we can use the theorem! We explain Rational Functions in the Real World with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. From anesthesia to economics, rational functions are used in multiple areas of study to help predict outcomes. From anesthesia to economics, rational functions are used in multiple areas of study to help predict outcomes. The general form of a rational function is p ( x ) q ( x ) , where p ( x ) and q ( x ) are polynomials and q ( x ) ≠ 0 . 3.For finding time etc.. Finally, check your solutions and throw out any that make the denominator zero. flashcard sets, {{courseNav.course.topics.length}} chapters | Chapter 3. Recall that you can solve equations containing fractions by using the least common denominator of all the fractions in the equation. Sciences, Culinary Arts and Personal This example may help clarify the idea of a vertical asymptote: We see there is a vertical asymptote when x = 1 since the function is approaching negative infinity as we approach 1 from the left, and the function is approaching positive infinity as we approach 1 from the right. Solving Rational Equations ©2001-2003www.beaconlearningcenter.com Rev.7/25/03 SOLVING RATIONAL EQUATIONS EXAMPLES 1. A rational function written in factored form will have an x-intercept where each factor of the numerator is equal to zero. Fortunately, the effect on the shape of the graph at those intercepts … Example 2. (a)x = 2 (b)x = 3 (c)x = 4, Another model of population growth that has been used to model salmon is the Bcvcrton-Holtz model given by f(x) = \frac{rx}{1+\frac{x}{b}} where r and b are positive constants. Learn all about them in this lesson! . + a-sub-n * x^n, where a-sub-0, a-sub-1, . Sociology 110: Cultural Studies & Diversity in the U.S. CPA Subtest IV - Regulation (REG): Study Guide & Practice, Properties & Trends in The Periodic Table, Solutions, Solubility & Colligative Properties, Creating Routines & Schedules for Your Child's Pandemic Learning Experience, How to Make the Hybrid Learning Model Effective for Your Child, Distance Learning Considerations for English Language Learner (ELL) Students, Roles & Responsibilities of Teachers in Distance Learning, Between Scylla & Charybdis in The Odyssey, Hermia & Helena in A Midsummer Night's Dream: Relationship & Comparison. Application of Rational Functions As I mentioned earlier, accuracy has a large role in medicine and very few mistakes can be made. Quiz & Worksheet - The Civil War West of the Mississippi River, Quiz & Worksheet - Melancholy Temperaments, What Is Asphyxia? The examples have detailed solutions in this page, the matched exercises have answers here. Learn how to apply the formula for rational functions in difference circumstances to provide a better understanding of a situation … Find the y-intercept, the x-intercept(s), the removable singularities, the vertical asymptotes and the horizontal asymptotes. Because there is a variable in both the numerator and denominator, there are effects on both the Log in here for access. Ex. What is the Main Frame Story of The Canterbury Tales? A rational function is an equation that takes the form y = N(x)/D(x) where N and D are polynomials.Attempting to sketch an accurate graph of one by hand can be a comprehensive review of many of the most important high school math topics from basic algebra to differential calculus. Also, note in the last example, we are dividing rationals, so we flip the second and multiply. Graphs of rational functions (old example) Our mission is to provide a free, world-class education to anyone, anywhere. Vertical asymptotes occur at x-values when the denominator of a rational function equals 0 and the numerator does not equal 0. flashcard set{{course.flashcardSetCoun > 1 ? Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. If a rational function does not have a constant in the denominator, the graph of the rational function features asymptotic behavior and … Rational Functions Word Problems - Work, Tank And Pipe Here are a few examples of work problems that are solved with rational equations. Answer: g (x) = 4 x 2, where x ≠ 0. Plus, get practice tests, quizzes, and personalized coaching to help you Hence, the only vertical asymptote occurs at x = -3. . x ↑, y ↑ Inverse Variation: =. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons We explain Rational Functions in the Real World with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Anyone can earn Again, the function never touches this line, but gets very close to it. 2 HA: because because approaches 0 as x increases. The cost C (in dollars) of removing p amount (percent) of the smokestack pollutants is given by:. Peter has taught Mathematics at the college level and has a master's degree in Mathematics. Richard Wright, Andrews Academy . a. Factor 2 out in the numerator. They are overwhelmed when presented with multiple pages of note Multiplying each side of the equation by the common denominator eliminates the fractions. Is it possible … When adding or subtracting rational functions, you must find a common denominator as you might do with regular fractions. ... 8.3 Graph General Rational Functions. y depends on both x and z a . Some examples and diagrams are taken from the textbook. Graphing Rational Functions: Introduction (page 1 of 4) Sections: Introduction, Examples , The special case with the "hole" To graph a rational function, you find the asymptotes and the intercepts , plot a few points, and then sketch in the graph. :) https://www.patreon.com/patrickjmt !! Example 2 Holes. Let f(x) = \frac{x^{2} -4}{x^{2} - z -6} Determine as the following points. \$1 per month helps!! Solve the equation. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Graphing Rational Functions: An Example (page 2 of 4) Sections: Introduction, Examples, The special case with the "hole" Graph the following: First I'll find any vertical asymptotes, by setting the denominator equal to zero and solving: x 2 + 1 = 0 x 2 = –1. Since our denominator is (x + 3)(x^2 + 1), we'll set it equal to 0 and solve for x. One of the most unique properties of a rational function is that it may have vertical asymptotes. Frequently, rationals can be simplified by factoring the numerator, denominator, or both, and crossing out factors. I thought I would post some of what we discussed, as a summary of how to analyze rational functions. Over 83,000 lessons in all major subjects, {{courseNav.course.mDynamicIntFields.lessonCount}}, How to Add and Subtract Rational Expressions, Practice Adding and Subtracting Rational Expressions, How to Multiply and Divide Rational Expressions, Multiplying and Dividing Rational Expressions: Practice Problems, Solving Rational Equations with Literal Coefficients, Biological and Biomedical Rational equations. Section 3.3 Rational Functions of the Form (˘)= #ˇˆ= >ˇˆ% In this section you will look at polynomial functions in which both the numerator and denominator are linear expressions. To graph rational functions. In this section, we’ll learn the different approaches we can use to find the limit of a given rational function. To unlock this lesson you must be a Study.com Member. Examples of rational functions Rational function of degree 3, with a graph of degree 3: y = x 3 − 2 x 2 ( x 2 − 5 ) {\displaystyle y={\frac {x^{3}-2x}{2(x^{2}-5)}}} Rational function of degree 2, with a graph of degree 3: y = x 2 − 3 x − 2 x 2 − 4 {\displaystyle y={\frac {x^{2}-3x-2}{x^{2}-4}}} Feel Free TO WATCH and LEARN! . The parent rational function is =1 . Application of Rational Functions Application 1. "Rational function" is the name given to a function which can be represented as the quotient of polynomials, just as a rational number is a number which can be expressed as a quotient of whole numbers. and career path that can help you find the school that's right for you. Slides created by . 99% of my students only want notes that fit on one page. Curious as to why these points remain unfilled? 99% of my students only want notes that fit on one page. credit by exam that is accepted by over 1,500 colleges and universities. Factor Numerators and Denominators. They can be multiplied and dividedlike regular fractions. If the quotient is constant, then y = this constant is the equation of a horizontal asymptote. Rational functions are ratios of polynomial functions, like the examples below. Let's use this theorem to find vertical asymptotes! Here are some examples. The complex analytic functions we construct will give conformal maps from The value of horizontal asymptotes depends on certain characteristics of the polynomials in the rational function. A rational function is a function which is the ratio of polynomial functions. The graph of function f is a horizontal line with a hole (function not defined) at x = -1 as shown below. Floater-Hormann's rational interpolation solves this problem. You might be thinking. Answer. Below is a simple example of a basic rational function, f (x). For example - 1.For finding pressure. Graphing Rational Functions. A rational function is a function thatcan be written as a ratio of two polynomials. We can use the following theorem: Theorem: Let R(x) be a rational function with no common factors between the numerator and the denominator. Topics. As with polynomials, factors of the numerator may have integer powers greater than one. Rational Functions Example November 12, 2020 / Suman Ganguli / 1 Comment. Were we supposed to? 6 x + 16 (b) f ( x ) = ? Factoring the left hand side, we get x(x + 1) = 0. Most rational functions will be made up of more than one piece. Rational Functions Graphing Cheat Sheet and Graph Paper. Rational expressions and rational equations can be useful tools for representing real life situations and for finding answers to real problems. A rational function is a ratio of polynomial functions. To transform the rational function , you can apply the general expression for function transformations. The function R(x) = (sqrt(x) + x^2) / (3x^2 - 9x + 2) is not a rational function since the numerator, sqrt(x) + x^2, is not a polynomial since the exponent of x is not an integer. To learn more, visit our Earning Credit Page. There is a common factor of x-5. A rational function is defined as the quotient of polynomials in which the denominator has a degree of at least 1 . Select a subject to preview related courses: First, we see that R(x) is indeed a rational function with no common factors between the numerator and denominator. Of x make our denominator equal to 0 an imaginary line x = 0 and solve for x \ne.... College you want to attend yet of x make our denominator is x^2 + x, we plot points test! Functions come from the textbook: g ( x ) is not the polynomial... When you cross out f… rational functions supply important examples and occur naturally in many contexts )... Of the numerator is equal to 0, we get x ( x ) = 24 4 x 6! Matched exercises have answers here questions, and personalized coaching to help succeed! Rev.7/25/03 Solving rational equations can be polynomials of any order, but the rational Expressions and rational inequalities m! The height h ( t ) can be expressed as a quadratic/cubic rational is! 3 ) ( 3 ) nonprofit organization recursive process, or steps for calculation from a.. Height h ( t ) can be calculated at any given time t after the object is.. Learn the different approaches we can note that the numerator and denominator are polynomials variable. Of their respective owners this lesson you must find a common denominator as you can test out the... Y coordinate of 2 to zero applied math the numerator may have vertical of... What points is the function y = \frac { x + 16 ( b ) f ( x ) 24. Like a ratio. real values of x make our denominator equal to 0 because approaches as... Horizontal line with a hole ( function not defined ) at x = 0 (! - functions which are ratios of polynomial functions make the denominator is identified as a of! A variable in the numerator and denominator are polynomials examples 1 function that can be useful tools for representing life. You must be a variable in the rational function R ( x ) is! Side of the Canterbury Tales the LCD of all, a rational is. And rational inequalities hole when x=5, so we have x^2 + 1 = 0, then =... ( 3 ) nonprofit organization … rational functions are typically identified by the numerator and denominator are polynomials h t... Of Solving the rational … example 2 Holes other trademarks and copyrights are the result a. X 2 6 x + 16 ( b ) f ( x ) = 24 x! ratio. the second and multiply functions supply important examples and diagrams taken... Such functions by considering asymptotes and intercepts and exams but we ’ learn. Given a rational function is a 501 ( c ) f ( x ) = 0 close to it and. Average cost, percentage and mixture problems since both the numerator and denominator are polynomials the . Years of teaching experience at high school and university level of their respective owners by D x... Create an account example, f ( x ) the largest exponent in the denominator up to add lesson. Recall that a rational function is defined as the quotient of two polynomials = 2x... Given time t after the object is launched the algebraic equivalent of rational functions examples function. Are common factors rational functions examples by the common factors between the numerator and denominator are.. To anyone, anywhere and nally the trigonometric functions and their inverses HA: because... To create our graph that 's great because that means we can use to what! Exponential functions, then continuing on to the exponential and loga-rithm functions, you can out... Multiplying each side of the first two years of college and save thousands off degree. Functions and their inverses apply the general expression for function transformations to attend yet each part rational functions examples to 0 graph! Functions to three formulas of such functions by considering asymptotes and intercepts because that we! Is Asphyxia singularities, the real values of x that make our denominator equal to 0 will have asymptotes... Determine an explicit expression, a rational function is a rational number multi-person work...., the height h ( t ) can be expressed as a ratio of two separate.. 1 = 0 has no real solutions, the only vertical asymptote at... Are both polynomials = this constant is the integral of a rational is! For representing real life situations and for finding answers to real problems function y this. Because that means we can use the denominator is identified as a summary of how to analyze rational are... To a Custom Course x^2 - 14x + 45 } continuous zero polynomial by considering and. The rational functions, and personalized coaching to help predict outcomes the examples below a common denominator as might. Judge Danforth in the last example, since both the numerator is to. To the exponential and loga-rithm functions, like the examples have detailed solutions in this video you will learn distinguish! Considering asymptotes and intercepts you will learn to distinguish rational functions, the! College level and has the property of their respective owners no real,. It may have integer powers greater than one piece: g ( x =! 2 ( x+1 ) } { x^2 - 14x + 45 } continuous also, note in the.. Function thatcan be written as the quotient is constant, then y = axz = x^3-4x^2-31x+70. A horizontal asymptote … rational functions are typically identified by the degrees of the rational functions examples may have integer powers than... The … below is a simple example of a basic rational function is pretty just!, factors of the numerator may have asymptotes we technically did use the,. To actually computing our vertical asymptotes occur at x-values when the denominator equals.! X ≠ 0 and multiply to zero ( in dollars ) of removing p amount ( percent ) removing... Following example: y = ( 3x 3 + …. this point on, most of the Canterbury?. Basic rational function is a function that is a non-negative integer: help and Review to. That make our denominator equal to 0 x=-1 as a ratio of integers: p/q in dollars ) of p... ), where a-sub-0, a-sub-1, x^2 + x, we should probably define a vertical asymptote y... Is made up of more than one piece West of the numerator is p ( )... Hand side, we get x ( x ) = \dfrac { 2 ( x+1 ) } x+1!, examples, solutions and important formulas research, mathematical modeling, mathematical modeling mathematical. We flip the second and multiply one of the function =1 has a master 's degree Mathematics. Our regions in order to create our graph denominator can be expressed as a summary of how to analyze functions. Given rational function from UW-Milwaukee in 2019 of such functions by considering asymptotes intercepts. A context expression, a quadratic for the … below is a function made up of than... \Ne -1 } continuous 2 ) points is the case in this section, should. Will learn to distinguish rational functions may have integer powers greater than one are cubic polynomials find out LCD. Is p ( x ) = 0 and the horizontal asymptotes this is a. 30 days, just create an account of 2 for function transformations and loga-rithm functions, you can,! Coal to generate electricity whose numerator and denominator can be calculated at any given time t after the is! In Mathematics hole when x=5 points and test our regions in order to create our!... Between the numerator and a horizontal asymptote of a rational function, examples, and. Gmat Prep: help and Review page to learn more, visit our Earning Credit page smokestack pollutants is by! & Worksheet - the Civil War West of the word rational '' is ratio. use. =0 when x=1, so we flip the second and multiply comes to actually computing our vertical asymptotes and horizontal... Peter has taught Mathematics at the point ( 1,2 ) horizontal line with a hole ( function not defined at... Calculated at any given time t after the object is launched 7 x 2 + 10 c. Where q ( x ) = 0 the page, the only vertical asymptote comes from x + x... ) f ( x ) andthedenominator is q ( x ) anyone can earn regardless... I would post some of what we discussed, as a quadratic/cubic rational function always a rational function defined. Situations and for finding answers to real problems help predict outcomes type of function f ( x ) \dfrac. Function =1 has a vertical asymptote occurs at x = -1 is q ( x.. You who support me on Patreon result is a simple example of Limits at!! Tests, quizzes, and nally the trigonometric functions and their inverses free, world-class education to anyone,.... From a context example of Limits at Infinity 1: Represent the speed = =. Rev.7/25/03 Solving rational equations examples 1 check your solutions and throw out any that make the is. Good for describing distance-speed-time questions, and personalized coaching to help you.. And the exponents of each x is a non-negative integer one is being divided by numerator! … rational functions, and personalized coaching to help predict outcomes domain: recall that you can out. Formulas of such functions by considering asymptotes and intercepts this theorem to find vertical asymptotes and intercepts Solving. With a hole ( function not defined ) at x = -3 is to a. Real numbers and the exponents of each x is a simple example of a rational function R ( )... Real solutions, the matched exercises have answers here Mathematics at the level! Time t after the object is launched = 0 need to find vertical.
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# Beamer+Xelatex+Hebrew produce empty presentation
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I am trying to compile some hebrew presentation with xelatex. The code is:
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\end{frame}
\end{document}
The compilation reaches to the ends but the slides are empty. Take a look in the attached files.
Ido
1. This is caused by polyglossia not by beamer itself. While the lack of RTL support in beamer is not ideal this can't be regarded as a bug in the class.
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# A particle of charge $-e$ orbits a particle of charge $Ze$, what is its orbital frequency?
A point particle $P$ of charge $Ze$ is fixed at the origin in 3-dimensions, while a point particle $E$ of mass $m$ and charge $-e$ moves in the electric field of $P$.
I have the Newtonian equation of motion as $$- \frac{Ze^2}{4 \pi \epsilon_{0}} \frac{\vec{r}}{r^3} = m \ddot{\vec{r}}$$ I derived this from Newton's law and Coulomb's law.
I then went on to show that the particle moves in a plane by showing it has a constant normal vector.
Can anyone help me find this if the orbit is circular about $P$ what it's orbital frequency is -in terms of the constants we have? I would be extremely grateful.
I have attempted this by parameterizing the orbit, differentiating with respect to time and substituting into the equation of motion, however all I get are two unsolvable differentials equations.
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An important point to note are the conserved quantities: you have the angular momentum and the energy conserved (which helps you to reduce the differential equation to a more manageable form). For the circular orbit this is all you need. There is however an additional conserved quantity called the Runge-Lenz vector. – Fabian Oct 9 '12 at 19:11
If the orbit is circular, then $r=R$, $\vec{r} = R(\cos \omega t,\sin \omega t)$, and $\ddot{\vec{r}} = -R \omega^2 (\cos \omega t, \sin \omega t)$. Popping this values into your equation, you get
$$\frac{Ze^2}{4\pi \varepsilon_0 m} = R^3 \omega^2$$.
So your orbital frequency depends on the radius of the orbit, as well as $e$, $Z$ and $m$. This is a variant of Kepler's Third Law of Planetary Motion.
You may want to look at Bohr's Model for a discussion on how this result can be linked with quantun mechanics in the simplest atomic model.
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Thank you! I was trying to solve the parameters as functions of $t$ hence the complicated differential equations, this is very clear now – Freeman Oct 9 '12 at 20:07
This is a standard exercise in Classical physics, Several text books deal with this problem in the context of Gravitational physics. It is same problem with different constants. for example see Landau Classical Mechanics.
The most general orbit is elliptical, Only if there are special initial conditions the particle moves in a circular orbit.
You can easily calculate The velocity required to maintain circular orbit, at a given radius. Apply the concept of centripetal force.
If you have to however solve the problem using quantum mechanics, everything changes.
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# How do I use glFramebufferTexture2DEXT() ?
This topic is 2267 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hello,
I want to do some render-to-texture functionality in my game. When I try to call glFramebufferTexture2DEXT(), I get the following error when I try to compile it:
[ 83%] Building CXX object Plugins/TextureSet/C64SpindizzyTextureSet/CMakeFiles/TextureSet_C64SpindizzyTextureSet.dir/C64SpindizzyTexture.cpp.o /home/martin/spindizzyremake/Plugins/TextureSet/C64SpindizzyTextureSet/C64SpindizzyTexture.cpp: In member function 'void C64SpindizzyTexture::setRenderTarget()': /home/martin/spindizzyremake/Plugins/TextureSet/C64SpindizzyTextureSet/C64SpindizzyTexture.cpp:25: error: 'glFramebufferTexture2DEXT' was not declared in this scope make[2]: *** [Plugins/TextureSet/C64SpindizzyTextureSet/CMakeFiles/TextureSet_C64SpindizzyTextureSet.dir/C64SpindizzyTexture.cpp.o] Error 1 make[1]: *** [Plugins/TextureSet/C64SpindizzyTextureSet/CMakeFiles/TextureSet_C64SpindizzyTextureSet.dir/all] Error 2 make: *** [all] Error 2
My function that calls this function simply looks like this:
void C64SpindizzyTexture::setRenderTarget() { glFramebufferTexture2DEXT(GL_FRAMEBUFFER_EXT, GL_COLOR_ATTACHMENT0_EXT, GL_TEXTURE_2D, cTextureID, 0); }
I've trying include both <GL/gl.h> and <GL/glext.h>, but still get the same error.
Surprisingly, I could not find any formal documentation on the use of this function on Google. Does anyone know how I can proceed to use this function so that I can perform render-to-texture in my game?
With Kind Regards,
MartinB
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I got the function from glew.h (google it), there its defined in like that:
#define glFramebufferTexture2DEXT GLEW_GET_FUN(__glewFramebufferTexture2DEXT)
Don't forget to call glewInit() once.
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Thanks. Turns out I had to change every <GL/gl.h> in my project to <GL/glew.h> too, otherwise it was failing for including <GL/gl.h> before <GL/glew.h>. Once I'd done that, it worked after calling glewInit().
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# Is the “field” I learned about in analysis different from the “field” I learned about in econometrics?
Today was my first day of econometrics, and real analysis, and in both courses the professor defined something called a "field". Unfortunately, the field I learned about in real analysis seems completely different from the field I learned about in econometrics.
The field from analysis is a set with addition and multiplication which obeys 11 familiar axioms. I had already been familiar with this definition of a field from linear algebra but here it was again.
But the field from econometrics was completely different! My notes say,
If $S$ is a sample space, a collection of subsets $\mathcal{S}$ of $S$ is called a field if:
1. $S\in\mathcal{S}$
2. Whenever $A\in \mathcal{S}$, $A^C\in \mathcal{S}$
3. Whenever $A$ and $B$ are in $\mathcal{S}$, $A\cup B\in \mathcal{S}$
Is the field that I learned about in econometrics a completely different thing? Or are they somehow related?
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Sounds like a $\sigma$-algebra to me. – Rahul Sep 7 '12 at 0:01
@RahulNarain $\sigma$-algebras are closed under countable union. – William Sep 7 '12 at 0:02
@Williamm and Rahul It is sometimes simply called an algebra. – M Turgeon Sep 7 '12 at 0:05
According to wikipedia, a $\sigma$-algebra is sometimes called a $\sigma$-field. So I guess, some people may call a regular algebra a field. – William Sep 7 '12 at 0:07
The definition you gave is that of a field of sets (en.wikipedia.org/wiki/Field_of_sets). I suppose in some contexts it might just be called a field. – Trevor Wilson Sep 7 '12 at 0:13
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Ask question
For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places. begin{array}{|c|c|}hline x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 hline f(x) & 409.4 & 260.7 & 170.4 & 110.6 & 74 & 44.7 & 32.4 & 19.5 & 12.7 & 8.1 hline end{array}
Question
Exponential models
asked 2021-02-18
For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{x}&{1}&{2}&{3}&{4}&{5}&{6}&{7}&{8}&{9}&{10}\backslash{h}{l}\in{e}{f{{\left({x}\right)}}}&{409.4}&{260.7}&{170.4}&{110.6}&{74}&{44.7}&{32.4}&{19.5}&{12.7}&{8.1}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Answers (1)
2021-02-19
Step 1
Remember that regression analysis is the process of looking for a best fit of model for a set of data. This can be done on a graphing utility as follows:
1. Press [STAT], the input corresponging x-values of data in L1, and y-values of data in L2.
2. Use [STATPLOT] to observe a scatterplot of the data.
3. Press [STAT], then [CALC] then [ExpReg]/[LnReg]/[Logistic].
This will show you a function in either the form of an exponential, a logarithmic or a logistic model.
4. Graph this equation on the same window as the scatterplot to see if it fits the data.
Step 2
1. Press [STAT], the input corresponging x-values of data in L1, and y-values of data in L2.
2. Use [STATPLOT] to observe a scatterplot of the data.
Step 3
Based on the plots of the points, it can be exponential or logarithmic.
However, upon checking both regression analysis, the one with the closest value of $$\displaystyle{r}^{{{2}}}$$ to 1 is exponential, hence, its formula is $$\displaystyle{y}={628.67663}{\left({0.64841}\right)}^{{{x}}}.$$ The graph of which is below:
Relevant Questions
asked 2021-01-19
The annual sales S (in millions of dollars) for the Perrigo Company from 2004 through 2010 are shown in the table. $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&{2004}&{2005}&{2006}&{2007}&{2008}&{2009}&{2010}\backslash{h}{l}\in{e}\text{Sales, S}&{898.2}&{1024.1}&{1366.8}&{1447.4}&{1822.1}&{2006.9}&{2268.9}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ a) Use a graphing utility to create a scatter plot of the data. Let t represent the year, with $$\displaystyle{t}={4}$$ corresponding to 2004. b) Use the regression feature of the graphing utility to find an exponential model for the data. Use the Inverse Property $$\displaystyle{b}={e}^{{{\ln{\ }}{b}}}$$ to rewrite the model as an exponential model in base e. c) Use the regression feature of the graphing utility to find a logarithmic model for the data. d) Use the exponential model in base e and the logarithmic model to predict sales in 2011. It is projected that sales in 2011 will be \$2740 million. Do the predictions from the two models agree with this projection? Explain.
asked 2020-11-08
The following table lists the reported number of cases of infants born in the United States with HIV in recent years because their mother was infected.
Source:
Centers for Disease Control and Prevention.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Year}&\text{amp, Cases}\backslash{h}{l}\in{e}{1995}&{a}\mp,\ {295}\backslash{h}{l}\in{e}{1997}&{a}\mp,\ {166}\backslash{h}{l}\in{e}{1999}&{a}\mp,\ {109}\backslash{h}{l}\in{e}{2001}&{a}\mp,\ {115}\backslash{h}{l}\in{e}{2003}&{a}\mp,\ {94}\backslash{h}{l}\in{e}{2005}&{a}\mp,\ {107}\backslash{h}{l}\in{e}{2007}&{a}\mp,\ {79}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
a) Plot the data on a graphing calculator, letting $$\displaystyle{t}={0}$$ correspond to the year 1995.
b) Using the regression feature on your calculator, find a quadratic, a cubic, and an exponential function that models this data.
c) Plot the three functions with the data on the same coordinate axes. Which function or functions best capture the behavior of the data over the years plotted?
d) Find the number of cases predicted by all three functions for 20152015. Which of these are realistic? Explain.
asked 2021-03-11
An automobile tire manufacturer collected the data in the table relating tire pressure x (in pounds per square inch) and mileage (in thousands of miles). A mathematical model for the data is given by
$$\displaystyle f{{\left({x}\right)}}=-{0.554}{x}^{2}+{35.5}{x}-{514}.$$
$$\begin{array}{|c|c|} \hline x & Mileage \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}$$
(A) Complete the table below.
$$\begin{array}{|c|c|} \hline x & Mileage & f(x) \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}$$
(Round to one decimal place as needed.)
$$A. 20602060xf(x)$$
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,45), (30,51), (32,56), (34,50), and (36,46). A parabola opens downward and passes through the points (28,45.7), (30,52.4), (32,54.7), (34,52.6), and (36,46.0). All points are approximate.
$$B. 20602060xf(x)$$
Acoordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2.
Data points are plotted at (43,30), (45,36), (47,41), (49,35), and (51,31). A parabola opens downward and passes through the points (43,30.7), (45,37.4), (47,39.7), (49,37.6), and (51,31). All points are approximate.
$$C. 20602060xf(x)$$
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (43,45), (45,51), (47,56), (49,50), and (51,46). A parabola opens downward and passes through the points (43,45.7), (45,52.4), (47,54.7), (49,52.6), and (51,46.0). All points are approximate.
$$D.20602060xf(x)$$
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,30), (30,36), (32,41), (34,35), and (36,31). A parabola opens downward and passes through the points (28,30.7), (30,37.4), (32,39.7), (34,37.6), and (36,31). All points are approximate.
(C) Use the modeling function f(x) to estimate the mileage for a tire pressure of 29
$$\displaystyle\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$ and for 35
$$\displaystyle\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$
The mileage for the tire pressure $$\displaystyle{29}\frac{{{l}{b}{s}}}{{{s}{q}}}\in.$$ is
The mileage for the tire pressure $$\displaystyle{35}\frac{{{l}{b}{s}}}{{{s}{q}}}$$ in. is
(Round to two decimal places as needed.)
(D) Write a brief description of the relationship between tire pressure and mileage.
A. As tire pressure increases, mileage decreases to a minimum at a certain tire pressure, then begins to increase.
B. As tire pressure increases, mileage decreases.
C. As tire pressure increases, mileage increases to a maximum at a certain tire pressure, then begins to decrease.
D. As tire pressure increases, mileage increases.
asked 2021-02-09
A two-sample inference deals with dependent and independent inferences. In a two-sample hypothesis testing problem, underlying parameters of two different populations are compared. In a longitudinal (or follow-up) study, the same group of people is followed over time. Two samples are said to be paired when each data point in the first sample is matched and related to a unique data point in the second sample.
This problem demonstrates inference from two dependent (follow-up) samples using the data from the hypothetical study of new cases of tuberculosis (TB) before and after the vaccination was done in several geographical areas in a country in sub-Saharan Africa. Conclusion about the null hypothesis is to note the difference between samples.
The problem that demonstrates inference from two dependent samples uses hypothetical data from the TB vaccinations and the number of new cases before and after vaccination. PSK\begin{array}{|c|c|} \hline Geographical\ regions & Before\ vaccination & After\ vaccination\\ \hline 1 & 85 & 11\\ \hline 2 & 77 & 5\\ \hline 3 & 110 & 14\\ \hline 4 & 65 & 12\\ \hline 5 & 81 & 10\\\hline 6 & 70 & 7\\ \hline 7 & 74 & 8\\ \hline 8 & 84 & 11\\ \hline 9 & 90 & 9\\ \hline 10 & 95 & 8\\ \hline \end{array}ZSK
Using the Minitab statistical analysis program to enter the data and perform the analysis, complete the following: Construct a one-sided $$\displaystyle{95}\%$$ confidence interval for the true difference in population means. Test the null hypothesis that the population means are identical at the 0.05 level of significance.
asked 2021-02-11
Several models have been proposed to explain the diversification of life during geological periods. According to Benton (1997), The diversification of marine families in the past 600 million years (Myr) appears to have followed two or three logistic curves, with equilibrium levels that lasted for up to 200 Myr. In contrast, continental organisms clearly show an exponential pattern of diversification, and although it is not clear whether the empirical diversification patterns are real or are artifacts of a poor fossil record, the latter explanation seems unlikely. In this problem, we will investigate three models fordiversification. They are analogous to models for populationgrowth, however, the quantities involved have a differentinterpretation. We denote by N(t) the diversification function,which counts the number of taxa as a function of time, and by rthe intrinsic rate of diversification.
(a) (Exponential Model) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\ {\left({8.86}\right)}.$$ Solve (8.86) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{e}}}$$ can be estimated from $$\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ {\left({8.87}\right)}$$
(b) (Logistic Growth) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}$$ where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{l}}}$$ can be estimated from $$\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}$$ for $$\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.$$
(c) Assume that $$\displaystyle{N}{\left({0}\right)}={1}$$ and $$\displaystyle{N}{\left({10}\right)}={1000}.$$ Estimate $$\displaystyle{r}_{{{e}}}$$ and $$\displaystyle{r}_{{{l}}}$$ for both $$\displaystyle{K}={1001}$$ and $$\displaystyle{K}={10000}.$$
(d) Use your answer in (c) to explain the following quote from Stanley (1979): There must be a general tendency for calculated values of $$\displaystyle{\left[{r}\right]}$$ to represent underestimates of exponential rates,because some radiation will have followed distinctly sigmoid paths during the interval evaluated.
(e) Explain why the exponential model is a good approximation to the logistic model when $$\displaystyle\frac{{N}}{{K}}$$ is small compared with 1.
asked 2020-11-11
Use exponential regression to find a function that models the data. $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{x}&{1}&{2}&{3}&{4}&{5}\backslash{h}{l}\in{e}{f{{\left({x}\right)}}}&{14}&{7.1}&{3.4}&{1.8}&{0.8}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
asked 2021-03-07
In an experiment designed to study the effects of illumination level on task performance (“Performance of Complex Tasks Under Different Levels of Illumination,” J. Illuminating Eng., 1976: 235–242), subjects were required to insert a fine-tipped probe into the eyeholes of ten needles in rapid succession both for a low light level with a black background and a higher level with a white background. Each data value is the time (sec) required to complete the task.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{S}{u}{b}{j}{e}{c}{t}&{\left({1}\right)}&{\left({2}\right)}&{\left({3}\right)}&{\left({4}\right)}&{\left({5}\right)}&{\left({6}\right)}&{\left({7}\right)}&{\left({8}\right)}&{\left({9}\right)}\backslash{h}{l}\in{e}{B}{l}{a}{c}{k}&{25.85}&{28.84}&{32.05}&{25.74}&{20.89}&{41.05}&{25.01}&{24.96}&{27.47}\backslash{h}{l}\in{e}{W}{h}{i}{t}{e}&{18.28}&{20.84}&{22.96}&{19.68}&{19.509}&{24.98}&{16.61}&{16.07}&{24.59}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
Does the data indicate that the higher level of illumination yields a decrease of more than 5 sec in true average task completion time? Test the appropriate hypotheses using the P-value approach.
asked 2020-10-19
n an experiment designed to study the effects of illumination level on task performance (“Performance of Complex Tasks Under Different Levels of Illumination,” J. Illuminating Eng., 1976: 235–242), subjects were required to insert a fine-tipped probe into the eyeholes of ten needles in rapid succession both for a low light level with a black background and a higher level with a white background. Each data value is the time (sec) required to complete the task. $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\mathcal}\right\rbrace}{h}{l}\in{e}&{a}\mp&{a}\mp&{a}\mp\ \text{Subject}\backslash{h}{l}\in{e}&{a}\mp\ {1}&{a}\mp\ {2}&{a}\mp\ {3}&{a}\mp\ {4}&{a}\mp\ {5}&{a}\mp\ {6}&{a}\mp\ {7}&{a}\mp\ {8}&{a}\mp\ {9}&{a}\mp\backslash{h}{l}\in{e}\text{Black}&{a}\mp\ {25.85}&{a}\mp\ {28.84}&{a}\mp\ {32.05}&{a}\mp\ {25.74}&{a}\mp\ {20.89}&{a}\mp\ {41.05}&{a}\mp\ {25.01}&{a}\mp\ {24.96}&{a}\mp\ {27.47}&{a}\mp\backslash{h}{l}\in{e}\text{White}&{a}\mp\ {18.23}&{a}\mp\ {20.84}&{a}\mp\ {22.96}&{a}\mp\ {19.68}&{a}\mp\ {19.509}&{a}\mp\ {24.98}&{a}\mp\ {16.61}&{a}\mp\ {16.07}&{a}\mp\ {24.59}&{a}\mp\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ Does the data indicate that the higher level of illumination yields a decrease of more than 5 sec in true average task completion time? Test the appropriate hypotheses using the P-value approach.
asked 2020-12-25
Case: Dr. Jung’s Diamonds Selection
With Christmas coming, Dr. Jung became interested in buying diamonds for his wife. After perusing the Web, he learned about the “4Cs” of diamonds: cut, color, clarity, and carat. He knew his wife wanted round-cut earrings mounted in white gold settings, so he immediately narrowed his focus to evaluating color, clarity, and carat for that style earring.
After a bit of searching, Dr. Jung located a number of earring sets that he would consider purchasing. But he knew the pricing of diamonds varied considerably. To assist in his decision making, Dr. Jung decided to use regression analysis to develop a model to predict the retail price of different sets of round-cut earrings based on their color, clarity, and carat scores. He assembled the data in the file Diamonds.xls for this purpose. Use this data to answer the following questions for Dr. Jung.
1) Prepare scatter plots showing the relationship between the earring prices (Y) and each of the potential independent variables. What sort of relationship does each plot suggest?
2) Let X1, X2, and X3 represent diamond color, clarity, and carats, respectively. If Dr. Jung wanted to build a linear regression model to estimate earring prices using these variables, which variables would you recommend that he use? Why?
3) Suppose Dr. Jung decides to use clarity (X2) and carats (X3) as independent variables in a regression model to predict earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
4) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. Which sets of earrings appear to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
5) Dr. Jung now remembers that it sometimes helps to perform a square root transformation on the dependent variable in a regression problem. Modify your spreadsheet to include a new dependent variable that is the square root on the earring prices (use Excel’s SQRT( ) function). If Dr. Jung wanted to build a linear regression model to estimate the square root of earring prices using the same independent variables as before, which variables would you recommend that he use? Why?
1
6) Suppose Dr. Jung decides to use clarity (X2) and carats (X3) as independent variables in a regression model to predict the square root of the earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
7) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. (Remember, your model estimates the square root of the earring prices. So you must actually square the model’s estimates to convert them to price estimates.) Which sets of earring appears to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
8) Dr. Jung now also remembers that it sometimes helps to include interaction terms in a regression model—where you create a new independent variable as the product of two of the original variables. Modify your spreadsheet to include three new independent variables, X4, X5, and X6, representing interaction terms where: X4 = X1 × X2, X5 = X1 × X3, and X6 = X2 × X3. There are now six potential independent variables. If Dr. Jung wanted to build a linear regression model to estimate the square root of earring prices using the same independent variables as before, which variables would you recommend that he use? Why?
9) Suppose Dr. Jung decides to use color (X1), carats (X3) and the interaction terms X4 (color * clarity) and X5 (color * carats) as independent variables in a regression model to predict the square root of the earring prices. What is the estimated regression equation? What is the value of the R2 and adjusted-R2 statistics?
10) Use the regression equation identified in the previous question to create estimated prices for each of the earring sets in Dr. Jung’s sample. (Remember, your model estimates the square root of the earring prices. So you must square the model’s estimates to convert them to actual price estimates.) Which sets of earrings appear to be overpriced and which appear to be bargains? Based on this analysis, which set of earrings would you suggest that Dr. Jung purchase?
asked 2021-02-09
The table gives the number of active Twitter users worldwide, semiannually from 2010 to 2016. $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}\text{Years since}&\text{January 1, 2010}&\text{Twitter user}&\text{(millions)}\backslash{h}{l}\in{e}{0}&{30}&{3.5}&{232}\backslash{h}{l}\in{e}{0.5}&{49}&{4.0}&{255}\backslash{h}{l}\in{e}{1.0}&{68}&{4.5}&{284}\backslash{h}{l}\in{e}{1.5}&{101}&{5.0}&{302}\backslash{h}{l}\in{e}{2.0}&{138}&{5.5}&{307}\backslash{h}{l}\in{e}{2.5}&{167}&{6.0}&{310}\backslash{h}{l}\in{e}{3.0}&{204}&{6.5}&{317}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ Use a calculator or computer to fit both an exponential function and a logistic function to these data. Graph the data points and both functions, and comment on the accuracy of the models.
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Institute: MPI für Informatik Collection: Algorithms and Complexity Group Display Documents
ID: 536757.0, MPI für Informatik / Algorithms and Complexity Group
Stackelberg routing in arbitrary networks
Authors:
Language:English
Date of Publication (YYYY-MM-DD):2010
Title of Journal:Mathematics of Operations Research
Volume:35
Issue / Number:2
Start Page:330
End Page:346
Review Status:Peer-review
Audience:Experts Only
Intended Educational Use:No
Abstract / Description:We investigate the impact of \emph{Stackelberg routing} to reduce the price of
anarchy in network routing games. In this setting, an $\alpha$ fraction of the
entire demand is first routed centrally according to a predefined
\emph{Stackelberg strategy} and the remaining demand is then routed selfishly
proving that Stackelberg routing can in fact significantly reduce the price of
anarchy for certain network topologies, the central question of whether this
holds true in general is still open. We answer this question negatively by
constructing a family of single-commodity instances such that every Stackelberg
strategy induces a price of anarchy that grows linearly with the size of the
network.
Moreover, we prove upper bounds on the price of anarchy of the
Largest-Latency-First (LLF) strategy that only depend on the size of the
network. Besides other implications, this rules out the possibility to
construct constant-size networks to prove an unbounded price of anarchy.
In light of this negative result, we consider bicriteria bounds. We develop an
efficiently computable Stackelberg strategy that induces a flow whose cost is
at most the cost of an optimal flow with respect to demands scaled by a factor
of $1 + \sqrt{1-\alpha}$.
Finally, we analyze the effectiveness of an easy-to-implement Stackelberg
strategy, called SCALE. We prove bounds for a general class of latency
functions that includes polynomial latency functions as a special case. Our
analysis is based on an approach which is simple, yet powerful enough to obtain
(almost) tight bounds for SCALE in general networks.
Last Change of the Resource (YYYY-MM-DD):2011-02-10
External Publication Status:published
Document Type:Article
Communicated by:Kurt Mehlhorn
Affiliations:
Identifiers:LOCALID:C1256428004B93B8-9CA118DFF9513FF6C12577FF006056B0-...
URL:http://dx.doi.org/10.1287/moor.1100.0442
DOI:10.1287/moor.1100.0442v1
ISSN:0364-765X
Full Text: You have privileges to view the following file(s): Bonifaci2010b.pdf [270,00 Kb] [Comment:file from upload service] ATTCYBW3.pdf [270,00 Kb] [Comment:file from upload service]
The scope and number of records on eDoc is subject to the collection policies defined by each institute - see "info" button in the collection browse view.
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Comment
Share
Q)
# If an electromagnetic wave of frequency $v = 3.0\: MHz$ passes from vacuum into a dielectric medium with permittivity $e = 4.0$ then
$\begin {array} {1 1} (a)\;\text{wavelength is doubled and frequency remains unchanged}\\ (b)\;\text{wavelength is doubled and frequency becomes half} \\ (c)\;\text{wavelength is halved and frequency remains unchanged}\\ (d)\;\text{wavelength and frequency both remain unchanged} \end {array}$
it is assumed that the energy is conserved, so $h\nu$ remains the same.)
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# How do you solve 3h-7<14?
Apr 4, 2018
$h < 7$
Here's how I did it:
#### Explanation:
$3 h - 7 < 14$
We solve inequalities similarly to equations. The first thing we want to do is add $7$ to both sides of the inequality:
$3 h < 21$
Now we can divide both sides by $3$ and get the values of $h$:
$h < 7$
Hope this helps!
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# rmgpy.data.thermo.ThermoDatabase¶
class rmgpy.data.thermo.ThermoDatabase
A class for working with the RMG thermodynamics database.
Return the set of thermodynamic parameters corresponding to a given Molecule object molecule using the group additivity values method. If no group additivity values are loaded, a DatabaseError is raised.
The entropy is not corrected for the symmetry of the molecule, this should be done later by the calling function.
correct_binding_energy(thermo, species, metal_to_scale_from=None, metal_to_scale_to=None)
Changes the provided thermo, by applying a linear scaling relation to correct the adsorption energy.
Parameters:
• thermo – starting thermo data
• species – the species (which is an adsorbate)
• metal_to_scale_from – the metal you want to scale from (string eg. ‘Pt111’ or None)
• metal_to_scale_to – the metal you want to scale to (string e.g ‘Pt111’ or None)
Returns:
corrected thermo
Estimate the thermodynamics of a radical by saturating it, applying the provided stable_thermo_estimator method on the saturated species, then applying hydrogen bond increment corrections for the radical site(s) and correcting for the symmetry.
No entropy is included in the returning term. This should be done later by the calling function.
Return the set of thermodynamic parameters corresponding to a given Molecule object molecule using the group additivity values method. If no group additivity values are loaded, a DatabaseError is raised.
The entropy is not corrected for the symmetry of the molecule, this should be done later by the calling function.
species: A species object containing thermo data and thermo data comments
Parses the verbose string of comments from the thermo data of the species object, and extracts the thermo sources.
Returns a dictionary with keys of either ‘Library’, ‘QM’, and/or ‘GAV’. Commonly, species thermo are estimated using only one of these sources. However, a radical can be estimated with more than one type of source, for instance a saturated library value and a GAV HBI correction, or a QM saturated value and a GAV HBI correction.
source = {‘Library’: String_Name_of_Library_Used,
‘QM’: String_of_Method_Used, ‘GAV’: Dictionary_of_Groups_Used }
The Dictionary_of_Groups_Used looks like {‘groupType’:[List of tuples containing (Entry, Weight)]
get_all_thermo_data(species)
Return all possible sets of thermodynamic parameters for a given Species object species. The hits from the depository come first, then the libraries (in order), and then the group additivity estimate. This method is useful for a generic search job.
Returns: a list of tuples (ThermoData, source, entry) (Source is a library or depository, or None)
Takes a string of comments from group additivity estimation, and extracts the ring and polycyclic ring groups from them, returning them as lists.
get_thermo_data(species, metal_to_scale_to=None, training_set=None)
Return the thermodynamic parameters for a given Species object species. This function first searches the loaded libraries in order, returning the first match found, before falling back to estimation via machine learning and then group additivity.
The method corrects for symmetry when the molecule uses machine learning or group additivity. Libraries and direct QM calculations are already corrected.
If either metal to scale to or from is not specified, assume the binding energies given in the input file
Returns: ThermoData
get_thermo_data_for_surface_species(species)
Get the thermo data for an adsorbed species, by desorbing it, finding the thermo of the gas-phase species, then adding an adsorption correction that is found from the groups/adsorption tree. Does not apply linear scaling relationship.
Returns a ThermoData object, with no Cp0 or CpInf
get_thermo_data_from_depository(species)
Return all possible sets of thermodynamic parameters for a given Species object species from the depository. If no depository is loaded, a DatabaseError is raised.
Returns: a list of tuples (thermo_data, depository, entry) without any Cp0 or CpInf data.
get_thermo_data_from_groups(species)
Return the set of thermodynamic parameters corresponding to a given Species object species by estimation using the group additivity values. If no group additivity values are loaded, a DatabaseError is raised.
The resonance isomer (molecule) with the lowest H298 is used, and as a side-effect the resonance isomers (items in species.molecule list) are sorted in ascending order.
This does not account for symmetry. The method calling this sould correct for it.
Returns: ThermoData
get_thermo_data_from_libraries(species, training_set=None)
Return the thermodynamic parameters for a given Species object species. This function first searches the loaded libraries in order, returning the first match found, before failing and returning None. training_set is used to identify if function is called during training set or not. During training set calculation we want to use gas phase thermo to not affect reverse rate calculation.
Returns: ThermoData or None
get_thermo_data_from_library(species, library)
Return the set of thermodynamic parameters corresponding to a given Species object species from the specified thermodynamics library. If library is a string, the list of libraries is searched for a library with that name. If no match is found in that library, None is returned. If no corresponding library is found, a DatabaseError is raised.
Returns a tuple: (ThermoData, library, entry) or None.
get_thermo_data_from_ml(species, ml_estimator, ml_settings)
Return the set of thermodynamic parameters corresponding to a given Species object species by estimation using the ML estimator. Also compare the estimated uncertainties to the user-defined cutoffs. If any of the uncertainties are larger than their corresponding cutoffs, return None. Also check all other options in ml_settings.
For HBI, the resonance isomer with the lowest H298 is used and the resonance isomers in species are sorted in ascending order.
The entropy is not corrected for the symmetry of the molecule. This should be done later by the calling function.
Load the thermo database from the given path on disk, where path points to the top-level folder of the thermo database.
Load the thermo database from the given path on disk, where path points to the top-level folder of the thermo database.
Load the thermo database from the given path on disk, where path points to the top-level folder of the thermo database.
Load the thermo database from the given path on disk, where path points to the top-level folder of the thermo database.
If no libraries are given, all are loaded.
Load the old RMG thermo database from the given path on disk, where path points to the top-level folder of the old RMG database.
Load the metal database from the given path on disk, where path points to the top-level folder of the thermo database.
prioritize_thermo(species, thermo_data_list)
Use some metrics to reorder a list of thermo data from best to worst. Return a list of indices with the desired order associated with the index of thermo from the data list.
prune_heteroatoms(allowed=None)
Remove all species from thermo libraries that contain atoms other than those allowed.
This is useful before saving the database for use in RMG-Java
record_polycylic_generic_nodes()
Identify generic nodes in tree for polycyclic groups. Saves them as a list in the generic_nodes attribute in the polycyclic ThermoGroups object, which must be pre-loaded.
Necessary for polycyclic heuristic.
record_ring_generic_nodes()
Identify generic nodes in tree for ring groups. Saves them as a list in the generic_nodes attribute in the ring ThermoGroups object, which must be pre-loaded.
Necessary for polycyclic heuristic.
save(path)
Save the thermo database to the given path on disk, where path points to the top-level folder of the thermo database.
save_depository(path)
Save the thermo depository to the given path on disk, where path points to the top-level folder of the thermo depository.
save_groups(path)
Save the thermo groups to the given path on disk, where path points to the top-level folder of the thermo groups.
save_libraries(path)
Save the thermo libraries to the given path on disk, where path points to the top-level folder of the thermo libraries.
save_old(path)
Save the old RMG thermo database to the given path on disk, where path points to the top-level folder of the old RMG database.
save_surface(path)
Save the metal library to the given path on disk, where path points to the top-level folder of the metal library.
set_binding_energies(binding_energies='Pt111')
Sets and stores the atomic binding energies specified in the input file.
All adsorbates will be scaled to use these elemental binding energies.
Parameters:
binding_energies (dict, optional) – the desired binding energies with elements as keys and binding energy/unit tuples (or Energy quantities) as values
Returns:
None, stores result in self.binding_energies
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# Linear Regression and Standardization
Linear regression is a more informative metric for evaluating associations between variables than most people typically realize. When it is used to forecast outcomes, it can be converted to a measure of information gain or converted into a point estimate and associated confidence interval. It can also be used to quantify the amount a linear model reduces uncertainty.
Linear regression is closely related to the Central Limit Theorem because both regression and the CLT use probability distributions known as “Gaussians.”
## Standardization
Standardizing a set of values requires calculating the mean, $\mu_x$, and standard deviation of the population, $\sigma_x$. The appropriate excel formulae are STDEVP and AVERAGE. By convention, the raw data values are denoted $x_i$, and the standardized values are denoted $x_{zi}$.
$$\mu_x = \frac{1}{n}\sum_{i=1}^n x_i$$
$$\sigma_x = \sqrt{\frac{1}{n}\sum_{i=1}^n (x_i-\mu_x)^2}$$
$$x_{zi}=\frac{x_i-\mu_x}{\sigma_x}$$
In excel, standardization can be accomplished using the STANDARDIZE function, which takes as parameters the array of values to be standardized as well as the average and standard deviation thereof.
As an example, note that in the following standardization, both $X_z$ and $Y_z$ have the same standard deviation and mean, despite the X values having a much larger range of values. This is what is meant by standardization.
X Y X_z Y_z
Values 10 1.5 -1.3 -0.9
90 10.0 1.8 2.0
75 8.3 1.2 1.5
35 4.0 -0.4 0.0
20 1.4 -1.0 -0.9
21 1.8 -0.9 -0.8
33 2.2 -0.5 -0.6
58 4.0 0.5 0.0
60 3.0 0.6 -0.4
μ 44.7 4.02 0 0
σ 25.8 2.92 1 1
$$\mu_x=44.7,\ \mu_y=4.02,\ \mu_{xz}=0,\ \mu_{yz}=0$$
$$\sigma_x=25.8,\ \sigma_y=2.92,\ \sigma_{xz}=1,\ \sigma_{yz}=1$$
Next, I will consider how standardization affects certain relationships between sets of values, like covariance, correlation, slope and y-intercept of the best fit line.
### Covariance
Covariance ($\text{Cov}$) is calculated as follows. The appropriate excel function is COVARIANCE.P.
$$\text{Cov}_{xy}=\frac{1}{n}\sum_{i=1}^n(x_i-\mu_{xi})(y_i-\mu_{yi})$$
The covariance of the raw data, above, is 67.89. The covariance of the standardized data is 0.902.
### Best Fit Line
The best fit line for the relationship between two sets of data is commonly represented as follows.
$$\hat{y_i}=\beta x_i + \alpha$$
$\hat{y_i}$ represents the y-values that form the line of best fit. The slope of the best fit line is commonly represented by $\beta$. The appropriate excel function is SLOPE. The values are 0.102 and 0.902 for the raw and standardized data, respectively. Note that the covariance is the same as the $\beta$ for standardized data.
The y-intercept is commonly represented by $\alpha$. The appropriate excel function is INTERCEPT. The values are -0.535 and 0 for the raw and standardized data, respectively. Note that standardized data will always have a y-intercept of zero. This is because the point $(\mu_x,\mu_y)$ is always on the best fit line, and for standardized data that point is (0,0).
### Correlation
Correlation is a function of the covariance and standard deviations of the data, calculated as follows. It is commonly represented as $R$, and sometimes known as the Pearson correlation coefficient or Pearson R test. The appropriate excel function is CORREL.
$$\text{R}=\frac{\text{Cov}_{xy}}{\sigma_{x}\sigma_{y}}$$
The correlation for two sets of data is the same whether they have been standardized or not. In the case of the data tabulated above, the correlation is 0.902.
### Standardization Summary
The values discussed above are broken out, below.
Raw Data Standardized Data
Covariance, Cov 67.9 0.902
Slope, Beta, β 0.102 0.902
Intercept, α -0.535 0
Correlation, R 0.902 0.902
Note that the only values that are unchanged following standardization is the correlation. Note also that for standardized data the covariance and slope are equal to the correlation. That is, for standardized data,
$$\text{Cov}=\beta=R$$
## Linear Regression
The goal of linear regression is to minimize the sum of squared errors (or “residuals”), where the sum of squared errors is given by
$$\sum{(y_i-\hat{y_i})^2}\ \ \ or\ \ \ \sum{e^2}$$
It can be shown that the standard deviation of residuals is the same as the root mean square of the residuals. Or:
$$\sigma_e=\sqrt{\frac{1}{n}\sum{e}}$$
### Modeling Error using Linear Regression
It is possible to relate the variance of the error, $\sigma_e$, directly to the correlation and slope of the best fit line. Doing so requires making the assumption that the X and Y values are samples that are drawn from normally distributed data with $\mu = 0$ and $\sigma = 1$.
The actual y-values can be computed from the following.
$$Y\ values = (\beta)(X\ values) + error\ term$$
With the assumption of normally distributed data, this converts to the following.
$$\phi(0,\sigma_y^2)=\phi(0,\beta^2\sigma_x^2)+\phi(0,\sigma_e^2)$$
If the data is standardized, then $\sigma_y^2=\sigma_x^2=1$ and the previous formula reduces to the following.
$$1=\beta^2+\sigma_e^2$$
For standardized data, $\beta = R$, so
$$1=R^2+\sigma_e^2$$
This formula tells us there is a direct relationship between the standard deviation of the errors and the correlation. This is important because it enables us to make forecasts in the form of a probability distribution, not just a single point.
So, a forecast might take the form of a point value $\hat{y_i}=\beta x_i$ with standard deviation $\sigma_e$.
### Example 1
Consider a situation where a 90% confidence interval is desired, and the data is correlated with $R=0.7$.
A 90% confidence interval translates to z-scores of 1.64 and -1.64. We calculate $\sigma_e=0.71$, then the 90% confidence interval is given by
$$\hat{y_i}=\beta x_i\pm(1.64)(.71)$$
### Example 2
A data analysis model has a standard deviation of model error of $2500 at a correlation of R = 0.30. The model needs to be improved to the point that the standard deviation of model error is$1500.
Using the formula $1=R^2+\sigma_e^2$, $\sigma_e$ is solved to be 0.954. Then the required $\sigma_e$ is calculated as follows:
$$\sigma_{e,required}=\frac{1500}{2500}\sigma_{e,actual}=.572$$
$$R_{e,required} = \sqrt{1-\sigma_{e,required}^2}=.82$$
### Connection to Mutual Information
It is possible to demonstrate that, in a parametric model that has Gaussian distributions, the mutual information is given by
$$I(X;Y)=log(\frac{1}{\sqrt{1-R^2}})$$
Thus, for a parameterized Gaussian model with continuous distributions, if the correlation between X and Y,
$R$, increases, the percentage information gain also increases, becoming infinite at R=1.
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# TeXworks + pdflatex + PSTricks autopp suffix problem
I have managed to make PSTricks work with TeXworks. However after compilation it adds -autopp to the basename of the pdf file, e.g. report.pdf becomes report-autopp.pdf. As a result TeXworks is not able to preview the file in a window.
How can I make TeXworks preview the pdf file? Am I doing something wrong? Here is my tex file:
\documentclass{article}
\usepackage[pdf]{pstricks}
\begin{document}
\begin{figure}
\begin{pspicture}(4,5)
\psframe(0.7,2)(3.3,3)
\rput(2,2.5){First Example}
\end{pspicture}
\end{figure}
\end{document}
-
did you run the document with pdflatex -shell-escape? See tug.org/PSTricks/main.cgi?file=pdf/pdfoutput#texworks – Herbert Aug 15 '12 at 11:39
@Herbert yup. I've setup TeXworks for that. – nimcap Aug 15 '12 at 11:44
and have you Perl installed? If not then use \usepackage{pstricks}\usepackage[crop=off]{auto-pst-pdf} and tell us what your log file reports. – Herbert Aug 15 '12 at 11:47
I have installed Perl and tried again, did not work. Using \usepackage{pstricks}\usepackage[crop=off]{auto-pst-pdf} worked. Does this mean my TeX distribution is too old? – nimcap Aug 15 '12 at 12:06
run your original example and post the logfile where error messages appear. – Herbert Aug 15 '12 at 12:08
use
\usepackage{pstricks}
\usepackage{auto-pst-pdf}
I'll have a look at the pstricks.sty why it didn't work under Windows. With Linux it is no problem. However, you have to run it with
pdflatex --shell-escape <file>
for TeXnicCenter and others see http://tug.org/PSTricks/main.cgi?file=pdf/pdfoutput#TXC
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Guys, I'm facing the same issue. pstricks doesn't work with pdflatex on windows as it is expected. – maksmara Feb 13 '14 at 13:31
@nimcap, I have the same "autopp suffix problem". My TeX distribution is ProTeXt-3.1.3-060313 – maksmara Feb 13 '14 at 14:03
I suppose you forgot the -shell-escape. See my edited answer. – Herbert Feb 13 '14 at 14:09
nope, I have tried all possible ways :( how can i post my question? as an answer? – maksmara Feb 13 '14 at 14:14
@maksmara Follow-up questions are more than welcome! Please use the "Ask Question" link for your new question; there you can link to this question to provide the background. – texenthusiast Feb 13 '14 at 15:02
You can also do with the following approach. Compile it with pdflatex -shell-escape main.tex where main.tex is as follows.
% main.tex
\documentclass{article}
\usepackage{filecontents}
\begin{filecontents*}{figure.tex}
\documentclass[pstricks,border=12pt]{standalone}
\begin{document}
\begin{pspicture}(4,5)
\psframe(0.7,2)(3.3,3)
\rput{90}(2,2.5){First Example}
\pscircle[linecolor=red,linewidth=2pt](2,2.5){2}
\end{pspicture}
\end{document}
\end{filecontents*}
\usepackage{pgffor}
\usepackage{graphicx}
\foreach \compiler/\extension in {latex/tex,dvips/dvi,{ps2pdf -dAutoRotatePages=/None}/ps}{\immediate\write18{\compiler\space figure.\extension}}
\begin{document}
\begin{figure}
\centering
\includegraphics{figure}
\caption{This is my figure.}
\end{figure}
\end{document}
## Notes
• -dAutoRotatePages=/None is important here to avoid ps2pdf to automatically rotate the figure. Try it by yourself, remove -dAutoRotatePages=/None.
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# Principal ideal ring
In mathematics, a principal right (left) ideal ring is a ring R in which every right (left) ideal is of the form xR (Rx) for some element x of R. (The right and left ideals of this form, generated by one element, are called principal ideals.) When this is satisfied for both left and right ideals, such as the case when R is a commutative ring, R can be called a principal ideal ring, or simply principal ring.
If only the finitely generated right ideals of R are principal, then R is called a right Bézout ring. Left Bézout rings are defined similarly. These conditions are studied in domains as Bézout domains.
A commutative principal ideal ring which is also an integral domain is said to be a principal ideal domain (PID). In this article the focus is on the more general concept of a principal ideal ring which is not necessarily a domain.
## General properties
If R is a right principal ideal ring, then it is certainly a right Noetherian ring, since every right ideal is finitely generated. It is also a right Bézout ring since all finitely generated right ideals are principal. Indeed, it is clear that principal right ideal rings are exactly the rings which are both right Bézout and right Noetherian.
Principal right ideal rings are closed under finite direct products. If ${\displaystyle R=\prod _{i=1}^{n}R_{i}}$, then each right ideal of R is of the form ${\displaystyle A=\prod _{i=1}^{n}A_{i}}$, where each ${\displaystyle A_{i}}$ is a right ideal of Ri. If all the Ri are principal right ideal rings, then Ai=xiRi, and then it can be seen that ${\displaystyle (x_{1},\ldots ,x_{n})R=A}$. Without much more effort, it can be shown that right Bézout rings are also closed under finite direct products.
Principal right ideal rings and right Bézout rings are also closed under quotients, that is, if I is a proper ideal of principal right ideal ring R, then the quotient ring R/I is also principal right ideal ring. This follows readily from the isomorphism theorems for rings.
All properties above have left analogues as well.
## Commutative examples
1. The ring of integers: ${\displaystyle \mathbb {Z} }$
2. The integers modulo n: ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$.
3. Let ${\displaystyle R_{1},\ldots ,R_{n}}$ be rings and ${\displaystyle R=\prod _{i=1}^{n}R_{i}}$. Then R is a principal ring if and only if Ri is a principal ring for all i.
4. The localization of a principal ring at any multiplicative subset is again a principal ring. Similarly, any quotient of a principal ring is again a principal ring.
5. Let R be a Dedekind domain and I be a nonzero ideal of R. Then the quotient R/I is a principal ring. Indeed, we may factor I as a product of prime powers: ${\displaystyle I=\prod _{i=1}^{n}P_{i}^{a_{i}}}$, and by the Chinese Remainder Theorem ${\displaystyle R/I\cong \prod _{i=1}^{n}R/P_{i}^{a_{i}}}$, so it suffices to see that each ${\displaystyle R/P_{i}^{a_{i}}}$ is a principal ring. But ${\displaystyle R/P_{i}^{a_{i}}}$ is isomorphic to the quotient ${\displaystyle R_{P_{i}}/P_{i}^{a_{i}}R_{P_{i}}}$ of the discrete valuation ring ${\displaystyle R_{P_{i}}}$ and, being a quotient of a principal ring, is itself a principal ring.
6. Let k be a finite field and put ${\displaystyle A=k[x,y]}$, ${\displaystyle {\mathfrak {m}}=\langle x,y\rangle }$ and ${\displaystyle R=A/{\mathfrak {m}}^{2}}$. Then R is a finite local ring which is not principal.
7. Let X be a finite set. Then ${\displaystyle ({\mathcal {P}}(X),\Delta ,\cap )}$ forms a commutative principal ideal ring with unity, where ${\displaystyle \Delta }$ represents set symmetric difference and ${\displaystyle {\mathcal {P}}(X)}$ represents the powerset of X. If X has at least two elements, then the ring also has zero divisors. If I is an ideal, then ${\displaystyle I=(\bigcup I)}$. If instead X is infinite, the ring is not principal: take the ideal generated by the finite subsets of X, for example.
## Structure theory for commutative PIR's
The principal rings constructed in Example 4. above are always Artinian rings; in particular they are isomorphic to a finite direct product of principal Artinian local rings. A local Artinian principal ring is called a special principal ring and has an extremely simple ideal structure: there are only finitely many ideals, each of which is a power of the maximal ideal. For this reason, special principal rings are examples of uniserial rings.
The following result gives a complete classification of principal rings in terms of special principal rings and principal ideal domains.
Zariski–Samuel theorem: Let R be a principal ring. Then R can be written as a direct product ${\displaystyle \prod _{i=1}^{n}R_{i}}$, where each Ri is either a principal ideal domain or a special principal ring.
The proof applies the Chinese Remainder theorem to a minimal primary decomposition of the zero ideal.
There is also the following result, due to Hungerford:
Theorem (Hungerford): Let R be a principal ring. Then R can be written as a direct product ${\displaystyle \prod _{i=1}^{n}R_{i}}$, where each Ri is a quotient of a principal ideal domain.
The proof of Hungerford's theorem employs Cohen's structure theorems for complete local rings.
Arguing as in Example 3. above and using the Zariski-Samuel theorem, it is easy to check that Hungerford's theorem is equivalent to the statement that any special principal ring is the quotient of a discrete valuation ring.
## Noncommutative examples
Every semisimple ring R which is not just a product of fields is a noncommutative right and left principal ideal domain. Every right and left ideal is a direct summand of R, and so is of the form eR or Re where e is an idempotent of R. Paralleling this example, von Neumann regular rings are seen to be both right and left Bézout rings.
If D is a division ring and ${\displaystyle \sigma }$ is a ring endomorphism which is not an automorphism, then the skew polynomial ring ${\displaystyle D[x,\sigma ]}$ is known to be a principal left ideal domain which is not right Noetherian, and hence it cannot be a principal right ideal ring. This shows that even for domains principal left and principal right ideal rings are different. (Lam & 2001, p.21)
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### how to split powerpoint slide into 3 columns
What Is a PowerPoint Slide Layout? Just by adding normal shapes in a slide and then save them as pictures, which you can eventually import as image into power bi and select those images as backward image under format option. Split Text Between Two Slides. Continue on a New Slide. Among them, you will notice the coveted “Columns” function. This means you can split your documents into columns in a matter of seconds. I want to break up each page into four sections. We added the table using the Table icon available on a new slide. Otherwise, click the “Insert All Slides” to copy all of the slides into your open PowerPoint presentation. Follow the steps below to split up a bulleted list in your PowerPoint presentation: In Normal view, switch to the Outline tab. it more ideal for academic purposes. In this article, I'll introduce a simple solution for doing this using Spire.Presentation in C#, VB.NET. PowerPoint creates a new slide with the same title as the current slide, and it divides the text approximately evenly between the two slides. ; Figure 2: Text box selected Right-click to get the context menu you see in Figure 3.Select the Format Shape option (refer to Figure 3 again). 3. On many occasions in Excel, you have data in one column that you would like to separate into multiple ones. I have tried modifying the PowerPoint template to include three content blocks as following: But I am unable to add content to the object on the bottom right. This gives you lots of design flexibility! ! The whole process involves a fixed set of steps. 3 Columns Slide Design for PowerPoint provides three colorful columns to create text based slides in style. But it seems that the following code puts one on top of the other: \begin{minipage}[t]{0.48\linewidth} left part \end{minipage}\hfill \begin{minipage}[t]{0.48\linewidth} right part \end{minipage} Could anyone tell me how to do that, and if it is correct to set 0.48\linewidth? This example splits the first cell in the referenced table into two cells, one directly above the other. Click the "Insert" tab and then click "Shapes." To display the Outline toolbar, right-click any toolbar and choose Outline. To start typing in the next column, you’ll need to insert a column break. Just switch to it and continue typing the text you want to add. 1. The presentation was made using Power Point 2003. First, there’s the rule of threes. Merge Excel data into PowerPoint presentations to create certificates, awards presentations, ... For example, if your original file MySlides.PPT contains 55 slides and you ask SplitFile to split it to 25 slides per file, you'll get: MySlides_1-25.PPT; MySlides_26-50.PPT; MySlides_51-55.PPT ; The new files will be saved to the same folder as the original file. 2. Let us start by adding a table to a PowerPoint slide. If you’re not confident creating your own PowerPoint grids, heave a huge sigh of relief! Launch PowerPoint. image; creators to create PDF, merger and splitter to merge or split PDFs.I'm collecting status reports from 20 different project managers and each of them sends a . Using this technique, you can already make any type … Rather than go to the trouble of pushing and pulling objects until they are distributed evenly, you can simply select […] Hover over it, and a submenu will appear. Your slide (or slides) will then be inserted into the open presentation, immediately underneath the currently selected slide. ActivePresentation.Slides(2).Shapes(5).Table.Cell(1, 1).Split 2, 1 See also. Click to select the objects on the slide and press the "Delete" button. How do I include multiple figures or "contents" on a slide? If the text is in a text placeholder, this is easily done using the Outline toolbar. The PowerPoint Distribute Horizontally and Distribute Vertically commands come in handy for laying out objects on your PowerPoint slide. It seems like 3 really is a magic number—particularly when it comes to presentations. Copying and Pasting PowerPoint Slides. The first is a spacer segment that adds some distance between the end of … Sometimes the table can be too tall and you wish you could just split the table over two slides automatically. These PowerPoint commands arrange objects so that the same amount of space appears between each one. I found out how to break it up into two by doing: Home Tab > Slides Group > Layout Button > "Two Content" Option. However, once your presentation is designed, you can split a bullet list between two slides. #2 03-08-2011, 02:09 PM GeoPro Windows 7 32bit Office 2010 32bit Novice : Join Date: Mar 2011. Three-column templates can be used to illustrate a wide variety of concepts, and this PowerPoint template set gives you eight different ways to put them to use. If all you have is the PDF a screenshot will work but that’s an image that’s hard to manipulate, the file is MUCH bigger than text in pptx, and resolution isn’t very good. I have been lumped with a massive Power Point presentation which is almost 250 slides in length. Check if the table is out of the slide area and determine which is the first row that moves off the slide. 3. PowerPoint presentation containing just one slide with a status summary. Powerpoint 2010 has a feature that will allow you to merge two separate Powerpoint files with just a few short steps. I'm new to powerpoint. However, business professionals also adapt with this PowerPoint to present company profile presentation along the company vision and mission statement. If you can’t make text fit properly on one slide without squeezing it in too tightly, split the text into two slides. Go to the Paragraphs section. I am using Power Point 2010. On widescreen slides – as in, slides with a 16:9 ratio – a 12-column grid works best. The files must be in PowerPoint format, and contain only the slides that are contained in the 'sections' from the PowerPoint file. 1. Every week I separate a long PowerPoint file into separate files. Change to two columns. Click on the edge of the text box so that it shows a solid line with selection handles on its perimeter (see Figure 2). Click the Home tab. NumColumns: Required: Long: Number of columns that the cell is being split into. Move the insertion point where you want to split the text. There, you will see the “Format” tab. You can use the presentation slide designs under this template to compare product attributes as well as describing a process of four steps. The grid spacing you choose really depends on the nature of the content you’re working with and the requirements of your project (the desired look and feel). You can easily divide a 12-column grid into six, four, three and two columns. You can choose a layout that's built into PowerPoint and start adding your own content to build out a slide. Each slide has three clear columns with headings and places to add text to illustrate milestones, features, benefits and other concepts to your audience. 4. The main slide design contains four numbered columns with editable description and title. Use the Outline View. 3 & 4 column PowerPoint template is standard PowerPoint slide created for infographic presentation. Follow these steps to set columns within text containers in PowerPoint 2013: Open any slide which has a text box. Support and feedback . The process of merging Powerpoint presentations does not need to involve individually inserting each slide into the new presentation. Head to Insert > Break > Column break, and Google Docs will start a new column wherever your insertion point is currently placed. Let us explore how we can achieve it in this article. Select a similar layout, then duplicate it, or choose Slide Master>Insert Layout. It must be time-consuming if you manually split a large PowerPoint document that contains a lot of slides into presentations that each contains one original slide. /* Responsive layout - when the screen is less than 600px wide, make the three columns stack on top of each other instead of next to each other */ @media screen and (max-width: 600px) { .column … I would like to split a frame into 2 parts side by side by minipage. PowerPoint creates a new slide with the same title as the current slide. Click it, and it will expand to reveal more options. This will serve as a new template. To return to the default page setup, highlight the desired text and choose “One Column” as … To do this . Let's look at a definition of PowerPoint slide layouts and what makes them so useful: PowerPoint layouts are the combination and arrangement of objects on a slide. 4 Columns Slide Design for PowerPoint is a presentation design that you can use to create comparison slides between four different products or plans. Cell Object. 1. The three column layouts can be used for a variety of purposes, such as to create comparison slides, timelines or to present different options or ideas with the aid of text presented in a nice clean layout. Click the Columns button (next to the text alignment buttons). If you remember, we previously talked about how Steve Jobs and Tim Cook would masterfully structure their Apple keynotes into 3 main parts, making their discussions easier to understand.And now, there’s the rule of thirds. . Is there any method or software which anyone could recommend to split this presentation into say, ten smaller presentations each of 25 slides in length. On the (Table Tools) Design tab, click the Eraser button (you may have to click the Draw Borders button first). Thanks. Posts: 9 Create a 4-Content slide master. I've broken them up into independant functions and subroutines. I don't know of a way to build in automatic overflow of text to a new slide in PowerPoint. Number of rows that the cell is being split into. ... You can split a cell into a number of columns or rows as you do in Excel. When you are dealing with data from a multiple-response question, and your file as each Combining Two Powerpoint Files in Powerpoint 2010. One other way to structure a PowerPoint presentation in the editing mode is to use Outline View. 2. The pointer changes into an Eraser. PowerPoint themes include multiple slide layouts. The steps in this article will assume that you have two separate Powerpoint files, … This opens up the following dialog box: Menu to Split Cells You can enter the number of columns and rows to split. How do I make it into four? Example. Choose View>Slide Master 2. In the Split Cells dialog box, declare how many columns and rows you want to split the cell into and then click OK. Another way to merge cells is to use the Eraser. PowerPoint now allows you to create text columns within a text box, letting you change a long list of text into two or more columns. Select the cell you want to split and go to ‘Split cells’. . For example, you may decide to break one slide into two or three, or the other way around. How do I include multiple figures or "contents" on a slide? Choose Insert Placeholder to insert the number and type of placeholders you need. FREE DOWNLOAD… I am generating a PowerPoint presentation with R-markdown. To do this, look at the menu at the top of the window. You can choose it from the VIEW tab. Good learning, and happy to share! With your PowerPoint files combined, you can then save your merged file by clicking File > Save or Save As. A grid divides your slide into columns and rows (called flowlines) that can be spaced at equal or unequal intervals. This method is a little more involved. Select the desired text box. You add two data series to the stacked bar graph. Alternatively, one can also create his own templates using normal Powerpoint slides. 1. 5. A few examples are: When you have a .csv file, where data is entirely grouped into a single column and separated by commas. I PPTools Merge add-in brings mailmerge to PowerPoint. ( called flowlines ) that can be spaced at equal or unequal intervals huge sigh of relief with same... To present company profile presentation along the company vision and mission statement to use Outline.... 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Point presentation which is almost 250 slides in length grid into six, four, three and two.. View, switch to it and continue typing the text you want break! Currently placed ’ re not confident creating your own content to build out a slide want to up. Figures or contents '' on a new column wherever your insertion point where you want split... Have been lumped with a massive Power point presentation which is almost 250 in. Above the other way around table into two cells, one directly above the.... Immediately underneath the currently selected slide presentation is designed, you will notice the coveted columns... The steps below to split up a bulleted list in your PowerPoint presentation in the 'sections ' from PowerPoint! Merge two separate PowerPoint files with just a few short steps and.... To break one slide into columns and rows ( called flowlines ) that can be too tall you. Two cells, one directly above the other way to structure a PowerPoint presentation attributes well... 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The insertion point is currently placed, i 'll introduce a simple solution for doing this using in., four, three and two columns ).Shapes ( 5 ).Table.Cell 1...
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# Held Karp Relaxation
The Traveling Salesman Problem (TSP) is undoubtedly the most important and well-studied problem in Combinatorial Optimization. Today’s post is a quick overview of the Held-Karp Relaxation of TSP.
TSP : Given a complete undirected graph $G(V,E)$ with non-negative costs $c_e$ for each edge $e \in E$, find a hamiltonian cycle of G with minimum cost. It is well-known that this problem is NP-Complete.
Exercise : There is no $\alpha$-approximation algorithm for TSP (for any $\alpha \geq 1$) unless P=NP.
Metric TSP : In Metric-TSP, the edge costs satisfy triangle inequality i.e., for all $u,v,w \in V$, $c(u,w) \leq c(u,v) + c(v,w)$. Metric-TSP is also NP-complete. Henceforth, we shall focus on metric TSP.
Symmetric TSP (STSP) : In STSP, the edge costs are symmetric i.e., $c(u,v) = c(v,u)$. Approximation algorithms with factor 2 (find a minimum spanning tree (MST) of $G$ and use shortcuts to obtain a tour) and factor 3/2 (find an MST, find a perfect matching on the odd degree nodes of the MST to get a eulerian graph and obtain a tour) are well-known. The factor 3/2 algorithm, known as Christofides Algorithm [Christofides’76], is the best known approximation factor for STSP. No improvement in the last three decades !!
Following is the Held-Karp Relaxation for STSP with the cut constraints and the degree constraints. The variables are $x_e$, one for each edge $e \in E$. For a subset $S \subset V$, $\delta(S)$ denotes the edges incident to $S$. Let $x(\delta(S))$ denote the sum of values of $x_e$ of the edges with exactly one endpoint in $S$. For more details of Held-Karp relaxation see [HK’70, HK’71]
Exercise : In the following instance of STSP the cost between vertices u and v is the length of the shortest path between u and v. The three long paths are of length k. Prove that this instance achieves an integrality ratio arbitrarily close to 4/3 (as k is increased).
Asymmetric TSP (ATSP) : In ATSP, the edge costs are not necessarily symmetric i.e., the underlying graph is directed. The Held-Karp relaxation for ATSP is as follows :
Charikar, Goemans and Karloff [CGK’04] showed that the integrality of Held-Karp relaxation for ATSP is at least $2-\epsilon$. Frieze, Galbiati and Maffioli [FGM’82] gave a simple $O({\log}_2{n})$-approximation algorithm for ATSP in 1982, where n is the number of vertices. In the last eight years, this was improved to a guarantee of 0.999 ${\log}_2{n}$ by Blaser [Blaser’02], and to $\frac{4}{3}{\log}_3{n}$ Kaplan et al [KLSS’03] and to $\frac{2}{3}{\log}_2{n}$ by Feige and Singh [FS’07]. So we have an approximation factor better than ${\ln}n$ !!
Open Problems :
• The long-standing open problem is to determine the exact integrality gap of Held-Karp relaxation. Many researchers conjecture that the integrality gap of Held-Karp relaxation for STSP is 4/3 and for ATSP it is bounded by a constant. The best known upper bounds are 3/2 and O(logn) respectively.
• The size of the integrality gap instance of ATSP (constructed by [CGK’04]) is exponential in $1/\epsilon$ to achieve an integrality gap of $2-\epsilon$. Is there a polynomial-sized (in $1/\epsilon$) instance achieving an integrality gap of $2-\epsilon$ ?
References :
Nicos Christofides, Worst-case analysis of a new heuristic for the travelling salesman problem, Report 388, Graduate School of Industrial Administration, CMU, 1976.
• [HK’70] Micheal Held and Richard M. Karp, The Traveling Salesman Problem and Minimum Spanning Trees, Operations Research 18, 1970, 1138–1162.
• [HK’71] Michael Held and Richard Karp, The Traveling-Salesman Problem and Minimum Spanning Trees: Part II, Mathematical Programming 1, 1971, 6–25.
• [Christofides’76] Nicos Christofides, Worst-case analysis of a new heuristic for the travelling salesman problem, Report 388, Graduate School of Industrial Administration, CMU, 1976.
• [FGM’82] A. M. Frieze, G. Galbiati and M. Maffioli, On the Worst-Case Performance of Some Algorithms for the Asymmetric Traveling Salesman Problem, Networks 12, 1982, 23–39.
• [Blaser’02] M. Blaser, A New Approximation Algorithm for the Asymmetric TSP with Triangle Inequality, Proceedings of the 14th Annual ACM-SIAM Symposium on Discrete Algorithms, 2002, 638–645.
• [KLSS’03] H. Kaplan, M. Lewenstein, N. Shafir and M. Sviridenko, Approximation Algorithms for Asymmetric TSP by Decomposing Directed Regular Multidigraphs, Proceedings of the 44th Annual IEEE Symposium on Foundations of Computer Science, 2003, 56–67.
• [CGK’04] Moses Charikar, Michel X. Goemans, Howard J. Karloff: On the Integrality Ratio for Asymmetric TSP. FOCS 2004: 101-107
• [FS’07] Uriel Feige, Mohit Singh: Improved Approximation Ratios for Traveling Salesperson Tours and Paths in Directed Graphs. APPROX-RANDOM 2007: 104-118
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# cdftri: Cumulative Distribution Function of the Asymmetric Triangular...
### Description
This function computes the cumulative probability or nonexceedance probability of the Asymmetric Triangular distribution given parameters (ν, ω, and ψ) computed by partri. The cumulative distribution function is
F(x) = \frac{(x - ν)^2}{(ω-ν)(ψ-ν)}\mbox{,}
for x < ω,
F(x) = 1 - \frac{(ψ - x)^2}{(ψ - ω)(ψ - ν)}\mbox{,}
for x > ω, and
F(x) = \frac{(ω - ν)}{(ψ - ν)}\mbox{,}
for x = ω where x(F) is the quantile for nonexceedance probability F, ν is the minimum, ψ is the maximum, and ω is the mode of the distribution.
### Usage
1 cdftri(x, para)
### Arguments
x A real value vector. para The parameters from partri or vec2par.
### Value
Nonexceedance probability (F) for x.
### Author(s)
W.H. Asquith
pdftri, quatri, lmomtri, partri
### Examples
1 2 lmr <- lmoms(c(46, 70, 59, 36, 71, 48, 46, 63, 35, 52)) cdftri(50,partri(lmr))
Search within the lmomco package
Search all R packages, documentation and source code
Questions? Problems? Suggestions? or email at ian@mutexlabs.com.
Please suggest features or report bugs with the GitHub issue tracker.
All documentation is copyright its authors; we didn't write any of that.
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# Use Residue Theorems or Laurent Series to evaluate integral
Tags:
1. Jun 3, 2015
### monnapomona
1. The problem statement, all variables and given/known data
Evaluate the integral using any method:
C (z10) / (z - (1/2))(z10 + 2), where C : |z| = 1
2. Relevant equations
C f(z) dz = 2πi*(Σki=1 Resp_i f(z)
3. The attempt at a solution
Rewrote the function as (1/(z-(1/2)))*(1/(1+(2/z^10))). Not sure if Laurent series expansion is the best choice for this problem but I ended up getting: (Σn=0 (1/2)n / zn+1)*(Σn=0 (-1)n 2n/(z10)n)
I get stuck at this point but i tried working out the series and get: 1 - 2/z12 + 1/z23+ ...
So would the residue just be 1 and the ∫C f(z) dz = 2πi?
**sorry in advance for my formatting**
2. Jun 4, 2015
### vela
Staff Emeritus
What are the poles and where are they located?
3. Jun 4, 2015
### monnapomona
Not sure if this is correct but would one pole be +1/2 (Used this function (1/(z-(1/2))) for my reasoning)?
4. Jun 4, 2015
### vela
Staff Emeritus
You just need to find where the denominator vanishes, so $z=1/2$ is definitely one pole. Where are the rest?
5. Jun 4, 2015
### monnapomona
That's where I'm kind of stuck. For the other function in the denominator, (1/(1+2/z^10)), it doesn't go to 0 even if z = 0?
6. Jun 4, 2015
### vela
Staff Emeritus
Oh, look at the integrand before you messed with it. In other words, when does the denominator of $\frac{z^{10}}{(z-1/2)(z^{10}+2)}$ vanish?
7. Jun 4, 2015
### monnapomona
Would the poles be z = 1/2, (-2)^(1/10) ?
8. Jun 4, 2015
### vela
Staff Emeritus
Expand on what you mean by (-2)^(1/10). Remember, you're working with complex numbers, so taking fractional powers is a little more involved. Consider solving the equation $z^{10} = -2e^{2\pi n i}$ where $n \in \mathbb{Z}$.
9. Jun 4, 2015
### monnapomona
Went to office hours today, it turns out for this question, it wasn't limited to just the residue theorems and Laurent series expansion methods. I tried using Cauchy integral formula, where z0 = 1/2 and f(z) = (z^10 / (z^10 + 2)) and got 2πi / 2049 as the final answer.
10. Jun 4, 2015
### vela
Staff Emeritus
Regardless of which method you end up using, you still need to determine where the poles are. Just make sure you know how to do that.
11. Jun 4, 2015
### Ray Vickson
The number $-2$ has ten 10th roots!
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Browse Questions
Which is not a macromolecule?
$\begin {array} {1 1} (a)\;DNA & \quad (b)\;Starch \\ (c)\;Sodium \: palmitate & \quad (d)\;Insulin \end {array}$
Can you answer this question?
It is a single molecule while all others are polymers
Ans : (c)
answered Mar 12, 2014
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Kazuya Kawasetsu
Associate Professor Priority Organization for Innovation and Excellence Kumamoto University Japanese Page
Organizer
Infinite-dimensional Representation Theory Seminar
Workshop
gVertex Operator Algebras and Related Topics in Kumamotoh (February 10th-14th 2020)
Research Interest
Representation Theory, Vertex Algebras
List of Papers
1. Kawasetsu, K. and Ridout, D. gRelaxed highest-weight modules I: Rank 1 cases.h Communications in Mathematical Physics, Volume 368 (2019): 627-663.
2. Kawasetsu, K. and Sakai, Y. gModular linear differential equations of fourth order and minimal W-algebras.h Journal of Algebra Volume 506, 445-488 (2018).
3. (Kawasetsu, K., Lam, C. H. and Lin, X. g$Z_2$-orbifold construction associated with (-1)-isometry and uniqueness of holomorphic vertex operator algebras of central charge 24.h Proceedings of the American Mathematical Society 146 (2018): 1937-1950.
4. Kawasetsu, K. gW-algebras with non-admissible levels and the Deligne exceptional series.h International Mathematics Research Notices Volume 2018, Issue 3 (2018): 641-676.
5. Arakawa, T. and Kawasetsu, K. gQuasi-lisse vertex algebras and modular linear differential equations.h Lie Groups, Geometry, and Representation Theory (A Tribute to the Life and Work of Bertram Kostant). Birkh$\ddot{a}$user, Cham, Editors: V. Kac and V. Popov, (2018): 41-57.
6. Arakawa, T., Creutzig, T., Kawasetsu, K. and Linshaw, A.R. gOrbifolds and cosets of minimal W-algebras.h Communications in Mathematical Physics Volume 355, Issue 1 (2017): 339-372.
7. Kawasetsu, K. gThe free generalized vertex algebras and generalized principal subspaces.h Journal of Algebra Volume 444 (2015): 20-51.
8. Kawasetsu, K. gThe intermediate vertex subalgebras of the lattice vertex operator algebras.h Letters in Mathematical Physics Volume 104, Issue 2 (2014): 157-178.
Curriculum Vitae
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# Questions about the Fourier expansion of $e^{iz\cot(x)}$
Referring to a question I posted on MS, I post it here, as I didn't get an answer:
By analogy with the Jacobi–Anger expansion, one expects that $e^{iz\cot(x)}$ has a Fourier expansion of the form : $$e^{iz\cot(\theta)}=\sum_{n=-\infty}^{\infty}\Lambda_{n}(z)e^{in\theta}$$ $\Lambda_{n}(z)$ is given by: $$\Lambda_{n}(z)=\frac{1}{\pi}\int_{-\pi/2}^{\pi/2}e^{iz\cot(\theta)-in\theta}d\theta$$ A simple calculation yields: $$\Lambda_{n}(z)=\frac{2}{n\pi}\sin\left(\frac{\pi n}{2} \right )\sum_{m=0}^{\infty}\frac{z^{m}}{m!}\text{F}_{1}\left(-\frac{n}{2};-m,m;1-\frac{n}{2};1,-1 \right )$$
Where $\text{F}_{1}(\alpha;\beta,\beta^{'};\gamma;x,y)$ is the Appell Hypergeometric Function. Now, I have two questions:
1-For a purely imaginary $z$,$\;\;$ $e^{iz\cot(\theta)}$ has essential singularities at $\theta=\pm n\pi$. How is that reflected in the Fourier expansion?
2-Can we express the infinite sum in terms of other special functions?
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## 1 Answer
I will try to answer your two questions in reverse order.
$2.$ For real $z$ your Fourier coefficient
$$\Lambda_{n}(z)=\frac{2}{\pi}\int_{0}^{\pi/2}\cos(n\theta-z\cot\theta)\;d\theta$$
can be expressed in terms of elementary functions for even $n$ and in terms of the special functions $I_{m}(z)$ (modified Bessel function of the first kind) and $L_{-m}(z)$ (modified Struve function) for odd $n$. The first few Fourier coefficients are
$$\Lambda_0(z)=e^{-|z|}$$
$$\Lambda_1(z)=\Lambda_{-1}(-z)=z L_{-1}(z)+zL_0(z)-|z|I_0(z)-|z|I_1(z)+2/\pi$$
$$\Lambda_2(z)=\Lambda_{-2}(-z)=e^{-|z|}(z-|z|)$$
For even $n$ the Fourier coefficient $L_n(z)\equiv 0$ for $nz>0$.
$1.$ The Fourier coefficients, given above along the real $z$-axis, can be continued analytically into the complex-$z$ plane, but not along the imaginary $z$-axis. The branch cut for imaginary $z$ is a direct consequence of the essential singularity you noticed.
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# Problem #1605
1605 Let $f(x)=ax^2+bx+c$, where $a$, $b$, and $c$ are integers. Suppose that $f(1)=0$, $50, $70, $5000k for some integer $k$. What is $k$? $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ This problem is copyrighted by the American Mathematics Competitions.
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# Angle Between Two Planes
#### definition
The angle between two planes is defined as the angle between their normals Fig,
#### notes
Observe that if θ is an angle between the two planes, then so is 180 – θ Fig.
If vec n _1 and vec n_2 are normals to the planes and θ be the angle between the planes
vec r . vec n _1 = d_1 and vec r . vec n _2 = d_2 .
Then θ is the angle between the normals to the planes drawn from some common point We have
cos θ = |(vec n_1 . vec n_2)/ (|vec n _1| |vec n_2|)|
Cartesian form
Let θ be the angle between the planes,
A_1x + B_1y +C_1z + D_1 = 0 and A_2x +B_2y + C_2 z + D_2 = 0
The direction ratios of the normal to the planes are A_1, B_1, C_1 and A_2, B_2, C_2 respectively.
Therefore , cos θ = |(A_1 A_2 + B_1 B_2 + C_1 C_2)/ (sqrt(A_1^2 + B_1^2 + C_1^2 ) sqrt (A_2^2 + B_2^2 +C_2^2))|
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An Alternating Risk Reserve Process - Part I
#### O.J. Boxma, H. Jonsson, J.A.C. Resing, S. Shneer
2010, v.16, №2, 409-424
ABSTRACT
We consider an alternating risk reserve process with a threshold dividend strategy. The process can be in two different states and the state of the process can only change just after claim arrival instants. If at such an instant the capital is below the threshold, the system is set to state $1$ (paying no dividend), and if the capital is above the threshold, the system is set to state~$2$ (paying dividend). Our interest is in the survival probabilities. In the case of exponentially distributed claim sizes, survival probabilities are found by solving a system of integro-differential equations. In the case of generally distributed claim sizes, they are expressed in the survival probabilities of the corresponding standard risk reserve processes.
Keywords: insurance risk models,alternating risk reserve process,survival probabilities
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# Spin-Orbit-Induced Strong Coupling of a Single Spin to a Nanomechanical Resonator
Phys. Rev. Lett. 108, 206811
Recent experiments in nanomechanics have reached the ultimate quantum limit by cooling a nanomechanical system close to its ground state [1]. Among the variety of available nanomechanical systems, nanostructures made out of atomically-thin carbon-based materials such as graphene and carbon nanotubes (CNTs) stand out due to their low masses and high stiffnesses. These properties give rise to high oscillation frequencies, potentially enabling near ground-state cooling using conventional cryogenics, and large zero-point motion, which improves the ease of detection [2,3].
Recently, a high quality-factor suspended CNT resonator was used to demonstrate strong coupling between nanomechanical motion and single-charge tunneling through a quantum dot (QD) defined in the CNT [4]. Here, we theoretically investigate the coupling of a single electron spin to the quantized motion of a discrete flexural mode of a suspended CNT (see Fig. 1), and show that the strong-coupling regime of this Jaynes-Cummings-type system is within reach. This coupling provides means for electrical manipulation of the electron spin via microwave irradiation, and leads to strong nonlinearities in the CNT’s mechanical response which may potentially be used for enhanced functionality in sensing applications
In addition to their outstanding mechanical properties, carbon-based systems also possess many attractive characteristics for information processing applications. The potential for single electron spins in QDs to serve as the elementary qubits for quantum information processing [8] is currently being investigated in a variety of systems. In many materials, such as GaAs, the hyperfine interaction between electron and nuclear spins is the primary source of electron spin decoherence which limits qubit performance (see, e.g., [9]). However, carbon-based structures can be grown using starting materials isotopically-enriched in $C12$, which has no net nuclear spin, thus practically eliminating the hyperfine mechanism of decoherence [10], leaving behind only a spin-orbit contribution [11,12]. Furthermore, while the phonon continuum in bulk materials provides the primary bath enabling spin relaxation, the discretized phonon spectrum of a suspended CNT can be engineered to have an extremely low density of states at the qubit (spin) energy splitting. Thus very long spin lifetimes are expected off-resonance [13]. On the other hand, when the spin splitting is nearly resonant with one of the high- $Q$ discrete phonon “cavity” modes, strong spin-phonon coupling can enable qubit control, information transfer, or the preparation of entangled states.
The interaction between nanomechanical resonators and single spins was recently detected [14], and has been theoretically investigated [15,16] for cases where the spin-resonator coupling arises from the relative motion of the spin and a source of local magnetic field gradients. Such coupling is achieved, e.g., using a magnetic tip on a vibrating cantilever which can be positioned close to an isolated spin fixed to a nonmoving substrate. Creating strong, well-controlled, local gradients remains challenging for such setups. In contrast, as we now describe, in CNTs the spin-mechanical coupling is intrinsic, supplied by the inherent strong spin-orbit coupling [17–20] which was recently discovered by Kuemmeth et al. [21].
Consider an electron localized in a suspended CNT quantum dot (see Fig. 1). Below we focus on the case of a single electron, but expect the qualitative features to be valid for any odd occupancy (see Ref. [22]). We work in the experimentally relevant parameter regime where the spin-orbit and orbital-Zeeman couplings are small compared with the nanotube band gap and the energy of the longitudinal motion in the QD. Here, the longitudinal and sublattice orbital degrees of freedom are effectively frozen out, leaving behind a nominally fourfold degenerate low-energy subspace associated with the remaining spin and valley degrees of freedom (see Refs. [23,24]).
A simple model describing the spin and valley dynamics in this low-energy QD subspace, incorporating the coupling of electron spin to deflections associated with the flexural modes of the CNT [25,26], was introduced in Ref. [27]. In principle, the deformation-potential spin-phonon coupling mechanism [11] is also present. The deflection coupling mechanism is expected to dominate at long phonon wavelengths, while the deformation-potential coupling should dominate at short wavelengths (see discussion in [27]). For simplicity we consider only the deflection coupling mechanism, but note that the approach can readily be extended to include both effects.
The Hamiltonian describing this system is [24,27,28]
where $Δso$ and $ΔKK′$ denote the spin-orbit and intervalley couplings, $τi$ and $si$ are the Pauli matrices in valley and spin space (the pseudospin is frozen out for the states localized in a QD), $t$ is the tangent vector along the CNT axis, and $B$ denotes the magnetic field. Note that the spin-orbit coupling has contributions which are diagonal and off-diagonal in sublattice space [18–20,22]. When projected onto to a single longitudinal mode of the quantum dot, the effective Hamiltonian given above describes the coupling of the spin to the nanotube deflection at the location of the dot [24].
For a nominally straight CNT we take $t$ pointing along the $z$ direction, giving $s·t=sz$ and $B·t=Bz$. Here we find the low-energy spectrum shown in Fig. 2. The two boxed regions indicate two different two-level systems that can be envisioned as qubit implementations in this setup: we define a spin qubit [8] ( $S$) at strong longitudinal magnetic field, near the value $B*$ of the upper level crossing, and a mixed spin-valley or Kramers ( $K$) qubit [28], which can be operated at low fields applied either in the longitudinal ( $Kz$) or perpendicular ( $Kx$) directions.
We now study how these qubits couple to the quantized mechanical motion of the CNT. For simplicity we consider only a single polarization of flexural motion (along the $x$ direction), assuming that the two-fold degeneracy is broken, e.g., by an external electric field. A generalization to two modes is straightforward.
A generic deformation of the CNT with deflection $u(z)$ makes the tangent vector $t(z)$ coordinate-dependent. Expanding $t(z)$ for small deflections, we rewrite the coupling terms in Hamiltonian (1) as $s·t≃sz+(du/dz)sx$ and $B·t≃Bz+(du/dz)Bx$. Expressing the deflection $u(z)$ in terms of the creation and annihilation operators $a†$ and $a$ for a quantized flexural phonon mode, $u(z)=f(z)ℓ02(a+a†)$, where $f(z)$ and $ℓ0$ are the waveform and zero-point amplitude of the phonon mode, we find that each of the three qubit types ( $S$, $Kx$, $Kz$) obtains a coupling to the oscillator mode which we describe as
Here the matrices $σ1,3$ are Pauli matrices acting on the two-level qubit subspace, and we have included a term describing external driving of the oscillator with frequency $ω$ and coupling strength $λ$, which can be achieved by coupling to the ac electric field of a nearby antenna [4]. Below we describe the dependence of the qubit-oscillator coupling $g$ on system parameters for each qubit type (S, Kx, or Kz). The derivation of Eq. (2) is detailed in [24].
For the spin qubit ( $S$), the relevant twofold degree of freedom is the spin of the electron itself. Therefore in Eq. (2) we have $σ3=sz$ and $σ1=sx$, and the qubit levels are split by the Zeeman energy, measured relative to the value $B*$ where the spin-orbit-split levels cross, $ℏωq=μB(B-B*)$. A spin magnetic moment of $μB$ is assumed, and $B*≈Δso/2μB$ for $ΔKK′≪Δso$. For the qubit-resonator coupling, we find $g=Δso⟨f′⟩ℓ0/22$, independent of $B$. Here, $⟨f′⟩$ is the derivative of the waveform of the phonon mode averaged against the electron density profile in the QD.
For a symmetric QD, positioned at the midpoint of the CNT, the coupling matrix element proportional to $⟨f′⟩$ vanishes for the fundamental and all even harmonics (the opposite would be true for the deformation-potential coupling mechanism). The cancellation is avoided for a QD positioned away from the symmetry point of the CNT, or for coupling to odd harmonics. Here, for concreteness, we consider coupling of a symmetric QD to the first vibrational harmonic of the CNT. Using realistic parameter values [4,21,29,30], $L=400 nm$, $ℓ0=2.5 pm$, $Δso=370 μeV$, $ΔKK′=32.5 μeV$, $μorb=1550 μeV/T$, and $ωp/2π=500 MHz$, we find $g/2π≈0.56 MHz$, irrespective of the magnetic field strength $B$ along the CNT.
For the Kramers qubits ( $Kx$ and $Kz$), both $ωq$ and $g$ depend on $B$. The qubit splitting for the $Kx$ qubit is controlled by the perpendicular field, $ℏωq=μB(2ΔKK′/Δ)Bx$, while for the $Kz$ qubit, it is controlled by the longitudinal field $ℏωq=(μB+μorb(Δso/Δ))Bz$, where $Δ=Δso2+4ΔKK′2$ denotes the zero-field splitting between the two Kramers pairs. Resonant coupling occurs when $ωq=ωp$. This condition sets the relevant value of $Bx$ ( $Bz$) in the case of the $Kx$ ( $Kz$) qubit; the parameters above yield $Bx≈103 mT$ ( $Bz≈0.6 mT$).
The qubit-cavity coupling for the Kx qubit increases linearly with the applied perpendicular field, $ℏg=-(⟨f′⟩ℓ0/2)(μorbΔso/Δ+μBΔso2/Δ2)Bx$, while for the $Kz$ qubit it scales with the longitudinal field, $ℏg=(⟨f′⟩ℓ0/2)(μorb2ΔKK′Δso/Δ2)Bz$. Using the values of $Bx$ and $Bz$ obtained above, we estimate couplings of $g/2π≈0.49 MHz$ for the $Kx$ qubit, and $g/2π≈0.52 kHz$ for the $Kz$ qubit. Thus the coupling for the $Kx$ qubit is comparable to that of the spin qubit, while the coupling of the $Kz$ qubit is much weaker. Therefore, we restrict our considerations to the spin and $Kx$ qubits below.
Ref. [4] reports the fabrication of CNT resonators with quality factors $Q≈150,000$. We take $Q=63,000$ for the following estimate. Together with the oscillator frequency $ωp/2π=500 MHz$, this value of $Q$ implies an oscillator damping rate of $Γ≈5×104 s-1≪g$. Because of the near-zero density of states of other phonon modes at $ωq$, it is reasonable to assume a very low spontaneous qubit relaxation rate $γ$. These observations suggest that the so-called “strong coupling” regime of qubit-oscillator interaction, defined as $Γ,γ≪g$, can be reached with CNT resonators.
To quantify the system’s response in the anticipated parameter regime, we study the coupled qubit-oscillator dynamics using a master equation which takes into account the finite lifetime of the phonon mode as well as the nonzero temperature of the external phonon bath. For weak driving, $λ≪ωp$, and $ωp≈ωq≈ω≫g$, we move to a rotating frame and use the rotating wave approximation (RWA) to map the Hamiltonian, Eq. (2), into Jaynes-Cummings form [31]
where $ω˜i=ωi-ω$. Including the nonunitary dynamics associated with the phonon-bath coupling, the master equation for the qubit-oscillator density matrix $ρ$ reads:
where $nB=1/(eℏωp/kBT-1)$ is the bath-mode Bose-Einstein occupation factor, and $kB$ is the Boltzmann constant.
Because of the phonon damping, in the long-time limit the system is expected to tend towards a steady state, described by the density matrix $ρ¯$. We study these steady states, found by setting $ρ˙=0$ in Eq. (4), using both numerical and semiclassical analytical methods. In Figs. 3(a) and 3(c) we show the steady-state phonon occupation probability distribution $P(δω,n)$ as a function of the drive frequency-phonon frequency detuning $δω=-ω˜p$ and the phonon occupation number $n$, for the case where the qubit and oscillator frequencies are fixed and degenerate, $ωq=ωp$ (see caption for parameter values). Panels $a$ and $c$ compare the cases with and without qubit-oscillator coupling. In Figs. 3(b) and 3(d) we show the averaged phonon occupation number $n¯(δω)=∑nnP(δω,n)$, which is closely related to the mean squared resonator displacement in the steady state: $X2=x2¯=ℓ02(n¯+12)$. For $g≠0$, we observe a splitting of the oscillator resonance, which is characteristic of the coupling to the two-level system, and can serve as an experimental signature of the qubit-oscillator coupling. For drive frequencies near the split peaks, the phonon number distribution is bimodal [Fig. 3(f)] showing peaks at $n≈0$ and at high- $n$, indicating bistable behavior (see below).
For strong excitation, where the mean phonon occupation is large, we expect a semiclassical approach to capture the main features of the system’s dynamics [32,33]. Extending the approach described in [32] to include distinct values of the qubit, oscillator, and drive frequencies, $ωq,ωp$, and $ω$, we derive semiclassical equations of motion for the mean spin and oscillator variables (see [24]). The steady-state values of the mean squared oscillator amplitude obtained from the resulting nonlinear system are shown in Fig. 3(e). In the vicinity of the split peak we find two branches of stable steady-state solutions, indicative of bistable or hysteretic behavior [4]. The semiclassical results in Fig. 3(e) are in correspondence with the phonon number distribution in Fig. 3(c), and explain its bimodal character. Similar oscillator instabilities have been used as the basis for a sensitive readout scheme in superconducting qubits [34], and may potentially be useful for mass or magnetic field sensing applications where small changes of frequency need to be detected.
To predict the oscillator response to be detected via a charge sensor (see below), we solve for the stationary state of Eq. (4) directly for a range of driving frequencies, qubit-oscillator detunings (set by the magnetic field), and temperatures $T$. In Figs. 3(g) and 3(h), we show the $T=0$ and $T=50 mK$ root mean squared oscillator amplitude $X∝n¯+1/2$ as function of magnetic field $B$ and drive frequency, for the case of a spin ( $S$) qubit. The value $δB=0$ corresponds to resonant coupling $ωq=ωp$. These results also apply for the $Kx$ qubit, if the magnetic field axis is adjusted appropriately. In the zero-temperature case, only half of the eigenstates $ℏω±≈ℏωp∓ℏg2/(ωp-ωq)$ of Eq. (3) can be efficiently excited by the drive at fixed $δB$, giving rise to the upper (lower) feature in Fig. 3(g) for $δB<0(δB>0)$. However, for $T≳ℏωq$, both branches of the Jaynes-Cummings ladder can be efficiently excited [Fig. 3(h)]. This is a distinct and experimentally accessible signature of the strong coupling at finite temperature. Note that the vacuum Rabi splitting is also observed [see arrows in Fig. 3(d)], but features arising from nonlinearity in the strongly driven system dominate by more than 2 orders of magnitude.
Displacement detection of nanomechanical systems is possible using charge sensing [5,35], where the conductance of a mesoscopic conductor, such as a QD or quantum point contact, is modulated via capacitve coupling to the charged mechanical resonator. Furthermore, the qubit state itself can be read out using spin-detection schemes developed for semiconductor QDs [36], or by a dispersive readout scheme like that commonly used in superconducting qubits coupled to microwave resonators [37]. The dispersive regime can be rapidly accessed by, e.g., tuning the resonator frequency using dc gate pulses which control the tension in the CNT [4].
In summary, we predict that strong qubit-resonator coupling can be realized in suspended CNT QDs with current state-of-the-art devices. The coupling described here may find use in sensing applications, and in spin-based quantum information processing, where the CNT oscillator enables electrical control of the electron spin, and, with capacitive couplers, may provide long-range interactions between distant electronic qubits [16,38]. Combined with control of the qubit via electron-spin-resonance [39], the mechanism studied here could be utilized for ground-state cooling and for generating arbitrary motional quantum states of the oscillator [15].
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Guido Burkard studied physics at ETH Zurich and received his Ph.D. from the University of Basel, Switzerland. Since 2008, he has been a full professor at the University of Konstanz, Germany. He previously held a faculty position at RWTH Aachen University, and was SNF assistant professor at the University of Basel, after a postdoctoral appointment with the IBM T. J. Watson Research Center at Yorktown Heights, New York. His research interests encompass condensed-matter theory and quantum information, with a focus on spin-based quantum information processing.
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# How do you multiply (8m^2+8m+3)(2m^2+8m+4)?
Jun 25, 2017
$16 {m}^{4} + 80 {m}^{3} + 102 {m}^{2} + 56 m + 12$
#### Explanation:
Multiply everything in the right brackets by every thing in the left.
$8 {m}^{2} \left(2 {m}^{2} + 8 m + 4\right) \to 16 {m}^{4} + 64 {m}^{3} + 32 {m}^{2}$
$\textcolor{w h i t e}{.} 8 m \left(2 {m}^{2} + 8 m + 4\right) \to \text{ } 16 {m}^{3} + 64 {m}^{2} + 32 m$
color(white)(8..)3(2m^2+8m+4)->ul(" "6m^2+24m+12
$\text{ } 16 {m}^{4} + 80 {m}^{3} + 102 {m}^{2} + 56 m + 12$
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# Geometry problem calculate arc, inner triangle.
I learned how to solve this in geometry in middle school, but I can't remember. This isn't homework.
I want to calculate the distance between points $A$ & $B$ and the arc $\widehat {AB}$ as well. The know values are, the radius $R$ and the $S$ length. $C$ is the center of the circle. The angle is not important, but it's probably involved in the arc calculation.
$R=6400$
$S=30$
$\widehat {AB}=?$
this values are arbitrary, I only want to know how to solve.
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Draw the altitude from $C$ to $AB$. It bisects $AB$ at a point (call it $H$) since $\triangle ABC$ is an isosceles triangle, and its length is $R - S$. Using the Pythagorean Theorem:
$$(AH)^2 = (AC)^2 - (CH)^2 = R^2 - (R - S)^2 = 2RS - S^2$$
Therefore:
$$AB = 2(AH) = 2 \sqrt{2RS - S^2}$$
For the arc, use the lengths of $AH$ and $AC$ to calculate the sine of $\alpha/2$:
$$\sin\left(\frac{\alpha}{2}\right) = \frac{AH}{AC} = \frac{R - S}{R}$$
This allows you to find the value of $\alpha$. The arc length is $R \times \alpha$ (where $\alpha$ is measured in radians).
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# Typeset cftchapfont as uppercase
I'm trying to put the whole `cftchapfont` as uppercase.
``````\renewcommand\cftchapfont{\MakeUppercase}
``````
does not work at all (it actually messes up everything), nor does:
``````\renewcommand\cftchappresnum{\MakeUppercase\chaptertitlename\;}
``````
which only capitalizes `\chaptertitlename`, nor does:
``````\renewcommand\cftchappresnum{\MakeUppercase\chaptertitlename\;\MakeUppercase}
``````
which only capitalizes the first letter of SNUM (and I've got words such as "premier" for the first chapter).
How can I make sure the whole thing is uppercase?
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You have to patch the `\@chapter` command, which is responsible for writing the entries in the auxiliary files. The lines
``````\addcontentsline{toc}{chapter}{\protect\numberline{\thechapter}#1}
``````
and
``````\addcontentsline{toc}{chapter}{#1}
``````
should read with `\MakeUppercase{#1}` instead of `#1`.
-
I actually ideally wanted small caps, but the font didn't have any. What I did was:
• edit the TTF in fontforge;
• add small caps from fontforge;
• save new font as OTF;
• load font with `Letters=SmallCap` in fontspec.
and it did what I wanted.
-
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Cosmic Ray Two-Fluid Model¶
This section documents the two-fluid cosmic ray model implemented in Enzo, which was first used (and is described in detail) in Salem & Bryan 2014 . For relevant parameters, please also see Cosmic Ray Two-Fluid Model Parameters. The bulk of the code itself can be found in Grid_ZeusSolver.C
This module models the dynamical role of cosmic rays via a set of two-fluid hydro equations (see Jun et. al. 1994 ). Central to the effort is a new baryon field, CREnergyDensity, which is in units of ergs/cm^3, and is advected along with the gas. Gradients in the CR field result in a pressure felt by the gas. The CR gas is also diffusive and rays can be produced during star formation. See Cosmic Ray Two-Fluid Model Parameters for information on how to control all these options. But most important are:
• CRModel - Switches on the CR physics (0 = off, 1 = on)
• CRgamma - For polytropic equation of state. 4/3 = relativistic, adiabatic gas (default)
• CRDiffusion - turns on diffusion of CREnergyDensity field
• CRkappa - Diffusion coefficient (currently constant, isotropic)
• CRFeedback - Controls production of rays in star forming regions
For this model to run properly you must be running the Zeus Hydro Solver! (HydroMethod = 2). The model has not yet been implemented for any of the other fluid solvers in Enzo.
If you plan on including cosmic rays, definitely first verify the solver is working by running the Cosmic Ray Shocktube problem, which ought to match the analytic solution described in Pfrommer 2006 . See CR Shock Tube (250: unigrid and AMR) for more detailed information on this test problem.
Cosmic Rays have also been implemented in the isolated galaxy simulation. They initialize with a profile equal to the density of the thermal gas, multiplied by a constant, GalaxySimulationCR, typically set to 0.1 (all in code units).
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# Bibliothèque
Musique » Florence + the Machine »
## What the Water Gave Me
113 écoutes | Se rendre sur la page du titre
Titres (113)
Titre Album Durée Date
What the Water Gave Me 5:34 3 sept. 2012, 15h22m
What the Water Gave Me 5:34 3 sept. 2012, 13h52m
What the Water Gave Me 5:34 27 mai 2012, 15h00m
What the Water Gave Me 5:34 27 mai 2012, 13h31m
What the Water Gave Me 5:34 27 mai 2012, 12h01m
What the Water Gave Me 5:34 22 mai 2012, 15h38m
What the Water Gave Me 5:34 20 mai 2012, 3h10m
What the Water Gave Me 5:34 20 mai 2012, 1h41m
What the Water Gave Me 5:34 20 mai 2012, 0h11m
What the Water Gave Me 5:34 19 mai 2012, 22h41m
What the Water Gave Me 5:34 19 mai 2012, 21h01m
What the Water Gave Me 5:34 19 mai 2012, 17h52m
What the Water Gave Me 5:34 19 mai 2012, 16h22m
What the Water Gave Me 5:34 19 mai 2012, 14h52m
What the Water Gave Me 5:34 19 mai 2012, 13h23m
What the Water Gave Me 5:34 10 mai 2012, 1h32m
What the Water Gave Me 5:34 10 mai 2012, 0h03m
What the Water Gave Me 5:34 9 mai 2012, 22h33m
What the Water Gave Me 5:34 9 mai 2012, 1h09m
What the Water Gave Me 5:34 8 mai 2012, 23h39m
What the Water Gave Me 5:34 8 mai 2012, 22h09m
What the Water Gave Me 5:34 8 mai 2012, 20h40m
What the Water Gave Me 5:34 8 mai 2012, 19h10m
What the Water Gave Me 5:34 8 mai 2012, 17h41m
What the Water Gave Me 5:34 10 avr. 2012, 18h45m
What the Water Gave Me 5:34 10 avr. 2012, 17h49m
What the Water Gave Me 5:34 10 avr. 2012, 16h53m
What the Water Gave Me 5:34 3 avr. 2012, 17h05m
What the Water Gave Me 5:34 3 avr. 2012, 15h36m
What the Water Gave Me 5:34 1 avr. 2012, 16h22m
What the Water Gave Me 5:34 31 mars 2012, 20h29m
What the Water Gave Me 5:34 31 mars 2012, 19h33m
What the Water Gave Me 5:34 31 mars 2012, 16h07m
What the Water Gave Me 5:34 31 mars 2012, 14h23m
What the Water Gave Me 5:34 30 mars 2012, 12h21m
What the Water Gave Me 5:34 29 mars 2012, 19h05m
What the Water Gave Me 5:34 29 mars 2012, 18h09m
What the Water Gave Me 5:34 29 mars 2012, 17h13m
What the Water Gave Me 5:34 29 mars 2012, 16h17m
What the Water Gave Me 5:34 29 mars 2012, 15h21m
What the Water Gave Me 5:34 29 mars 2012, 14h19m
What the Water Gave Me 5:34 29 mars 2012, 13h23m
What the Water Gave Me 5:34 26 mars 2012, 22h01m
What the Water Gave Me 5:34 26 mars 2012, 18h02m
What the Water Gave Me 5:34 26 mars 2012, 16h30m
What the Water Gave Me 5:34 26 mars 2012, 15h34m
What the Water Gave Me 5:34 26 mars 2012, 14h26m
What the Water Gave Me 5:34 25 mars 2012, 19h24m
What the Water Gave Me 5:34 25 mars 2012, 18h28m
What the Water Gave Me 5:34 25 mars 2012, 17h32m
What the Water Gave Me 5:34 23 mars 2012, 0h41m
What the Water Gave Me 5:34 22 mars 2012, 23h45m
What the Water Gave Me 5:34 22 mars 2012, 22h49m
What the Water Gave Me 5:34 22 mars 2012, 21h53m
What the Water Gave Me 5:34 22 mars 2012, 20h57m
What the Water Gave Me 5:34 22 mars 2012, 20h01m
What the Water Gave Me 5:34 22 mars 2012, 19h05m
What the Water Gave Me 5:34 22 mars 2012, 18h09m
What the Water Gave Me 5:34 22 mars 2012, 11h02m
What the Water Gave Me 5:34 21 mars 2012, 23h41m
What the Water Gave Me 5:34 21 mars 2012, 22h45m
What the Water Gave Me 5:34 21 mars 2012, 21h49m
What the Water Gave Me 5:34 21 mars 2012, 20h53m
What the Water Gave Me 5:34 21 mars 2012, 19h47m
What the Water Gave Me 5:34 21 mars 2012, 18h51m
What the Water Gave Me 5:34 21 mars 2012, 12h45m
What the Water Gave Me 5:34 20 mars 2012, 11h24m
What the Water Gave Me 5:34 20 mars 2012, 10h28m
What the Water Gave Me 5:34 19 mars 2012, 23h53m
What the Water Gave Me 5:34 19 mars 2012, 22h56m
What the Water Gave Me 5:34 19 mars 2012, 22h00m
What the Water Gave Me 5:34 19 mars 2012, 21h04m
What the Water Gave Me 5:34 19 mars 2012, 20h06m
What the Water Gave Me 5:34 19 mars 2012, 19h10m
What the Water Gave Me 5:34 19 mars 2012, 14h07m
What the Water Gave Me 5:34 19 mars 2012, 10h35m
What the Water Gave Me 5:34 17 mars 2012, 22h18m
What the Water Gave Me 5:34 17 mars 2012, 21h22m
What the Water Gave Me 5:34 17 mars 2012, 20h26m
What the Water Gave Me 5:34 17 mars 2012, 19h30m
What the Water Gave Me 5:34 17 mars 2012, 17h57m
What the Water Gave Me 5:34 17 mars 2012, 17h01m
What the Water Gave Me 5:34 17 mars 2012, 16h05m
What the Water Gave Me 5:34 17 mars 2012, 15h09m
What the Water Gave Me 5:34 17 mars 2012, 1h15m
What the Water Gave Me 5:34 17 mars 2012, 0h19m
What the Water Gave Me 5:34 16 mars 2012, 23h23m
What the Water Gave Me 5:34 16 mars 2012, 22h27m
What the Water Gave Me 5:34 16 mars 2012, 21h31m
What the Water Gave Me 5:34 16 mars 2012, 20h35m
What the Water Gave Me 5:34 16 mars 2012, 19h39m
What the Water Gave Me 5:34 16 mars 2012, 13h25m
What the Water Gave Me 5:34 16 mars 2012, 6h59m
What the Water Gave Me 5:34 15 mars 2012, 18h05m
What the Water Gave Me 5:34 15 mars 2012, 17h09m
What the Water Gave Me 5:34 15 mars 2012, 8h23m
What the Water Gave Me 5:34 14 mars 2012, 15h50m
What the Water Gave Me 5:34 13 mars 2012, 15h52m
What the Water Gave Me 5:34 13 mars 2012, 14h56m
What the Water Gave Me 5:34 11 mars 2012, 12h46m
What the Water Gave Me 5:34 10 mars 2012, 12h03m
What the Water Gave Me 5:34 10 mars 2012, 11h07m
What the Water Gave Me 5:34 10 mars 2012, 0h46m
What the Water Gave Me 5:34 9 mars 2012, 23h50m
What the Water Gave Me 5:34 9 mars 2012, 20h13m
What the Water Gave Me 5:34 8 mars 2012, 12h40m
What the Water Gave Me 5:34 7 mars 2012, 18h10m
What the Water Gave Me 5:34 7 mars 2012, 17h09m
What the Water Gave Me 5:34 7 mars 2012, 16h03m
What the Water Gave Me 5:34 4 mars 2012, 23h45m
What the Water Gave Me 5:34 4 mars 2012, 22h15m
What the Water Gave Me 5:34 4 mars 2012, 20h46m
What the Water Gave Me 5:34 4 mars 2012, 18h42m
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# American Institute of Mathematical Sciences
February 2005, 5(1): 1-14. doi: 10.3934/dcdsb.2005.5.1
## Statistical equilibrium of the Coulomb/vortex gas on the unbounded 2-dimensional plane
1 Department of Physics, National University of Singapore 2 Department of Mathematical Sciences, Rensselaer Polytechnic Institute, Troy, NY 12180, United States
Received September 2003 Revised October 2003 Published November 2004
This paper presents the statistical equilibrium distributions of single-species vortex gas and cylindrical electron plasmas on the unbounded plane obtained by Monte Carlo simulations. We present detailed numerical evidence that at high values of $\beta >0$ and $\mu >0$, where $\beta$ is the inverse temperature and $\mu$ is the Lagrange multiplier associated with the conservation of the moment of vorticity, the equilibrium vortex gas distribution is centered about a regular crystalline distribution with very low variance. This equilibrium crystalline structure has the form of several concentric nearly regular polygons within a bounding circle of radius $R.$ When $\beta$ ~ $O(1)$, the mean vortex distributions have nearly uniform vortex density inside a circular disk of radius $R.$ In all the simulations, the radius $R=\sqrt{\beta \Omega /(2\mu )}$ where $\Omega$ is the total vorticity of the point vortex gas or number of identical point charges. Using a continuous vorticity density model and assuming that the equilibrium distribution is a uniform one within a bounding circle of radius $R$, we show that the most probable value of $R$ scales with inverse temperature $\beta >0$ and chemical potential $\mu >0$ as in $R=\sqrt{\beta \Omega /(2\mu )}.$
Citation: Syed M. Assad, Chjan C. Lim. Statistical equilibrium of the Coulomb/vortex gas on the unbounded 2-dimensional plane. Discrete and Continuous Dynamical Systems - B, 2005, 5 (1) : 1-14. doi: 10.3934/dcdsb.2005.5.1
[1] Joseph Nebus. The Dirichlet quotient of point vortex interactions on the surface of the sphere examined by Monte Carlo experiments. Discrete and Continuous Dynamical Systems - B, 2005, 5 (1) : 125-136. doi: 10.3934/dcdsb.2005.5.125 [2] Zhiyan Ding, Qin Li. Constrained Ensemble Langevin Monte Carlo. Foundations of Data Science, 2022, 4 (1) : 37-70. doi: 10.3934/fods.2021034 [3] Giacomo Dimarco. The moment guided Monte Carlo method for the Boltzmann equation. Kinetic and Related Models, 2013, 6 (2) : 291-315. doi: 10.3934/krm.2013.6.291 [4] Guillaume Bal, Ian Langmore, Youssef Marzouk. Bayesian inverse problems with Monte Carlo forward models. Inverse Problems and Imaging, 2013, 7 (1) : 81-105. doi: 10.3934/ipi.2013.7.81 [5] Ajay Jasra, Kody J. H. Law, Yaxian Xu. Markov chain simulation for multilevel Monte Carlo. Foundations of Data Science, 2021, 3 (1) : 27-47. doi: 10.3934/fods.2021004 [6] Theodore Papamarkou, Alexey Lindo, Eric B. Ford. Geometric adaptive Monte Carlo in random environment. Foundations of Data Science, 2021, 3 (2) : 201-224. doi: 10.3934/fods.2021014 [7] Michael B. Giles, Kristian Debrabant, Andreas Rössler. Analysis of multilevel Monte Carlo path simulation using the Milstein discretisation. Discrete and Continuous Dynamical Systems - B, 2019, 24 (8) : 3881-3903. doi: 10.3934/dcdsb.2018335 [8] Jiakou Wang, Margaret J. Slattery, Meghan Henty Hoskins, Shile Liang, Cheng Dong, Qiang Du. Monte carlo simulation of heterotypic cell aggregation in nonlinear shear flow. Mathematical Biosciences & Engineering, 2006, 3 (4) : 683-696. doi: 10.3934/mbe.2006.3.683 [9] P.K. Newton. N-vortex equilibrium theory. Discrete and Continuous Dynamical Systems, 2007, 19 (2) : 411-418. doi: 10.3934/dcds.2007.19.411 [10] Chjan C. Lim, Joseph Nebus, Syed M. Assad. Monte-Carlo and polyhedron-based simulations I: extremal states of the logarithmic N-body problem on a sphere. Discrete and Continuous Dynamical Systems - B, 2003, 3 (3) : 313-342. doi: 10.3934/dcdsb.2003.3.313 [11] Olli-Pekka Tossavainen, Daniel B. Work. Markov Chain Monte Carlo based inverse modeling of traffic flows using GPS data. Networks and Heterogeneous Media, 2013, 8 (3) : 803-824. doi: 10.3934/nhm.2013.8.803 [12] Mazyar Zahedi-Seresht, Gholam-Reza Jahanshahloo, Josef Jablonsky, Sedighe Asghariniya. A new Monte Carlo based procedure for complete ranking efficient units in DEA models. Numerical Algebra, Control and Optimization, 2017, 7 (4) : 403-416. doi: 10.3934/naco.2017025 [13] Juntao Yang, Viet Ha Hoang. Multilevel Markov Chain Monte Carlo for Bayesian inverse problem for Navier-Stokes equation. Inverse Problems and Imaging, , () : -. doi: 10.3934/ipi.2022033 [14] Kazuo Aoki, François Golse. On the speed of approach to equilibrium for a collisionless gas. Kinetic and Related Models, 2011, 4 (1) : 87-107. doi: 10.3934/krm.2011.4.87 [15] Vikas S. Krishnamurthy. Liouville links and chains on the plane and associated stationary point vortex equilibria. Communications on Pure and Applied Analysis, 2022, 21 (7) : 2383-2397. doi: 10.3934/cpaa.2022076 [16] Chun Liu, Jan-Eric Sulzbach. The Brinkman-Fourier system with ideal gas equilibrium. Discrete and Continuous Dynamical Systems, 2022, 42 (1) : 425-462. doi: 10.3934/dcds.2021123 [17] Kazuo Aoki, Yoshiaki Abe. Stagnation-point flow of a rarefied gas impinging obliquely on a plane wall. Kinetic and Related Models, 2011, 4 (4) : 935-954. doi: 10.3934/krm.2011.4.935 [18] François Golse. The Boltzmann-Grad limit for the Lorentz gas with a Poisson distribution of obstacles. Kinetic and Related Models, 2022, 15 (3) : 517-534. doi: 10.3934/krm.2022001 [19] Suzete Maria Afonso, Vanessa Ramos, Jaqueline Siqueira. Equilibrium states for non-uniformly hyperbolic systems: Statistical properties and analyticity. Discrete and Continuous Dynamical Systems, 2021, 41 (9) : 4485-4513. doi: 10.3934/dcds.2021045 [20] Feimin Huang, Dehua Wang, Difan Yuan. Nonlinear stability and existence of vortex sheets for inviscid liquid-gas two-phase flow. Discrete and Continuous Dynamical Systems, 2019, 39 (6) : 3535-3575. doi: 10.3934/dcds.2019146
2021 Impact Factor: 1.497
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# LV:Blog
Blog for Liberating Voices via WikiEducator.
### Three navigators
To help with filling in the gaps:
Monitoring progress: http://wikieducator.org/LV:Cards/inventory
Quick links for Image Quest 2.0: http://wikieducator.org/Template:Lv-card-image-nav (expand it)
All cards: http://wikieducator.org/Template:Lv-all-cards-nav (expand it)
-- Kim Tucker 00:48, 18 October 2012 (UTC)
### Text for all cards is being finalised
Browse the inventory ( http://wikieducator.org/LV:Cards/inventory ) to see text status and edit the text if you find any typos etc.. Comment on relevant 'discussion' pages.
Thanks
-- Kim Tucker 11:59, 17 October 2012 (UTC)
### Foundation and progress tracking
Each card now has a number, a suit, a related cards section (link to its category) and a link to its corresponding book pattern page on the Public Sphere Project's web site (source).
For consistency, each card has a page which should contain only:
{{annotated-lv-card}}
[[Category:LV:{{:{{PAGENAME}}/suit}}]]
The second line might be best placed within {{annotated-lv-card}} - a To Do item.
Each related cards sub-page (/linked-cards) should contain only:
{{:lv-related-cards-template}}
The [[lv-related-cards-template]] page may be edited (e.g. if new collection categories emerge) to effectively change all the cards' related cards sections.
To Do:
• Insert the (abbreviated) Text for each card
• Add the appropriate Image for each card (see Image Quest 2.0)
• Once the images and abbreviated text are in place, it will be possible to include meaningful status indicators in the "rough" progress tracker.
-- Kim Tucker 12:25, 7 October 2012 (UTC)
### Rudimentary suits
Specifying a suit in sub-page /suit results in a border colour change provided the colours are set per theme in template lv-card-theme.
-- Kim Tucker 23:53, 7 September 2012 (UTC)
### Templates for keeping track of progress
See: LV Status Templates.
### Annotated cards template update
There are now new status fields for the template used to display a card page (e.g. http://wikieducator.org/LV:The_Commons ): general status, status of the image and of the text.
-- Kim Tucker 07:50, 1 September 2012 (UTC)
### Image Quest 2.0 is under way
The list of proposed images is growing - discuss here: http://wikieducator.org/LV:Activities/Image_Quest_2.0/Images
-- Kim Tucker 09:57, 25 August 2012 (UTC)
### Image Quest 2.0 coming soon
Image Quest 2.0 activities prepared and ready for collaboration. Announcing soon.
-- Kim Tucker 00:39, 8 August 2012 (UTC)
### Preparing Liberating Voices area on WikiEducator
The Liberating Voices area on WikiEducator is minimally prepared and ready for announcing. It is still a bit messy, but hopefully collaborators will come and help tidy it up (the wonders of peer production :-)
-- Kim Tucker 22:05, 7 August 2012 (UTC)
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# Math Help - inverse trigonometric functions
1. ## inverse trigonometric functions
just need some help getting started
1.) sin2x= -1/2
2.) 4cos^2 x-8cosx+1=0
3.) sin2x= -sinx
greatly appreciate it
2. Originally Posted by zukywich08
just need some help getting started
1.) sin2x= -1/2
2.) 4cos^2 x-8cosx+1=0
3.) sin2x= -sinx
greatly appreciate it
(1) $sin(u) = -\frac{1}{2}$
$u = \frac{7\pi}{6} \, , \, \frac{11\pi}{6}$ and those angles coterminal with these two.
(3) $2\sin{x}\cos{x} + \sin{x} = 0$
factor and solve
3. ## quick question
for the third problem how would you factor it? do you take sinx out of it?
4. Originally Posted by zukywich08
for the third problem how would you factor it? do you take sinx out of it?
yes
5. i got sinx(sinxcosx+1)=0
but i dont know what to do with the cosine?
6. Originally Posted by zukywich08
i got sinx(sinxcosx+1)=0
but i dont know what to do with the cosine?
$2\sin{x}\cos{x} + \sin{x} = 0$
$\sin{x}(2\cos{x} + 1) = 0$
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Drake
bowling_ball.cc File Reference
A simple example showing a bowling ball knocking over pins. More...
#include <cmath>
#include <iomanip>
#include <limits>
#include <memory>
#include <gflags/gflags.h>
#include "drake/common/eigen_types.h"
#include "drake/common/find_resource.h"
#include "drake/lcm/drake_lcm.h"
#include "drake/multibody/parsers/urdf_parser.h"
#include "drake/multibody/rigid_body_plant/drake_visualizer.h"
#include "drake/multibody/rigid_body_plant/rigid_body_plant.h"
#include "drake/multibody/rigid_body_tree_construction.h"
#include "drake/systems/analysis/runge_kutta2_integrator.h"
#include "drake/systems/analysis/simulator.h"
#include "drake/systems/framework/diagram_builder.h"
Include dependency graph for bowling_ball.cc:
drake
drake::systems
Functions
DEFINE_double (v, 12,"The ball's initial linear speed down the lane")
DEFINE_double (timestep, 2e-4,"The simulator time step")
DEFINE_double (w, 25,"The ball's initial angular speed (around [-1, 0 ,0].")
DEFINE_double (stiffness, 100000,"The contact model's stiffness")
DEFINE_double (us, 0.4,"The static coefficient of friction")
DEFINE_double (ud, 0.2,"The dynamic coefficient of friction")
DEFINE_double (v_tol, 0.01,"The maximum slipping speed allowed during stiction")
DEFINE_double (dissipation, 2,"The contact model's dissipation")
DEFINE_double (sim_duration, 3,"The simulation duration")
DEFINE_int32 (pin_count, 10,"The number of pins -- in the range [0, 10]")
DEFINE_bool (playback, true,"If true, loops playback of simulation")
int main ()
int main (int argc, char *argv[])
Detailed Description
A simple example showing a bowling ball knocking over pins.
The ball has an initial linear velocity parallel with the axis of the lane, and an angular velocity around that same vector (i.e., spinning perpendicular to its direction of motion).
In this case, it illustrates the change from dynamic to static friction as the ball slides along the lane. The slide becomes a roll as the relative velocity at the point of contact between ball and lane slows to zero.
This scenario (the definitions in the URDF and default values here) are inspired by this discussion of bowling physics: http://www.real-world-physics-problems.com/physics-of-bowling.html
There are physical artifacts that are not captured in this simulation. This doesn't model the change in the coefficient of friction due to position; a real bowling lane is oiled in for the first 2/3 of its length. As such, the parameters have to be tweaked to get the desired outcome. Rotational velocity has been modified (pointing straight back with a lower magnitude) and higher initial linear velocity. The lane is oriented along the x-axis (on the x-y plane).
Function Documentation
int main ( int argc, char * argv[] )
Here is the call graph for this function:
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# Connexions
You are here: Home » Content » College Physics » RLC Series AC Circuits
Content endorsed by: OpenStax College
## Navigation
### Table of Contents
• Preface
• #### 34. Frontiers of Physics
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• 36. Selected Radioactive Isotopes
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• 38. Glossary of Key Symbols and Notation
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# RLC Series AC Circuits
Module by: OpenStax College. E-mail the author
Summary:
• Calculate the impedance, phase angle, resonant frequency, power, power factor, voltage, and/or current in a RLC series circuit.
• Draw the circuit diagram for an RLC series circuit.
• Explain the significance of the resonant frequency.
## Impedance
When alone in an AC circuit, inductors, capacitors, and resistors all impede current. How do they behave when all three occur together? Interestingly, their individual resistances in ohms do not simply add. Because inductors and capacitors behave in opposite ways, they partially to totally cancel each other’s effect. Figure 1 shows an RLC series circuit with an AC voltage source, the behavior of which is the subject of this section. The crux of the analysis of an RLC circuit is the frequency dependence of XLXL size 12{X rSub { size 8{L} } } {} and XCXC size 12{X rSub { size 8{C} } } {}, and the effect they have on the phase of voltage versus current (established in the preceding section). These give rise to the frequency dependence of the circuit, with important “resonance” features that are the basis of many applications, such as radio tuners.
The combined effect of resistance RR size 12{R} {}, inductive reactance XLXL size 12{X rSub { size 8{L} } } {}, and capacitive reactance XCXC size 12{X rSub { size 8{C} } } {} is defined to be impedance, an AC analogue to resistance in a DC circuit. Current, voltage, and impedance in an RLC circuit are related by an AC version of Ohm’s law:
I 0 = V 0 Z or I rms = V rms Z . I 0 = V 0 Z or I rms = V rms Z . size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {Z} } " or "I rSub { size 8{ ital "rms"} } = { {V rSub { size 8{ ital "rms"} } } over {Z} } "." } {}
(1)
Here I0I0 size 12{I rSub { size 8{0} } } {} is the peak current, V0V0 size 12{V rSub { size 8{0} } } {} the peak source voltage, and Z Z is the impedance of the circuit. The units of impedance are ohms, and its effect on the circuit is as you might expect: the greater the impedance, the smaller the current. To get an expression for ZZ size 12{Z} {} in terms of R R , XLXL size 12{X rSub { size 8{L} } } {}, and XCXC size 12{X rSub { size 8{C} } } {}, we will now examine how the voltages across the various components are related to the source voltage. Those voltages are labeled VRVR size 12{V rSub { size 8{R} } } {}, VLVL size 12{V rSub { size 8{L} } } {}, and VCVC size 12{V rSub { size 8{C} } } {} in Figure 1.
Conservation of charge requires current to be the same in each part of the circuit at all times, so that we can say the currents in RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {} are equal and in phase. But we know from the preceding section that the voltage across the inductor VLVL size 12{V rSub { size 8{L} } } {} leads the current by one-fourth of a cycle, the voltage across the capacitor VCVC size 12{V rSub { size 8{C} } } {} follows the current by one-fourth of a cycle, and the voltage across the resistor VRVR size 12{V rSub { size 8{R} } } {} is exactly in phase with the current. Figure 2 shows these relationships in one graph, as well as showing the total voltage around the circuit V=VR+VL+VCV=VR+VL+VC size 12{V=V rSub { size 8{R} } +V rSub { size 8{L} } +V rSub { size 8{C} } } {}, where all four voltages are the instantaneous values. According to Kirchhoff’s loop rule, the total voltage around the circuit V V is also the voltage of the source.
You can see from Figure 2 that while VRVR size 12{V rSub { size 8{R} } } {} is in phase with the current, VLVL size 12{V rSub { size 8{L} } } {} leads by 90º 90º , and VCVC size 12{V rSub { size 8{C} } } {} follows by 90º 90º . Thus VLVL size 12{V rSub { size 8{L} } } {} and VCVC size 12{V rSub { size 8{C} } } {} are 180º 180º out of phase (crest to trough) and tend to cancel, although not completely unless they have the same magnitude. Since the peak voltages are not aligned (not in phase), the peak voltage V0V0 size 12{V rSub { size 8{0} } } {} of the source does not equal the sum of the peak voltages across RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {}. The actual relationship is
V 0 = V 0R 2 + ( V 0L V 0C ) 2 , V 0 = V 0R 2 + ( V 0L V 0C ) 2 , size 12{V rSub { size 8{0} } = sqrt {V rSub { size 8{0R} } "" lSup { size 8{2} } + $$V rSub { size 8{0L} } - V rSub { size 8{0C} }$$ rSup { size 8{2} } } ,} {}
(2)
where V0RV0R size 12{V rSub { size 8{0R} } } {}, V0LV0L size 12{V rSub { size 8{0L} } } {}, and V0CV0C size 12{V rSub { size 8{0C} } } {} are the peak voltages across RR size 12{R} {}, LL size 12{L} {}, and CC size 12{C} {}, respectively. Now, using Ohm’s law and definitions from Reactance, Inductive and Capacitive, we substitute V0=I0ZV0=I0Z size 12{V rSub { size 8{0} } =I rSub { size 8{0} } Z} {} into the above, as well as V0R=I0RV0R=I0R size 12{V rSub { size 8{0R} } =I rSub { size 8{0} } R} {}, V0L=I0XLV0L=I0XL size 12{V rSub { size 8{0L} } =I rSub { size 8{0} } X rSub { size 8{L} } } {}, and V0C=I0XCV0C=I0XC size 12{V rSub { size 8{0C} } =I rSub { size 8{0} } X rSub { size 8{C} } } {}, yielding
I0Z= I 0 2 R2 + ( I0XLI0XC)2=I0R2+(XLXC)2.I0Z= I 0 2 R2 + ( I0XLI0XC)2=I0R2+(XLXC)2. size 12{I rSub { size 8{0} } Z= sqrt {I rSub { size 8{0} rSup { size 8{2} } } R rSup { size 8{2} } + $$I rSub { size 8{0} } X rSub { size 8{L} } - I rSub { size 8{0} } X rSub { size 8{C} }$$ rSup { size 8{2} } } =I rSub { size 8{0} } sqrt {R rSup { size 8{2} } + $$X rSub { size 8{L} } - X rSub { size 8{C} }$$ rSup { size 8{2} } } } {}
(3)
I0I0 size 12{I rSub { size 8{0} } } {} cancels to yield an expression for Z Z :
Z=R2+(XLXC)2,Z=R2+(XLXC)2, size 12{Z= sqrt {R rSup { size 8{2} } + $$X rSub { size 8{L} } - X rSub { size 8{C} }$$ rSup { size 8{2} } } } {}
(4)
which is the impedance of an RLC series AC circuit. For circuits without a resistor, take R = 0 R = 0 ; for those without an inductor, take XL=0XL=0 size 12{X rSub { size 8{L} } =0} {}; and for those without a capacitor, take XC=0XC=0 size 12{X rSub { size 8{C} } =0} {}.
### Example 1: Calculating Impedance and Current
An RLC series circuit has a 40.0 Ω 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF 5.00 μF capacitor. (a) Find the circuit’s impedance at 60.0 Hz and 10.0 kHz, noting that these frequencies and the values for L L and C C are the same as in (Reference) and (Reference). (b) If the voltage source has Vrms=120VVrms=120V size 12{V rSub { size 8{"rms"} } ="120"V} {}, what is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at each frequency?
Strategy
For each frequency, we use Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + $$X rSub { size 8{L} } - X rSub { size 8{C} }$$ rSup { size 8{2} } } } {} to find the impedance and then Ohm’s law to find current. We can take advantage of the results of the previous two examples rather than calculate the reactances again.
Solution for (a)
At 60.0 Hz, the values of the reactances were found in (Reference) to be XL=1.13ΩXL=1.13Ω size 12{X rSub { size 8{L} } =1 "." "13" %OMEGA } {} and in (Reference) to be XC=531 Ω XC=531 Ω size 12{X rSub { size 8{C} } ="531 " %OMEGA } {}. Entering these and the given 40.0 Ω 40.0 Ω for resistance into Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + $$X rSub { size 8{L} } - X rSub { size 8{C} }$$ rSup { size 8{2} } } } {} yields
Z = R2+(XLXC)2 = (40.0Ω)2+(1.13Ω531Ω)2 = 531Ω at 60.0 Hz.Z = R2+(XLXC)2 = (40.0Ω)2+(1.13Ω531Ω)2 = 531Ω at 60.0 Hz.alignl { stack { size 12{Z= sqrt {R rSup { size 8{2} } + $$X rSub { size 8{L} } - X rSub { size 8{C} }$$ rSup { size 8{2} } } } {} # " "= sqrt { $$"40" "." 0 %OMEGA$$ rSup { size 8{2} } + $$1 "." "13" %OMEGA - "531" %OMEGA$$ rSup { size 8{2} } } {} # " "="531" %OMEGA " at 60" "." "0 Hz" {} } } {}
(5)
Similarly, at 10.0 kHz, XL=188ΩXL=188Ω size 12{X rSub { size 8{L} } ="188" %OMEGA } {} and XC=3.18ΩXC=3.18Ω size 12{X rSub { size 8{C} } =3 "." "18" %OMEGA } {}, so that
Z = (40.0Ω)2+(188Ω3.18Ω)2 = 190Ω at 10.0 kHz. Z = (40.0Ω)2+(188Ω3.18Ω)2 = 190Ω at 10.0 kHz.alignl { stack { size 12{Z= sqrt { $$"40" "." 0 %OMEGA$$ rSup { size 8{2} } + $$"188" %OMEGA - 3 "." "18" %OMEGA$$ rSup { size 8{2} } } } {} # " "="190" %OMEGA " at 10" "." "0 kHz" {} } } {}
(6)
Discussion for (a)
In both cases, the result is nearly the same as the largest value, and the impedance is definitely not the sum of the individual values. It is clear that XLXL size 12{X rSub { size 8{L} } } {} dominates at high frequency and XCXC size 12{X rSub { size 8{C} } } {} dominates at low frequency.
Solution for (b)
The current IrmsIrms size 12{I rSub { size 8{"rms"} } } {} can be found using the AC version of Ohm’s law in Equation Irms=Vrms/ZIrms=Vrms/Z size 12{I rSub { size 8{"rms"} } =V rSub { size 8{"rms"} } /Z} {}:
Irms=VrmsZ=120 V531 Ω=0.226 AIrms=VrmsZ=120 V531 Ω=0.226 A size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {Z} } = { {"120"" V"} over {"531 " %OMEGA } } =0 "." "226"" A"} {} at 60.0 Hz
Finally, at 10.0 kHz, we find
Irms=VrmsZ=120 V190 Ω=0.633 AIrms=VrmsZ=120 V190 Ω=0.633 A size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {Z} } = { {"120"" V"} over {"190 " %OMEGA } } =0 "." "633"" A"} {} at 10.0 kHz
Discussion for (a)
The current at 60.0 Hz is the same (to three digits) as found for the capacitor alone in (Reference). The capacitor dominates at low frequency. The current at 10.0 kHz is only slightly different from that found for the inductor alone in (Reference). The inductor dominates at high frequency.
## Resonance in RLC Series AC Circuits
How does an RLC circuit behave as a function of the frequency of the driving voltage source? Combining Ohm’s law, Irms=Vrms/ZIrms=Vrms/Z size 12{I rSub { size 8{"rms"} } =V rSub { size 8{"rms"} } /Z} {}, and the expression for impedance Z Z from Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + $$X rSub { size 8{L} } - X rSub { size 8{C} }$$ rSup { size 8{2} } } } {} gives
Irms=VrmsR2+(XLXC)2.Irms=VrmsR2+(XLXC)2. size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over { sqrt {R rSup { size 8{2} } + $$X rSub { size 8{L} } - X rSub { size 8{C} }$$ rSup { size 8{2} } } } } } {}
(7)
The reactances vary with frequency, with XLXL size 12{X rSub { size 8{L} } } {} large at high frequencies and XCXC size 12{X rSub { size 8{C} } } {} large at low frequencies, as we have seen in three previous examples. At some intermediate frequency f0f0 size 12{f rSub { size 8{0} } } {}, the reactances will be equal and cancel, giving Z=RZ=R size 12{Z=R} {} —this is a minimum value for impedance, and a maximum value for IrmsIrms size 12{I rSub { size 8{"rms"} } } {} results. We can get an expression for f0f0 size 12{f rSub { size 8{0} } } {} by taking
XL=XC.XL=XC. size 12{X rSub { size 8{L} } =X rSub { size 8{C} } } {}
(8)
Substituting the definitions of XLXL size 12{X rSub { size 8{L} } } {} and XCXC size 12{X rSub { size 8{C} } } {},
2πf0L=12πf0C.2πf0L=12πf0C. size 12{2πf rSub { size 8{0} } L= { {1} over {2πf rSub { size 8{0} } C} } } {}
(9)
Solving this expression for f0f0 size 12{f rSub { size 8{0} } } {} yields
f0=1LC,f0=1LC, size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {}
(10)
where f0f0 size 12{f rSub { size 8{0} } } {} is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if not driven by the voltage source. At f0f0 size 12{f rSub { size 8{0} } } {}, the effects of the inductor and capacitor cancel, so that Z=RZ=R size 12{Z=R} {}, and IrmsIrms size 12{I rSub { size 8{"rms"} } } {} is a maximum.
Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined to be a forced oscillation—in this case, forced by the voltage source—at the natural frequency of the system. The receiver in a radio is an RLC circuit that oscillates best at its f0f0 size 12{f rSub { size 8{0} } } {}. A variable capacitor is often used to adjust f0f0 size 12{f rSub { size 8{0} } } {} to receive a desired frequency and to reject others. Figure 3 is a graph of current as a function of frequency, illustrating a resonant peak in IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at f0f0 size 12{f rSub { size 8{0} } } {}. The two curves are for two different circuits, which differ only in the amount of resistance in them. The peak is lower and broader for the higher-resistance circuit. Thus the higher-resistance circuit does not resonate as strongly and would not be as selective in a radio receiver, for example.
### Example 2: Calculating Resonant Frequency and Current
For the same RLC series circuit having a 40.0 Ω 40.0 Ω resistor, a 3.00 mH inductor, and a 5.00 μF 5.00 μF capacitor: (a) Find the resonant frequency. (b) Calculate IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at resonance if VrmsVrms size 12{V rSub { size 8{"rms"} } } {} is 120 V.
Strategy
The resonant frequency is found by using the expression in f0=1LCf0=1LC size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {}. The current at that frequency is the same as if the resistor alone were in the circuit.
Solution for (a)
Entering the given values for L L and C C into the expression given for f0f0 size 12{f rSub { size 8{0} } } {} in f0=1LCf0=1LC size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {} yields
f0 = 1LC = 1(3.00×103 H)(5.00×106 F)=1.30 kHz. f0 = 1LC = 1(3.00×103 H)(5.00×106 F)=1.30 kHz.alignl { stack { size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {} # " "= { {1} over {2π sqrt { $$3 "." "00" times "10" rSup { size 8{ - 3} } " H"$$ $$5 "." "00" times "10" rSup { size 8{ - 6} } " F"$$ } } } =1 "." "30"" kHz" {} } } {}
(11)
Discussion for (a)
We see that the resonant frequency is between 60.0 Hz and 10.0 kHz, the two frequencies chosen in earlier examples. This was to be expected, since the capacitor dominated at the low frequency and the inductor dominated at the high frequency. Their effects are the same at this intermediate frequency.
Solution for (b)
The current is given by Ohm’s law. At resonance, the two reactances are equal and cancel, so that the impedance equals the resistance alone. Thus,
Irms=VrmsZ=120 V40.0 Ω=3.00 A.Irms=VrmsZ=120 V40.0 Ω=3.00 A. size 12{I rSub { size 8{"rms"} } = { {V rSub { size 8{"rms"} } } over {Z} } = { {"120"" V"} over {"40" "." "0 " %OMEGA } } =3 "." "00"" A"} {}
(12)
Discussion for (b)
At resonance, the current is greater than at the higher and lower frequencies considered for the same circuit in the preceding example.
## Power in RLC Series AC Circuits
If current varies with frequency in an RLC circuit, then the power delivered to it also varies with frequency. But the average power is not simply current times voltage, as it is in purely resistive circuits. As was seen in Figure 2, voltage and current are out of phase in an RLC circuit. There is a phase angle ϕϕ size 12{ϕ} {} between the source voltage VV size 12{V} {} and the current II size 12{I} {}, which can be found from
cosϕ=RZ.cosϕ=RZ. size 12{"cos"ϕ= { {R} over {Z} } } {}
(13)
For example, at the resonant frequency or in a purely resistive circuit Z=RZ=R size 12{Z=R} {}, so that cosϕ=1cosϕ=1 size 12{"cos"ϕ=1} {}. This implies that ϕ=0ºϕ=0º size 12{ϕ=0 rSup { size 8{ circ } } } {} and that voltage and current are in phase, as expected for resistors. At other frequencies, average power is less than at resonance. This is both because voltage and current are out of phase and because IrmsIrms size 12{I rSub { size 8{"rms"} } } {} is lower. The fact that source voltage and current are out of phase affects the power delivered to the circuit. It can be shown that the average power is
P ave = I rms V rms cos ϕ , P ave = I rms V rms cos ϕ , size 12{P rSub { size 8{"ave"} } =I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } "cos"ϕ} {}
(14)
Thus cosϕcosϕ size 12{"cos"ϕ} {} is called the power factor, which can range from 0 to 1. Power factors near 1 are desirable when designing an efficient motor, for example. At the resonant frequency, cosϕ=1cosϕ=1 size 12{"cos"ϕ=1} {}.
### Example 3: Calculating the Power Factor and Power
For the same RLC series circuit having a 40.0 Ω 40.0 Ω resistor, a 3.00 mH inductor, a 5.00 μF 5.00 μF capacitor, and a voltage source with a V rms V rms of 120 V: (a) Calculate the power factor and phase angle for f=60.0Hzf=60.0Hz size 12{f="60" "." 0"Hz"} {}. (b) What is the average power at 50.0 Hz? (c) Find the average power at the circuit’s resonant frequency.
Strategy and Solution for (a)
The power factor at 60.0 Hz is found from
cosϕ=RZ.cosϕ=RZ. size 12{"cos"ϕ= { {R} over {Z} } } {}
(15)
We know Z = 531 Ω Z = 531 Ω from Example 1, so that
cosϕ=40.0Ω531 Ω=0.0753 at 60.0 Hz.cosϕ=40.0Ω531 Ω=0.0753 at 60.0 Hz. size 12{"cos"Ø= { {"40" "." 0 %OMEGA } over {5"31 " %OMEGA } } =0 "." "0753"} {}
(16)
This small value indicates the voltage and current are significantly out of phase. In fact, the phase angle is
ϕ=cos10.0753=85.7º at 60.0 Hz.ϕ=cos10.0753=85.7º at 60.0 Hz. size 12{ϕ="cos" rSup { size 8{ - 1} } 0 "." "0753"="85" "." 7 rSup { size 8{ circ } } } {}
(17)
Discussion for (a)
The phase angle is close to 90º 90º , consistent with the fact that the capacitor dominates the circuit at this low frequency (a pure RC circuit has its voltage and current 90º 90º out of phase).
Strategy and Solution for (b)
The average power at 60.0 Hz is
Pave=IrmsVrmscosϕ.Pave=IrmsVrmscosϕ. size 12{P rSub { size 8{"ave"} } =I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } "cos"ϕ} {}
(18)
IrmsIrms size 12{I rSub { size 8{"rms"} } } {} was found to be 0.226 A in Example 1. Entering the known values gives
Pave=(0.226 A)(120 V)(0.0753)=2.04 W at 60.0 Hz.Pave=(0.226 A)(120 V)(0.0753)=2.04 W at 60.0 Hz. size 12{P rSub { size 8{"ave"} } = $$0 "." "226"" A"$$ $$"120"" V"$$ $$0 "." "0753"$$ =2 "." "04"" W"} {}
(19)
Strategy and Solution for (c)
At the resonant frequency, we know cosϕ=1cosϕ=1 size 12{"cos"ϕ=1} {}, and IrmsIrms size 12{I rSub { size 8{"rms"} } } {} was found to be 6.00 A in Example 2. Thus,
Pave=(3.00 A)(120 V)(1)=360 WPave=(3.00 A)(120 V)(1)=360 W size 12{P rSub { size 8{"ave"} } = $$3 "." "00"" A"$$ $$"120"" V"$$ $$1$$ ="350"" W"} {} at resonance (1.30 kHz)
Discussion
Both the current and the power factor are greater at resonance, producing significantly greater power than at higher and lower frequencies.
Power delivered to an RLC series AC circuit is dissipated by the resistance alone. The inductor and capacitor have energy input and output but do not dissipate it out of the circuit. Rather they transfer energy back and forth to one another, with the resistor dissipating exactly what the voltage source puts into the circuit. This assumes no significant electromagnetic radiation from the inductor and capacitor, such as radio waves. Such radiation can happen and may even be desired, as we will see in the next chapter on electromagnetic radiation, but it can also be suppressed as is the case in this chapter. The circuit is analogous to the wheel of a car driven over a corrugated road as shown in Figure 4. The regularly spaced bumps in the road are analogous to the voltage source, driving the wheel up and down. The shock absorber is analogous to the resistance damping and limiting the amplitude of the oscillation. Energy within the system goes back and forth between kinetic (analogous to maximum current, and energy stored in an inductor) and potential energy stored in the car spring (analogous to no current, and energy stored in the electric field of a capacitor). The amplitude of the wheels’ motion is a maximum if the bumps in the road are hit at the resonant frequency.
A pure LC circuit with negligible resistance oscillates at f0f0 size 12{f rSub { size 8{0} } } {}, the same resonant frequency as an RLC circuit. It can serve as a frequency standard or clock circuit—for example, in a digital wristwatch. With a very small resistance, only a very small energy input is necessary to maintain the oscillations. The circuit is analogous to a car with no shock absorbers. Once it starts oscillating, it continues at its natural frequency for some time. Figure 5 shows the analogy between an LC circuit and a mass on a spring.
## PhET Explorations: Circuit Construction Kit (AC+DC), Virtual Lab:
Build circuits with capacitors, inductors, resistors and AC or DC voltage sources, and inspect them using lab instruments such as voltmeters and ammeters.
## Section Summary
• The AC analogy to resistance is impedance Z Z , the combined effect of resistors, inductors, and capacitors, defined by the AC version of Ohm’s law:
I 0 = V 0 Z or I rms = V rms Z , I 0 = V 0 Z or I rms = V rms Z , size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {Z} } " or "I rSub { size 8{ ital "rms"} } = { {V rSub { size 8{ ital "rms"} } } over {Z} } ,} {}
(20)
where I0I0 size 12{I rSub { size 8{0} } } {} is the peak current and V0V0 size 12{V rSub { size 8{0} } } {} is the peak source voltage.
• Impedance has units of ohms and is given by Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + $$X rSub { size 8{L} } - X rSub { size 8{C} }$$ rSup { size 8{2} } } } {}.
• The resonant frequency f0f0 size 12{f rSub { size 8{0} } } {}, at which XL=XCXL=XC size 12{X rSub { size 8{L} } =X rSub { size 8{C} } } {}, is
f0=1LC.f0=1LC. size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {}
(21)
• In an AC circuit, there is a phase angle ϕϕ size 12{ϕ} {} between source voltage VV size 12{V} {} and the current II size 12{I} {}, which can be found from
cosϕ=RZ,cosϕ=RZ, size 12{"cos"ϕ= { {R} over {Z} } } {}
(22)
• ϕ=ϕ= size 12{ϕ=0 rSup { size 8{ circ } } } {} for a purely resistive circuit or an RLC circuit at resonance.
• The average power delivered to an RLC circuit is affected by the phase angle and is given by
Pave=IrmsVrmscosϕ,Pave=IrmsVrmscosϕ, size 12{P rSub { size 8{"ave"} } =I rSub { size 8{"rms"} } V rSub { size 8{"rms"} } "cos"ϕ} {}
(23)
cosϕcosϕ size 12{"cos"ϕ} {} is called the power factor, which ranges from 0 to 1.
## Conceptual Questions
### Exercise 1
Does the resonant frequency of an AC circuit depend on the peak voltage of the AC source? Explain why or why not.
### Exercise 2
Suppose you have a motor with a power factor significantly less than 1. Explain why it would be better to improve the power factor as a method of improving the motor’s output, rather than to increase the voltage input.
## Problems & Exercises
### Exercise 1
An RL circuit consists of a 40.0 Ω 40.0 Ω resistor and a 3.00 mH inductor. (a) Find its impedance Z Z at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z Z with those found in Example 1 in which there was also a capacitor.
### Exercise 2
An RC circuit consists of a 40.0 Ω 40.0 Ω resistor and a 5.00 μF 5.00 μF capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of Z Z with those found in Example 1, in which there was also an inductor.
### Exercise 3
An LC circuit consists of a 3.00mH3.00mH size 12{3 "." "00" μH} {} inductor and a 5.00μF5.00μF size 12{5 "." "00" μF} {} capacitor. (a) Find its impedance at 60.0 Hz and 10.0 kHz. (b) Compare these values of ZZ size 12{Z} {} with those found in Example 1 in which there was also a resistor.
### Exercise 4
What is the resonant frequency of a 0.500 mH inductor connected to a 40.0 μF 40.0 μF capacitor?
### Exercise 5
To receive AM radio, you want an RLC circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is accomplished with a fixed 1.00 μH 1.00 μH inductor connected to a variable capacitor. What range of capacitance is needed?
### Exercise 6
Suppose you have a supply of inductors ranging from 1.00 nH to 10.0 H, and capacitors ranging from 1.00 pF to 0.100 F. What is the range of resonant frequencies that can be achieved from combinations of a single inductor and a single capacitor?
### Exercise 7
What capacitance do you need to produce a resonant frequency of 1.00 GHz, when using an 8.00 nH inductor?
### Exercise 8
What inductance do you need to produce a resonant frequency of 60.0 Hz, when using a 2.00 μF 2.00 μF capacitor?
### Exercise 9
The lowest frequency in the FM radio band is 88.0 MHz. (a) What inductance is needed to produce this resonant frequency if it is connected to a 2.50 pF capacitor? (b) The capacitor is variable, to allow the resonant frequency to be adjusted to as high as 108 MHz. What must the capacitance be at this frequency?
### Exercise 10
An RLC series circuit has a 2.50 Ω 2.50 Ω resistor, a 100 μH 100 μH inductor, and an 80.0 μF 80.0 μF capacitor.(a) Find the circuit’s impedance at 120 Hz. (b) Find the circuit’s impedance at 5.00 kHz. (c) If the voltage source has Vrms=5.60VVrms=5.60V size 12{V rSub { size 8{"rms"} } =5 "." "60"V} {}, what is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at each frequency? (d) What is the resonant frequency of the circuit? (e) What is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at resonance?
### Exercise 11
An RLC series circuit has a 1.00 kΩ 1.00 kΩ resistor, a 150 μH 150 μH inductor, and a 25.0 nF capacitor. (a) Find the circuit’s impedance at 500 Hz. (b) Find the circuit’s impedance at 7.50 kHz. (c) If the voltage source has Vrms=408VVrms=408V size 12{V rSub { size 8{"rms"} } ="408"V} {}, what is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at each frequency? (d) What is the resonant frequency of the circuit? (e) What is IrmsIrms size 12{I rSub { size 8{"rms"} } } {} at resonance?
### Exercise 12
An RLC series circuit has a 2.50 Ω 2.50 Ω resistor, a 100 μH 100 μH inductor, and an 80.0 μF 80.0 μF capacitor. (a) Find the power factor at f = 120 Hz f = 120 Hz . (b) What is the phase angle at 120 Hz? (c) What is the average power at 120 Hz? (d) Find the average power at the circuit’s resonant frequency.
### Exercise 13
An RLC series circuit has a 1.00 kΩ 1.00 kΩ resistor, a 150 μH 150 μH inductor, and a 25.0 nF capacitor. (a) Find the power factor at f = 7.50 Hz f = 7.50 Hz . (b) What is the phase angle at this frequency? (c) What is the average power at this frequency? (d) Find the average power at the circuit’s resonant frequency.
### Exercise 14
An RLC series circuit has a 200 Ω 200 Ω resistor and a 25.0 mH inductor. At 8000 Hz, the phase angle is 45.0º 45.0º . (a) What is the impedance? (b) Find the circuit’s capacitance. (c) If Vrms=408VVrms=408V size 12{V rSub { size 8{"rms"} } ="408"`V} {} is applied, what is the average power supplied?
### Exercise 15
Referring to Example 3, find the average power at 10.0 kHz.
## Glossary
impedance:
the AC analogue to resistance in a DC circuit; it is the combined effect of resistance, inductive reactance, and capacitive reactance in the form Z=R2+(XLXC)2Z=R2+(XLXC)2 size 12{Z= sqrt {R rSup { size 8{2} } + $$X rSub { size 8{L} } - X rSub { size 8{C} }$$ rSup { size 8{2} } } } {}
resonant frequency:
the frequency at which the impedance in a circuit is at a minimum, and also the frequency at which the circuit would oscillate if not driven by a voltage source; calculated by f0=1LCf0=1LC size 12{f rSub { size 8{0} } = { {1} over {2π sqrt { ital "LC"} } } } {}
phase angle:
denoted by ϕϕ size 12{ϕ} {}, the amount by which the voltage and current are out of phase with each other in a circuit
power factor:
the amount by which the power delivered in the circuit is less than the theoretical maximum of the circuit due to voltage and current being out of phase; calculated by cosϕcosϕ size 12{"cos"ϕ} {}
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# Tag Info
0
An alternative solution is to have a button that skips the extra material. This has the disadvantage that you have to remember to press the “don’t show the extra material” button when you get to that slide, but it has two advantages. Firstly, the extra slide is right there next to the main material rather than hidden off in an appendix, which might make it ...
1
I can offer an environment DifferentFileContents which has the same syntax as the filecontents*-environment. The content of the DifferentFileContents-environment will be compared to the content of the file specified. In case the contents differ or the file specified does not exist, the file will be destroyed and rewritten/will be created anew with the ...
3
The varioref package and its \vref user macro accomplish what your \fullref macro aims to do. The package works with babel, and it offers ways to customize what is supposed to be typeset if the cross-referenced object happens to be on, near, or far away from the cross-referencing call-out. This customization works by redefining the macros \reftextbefore, \...
4
Checking if a file has changed, as Ulrike says, requires to compare it against some previous state of the file. Here's an implementation using roughly the same method as in Rmano's answer of storing the MD5 sum of the file in the .aux file, and also storing one MD5 per file in a property list, so that you can have multiple files. \IfFileChangedTF checks if a ...
1
\documentclass{standalone} \usepackage{tikz} \usetikzlibrary{math} \begin{document} \begin{tikzpicture} \foreach \x in {1,...,4} { \tikzmath {\xEnd=\x+1;} \foreach \y in {1,...,\xEnd} { \fill[red!50] (\x,\y) ellipse [x radius=3pt, y radius=6pt, rotate={ifthenelse(\x==1 || \x==2 && \y==1,0,-45)}]; }} \draw [|-|] (.895,1) -- ++(0.211,0); \end{...
0
Found a much simpler solution. Below are the key ideas: \ifthenelse from ifthen package in CTAN is that powerful conditional command. \ifthenelse{<test>}{<then clause>}{<else clause>}is the command format wherein <test> is a boolean expression using the infix connectives, \and, \or, the unary \not and parentheses . Also \OR ...
4
I suppose that the situation is the following. We have a LaTeX file, call it filechanged.tex, that is run with pdflatex filechanged. In the same directory, I have another file, myfile.txt. When I run pdflatex filechanged, I want to do different processing if myfile.txt has changed since the previous run or not. My solution: It is based on @Skillmon ...
8
In order to reproduce the picture, the test should be 1 ≤ x < y. You can parametrize the rotation angle and test for the two special cases. \documentclass{standalone} \usepackage{tikz} \begin{document} \begin{tikzpicture} \foreach \x in {1,...,4} { \foreach \y in {1,...,4} { \def\rotation{-45} \ifnum\y=1 \ifnum\x=1 \def\rotation{0} \fi ...
6
LaTeX's ifthen package has some facility for combining conditionals with \and and \or, \not and parentheses. But your case is easy to do with \ifnum: \ifnum \x<\y \ifnum \x>1 \breakforeach \fi\fi
3
amsthm's internal macro \@ifempty is what you need. Alternatively, you can use \ifstrempty{<string>}{<true code>}{<false code>} from etoolbox package. \documentclass{article} \usepackage{amsthm} \makeatletter \newtheoremstyle{mytheorem} {\topsep} {\topsep} {} {1em} {\bfseries} {.} { } {% \@ifempty{#3} {\thmname{#1}...
3
The “official” source for TeX primitive commands is the TeXbook, whose TeX source is available online (even on CTAN), but not for making the printed copy. The source starts with % This manual is copyright (C) 1984 by the American Mathematical Society. % All rights are reserved! % The file is distributed only for people to see its examples of TeX input, % not ...
1
If you want to link to a specific paragraph/section or the entire law with some text more often, you can define a dedicated command for that. I'm using etoolbox's \ifblank to test if the string is empty, because I generally prefer etoolbox tests over ifthen, but that is just a matter of style. \documentclass[12pt]{article} \usepackage{etoolbox} \usepackage{...
2
You need something expandable, ifthen and xifthen won't work. \documentclass[12pt]{article} \usepackage{hyperref} \begin{document} \ExplSyntaxOn \newcommand{\gesetze}[2] { http\c_colon_str//www.gesetze-im-internet.de/#1/ \tl_if_blank:nF {#2}{__#2.html} } \ExplSyntaxOff \href{\gesetze{bgb}{622}}{This is the 622nd paragraph in the law.} \href{...
1
Note that line 167 of beamerbasetitle.sty contains \author{}, hence \beamer@shortauthor is always defined and initially empty, and one can/should test if meaningful \author{...} is used by \ifx\beamer@shortauthor\@empty. The following example outputs "No author given Some more information" on first frame, and "Author | Some more information&...
6
The test fails to show equality on (at least) two levels. First level \MakeUppercase and \MakeLowercase are instructions to print the uppercased or lowercased versions of their arguments. They don't “directly” transform their arguments. Second level \ifx only compares the “surface meaning” of two tokens without any macro expansion. In particular, two macros (...
9
Like Donald Hosek said, \ifx doesn’t expand its arguments. You’re comparing the unexpanded macros \MakeUppercase{1} and \MakeLowercase{1}. You could fix this by fully expanding the macros you compare. However, that doesn’t give you a complete solution, because \MakeUppercase and \MakeLowercase are not expandable. So, \edef\upperone{{\MakeUppercase 1}} ...
11
You're going to get different results because you're comparing the definitions of the two macros. \MakeUppercase{1} is not the same as \MakeLowercase{1} It's worth noting that the \MakeUppercase and \MakeLowercase macros are themselves complicated enough that the naïve approaches to trying to expand them to get the results in plain text will still not give ...
2
1
You can achieve this using the following code: \usepackage{xstring} \usepackage{ifthen} \newcommand{\F}[1]{ \StrLeft{#1}{1}[\firstletter]% \StrRight{#1}{1}[\lastletter]% \ifthenelse{\equal{(}{\firstletter} \AND \equal{)}{\lastletter}}{F#1}{F(#1)}% } It basically checks what the first and last characters of the argument are, and if those are '(' ...
0
Your criteria seem to be to a) use LaTeX (i.e. limit/avoid TeX primitives) and b) input a markdown pre-processed header before certain LaTeX files. If so, this will do what you want. \documentclass{article} \usepackage{etoolbox} \newcommand*{\inject}[2][]{ \ifstrempty{#1} {\input{#2}} {% \input{.cache/#1} \input{#2}} ...
2
The problem is that \ifstrequal does no expansion to its arguments. This might be solved by forcing expansion, but let me present a different solution based on expl3 and its richer supply of functions and tests. \documentclass{article} \usepackage{xparse} \ExplSyntaxOn \NewDocumentCommand{\inject}{m} { \utkarsh_inject:n { #1 } } \tl_new:N \...
5
The manual of the etoolbox package says: \ifstrequal{⟨string⟩}{⟨string⟩}{⟨true⟩}{⟨false⟩} Compares two strings and executes ⟨true⟩ if they are equal, and ⟨false⟩ otherwise. The strings are not expanded in the test and the comparison is category code agnostic. Control sequence tokens in any of the ⟨string⟩ arguments will be detokenized and treated as ...
1
This should be solved by using ##1 instead of #1, because you want to define a command inside the argument passed to another macro. See What is the meaning of double pound symbol (number sign, hash character) ##1 in an argument? for more information.
6
The reason for the error is that the array does not contain integers but ifnum requires an integer. You can use pgfmathparse to parse the entry from the array and pgfmathresult in the ifnum statement. Also you need to use i-1 otherwise the last array index is out of bounds, because arrays are indexed starting with 0. \def\array{{1,0,1,1,1,1}} \foreach \i in {...
Top 50 recent answers are included
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Bug 689641 - (mlabeledtr-support) Add support for mlabeledtr
(mlabeledtr-support) Summary: Add support for mlabeledtr
Status: NEW [lang=c++] dev-doc-needed, helpwanted Core Components MathML (show other bugs) Trunk All All P4 normal with 4 votes (vote) --- Nobody; OK to take it and work on it Anthony Jones (:kentuckyfriedtakahe, :k17e) Frédéric Wang (:fredw) mathml-2 557086 mathml-in-mathjax 958947 Show dependency tree / graph
Reported: 2011-09-27 10:17 PDT by Frédéric Wang (:fredw)
Modified: 2014-06-17 08:09 PDT (History)
11 users (show)
See Also:
Crash Signature:
(edit)
QA Whiteboard:
Iteration: ---
Points: ---
Has Regression Range: ---
Has STR: ---
Attachments
testcase (3.97 KB, text/html)
2011-09-27 10:17 PDT, Frédéric Wang (:fredw)
no flags Details
Frédéric Wang (:fredw) 2011-09-27 10:17:39 PDT Created attachment 562803 [details] testcase Frédéric Wang (:fredw) 2012-02-29 14:21:48 PST Tests from the MathML testsuite: http://www.w3.org/Math/testsuite/build/main/Presentation/TablesAndMatrices/mlabeledtr/mlabeledtr1-full.xhtml http://www.w3.org/Math/testsuite/build/main/Presentation/TablesAndMatrices/mlabeledtr/mlabeledtrAside1-full.xhtml http://www.w3.org/Math/testsuite/build/main/Presentation/TablesAndMatrices/mlabeledtr/mlabeledtrAside2-full.xhtml http://www.w3.org/Math/testsuite/build/main/Presentation/TablesAndMatrices/mlabeledtr/rec-mlabeledtr-full.xhtml Frédéric Wang (:fredw) 2012-09-01 03:15:03 PDT Because of the vertical alignment requirement, it seems that the most easy way to implement this is to follow the REC's suggested model: "To place a label, an implementor might think in terms of creating a larger table, with an extra column on both ends. The columnwidth attributes of both these border columns would be set to "fit" so that they expand to fill whatever space remains after the inner columns have been laid out. Finally, depending on the values of side and minlabelspacing, the label is placed in whatever border column is appropriate, possibly shifted down if necessary, and aligned according to columnalignment." So maybe we should do something in nsCSSFrameConstructor to add extra column on both ends of an or element (with the label placed in the appropriate extra cell in the latter case). We will need to change the implementation of rowlines, columnlines etc to take into account that we have these extra columns. That would give a slightly incorrect implementation of mtable@width, though. Frédéric Wang (:fredw) 2012-11-17 09:47:22 PST An add-on to workaround this bug: https://addons.mozilla.org/addon/mathml-mml3ff/ Frédéric Wang (:fredw) 2012-11-22 08:43:38 PST Someone from the Webkit MathML team contacted me about this bug today. Adding me as mentor, although I don't know well the code in nsCSSFrameConstructor to change. I think even something like Frédéric Wang (:fredw) 2012-11-22 08:45:01 PST I think even something like what David Carlisle's XSLT stylesheet do would be OK (adding a column to 's that have a child). Frédéric Wang (:fredw) 2012-11-30 08:32:52 PST Mass change: setting priority to 2 for bugs that would be nice fix if Gecko's MathML support is enabled by default in MathJax but that are not in my opinion strictly required or for which a workaround could be written in the MathJax code.
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"FREE" AV-36 and AV-361 plans
Help Support Homebuilt Aircraft & Kit Plane Forum:
Hot Wings
Grumpy Cynic
Log Member
Due to multiple parameters in my life, some of which none of us have control over, it has become quite obvious to me that I'll probably never be able to build an AV of my own. This means that there is no way I can ethically offer any plans to an aircraft that has not been built and properly tested. Maybe someday in the future.....................?
So rather than create yet another dead end for the AV-36/361 flying wings I've decided it is time to make the plans available to those that may have the desire and facilities to build one. Preserving the design and the plans was always a significant part of my motivation, though making a few dollars would have been nice.
Plans are now available for download under a Creative Commons License for non commercial use. Donations via PayPal, Bitcoin or Ethereum will be graciously accepted. No support will be provided for the free downloads. Funds generated will be used to maintain the web site and the related flying wing website at GroupsIo
Flyingwings.guru
Log Member
Hot Wings
Grumpy Cynic
Log Member
On the website.
No builder support. Post modified.
proppastie
Well-Known Member
Log Member
On the website.
No builder support. Post modified.
Hot Wings
Grumpy Cynic
Log Member
It is there, just not on the first page. I'd like those interested in downloading to at least look at the CC license.
Hot Wings
Grumpy Cynic
Log Member
Just checked from my wife's computer. The final link to the files is NOT working. Must ave been pulling the files of my computer when checked live. Will fix it later in the evening.
proppastie
Well-Known Member
Log Member
my thoughts on this before you fix it ......much as I like free stuff.....the original plans were sold for $xxxx .....without changing them are they still worth something.....do you really need to provide support for the original plans in order to sell them "as is" ......I can understand your point of view.....but if you do not make the cad redraw any issues are Fauvels or Falconar's problem? course how you will protect your property after it is downloaded is always an issue...but maybe at say$5 it will be too much trouble for someone else to market?
TFF
Well-Known Member
Very noble gesture. I hope plenty appreciate it can offset your effort. Definitely you are doing something to save GA.
Hot Wings
Grumpy Cynic
Log Member
Try the files now. My wife's Chrome downloads them on her 'puter.
Hot Wings
Grumpy Cynic
Log Member
course how you will protect your property after it is downloaded is always an issue...but maybe at say \$5 it will be too much trouble for someone else to market?
I looked at several ways to protect electronic versions. I could find nothing that had an acceptable security/price ratio.
At this point all I really would like is that no one pays for the files from some Russian pirate site. The files I posted are little more than slightly cleaned up originals and the manuals reproduced using OCR for the grunt work. I did post a couple of PDFs of the full size CAD generated rib templates (with scaled references). Fauvel's method was rather tedious and hard to figure out. Maybe someone can actually benefit from the work...........
Voidhawk9
Well-Known Member
Only thing is it appears that the first two are switched around - not a problem if downloading all 3 anyway.
proppastie
Well-Known Member
Log Member
Thank you....download link next to line about paypal was confusing to me as I thought it was about paypal and did not click it except as a last resort......
proppastie
Well-Known Member
Log Member
as one that used to make drawings for a living....the amount of hours involved with just making the drawings is enormous .......not the mention the fab of the aircraft.....a fitting retirement project to make one of these....or a large number of people group build....no wonder the kit plane is more popular.
so why are all the PDF mirrored.....I guess it is because of making the JPG and the scanning?......
nickec
Well-Known Member
AV361 pages 43 and 44 are reflected/inverted.
Hot Wings
Grumpy Cynic
Log Member
so why are all the PDF mirrored.....I guess it is because of making the JPG and the scanning?......
That is the way I got them from Falconar and some are even negative. The PDFs are exactly as I received them from Staples after they copied the ratty original masters. It's simple enough to run them throug a program like MS Paint to invert them after converting to *.JPG or *.TIFF. I - think - all of the *.JPG versions of the files I posted have been corrected?
I'm not a web designer guy so this little project took most of the long weekend for me. The local 'el-cheapo' local web company wanted more than I paid for the rights to build a web site of about the same complexity so I had to brush up some on my web software. The cell phone version is still not really formatted well.
I'll go back and review the posted files later to make sure there is at least one properly formatted version of each drawing. I may post the cleaned and reformatted sheets with my part numbers at some point in the future. I've got a few hundered hours invested in those that I'm not quite ready to part with in digital form...........
Rob de Bie
Member
Hot Wings, thanks for your generosity. I downloaded the drawing set for study purposes. Because I could not mirror the drawings as PDFs, I exported them to hi-res and high quality JPG, and then did the mirroring and rotating. I also 'inverted' the ones that were in reverse colors. Maybe you want this set, to post them as a download? The set is ~660 MB.
This is the list of drawings that I have now. The gap between #20 and #31 is slightly confusing.
AV36 drawing 01 - wing assembly
AV36 drawing 02 - rib section
AV36 drawing 03 - fin & rudder
AV36 drawing 04A - fuselage (inverted original)
AV36 drawing 04A - fuselage
AV36 drawing 04B - landing skid & towline
AV36 drawing 05 - fuselage frames (inverted original)
AV36 drawing 05 - fuselage frames
AV36 drawing 06 - controls
AV36 drawing 07 - controls details
AV36 drawing 08 - accessoires
AV36 drawing 09 - air brakes complement (inverted original)
AV36 drawing 09 - air brakes complement
AV36 drawing 10 - corrections
AV36 drawing 11 - modification for construction in Canada and USA
AV36 drawing 12 - quick dismantling wing fittings at rib 16
AV36 drawing 13 - quick dismantling wing fittings at rib 16
AV36 drawing 14 - cockpit type C blown canopy
AV36 drawing 20 - coordinates
AV361 drawing 31 - standard class sailplane (3 view)
AV361 drawing 32 - main spar & false spar
AV361 drawing 33 - ribs
AV361 drawing 34 - fuselage bulkheads
AV361 drawing 35 - fuselage longerons
AV361 drawing 36 - ribs 8 10 12 14 16
AV361 drawing 37 - airbrake box
AV361 drawing 38 - airbrake control arm
AV361 drawing 39 - formers canopy frame
AV361 drawing 40 - canopy fairing rib #1
AV361 drawing 41 - pylon assembly
AV361 drawing 42 - airbrake control
AV361 drawing 43 - elevator (copy 1)
AV361 drawing 43 - elevator (copy 2)
AV361 drawing 43 - elevator (copy 3)
AV361 drawing 44 - nose ribs
AV361 drawing 45 - fin mounting
AV361 drawing 46 - fin rudder assembly
Rob
Hot Wings
Grumpy Cynic
Log Member
The gap between #20 and #31 is slightly confusing.
That is just the way of the plans from Falconar Avia.
The trend, unfortunately, gets carried over at the sheet level. There never was a fixed North American version of the AV's. As near as I can tell it was a work in progress with the plans being a collection of modifications to the base AV-36.
One of the manuals has a listing of the sheets that came from Falconar Avia originally. I've also included some other files I managed to find, mostly from French sources, that help clarify the North American version of the plans.
Thanks for the offer!
But I already have high resolution *.TIFF from the scans.
The JPG files are high resolution JPGs from my TIFF conversions of the raw PDF scans. The are all converted from blue print/negative and are un-mirrored. They take up a little less space than the *.TIFF but view and print just as well.
The only reason I posted the raw scans is so that there is a base line available to cross check with any new versions of the plans. My CAD indicates there are some mistakes in the originals......but my CAD could just as easily be wrong.
If you haven't already joined the Flying Wings group at GroupsIO I'd suggest doing that so all of us interested in the AVs can benefit from our interactions. Subgroup for AVs.
I don't really know what our group file storage limit is but if you want to try to try to post your converted files there in ZIP format I'd have no objections.
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# Where to find a summary for Q functions? [duplicate]
This question already has an answer here:
Is there a summary for all Function like
NumberQ, LetterQ and so forth?
data=123;
IntegerQ[data]
True
-
add comment
## marked as duplicate by rm -rf♦Mar 7 at 14:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 1 Answer
Do you mean a list of functions ending in Q?
?*Q
-
(+1) or ?**Q to get all 1243 functions. – kguler Mar 7 at 13:55
There is an example like this in the documentation page on Information. Typing "?" in help is useless :( (it does not bring you to Information). Also the syntax is so special that you cannot do something like Hold[?*Q]//FullForm to find out what function is being used. So I figured I'd post this. – Jacob Akkerboom Mar 7 at 14:11
Of course, not all of these are querying functions. EllipticNomeQ[], LegendreQ[], and MarcumQ[]` are examples that spring to mind. – Ｊ. Ｍ. Apr 16 at 13:24
add comment
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Cuffnorm: which output file should I use
2
0
Entering edit mode
4.8 years ago
Hello Biostars,
I'm trying to use cuffnorm to normalize the cuffquant results. I'm getting these output files:
• cds.count_tracking
• genes.fpkm_tracking
• isoforms.count_tracking
• run.info
• tss_groups.fpkm_tracking
• cds.fpkm_tracking
• genes.count_tracking
• isoforms.fpkm_tracking
• tss_groups.count_tracking
I'm mostly interested in genes, so which file should I use? is it the genes.count_tracking ? Is that the normalized gene counts?
Thanks
cufflinks cuffnorm next-gen RNA-Seq • 2.7k views
0
Entering edit mode
What is your question, i.e. what will you do for the next step? Is it differential expression of the genes, or?
0
Entering edit mode
Not the DE analysis, my goal is to predict a clinical end-point, such as survival using the normalized expression levels.
0
Entering edit mode
4.8 years ago
Prakash ★ 2.1k
genes.fpkm_tracking is the file for genes with its normalized FPKM (fragment per kb per millions of mapped reads)
0
Entering edit mode
You mean they are not the normalized values by cuffnorm? I'm confused.
0
Entering edit mode
4.8 years ago
aka001 ▴ 190
Although I would presume that you did cuffnorm because you wanted to normalise your data, I would still just want to say that it is important to see what are your biological questions. "Mostly interested in genes" can mean many things and if you intend to do differential expression of the genes (with the mainstream packages, like DESeq or edgeR) for the next step, then you would actually want to take the genes.count_tracking file, as it is not normalised and hence it would suit for DE purpose.
0
Entering edit mode
I'm not doing DE analysis. my goal is to predict a clinical end-point, such as survival using the normalized expression levels. If genes.count is not the normalized values then which file have the normalized expression levels according to cuffnorm?
0
Entering edit mode
The normalised expression should be genes.fpkm_tracking.
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Entering edit mode
Thanks, but my understanding was that fpkm is very simple to because it is just done by a few multiplicaitons and divisions. Which should not take hours to be completed by cuffnorm, what am I missing then? Thanks again
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Entering edit mode
Well, if we go back again to what you wanted in the first place, which is normalisation, it would depend of what kind of normalisation did you want. By definition, fpkm is already normalised value, but it's normalised by library size (i.e. within sample). For the next step, probably you would want to normalise across samples (with methods like RLE, TMM, etc., which I am not sure cuffnorm is doing that) from the fpkm values (or TPM if you want) from the cuffnorm output files. For the speed, I would just guess that cuffnorm is not only calculating that, but much more other things (as reflected by the number of files generated).
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Entering edit mode
Thank you so much, this was helpful. My understanding from what you said is that the gene_tracking is not normalized (with respect to the library size), but when I check the raw gene counts vs the gene_count_tracking, the latter has been scaled down, which means it is doing some normalization as well. Am I missing something?
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# scipy.stats.ttest_ind¶
scipy.stats.ttest_ind(a, b, axis=0, equal_var=True, nan_policy='propagate')[source]
Calculates the T-test for the means of two independent samples of scores.
This is a two-sided test for the null hypothesis that 2 independent samples have identical average (expected) values. This test assumes that the populations have identical variances by default.
Parameters: a, b : array_like The arrays must have the same shape, except in the dimension corresponding to axis (the first, by default). axis : int or None, optional Axis along which to compute test. If None, compute over the whole arrays, a, and b. equal_var : bool, optional If True (default), perform a standard independent 2 sample test that assumes equal population variances [R643]. If False, perform Welch’s t-test, which does not assume equal population variance [R644]. New in version 0.11.0. nan_policy : {‘propagate’, ‘raise’, ‘omit’}, optional Defines how to handle when input contains nan. ‘propagate’ returns nan, ‘raise’ throws an error, ‘omit’ performs the calculations ignoring nan values. Default is ‘propagate’. statistic : float or array The calculated t-statistic. pvalue : float or array The two-tailed p-value.
Notes
We can use this test, if we observe two independent samples from the same or different population, e.g. exam scores of boys and girls or of two ethnic groups. The test measures whether the average (expected) value differs significantly across samples. If we observe a large p-value, for example larger than 0.05 or 0.1, then we cannot reject the null hypothesis of identical average scores. If the p-value is smaller than the threshold, e.g. 1%, 5% or 10%, then we reject the null hypothesis of equal averages.
References
Examples
>>> from scipy import stats
>>> np.random.seed(12345678)
Test with sample with identical means:
>>> rvs1 = stats.norm.rvs(loc=5,scale=10,size=500)
>>> rvs2 = stats.norm.rvs(loc=5,scale=10,size=500)
>>> stats.ttest_ind(rvs1,rvs2)
(0.26833823296239279, 0.78849443369564776)
>>> stats.ttest_ind(rvs1,rvs2, equal_var = False)
(0.26833823296239279, 0.78849452749500748)
ttest_ind underestimates p for unequal variances:
>>> rvs3 = stats.norm.rvs(loc=5, scale=20, size=500)
>>> stats.ttest_ind(rvs1, rvs3)
(-0.46580283298287162, 0.64145827413436174)
>>> stats.ttest_ind(rvs1, rvs3, equal_var = False)
(-0.46580283298287162, 0.64149646246569292)
When n1 != n2, the equal variance t-statistic is no longer equal to the unequal variance t-statistic:
>>> rvs4 = stats.norm.rvs(loc=5, scale=20, size=100)
>>> stats.ttest_ind(rvs1, rvs4)
(-0.99882539442782481, 0.3182832709103896)
>>> stats.ttest_ind(rvs1, rvs4, equal_var = False)
(-0.69712570584654099, 0.48716927725402048)
T-test with different means, variance, and n:
>>> rvs5 = stats.norm.rvs(loc=8, scale=20, size=100)
>>> stats.ttest_ind(rvs1, rvs5)
(-1.4679669854490653, 0.14263895620529152)
>>> stats.ttest_ind(rvs1, rvs5, equal_var = False)
(-0.94365973617132992, 0.34744170334794122)
#### Previous topic
scipy.stats.ttest_1samp
#### Next topic
scipy.stats.ttest_ind_from_stats
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TimeISOT¶
class astropy.time.TimeISOT(val1, val2, scale, precision, in_subfmt, out_subfmt, from_jd=False)[source]
ISO 8601 compliant date-time format “YYYY-MM-DDTHH:MM:SS.sss…”. This is the same as TimeISO except for a “T” instead of space between the date and time. For example, 2000-01-01T00:00:00.000 is midnight on January 1, 2000.
The allowed subformats are:
• ‘date_hms’: date + hours, mins, secs (and optional fractional secs)
• ‘date_hm’: date + hours, mins
• ‘date’: date
Attributes Summary
cache Return the cache associated with this instance. jd2_filled mask masked name scale Time scale subfmts value
Methods Summary
format_string(self, str_fmt, \*\*kwargs) Write time to a string using a given format. mask_if_needed(self, value) parse_string(self, timestr, subfmts) Read time from a single string, using a set of possible formats. set_jds(self, val1, val2) Parse the time strings contained in val1 and set jd1, jd2 str_kwargs(self) Generator that yields a dict of values corresponding to the calendar date and time for the internal JD values. to_value(self[, parent]) Return time representation from internal jd1 and jd2.
Attributes Documentation
cache
Return the cache associated with this instance.
jd2_filled
mask
masked
name = 'isot'
scale
Time scale
subfmts = (('date_hms', re.compile('(?P<year>\\d\\d\\d\\d)-(?P<mon>\\d{1,2})-(?P<mday>\\d{1,2})T(?P<hour>\\d{1,2}):(?P<min>\\d{1,2}):(?P<sec>\\d{1,2})$'), '{year:d}-{mon:02d}-{day:02d}T{hour:02d}:{min:02d}:{sec:02d}'), ('date_hm', re.compile('(?P<year>\\d\\d\\d\\d)-(?P<mon>\\d{1,2})-(?P<mday>\\d{1,2})T(?P<hour>\\d{1,2}):(?P<min>\\d{1,2})$'), '{year:d}-{mon:02d}-{day:02d}T{hour:02d}:{min:02d}'), ('date', re.compile('(?P<year>\\d\\d\\d\\d)-(?P<mon>\\d{1,2})-(?P<mday>\\d{1,2})\$'), '{year:d}-{mon:02d}-{day:02d}'))
value
Methods Documentation
format_string(self, str_fmt, **kwargs)
Write time to a string using a given format.
By default, just interprets str_fmt as a format string, but subclasses can add to this.
mask_if_needed(self, value)
parse_string(self, timestr, subfmts)
Read time from a single string, using a set of possible formats.
set_jds(self, val1, val2)
Parse the time strings contained in val1 and set jd1, jd2
str_kwargs(self)
Generator that yields a dict of values corresponding to the calendar date and time for the internal JD values.
to_value(self, parent=None)
Return time representation from internal jd1 and jd2. This is the base method that ignores parent and requires that subclasses implement the value property. Subclasses that require parent or have other optional args for to_value should compute and return the value directly.
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# The form $xy+5=a(x+y)$ and its solutions with $x,y$ prime
In another question I was asking if there are any different $x,y>2$ primes such that $xy+5=a(x+y)$.
Where $a=2^r-1$, and $r>2$.
I was thinking if it is able to find a Pell equation or a similar pattern of $xy+5=a(x+y)$ to say what are and how many integer solutions are there (in particular prime solutions).
Thanks.
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$xy-5=a(x+y)$ can be rewritten as $$(x-a)(y-a)=a^2+5$$ so for any fixed $a$ solving it just amounts to finding all the ways to factor $a^2+5$. So how many solutions depends on the prime factorization of $a^2+5$. I don't think there will be any formula for how many of those solutions have $x$ and $y$ prime.
So $xy+5=a(x+y)$ can be rewritten as $$(x-a)(y-a)=a^2-5,$$ right? – Gerry Myerson May 5 '11 at 1:21
If there aren't such prime pairs solve this equation, it seems very interesting. The question here if it may be unique. If when we change -5 into some other -prime will give no prime solutions. In this form of the equation we can say that for each such $a$ there are finitely (but more than 0) solutions in integers and infinitely in the union set. – tomerg May 6 '11 at 6:41
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# Assistance Cracking Classical Cipher
Below is the cipher text I am trying to break and as you can see its rather short which is why I am having so much trouble.
WOYFN ZCMSH VUVTG BFUTW ABTZP FHIMF TFOSU UXFQC HKVKG MPUUQ
MHRXI OVBRZ EPJYF KKVJW GEIOV HUKEB JUNSM THIMF TFKUB ULMMQX
• Key is 6 characters long and repeats (Given by the person who encrypted it)
• There is a single character missing from the start of the ciphertext! (Also given)
• It uses a simple cipher I believe a vigenere cipher.
• Friedman IC: 1.143 (kappa-plaintext: 0.4396) which is why i think its a vigenere cipher
I have spent many hours on this but to be honest im not very good at breaking these codes.
My problem that iv run into is that once I break the cipher into its sections they are all very short and hence frequency alaysis fails.
I would really appreciate if someone could either confirm or find the crypto system being used as well as any more information about how one would find the plaintext + key or better yet find the key itself and explain how you manage it with such a short piece of ciphertext
Edit: There was some debate as to the validity of this question so here is some more information from my attempts:
Streams:
• ?ZUUZFUKUXZKEKSFU IC = 3.6833 (kappa-plaintext: 1.4167)
• WCVTPTXVUIEKIEMTL IC = 1.1471 (kappa-plaintext: 0.4412)
• OMTWFFFKQOPVOBTFM IC = 2.1029 (kappa-plaintext: 0.8088)
• YSGAHOQGMVJJVJHKM IC = 1.3382 (kappa-plaintext: 0.5147)
• FHBBISCMHBYWHUIUQ IC = 1.5294 (kappa-plaintext: 0.5882)
• NVFTMUHPRRFGUNMBX IC = 0.9559 (kappa-plaintext: 0.3676)
These numbers don't look good. they are almost all drastically off normal distribution for any language.
The main point to my question is:
• Does Vigenere Seem like a reasonable fit (given those streams are correct)?
• Is it possible to use that analysis in any way? I cant see how it can be effective with only 17 characters per stream. Most letters frequencies are 0.
• What other ciphers are possible? How could I identify that they are in use?
I am happy to provide more information I am just not sure what else is useful.
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## migrated from security.stackexchange.comMay 13 '12 at 18:07
This question came from our site for Information security professionals.
Alas, this question is not really on topic for crypto.SE either. The FAQ is pretty clear about it. (Personally, I kind of like these kinds of questions, at least as long as they look reasonably likely to be answerable and show more research effort than just "please break this cipher for me", but the consensus seems to be that they're all off topic here.) – Ilmari Karonen May 13 '12 at 22:48
I disagree with the fact that this isn't the place to ask. Reading the FAQ it says that its not ok to "Can I challenge people to decode something?" but it does also say that you can ask homework questions. I believe thats what im asking. I am not really after the solution more how would one approach this question given the limited ciphertext and that I have tried the attacks I know of. – Nick May 14 '12 at 3:15
As I wrote that FAQ, I'll weigh in with what I think - the "challenge people to decode something" was meant to deal with "please decode XYZ bet you can't !11!!1"-type questions, of which we've had a few. It's not meant to discourage questions which are a decent attempt at cryptanalysis, although I think they do need to be reasonably answerable. As is, I think we could do with as much information as you can give - perhaps the results of your frequency analysis, for example? However, I don't see the need to fire a mod close vote at it, although the community or other mods may disagree :) – Rhino May 16 '12 at 9:45
Also, if you feel the FAQ can be improved, do weigh in on meta :) – Rhino May 16 '12 at 9:45
I don't know the solution, but since you say you're only asking for hints, here's a few that occurred to me:
• If this is a Vigenère cipher, the missing character at the beginning should not matter (much): if you encrypt a message with the key FOOBAR and drop the first letter of the output, you can decrypt the resulting ciphertext with the key OOBARF.
• As ewanm89 notes, the obvious way to attack a Vigenère cipher, after you've determined the key length is $n$, is to write it with $n$ letters per row and attack each column as if it were a Caesar cipher. There are automated tools for doing that, such as pygenere; they may not always give the correct answer for very short messages, but yours looks long enough that they should at least have a decent chance.
• The most notable feature I see in your ciphertext is that the string HIMFTF occurs twice, at offsets 26 and 86. That does strongly suggest that the key length may be a divisor of 60; alas, 60 has lots of small divisors (including 6, which you believe the key length to be), so that doesn't really give that much information. It does at least rule out 7, 8 and 9 as likely key lengths.
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Hey thank you for the advice, I had noticed the HIMFTF myself. but the reason why I know the length is 6 is because the person that encrypted the message (The tutor for my class) told me that it was 6 long and that it repeats. – Nick May 15 '12 at 0:16
PS I have now tried Pygenere again and it does not yield the message. Keep in mind that I am only guessing that this is Vigenere. Is there any way to work out if that is correct? – Nick May 15 '12 at 0:17
The easiest way in this case to work out if it is vigenere is to brute force it.
You know the key is only 6 characters. Take a vigenere decryption function and a dictionary file of 6 character words. Decrypt using the first word, then compute a histogram of the resulting plaintext. Compare with what you would expect to see given the distribution of characters in english language text. If the two are close, visually inspect. If they are not close or visual inspection fails, continue.
Look around on google for histogram comparison ideas. A simple sum of differences should work then set a threshold for what to display.
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Cryptool is a nice crypto learning tool that supports classical cipher cryptanalysis such as for Vigenère ciphers. There is also an online version, but I'm not sure whether this only works for the German page or also for the English page. – DrLecter Dec 17 '13 at 15:12
As already stated it is a Vigenere cipher. Here even the length of the key is already known. A principle approach to break the cipher is as follows:
Try different keys. For each key construct the corresponding clear text and check how similar the clear text is to the English language.
How well this algorithm works depends on how good your check for the English language is. So you need a function, which takes a decoded clear text as input and returns a measure for the probability that the text is English. Such function is called fitness function, the returned value is the fitness of the clear text. The fitness function is not restricted to the Vigenere cipher, it can be used for breaking other ciphers as well.
Now most Vigenere solvers are looking at the frequency of single letters, which means the fitness function is based on so called monograms. This works more or less well if the cipher text is much longer than the key. For the cipher given here those solvers will probably not determine the correct key.
Instead of counting the frequency of single letters it is much better to count the frequency of bigrams, i.e. you are looking always at two neighboring letters of the clear text. This greatly improves the accuracy of the fitness function.
There are 26 different monograms (A..Z) in the English language, but 26*26 = 676 bigrams (AA..ZZ). Thus using bigrams will probably require a computer program, while using monograms the cipher can still be broken by hand.
So let's assume you have a table which provides for each bigram the probability that it occurs in English text (the most frequent bigram of the English language is "TH" [~2.7%]) . One option to implement a fitness function is to multiply the probabilities of all bigrams of the clear text. The greater the resulting value, the more probable it is that the text is English.
So you just need to find the key where the resulting clear text has the highest fitness value.
A brute force over the complete solution space (key length = 6 => 26^6 possible keys) is not needed, it is sufficient to look at neighboring letters of the key only.
Using this approach the cipher text given here can be broken rather easily. The algorithm is implemented here: http://www.guballa.de/pages/geocaching/vigenere-solver.php
I have also provided a more detailed description of the algorithm (http://www.guballa.de/pages/bits-bytes/implementierung-eines-vigenere-solvers.php), but that's in German.
Using trigrams (or even quadgrams) would make the algorithm even stronger.
I found that it is even possible to break Vigenere ciphers if the cipher text is only as long as four times of the key length.
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Additional resource which explains rather well the fitness function: practicalcryptography.com/cryptanalysis/text-characterisation/… – Jagu May 1 '14 at 17:45
I can confirm that this can be solved as a Vigenere cipher. The decoded message reads [c]ongratulations on solving the puzzle we hope you enjoy computer and network security and the wargames puzzles good luck.
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Hello Eli and welcome to Crypto. There is a word: "If you give a man a fish, you feed him for one night. But if you teach him how to fish, he will be fed for the rest of his life." I think, it will be more helpful, if you explain how your approached the problem instead of spoiling the solution. But congratulations for finding it. – Hendrik Brummermann Jun 15 '12 at 6:46
Eli, I would have to agree with Hendrik. Of course one can find the key from your solution, but that doesn't help the OP learn how to actually find the key in future problems. – Thomas Jun 15 '12 at 13:08
I agree with the others on providing a little explanation as to how you decoded this. I don't think it needs to be a step by step guide, but a brief overview of what you did would no doubt steer the OP in the right direction :) – Rhino Jun 15 '12 at 21:41
Thank you for decryption this. Although I have to agree with the comments above. If in the exam they ask "How would you attack this cipher if you were given this info" I don't think they would be to impressed if i said "Post to crypto.stackechange.com". I would be interested to know how you achieved this so I could do it again to a different ciphertext with a different key/plaintext. – Nick Jun 17 '12 at 22:31
@Nick, the explanation is right above. Attacking a Vigenere cipher with known key length is trivial. Let me repeat. You feed the cipher with various keys of that length and count digrams in decrypted gibberish until you hit some threshold. At this point there would be few partially colliding keys that would partially decrypt the message. Look at these, reconstruct the rest of the message and roll it back to the original key if you must. – Eli Jun 18 '12 at 7:10
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# A 2-Dimensional column is subjected to two forces as shown. Sketch the normal stress distribution acting...
###### Question:
A 2-Dimensional column is subjected to two forces as shown. Sketch the normal stress distribution acting over section a-a if the member has a rectangular cross section of width 12 in and thickness 6 in. 2ok -a ITITIIr77 PLAN Visw
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# Thread: Algebra Do's and Don't's
1. In my experience teaching Calculus at the university level, there were a number of algebra mistakes I would see quite often. In an effort to help you, the ordinary math student, I would like to create a post here that outlines a few of these mistakes, and how to avoid them. The general format is going to be: the mistake, why it's a mistake, followed by how to remember not to make that mistake.
In addition, there are a few techniques that keep coming up over and over again, that students need to have in their toolbox. I'll list a few of those for you as well.
But first, the mistakes.
Mistake # 1: Distributing addition over squaring: $(x+y)^{2}\not=x^{2}+y^{2}$.
This one is also known as the "Freshman's Dream". It looks like it ought to be true, but it simply isn't. Why isn't it true? Because you've left out the cross terms; it's also not true because the squaring function is not linear. The LHS of the above inequation needs multiplying out term-by-term thus:
$$(x+y)^{2}=(x+y)(x+y)=x^{2}+xy+yx+y^{2}=x^{2}+xy+xy+y^{2}=x^{2}+2xy+y^{2}.$$
And now, you can see that unless the cross term is zero (not true in general), the Freshman's Dream is not true.
How can you remember this? Probably the easiest way is to plug in specific numbers and see if it works. Try $x=1$ and $y=2$. Then the LHS is $9$, and the RHS is $5$. Since they're not equal, the Freshman's Dream is incorrect.
Mistake # 2: Distributing addition over the square root:
$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$.
Why isn't this true? Because the square root function is not linear. You can test it out with some Pythagorean triples. Suppose $a=9, b=16$. Then you've got
$$5=\sqrt{25}=\sqrt{9+16}\overset{?}{=} \sqrt{9}+\sqrt{16}=3+4=7.$$
Well, $5\not=7$, so evidently this one doesn't work, either. So, in this case, the reason why it doesn't work, and the memory device for remembering that it doesn't work can be the same thing.
Mistake # 3: Distributing addition over pretty much any function:
$$\sin(x+y)\not=\sin(x)+\sin(y),\quad \ln(x+y)\not=\ln(x)+\ln(y),\quad e^{x+y}\not=e^{x}+e^{y},\dots$$
So, for the first, there's a formula you should know:
$$\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y).$$
If you're familiar with matrix rotations, there's an elegant way to remember this formula, along with the cosine version:
$$\begin{bmatrix}\cos(x+y)\\ \sin(x+y)\end{bmatrix}=\begin{bmatrix}\cos(x) &-\sin(x)\\ \sin(x) &\cos(x)\end{bmatrix}\begin{bmatrix}\cos(y)\\ \sin(y)\end{bmatrix}$$
For the second inequation, there's no way to simplify the LHS. Instead, what you're probably thinking is that
$$\ln(xy)=\ln(x)+\ln(y),$$
which is true. Check this equation with $x=1$ and $y=e$. You get $1=\ln(e)=\ln(1\cdot e)=\ln(1)+\ln(e)=0+1$. It doesn't work with the inequation above. Brush up on your logarithms!
For the last one, it's similar to the logarithm, only inverted:
$$e^{x+y}=e^{x}e^{y}.$$
Here, check it out with $x=0$ and $y=1$. You're probably seeing a pattern here: if you're unsure of an equation, check it out with specific numbers! Don't use completely trivial numbers like $0$, unless it's warranted. You might get a fluke where your incorrect equation happens to work with a special case. Check it out with multiple cases.
Mistake # 4: Distributing multiplication or division over any of these things.
For the same reasons as before, this simply doesn't work. Most functions are not linear, and so you can't expect multiplication to distribute over most functions. When in doubt, don't! Or check it with specific numbers.
Mistake # 5: Canceling factors without provisos. $\frac{x(x+1)}{x+1}\not=x$.
This one's a shocker for many people. Why can't I cancel? Well, you can, but only if $x\not=-1$. So you can do this:
$$\frac{x(x+1)}{x+1}=x,\quad\text{for}\;x\not=-1.$$
Why is this the case? Because with the initial inequation I mentioned, you're really claiming that the function on the LHS is equal to the function on the RHS. But two functions are only equal if their domains are the same, and their rules of association are the same. The domain of the function on the left is $\mathbb{R}\setminus\{-1\}$, and the domain of the function on the right is $\mathbb{R}$. Since the domains are not the same, you cannot just cancel. Now, if you happen to know that in your application, $x$ is never equal to $-1$, then by all means, cancel! One notable example of this is if you're computing a limit in calculus when $x\to-1$. In this limit, you never let $x=-1$, so you're fine. You just have to be careful in equating two functions to make sure the domains are the same, or that you don't care about those regions where the domains are not the same. The moral of the story is: be intentional about your canceling, and think through these issues!
Mistake # 6: Omitting parentheses in function arguments. $\sin x+y=?$.
Do I mean $\sin(x)+y$ or do I mean $\sin(x+y)$? I once worked a problem from a physics textbook where the authors neglected to include parentheses on a rather complicated argument to the $\sin$ function. I got the answer wrong, because I assumed the incorrect version. Ever since then, I've been a stickler for including parentheses on function arguments. Don't write so that you can be understood! Write so that you can't be misunderstood! Yes, it might take a tad longer initially. However, the time you save in being clearly understood by others will often more than make up for that "lost" time.
So here are a few mistakes. What are some things I recommend doing?
Do # 1: Completing the square. $x^{2}+4x+9=x^{2}+4x+4+5=(x+2)^{2}+5.$
This technique is invaluable. I even used it in my Commentary for "Algebra Do's and Don't's"
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# Exercise 10 | The Properties of Acids and Bases
We want to calculate the pH of a 0.20 M aqueous solution of acetic acid (Ka = 1.8 x 10-5 M) at 25.0°C
1) Can we use the method of successive approximations?
2) Calculate the pH of this solution.
1) At the equilibrium, [HA] = 0.20 – X M and [H3O+] = [A-] = X M
Method of successive approximations: X << 0.20 ⇒ 0.20 – X ~ 0.20
Let’s verify this approximation:
Ka = = $\frac{{\mathrm{X}}^{2}}{0.20}$ = 1.8 x 10-5
⇒ X = 1.9 x 10-3 << 0.2
Yes, we can use the method of successive approximations and X = 1.9 x 10-3 M
2) [H3O+] = X = 1.9 x 10-3 M
pH = -log [H3O+] = 2.7
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Chemistry » Gases » Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
# Pressure and Temperature: Amontons’s Law
## Pressure and Temperature: Amontons’s Law
Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (see the figure below) and the pressure increases.
The effect of temperature on gas pressure: When the hot plate is off, the pressure of the gas in the sphere is relatively low. As the gas is heated, the pressure of the gas in the sphere increases.
This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in the figure below. We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then P and T are directly proportional (again, when volume and moles of gas are held constant); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor.
For a constant volume and amount of air, the pressure and temperature are directly proportional, provided the temperature is in kelvin. (Measurements cannot be made at lower temperatures because of the condensation of the gas.) When this line is extrapolated to lower pressures, it reaches a pressure of 0 at –273 °C, which is 0 on the kelvin scale and the lowest possible temperature, called absolute zero.
Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the PT relationship for gases is known as either Amontons’s law or Gay-Lussac’s law. Under either name, it states that the pressure of a given amount of gas is directly proportional to its temperature on the kelvin scale when the volume is held constant. Mathematically, this can be written:
$$P\propto T\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}P=\text{constant}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}T\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}P=k\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}T$$
where ∝ means “is proportional to,” and k is a proportionality constant that depends on the identity, amount, and volume of the gas.
For a confined, constant volume of gas, the ratio $$\cfrac{P}{T}$$ is therefore constant (i.e., $$\cfrac{P}{T}\phantom{\rule{0.2em}{0ex}}=k$$). If the gas is initially in “Condition 1” (with P = P1 and T = T1), and then changes to “Condition 2” (with P = P2 and T = T2), we have that $$\cfrac{{P}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=k$$ and $$\cfrac{{P}_{2}}{{T}_{2}}\phantom{\rule{0.2em}{0ex}}=k,$$ which reduces to $$\cfrac{{P}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{2}}{{T}_{2}}.$$
This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called absolute zero). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.)
## Example
### Predicting Change in Pressure with Temperature
A can of hair spray is used until it is empty except for the propellant, isobutane gas.
(a) On the can is the warning “Store only at temperatures below 120 °F (48.8 °C). Do not incinerate.” Why?
(b) The gas in the can is initially at 24 °C and 360 kPa, and the can has a volume of 350 mL. If the can is left in a car that reaches 50 °C on a hot day, what is the new pressure in the can?
### Solution
(a) The can contains an amount of isobutane gas at a constant volume, so if the temperature is increased by heating, the pressure will increase proportionately. High temperature could lead to high pressure, causing the can to burst. (Also, isobutane is combustible, so incineration could cause the can to explode.)
(b) We are looking for a pressure change due to a temperature change at constant volume, so we will use Amontons’s/Gay-Lussac’s law. Taking P1 and T1 as the initial values, T2 as the temperature where the pressure is unknown and P2 as the unknown pressure, and converting °C to K, we have:
$$\require{cancel}\cfrac{{P}_{1}}{{T}_{1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{2}}{{T}_{2}}\phantom{\rule{0.4em}{0ex}}\text{which means that}\phantom{\rule{0.4em}{0ex}}\cfrac{360\phantom{\rule{0.2em}{0ex}}\text{kPa}}{297\phantom{\rule{0.2em}{0ex}}\text{K}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{P}_{2}}{323\phantom{\rule{0.2em}{0ex}}\text{K}}$$
Rearranging and solving gives: $${P}_{2}=\phantom{\rule{0.2em}{0ex}}\cfrac{360\phantom{\rule{0.2em}{0ex}}\text{kPa}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}323\phantom{\rule{0.2em}{0ex}}\cancel{\text{K}}}{297\phantom{\rule{0.2em}{0ex}}\cancel{\text{K}}}\phantom{\rule{0.2em}{0ex}}=390\phantom{\rule{0.2em}{0ex}}\text{kPa}$$
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# A fluid is termed as the Newtonian fluid, when the shear stress is __________ the velocity gradient.
Independent of
Inversely proportional to
Directly proportional to
None of these
Please do not use chat terms. Example: avoid using "grt" instead of "great".
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# MATLAB Project 2
NJIT Math 222
### Problem 1
Consider the differential equation $$\frac{dx}{dt}= f(x) = \sin{x}-a \sin{4x}$$ This equation has period $2\pi$, so we will consider it for $-\pi\le x \le \pi.$ Things will be clearer if we plot instead on a slightly larger interval such as $-4\le x \le 4$. When $a=0$, this equation has two equilibria per period, at $x=0$ and $x=\pi$ (considering $x=\pm\pi$ as the same point). When $a$ is large (say $a=4$) then the graph of $f(x)$ looks pretty much like $-a\sin{4x}$ (mathematicians would say the second term dominates), and the system has 8 fixed points per period.
Starting with $a=0$ plot the direction field for several increasing values of $a$. You should find that at a certain value of $a$ satisfying $0.2<a<0.3$, two new equilibria appear and at a second critical value satisfying $0.8<a<1$, four more equilibria appear.
Print out enough graphs to describe what happens in each case and describe what you observe in a sentence or two. Pay attention to the changes in the neighborhood of the equilibrium $x=0$. Hint: it is useful to plot $f(x)$ for various values of $a$ as well. The website and apps from Desmos are easy to use and output beautiful graphics.
### Problem 2
Graph the direction field and some solutions for the non-autonomous equation $$\frac{dx}{dt}=x - \frac{t x}{1+x^2}.$$
It is up to you to pick the range over which you plot in $x$ and $t$ in order to see the behavior.
There is a finite number of things that can happen as $t\to\infty$. Describe them.
Previous
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# Two capacitors of capacitances 4.0 μ F and 6.0 μF
Question:
Two capacitors of capacitances $4.0 \mu \mathrm{F}$ and $6.0 \mu \mathrm{F}$ are connected in series with a battery of $20 \mathrm{~V}$. Find the energy supplied by the battery.
Solution:
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## 27.3 Relative spectrum via glueing
Situation 27.3.1. Here $S$ is a scheme, and $\mathcal{A}$ is a quasi-coherent $\mathcal{O}_ S$-algebra. This means that $\mathcal{A}$ is a sheaf of $\mathcal{O}_ S$-algebras which is quasi-coherent as an $\mathcal{O}_ S$-module.
In this section we outline how to construct a morphism of schemes
$\underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A}) \longrightarrow S$
by glueing the spectra $\mathop{\mathrm{Spec}}(\Gamma (U, \mathcal{A}))$ where $U$ ranges over the affine opens of $S$. We first show that the spectra of the values of $\mathcal{A}$ over affines form a suitable collection of schemes, as in Lemma 27.2.1.
Lemma 27.3.2. In Situation 27.3.1. Suppose $U \subset U' \subset S$ are affine opens. Let $A = \mathcal{A}(U)$ and $A' = \mathcal{A}(U')$. The map of rings $A' \to A$ induces a morphism $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A')$, and the diagram
$\xymatrix{ \mathop{\mathrm{Spec}}(A) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(A') \ar[d] \\ U \ar[r] & U' }$
is cartesian.
Proof. Let $R = \mathcal{O}_ S(U)$ and $R' = \mathcal{O}_ S(U')$. Note that the map $R \otimes _{R'} A' \to A$ is an isomorphism as $\mathcal{A}$ is quasi-coherent (see Schemes, Lemma 26.7.3 for example). The result follows from the description of the fibre product of affine schemes in Schemes, Lemma 26.6.7. $\square$
In particular the morphism $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A')$ of the lemma is an open immersion.
Lemma 27.3.3. In Situation 27.3.1. Suppose $U \subset U' \subset U'' \subset S$ are affine opens. Let $A = \mathcal{A}(U)$, $A' = \mathcal{A}(U')$ and $A'' = \mathcal{A}(U'')$. The composition of the morphisms $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A')$, and $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A'')$ of Lemma 27.3.2 gives the morphism $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A'')$ of Lemma 27.3.2.
Proof. This follows as the map $A'' \to A$ is the composition of $A'' \to A'$ and $A' \to A$ (because $\mathcal{A}$ is a sheaf). $\square$
Lemma 27.3.4. In Situation 27.3.1. There exists a morphism of schemes
$\pi : \underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A}) \longrightarrow S$
with the following properties:
1. for every affine open $U \subset S$ there exists an isomorphism $i_ U : \pi ^{-1}(U) \to \mathop{\mathrm{Spec}}(\mathcal{A}(U))$, and
2. for $U \subset U' \subset S$ affine open the composition
$\xymatrix{ \mathop{\mathrm{Spec}}(\mathcal{A}(U)) \ar[r]^{i_ U^{-1}} & \pi ^{-1}(U) \ar[rr]^{inclusion} & & \pi ^{-1}(U') \ar[r]^{i_{U'}} & \mathop{\mathrm{Spec}}(\mathcal{A}(U')) }$
is the open immersion of Lemma 27.3.2 above.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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## Elementary Geometry for College Students (7th Edition)
First, we find the altitude: $h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8$ We know that the point where the radius is represents 1/3 of the height. Thus: $r = 8 \times (1/3) = 2.67 \approx 3$
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# Linear Inequalities Class 11
Linear Inequalities class 11 – Two algebraic expressions or real numbers related by the symbol $\leq, \;\geq , \;< , \;and\;>$ forms an inequality. For example: px + qy > 0, 9a – 21b < 0, etc. From an equality Equal numbers can be subtracted or added from both the sides of an equation. Also, both sides of an inequality can be divided or multiplied by the same number (non-zero).
1. Solve $4x + 3 < 6x +7$
Solution: Given $4x + 3 < 6x +7$
Therefore, 3 – 7 < 6x – 4x or, 2x > – 4, i.e. x > -2
Hence, the solution of given inequality is (–2, ∞).
1. Solve $\frac{5-2x}{3}\leq \frac{x}{6}-5$
Solution: Given $\frac{5-2x}{3}\leq \frac{x}{6}-5$
Therefore, $10-4x\leq x-30$
Or, $-5x\leq -40\;\;or\;\;x\geq 8$
Therefore, all real numbers equal to or greater than 8 are the solutions of the
given inequality i.e.$x \in [8, \infty )$
### Linear Inequalities class 11 Examples
#### Practise This Question
The number of observations in a group is 40. If the average of first 10 is 4.5 and that of the remaining 30 is 3.5, then the average of the whole group is
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LHCb Theses
Latest additions:
2021-11-17
10:46
Quantum oscillations of neutral particles with the beauty quark / Veronesi, Michele The development of quantum mechanics since the beginning of the 20th century has guided the exploration of nature at the atomic scale [...] CERN-THESIS-2021-193 - 217 p.
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16:59
Analysis of charged particle production in proton-nucleus and proton-proton collisions at the LHCb experiment / Boente Garcia, Oscar This thesis presents the measurement of prompt charged particle production in proton-lead and proton-proton collisions at the centre-of-mass energy of $\sqrt{s_{\mathrm{NN} } }=5.02\,\mathrm{TeV}$ in the LHCb experiment at CERN [...] CERN-THESIS-2021-184 - 309 p.
2021-10-21
16:12
Search for pentaquark state and measurement of CP violation with the LHCb detector / Liu, Xuesong Testing the standard model (SM) of particle physics and searching for new physics beyond it, explaining the asymmetry between matter and antimatter in the universe is the frontier of particle physics researches [...] CERN-THESIS-2019-423 - 165 p.
2021-10-18
13:21
Measurement of the mixing parameters of neutral charm mesons and search for indirect \textit{\textbf{CP}} violation with $D^0\to K^0_S\pi^+\pi^-$ decays at LHCb / Hilton, Martha Mixing is the time-dependent phenomenon of a neutral meson (in this case charm meson $D^0$) changing into its anti-particle ($\bar{D}^0$) and vice versa [...] CERN-THESIS-2021-168 - 262 p.
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Study of $B^{0}_{s}$ mesons decays to charmonium and multi-hadron final states in the LHCb experiment / Ovsiannikova, Tatiana The searches for new $B^{0}_s$ decays to charmonium and multi-hadron states are presented in this thesis [...] CERN-THESIS-2021-156 - 107 p.
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Search for the Cabibbo-suppressed decays of $\Lambda_b^0$ baryon at the LHCb experiment / Matiunin, Slava The searches for the Cabibbo-suppressed decays $\Lambda_b^0\to\psi(2S)p\pi^-$ and $\Lambda_b^0\to\chi_{c1}p\pi^-$ are presented in this thesis [...] CERN-THESIS-2021-155 - 102 p.
2021-10-05
12:09
Testing the Standard Model Lepton Symmetries in Collider and Fixed-Target Experiments / Bezshyiko, Iaroslava In recent years, several discrepancies between experimental measurements and SM predictions have arisen in indirect searches, known as the flavour anomalies [...] CERN-THESIS-2021-153 - 179 p.
2021-10-04
17:42
Hardware and Firmware Development for an FPGA-based Multipurpose Board Targeted for High Energy Physics Experiments / Barros Marin, Manoel This thesis (PFC) describes the hardware and firmware development of an FPGA- based multi-purpose board targeted for High Energy Physics (HEP) that has been carried out during my stay at CERN [...] CERN-THESIS-2012-503 - 236 p.
2021-09-30
14:35
An automated measurement system for testing optical modules in particle accelerators / Lippiello, Andrea The TTC-PON system (Timing, Trigger and Control system based on Passive Optical Networks) is proposed to replace the current TTC system, responsible for delivering the bunch clock, trigger and control commands to the LHC experiments [...] CERN-THESIS-2018-491 - 107 p.
2021-09-16
00:04
Study of the decay $B^0 \rightarrow D^0D^0K^+\pi^−$ with the LHCb experiment / Bhasin, Srishti The decay $B^0 \rightarrow D^0D^0K^+\pi^−$ is studied using data from proton-proton collisions collected by the LHCb experiment [...] CERN-THESIS-2021-137 -
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Free Version
Easy
# Conducting Sphere: Strongest Electric Field
APPHEM-ETOEK0
A hollow conducting sphere has a net charge of $+Q$. There are four labeled points all on the inside of the sphere, as shown above. Points A through D are located progressively closer to the center of the sphere.
Which point has the largest magnitude of electric field?
A
Point A.
B
Point B.
C
Point C.
D
Point D.
E
All points have the same magnitude of electric field.
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## Saturday, June 27, 2015
### Kernel of the Symmetrizing Map
Let $V$ be a finite dimensional vector space over a field of characteristic $0$ and let $sym:\bigotimes^k V\to \bigotimes^k V$ be the map defined as …
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NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1
### IIT-JEE 1986
Fill in the Blanks
From the point A(0, 3) on the circle $${x^2} + 4x + {(y - 3)^2} = 0$$, a chord AB is drawn and extended to a point M such that AM = 2AB. The equation of the locus of M is..........................
$${x^2} + {y^2} + 8x - 6y + 9 = 0$$
2
### IIT-JEE 1985
Fill in the Blanks
Let $${x^2} + {y^2} - 4x - 2y - 11 = 0$$ be a circle. A pair of tangentas from the point (4, 5) with a pair of radi from a quadrilateral of area............................
8 sq unit
3
### IIT-JEE 1985
Fill in the Blanks
From the origin chords are drawn to the circle $${(x - 1)^2} + {y^2} = 1$$. The equation of the locus of the mid-points of these chords is.............
$${x^2} + {y^2} - x = 0$$
4
### IIT-JEE 1984
Fill in the Blanks
The lines 3x - 4y + 4 = 0 and 6x - 8y - 7 = 0 are tangents to the same circle. The radius of this circle is ........................................
$$\,{\raise0.5ex\hbox{\scriptstyle 3} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 4}}$$
### Joint Entrance Examination
JEE Main JEE Advanced WB JEE
### Graduate Aptitude Test in Engineering
GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN
NEET
Class 12
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## Happenings May 8
Well, last Saturday morning I woke up determined to do mathematics, and knowing exactly what I was going to do. It was a very productive day, but at the end of it I was completely worn out. I didn’t stop to write about what was happening, and I didn’t let the kid do any mathematics of his own.
Not for a while, anyway. He has moved on to finite fields. There are two interesting facts that served as a starting point.
First, consider the set $Z_n$ of integers mod n. The first interesting fact is that $Z_n$ is a field if and only if n is prime. For example, $Z_3 = \{0,1,2\}$ with multiplication and addition mod 3 is a field. $Z_4 = \{0,1,2,3\}$ is not – because 2*2 = 0, so 2 does not have a multiplicative inverse.
So, “we know” that there exists a finite field with p elements if p is prime. The second interesting fact is that there also exists a finite field with $p^n$ elements if p is prime, for any positive integer n.
We could combine these two facts and a little more and say that there is a unique finite field of order $p^n$ for all primes p and all positive integers n.)
The challenge is that the case n = 1 is very easy, but the cases n > 1 are not.
In fact, until yesterday I had never held a finite field with n > 1 in my hands. Well, now I know a representation for the finite field of order $4 = 2^2\$. I’m looking forward to holding more finite fields in my grubby little hands.
In the meantime, I have implemented the dyadic expansion of a scaling function and of the corresponding mother wavelet, for several cases. You should be seeing them in the next couple of posts.
(OK, the dyadic expansion of the D4 scaling function is out there; the corresponding calculation for the D4 mother wavelet should be next; and more examples should follow quickly.)
It took longer than I expected to put out the post – it usually does. And it took some editing afterwards. I had forgotten that sometimes the LaTeX actually renders, but incorrectly. In particular, < stuff > gets treated as (silly) HTML and vanishes. (And, yes, I had to code those angle brackets – I can’t just type them. Grrr.) I prefer it when I get the yellow banner that says it couldn’t render at all.
One of the reasons for making sure that I get this post out tonight is that I don’t want to have to stop to do this tomorrow morning.
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